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6,701
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What is the relationship between $Y$ and $X$ in this plot?
|
Let me describe what I see as soon as I look at it:
If we're interested in the conditional distribution of $y$ (which if often where interest focuses if we see $x$ as IV and $y$ as DV),
then for $x\leq 0.5$ the conditional distribution of $Y|x$ appears bimodal with an upper group (between about 70 and 125, with mean a bit below 100) and a lower group (between 0 and about 70, with mean around 30 or so). Within each modal group, the relationship with $x$ is nearly flat. (See red and blue lines below drawn roughly
where I guess some rough sense of location to be)
Then by looking at where those two groups are more or less dense in $X$, we can go on to say more:
For $x>0.5$ the upper group disappears completely, which makes the overall mean of $x$ fall, and below about 0.2, the lower group is much less dense than above it, making the overall average higher.
Between these two effects, it induces an apparent negative (but nonlinear) relationship between the two, as $E(Y|X=x)$ seems to be decreasing against $x$ but with a broad, mostly flat region in the center. (See purple dashed line)
No doubt it would be important to know what $Y$ and $X$ were, because then it might be clearer why the conditional distribution for $Y$ might be bimodal over much of its range (indeed, it might even become clear that there are indeed two groups, whose distributions in $X$ induce the apparent decreasing relationship in $Y|x$).
This what I saw based on purely "by-eye" inspection. With a bit of playing around in something like a basic image manipulation program (like the one I drew the lines with) we could start to figure out some more accurate numbers. If we digitize the data (which is pretty simple with decent tools, if sometimes a little tedious to get right), then we can undertake more sophisticated analyses of that sort of impression.
This kind of exploratory analysis can lead to some important questions (sometimes ones that surprise the person who has the data but has only shown a plot), but we must take some care over the extent to which our models are chosen by such inspections - if we apply models chosen on the basis of the appearance of a plot and then estimate those models on the same data, we'll tend to encounter the same problems we get when we use more formal model-selection and estimation on the same data. [This is not to deny the importance of exploratory analysis at all - it's just we must be careful of the consequences of doing it without regard to how we go about it. ]
Response to Russ' comments:
[later edit: To clarify -- I broadly agree with Russ' criticisms taken as a general precaution, and there's certainly some possibility I've seen more than is really there. I plan to come back and edit these into a more extensive commentary on spurious patterns we commonly identify by eye and ways we might start to avoid the worst of that. I believe I'll also be able to add some justification about why I think it's probably not just spurious in this specific case (e.g. via a regressogram or 0-order kernel smooth, though of course, absent more data to test against, there's only so far that can go; for example, if our sample is unrepresentative, even resampling only gets us so far.]
I completely agree we have a tendency to see spurious patterns; it's a point I make frequently both here and elsewhere.
One thing I suggest, for example, when looking at residual plots or Q-Q plots is to generate many plots where the situation is known (both as things should be and where assumptions don't hold) to get a clear idea how much pattern should be ignored.
Here's an example where a Q-Q plot is placed among 24 others (which satisfy the assumptions), in order for us to see how unusual the plot is. This kind of exercise is important because it helps us avoid fooling ourselves by interpreting every little wiggle, most of which will be simple noise.
I often point out that if you can change an impression by covering a few points, we may be relying on an impression generated by nothing more than noise.
[However, when it's apparent from many points rather than few, it's harder to maintain that it's not there.]
The displays in whuber's answer supports my impression, the Gaussian blur plot seems to pick up the same tendency to bimodality in $Y$.
When we don't have more data to check, we can at least look at whether the impression tends to survive resampling (bootstrap the bivariate distribution and see if it's nearly always still present), or other manipulations where the impression shouldn't be apparent if it's simple noise.
1) Here's one way to see if the apparent bimodality is more than just skewness plus noise - does it show up in a kernel density estimate? Is it still visible if we plot kernel density estimates under a variety of transformations? Here I transform it toward greater symmetry, at 85% of default bandwidth (since we're trying to identify a relatively small mode, and the default bandwidth is not optimized for that task):
The plots are $Y$, $\sqrt{Y}$ and $\log(Y)$. The vertical lines are at $68$, $\sqrt{68}$ and $\log(68)$. The bimodality is diminished, but still quite visible. Since it's very clear in the original KDE it seems to confirm it's there - and the second and third plots suggest its at least somewhat robust to transformation.
2) Here's another basic way to see if it's more than just "noise":
Step 1: perform clustering on Y
Step 2: Split into two groups on $X$, and cluster the two groups separately, and see if it's quite similar. If there's nothing going on the two halves shouldn't be expected to split all that much alike.
The points with dots were clustered differently from the "all in one set" cluster in the previous plot. I'll do some more later, but it seems like perhaps there really might be a horizontal "split" near that position.
I'm going to try a regressogram or Nadaraya-Watson estimator (both being local estimates of the regression function, $E(Y|x)$). I haven't generated either yet, but we'll see how they go. I'd probably exclude the very ends where there's little data.
3) Edit: Here's the regressogram, for bins of width 0.1 (excluding the very ends, as I suggested earlier):
This is entirely consistent with the original impression I had of the plot; it doesn't prove my reasoning was correct, but my conclusions arrived at the same result the regressogram does.
If what I saw in the plot - and the resulting reasoning - was spurious, I probably should not have succeeded at discerning $E(Y|x)$ like this.
(Next thing to try would be a Nadayara-Watson estimator. Then I might see how it goes under resampling if I have time.)
4) Later edit:
Nadarya-Watson, Gaussian kernel, bandwidth 0.15:
Again, this is surprisingly consistent with my initial impression. Here's The NW estimators based on ten bootstrap resamples:
The broad pattern is there, though a couple of the resamples don't as clearly follow the description based on the whole of the data. We see that the case of the level of the left is less certain than on the right - the level of noise (partly from few observations, partly from the wide spread) is such that it's less easy to claim the mean is really higher at the left than at the center.
My overall impression is that I probably wasn't simply fooling myself, because the various aspects stand up moderately well to a variety of challenges (smoothing, transformation, splitting into subgroups, resampling) that would tend to obscure them if they were simply noise. On the other hand, the indications are that the effects, while broadly consistent with my initial impression, are relatively weak, and it may be too much to claim any real change in expectation moving from the left side to the center.
|
What is the relationship between $Y$ and $X$ in this plot?
|
Let me describe what I see as soon as I look at it:
If we're interested in the conditional distribution of $y$ (which if often where interest focuses if we see $x$ as IV and $y$ as DV),
then for $x\l
|
What is the relationship between $Y$ and $X$ in this plot?
Let me describe what I see as soon as I look at it:
If we're interested in the conditional distribution of $y$ (which if often where interest focuses if we see $x$ as IV and $y$ as DV),
then for $x\leq 0.5$ the conditional distribution of $Y|x$ appears bimodal with an upper group (between about 70 and 125, with mean a bit below 100) and a lower group (between 0 and about 70, with mean around 30 or so). Within each modal group, the relationship with $x$ is nearly flat. (See red and blue lines below drawn roughly
where I guess some rough sense of location to be)
Then by looking at where those two groups are more or less dense in $X$, we can go on to say more:
For $x>0.5$ the upper group disappears completely, which makes the overall mean of $x$ fall, and below about 0.2, the lower group is much less dense than above it, making the overall average higher.
Between these two effects, it induces an apparent negative (but nonlinear) relationship between the two, as $E(Y|X=x)$ seems to be decreasing against $x$ but with a broad, mostly flat region in the center. (See purple dashed line)
No doubt it would be important to know what $Y$ and $X$ were, because then it might be clearer why the conditional distribution for $Y$ might be bimodal over much of its range (indeed, it might even become clear that there are indeed two groups, whose distributions in $X$ induce the apparent decreasing relationship in $Y|x$).
This what I saw based on purely "by-eye" inspection. With a bit of playing around in something like a basic image manipulation program (like the one I drew the lines with) we could start to figure out some more accurate numbers. If we digitize the data (which is pretty simple with decent tools, if sometimes a little tedious to get right), then we can undertake more sophisticated analyses of that sort of impression.
This kind of exploratory analysis can lead to some important questions (sometimes ones that surprise the person who has the data but has only shown a plot), but we must take some care over the extent to which our models are chosen by such inspections - if we apply models chosen on the basis of the appearance of a plot and then estimate those models on the same data, we'll tend to encounter the same problems we get when we use more formal model-selection and estimation on the same data. [This is not to deny the importance of exploratory analysis at all - it's just we must be careful of the consequences of doing it without regard to how we go about it. ]
Response to Russ' comments:
[later edit: To clarify -- I broadly agree with Russ' criticisms taken as a general precaution, and there's certainly some possibility I've seen more than is really there. I plan to come back and edit these into a more extensive commentary on spurious patterns we commonly identify by eye and ways we might start to avoid the worst of that. I believe I'll also be able to add some justification about why I think it's probably not just spurious in this specific case (e.g. via a regressogram or 0-order kernel smooth, though of course, absent more data to test against, there's only so far that can go; for example, if our sample is unrepresentative, even resampling only gets us so far.]
I completely agree we have a tendency to see spurious patterns; it's a point I make frequently both here and elsewhere.
One thing I suggest, for example, when looking at residual plots or Q-Q plots is to generate many plots where the situation is known (both as things should be and where assumptions don't hold) to get a clear idea how much pattern should be ignored.
Here's an example where a Q-Q plot is placed among 24 others (which satisfy the assumptions), in order for us to see how unusual the plot is. This kind of exercise is important because it helps us avoid fooling ourselves by interpreting every little wiggle, most of which will be simple noise.
I often point out that if you can change an impression by covering a few points, we may be relying on an impression generated by nothing more than noise.
[However, when it's apparent from many points rather than few, it's harder to maintain that it's not there.]
The displays in whuber's answer supports my impression, the Gaussian blur plot seems to pick up the same tendency to bimodality in $Y$.
When we don't have more data to check, we can at least look at whether the impression tends to survive resampling (bootstrap the bivariate distribution and see if it's nearly always still present), or other manipulations where the impression shouldn't be apparent if it's simple noise.
1) Here's one way to see if the apparent bimodality is more than just skewness plus noise - does it show up in a kernel density estimate? Is it still visible if we plot kernel density estimates under a variety of transformations? Here I transform it toward greater symmetry, at 85% of default bandwidth (since we're trying to identify a relatively small mode, and the default bandwidth is not optimized for that task):
The plots are $Y$, $\sqrt{Y}$ and $\log(Y)$. The vertical lines are at $68$, $\sqrt{68}$ and $\log(68)$. The bimodality is diminished, but still quite visible. Since it's very clear in the original KDE it seems to confirm it's there - and the second and third plots suggest its at least somewhat robust to transformation.
2) Here's another basic way to see if it's more than just "noise":
Step 1: perform clustering on Y
Step 2: Split into two groups on $X$, and cluster the two groups separately, and see if it's quite similar. If there's nothing going on the two halves shouldn't be expected to split all that much alike.
The points with dots were clustered differently from the "all in one set" cluster in the previous plot. I'll do some more later, but it seems like perhaps there really might be a horizontal "split" near that position.
I'm going to try a regressogram or Nadaraya-Watson estimator (both being local estimates of the regression function, $E(Y|x)$). I haven't generated either yet, but we'll see how they go. I'd probably exclude the very ends where there's little data.
3) Edit: Here's the regressogram, for bins of width 0.1 (excluding the very ends, as I suggested earlier):
This is entirely consistent with the original impression I had of the plot; it doesn't prove my reasoning was correct, but my conclusions arrived at the same result the regressogram does.
If what I saw in the plot - and the resulting reasoning - was spurious, I probably should not have succeeded at discerning $E(Y|x)$ like this.
(Next thing to try would be a Nadayara-Watson estimator. Then I might see how it goes under resampling if I have time.)
4) Later edit:
Nadarya-Watson, Gaussian kernel, bandwidth 0.15:
Again, this is surprisingly consistent with my initial impression. Here's The NW estimators based on ten bootstrap resamples:
The broad pattern is there, though a couple of the resamples don't as clearly follow the description based on the whole of the data. We see that the case of the level of the left is less certain than on the right - the level of noise (partly from few observations, partly from the wide spread) is such that it's less easy to claim the mean is really higher at the left than at the center.
My overall impression is that I probably wasn't simply fooling myself, because the various aspects stand up moderately well to a variety of challenges (smoothing, transformation, splitting into subgroups, resampling) that would tend to obscure them if they were simply noise. On the other hand, the indications are that the effects, while broadly consistent with my initial impression, are relatively weak, and it may be too much to claim any real change in expectation moving from the left side to the center.
|
What is the relationship between $Y$ and $X$ in this plot?
Let me describe what I see as soon as I look at it:
If we're interested in the conditional distribution of $y$ (which if often where interest focuses if we see $x$ as IV and $y$ as DV),
then for $x\l
|
6,702
|
What is the relationship between $Y$ and $X$ in this plot?
|
OK folks, I followed Alexis's lead and captured the data. Here is a plot of $\log y$ versus $x$.
And the correlations:
> cor.test(~ x + y, data = data)
Pearson's product-moment correlation
data: x and y
t = -2.6311, df = 169, p-value = 0.009298
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.33836844 -0.04977867
sample estimates:
cor
-0.1983692
> cor.test(~ x + log(y), data = data)
Pearson's product-moment correlation
data: x and log(y)
t = -2.8901, df = 169, p-value = 0.004356
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.35551268 -0.06920015
sample estimates:
cor
-0.2170188
The correlation test does indicate a likely negative dependence. I remain unconvinced of any bimodality (but also unconvinced that it's absent).
[I removed a residual plot I had in an earlier version because I overlooked the point that @whuber was trying to predict $X|Y$.]
|
What is the relationship between $Y$ and $X$ in this plot?
|
OK folks, I followed Alexis's lead and captured the data. Here is a plot of $\log y$ versus $x$.
And the correlations:
> cor.test(~ x + y, data = data)
Pearson's product-moment correlation
data
|
What is the relationship between $Y$ and $X$ in this plot?
OK folks, I followed Alexis's lead and captured the data. Here is a plot of $\log y$ versus $x$.
And the correlations:
> cor.test(~ x + y, data = data)
Pearson's product-moment correlation
data: x and y
t = -2.6311, df = 169, p-value = 0.009298
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.33836844 -0.04977867
sample estimates:
cor
-0.1983692
> cor.test(~ x + log(y), data = data)
Pearson's product-moment correlation
data: x and log(y)
t = -2.8901, df = 169, p-value = 0.004356
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.35551268 -0.06920015
sample estimates:
cor
-0.2170188
The correlation test does indicate a likely negative dependence. I remain unconvinced of any bimodality (but also unconvinced that it's absent).
[I removed a residual plot I had in an earlier version because I overlooked the point that @whuber was trying to predict $X|Y$.]
|
What is the relationship between $Y$ and $X$ in this plot?
OK folks, I followed Alexis's lead and captured the data. Here is a plot of $\log y$ versus $x$.
And the correlations:
> cor.test(~ x + y, data = data)
Pearson's product-moment correlation
data
|
6,703
|
What is the relationship between $Y$ and $X$ in this plot?
|
Russ Lenth wondered how the graph would look if the Y axis were logarithmic. Alexis scraped the data, so it is easy to plot on with a log axis:
On a log scale, there is no hint of bimodality or trend. Whether a log scale makes sense or not depends, of course, on the details of what the data represent. Similarly, whether it makes sense to think that the data represent sampling from two populations as whuber suggests depends on the details.
Addendum: Based on the comments below, here is a revised version:
|
What is the relationship between $Y$ and $X$ in this plot?
|
Russ Lenth wondered how the graph would look if the Y axis were logarithmic. Alexis scraped the data, so it is easy to plot on with a log axis:
On a log scale, there is no hint of bimodality or trend
|
What is the relationship between $Y$ and $X$ in this plot?
Russ Lenth wondered how the graph would look if the Y axis were logarithmic. Alexis scraped the data, so it is easy to plot on with a log axis:
On a log scale, there is no hint of bimodality or trend. Whether a log scale makes sense or not depends, of course, on the details of what the data represent. Similarly, whether it makes sense to think that the data represent sampling from two populations as whuber suggests depends on the details.
Addendum: Based on the comments below, here is a revised version:
|
What is the relationship between $Y$ and $X$ in this plot?
Russ Lenth wondered how the graph would look if the Y axis were logarithmic. Alexis scraped the data, so it is easy to plot on with a log axis:
On a log scale, there is no hint of bimodality or trend
|
6,704
|
What is the relationship between $Y$ and $X$ in this plot?
|
Well, you are right, the relationship is weak, but not zero. I would guess positive. However, don't guess, just run a simple linear regression (OLS regression) and find out! There you will get a slope of xxx which tells you what the relationship is. And yes, you do have outliers that might bias the results. That can be dealt with. You could use Cook's distance or create a leverage plot to estimate the outliers' effect on the relationship.
Good luck
|
What is the relationship between $Y$ and $X$ in this plot?
|
Well, you are right, the relationship is weak, but not zero. I would guess positive. However, don't guess, just run a simple linear regression (OLS regression) and find out! There you will get a slope
|
What is the relationship between $Y$ and $X$ in this plot?
Well, you are right, the relationship is weak, but not zero. I would guess positive. However, don't guess, just run a simple linear regression (OLS regression) and find out! There you will get a slope of xxx which tells you what the relationship is. And yes, you do have outliers that might bias the results. That can be dealt with. You could use Cook's distance or create a leverage plot to estimate the outliers' effect on the relationship.
Good luck
|
What is the relationship between $Y$ and $X$ in this plot?
Well, you are right, the relationship is weak, but not zero. I would guess positive. However, don't guess, just run a simple linear regression (OLS regression) and find out! There you will get a slope
|
6,705
|
What is the relationship between $Y$ and $X$ in this plot?
|
You already provided some intuition to your question by looking at the orientation of the X/Y data points and their dispersion. In short you're correct.
In formal terms orientation can be referred to as correlation sign and dispersion as variance. These two links will give you more information on how to interpret the linear relationship between two variables.
|
What is the relationship between $Y$ and $X$ in this plot?
|
You already provided some intuition to your question by looking at the orientation of the X/Y data points and their dispersion. In short you're correct.
In formal terms orientation can be referred to
|
What is the relationship between $Y$ and $X$ in this plot?
You already provided some intuition to your question by looking at the orientation of the X/Y data points and their dispersion. In short you're correct.
In formal terms orientation can be referred to as correlation sign and dispersion as variance. These two links will give you more information on how to interpret the linear relationship between two variables.
|
What is the relationship between $Y$ and $X$ in this plot?
You already provided some intuition to your question by looking at the orientation of the X/Y data points and their dispersion. In short you're correct.
In formal terms orientation can be referred to
|
6,706
|
What is the relationship between $Y$ and $X$ in this plot?
|
This is a home work. So, the answer to your question is simple. Run a linear regression of Y on X, you'll get something like this:
Coefficient Standard Er t Stat
C 53.14404163 6.522516463 8.147781908
X -44.8798926 16.80565866 -2.670522684
So, the t-statistics is significant on X variable at 99% confidence. Hence, you can declare the variables as having some kind of relationship.
Is it linear? Add a variable X2 = (X-mean(X))^2, and regress again.
Coefficient Stand Err t Stat
C 53.46173893 6.58938281 8.11331508
X -43.9503443 17.01532569 -2.582985779
X2 -44.601130 114.1461801 -0.390736951
The coefficient at X is still significant, but X2 is not. X2 represents nonlinearity. So, you declare that teh relationship appears to be linear.
The above was for a home work.
In real life, things are more complicated. Imagine, that this was the data on a class of students. Y - bench press in pounds, X - time in minutes of holding one's breath before the bench press. I'd ask for the gender of the students. Just for fun of it, let;s add another variable, Z, and let's say that Z=1 (girls) for all Y<60, and Z=0 (boys) when Y>=60. Run the regression with three variables:
Coefficient Stand Error t Stat
C 92.93031357 3.877092841 23.969071
X -6.55246715 8.977138488 -0.72990599
X2 -43.6291362 59.06955097 -0.738606194
Z -63.3231270 2.960160265 -21.39179009
What happened?! The "relationship" between X and Y has disappeared! Oh, it seems that the relationship was spurious due to confounding variable, gender.
What is the moral of the story? You need to know what is the data to "explain" the "relationship", or even to establish it in the first place. In this case, the moment I'm told that the data on students' physical activity, I'll immediately ask for their gender, and will not even bother analyzing the data without getting the gender variable.
On the other hand, if you're asked to "describe" the scatter plot, then anything goes. Correlations, linear fits etc. For your home work, first two steps above should be sufficient: look at coefficient of X (relationship), then X^2 (linearity). Make sure you de-mean the X variable (subtract the mean).
|
What is the relationship between $Y$ and $X$ in this plot?
|
This is a home work. So, the answer to your question is simple. Run a linear regression of Y on X, you'll get something like this:
Coefficient Standard Er t Stat
C 53.14404163 6.522516463 8.1477
|
What is the relationship between $Y$ and $X$ in this plot?
This is a home work. So, the answer to your question is simple. Run a linear regression of Y on X, you'll get something like this:
Coefficient Standard Er t Stat
C 53.14404163 6.522516463 8.147781908
X -44.8798926 16.80565866 -2.670522684
So, the t-statistics is significant on X variable at 99% confidence. Hence, you can declare the variables as having some kind of relationship.
Is it linear? Add a variable X2 = (X-mean(X))^2, and regress again.
Coefficient Stand Err t Stat
C 53.46173893 6.58938281 8.11331508
X -43.9503443 17.01532569 -2.582985779
X2 -44.601130 114.1461801 -0.390736951
The coefficient at X is still significant, but X2 is not. X2 represents nonlinearity. So, you declare that teh relationship appears to be linear.
The above was for a home work.
In real life, things are more complicated. Imagine, that this was the data on a class of students. Y - bench press in pounds, X - time in minutes of holding one's breath before the bench press. I'd ask for the gender of the students. Just for fun of it, let;s add another variable, Z, and let's say that Z=1 (girls) for all Y<60, and Z=0 (boys) when Y>=60. Run the regression with three variables:
Coefficient Stand Error t Stat
C 92.93031357 3.877092841 23.969071
X -6.55246715 8.977138488 -0.72990599
X2 -43.6291362 59.06955097 -0.738606194
Z -63.3231270 2.960160265 -21.39179009
What happened?! The "relationship" between X and Y has disappeared! Oh, it seems that the relationship was spurious due to confounding variable, gender.
What is the moral of the story? You need to know what is the data to "explain" the "relationship", or even to establish it in the first place. In this case, the moment I'm told that the data on students' physical activity, I'll immediately ask for their gender, and will not even bother analyzing the data without getting the gender variable.
On the other hand, if you're asked to "describe" the scatter plot, then anything goes. Correlations, linear fits etc. For your home work, first two steps above should be sufficient: look at coefficient of X (relationship), then X^2 (linearity). Make sure you de-mean the X variable (subtract the mean).
|
What is the relationship between $Y$ and $X$ in this plot?
This is a home work. So, the answer to your question is simple. Run a linear regression of Y on X, you'll get something like this:
Coefficient Standard Er t Stat
C 53.14404163 6.522516463 8.1477
|
6,707
|
Variance of a bounded random variable
|
You can prove Popoviciu's inequality as follows. Use the notation $m=\inf X$ and $M=\sup X$. Define a function $g$ by
$$
g(t)=\mathbb{E}\!\left[\left(X-t\right)^2\right] \, .
$$
Computing the derivative $g'$, and solving
$$
g'(t) = -2\mathbb{E}[X] +2t=0 \, ,
$$
we find that $g$ achieves its minimum at $t=\mathbb{E}[X]$ (note that $g''>0$).
Now, consider the value of the function $g$ at the special point $t=\frac{M+m}{2}$. It must be the case that
$$
\mathbb{Var}[X]=g(\mathbb{E}[X])\leq g\left(\frac{M+m}{2}\right) \, .
$$
But
$$
g\left(\frac{M+m}{2}\right) = \mathbb{E}\!\left[\left(X - \frac{M+m}{2}\right)^2 \right] = \frac{1}{4}\mathbb{E}\!\left[\left((X-m) + (X-M)\right)^2 \right] \, .
$$
Since $X-m\geq 0$ and $X-M\leq 0$, we have
$$
\left((X-m)+(X-M)\right)^2\leq\left((X-m)-(X-M)\right)^2=\left(M-m\right)^2 \, ,
$$
implying that
$$
\frac{1}{4}\mathbb{E}\!\left[\left((X-m) + (X-M)\right)^2 \right] \leq \frac{1}{4}\mathbb{E}\!\left[\left((X-m) - (X-M)\right)^2 \right] = \frac{(M-m)^2}{4} \, .
$$
Therefore, we proved Popoviciu's inequality
$$
\mathbb{Var}[X]\leq \frac{(M-m)^2}{4} \, .
$$
|
Variance of a bounded random variable
|
You can prove Popoviciu's inequality as follows. Use the notation $m=\inf X$ and $M=\sup X$. Define a function $g$ by
$$
g(t)=\mathbb{E}\!\left[\left(X-t\right)^2\right] \, .
$$
Computing the deriva
|
Variance of a bounded random variable
You can prove Popoviciu's inequality as follows. Use the notation $m=\inf X$ and $M=\sup X$. Define a function $g$ by
$$
g(t)=\mathbb{E}\!\left[\left(X-t\right)^2\right] \, .
$$
Computing the derivative $g'$, and solving
$$
g'(t) = -2\mathbb{E}[X] +2t=0 \, ,
$$
we find that $g$ achieves its minimum at $t=\mathbb{E}[X]$ (note that $g''>0$).
Now, consider the value of the function $g$ at the special point $t=\frac{M+m}{2}$. It must be the case that
$$
\mathbb{Var}[X]=g(\mathbb{E}[X])\leq g\left(\frac{M+m}{2}\right) \, .
$$
But
$$
g\left(\frac{M+m}{2}\right) = \mathbb{E}\!\left[\left(X - \frac{M+m}{2}\right)^2 \right] = \frac{1}{4}\mathbb{E}\!\left[\left((X-m) + (X-M)\right)^2 \right] \, .
$$
Since $X-m\geq 0$ and $X-M\leq 0$, we have
$$
\left((X-m)+(X-M)\right)^2\leq\left((X-m)-(X-M)\right)^2=\left(M-m\right)^2 \, ,
$$
implying that
$$
\frac{1}{4}\mathbb{E}\!\left[\left((X-m) + (X-M)\right)^2 \right] \leq \frac{1}{4}\mathbb{E}\!\left[\left((X-m) - (X-M)\right)^2 \right] = \frac{(M-m)^2}{4} \, .
$$
Therefore, we proved Popoviciu's inequality
$$
\mathbb{Var}[X]\leq \frac{(M-m)^2}{4} \, .
$$
|
Variance of a bounded random variable
You can prove Popoviciu's inequality as follows. Use the notation $m=\inf X$ and $M=\sup X$. Define a function $g$ by
$$
g(t)=\mathbb{E}\!\left[\left(X-t\right)^2\right] \, .
$$
Computing the deriva
|
6,708
|
Variance of a bounded random variable
|
If the random variable is restricted to $[a,b]$ and we know the mean $\mu=E[X]$, the variance is bounded by $(b-\mu)(\mu-a)$.
Let us first consider the case $a=0, b=1$. Note that for all $x\in [0,1]$, $x^2\leq x$, wherefore also $E[X^2]\leq E[X]$. Using this result,
\begin{equation}
\sigma^2 = E[X^2] - (E[X]^2) = E[X^2] - \mu^2 \leq \mu - \mu^2 = \mu(1-\mu).
\end{equation}
To generalize to intervals $[a,b]$ with $b>a$, consider $Y$ restricted to $[a,b]$. Define $X=\frac{Y-a}{b-a}$, which is restricted in $[0,1]$. Equivalently, $Y = (b-a)X + a$, and thus
\begin{equation}
Var[Y] = (b-a)^2Var[X] \leq (b-a)^2\mu_X (1-\mu_X).
\end{equation}
where the inequality is based on the first result.
Now, by substituting $\mu_X = \frac{\mu_Y - a}{b-a}$, the bound equals
\begin{equation}
(b-a)^2\, \frac{\mu_Y - a}{b-a}\,\left(1- \frac{\mu_Y - a}{b-a}\right) = (b-a)^2 \frac{\mu_Y -a}{b-a}\,\frac{b - \mu_Y}{b-a} = (\mu_Y - a)(b- \mu_Y),
\end{equation}
which is the desired result.
|
Variance of a bounded random variable
|
If the random variable is restricted to $[a,b]$ and we know the mean $\mu=E[X]$, the variance is bounded by $(b-\mu)(\mu-a)$.
Let us first consider the case $a=0, b=1$. Note that for all $x\in [0,1]$,
|
Variance of a bounded random variable
If the random variable is restricted to $[a,b]$ and we know the mean $\mu=E[X]$, the variance is bounded by $(b-\mu)(\mu-a)$.
Let us first consider the case $a=0, b=1$. Note that for all $x\in [0,1]$, $x^2\leq x$, wherefore also $E[X^2]\leq E[X]$. Using this result,
\begin{equation}
\sigma^2 = E[X^2] - (E[X]^2) = E[X^2] - \mu^2 \leq \mu - \mu^2 = \mu(1-\mu).
\end{equation}
To generalize to intervals $[a,b]$ with $b>a$, consider $Y$ restricted to $[a,b]$. Define $X=\frac{Y-a}{b-a}$, which is restricted in $[0,1]$. Equivalently, $Y = (b-a)X + a$, and thus
\begin{equation}
Var[Y] = (b-a)^2Var[X] \leq (b-a)^2\mu_X (1-\mu_X).
\end{equation}
where the inequality is based on the first result.
Now, by substituting $\mu_X = \frac{\mu_Y - a}{b-a}$, the bound equals
\begin{equation}
(b-a)^2\, \frac{\mu_Y - a}{b-a}\,\left(1- \frac{\mu_Y - a}{b-a}\right) = (b-a)^2 \frac{\mu_Y -a}{b-a}\,\frac{b - \mu_Y}{b-a} = (\mu_Y - a)(b- \mu_Y),
\end{equation}
which is the desired result.
|
Variance of a bounded random variable
If the random variable is restricted to $[a,b]$ and we know the mean $\mu=E[X]$, the variance is bounded by $(b-\mu)(\mu-a)$.
Let us first consider the case $a=0, b=1$. Note that for all $x\in [0,1]$,
|
6,709
|
Variance of a bounded random variable
|
Let $F$ be a distribution on $[0,1]$. We will show that if the variance of $F$ is maximal, then $F$ can have no support in the interior, from which it follows that $F$ is Bernoulli and the rest is trivial.
As a matter of notation, let $\mu_k = \int_0^1 x^k dF(x)$ be the $k$th raw moment of $F$ (and, as usual, we write $\mu = \mu_1$ and $\sigma^2 = \mu_2 - \mu^2$ for the variance).
We know $F$ does not have all its support at one point (the variance is minimal in that case). Among other things, this implies $\mu$ lies strictly between $0$ and $1$. In order to argue by contradiction, suppose there is some measurable subset $I$ in the interior $(0,1)$ for which $F(I)\gt 0$. Without any loss of generality we may assume (by changing $X$ to $1-X$ if need be) that $F(J = I \cap (0, \mu]) \gt 0$: in other words, $J$ is obtained by cutting off any part of $I$ above the mean and $J$ has positive probability.
Let us alter $F$ to $F'$ by taking all the probability out of $J$ and placing it at $0$. In so doing, $\mu_k$ changes to
$$\mu'_k = \mu_k - \int_J x^k dF(x).$$
As a matter of notation, let us write $[g(x)] = \int_J g(x) dF(x)$ for such integrals, whence
$$\mu'_2 = \mu_2 - [x^2], \quad \mu' = \mu - [x].$$
Calculate
$$\sigma'^2 = \mu'_2 - \mu'^2 = \mu_2 - [x^2] - (\mu - [x])^2 = \sigma^2 + \left((\mu[x] - [x^2]) + (\mu[x] - [x]^2)\right).$$
The second term on the right, $(\mu[x] - [x]^2)$, is non-negative because $\mu \ge x$ everywhere on $J$. The first term on the right can be rewritten
$$\mu[x] - [x^2] = \mu(1 - [1]) + ([\mu][x] - [x^2]).$$
The first term on the right is strictly positive because (a) $\mu \gt 0$ and (b) $[1] = F(J) \lt 1$ because we assumed $F$ is not concentrated at a point. The second term is non-negative because it can be rewritten as $[(\mu-x)(x)]$ and this integrand is nonnegative from the assumptions $\mu \ge x$ on $J$ and $0 \le x \le 1$. It follows that $\sigma'^2 - \sigma^2 \gt 0$.
We have just shown that under our assumptions, changing $F$ to $F'$ strictly increases its variance. The only way this cannot happen, then, is when all the probability of $F'$ is concentrated at the endpoints $0$ and $1$, with (say) values $1-p$ and $p$, respectively. Its variance is easily calculated to equal $p(1-p)$ which is maximal when $p=1/2$ and equals $1/4$ there.
Now when $F$ is a distribution on $[a,b]$, we recenter and rescale it to a distribution on $[0,1]$. The recentering does not change the variance whereas the rescaling divides it by $(b-a)^2$. Thus an $F$ with maximal variance on $[a,b]$ corresponds to the distribution with maximal variance on $[0,1]$: it therefore is a Bernoulli$(1/2)$ distribution rescaled and translated to $[a,b]$ having variance $(b-a)^2/4$, QED.
|
Variance of a bounded random variable
|
Let $F$ be a distribution on $[0,1]$. We will show that if the variance of $F$ is maximal, then $F$ can have no support in the interior, from which it follows that $F$ is Bernoulli and the rest is tr
|
Variance of a bounded random variable
Let $F$ be a distribution on $[0,1]$. We will show that if the variance of $F$ is maximal, then $F$ can have no support in the interior, from which it follows that $F$ is Bernoulli and the rest is trivial.
As a matter of notation, let $\mu_k = \int_0^1 x^k dF(x)$ be the $k$th raw moment of $F$ (and, as usual, we write $\mu = \mu_1$ and $\sigma^2 = \mu_2 - \mu^2$ for the variance).
We know $F$ does not have all its support at one point (the variance is minimal in that case). Among other things, this implies $\mu$ lies strictly between $0$ and $1$. In order to argue by contradiction, suppose there is some measurable subset $I$ in the interior $(0,1)$ for which $F(I)\gt 0$. Without any loss of generality we may assume (by changing $X$ to $1-X$ if need be) that $F(J = I \cap (0, \mu]) \gt 0$: in other words, $J$ is obtained by cutting off any part of $I$ above the mean and $J$ has positive probability.
Let us alter $F$ to $F'$ by taking all the probability out of $J$ and placing it at $0$. In so doing, $\mu_k$ changes to
$$\mu'_k = \mu_k - \int_J x^k dF(x).$$
As a matter of notation, let us write $[g(x)] = \int_J g(x) dF(x)$ for such integrals, whence
$$\mu'_2 = \mu_2 - [x^2], \quad \mu' = \mu - [x].$$
Calculate
$$\sigma'^2 = \mu'_2 - \mu'^2 = \mu_2 - [x^2] - (\mu - [x])^2 = \sigma^2 + \left((\mu[x] - [x^2]) + (\mu[x] - [x]^2)\right).$$
The second term on the right, $(\mu[x] - [x]^2)$, is non-negative because $\mu \ge x$ everywhere on $J$. The first term on the right can be rewritten
$$\mu[x] - [x^2] = \mu(1 - [1]) + ([\mu][x] - [x^2]).$$
The first term on the right is strictly positive because (a) $\mu \gt 0$ and (b) $[1] = F(J) \lt 1$ because we assumed $F$ is not concentrated at a point. The second term is non-negative because it can be rewritten as $[(\mu-x)(x)]$ and this integrand is nonnegative from the assumptions $\mu \ge x$ on $J$ and $0 \le x \le 1$. It follows that $\sigma'^2 - \sigma^2 \gt 0$.
We have just shown that under our assumptions, changing $F$ to $F'$ strictly increases its variance. The only way this cannot happen, then, is when all the probability of $F'$ is concentrated at the endpoints $0$ and $1$, with (say) values $1-p$ and $p$, respectively. Its variance is easily calculated to equal $p(1-p)$ which is maximal when $p=1/2$ and equals $1/4$ there.
Now when $F$ is a distribution on $[a,b]$, we recenter and rescale it to a distribution on $[0,1]$. The recentering does not change the variance whereas the rescaling divides it by $(b-a)^2$. Thus an $F$ with maximal variance on $[a,b]$ corresponds to the distribution with maximal variance on $[0,1]$: it therefore is a Bernoulli$(1/2)$ distribution rescaled and translated to $[a,b]$ having variance $(b-a)^2/4$, QED.
|
Variance of a bounded random variable
Let $F$ be a distribution on $[0,1]$. We will show that if the variance of $F$ is maximal, then $F$ can have no support in the interior, from which it follows that $F$ is Bernoulli and the rest is tr
|
6,710
|
Variance of a bounded random variable
|
At @user603's request....
A useful upper bound on the variance $\sigma^2$ of a random variable that takes on values in $[a,b]$ with probability $1$ is $\sigma^2 \leq \frac{(b−a)^2}{4}$. A proof for the
special case $a=0, b=1$ (which is what the OP asked about) can be found
here on math.SE, and
it is easily adapted to
the more general case. As noted in my comment above and also in the answer referenced
herein, a discrete random variable that takes on values $a$ and $b$ with equal
probability $\frac{1}{2}$ has variance $\frac{(b−a)^2}{4}$ and thus no tighter
general bound can be found.
Another point to keep in mind is that a bounded random variable has finite
variance, whereas for an unbounded random variable, the variance might not
be finite, and in some cases might not even be definable. For example,
the mean cannot be defined for Cauchy random variables,
and so one cannot
define the variance (as the expectation of the squared deviation from the mean).
|
Variance of a bounded random variable
|
At @user603's request....
A useful upper bound on the variance $\sigma^2$ of a random variable that takes on values in $[a,b]$ with probability $1$ is $\sigma^2 \leq \frac{(b−a)^2}{4}$. A proof for th
|
Variance of a bounded random variable
At @user603's request....
A useful upper bound on the variance $\sigma^2$ of a random variable that takes on values in $[a,b]$ with probability $1$ is $\sigma^2 \leq \frac{(b−a)^2}{4}$. A proof for the
special case $a=0, b=1$ (which is what the OP asked about) can be found
here on math.SE, and
it is easily adapted to
the more general case. As noted in my comment above and also in the answer referenced
herein, a discrete random variable that takes on values $a$ and $b$ with equal
probability $\frac{1}{2}$ has variance $\frac{(b−a)^2}{4}$ and thus no tighter
general bound can be found.
Another point to keep in mind is that a bounded random variable has finite
variance, whereas for an unbounded random variable, the variance might not
be finite, and in some cases might not even be definable. For example,
the mean cannot be defined for Cauchy random variables,
and so one cannot
define the variance (as the expectation of the squared deviation from the mean).
|
Variance of a bounded random variable
At @user603's request....
A useful upper bound on the variance $\sigma^2$ of a random variable that takes on values in $[a,b]$ with probability $1$ is $\sigma^2 \leq \frac{(b−a)^2}{4}$. A proof for th
|
6,711
|
Variance of a bounded random variable
|
Here's a really simple proof I found in Sheldon Ross's A first course in probability 10th ed., "theory problem 5.8".
Suppose we have a random variable $X$ between $0$ and $c$. Then $X \leq c$ and thus $E(X^2) \leq c E(X)$. We thus have
$$\mathrm{Var}(X) \leq c E(X) - E(X)^2 = c^2 \alpha (1 - \alpha) \leq c^2 / 4$$
where $\alpha \equiv E(X) / c$ and the final step follows from straightforward differentiation to see that $\alpha (1 - \alpha)$ is minimized at $\alpha = 1/2$.
|
Variance of a bounded random variable
|
Here's a really simple proof I found in Sheldon Ross's A first course in probability 10th ed., "theory problem 5.8".
Suppose we have a random variable $X$ between $0$ and $c$. Then $X \leq c$ and thus
|
Variance of a bounded random variable
Here's a really simple proof I found in Sheldon Ross's A first course in probability 10th ed., "theory problem 5.8".
Suppose we have a random variable $X$ between $0$ and $c$. Then $X \leq c$ and thus $E(X^2) \leq c E(X)$. We thus have
$$\mathrm{Var}(X) \leq c E(X) - E(X)^2 = c^2 \alpha (1 - \alpha) \leq c^2 / 4$$
where $\alpha \equiv E(X) / c$ and the final step follows from straightforward differentiation to see that $\alpha (1 - \alpha)$ is minimized at $\alpha = 1/2$.
|
Variance of a bounded random variable
Here's a really simple proof I found in Sheldon Ross's A first course in probability 10th ed., "theory problem 5.8".
Suppose we have a random variable $X$ between $0$ and $c$. Then $X \leq c$ and thus
|
6,712
|
Variance of a bounded random variable
|
Given random variable $D$ with mean $E D=\mu$, when $a\le D\le b$, we have
\begin{eqnarray*}
E (D-\mu)^2&\le& E[(D-\mu)^2-(D-a)(D-b)]\\
&=& E[\mu^2-2\mu D+(a+b)D-ab]\\
&=& (a+b)\mu-\mu^2-ab\\
&=&(\mu-a)(b-\mu),
\end{eqnarray*}
where the equation holds if and only if $D\in\{a,b\}$ with probability $1$.
Moreover, if $\mu$ can be arbitrary in the support set $[a,b]$, then clearly
$$E(D-\mu)^2\le \left(\tfrac{a+b}{2}-a\right)\left(\tfrac{a+b}{2}-b\right)=\left(\tfrac{b-a}{2}\right)^2.$$
|
Variance of a bounded random variable
|
Given random variable $D$ with mean $E D=\mu$, when $a\le D\le b$, we have
\begin{eqnarray*}
E (D-\mu)^2&\le& E[(D-\mu)^2-(D-a)(D-b)]\\
&=& E[\mu^2-2\mu D+(a+b)D-ab]\\
&=& (a+b)\mu-\mu^2-ab\\
&=&(\mu-
|
Variance of a bounded random variable
Given random variable $D$ with mean $E D=\mu$, when $a\le D\le b$, we have
\begin{eqnarray*}
E (D-\mu)^2&\le& E[(D-\mu)^2-(D-a)(D-b)]\\
&=& E[\mu^2-2\mu D+(a+b)D-ab]\\
&=& (a+b)\mu-\mu^2-ab\\
&=&(\mu-a)(b-\mu),
\end{eqnarray*}
where the equation holds if and only if $D\in\{a,b\}$ with probability $1$.
Moreover, if $\mu$ can be arbitrary in the support set $[a,b]$, then clearly
$$E(D-\mu)^2\le \left(\tfrac{a+b}{2}-a\right)\left(\tfrac{a+b}{2}-b\right)=\left(\tfrac{b-a}{2}\right)^2.$$
|
Variance of a bounded random variable
Given random variable $D$ with mean $E D=\mu$, when $a\le D\le b$, we have
\begin{eqnarray*}
E (D-\mu)^2&\le& E[(D-\mu)^2-(D-a)(D-b)]\\
&=& E[\mu^2-2\mu D+(a+b)D-ab]\\
&=& (a+b)\mu-\mu^2-ab\\
&=&(\mu-
|
6,713
|
Variance of a bounded random variable
|
are you sure that this is true in general - for continuous as well as discrete distributions? Can you provide a link to the other pages?
For a general distibution on $[a,b]$ it is trivial to show that
$$
Var(X) = E[(X-E[X])^2] \le E[(b-a)^2] = (b-a)^2.
$$
I can imagine that sharper inequalities exist ...
Do you need the factor $1/4$ for your result?
On the other hand one can find it with the factor $1/4$ under the name Popoviciu's_inequality on wikipedia.
This article looks better than the wikipedia article ...
For a uniform distribution it holds that
$$
Var(X) = \frac{(b-a)^2}{12}.
$$
|
Variance of a bounded random variable
|
are you sure that this is true in general - for continuous as well as discrete distributions? Can you provide a link to the other pages?
For a general distibution on $[a,b]$ it is trivial to show that
|
Variance of a bounded random variable
are you sure that this is true in general - for continuous as well as discrete distributions? Can you provide a link to the other pages?
For a general distibution on $[a,b]$ it is trivial to show that
$$
Var(X) = E[(X-E[X])^2] \le E[(b-a)^2] = (b-a)^2.
$$
I can imagine that sharper inequalities exist ...
Do you need the factor $1/4$ for your result?
On the other hand one can find it with the factor $1/4$ under the name Popoviciu's_inequality on wikipedia.
This article looks better than the wikipedia article ...
For a uniform distribution it holds that
$$
Var(X) = \frac{(b-a)^2}{12}.
$$
|
Variance of a bounded random variable
are you sure that this is true in general - for continuous as well as discrete distributions? Can you provide a link to the other pages?
For a general distibution on $[a,b]$ it is trivial to show that
|
6,714
|
Variance of a bounded random variable
|
The key elements here are that $f(x) = x^2$ is convex, $EX$ minimises $E(X-t)^2$ and $X(\omega) \in [a,b]$.
Let $x \in [a,b]$, then
$f(x-{1 \over 2}(a+b)) \le {1 \over 2} (f({x-a \over 2}) + f({x-b \over 2}))$, or
$(x-{1 \over 2}(a+b))^2 \le {1 \over 2} (({x-a \over 2})^2 + ({x-b \over 2})^2 ) \le ({b-a \over 2})^2$.
Since $E(X-t)^2 = E(X-EX + EX-t)^2 = \operatorname{var} X + (EX-t)^2$ for any $t$, setting $t={1 \over 2}(a+b)$ shows that
$ \operatorname{var} X \le E(X-{1 \over 2}(a+b))^2 \le ({b-a \over 2})^2$.
|
Variance of a bounded random variable
|
The key elements here are that $f(x) = x^2$ is convex, $EX$ minimises $E(X-t)^2$ and $X(\omega) \in [a,b]$.
Let $x \in [a,b]$, then
$f(x-{1 \over 2}(a+b)) \le {1 \over 2} (f({x-a \over 2}) + f({x-b \o
|
Variance of a bounded random variable
The key elements here are that $f(x) = x^2$ is convex, $EX$ minimises $E(X-t)^2$ and $X(\omega) \in [a,b]$.
Let $x \in [a,b]$, then
$f(x-{1 \over 2}(a+b)) \le {1 \over 2} (f({x-a \over 2}) + f({x-b \over 2}))$, or
$(x-{1 \over 2}(a+b))^2 \le {1 \over 2} (({x-a \over 2})^2 + ({x-b \over 2})^2 ) \le ({b-a \over 2})^2$.
Since $E(X-t)^2 = E(X-EX + EX-t)^2 = \operatorname{var} X + (EX-t)^2$ for any $t$, setting $t={1 \over 2}(a+b)$ shows that
$ \operatorname{var} X \le E(X-{1 \over 2}(a+b))^2 \le ({b-a \over 2})^2$.
|
Variance of a bounded random variable
The key elements here are that $f(x) = x^2$ is convex, $EX$ minimises $E(X-t)^2$ and $X(\omega) \in [a,b]$.
Let $x \in [a,b]$, then
$f(x-{1 \over 2}(a+b)) \le {1 \over 2} (f({x-a \over 2}) + f({x-b \o
|
6,715
|
Variance of a bounded random variable
|
A simple proof of Popoviciu's inequality is the following, where $X\in [m, M]$:
\begin{equation}
\color{blue}{\operatorname{Var}[X] \le \operatorname{Var}[X] + \mathbb{E}[(M-X)(X-m)] = \frac{(M-m)^2}{4} - \left(\mathbb{E}[X] - \frac{M+m}{2}\right)^2 \le \frac{(M-m)^2}{4}.}
\end{equation}
Source: https://math.stackexchange.com/a/4264325/31498
(Check this answer out for a one-liner proof of Bhatia—Davis's inequality as well.)
|
Variance of a bounded random variable
|
A simple proof of Popoviciu's inequality is the following, where $X\in [m, M]$:
\begin{equation}
\color{blue}{\operatorname{Var}[X] \le \operatorname{Var}[X] + \mathbb{E}[(M-X)(X-m)] = \frac{(M-m)^2}{
|
Variance of a bounded random variable
A simple proof of Popoviciu's inequality is the following, where $X\in [m, M]$:
\begin{equation}
\color{blue}{\operatorname{Var}[X] \le \operatorname{Var}[X] + \mathbb{E}[(M-X)(X-m)] = \frac{(M-m)^2}{4} - \left(\mathbb{E}[X] - \frac{M+m}{2}\right)^2 \le \frac{(M-m)^2}{4}.}
\end{equation}
Source: https://math.stackexchange.com/a/4264325/31498
(Check this answer out for a one-liner proof of Bhatia—Davis's inequality as well.)
|
Variance of a bounded random variable
A simple proof of Popoviciu's inequality is the following, where $X\in [m, M]$:
\begin{equation}
\color{blue}{\operatorname{Var}[X] \le \operatorname{Var}[X] + \mathbb{E}[(M-X)(X-m)] = \frac{(M-m)^2}{
|
6,716
|
Why are survival times assumed to be exponentially distributed?
|
Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. This is because they are memoryless, and thus the hazard function is constant w/r/t time, which makes analysis very simple. This kind of assumption may be valid, for example, for some kinds of electronic components like high-quality integrated circuits. I'm sure you can think of more examples where the effect of time on hazard can safely be assumed to be negligible.
However, you are correct to observe that this would not be an appropriate assumption to make in many cases. Normal distributions can be alright in some situations, though obviously negative survival times are meaningless. For this reason, lognormal distributions are often considered. Other common choices include Weibull, Smallest Extreme Value, Largest Extreme Value, Log-logistic, etc. A sensible choice for model would be informed by subject-area experience and probability plotting. You can also, of course, consider non-parametric modeling.
A good reference for classical parametric modeling in survival analysis is:
William Q. Meeker and Luis A. Escobar (1998). Statistical Methods for
Reliability Data, Wiley
|
Why are survival times assumed to be exponentially distributed?
|
Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. This is because they are memo
|
Why are survival times assumed to be exponentially distributed?
Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. This is because they are memoryless, and thus the hazard function is constant w/r/t time, which makes analysis very simple. This kind of assumption may be valid, for example, for some kinds of electronic components like high-quality integrated circuits. I'm sure you can think of more examples where the effect of time on hazard can safely be assumed to be negligible.
However, you are correct to observe that this would not be an appropriate assumption to make in many cases. Normal distributions can be alright in some situations, though obviously negative survival times are meaningless. For this reason, lognormal distributions are often considered. Other common choices include Weibull, Smallest Extreme Value, Largest Extreme Value, Log-logistic, etc. A sensible choice for model would be informed by subject-area experience and probability plotting. You can also, of course, consider non-parametric modeling.
A good reference for classical parametric modeling in survival analysis is:
William Q. Meeker and Luis A. Escobar (1998). Statistical Methods for
Reliability Data, Wiley
|
Why are survival times assumed to be exponentially distributed?
Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. This is because they are memo
|
6,717
|
Why are survival times assumed to be exponentially distributed?
|
To add a bit of mathematical intuition behind how exponents pop up in survival distributions:
The probability density of a survival variable is $f(t) = h(t)S(t)$, where $h(t)$ is the current hazard (risk for a person to "die" this day) and $S(t)$ is the probability that a person survived until $t$. $S(t)$ can be expanded as the probability that a person survived day 1, and survived day 2, ... up to day $t$. Then:
$$ P(survived\ day\ t)=1-h(t)$$
$$ P(survived\ days\ 1, 2, ..., t) = (1-h(t))^t$$
With constant and small hazard $\lambda$, we can use:
$$ e^{-\lambda} \approx 1-\lambda$$
to approximate $S(t)$ as simply
$$ (1-\lambda)^t \approx e^{-\lambda t} $$
, and the probability density is then
$$ f(t) = h(t)S(t) = \lambda e^{-\lambda t}$$
Disclaimer: this is in no way an attempt at a proper derivation of the pdf - I just figured this is a neat coincidence, and welcome any comments on why this is correct/incorrect.
EDIT: changed the approximation per advice by @SamT, see comments for discussion.
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Why are survival times assumed to be exponentially distributed?
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To add a bit of mathematical intuition behind how exponents pop up in survival distributions:
The probability density of a survival variable is $f(t) = h(t)S(t)$, where $h(t)$ is the current hazard
|
Why are survival times assumed to be exponentially distributed?
To add a bit of mathematical intuition behind how exponents pop up in survival distributions:
The probability density of a survival variable is $f(t) = h(t)S(t)$, where $h(t)$ is the current hazard (risk for a person to "die" this day) and $S(t)$ is the probability that a person survived until $t$. $S(t)$ can be expanded as the probability that a person survived day 1, and survived day 2, ... up to day $t$. Then:
$$ P(survived\ day\ t)=1-h(t)$$
$$ P(survived\ days\ 1, 2, ..., t) = (1-h(t))^t$$
With constant and small hazard $\lambda$, we can use:
$$ e^{-\lambda} \approx 1-\lambda$$
to approximate $S(t)$ as simply
$$ (1-\lambda)^t \approx e^{-\lambda t} $$
, and the probability density is then
$$ f(t) = h(t)S(t) = \lambda e^{-\lambda t}$$
Disclaimer: this is in no way an attempt at a proper derivation of the pdf - I just figured this is a neat coincidence, and welcome any comments on why this is correct/incorrect.
EDIT: changed the approximation per advice by @SamT, see comments for discussion.
|
Why are survival times assumed to be exponentially distributed?
To add a bit of mathematical intuition behind how exponents pop up in survival distributions:
The probability density of a survival variable is $f(t) = h(t)S(t)$, where $h(t)$ is the current hazard
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6,718
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Why are survival times assumed to be exponentially distributed?
|
You'll almost certainly want to look at reliability engineering and predictions for thorough analyses of survival times. Within that, there are a few distributions which get used often:
The Weibull (or "bathtub") distribution is the most complex. It accounts for three types of failure modes, which dominate at different ages: infant mortality (where defective parts break early on), induced failures (where parts break randomly throughout the life of the system), and wear out (where parts break down from use). As used, it has a PDF which looks like "\__/". For some electronics especially, you might hear about "burn in" times, which means those parts have already been operated through the "\" part of the curve, and early failures have been screened out (ideally). Unfortunately, Weibull analysis breaks down fast if your parts aren't homogeneous (including use environment!) or if you are using them at different time scales (e.g. if some parts go directly into use, and other parts go into storage first, the "random failure" rate is going to be significantly different, due to blending two measurements of time (operating hours vs. use hours).
Normal distributions are almost always wrong. Every normal distribution has negative values, no reliability distribution does. They can sometimes be a useful approximation, but the times when that's true, you're almost always looking at a log-normal anyway, so you may as well just use the right distribution. Log-normal distributions are correctly used when you have some sort of wear-out and negligible random failures, and in no other circumstances! Like the Normal distribution, they're flexible enough that you can force them to fit most data; you need to resist that urge and check that the circumstances make sense.
Finally, the exponential distribution is the real workhorse. You often don't know how old parts are (for example, when parts aren't serialized and have different times when they entered into service), so any memory-based distribution is out. Additionally, many parts have a wearout time that is so arbitrarily long that it's either completely dominated by induced failures or outside the useful time-frame of the analysis. So while it may not be as perfect a model as other distributions, it just doesn't care about things which trip them up. If you have an MTTF (population time/failure count), you have an exponential distribution. On top of that, you don't need any physical understanding of your system. You can do exponential estimates just based on observed part MTTFs (assuming a large enough sample), and they come out pretty dang close. It's also resilient to causes: if every other month, someone gets bored and plays croquet with some part until it breaks, exponential accounts for that (it rolls into the MTTF). Exponential is also simple enough that you can do back-of-the-envelope calculations for availability of redundant systems and such, which significantly increases its usefulness.
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Why are survival times assumed to be exponentially distributed?
|
You'll almost certainly want to look at reliability engineering and predictions for thorough analyses of survival times. Within that, there are a few distributions which get used often:
The Weibull (
|
Why are survival times assumed to be exponentially distributed?
You'll almost certainly want to look at reliability engineering and predictions for thorough analyses of survival times. Within that, there are a few distributions which get used often:
The Weibull (or "bathtub") distribution is the most complex. It accounts for three types of failure modes, which dominate at different ages: infant mortality (where defective parts break early on), induced failures (where parts break randomly throughout the life of the system), and wear out (where parts break down from use). As used, it has a PDF which looks like "\__/". For some electronics especially, you might hear about "burn in" times, which means those parts have already been operated through the "\" part of the curve, and early failures have been screened out (ideally). Unfortunately, Weibull analysis breaks down fast if your parts aren't homogeneous (including use environment!) or if you are using them at different time scales (e.g. if some parts go directly into use, and other parts go into storage first, the "random failure" rate is going to be significantly different, due to blending two measurements of time (operating hours vs. use hours).
Normal distributions are almost always wrong. Every normal distribution has negative values, no reliability distribution does. They can sometimes be a useful approximation, but the times when that's true, you're almost always looking at a log-normal anyway, so you may as well just use the right distribution. Log-normal distributions are correctly used when you have some sort of wear-out and negligible random failures, and in no other circumstances! Like the Normal distribution, they're flexible enough that you can force them to fit most data; you need to resist that urge and check that the circumstances make sense.
Finally, the exponential distribution is the real workhorse. You often don't know how old parts are (for example, when parts aren't serialized and have different times when they entered into service), so any memory-based distribution is out. Additionally, many parts have a wearout time that is so arbitrarily long that it's either completely dominated by induced failures or outside the useful time-frame of the analysis. So while it may not be as perfect a model as other distributions, it just doesn't care about things which trip them up. If you have an MTTF (population time/failure count), you have an exponential distribution. On top of that, you don't need any physical understanding of your system. You can do exponential estimates just based on observed part MTTFs (assuming a large enough sample), and they come out pretty dang close. It's also resilient to causes: if every other month, someone gets bored and plays croquet with some part until it breaks, exponential accounts for that (it rolls into the MTTF). Exponential is also simple enough that you can do back-of-the-envelope calculations for availability of redundant systems and such, which significantly increases its usefulness.
|
Why are survival times assumed to be exponentially distributed?
You'll almost certainly want to look at reliability engineering and predictions for thorough analyses of survival times. Within that, there are a few distributions which get used often:
The Weibull (
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6,719
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Why are survival times assumed to be exponentially distributed?
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Some ecology might help answer the "Why" behind this question.
The reason why exponential distribution is used for modeling survival is due to the life strategies involved in organisms living in nature. There's essentially two extremes with regard to survival strategy with some room for the middle ground.
Here's an image that illustrates what I mean (courtesy of Khan Academy):
This graph plots surviving individuals on the Y axis, and "percentage of maximum life expectancy" (a.k.a. approximation of the individual's age) on the X axis.
Type I is humans, which model organisms which have an extreme level of care of their offspring ensuring very low infant mortality. Often these species have very few offspring because each one takes a large amount of the parents time and effort. The majority of what kills Type I organisms is the type of complications that arise in old age. The strategy here is high investment for high payoff in long, productive lives, if at the cost of sheer numbers.
Conversely, Type III is modeled by trees (but could also be plankton, corals, spawning fish, many types of insects, etc) where the parent invests relatively little in each offspring, but produces a ton of them in the hopes that a few will survive. The strategy here is "spray and pray" hoping that while most offspring will be destroyed relatively quickly by predators taking advantage of easy pickings, the few that survive long enough to grow will become increasingly difficult to kill, eventually becoming (practically) impossible to be eaten. All the while these individuals produce huge numbers of offspring hoping that a few will likewise survive to their own age.
Type II is a middling strategy with moderate parental investment for moderate survivability at all ages.
I had an ecology professor who put it this way:
"Type III (trees) is the 'Curve of Hope', because the longer an individual survives, the more likely it becomes that it will continue to survive. Meanwhile Type I (humans) is the 'Curve of Despair', because the longer you live, the more likely it becomes that you will die."
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Why are survival times assumed to be exponentially distributed?
|
Some ecology might help answer the "Why" behind this question.
The reason why exponential distribution is used for modeling survival is due to the life strategies involved in organisms living in natur
|
Why are survival times assumed to be exponentially distributed?
Some ecology might help answer the "Why" behind this question.
The reason why exponential distribution is used for modeling survival is due to the life strategies involved in organisms living in nature. There's essentially two extremes with regard to survival strategy with some room for the middle ground.
Here's an image that illustrates what I mean (courtesy of Khan Academy):
This graph plots surviving individuals on the Y axis, and "percentage of maximum life expectancy" (a.k.a. approximation of the individual's age) on the X axis.
Type I is humans, which model organisms which have an extreme level of care of their offspring ensuring very low infant mortality. Often these species have very few offspring because each one takes a large amount of the parents time and effort. The majority of what kills Type I organisms is the type of complications that arise in old age. The strategy here is high investment for high payoff in long, productive lives, if at the cost of sheer numbers.
Conversely, Type III is modeled by trees (but could also be plankton, corals, spawning fish, many types of insects, etc) where the parent invests relatively little in each offspring, but produces a ton of them in the hopes that a few will survive. The strategy here is "spray and pray" hoping that while most offspring will be destroyed relatively quickly by predators taking advantage of easy pickings, the few that survive long enough to grow will become increasingly difficult to kill, eventually becoming (practically) impossible to be eaten. All the while these individuals produce huge numbers of offspring hoping that a few will likewise survive to their own age.
Type II is a middling strategy with moderate parental investment for moderate survivability at all ages.
I had an ecology professor who put it this way:
"Type III (trees) is the 'Curve of Hope', because the longer an individual survives, the more likely it becomes that it will continue to survive. Meanwhile Type I (humans) is the 'Curve of Despair', because the longer you live, the more likely it becomes that you will die."
|
Why are survival times assumed to be exponentially distributed?
Some ecology might help answer the "Why" behind this question.
The reason why exponential distribution is used for modeling survival is due to the life strategies involved in organisms living in natur
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6,720
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Why are survival times assumed to be exponentially distributed?
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To answer your explicit question, you cannot use the normal distribution for survival because the normal distribution goes to negative infinity, and survival is strictly non-negative. Moreover, I don't think it's true that "survival times are assumed to be exponentially distributed" by anyone in reality.
When survival times are modeled parametrically (i.e., when any named distribution is invoked), the Weibull distribution is the typical starting place. Note that the Weibull has two parameters, shape and scale, and that when shape = 1, the Weibull simplifies to the exponential distribution. A way of thinking about this is that the exponential distribution the simplest possible parametric distribution for survival times, which is why it is often discussed first when survival analysis is being taught. (By analogy, consider that we often begin teaching hypothesis testing by going over the one-sample $z$-test, where we pretend to know the population SD a-priori, and then work up to the $t$-test.)
The exponential distribution assumes that the hazard is always exactly the same, no matter how long a unit has survived (consider the figure in @CaffeineConnoisseur's answer). In contrast, when the shape is $>1$ in the Weibull distribution, it implies that hazards increase the longer you survive (like the 'human curve'); and when it is $<1$, it implies hazards decrease (the 'tree').
Most commonly, survival distributions are complex and not well fit by any named distribution. People typically don't even bother trying to figure out what distribution it might be. That's what makes the Cox proportional hazards model so popular: it is semi-parametric in that the baseline hazard can be left completely unspecified but the rest of the model can be parametric in terms of its relationship to the unspecified baseline.
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Why are survival times assumed to be exponentially distributed?
|
To answer your explicit question, you cannot use the normal distribution for survival because the normal distribution goes to negative infinity, and survival is strictly non-negative. Moreover, I don
|
Why are survival times assumed to be exponentially distributed?
To answer your explicit question, you cannot use the normal distribution for survival because the normal distribution goes to negative infinity, and survival is strictly non-negative. Moreover, I don't think it's true that "survival times are assumed to be exponentially distributed" by anyone in reality.
When survival times are modeled parametrically (i.e., when any named distribution is invoked), the Weibull distribution is the typical starting place. Note that the Weibull has two parameters, shape and scale, and that when shape = 1, the Weibull simplifies to the exponential distribution. A way of thinking about this is that the exponential distribution the simplest possible parametric distribution for survival times, which is why it is often discussed first when survival analysis is being taught. (By analogy, consider that we often begin teaching hypothesis testing by going over the one-sample $z$-test, where we pretend to know the population SD a-priori, and then work up to the $t$-test.)
The exponential distribution assumes that the hazard is always exactly the same, no matter how long a unit has survived (consider the figure in @CaffeineConnoisseur's answer). In contrast, when the shape is $>1$ in the Weibull distribution, it implies that hazards increase the longer you survive (like the 'human curve'); and when it is $<1$, it implies hazards decrease (the 'tree').
Most commonly, survival distributions are complex and not well fit by any named distribution. People typically don't even bother trying to figure out what distribution it might be. That's what makes the Cox proportional hazards model so popular: it is semi-parametric in that the baseline hazard can be left completely unspecified but the rest of the model can be parametric in terms of its relationship to the unspecified baseline.
|
Why are survival times assumed to be exponentially distributed?
To answer your explicit question, you cannot use the normal distribution for survival because the normal distribution goes to negative infinity, and survival is strictly non-negative. Moreover, I don
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6,721
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Why are survival times assumed to be exponentially distributed?
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This doesn't directly answer the question, but I think it's very important to note, and does not fit nicely into a single comment.
While the exponential distribution has a very nice theoretical derivation, and thus assuming the data produced follows the mechanisms assumed in the exponential distribution, it should theoretically give optimal estimates, in practice I've yet to run into a dataset where the exponential
distribution produces even close to acceptable results (of course, this is dependent on the data types I've analyzed, almost all biological data). For example, I just looked at fitting a model with a variety of distributions using the first data set I could find in my R-package. For model checking of the baseline distribution, we typically compare against the semi-parametric model. Take a look at the results.
Of the Weibull, log-logistic and log-normal distribution, there's not an absolute clear victor in terms of appropriate fit. But there's a clear loser: the exponential distribution! It's been my experience that this magnitude of mis-fitting is not exceptional, but rather the norm for the exponential distribution.
Why? Because the exponential distribution is a single parameter family. Thus, if I specify the mean of this distribution, I've specified all other moments of the distribution. These other families are all two parameter families. Thus, there's a lot more flexibility in those families to adapt to the data itself.
Now keep in mind that the Weibull distribution has the exponential distribution as a special case (i.e. when the shape parameter = 1). So even if the data truly is exponential, we only add a little more noise to our estimates by using a Weibull distribution over an exponential distribution.
As such, I would just about never recommend using the exponential distribution to model real data (and I'm curious to hear if any readers have an example of when it's actually a good idea).
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Why are survival times assumed to be exponentially distributed?
|
This doesn't directly answer the question, but I think it's very important to note, and does not fit nicely into a single comment.
While the exponential distribution has a very nice theoretical deriv
|
Why are survival times assumed to be exponentially distributed?
This doesn't directly answer the question, but I think it's very important to note, and does not fit nicely into a single comment.
While the exponential distribution has a very nice theoretical derivation, and thus assuming the data produced follows the mechanisms assumed in the exponential distribution, it should theoretically give optimal estimates, in practice I've yet to run into a dataset where the exponential
distribution produces even close to acceptable results (of course, this is dependent on the data types I've analyzed, almost all biological data). For example, I just looked at fitting a model with a variety of distributions using the first data set I could find in my R-package. For model checking of the baseline distribution, we typically compare against the semi-parametric model. Take a look at the results.
Of the Weibull, log-logistic and log-normal distribution, there's not an absolute clear victor in terms of appropriate fit. But there's a clear loser: the exponential distribution! It's been my experience that this magnitude of mis-fitting is not exceptional, but rather the norm for the exponential distribution.
Why? Because the exponential distribution is a single parameter family. Thus, if I specify the mean of this distribution, I've specified all other moments of the distribution. These other families are all two parameter families. Thus, there's a lot more flexibility in those families to adapt to the data itself.
Now keep in mind that the Weibull distribution has the exponential distribution as a special case (i.e. when the shape parameter = 1). So even if the data truly is exponential, we only add a little more noise to our estimates by using a Weibull distribution over an exponential distribution.
As such, I would just about never recommend using the exponential distribution to model real data (and I'm curious to hear if any readers have an example of when it's actually a good idea).
|
Why are survival times assumed to be exponentially distributed?
This doesn't directly answer the question, but I think it's very important to note, and does not fit nicely into a single comment.
While the exponential distribution has a very nice theoretical deriv
|
6,722
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Why are survival times assumed to be exponentially distributed?
|
Another reason why the exponential distribution crops up often to model interval between events is the following.
It is well known that, under some assumptions, the sum of a large number of independent random variables will be close to a Gaussian distribution. A similar theorem holds for renewal processes, i.e. stochastic models for events that occur randomly in time with I.I.D. inter-event intervals. In fact, the Palm–Khintchine theorem states that the superposition of a large number of (not necessarily Poissonian) renewal processes behaves asymptotically like a Poisson process. The inter-event intervals of a Poisson process are exponentially distributed.
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Why are survival times assumed to be exponentially distributed?
|
Another reason why the exponential distribution crops up often to model interval between events is the following.
It is well known that, under some assumptions, the sum of a large number of independen
|
Why are survival times assumed to be exponentially distributed?
Another reason why the exponential distribution crops up often to model interval between events is the following.
It is well known that, under some assumptions, the sum of a large number of independent random variables will be close to a Gaussian distribution. A similar theorem holds for renewal processes, i.e. stochastic models for events that occur randomly in time with I.I.D. inter-event intervals. In fact, the Palm–Khintchine theorem states that the superposition of a large number of (not necessarily Poissonian) renewal processes behaves asymptotically like a Poisson process. The inter-event intervals of a Poisson process are exponentially distributed.
|
Why are survival times assumed to be exponentially distributed?
Another reason why the exponential distribution crops up often to model interval between events is the following.
It is well known that, under some assumptions, the sum of a large number of independen
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6,723
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Why are survival times assumed to be exponentially distributed?
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tl;dr- An expontential distribution is equivalent to assuming that individuals are as likely to die at any given moment as any other.
Derivation
Assume that a living individual is as likely to die at any given moment as at any other.
So, the death rate $-\frac{\text{d}P}{\text{d}t}$ is proportional to the population, $P$.
$$-\frac{\text{d}P}{\text{d}t}{\space}{\propto}{\space}P$$
Solving on WolframAlpha shows:
$$P\left(t\right)={c_1}{e^{-t}}$$
So, the population follows an exponential distribution.
Math note
The above math is a reduction of a first-order ordinary differential equation (ODE). Normally, we would also solve for $c_0$ by noting the boundary condition that population starts at some given value, $P\left(t_0\right)$, at start-time $t_0$.
Then the equation becomes:
$$P\left(t\right)={e^{-t}}P\left({t_0}\right).$$
Reality check
The exponential distribution assumes that people in the population tend to die at the same rate over time. In reality, death rates will tend to vary for finite populations.
Coming up with better distributions involves stochastic differential equations. Then, we can't say that there's a constant death likelihood; rather, we have to come up with a distribution for each individual's odds of dying at any given moment, then combine those various possibility trees together for the entire population, then solve that differential equation over time.
I can't recall having seen this done in anything online before, so you probably won't run into it; but, that's the next modeling step if you want to improve upon the exponential distribution.
|
Why are survival times assumed to be exponentially distributed?
|
tl;dr- An expontential distribution is equivalent to assuming that individuals are as likely to die at any given moment as any other.
Derivation
Assume that a living individual is as likely to die a
|
Why are survival times assumed to be exponentially distributed?
tl;dr- An expontential distribution is equivalent to assuming that individuals are as likely to die at any given moment as any other.
Derivation
Assume that a living individual is as likely to die at any given moment as at any other.
So, the death rate $-\frac{\text{d}P}{\text{d}t}$ is proportional to the population, $P$.
$$-\frac{\text{d}P}{\text{d}t}{\space}{\propto}{\space}P$$
Solving on WolframAlpha shows:
$$P\left(t\right)={c_1}{e^{-t}}$$
So, the population follows an exponential distribution.
Math note
The above math is a reduction of a first-order ordinary differential equation (ODE). Normally, we would also solve for $c_0$ by noting the boundary condition that population starts at some given value, $P\left(t_0\right)$, at start-time $t_0$.
Then the equation becomes:
$$P\left(t\right)={e^{-t}}P\left({t_0}\right).$$
Reality check
The exponential distribution assumes that people in the population tend to die at the same rate over time. In reality, death rates will tend to vary for finite populations.
Coming up with better distributions involves stochastic differential equations. Then, we can't say that there's a constant death likelihood; rather, we have to come up with a distribution for each individual's odds of dying at any given moment, then combine those various possibility trees together for the entire population, then solve that differential equation over time.
I can't recall having seen this done in anything online before, so you probably won't run into it; but, that's the next modeling step if you want to improve upon the exponential distribution.
|
Why are survival times assumed to be exponentially distributed?
tl;dr- An expontential distribution is equivalent to assuming that individuals are as likely to die at any given moment as any other.
Derivation
Assume that a living individual is as likely to die a
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6,724
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Why are survival times assumed to be exponentially distributed?
|
(Note that in the part you quoted, the statement was conditional; the sentence itself didn't assume exponential survival, it explained a consequence of doing so. Nevertheless assumption of exponential survival are common, so it's worth dealing with the question of "why exponential" and "why not normal" -- since the first is pretty well covered already I'll focus more on the second thing)
Normally distributed survival times don't make sense because they have a non-zero probability of the survival time being negative.
If you then restrict your consideration to normal distributions that have almost no chance of being near zero, you can't model survival data that has a reasonable probability of a short survival time:
Maybe once in a while survival times which have almost no chance of short survival times would be reasonable, but you need distributions that make sense in practice -- usually you observe short and long survival times (and anything in between), with typically a skewed distribution of survival times). An unmodified normal distribution will rarely be useful in practice.
[A truncated normal might more often be a reasonable rough approximation than a normal, but other distributions will often do better.]
The constant-hazard of the exponential is sometimes a reasonable approximation for survival times.. For example, if "random events" like accident are a major contributor to death-rate, exponential survival will work fairly well. (Among animal populations for example, sometimes both predation and disease can act at least roughly like a chance process, leaving something like an exponential as a reasonable first approximation to survival times.)
One additional question related truncated normal: if normal is not appropriate why not normal squared (chi sq with df 1)?
Indeed that might be a little better ... but note that that would correspond to an infinite hazard at 0, so it would only occasionally be useful. While it can model cases with a very high proportion of very short times, it has the converse problem of only being able to model cases with typically much shorter than average survival (25% of survival times are below 10.15% of the mean survival time and half of the survival times are less than 45.5% of the mean; that is median survival is less than half the mean.)
Let's look at a scaled $χ^2_1$ (i.e. a gamma with shape parameter $\frac12$):
[Maybe if you sum two of those $χ^2_1$ variates... or maybe if you considered noncentral $χ^2$ you would get some suitable possibilities. Outside of the exponential, common choices of parametric distributions for survival times include Weibull, lognormal, gamma, log-logistic among many others ... note that the Weibull and the gamma include the exponential as a special case]
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Why are survival times assumed to be exponentially distributed?
|
(Note that in the part you quoted, the statement was conditional; the sentence itself didn't assume exponential survival, it explained a consequence of doing so. Nevertheless assumption of exponential
|
Why are survival times assumed to be exponentially distributed?
(Note that in the part you quoted, the statement was conditional; the sentence itself didn't assume exponential survival, it explained a consequence of doing so. Nevertheless assumption of exponential survival are common, so it's worth dealing with the question of "why exponential" and "why not normal" -- since the first is pretty well covered already I'll focus more on the second thing)
Normally distributed survival times don't make sense because they have a non-zero probability of the survival time being negative.
If you then restrict your consideration to normal distributions that have almost no chance of being near zero, you can't model survival data that has a reasonable probability of a short survival time:
Maybe once in a while survival times which have almost no chance of short survival times would be reasonable, but you need distributions that make sense in practice -- usually you observe short and long survival times (and anything in between), with typically a skewed distribution of survival times). An unmodified normal distribution will rarely be useful in practice.
[A truncated normal might more often be a reasonable rough approximation than a normal, but other distributions will often do better.]
The constant-hazard of the exponential is sometimes a reasonable approximation for survival times.. For example, if "random events" like accident are a major contributor to death-rate, exponential survival will work fairly well. (Among animal populations for example, sometimes both predation and disease can act at least roughly like a chance process, leaving something like an exponential as a reasonable first approximation to survival times.)
One additional question related truncated normal: if normal is not appropriate why not normal squared (chi sq with df 1)?
Indeed that might be a little better ... but note that that would correspond to an infinite hazard at 0, so it would only occasionally be useful. While it can model cases with a very high proportion of very short times, it has the converse problem of only being able to model cases with typically much shorter than average survival (25% of survival times are below 10.15% of the mean survival time and half of the survival times are less than 45.5% of the mean; that is median survival is less than half the mean.)
Let's look at a scaled $χ^2_1$ (i.e. a gamma with shape parameter $\frac12$):
[Maybe if you sum two of those $χ^2_1$ variates... or maybe if you considered noncentral $χ^2$ you would get some suitable possibilities. Outside of the exponential, common choices of parametric distributions for survival times include Weibull, lognormal, gamma, log-logistic among many others ... note that the Weibull and the gamma include the exponential as a special case]
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Why are survival times assumed to be exponentially distributed?
(Note that in the part you quoted, the statement was conditional; the sentence itself didn't assume exponential survival, it explained a consequence of doing so. Nevertheless assumption of exponential
|
6,725
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Why are survival times assumed to be exponentially distributed?
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If we want time to be strictly positive, why not make normal distribution with higher mean and very small variance (will have almost no chance to get negative number.)?
Because
that still has a nonzero probability of being negative, so it's not strictly positive;
the mean and variance are something that you can measure from the population you're trying to model. If your population has mean 2 and variance 1, and you model it with a normal distribution, that normal distribution will have substantial mass below zero; if you model it with a normal distribution with mean 5 and variance 0.1, your model obviously has very differnt properties to the thing it's supposed to model.
The normal distribution has a particular shape, and that shape is symmetrical about the mean. The only way to adjust the shape are to move it right and left (increase or decrease the mean) or to make it more or less spread out (increase or decrease the variance). This means that the only way to get a normal distribution where most of the mass is between two and ten and only a tiny amount of the mass is below zero, you need to put your mean at, say, six (the middle of the range) and set the variance small enough that only a tiny fraction of samples are negative. But then you'll probably find that most of your samples are 5, 6 or 7, whereas you were supposed to have quite a lot of 2s, 3s, 4s, 8s, 9s and 10s.
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Why are survival times assumed to be exponentially distributed?
|
If we want time to be strictly positive, why not make normal distribution with higher mean and very small variance (will have almost no chance to get negative number.)?
Because
that still has a non
|
Why are survival times assumed to be exponentially distributed?
If we want time to be strictly positive, why not make normal distribution with higher mean and very small variance (will have almost no chance to get negative number.)?
Because
that still has a nonzero probability of being negative, so it's not strictly positive;
the mean and variance are something that you can measure from the population you're trying to model. If your population has mean 2 and variance 1, and you model it with a normal distribution, that normal distribution will have substantial mass below zero; if you model it with a normal distribution with mean 5 and variance 0.1, your model obviously has very differnt properties to the thing it's supposed to model.
The normal distribution has a particular shape, and that shape is symmetrical about the mean. The only way to adjust the shape are to move it right and left (increase or decrease the mean) or to make it more or less spread out (increase or decrease the variance). This means that the only way to get a normal distribution where most of the mass is between two and ten and only a tiny amount of the mass is below zero, you need to put your mean at, say, six (the middle of the range) and set the variance small enough that only a tiny fraction of samples are negative. But then you'll probably find that most of your samples are 5, 6 or 7, whereas you were supposed to have quite a lot of 2s, 3s, 4s, 8s, 9s and 10s.
|
Why are survival times assumed to be exponentially distributed?
If we want time to be strictly positive, why not make normal distribution with higher mean and very small variance (will have almost no chance to get negative number.)?
Because
that still has a non
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6,726
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Is there a name for the opposite of the gambler's fallacy?
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That seems to be a typical example of the Ludic Fallacy: https://en.wikipedia.org/wiki/Ludic_fallacy#Example:_Suspicious_coin
|
Is there a name for the opposite of the gambler's fallacy?
|
That seems to be a typical example of the Ludic Fallacy: https://en.wikipedia.org/wiki/Ludic_fallacy#Example:_Suspicious_coin
|
Is there a name for the opposite of the gambler's fallacy?
That seems to be a typical example of the Ludic Fallacy: https://en.wikipedia.org/wiki/Ludic_fallacy#Example:_Suspicious_coin
|
Is there a name for the opposite of the gambler's fallacy?
That seems to be a typical example of the Ludic Fallacy: https://en.wikipedia.org/wiki/Ludic_fallacy#Example:_Suspicious_coin
|
6,727
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Is there a name for the opposite of the gambler's fallacy?
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N Blake's answer is great, using the language of cognitive psychology. If you're up for a Bayesian slant, you could treat this as violating Cromwell's rule.
You have a prior certain belief that the coin's probability of giving tails is $f=0.5$. In other words, you believe that $p(f=0.5) = 1$, which is the problem. This gives you no flexibility to update this as the experiment continues. Your posterior $p(f \mid \text{100 heads})$ is now proportional to
$$
\cases{
p(\mathrm{H} \mid f)^{100} \times 1 & if $f=0.5$ \\
p(\mathrm{H} \mid f)^{100} \times 0 \color{red}{= 0} & otherwise
}
$$
Despite this long run of heads, your posterior remains unchanged from the prior! You didn't update your mental model of the coin despite the consilience of evidence against it being a fair coin.
|
Is there a name for the opposite of the gambler's fallacy?
|
N Blake's answer is great, using the language of cognitive psychology. If you're up for a Bayesian slant, you could treat this as violating Cromwell's rule.
You have a prior certain belief that the co
|
Is there a name for the opposite of the gambler's fallacy?
N Blake's answer is great, using the language of cognitive psychology. If you're up for a Bayesian slant, you could treat this as violating Cromwell's rule.
You have a prior certain belief that the coin's probability of giving tails is $f=0.5$. In other words, you believe that $p(f=0.5) = 1$, which is the problem. This gives you no flexibility to update this as the experiment continues. Your posterior $p(f \mid \text{100 heads})$ is now proportional to
$$
\cases{
p(\mathrm{H} \mid f)^{100} \times 1 & if $f=0.5$ \\
p(\mathrm{H} \mid f)^{100} \times 0 \color{red}{= 0} & otherwise
}
$$
Despite this long run of heads, your posterior remains unchanged from the prior! You didn't update your mental model of the coin despite the consilience of evidence against it being a fair coin.
|
Is there a name for the opposite of the gambler's fallacy?
N Blake's answer is great, using the language of cognitive psychology. If you're up for a Bayesian slant, you could treat this as violating Cromwell's rule.
You have a prior certain belief that the co
|
6,728
|
Is there a name for the opposite of the gambler's fallacy?
|
Belief perseverance seems to fit - maintaining a belief (that it's a fair coin), despite contrary evidence (100/100 heads).
|
Is there a name for the opposite of the gambler's fallacy?
|
Belief perseverance seems to fit - maintaining a belief (that it's a fair coin), despite contrary evidence (100/100 heads).
|
Is there a name for the opposite of the gambler's fallacy?
Belief perseverance seems to fit - maintaining a belief (that it's a fair coin), despite contrary evidence (100/100 heads).
|
Is there a name for the opposite of the gambler's fallacy?
Belief perseverance seems to fit - maintaining a belief (that it's a fair coin), despite contrary evidence (100/100 heads).
|
6,729
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Is there a name for the opposite of the gambler's fallacy?
|
It's called a strong prior.
The more enduring your belief in the coin's fairness is, the stronger the prior would be said to be. Might call it an absolute prior if it'd never change.
It's not necessarily a fallacy so much as an approximation, especially if the prior is strong enough that it'd be difficult to meaningfully distinguish its strength from absolution.
|
Is there a name for the opposite of the gambler's fallacy?
|
It's called a strong prior.
The more enduring your belief in the coin's fairness is, the stronger the prior would be said to be. Might call it an absolute prior if it'd never change.
It's not necessa
|
Is there a name for the opposite of the gambler's fallacy?
It's called a strong prior.
The more enduring your belief in the coin's fairness is, the stronger the prior would be said to be. Might call it an absolute prior if it'd never change.
It's not necessarily a fallacy so much as an approximation, especially if the prior is strong enough that it'd be difficult to meaningfully distinguish its strength from absolution.
|
Is there a name for the opposite of the gambler's fallacy?
It's called a strong prior.
The more enduring your belief in the coin's fairness is, the stronger the prior would be said to be. Might call it an absolute prior if it'd never change.
It's not necessa
|
6,730
|
Is there a name for the opposite of the gambler's fallacy?
|
This belief was examined in a series of papers on the "gambler's fallacy" and broader methods of binomial prediction under the Bayesian paradigm (see O'Neill and Puza 2005; O'Neill 2012; O'Neill 2015). These papers argue in favour of your view here – that observing more heads in a series of coin-flips should shift your belief somewhat towards having more heads in the future. Those papers referred to this (correct) belief as the "frequent outcome approach", and noted (just as you have) that the persistent belief in fairness is problematic. Those papers do not give any name to the latter error, but I agree with some of the names suggested by other commentators ("stubbornness", "belief persistence", etc.).
|
Is there a name for the opposite of the gambler's fallacy?
|
This belief was examined in a series of papers on the "gambler's fallacy" and broader methods of binomial prediction under the Bayesian paradigm (see O'Neill and Puza 2005; O'Neill 2012; O'Neill 2015)
|
Is there a name for the opposite of the gambler's fallacy?
This belief was examined in a series of papers on the "gambler's fallacy" and broader methods of binomial prediction under the Bayesian paradigm (see O'Neill and Puza 2005; O'Neill 2012; O'Neill 2015). These papers argue in favour of your view here – that observing more heads in a series of coin-flips should shift your belief somewhat towards having more heads in the future. Those papers referred to this (correct) belief as the "frequent outcome approach", and noted (just as you have) that the persistent belief in fairness is problematic. Those papers do not give any name to the latter error, but I agree with some of the names suggested by other commentators ("stubbornness", "belief persistence", etc.).
|
Is there a name for the opposite of the gambler's fallacy?
This belief was examined in a series of papers on the "gambler's fallacy" and broader methods of binomial prediction under the Bayesian paradigm (see O'Neill and Puza 2005; O'Neill 2012; O'Neill 2015)
|
6,731
|
Is there a name for the opposite of the gambler's fallacy?
|
I would call that a Type II error, which is anytime in science you fail to reject a null hypothesis even though it is false. If you had no other reason to think the coin is fair other than the flips you've seen, you could calculate draw a conclusion on evidence alone/ If you conclude it is fair just because you think you have insufficient evidence to conclude the proportion is different than 50/50 in the infinite limit, and if it is in fact not 50/50, that is a Type II error.
|
Is there a name for the opposite of the gambler's fallacy?
|
I would call that a Type II error, which is anytime in science you fail to reject a null hypothesis even though it is false. If you had no other reason to think the coin is fair other than the flips y
|
Is there a name for the opposite of the gambler's fallacy?
I would call that a Type II error, which is anytime in science you fail to reject a null hypothesis even though it is false. If you had no other reason to think the coin is fair other than the flips you've seen, you could calculate draw a conclusion on evidence alone/ If you conclude it is fair just because you think you have insufficient evidence to conclude the proportion is different than 50/50 in the infinite limit, and if it is in fact not 50/50, that is a Type II error.
|
Is there a name for the opposite of the gambler's fallacy?
I would call that a Type II error, which is anytime in science you fail to reject a null hypothesis even though it is false. If you had no other reason to think the coin is fair other than the flips y
|
6,732
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Should parsimony really still be the gold standard?
|
@Matt's original answer does a great job of describing one of the benefits of parsimony but I don't think it actually answers your question. In reality, parsimony isn't the gold standard. Not now nor has it ever been. A "gold standard" related to parsimony is generalization error. We would like to develop models that don't overfit. That are as useful for prediction (or as interpretable or with minimum error) out of sample as they are in sample. It turns out (because of things laid out above) that parsimony is actually quite a good proxy for generalization error but it's by no means the only one.
Really, think about why we use cross validation or bootstrapping or train/test sets. The goal is to create models with good generalization accuracy. A lot of the time, these ways of estimating out of sample performance do end up choosing models with lower complexity but not always. As an extreme example imagine the oracle hands us the true but extremely complex model and a poor but parsimonious model. If parsimony was really our goal then we would choose the second but in reality, the first is what we would like to learn if we could. Unfortunately a lot of the time that last sentence is the kicker, "if we could".
|
Should parsimony really still be the gold standard?
|
@Matt's original answer does a great job of describing one of the benefits of parsimony but I don't think it actually answers your question. In reality, parsimony isn't the gold standard. Not now nor
|
Should parsimony really still be the gold standard?
@Matt's original answer does a great job of describing one of the benefits of parsimony but I don't think it actually answers your question. In reality, parsimony isn't the gold standard. Not now nor has it ever been. A "gold standard" related to parsimony is generalization error. We would like to develop models that don't overfit. That are as useful for prediction (or as interpretable or with minimum error) out of sample as they are in sample. It turns out (because of things laid out above) that parsimony is actually quite a good proxy for generalization error but it's by no means the only one.
Really, think about why we use cross validation or bootstrapping or train/test sets. The goal is to create models with good generalization accuracy. A lot of the time, these ways of estimating out of sample performance do end up choosing models with lower complexity but not always. As an extreme example imagine the oracle hands us the true but extremely complex model and a poor but parsimonious model. If parsimony was really our goal then we would choose the second but in reality, the first is what we would like to learn if we could. Unfortunately a lot of the time that last sentence is the kicker, "if we could".
|
Should parsimony really still be the gold standard?
@Matt's original answer does a great job of describing one of the benefits of parsimony but I don't think it actually answers your question. In reality, parsimony isn't the gold standard. Not now nor
|
6,733
|
Should parsimony really still be the gold standard?
|
Parsimonious models are desirable not just due to computing requirements, but also for generalization performance. It's impossible to achieve the ideal of infinite data that completely and accurately covers the sample space, meaning that non-parsimonious models have the potential to overfit and model noise or idiosyncrasies in the sample population.
It's certainly possible to build a model with millions of variables, but you'd be using variables that have no impact on the output to model the system. You could achieve great predictive performance on your training dataset, but those irrelevant variables will more than likely decrease your performance on an unseen test set.
If an output variable truly is the result of a million input variables, then you would do well to put them all in your predictive model, but only if you have enough data. To accurately build a model of this size, you'd need several million data points, at minimum. Parsimonious models are nice because in many real-world systems, a dataset of this size simply isn't available, and furthermore, the output is largely determined by a relatively small number of variables.
|
Should parsimony really still be the gold standard?
|
Parsimonious models are desirable not just due to computing requirements, but also for generalization performance. It's impossible to achieve the ideal of infinite data that completely and accurately
|
Should parsimony really still be the gold standard?
Parsimonious models are desirable not just due to computing requirements, but also for generalization performance. It's impossible to achieve the ideal of infinite data that completely and accurately covers the sample space, meaning that non-parsimonious models have the potential to overfit and model noise or idiosyncrasies in the sample population.
It's certainly possible to build a model with millions of variables, but you'd be using variables that have no impact on the output to model the system. You could achieve great predictive performance on your training dataset, but those irrelevant variables will more than likely decrease your performance on an unseen test set.
If an output variable truly is the result of a million input variables, then you would do well to put them all in your predictive model, but only if you have enough data. To accurately build a model of this size, you'd need several million data points, at minimum. Parsimonious models are nice because in many real-world systems, a dataset of this size simply isn't available, and furthermore, the output is largely determined by a relatively small number of variables.
|
Should parsimony really still be the gold standard?
Parsimonious models are desirable not just due to computing requirements, but also for generalization performance. It's impossible to achieve the ideal of infinite data that completely and accurately
|
6,734
|
Should parsimony really still be the gold standard?
|
I think the previous answers do a good job of making important points:
Parsimonious models tend to have better generalization characteristics.
Parsimony is not truly a gold standard, but just a consideration.
I want to add a few comments that come out of my day to day job experience.
The generalization of predictive accuracy argument is, of course, strong, but is academically bias in its focus. In general, when producing a statistical model, the economies are not such that predictive performance is a completely dominant consideration. Very often there are large outside constraints on what a useful model looks like for a given application:
The model must be implementable within an existing framework or system.
The model must be understandable by a non-technical entity.
The model must be efficient computationally.
The model must be documentable.
The model must pass regulatory constraints.
In real application domains, many if not all of these considerations come before, not after, predictive performance - and the optimization of model form and parameters is constrained by these desires. Each of these constraints biases the scientist towards parsimony.
It may be true that in many domains these constraints are being gradually lifted. But it is the lucky scientist indeed that gets to ignore them are focus purely on minimizing generalization error.
This can be very frustrating for the first time scientist, fresh out of school (it definitely was for me, and continues to be when I feel that the constraints placed on my work are not justified). But in the end, working hard to produce an unacceptable product is a waste, and that feels worse than the sting to your scientific pride.
|
Should parsimony really still be the gold standard?
|
I think the previous answers do a good job of making important points:
Parsimonious models tend to have better generalization characteristics.
Parsimony is not truly a gold standard, but just a consi
|
Should parsimony really still be the gold standard?
I think the previous answers do a good job of making important points:
Parsimonious models tend to have better generalization characteristics.
Parsimony is not truly a gold standard, but just a consideration.
I want to add a few comments that come out of my day to day job experience.
The generalization of predictive accuracy argument is, of course, strong, but is academically bias in its focus. In general, when producing a statistical model, the economies are not such that predictive performance is a completely dominant consideration. Very often there are large outside constraints on what a useful model looks like for a given application:
The model must be implementable within an existing framework or system.
The model must be understandable by a non-technical entity.
The model must be efficient computationally.
The model must be documentable.
The model must pass regulatory constraints.
In real application domains, many if not all of these considerations come before, not after, predictive performance - and the optimization of model form and parameters is constrained by these desires. Each of these constraints biases the scientist towards parsimony.
It may be true that in many domains these constraints are being gradually lifted. But it is the lucky scientist indeed that gets to ignore them are focus purely on minimizing generalization error.
This can be very frustrating for the first time scientist, fresh out of school (it definitely was for me, and continues to be when I feel that the constraints placed on my work are not justified). But in the end, working hard to produce an unacceptable product is a waste, and that feels worse than the sting to your scientific pride.
|
Should parsimony really still be the gold standard?
I think the previous answers do a good job of making important points:
Parsimonious models tend to have better generalization characteristics.
Parsimony is not truly a gold standard, but just a consi
|
6,735
|
Should parsimony really still be the gold standard?
|
I think this is a very good question. In my opinion parsimony is overrated. Nature is rarely parsimonious, and so we shouldn't necessarily expect accurate predictive or descriptive models to be so either. Regarding the question of interpretability, if you choose a simpler model that only modestly conforms to reality merely because you can understand it, what exactly are you understanding? Assuming a more complex model had better predictive power, it would appear to be closer to the actual facts anyways.
|
Should parsimony really still be the gold standard?
|
I think this is a very good question. In my opinion parsimony is overrated. Nature is rarely parsimonious, and so we shouldn't necessarily expect accurate predictive or descriptive models to be so e
|
Should parsimony really still be the gold standard?
I think this is a very good question. In my opinion parsimony is overrated. Nature is rarely parsimonious, and so we shouldn't necessarily expect accurate predictive or descriptive models to be so either. Regarding the question of interpretability, if you choose a simpler model that only modestly conforms to reality merely because you can understand it, what exactly are you understanding? Assuming a more complex model had better predictive power, it would appear to be closer to the actual facts anyways.
|
Should parsimony really still be the gold standard?
I think this is a very good question. In my opinion parsimony is overrated. Nature is rarely parsimonious, and so we shouldn't necessarily expect accurate predictive or descriptive models to be so e
|
6,736
|
Should parsimony really still be the gold standard?
|
Perhaps have a review of the Akaike Information Criterion, a concept that I only discovered by serendipity yesterday. The AIC seeks to identify which model and how many parameters are the best explanation for the observations at hand, rather than any basic Occam's Razor, or parsimony approach.
|
Should parsimony really still be the gold standard?
|
Perhaps have a review of the Akaike Information Criterion, a concept that I only discovered by serendipity yesterday. The AIC seeks to identify which model and how many parameters are the best explana
|
Should parsimony really still be the gold standard?
Perhaps have a review of the Akaike Information Criterion, a concept that I only discovered by serendipity yesterday. The AIC seeks to identify which model and how many parameters are the best explanation for the observations at hand, rather than any basic Occam's Razor, or parsimony approach.
|
Should parsimony really still be the gold standard?
Perhaps have a review of the Akaike Information Criterion, a concept that I only discovered by serendipity yesterday. The AIC seeks to identify which model and how many parameters are the best explana
|
6,737
|
Should parsimony really still be the gold standard?
|
Parsimony is not a golden start. It's an aspect in modeling. Modeling and especially forecasting can not be scripted, i.e. you can't just hand a script to a modeler to follow. You rather define principles upon which the modeling process must be based. So, the parsimony is one of these principles, application of which can not be scripted (again!). A modeler will consider the complexity when a selecting model.
Computational power has little to do with this. If you're in the industry your models will be consumed by business folks, product people, whoever you call them. You have to explain your model to them, it should make a sense to them. Having parsimonious models helps in this regard.
For instance, you're forecasting product sales. You should be able to describe what are the drivers of sales, and how they work. These must be related to concepts with which business operates, and the correlations must be understood and accepted by business. With complex models it could be very difficult to interpret the results of the model or attribute the differences with actuals. If you can't explain your models to business, you will not be valued by it.
One more thing that is particularly important for forecasting. Let's say your model is dependent on N exogenous variables. This means that you have to first obtain the forecasts of these variables in order to forecast your dependent variable. Having smaller N makes your life easier, so a simpler model is easier to use.
|
Should parsimony really still be the gold standard?
|
Parsimony is not a golden start. It's an aspect in modeling. Modeling and especially forecasting can not be scripted, i.e. you can't just hand a script to a modeler to follow. You rather define princi
|
Should parsimony really still be the gold standard?
Parsimony is not a golden start. It's an aspect in modeling. Modeling and especially forecasting can not be scripted, i.e. you can't just hand a script to a modeler to follow. You rather define principles upon which the modeling process must be based. So, the parsimony is one of these principles, application of which can not be scripted (again!). A modeler will consider the complexity when a selecting model.
Computational power has little to do with this. If you're in the industry your models will be consumed by business folks, product people, whoever you call them. You have to explain your model to them, it should make a sense to them. Having parsimonious models helps in this regard.
For instance, you're forecasting product sales. You should be able to describe what are the drivers of sales, and how they work. These must be related to concepts with which business operates, and the correlations must be understood and accepted by business. With complex models it could be very difficult to interpret the results of the model or attribute the differences with actuals. If you can't explain your models to business, you will not be valued by it.
One more thing that is particularly important for forecasting. Let's say your model is dependent on N exogenous variables. This means that you have to first obtain the forecasts of these variables in order to forecast your dependent variable. Having smaller N makes your life easier, so a simpler model is easier to use.
|
Should parsimony really still be the gold standard?
Parsimony is not a golden start. It's an aspect in modeling. Modeling and especially forecasting can not be scripted, i.e. you can't just hand a script to a modeler to follow. You rather define princi
|
6,738
|
Should parsimony really still be the gold standard?
|
Regarding Neural Networks, this topic has been very nicely covered in a recent NeurIPS 2020 paper entitled
The Pitfalls of Simplicity Bias in Neural Networks (Shah et al.).
I totally recommend reading the paper, I think it is very nicely structured and rigorous. Here is an attempt to summarize its main ideas:
The core of the problem is that the field seems to associate simplicity bias to "justify why NNs work well", but not to their "lack of robustness".
The authors tackle this problem via the design of datasets that comprise multiple predictive features with varying levels of simplicity. These will serve as control variables.
Following a series of theoretical analyses and targeted experiments, they found that the simplicity bias given by Stochastic Gradient Descent can be extreme, that common approaches to overcome this can fail, and in general, that there is still quite a way to go in order to better understand SB in NNs.
The proposed datasets and methods can serve as a basis for future work in that direction.
This principle of SB for NNs is very much embedded in the Deep Learning culture: See e.g. this video from the Udacity "Deep Learning" course (Vanhoucke et al.), where they use the "stretch pants" analogy. The idea is that instead of trying to guess the right pant size, you start with several extra sizes and "stretch them down" until they fit perfectly. This way, training is seen as reducing the capacity of the model wherever there is room to do so.
The paper above questions the general validity of this principle, showing that when trained with SGD, the pants tend to fit the simpler legs, no matter how many dimensions they have.
So whereas the original question is from 2015, this topic is still extremely current. Particularly, the following OP sentence is right on point:
Today's computing power enables us to build increasingly complex models with ever-greater ability for prediction. As a result of this increasing ceiling in computing power, do we really still need to gravitate toward simplicity?
Since the paper above shows that the ever-growing NN architectures still invariably gravitate towards simpler features when trained with SGD.
I would also like to finish commenting about another bias related to SB, that is Inductive Bias, "the set of assumptions that the learner uses to predict outputs of given inputs that it has not encountered".
Instead of assuming simplicity, inductive bias assumes any other arbitrary property. For example, when training a NN for transcription of piano music, it is safe to assume that all keyboard strokes generate mainly harmonic waveforms that have an onset and decay exponentially. In terms of "no free lunch", correctly applying an inductive bias would translate to getting lunch for a good prize. Physical models are known for their extremely precise and rigorous application of inductive biases.
Another amazing ICLR 2020 paper that treats the systematic introduction of Inductive Bias into NN design and training is
DDSP: DIFFERENTIABLE DIGITAL SIGNAL PROCESSING (Engel et al.)
With application of DDSP techniques it can be observed that introducing inductive biases helps the models satisfactorily solve complex tasks with less capacity, also the SGD convergence is faster and requires less training data.
So again coming back to the OP: While the No Free Lunch theorem states that no algorithm will work better than other for all kinds of problems, inductive biases are a good mechanism to gain performance without sacrificing much interpretability, if you are willing to reduce the set of targeted problems.
|
Should parsimony really still be the gold standard?
|
Regarding Neural Networks, this topic has been very nicely covered in a recent NeurIPS 2020 paper entitled
The Pitfalls of Simplicity Bias in Neural Networks (Shah et al.).
I totally recommend reading
|
Should parsimony really still be the gold standard?
Regarding Neural Networks, this topic has been very nicely covered in a recent NeurIPS 2020 paper entitled
The Pitfalls of Simplicity Bias in Neural Networks (Shah et al.).
I totally recommend reading the paper, I think it is very nicely structured and rigorous. Here is an attempt to summarize its main ideas:
The core of the problem is that the field seems to associate simplicity bias to "justify why NNs work well", but not to their "lack of robustness".
The authors tackle this problem via the design of datasets that comprise multiple predictive features with varying levels of simplicity. These will serve as control variables.
Following a series of theoretical analyses and targeted experiments, they found that the simplicity bias given by Stochastic Gradient Descent can be extreme, that common approaches to overcome this can fail, and in general, that there is still quite a way to go in order to better understand SB in NNs.
The proposed datasets and methods can serve as a basis for future work in that direction.
This principle of SB for NNs is very much embedded in the Deep Learning culture: See e.g. this video from the Udacity "Deep Learning" course (Vanhoucke et al.), where they use the "stretch pants" analogy. The idea is that instead of trying to guess the right pant size, you start with several extra sizes and "stretch them down" until they fit perfectly. This way, training is seen as reducing the capacity of the model wherever there is room to do so.
The paper above questions the general validity of this principle, showing that when trained with SGD, the pants tend to fit the simpler legs, no matter how many dimensions they have.
So whereas the original question is from 2015, this topic is still extremely current. Particularly, the following OP sentence is right on point:
Today's computing power enables us to build increasingly complex models with ever-greater ability for prediction. As a result of this increasing ceiling in computing power, do we really still need to gravitate toward simplicity?
Since the paper above shows that the ever-growing NN architectures still invariably gravitate towards simpler features when trained with SGD.
I would also like to finish commenting about another bias related to SB, that is Inductive Bias, "the set of assumptions that the learner uses to predict outputs of given inputs that it has not encountered".
Instead of assuming simplicity, inductive bias assumes any other arbitrary property. For example, when training a NN for transcription of piano music, it is safe to assume that all keyboard strokes generate mainly harmonic waveforms that have an onset and decay exponentially. In terms of "no free lunch", correctly applying an inductive bias would translate to getting lunch for a good prize. Physical models are known for their extremely precise and rigorous application of inductive biases.
Another amazing ICLR 2020 paper that treats the systematic introduction of Inductive Bias into NN design and training is
DDSP: DIFFERENTIABLE DIGITAL SIGNAL PROCESSING (Engel et al.)
With application of DDSP techniques it can be observed that introducing inductive biases helps the models satisfactorily solve complex tasks with less capacity, also the SGD convergence is faster and requires less training data.
So again coming back to the OP: While the No Free Lunch theorem states that no algorithm will work better than other for all kinds of problems, inductive biases are a good mechanism to gain performance without sacrificing much interpretability, if you are willing to reduce the set of targeted problems.
|
Should parsimony really still be the gold standard?
Regarding Neural Networks, this topic has been very nicely covered in a recent NeurIPS 2020 paper entitled
The Pitfalls of Simplicity Bias in Neural Networks (Shah et al.).
I totally recommend reading
|
6,739
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
|
The Dirichlet distribution is a conjugate prior for the multinomial distribution. This means that if the prior distribution of the multinomial parameters is Dirichlet then the posterior distribution is also a Dirichlet distribution (with parameters different from those of the prior). The benefit of this is that (a) the posterior distribution is easy to compute and (b) it in some sense is possible to quantify how much our beliefs have changed after collecting the data.
It can certainly be discussed whether these are good reasons to choose a particular prior, as these criteria are unrelated to actual prior beliefs... Nevertheless, conjugate priors are popular, as they often are reasonably flexible and convenient to use for the reasons stated above.
For the special case of the multinomial distribution, let $(p_1,\ldots,p_k)$ be the vector of multinomial parameters (i.e. the probabilities for the different categories). If $$(p_1,\ldots,p_k)\sim \mbox{Dirichlet}(\alpha_1,\ldots,\alpha_k)$$ prior to collecting the data, then, given observations $(x_1,\ldots,x_k)$ in the different categories,
$$(p_1,\ldots,p_k)\Big|(x_1,\ldots,x_k)\sim \mbox{Dirichlet}(\alpha_1+x_1,\ldots,\alpha_k+x_k).$$
The uniform distribution is actually a special case of the Dirichlet distribution, corresponding to the case $\alpha_1=\alpha_2=\cdots=\alpha_k=1$. So is the least-informative Jeffreys prior, for which $\alpha_1=\cdots=\alpha_k=1/2$. The fact that the Dirichlet class includes these natural "non-informative" priors is another reason for using it.
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Why is the Dirichlet distribution the prior for the multinomial distribution?
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The Dirichlet distribution is a conjugate prior for the multinomial distribution. This means that if the prior distribution of the multinomial parameters is Dirichlet then the posterior distribution i
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
The Dirichlet distribution is a conjugate prior for the multinomial distribution. This means that if the prior distribution of the multinomial parameters is Dirichlet then the posterior distribution is also a Dirichlet distribution (with parameters different from those of the prior). The benefit of this is that (a) the posterior distribution is easy to compute and (b) it in some sense is possible to quantify how much our beliefs have changed after collecting the data.
It can certainly be discussed whether these are good reasons to choose a particular prior, as these criteria are unrelated to actual prior beliefs... Nevertheless, conjugate priors are popular, as they often are reasonably flexible and convenient to use for the reasons stated above.
For the special case of the multinomial distribution, let $(p_1,\ldots,p_k)$ be the vector of multinomial parameters (i.e. the probabilities for the different categories). If $$(p_1,\ldots,p_k)\sim \mbox{Dirichlet}(\alpha_1,\ldots,\alpha_k)$$ prior to collecting the data, then, given observations $(x_1,\ldots,x_k)$ in the different categories,
$$(p_1,\ldots,p_k)\Big|(x_1,\ldots,x_k)\sim \mbox{Dirichlet}(\alpha_1+x_1,\ldots,\alpha_k+x_k).$$
The uniform distribution is actually a special case of the Dirichlet distribution, corresponding to the case $\alpha_1=\alpha_2=\cdots=\alpha_k=1$. So is the least-informative Jeffreys prior, for which $\alpha_1=\cdots=\alpha_k=1/2$. The fact that the Dirichlet class includes these natural "non-informative" priors is another reason for using it.
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
The Dirichlet distribution is a conjugate prior for the multinomial distribution. This means that if the prior distribution of the multinomial parameters is Dirichlet then the posterior distribution i
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6,740
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Why is the Dirichlet distribution the prior for the multinomial distribution?
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In addition rather than contradiction to Måns T's answer, I simply point out that there is no such thing as "the prior" in Bayesian modelling! The Dirichlet distribution is a convenient choice because of (a) conjugacy, (b) computing, and (c) connection with non-parametric statistics (since this is the discretised version of the Dirichlet process).
However, (i) whatever prior you put on the weights of the multinomial is a legitimate answer at the subjective Bayes level and (ii) in case of prior information being available there is no reason it simplifies into a Dirichlet distribution. Note also that mixtures and convolutions of Dirichlet distributions can be used as priors.
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
|
In addition rather than contradiction to Måns T's answer, I simply point out that there is no such thing as "the prior" in Bayesian modelling! The Dirichlet distribution is a convenient choice because
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
In addition rather than contradiction to Måns T's answer, I simply point out that there is no such thing as "the prior" in Bayesian modelling! The Dirichlet distribution is a convenient choice because of (a) conjugacy, (b) computing, and (c) connection with non-parametric statistics (since this is the discretised version of the Dirichlet process).
However, (i) whatever prior you put on the weights of the multinomial is a legitimate answer at the subjective Bayes level and (ii) in case of prior information being available there is no reason it simplifies into a Dirichlet distribution. Note also that mixtures and convolutions of Dirichlet distributions can be used as priors.
|
Why is the Dirichlet distribution the prior for the multinomial distribution?
In addition rather than contradiction to Måns T's answer, I simply point out that there is no such thing as "the prior" in Bayesian modelling! The Dirichlet distribution is a convenient choice because
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6,741
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XGBoost Loss function Approximation With Taylor Expansion
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This is a very interesting question. In order to fully understand what was going on, I had to go through what XGBoost is trying to do, and what other methods we had in our toolbox to deal with it. My answer goes over traditional methods, and how/why XGBoost is an improvement. If you want only the bullet points, there is a summary at the end.
Traditional Gradient Boosting
Consider the traditional Gradient Boosting Algorithm (Wikipedia):
Compute base model $H_0$
For $m \leftarrow 1:M$
Compute pseudo-residuals $r_{im} = -\frac{\partial \ell(y_i, H_{m-1}(x_i))}{\partial H_{m-1}(x_i)}$
Fit a base learner $h_m(x)$ to the pseudo-residuals
Compute the multiplier $\gamma$ that minimizes the cost, $\gamma = \arg \min_\gamma \sum_{i=1}^N \ell(y_i, H_{m-1}(x_i) + \gamma h_m(x_i))$, (using line search)
Update the model $H_m(x) = H_{m-1}(x) + \gamma h_m(x)$.
You get your boosted model $H_M(x)$.
The function approximation is important is for the following part,
Fit a base learner $h_m(x)$ to the pseudo-residuals.
Imagine you where to construct your Gradient Boosting Algorithm naively. You would build the algorithm above using existing regression trees as weak learners. Let assume you are not allowed to tweak the existing implementation of the weak learners. In Matlab, the default split criterion is the Mean Square Error. The same goes for scikit learn.
You are trying to find the best model $h_m(x)$ that minimize the cost $\ell(y_i, H_{m-1}(x_i) + h_m(x_i))$. But to do so, you are fitting a simple regression model to the residuals using the MSE as objective function. Notice that you are not directly minimizing what you want, but using the residuals and the MSE as a proxy to do so. The bad part is that it does not necessarily yields the optimal solution. The good part is that it work.
Traditional Gradient Descent
This is analogous to the traditional Gradient Descent (Wikipedia), where you are trying to minimize a cost function $f(x)$ by following the (negative of the) gradient of the function, $-\nabla f(x)$ at every step.
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
It does not allow you to find the exact minimum after one step, but each step gets you closer to the minimum (if the function is convex). This is an approximation, but it works very well and it is the algorithm we traditionally use to do a logistic regression, for example.
Interlude
At this point, the thing to understand is that the general gradient boosting algorithm does not compute the cost function $\ell$ for each possible splits, it uses the cost function of the regression weak learner to fit the residuals.
What your question seems to imply is that the "true XGBoost" should compute the cost function for each split, and that the "approximate XGBoost" is using a heuristic to approximate it. You can see it that way, but historically, we have had the general gradient boosting algorithm, which does not use information about the cost function, except the derivative at the current point. XGBoost is an extension to Gradient Boosting that tries to be smarter about growing the weak regression trees by using a more accurate approximation than just the gradient.
Other ways to choose the best model $h_m(x)$
If we take a look at AdaBoost as special case of gradient boosting, it does not selects regressors but classifiers as weak learners. If we set $h_m(x) \in \{-1,1\}$, the way AdaBoost selects the best model is by finding
$$h_m = \arg \max_{h_m} \sum_{i=1}^N w_i h_m(x_i)$$
where $w_i$ are the residuals (source, starts at slide 20). The reasoning for the use of this objective function is that if $w_i$ and $h_m(x_i)$ go in the same direction/have the same sign, the point is moving to the right direction, and you are trying to maximize the maximum amount of movement in the right direction.
But once again, this is not directly measuring which $h_m$ minimizes $\ell(y_i, H_{m-1}(x_i) + h_m(x_i))$. It is measuring how good the move $h_m$ is, with respect with the overall direction you should go, as measured with the residuals $w_i$, which are also an approximation. The residuals tell you in which direction you should be moving by their sign, and roughly by how much by their magnitude, but they do not tell you exactly where you should stop.
Better Gradient Descent
The next three examples are not essential to the explanation and are just here to present some ways to do better than a vanilla gradient descent, to support the idea that what XGBoost does is just another way of improving on gradient descent. In a traditional gradient descent setting, when trying to minimize $f(x)$, it is possible to do better than just following the gradient. Many extensions have been proposed (Wikipedia). Here are some of them, to show that it is possible to do better, given more computation time or more properties of the function $f$.
Line search/Backtracking: In Gradient Descent, once the gradient $-\nabla f(x^{(i)})$ is computed, the next point should be
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
But the gradient gives only the direction in which one should move, not really by "how much", so another procedure can be used, to find the best $c > 0$ such that
$$x_c^{(i+1)} = x^{(i)} - c \nabla f(x^{(i)})$$
minimizes the cost function. This is done be evaluating $f(x_c^{(i+1)})$ for some $c$, and since the function $f$ should be convex, it is relatively easy to do through Line Search (Wikipedia) or Backtracking Line Search (Wikipedia). Here, the main cost is evaluation $f(x)$. So this extension works best if $f$ is easy to compute. Note that the general algorithm for gradient boosting uses line search, as shown in the beginning of my answer.
Fast proximal gradient method: If the function to minimize is strongly convex, and its gradient is smooth (Lipschitz (Wikipedia)), then there is some trick using those properties that speeds up the convergence.
Stochastic Gradient Descent and the Momentum method: In Stochastic Gradient Descent, you do not evaluate the gradient on all points, but only on a subset of those points. You take a step, then compute the gradient on another batch, and continue on. Stochastic Gradient Descent may be used because the computation on all points is very expensive, or maybe all those points do not even fit into memory. This allows you to take more steps, more quickly, but less accurately.
When doing so, the direction of the gradient might change depending on which points are sampled. To counteract this effect, the momentum methods keeps a moving average of the direction for each dimension, reducing the variance in each move.
The most relevant extension to gradient descent in our discussion of XGBoost is Newton's method (Wikipedia). Instead of just computing the gradient and following it, it uses the second order derivative to gather more information about the direction it should go into. If we use gradient descent, we have that at each iteration, we update our point $x^{(i)}$ as follow,
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
And since the gradient $\nabla f(x^{(i)})$ points to the direction of highest increase in $f$, its negative points in the direction of highest decrease, and we hope that $f(x^{(i+1)}) < f(x^{(i)})$. This might not hold, as we might go too far in the direction of the gradient (hence the line search extension), but it is a good approximation. In Newton's method, we update $x^{(i)}$ as follow,
$$x^{(i+1)} = x^{(i)} - \frac{\nabla f(x^{(i)})}{\text{Hess} f(x^{(i)})}$$
Where $\text{Hess} f(x)$ is the Hessian of $f$ in $x$. This update takes into account second order information, so the direction is no longer the direction of highest decrease, but should point more precisely towards the $x^{(i+1)}$ such that $f(x^{(i+1)}) = 0$ (or the point where $f$ is minimal, if there is no zero). If $f$ is a second order polynomial, then Newton's method coupled with a line search should be able to find the minimum in one step.
Newton's method contrasts with Stochastic gradient descent. In Stochastic Gradient Descent, we use less point to take less time to compute the direction we should go towards, in order to make more of them, in the hope we go there quicker. In Newton's method, we take more time to compute the direction we want to go into, in the hope we have to take less steps in order to get there.
Now, the reason why Newton's method works is the same as to why the XGBoost approximation works, and it relies on Taylor's expansion (Wikipedia) and Taylor's theorem (Wikipedia). The Taylor expansion (or Taylor series) of a function at a point $f(x + a)$ is
$$f(x) + \frac{\partial f(x)}{\partial x}a + \frac{1}{2}\frac{\partial^2 f(x)}{\partial x^2}a^2 + \cdots = \sum_{n=0} ^\infty \frac{1}{n!} \frac{\partial^n f(x)}{\partial x^n}a^n.$$
Note the similarity between this expression and the approximation XGBoost is using. Taylor's Theorem states that if you stop the expansion at the order $k$, then the error, or the difference between $f(x+a)$ and $\sum_{n=0}^k \frac{1}{n!}\frac{\partial^n f(x)}{\partial x^n}a^n$, is at most $h_k(x) a^k$, where $h_k$ is a function with the nice property that it goes to zero as $a$ goes to zero.
If you want some visualisation of how well it approximate some functions, take a look at the wikipedia pages, they have some graphs for the approximation of non-polynomial function such as $e^x$, $\log(x)$.
The thing to note is that approximation works very well if you want to compute the value of $f$ in the neighbourhood of $x$, that is, for very small changes $a$. This is what we want to do in Boosting. Of course we would like to find the tree that makes the biggest change. If the weak learners we build are very good and want to make a very big change, then we can arbitrarily hinder it by only applying $0.1$ or $0.01$ of its effect. This is the step-size or the learning rate of the gradient descent. This is acceptable, because if our weak learners are getting very good solutions, this means that either the problem is easy, in which case we are going to end up with a good solution anyway, or we are overfitting, so going a little or very much in this bad direction does not change the underlying problem.
So what is XGBoost doing, and why does it work?
XGBoost is a Gradient Boosting algorithm that build regression trees as weak learners. The traditional Gradient Boosting algorithm is very similar to a gradient descent with a line search, where the direction in which to go is drawn from the available weak learners. The naïve implementation of Gradient Boosting would use the cost function of the weak learner to fit it to the residual. This is a proxy to minimize the cost of the new model, which is expensive to compute. What XGBoost is doing is building a custom cost function to fit the trees, using the Taylor series of order two as an approximation for the true cost function, such that it can be more sure that the tree it picks is a good one. In this respect, and as a simplification, XGBoost is to Gradient Boosting what Newton's Method is to Gradient Descent.
Why did they build it that way
Your question as to why using this approximation comes to a cost/performance tradeoff. This cost function is used to compare potential splits for regression trees, so if our points have say 50 features, with an average of 10 different values, each node has 500 potential splits, so 500 evaluation of the function. If you drop a continuous feature, the number of splits explode, and the evaluation of the split is called more and more (XGBoost has another trick to deal with continuous features, but that's out of the scope). As the algorithm will spend most of its time evaluating splits, the way to speed up the algorithm is to speed up tree evaluation.
If you evaluated the tree with the full cost function, $\ell$, it is a new computation for every new split. In order to do optimization in the computation of the cost function, you would need to have information about the cost function, which is the whole point of Gradient Boosting: It should work for every cost function.
The second order approximation is computationally nice, because most terms are the same in a given iteration. For a given iteration, most of the expression can be computed once, and reused as constant for all splits:
$$\mathcal{L}^{(t)}\approx \sum_{i=1}^n \underbrace{\ell(y_i,\hat{y}_i^{(t-1)})}_{\text{constant}}+\underbrace{g_i}_{\text{constant}}f_t(\mathbf{x}_i)+\frac{1}{2}\underbrace{h_i}_{\text{constant}}f_t^2(\mathbf{x}_i)+\Omega(f_t),$$
So the only thing you have to compute is $f_t(x_i)$ and $\Omega(f_t)$, and then what is left is mostly additions, and some multiplications. Moreover, if you take a look at the XGBoost paper (arxiv), you will see that they use the fact that they are building a tree to further simplify the expression down to a bunch of summation of indexes, which is very, very quick.
Summary
You can see XGBoost (with approximation) as a regression from the exact solution, an approximation of the "true XGBoost", with exact evaluation. But since the exact evaluation is so costly, another way to see it is that on huge datasets, the approximation is all we can realistically do, and this approximation is more accurate than the first-order approximation a "naïve" gradient boosting algorithm would do.
The approximation in use is similar to Newton's Method, and is justified by Taylor Series (Wikipedia) and Taylor Theorem (Wikipedia).
Higher order information is indeed not completely used, but it is not necessary, because we want a good approximation in the neighbourhood of our starting point.
For visualisation, check the Wikipedia page of Taylor Series/Taylor's Theorem, or Khan Academy on Taylor Series approximation, or MathDemo page on polynomial approximation of non-polynomials
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XGBoost Loss function Approximation With Taylor Expansion
|
This is a very interesting question. In order to fully understand what was going on, I had to go through what XGBoost is trying to do, and what other methods we had in our toolbox to deal with it. My
|
XGBoost Loss function Approximation With Taylor Expansion
This is a very interesting question. In order to fully understand what was going on, I had to go through what XGBoost is trying to do, and what other methods we had in our toolbox to deal with it. My answer goes over traditional methods, and how/why XGBoost is an improvement. If you want only the bullet points, there is a summary at the end.
Traditional Gradient Boosting
Consider the traditional Gradient Boosting Algorithm (Wikipedia):
Compute base model $H_0$
For $m \leftarrow 1:M$
Compute pseudo-residuals $r_{im} = -\frac{\partial \ell(y_i, H_{m-1}(x_i))}{\partial H_{m-1}(x_i)}$
Fit a base learner $h_m(x)$ to the pseudo-residuals
Compute the multiplier $\gamma$ that minimizes the cost, $\gamma = \arg \min_\gamma \sum_{i=1}^N \ell(y_i, H_{m-1}(x_i) + \gamma h_m(x_i))$, (using line search)
Update the model $H_m(x) = H_{m-1}(x) + \gamma h_m(x)$.
You get your boosted model $H_M(x)$.
The function approximation is important is for the following part,
Fit a base learner $h_m(x)$ to the pseudo-residuals.
Imagine you where to construct your Gradient Boosting Algorithm naively. You would build the algorithm above using existing regression trees as weak learners. Let assume you are not allowed to tweak the existing implementation of the weak learners. In Matlab, the default split criterion is the Mean Square Error. The same goes for scikit learn.
You are trying to find the best model $h_m(x)$ that minimize the cost $\ell(y_i, H_{m-1}(x_i) + h_m(x_i))$. But to do so, you are fitting a simple regression model to the residuals using the MSE as objective function. Notice that you are not directly minimizing what you want, but using the residuals and the MSE as a proxy to do so. The bad part is that it does not necessarily yields the optimal solution. The good part is that it work.
Traditional Gradient Descent
This is analogous to the traditional Gradient Descent (Wikipedia), where you are trying to minimize a cost function $f(x)$ by following the (negative of the) gradient of the function, $-\nabla f(x)$ at every step.
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
It does not allow you to find the exact minimum after one step, but each step gets you closer to the minimum (if the function is convex). This is an approximation, but it works very well and it is the algorithm we traditionally use to do a logistic regression, for example.
Interlude
At this point, the thing to understand is that the general gradient boosting algorithm does not compute the cost function $\ell$ for each possible splits, it uses the cost function of the regression weak learner to fit the residuals.
What your question seems to imply is that the "true XGBoost" should compute the cost function for each split, and that the "approximate XGBoost" is using a heuristic to approximate it. You can see it that way, but historically, we have had the general gradient boosting algorithm, which does not use information about the cost function, except the derivative at the current point. XGBoost is an extension to Gradient Boosting that tries to be smarter about growing the weak regression trees by using a more accurate approximation than just the gradient.
Other ways to choose the best model $h_m(x)$
If we take a look at AdaBoost as special case of gradient boosting, it does not selects regressors but classifiers as weak learners. If we set $h_m(x) \in \{-1,1\}$, the way AdaBoost selects the best model is by finding
$$h_m = \arg \max_{h_m} \sum_{i=1}^N w_i h_m(x_i)$$
where $w_i$ are the residuals (source, starts at slide 20). The reasoning for the use of this objective function is that if $w_i$ and $h_m(x_i)$ go in the same direction/have the same sign, the point is moving to the right direction, and you are trying to maximize the maximum amount of movement in the right direction.
But once again, this is not directly measuring which $h_m$ minimizes $\ell(y_i, H_{m-1}(x_i) + h_m(x_i))$. It is measuring how good the move $h_m$ is, with respect with the overall direction you should go, as measured with the residuals $w_i$, which are also an approximation. The residuals tell you in which direction you should be moving by their sign, and roughly by how much by their magnitude, but they do not tell you exactly where you should stop.
Better Gradient Descent
The next three examples are not essential to the explanation and are just here to present some ways to do better than a vanilla gradient descent, to support the idea that what XGBoost does is just another way of improving on gradient descent. In a traditional gradient descent setting, when trying to minimize $f(x)$, it is possible to do better than just following the gradient. Many extensions have been proposed (Wikipedia). Here are some of them, to show that it is possible to do better, given more computation time or more properties of the function $f$.
Line search/Backtracking: In Gradient Descent, once the gradient $-\nabla f(x^{(i)})$ is computed, the next point should be
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
But the gradient gives only the direction in which one should move, not really by "how much", so another procedure can be used, to find the best $c > 0$ such that
$$x_c^{(i+1)} = x^{(i)} - c \nabla f(x^{(i)})$$
minimizes the cost function. This is done be evaluating $f(x_c^{(i+1)})$ for some $c$, and since the function $f$ should be convex, it is relatively easy to do through Line Search (Wikipedia) or Backtracking Line Search (Wikipedia). Here, the main cost is evaluation $f(x)$. So this extension works best if $f$ is easy to compute. Note that the general algorithm for gradient boosting uses line search, as shown in the beginning of my answer.
Fast proximal gradient method: If the function to minimize is strongly convex, and its gradient is smooth (Lipschitz (Wikipedia)), then there is some trick using those properties that speeds up the convergence.
Stochastic Gradient Descent and the Momentum method: In Stochastic Gradient Descent, you do not evaluate the gradient on all points, but only on a subset of those points. You take a step, then compute the gradient on another batch, and continue on. Stochastic Gradient Descent may be used because the computation on all points is very expensive, or maybe all those points do not even fit into memory. This allows you to take more steps, more quickly, but less accurately.
When doing so, the direction of the gradient might change depending on which points are sampled. To counteract this effect, the momentum methods keeps a moving average of the direction for each dimension, reducing the variance in each move.
The most relevant extension to gradient descent in our discussion of XGBoost is Newton's method (Wikipedia). Instead of just computing the gradient and following it, it uses the second order derivative to gather more information about the direction it should go into. If we use gradient descent, we have that at each iteration, we update our point $x^{(i)}$ as follow,
$$x^{(i+1)} = x^{(i)} - \nabla f(x^{(i)})$$
And since the gradient $\nabla f(x^{(i)})$ points to the direction of highest increase in $f$, its negative points in the direction of highest decrease, and we hope that $f(x^{(i+1)}) < f(x^{(i)})$. This might not hold, as we might go too far in the direction of the gradient (hence the line search extension), but it is a good approximation. In Newton's method, we update $x^{(i)}$ as follow,
$$x^{(i+1)} = x^{(i)} - \frac{\nabla f(x^{(i)})}{\text{Hess} f(x^{(i)})}$$
Where $\text{Hess} f(x)$ is the Hessian of $f$ in $x$. This update takes into account second order information, so the direction is no longer the direction of highest decrease, but should point more precisely towards the $x^{(i+1)}$ such that $f(x^{(i+1)}) = 0$ (or the point where $f$ is minimal, if there is no zero). If $f$ is a second order polynomial, then Newton's method coupled with a line search should be able to find the minimum in one step.
Newton's method contrasts with Stochastic gradient descent. In Stochastic Gradient Descent, we use less point to take less time to compute the direction we should go towards, in order to make more of them, in the hope we go there quicker. In Newton's method, we take more time to compute the direction we want to go into, in the hope we have to take less steps in order to get there.
Now, the reason why Newton's method works is the same as to why the XGBoost approximation works, and it relies on Taylor's expansion (Wikipedia) and Taylor's theorem (Wikipedia). The Taylor expansion (or Taylor series) of a function at a point $f(x + a)$ is
$$f(x) + \frac{\partial f(x)}{\partial x}a + \frac{1}{2}\frac{\partial^2 f(x)}{\partial x^2}a^2 + \cdots = \sum_{n=0} ^\infty \frac{1}{n!} \frac{\partial^n f(x)}{\partial x^n}a^n.$$
Note the similarity between this expression and the approximation XGBoost is using. Taylor's Theorem states that if you stop the expansion at the order $k$, then the error, or the difference between $f(x+a)$ and $\sum_{n=0}^k \frac{1}{n!}\frac{\partial^n f(x)}{\partial x^n}a^n$, is at most $h_k(x) a^k$, where $h_k$ is a function with the nice property that it goes to zero as $a$ goes to zero.
If you want some visualisation of how well it approximate some functions, take a look at the wikipedia pages, they have some graphs for the approximation of non-polynomial function such as $e^x$, $\log(x)$.
The thing to note is that approximation works very well if you want to compute the value of $f$ in the neighbourhood of $x$, that is, for very small changes $a$. This is what we want to do in Boosting. Of course we would like to find the tree that makes the biggest change. If the weak learners we build are very good and want to make a very big change, then we can arbitrarily hinder it by only applying $0.1$ or $0.01$ of its effect. This is the step-size or the learning rate of the gradient descent. This is acceptable, because if our weak learners are getting very good solutions, this means that either the problem is easy, in which case we are going to end up with a good solution anyway, or we are overfitting, so going a little or very much in this bad direction does not change the underlying problem.
So what is XGBoost doing, and why does it work?
XGBoost is a Gradient Boosting algorithm that build regression trees as weak learners. The traditional Gradient Boosting algorithm is very similar to a gradient descent with a line search, where the direction in which to go is drawn from the available weak learners. The naïve implementation of Gradient Boosting would use the cost function of the weak learner to fit it to the residual. This is a proxy to minimize the cost of the new model, which is expensive to compute. What XGBoost is doing is building a custom cost function to fit the trees, using the Taylor series of order two as an approximation for the true cost function, such that it can be more sure that the tree it picks is a good one. In this respect, and as a simplification, XGBoost is to Gradient Boosting what Newton's Method is to Gradient Descent.
Why did they build it that way
Your question as to why using this approximation comes to a cost/performance tradeoff. This cost function is used to compare potential splits for regression trees, so if our points have say 50 features, with an average of 10 different values, each node has 500 potential splits, so 500 evaluation of the function. If you drop a continuous feature, the number of splits explode, and the evaluation of the split is called more and more (XGBoost has another trick to deal with continuous features, but that's out of the scope). As the algorithm will spend most of its time evaluating splits, the way to speed up the algorithm is to speed up tree evaluation.
If you evaluated the tree with the full cost function, $\ell$, it is a new computation for every new split. In order to do optimization in the computation of the cost function, you would need to have information about the cost function, which is the whole point of Gradient Boosting: It should work for every cost function.
The second order approximation is computationally nice, because most terms are the same in a given iteration. For a given iteration, most of the expression can be computed once, and reused as constant for all splits:
$$\mathcal{L}^{(t)}\approx \sum_{i=1}^n \underbrace{\ell(y_i,\hat{y}_i^{(t-1)})}_{\text{constant}}+\underbrace{g_i}_{\text{constant}}f_t(\mathbf{x}_i)+\frac{1}{2}\underbrace{h_i}_{\text{constant}}f_t^2(\mathbf{x}_i)+\Omega(f_t),$$
So the only thing you have to compute is $f_t(x_i)$ and $\Omega(f_t)$, and then what is left is mostly additions, and some multiplications. Moreover, if you take a look at the XGBoost paper (arxiv), you will see that they use the fact that they are building a tree to further simplify the expression down to a bunch of summation of indexes, which is very, very quick.
Summary
You can see XGBoost (with approximation) as a regression from the exact solution, an approximation of the "true XGBoost", with exact evaluation. But since the exact evaluation is so costly, another way to see it is that on huge datasets, the approximation is all we can realistically do, and this approximation is more accurate than the first-order approximation a "naïve" gradient boosting algorithm would do.
The approximation in use is similar to Newton's Method, and is justified by Taylor Series (Wikipedia) and Taylor Theorem (Wikipedia).
Higher order information is indeed not completely used, but it is not necessary, because we want a good approximation in the neighbourhood of our starting point.
For visualisation, check the Wikipedia page of Taylor Series/Taylor's Theorem, or Khan Academy on Taylor Series approximation, or MathDemo page on polynomial approximation of non-polynomials
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XGBoost Loss function Approximation With Taylor Expansion
This is a very interesting question. In order to fully understand what was going on, I had to go through what XGBoost is trying to do, and what other methods we had in our toolbox to deal with it. My
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6,742
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Neural network with skip-layer connections
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I am very late to the game, but I wanted to post to reflect some current developments in convolutional neural networks with respect to skip connections.
A Microsoft Research team recently won the ImageNet 2015 competition and released a technical report Deep Residual Learning for Image Recognition describing some of their main ideas.
One of their main contributions is this concept of deep residual layers. These deep residual layers use skip connections. Using these deep residual layers, they were able to train a 152 layer conv net for ImageNet 2015. They even trained a 1000+ layer conv net for the CIFAR-10.
The problem that motivated them is the following:
When deeper networks are able to start converging, a degradation problem has been exposed: with the network depth increasing, accuracy gets saturated (which might be unsurprising) and then degrades rapidly. Unexpectedly, such degradation is not caused by overfitting, and adding more layers to a suitably deep model leads to higher training error...
The idea is if that if you take a "shallow" network and just stack on more layers to create a deeper network, the performance of the deeper network should be at least as good as the shallow network as the deeper network could learn the exact shallow network by setting the new stacked layers to identity layers (in reality we know this is probably highly unlikely to happen using no architectural priors or current optimization methods). They observed that this was not the case and that training error sometimes got worse when they stacked more layers on top of a shallower model.
So this motivated them to use skip connections and use so-called deep residual layers to allow their network to learn deviations from the identity layer, hence the term residual, residual here referring to difference from the identity.
They implement skip connections in the following manner:
So they view the map $\mathcal{F}(x) := \mathcal{H}(x) - x$ as some residual map. They use a skip layer connection to cast this mapping into $\mathcal{F}(x) + x = \mathcal{H}(x)$. So if the residual $\mathcal{F}(x)$ is "small", the map $\mathcal{H}(x)$ is roughly the identity.
In this manner the use of deep residual layers via skip connections allows their deep nets to learn approximate identity layers, if that is indeed what is optimal, or locally optimal. Indeed they claim that their residual layers:
We show by experiments (Fig. 7) that the learned residual functions in general have small responses
As to why exactly this works they don't have an exact answer. It is highly unlikely that identity layers are optimal, but they believe that using these residual layers helps precondition the problem and that it's easier to learn a new function given a reference/baseline of comparison to the identity mapping than to learn one "from scratch" without using the identity baseline. Who knows. But I thought this would be a nice answer to your question.
By the way, in hindsight: sashkello's answer is even better isn't it?
|
Neural network with skip-layer connections
|
I am very late to the game, but I wanted to post to reflect some current developments in convolutional neural networks with respect to skip connections.
A Microsoft Research team recently won the Ima
|
Neural network with skip-layer connections
I am very late to the game, but I wanted to post to reflect some current developments in convolutional neural networks with respect to skip connections.
A Microsoft Research team recently won the ImageNet 2015 competition and released a technical report Deep Residual Learning for Image Recognition describing some of their main ideas.
One of their main contributions is this concept of deep residual layers. These deep residual layers use skip connections. Using these deep residual layers, they were able to train a 152 layer conv net for ImageNet 2015. They even trained a 1000+ layer conv net for the CIFAR-10.
The problem that motivated them is the following:
When deeper networks are able to start converging, a degradation problem has been exposed: with the network depth increasing, accuracy gets saturated (which might be unsurprising) and then degrades rapidly. Unexpectedly, such degradation is not caused by overfitting, and adding more layers to a suitably deep model leads to higher training error...
The idea is if that if you take a "shallow" network and just stack on more layers to create a deeper network, the performance of the deeper network should be at least as good as the shallow network as the deeper network could learn the exact shallow network by setting the new stacked layers to identity layers (in reality we know this is probably highly unlikely to happen using no architectural priors or current optimization methods). They observed that this was not the case and that training error sometimes got worse when they stacked more layers on top of a shallower model.
So this motivated them to use skip connections and use so-called deep residual layers to allow their network to learn deviations from the identity layer, hence the term residual, residual here referring to difference from the identity.
They implement skip connections in the following manner:
So they view the map $\mathcal{F}(x) := \mathcal{H}(x) - x$ as some residual map. They use a skip layer connection to cast this mapping into $\mathcal{F}(x) + x = \mathcal{H}(x)$. So if the residual $\mathcal{F}(x)$ is "small", the map $\mathcal{H}(x)$ is roughly the identity.
In this manner the use of deep residual layers via skip connections allows their deep nets to learn approximate identity layers, if that is indeed what is optimal, or locally optimal. Indeed they claim that their residual layers:
We show by experiments (Fig. 7) that the learned residual functions in general have small responses
As to why exactly this works they don't have an exact answer. It is highly unlikely that identity layers are optimal, but they believe that using these residual layers helps precondition the problem and that it's easier to learn a new function given a reference/baseline of comparison to the identity mapping than to learn one "from scratch" without using the identity baseline. Who knows. But I thought this would be a nice answer to your question.
By the way, in hindsight: sashkello's answer is even better isn't it?
|
Neural network with skip-layer connections
I am very late to the game, but I wanted to post to reflect some current developments in convolutional neural networks with respect to skip connections.
A Microsoft Research team recently won the Ima
|
6,743
|
Neural network with skip-layer connections
|
In theory, skip-layer connections should not improve on the network performance. But, since complex networks are hard to train and easy to overfit it may be very useful to explicitly add this as a linear regression term, when you know that your data has a strong linear component. This hints the model in a right direction... In addition, this is more interpretable since it presents your model as linear + perturbations, unraveling a bit of a structure behind the network, which is usually seen merely as a black box.
|
Neural network with skip-layer connections
|
In theory, skip-layer connections should not improve on the network performance. But, since complex networks are hard to train and easy to overfit it may be very useful to explicitly add this as a lin
|
Neural network with skip-layer connections
In theory, skip-layer connections should not improve on the network performance. But, since complex networks are hard to train and easy to overfit it may be very useful to explicitly add this as a linear regression term, when you know that your data has a strong linear component. This hints the model in a right direction... In addition, this is more interpretable since it presents your model as linear + perturbations, unraveling a bit of a structure behind the network, which is usually seen merely as a black box.
|
Neural network with skip-layer connections
In theory, skip-layer connections should not improve on the network performance. But, since complex networks are hard to train and easy to overfit it may be very useful to explicitly add this as a lin
|
6,744
|
Neural network with skip-layer connections
|
My old neural network toolbox (I mostly use kernel machines these days) used L1 regularisation to prune away redundant weights and hidden units, and also had skip-layer connections. This has the advantage that if the problem is essentially linear, the hidden units tend to get pruned and you are left with a linear model, which clearly tells you that the problem is linear.
As sashkello (+1) suggests, MLPs are universal approximators, so skip layer connections won't improve results in the limit of infinite data and an infinite number of hidden units (but when do we ever approach that limit?). The real advantage is that it makes estimating good values for the weights easier if the network architecture is well matched to the problem, and you may be able to use a smaller network and obtain better generalisation performance.
However, as with most neural network questions, generally the only way to find out if it will be helpful or harmful for a particular dataset is to try it and see (using a reliable performance evaluation procedure).
|
Neural network with skip-layer connections
|
My old neural network toolbox (I mostly use kernel machines these days) used L1 regularisation to prune away redundant weights and hidden units, and also had skip-layer connections. This has the adva
|
Neural network with skip-layer connections
My old neural network toolbox (I mostly use kernel machines these days) used L1 regularisation to prune away redundant weights and hidden units, and also had skip-layer connections. This has the advantage that if the problem is essentially linear, the hidden units tend to get pruned and you are left with a linear model, which clearly tells you that the problem is linear.
As sashkello (+1) suggests, MLPs are universal approximators, so skip layer connections won't improve results in the limit of infinite data and an infinite number of hidden units (but when do we ever approach that limit?). The real advantage is that it makes estimating good values for the weights easier if the network architecture is well matched to the problem, and you may be able to use a smaller network and obtain better generalisation performance.
However, as with most neural network questions, generally the only way to find out if it will be helpful or harmful for a particular dataset is to try it and see (using a reliable performance evaluation procedure).
|
Neural network with skip-layer connections
My old neural network toolbox (I mostly use kernel machines these days) used L1 regularisation to prune away redundant weights and hidden units, and also had skip-layer connections. This has the adva
|
6,745
|
Neural network with skip-layer connections
|
An in-depth explanation of skip connections from multiple perspectives can be found here:
https://theaisummer.com/skip-connections/
Here I provide the main point from the article:
Basically, skip connection is a standard module in many convolutional architectures. By using a skip connection, we provide an alternative path for the gradient (with backpropagation). It is experimentally validated that these additional paths are often beneficial for the model convergence. Skip connections in deep architectures, as the name suggests, skip some layer in the neural network and feeds the output of one layer as the input to the next layers (instead of only the next one).
As previously explained, using the chain rule, we must keep multiplying terms with the error gradient as we go backwards. However, in the long chain of multiplication, if we multiply many things together that are less than one, then the resulting gradient will be very small. Thus, the gradient becomes very small as we approach the earlier layers in a deep architecture. In some cases, the gradient becomes zero, meaning that we do not update the early layers at all.
In general, there are two fundamental ways that one could use skip connections through different non-sequential layers:
a) addition as in residual architectures,
b) concatenation as in densely connected architectures.
I hope it clarifies your understanding!
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Neural network with skip-layer connections
|
An in-depth explanation of skip connections from multiple perspectives can be found here:
https://theaisummer.com/skip-connections/
Here I provide the main point from the article:
Basically, skip conn
|
Neural network with skip-layer connections
An in-depth explanation of skip connections from multiple perspectives can be found here:
https://theaisummer.com/skip-connections/
Here I provide the main point from the article:
Basically, skip connection is a standard module in many convolutional architectures. By using a skip connection, we provide an alternative path for the gradient (with backpropagation). It is experimentally validated that these additional paths are often beneficial for the model convergence. Skip connections in deep architectures, as the name suggests, skip some layer in the neural network and feeds the output of one layer as the input to the next layers (instead of only the next one).
As previously explained, using the chain rule, we must keep multiplying terms with the error gradient as we go backwards. However, in the long chain of multiplication, if we multiply many things together that are less than one, then the resulting gradient will be very small. Thus, the gradient becomes very small as we approach the earlier layers in a deep architecture. In some cases, the gradient becomes zero, meaning that we do not update the early layers at all.
In general, there are two fundamental ways that one could use skip connections through different non-sequential layers:
a) addition as in residual architectures,
b) concatenation as in densely connected architectures.
I hope it clarifies your understanding!
|
Neural network with skip-layer connections
An in-depth explanation of skip connections from multiple perspectives can be found here:
https://theaisummer.com/skip-connections/
Here I provide the main point from the article:
Basically, skip conn
|
6,746
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Neural network with skip-layer connections
|
Based on Bishop 5.1. Feed-forward Network Functions:
[A way to generalize] of the network architecture is to include skip-layer connections, each of which is associated with a corresponding adaptive parameter. For instance, in a two-layer network these would go directly from inputs to outputs. In
principle, a network with sigmoidal hidden units can always mimic skip layer connections (for bounded input values) by using a sufficiently small first-layer weight that, over its operating range, the hidden unit is effectively linear, and then compensating with a large weight value from the hidden unit to the output.
In practice,
however, it may be advantageous to include skip-layer connections explicitly.
|
Neural network with skip-layer connections
|
Based on Bishop 5.1. Feed-forward Network Functions:
[A way to generalize] of the network architecture is to include skip-layer connections, each of which is associated with a corresponding adaptive
|
Neural network with skip-layer connections
Based on Bishop 5.1. Feed-forward Network Functions:
[A way to generalize] of the network architecture is to include skip-layer connections, each of which is associated with a corresponding adaptive parameter. For instance, in a two-layer network these would go directly from inputs to outputs. In
principle, a network with sigmoidal hidden units can always mimic skip layer connections (for bounded input values) by using a sufficiently small first-layer weight that, over its operating range, the hidden unit is effectively linear, and then compensating with a large weight value from the hidden unit to the output.
In practice,
however, it may be advantageous to include skip-layer connections explicitly.
|
Neural network with skip-layer connections
Based on Bishop 5.1. Feed-forward Network Functions:
[A way to generalize] of the network architecture is to include skip-layer connections, each of which is associated with a corresponding adaptive
|
6,747
|
How did scientists figure out the shape of the normal distribution probability density function?
|
"The Evolution of the Normal Distribution" by SAUL STAHL is the best source of information to answer pretty much all the questions in your post. I'll recite a few points for your convenience only, because you'll find the detailed discussion inside the paper.
This is probably an amateur question
No, it's an interesting question to anyone who uses statistics, because this is not covered in detail anywhere in standard courses.
Basically what bugs me is that for someone it would perhaps be more intuitive that the probability function of normally distributed data has a shape of an isosceles triangle rather than a bell curve, and how would you prove to such a person that the probability density function of all normally distributed data has a bell shape?
Look at this picture from the paper. It shows the error curves that Simpson came up with before Gaussian (Normal) was discovered to analyze experimental data. So, your intuition is spot on.
By experiment?
Yes, that's why they were called "error curves". The experiment was astronomical measurements. Astronomers struggled with measurement errors for centuries.
Or by some mathematical derivation?
Again, YES! Long story short: the analysis of errors in astronomical data led Gauss to his (aka Normal) distribution. These are the assumptions he used:
By the way, Laplace used a few different approaches, and also came up with his distribution too while working with astronomical data:
As to why normal distribution shows in experiment as measurement errors, here's a typical "hand-wavy" explanation physicist are used to give (a quote from Gerhard Bohm, Günter Zech, Introduction to Statistics and Data
Analysis for Physicists p.85):
Many experimental signals follow to a very good approximation a normal
distribution. This is due to the fact that they consist of the sum of
many contributions and a consequence of the central limit theorem.
|
How did scientists figure out the shape of the normal distribution probability density function?
|
"The Evolution of the Normal Distribution" by SAUL STAHL is the best source of information to answer pretty much all the questions in your post. I'll recite a few points for your convenience only, bec
|
How did scientists figure out the shape of the normal distribution probability density function?
"The Evolution of the Normal Distribution" by SAUL STAHL is the best source of information to answer pretty much all the questions in your post. I'll recite a few points for your convenience only, because you'll find the detailed discussion inside the paper.
This is probably an amateur question
No, it's an interesting question to anyone who uses statistics, because this is not covered in detail anywhere in standard courses.
Basically what bugs me is that for someone it would perhaps be more intuitive that the probability function of normally distributed data has a shape of an isosceles triangle rather than a bell curve, and how would you prove to such a person that the probability density function of all normally distributed data has a bell shape?
Look at this picture from the paper. It shows the error curves that Simpson came up with before Gaussian (Normal) was discovered to analyze experimental data. So, your intuition is spot on.
By experiment?
Yes, that's why they were called "error curves". The experiment was astronomical measurements. Astronomers struggled with measurement errors for centuries.
Or by some mathematical derivation?
Again, YES! Long story short: the analysis of errors in astronomical data led Gauss to his (aka Normal) distribution. These are the assumptions he used:
By the way, Laplace used a few different approaches, and also came up with his distribution too while working with astronomical data:
As to why normal distribution shows in experiment as measurement errors, here's a typical "hand-wavy" explanation physicist are used to give (a quote from Gerhard Bohm, Günter Zech, Introduction to Statistics and Data
Analysis for Physicists p.85):
Many experimental signals follow to a very good approximation a normal
distribution. This is due to the fact that they consist of the sum of
many contributions and a consequence of the central limit theorem.
|
How did scientists figure out the shape of the normal distribution probability density function?
"The Evolution of the Normal Distribution" by SAUL STAHL is the best source of information to answer pretty much all the questions in your post. I'll recite a few points for your convenience only, bec
|
6,748
|
How did scientists figure out the shape of the normal distribution probability density function?
|
You seem to assume in your question that the concept of the normal distribution was around before the distribution was identified, and people tried to figure out what it was. It's not clear to me how that would work. [Edit: there is at least one sense it which we might consider there being a "search for a distribution" but it's not "a search for a distribution that describes lots and lots of phenomena"]
This is not the case; the distribution was known about before it was called the normal distribution.
how would you prove to such a person that the probability density function of all normally distributed data has a bell shape
The normal distribution function is the thing that has what is usually called a "bell shape" -- all normal distributions have the same "shape" (in the sense that they only differ in scale and location).
Data can look more or less "bell-shaped" in distribution but that doesn't make it normal. Lots of non-normal distributions look similarly "bell-shaped".
The actual population distributions that data are drawn from are likely never actually normal, though it's sometimes quite a reasonable approximation.
This is typically true of almost all the distributions we apply to things in the real world -- they're models, not facts about the world. [As an example, if we make certain assumptions (those for a Poisson process), we can derive the Poisson distribution -- a widely used distribution. But are those assumptions ever exactly satisfied? Generally the best we can say (in the right situations) is that they're very nearly true.]
what do we actually consider normally distributed data? Data that follows the probability pattern of a normal distribution, or something else?
Yes, to actually be normally distributed, the population the sample was drawn from would have to have a distribution that has the exact functional form of a normal distribution. As a result, any finite population cannot be normal. Variables that necessarily bounded cannot be normal (for example, times taken for particular tasks, lengths of particular things cannot be negative, so they cannot actually be normally distributed).
it would perhaps be more intuitive that the probability function of normally distributed data has a shape of an isosceles triangle
I don't see why this is necessarily more intuitive. It's certainly simpler.
When first developing models for error distributions (specifically for astronomy in the early period), mathematicians considered a variety of shapes in relation to error distributions (including at one early point a triangular distribution), but in much of this work it was mathematics (rather than intuition) that was used. Laplace looked at double exponential and normal distributions (among several others), for example. Similarly Gauss used mathematics to derive it at around the same time, but in relation to a different set of considerations than Laplace did.
In the narrow sense that Laplace and Gauss were considering "distributions of errors", we could regard there as being a "search for a distribution", at least for a time. Both postulated some properties for a distribution of errors they considered important (Laplace considered a sequence of somewhat different criteria over time) led to different distributions.
Basically my question is why does the normal distribution probability density function has a bell shape and not any other?
The functional form of the thing that is called the normal density function gives it that shape. Consider the standard normal (for simplicity; every other normal has the same shape, differing only in scale and location):
$$f_Z(z) = k \cdot e^{-\frac12 z^2};\;-\infty<z<\infty$$
(where $k$ is simply a constant chosen to make the total area 1)
this defines the value of the density at every value of $x$, so it completely describes the shape of the density. That mathematical object is the thing we attach the label "normal distribution" to. There's nothing special about the name; it's just a label we attach to the distribution. It's had many names (and is still called different things by different people).
While some people have regarded the normal distribution as somehow "usual" it's really only in particular sets of situations that you even tend to see it as an approximation.
The discovery of the distribution is usually credited to de Moivre (as an approximation to the binomial). He in effect derived the functional form when trying to approximate binomial coefficients (/binomial probabilities) to approximate otherwise tedious calculations but - while he does effectively derive the form of the normal distribution - he doesn't seem to have thought about his approximation as a probability distribution, though some authors do suggest that he did. A certain amount of interpretation is required so there's scope for differences in that interpretation.
Gauss and Laplace did work on it in the early 1800s; Gauss wrote about it in 1809 (in connection with it being the distribution for which the mean is the MLE of the center) and Laplace in 1810, as an approximation to the distribution of sums of symmetric random variables. A decade later Laplace gives an early form of central limit theorem, for discrete and for continuous variables.
Early names for the distribution include the law of error, the law of frequency of errors, and it was also named after both Laplace and Gauss, sometimes jointly.
The term "normal" was used to describe the distribution independently by three different authors in the 1870s (Peirce, Lexis and Galton), the first in 1873 and the other two in 1877. This is more than sixty years after the work by Gauss and Laplace and more than twice that since de Moivre's approximation. Galton's use of it was probably most influential but he used the term "normal" in relation to it only once in that 1877 work (mostly calling it "the law of deviation").
However, in the 1880s Galton used the adjective "normal" in relation to the distribution numerous times (e.g. as the "normal curve" in 1889), and he in turn had a lot of influence on later statisticians in the UK (especially Karl Pearson). He didn't say why he used the term "normal" in this way, but presumably meant it in the sense of "typical" or "usual".
The first explicit use of the phrase "normal distribution" appears to be by Karl Pearson; he certainly uses it in 1894, though he claims to have used it long before (a claim I would view with some caution).
References:
Miller, Jeff
"Earliest Known Uses of Some of the Words of Mathematics:"
Normal distribution (Entry by John Aldrich)
http://jeff560.tripod.com/n.html
(alternate: https://mathshistory.st-andrews.ac.uk/Miller/mathword/n/)
Stahl, Saul (2006),
"The Evolution of the Normal Distribution",
Mathematics Magazine, Vol. 79, No. 2 (April), pp 96-113
https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/stahl96.pdf
Normal distribution, (2016, August 1).
In Wikipedia, The Free Encyclopedia.
Retrieved 12:02, August 3, 2016, from
https://en.wikipedia.org/w/index.php?title=Normal_distribution&oldid=732559095#History
Hald, A (2007),
"De Moivre’s Normal Approximation to the Binomial, 1733, and Its Generalization",
In: A History of Parametric Statistical Inference from Bernoulli to Fisher, 1713–1935; pp 17-24
[You may note substantial discrepancies between these sources in relation to their account of de Moivre]
|
How did scientists figure out the shape of the normal distribution probability density function?
|
You seem to assume in your question that the concept of the normal distribution was around before the distribution was identified, and people tried to figure out what it was. It's not clear to me how
|
How did scientists figure out the shape of the normal distribution probability density function?
You seem to assume in your question that the concept of the normal distribution was around before the distribution was identified, and people tried to figure out what it was. It's not clear to me how that would work. [Edit: there is at least one sense it which we might consider there being a "search for a distribution" but it's not "a search for a distribution that describes lots and lots of phenomena"]
This is not the case; the distribution was known about before it was called the normal distribution.
how would you prove to such a person that the probability density function of all normally distributed data has a bell shape
The normal distribution function is the thing that has what is usually called a "bell shape" -- all normal distributions have the same "shape" (in the sense that they only differ in scale and location).
Data can look more or less "bell-shaped" in distribution but that doesn't make it normal. Lots of non-normal distributions look similarly "bell-shaped".
The actual population distributions that data are drawn from are likely never actually normal, though it's sometimes quite a reasonable approximation.
This is typically true of almost all the distributions we apply to things in the real world -- they're models, not facts about the world. [As an example, if we make certain assumptions (those for a Poisson process), we can derive the Poisson distribution -- a widely used distribution. But are those assumptions ever exactly satisfied? Generally the best we can say (in the right situations) is that they're very nearly true.]
what do we actually consider normally distributed data? Data that follows the probability pattern of a normal distribution, or something else?
Yes, to actually be normally distributed, the population the sample was drawn from would have to have a distribution that has the exact functional form of a normal distribution. As a result, any finite population cannot be normal. Variables that necessarily bounded cannot be normal (for example, times taken for particular tasks, lengths of particular things cannot be negative, so they cannot actually be normally distributed).
it would perhaps be more intuitive that the probability function of normally distributed data has a shape of an isosceles triangle
I don't see why this is necessarily more intuitive. It's certainly simpler.
When first developing models for error distributions (specifically for astronomy in the early period), mathematicians considered a variety of shapes in relation to error distributions (including at one early point a triangular distribution), but in much of this work it was mathematics (rather than intuition) that was used. Laplace looked at double exponential and normal distributions (among several others), for example. Similarly Gauss used mathematics to derive it at around the same time, but in relation to a different set of considerations than Laplace did.
In the narrow sense that Laplace and Gauss were considering "distributions of errors", we could regard there as being a "search for a distribution", at least for a time. Both postulated some properties for a distribution of errors they considered important (Laplace considered a sequence of somewhat different criteria over time) led to different distributions.
Basically my question is why does the normal distribution probability density function has a bell shape and not any other?
The functional form of the thing that is called the normal density function gives it that shape. Consider the standard normal (for simplicity; every other normal has the same shape, differing only in scale and location):
$$f_Z(z) = k \cdot e^{-\frac12 z^2};\;-\infty<z<\infty$$
(where $k$ is simply a constant chosen to make the total area 1)
this defines the value of the density at every value of $x$, so it completely describes the shape of the density. That mathematical object is the thing we attach the label "normal distribution" to. There's nothing special about the name; it's just a label we attach to the distribution. It's had many names (and is still called different things by different people).
While some people have regarded the normal distribution as somehow "usual" it's really only in particular sets of situations that you even tend to see it as an approximation.
The discovery of the distribution is usually credited to de Moivre (as an approximation to the binomial). He in effect derived the functional form when trying to approximate binomial coefficients (/binomial probabilities) to approximate otherwise tedious calculations but - while he does effectively derive the form of the normal distribution - he doesn't seem to have thought about his approximation as a probability distribution, though some authors do suggest that he did. A certain amount of interpretation is required so there's scope for differences in that interpretation.
Gauss and Laplace did work on it in the early 1800s; Gauss wrote about it in 1809 (in connection with it being the distribution for which the mean is the MLE of the center) and Laplace in 1810, as an approximation to the distribution of sums of symmetric random variables. A decade later Laplace gives an early form of central limit theorem, for discrete and for continuous variables.
Early names for the distribution include the law of error, the law of frequency of errors, and it was also named after both Laplace and Gauss, sometimes jointly.
The term "normal" was used to describe the distribution independently by three different authors in the 1870s (Peirce, Lexis and Galton), the first in 1873 and the other two in 1877. This is more than sixty years after the work by Gauss and Laplace and more than twice that since de Moivre's approximation. Galton's use of it was probably most influential but he used the term "normal" in relation to it only once in that 1877 work (mostly calling it "the law of deviation").
However, in the 1880s Galton used the adjective "normal" in relation to the distribution numerous times (e.g. as the "normal curve" in 1889), and he in turn had a lot of influence on later statisticians in the UK (especially Karl Pearson). He didn't say why he used the term "normal" in this way, but presumably meant it in the sense of "typical" or "usual".
The first explicit use of the phrase "normal distribution" appears to be by Karl Pearson; he certainly uses it in 1894, though he claims to have used it long before (a claim I would view with some caution).
References:
Miller, Jeff
"Earliest Known Uses of Some of the Words of Mathematics:"
Normal distribution (Entry by John Aldrich)
http://jeff560.tripod.com/n.html
(alternate: https://mathshistory.st-andrews.ac.uk/Miller/mathword/n/)
Stahl, Saul (2006),
"The Evolution of the Normal Distribution",
Mathematics Magazine, Vol. 79, No. 2 (April), pp 96-113
https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/stahl96.pdf
Normal distribution, (2016, August 1).
In Wikipedia, The Free Encyclopedia.
Retrieved 12:02, August 3, 2016, from
https://en.wikipedia.org/w/index.php?title=Normal_distribution&oldid=732559095#History
Hald, A (2007),
"De Moivre’s Normal Approximation to the Binomial, 1733, and Its Generalization",
In: A History of Parametric Statistical Inference from Bernoulli to Fisher, 1713–1935; pp 17-24
[You may note substantial discrepancies between these sources in relation to their account of de Moivre]
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How did scientists figure out the shape of the normal distribution probability density function?
You seem to assume in your question that the concept of the normal distribution was around before the distribution was identified, and people tried to figure out what it was. It's not clear to me how
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6,749
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How did scientists figure out the shape of the normal distribution probability density function?
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The "normal" distribution is defined to be that particular distribution.
The question is why would we expect this particular distribution to be common in nature, and why is it so often used as an approximation even when the real data does not exactly follow that distribution? (Real data is often found to have a "fat tail", i.e. values far from the mean are much more common than the normal distribution would predict).
To put it another way, what is special about the normal distribution?
The normal has a lot of "nice" statistical properties, (see e.g. https://en.wikipedia.org/wiki/Central_limit_theorem), but the most relevant IMO is the fact that is the "maximum entropy" function for any distribution with a given mean and variance. https://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution
To express this in ordinary language, if you are given only the mean (central point) and variance (width) of a distribution, and you assume nothing else whatsoever about it, you will be forced to draw a normal distribution. Anything else requires additional information (in the sense of Shannon information theory), for example skewness, to determine it.
The principle of maximum entropy was introduced by E.T. Jaynes as a way of determining reasonable priors in Bayesian inference, and I think he was the first to draw attention to this property.
See this for further discussion: http://www.inf.fu-berlin.de/inst/ag-ki/rojas_home/documents/tutorials/Gaussian-distribution.pdf
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How did scientists figure out the shape of the normal distribution probability density function?
|
The "normal" distribution is defined to be that particular distribution.
The question is why would we expect this particular distribution to be common in nature, and why is it so often used as an appr
|
How did scientists figure out the shape of the normal distribution probability density function?
The "normal" distribution is defined to be that particular distribution.
The question is why would we expect this particular distribution to be common in nature, and why is it so often used as an approximation even when the real data does not exactly follow that distribution? (Real data is often found to have a "fat tail", i.e. values far from the mean are much more common than the normal distribution would predict).
To put it another way, what is special about the normal distribution?
The normal has a lot of "nice" statistical properties, (see e.g. https://en.wikipedia.org/wiki/Central_limit_theorem), but the most relevant IMO is the fact that is the "maximum entropy" function for any distribution with a given mean and variance. https://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution
To express this in ordinary language, if you are given only the mean (central point) and variance (width) of a distribution, and you assume nothing else whatsoever about it, you will be forced to draw a normal distribution. Anything else requires additional information (in the sense of Shannon information theory), for example skewness, to determine it.
The principle of maximum entropy was introduced by E.T. Jaynes as a way of determining reasonable priors in Bayesian inference, and I think he was the first to draw attention to this property.
See this for further discussion: http://www.inf.fu-berlin.de/inst/ag-ki/rojas_home/documents/tutorials/Gaussian-distribution.pdf
|
How did scientists figure out the shape of the normal distribution probability density function?
The "normal" distribution is defined to be that particular distribution.
The question is why would we expect this particular distribution to be common in nature, and why is it so often used as an appr
|
6,750
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How did scientists figure out the shape of the normal distribution probability density function?
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The Normal Distribution (aka "Gaussian Distribution") has a firm mathematical foundation. The Central Limit Theorem says that if you have a finite set of n independent and identically distributed random variables having a specific mean and variance, and you take the average of those random variables, the distribution of the result will converge to a Gaussian Distribution as n goes to infinity. There is no guesswork here, since the mathematical derivation leads to this specific distribution function and no other.
To put this into more tangible terms, consider a single random variable, such as flipping a fair coin (2 equally possible outcomes). The odds of getting a particular outcome is 1/2 for heads and 1/2 for tails.
If you increase the number of coins and keep track of the total number of heads obtained with each trial, then you will get a Binomial Distribution, which has a roughly bell shape. Just graph with the number of heads along the x-axis, and the number of times you flipped that many heads along the y-axis.
The more coins you use, and the more times you flip the coins, the closer the graph will come to looking like a Gaussian bell curve. That's what the Central Limit Theorem asserts.
The amazing thing is that the theorem does not depend on how the random variables are actually distributed, just so long as each of the random variables has the same distribution. One key idea in the theorem is that you are adding or averaging the random variables. Another key concept is that the theorem is describing the mathematical limit as the number of random variables becomes larger and larger. The more variables you use, the closer the distribution will approach a Normal Distribution.
I recommend you take a class in Mathematical Statistics if you want to see how mathematicians determined that the Normal Distribution is actually the mathematically correct function for the bell curve.
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How did scientists figure out the shape of the normal distribution probability density function?
|
The Normal Distribution (aka "Gaussian Distribution") has a firm mathematical foundation. The Central Limit Theorem says that if you have a finite set of n independent and identically distributed ran
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How did scientists figure out the shape of the normal distribution probability density function?
The Normal Distribution (aka "Gaussian Distribution") has a firm mathematical foundation. The Central Limit Theorem says that if you have a finite set of n independent and identically distributed random variables having a specific mean and variance, and you take the average of those random variables, the distribution of the result will converge to a Gaussian Distribution as n goes to infinity. There is no guesswork here, since the mathematical derivation leads to this specific distribution function and no other.
To put this into more tangible terms, consider a single random variable, such as flipping a fair coin (2 equally possible outcomes). The odds of getting a particular outcome is 1/2 for heads and 1/2 for tails.
If you increase the number of coins and keep track of the total number of heads obtained with each trial, then you will get a Binomial Distribution, which has a roughly bell shape. Just graph with the number of heads along the x-axis, and the number of times you flipped that many heads along the y-axis.
The more coins you use, and the more times you flip the coins, the closer the graph will come to looking like a Gaussian bell curve. That's what the Central Limit Theorem asserts.
The amazing thing is that the theorem does not depend on how the random variables are actually distributed, just so long as each of the random variables has the same distribution. One key idea in the theorem is that you are adding or averaging the random variables. Another key concept is that the theorem is describing the mathematical limit as the number of random variables becomes larger and larger. The more variables you use, the closer the distribution will approach a Normal Distribution.
I recommend you take a class in Mathematical Statistics if you want to see how mathematicians determined that the Normal Distribution is actually the mathematically correct function for the bell curve.
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How did scientists figure out the shape of the normal distribution probability density function?
The Normal Distribution (aka "Gaussian Distribution") has a firm mathematical foundation. The Central Limit Theorem says that if you have a finite set of n independent and identically distributed ran
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6,751
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How did scientists figure out the shape of the normal distribution probability density function?
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There are some excellent answers on this thread. I can't help feeling the OP wasn't asking the same question as everyone wants to answer. I get that, though, because this is close to being one of the most exciting questions to answer - I actually found it because I was hoping someone had the question "How do we know the normal PDF is a PDF?" and I searched for it. But I think the answer to the question may be to demonstrate the origin of the normal distribution.
The normal distribution was first designed to be used to approximate the binomial distribution for very large $n$. In 1744, a mathematician named De Moivre showed that the binomial distribution, for large $n$, has very similar probabilities to a normal distribution with mean $np$ and variance $np(1-p)$. The proof of this follows pretty naturally from taking the limit of the binomial pdf as $n\to\infty$, and replacing the factorial values with Stirling's approximation.
But I am again tempted to get very deep into the proof that this happens, and I don't know that is what the OP wanted. If interested, it is explained here. Just know that we can "easily" prove that the limit of the binomial distribution as $n\to\infty$ and $p\to0$ such that $np=1$ is a normal distribution.
Taking that knowledge, we can see why the normal distribution is bell shaped if we can see why the binomial distribution is bell shaped, which is much easier to see. Go ahead and try it for yourself - make a discrete graph of the binomial probabilities for $n=10$ and $p=0.5$. How is it shaped? What about a discrete graph of the binomial probabilities for $n=100$ and $p=0.5$? Indeed, do it empirically, generate some random data distributed Binomially and see how the histogram looks! Of course, it's a pretty blocky looking bell, but it gets more curvy the higher $n$ is. But why is it bell-shaped at all?
If I dump 100 coins on the ground right now and count how many heads I get, I might count 0 heads, or I might count 100 heads, but I'm way more likely to count a number somewhere in between. Do you see why this histogram should be bell shaped?
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How did scientists figure out the shape of the normal distribution probability density function?
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There are some excellent answers on this thread. I can't help feeling the OP wasn't asking the same question as everyone wants to answer. I get that, though, because this is close to being one of th
|
How did scientists figure out the shape of the normal distribution probability density function?
There are some excellent answers on this thread. I can't help feeling the OP wasn't asking the same question as everyone wants to answer. I get that, though, because this is close to being one of the most exciting questions to answer - I actually found it because I was hoping someone had the question "How do we know the normal PDF is a PDF?" and I searched for it. But I think the answer to the question may be to demonstrate the origin of the normal distribution.
The normal distribution was first designed to be used to approximate the binomial distribution for very large $n$. In 1744, a mathematician named De Moivre showed that the binomial distribution, for large $n$, has very similar probabilities to a normal distribution with mean $np$ and variance $np(1-p)$. The proof of this follows pretty naturally from taking the limit of the binomial pdf as $n\to\infty$, and replacing the factorial values with Stirling's approximation.
But I am again tempted to get very deep into the proof that this happens, and I don't know that is what the OP wanted. If interested, it is explained here. Just know that we can "easily" prove that the limit of the binomial distribution as $n\to\infty$ and $p\to0$ such that $np=1$ is a normal distribution.
Taking that knowledge, we can see why the normal distribution is bell shaped if we can see why the binomial distribution is bell shaped, which is much easier to see. Go ahead and try it for yourself - make a discrete graph of the binomial probabilities for $n=10$ and $p=0.5$. How is it shaped? What about a discrete graph of the binomial probabilities for $n=100$ and $p=0.5$? Indeed, do it empirically, generate some random data distributed Binomially and see how the histogram looks! Of course, it's a pretty blocky looking bell, but it gets more curvy the higher $n$ is. But why is it bell-shaped at all?
If I dump 100 coins on the ground right now and count how many heads I get, I might count 0 heads, or I might count 100 heads, but I'm way more likely to count a number somewhere in between. Do you see why this histogram should be bell shaped?
|
How did scientists figure out the shape of the normal distribution probability density function?
There are some excellent answers on this thread. I can't help feeling the OP wasn't asking the same question as everyone wants to answer. I get that, though, because this is close to being one of th
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6,752
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How did scientists figure out the shape of the normal distribution probability density function?
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Would also mention Maxwell-Herschel derivation of independent multivariate normal distribution from two assumptions:
Distribution is not affected by rotation of the vector.
Components of the vector are independent.
Here is the exposition by Jaynes
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How did scientists figure out the shape of the normal distribution probability density function?
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Would also mention Maxwell-Herschel derivation of independent multivariate normal distribution from two assumptions:
Distribution is not affected by rotation of the vector.
Components of the vector a
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How did scientists figure out the shape of the normal distribution probability density function?
Would also mention Maxwell-Herschel derivation of independent multivariate normal distribution from two assumptions:
Distribution is not affected by rotation of the vector.
Components of the vector are independent.
Here is the exposition by Jaynes
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How did scientists figure out the shape of the normal distribution probability density function?
Would also mention Maxwell-Herschel derivation of independent multivariate normal distribution from two assumptions:
Distribution is not affected by rotation of the vector.
Components of the vector a
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6,753
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Why is the null hypothesis often sought to be rejected?
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The purpose of statistical hypothesis testing is largely to impose self-skepticism, making us cautious about promulgating our hypothesis unless there is reasonable evidence to support it. Thus in the usual form of hypothesis testing the null hypothesis provides a "devils advocate", arguing against us, and only promulgate our hypothesis if we can show that the observations mean that it is unlikely that the advocate's argument is sound. So we take $H_0$ to be the thing we don't want to be true and then see if we are able to reject it. If we can reject it, it doesn't mean that our hypothesis is likely to be correct, just that it has passed this basic hurdle and so is worthy of consideration. If we can't, it doesn't mean that our hypothesis is false, it may be that we just don't have enough data to provide suffcient evidence. As @Bahgat rightly suggests (+1) this is very much the idea of Popper's falsificationism idea.
However, it is possible to have a test where $H_0$ is the thing you want to be true, but in order for that to work, you need to show that the test has sufficiently high statistical power in order to be confident of rejecting the null if it actually is false. Computing statistical power is rather more difficult that performing the test, which is why this form of testing is rarely used and the alternative where $H_0$ is what you don't want to be true is normally used instead.
So you don't have to take $H_0$ to oppose your hypothesis, but it does make the testing procedure much easier.
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Why is the null hypothesis often sought to be rejected?
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The purpose of statistical hypothesis testing is largely to impose self-skepticism, making us cautious about promulgating our hypothesis unless there is reasonable evidence to support it. Thus in the
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Why is the null hypothesis often sought to be rejected?
The purpose of statistical hypothesis testing is largely to impose self-skepticism, making us cautious about promulgating our hypothesis unless there is reasonable evidence to support it. Thus in the usual form of hypothesis testing the null hypothesis provides a "devils advocate", arguing against us, and only promulgate our hypothesis if we can show that the observations mean that it is unlikely that the advocate's argument is sound. So we take $H_0$ to be the thing we don't want to be true and then see if we are able to reject it. If we can reject it, it doesn't mean that our hypothesis is likely to be correct, just that it has passed this basic hurdle and so is worthy of consideration. If we can't, it doesn't mean that our hypothesis is false, it may be that we just don't have enough data to provide suffcient evidence. As @Bahgat rightly suggests (+1) this is very much the idea of Popper's falsificationism idea.
However, it is possible to have a test where $H_0$ is the thing you want to be true, but in order for that to work, you need to show that the test has sufficiently high statistical power in order to be confident of rejecting the null if it actually is false. Computing statistical power is rather more difficult that performing the test, which is why this form of testing is rarely used and the alternative where $H_0$ is what you don't want to be true is normally used instead.
So you don't have to take $H_0$ to oppose your hypothesis, but it does make the testing procedure much easier.
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Why is the null hypothesis often sought to be rejected?
The purpose of statistical hypothesis testing is largely to impose self-skepticism, making us cautious about promulgating our hypothesis unless there is reasonable evidence to support it. Thus in the
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6,754
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Why is the null hypothesis often sought to be rejected?
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Karl Popper says "We cannot conclusively affirm a hypothesis, but we can conclusively negate it". So when we do hypothesis testing in statistics, we try to negate (reject) the opposite hypothesis (the null hypothesis) of the hypothesis we are interested in (the alternative hypothesis) and which we can not affirm. Since we can specify the null hypothesis easily, but we don't know what exactly the alternative hypothesis is. We may hypothesize for example that there is a mean difference between the two populations, but we cannot point out how wide the gap would be.
See also Don't believe in the null hypothesis?
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Why is the null hypothesis often sought to be rejected?
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Karl Popper says "We cannot conclusively affirm a hypothesis, but we can conclusively negate it". So when we do hypothesis testing in statistics, we try to negate (reject) the opposite hypothesis (the
|
Why is the null hypothesis often sought to be rejected?
Karl Popper says "We cannot conclusively affirm a hypothesis, but we can conclusively negate it". So when we do hypothesis testing in statistics, we try to negate (reject) the opposite hypothesis (the null hypothesis) of the hypothesis we are interested in (the alternative hypothesis) and which we can not affirm. Since we can specify the null hypothesis easily, but we don't know what exactly the alternative hypothesis is. We may hypothesize for example that there is a mean difference between the two populations, but we cannot point out how wide the gap would be.
See also Don't believe in the null hypothesis?
|
Why is the null hypothesis often sought to be rejected?
Karl Popper says "We cannot conclusively affirm a hypothesis, but we can conclusively negate it". So when we do hypothesis testing in statistics, we try to negate (reject) the opposite hypothesis (the
|
6,755
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Why is the null hypothesis often sought to be rejected?
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It is not taken for granted that the null is always to be rejected. In model fit testing, the null is usually that the model fits well, and that is something desirable that we would hate to reject. It is, however, usually true that the sampling distribution of the test statistic is easier to derive under the null, which is usually far more restrictive than the alternative. The null that the mean difference between two groups is zero leads to a $t$-test; the null that the two distributions are the same leading to Kolmogorov-Smirnov test; the null that the linear regression model does not need nonlinear terms via Ramsey RESET test; the null that a latent variable model describes the observed covariance matrix adequately leads to a parametric space of lower dimension than an unrestricted alternative, and an asymptotic chi-square test of the distance to the smooth parametric surface in the space of $p(p+1)/2$ covariances defined by the model. So my take on this is that, as @whuber put it in the comment below, the null is usually a crucial albeit convenient technical assumption. The null is either a point (potentially multivariate) in the parametric space, so that the sampling distribution is fully specified; or a restricted parametric space, with the alternative that can be formulated to be complementary in that space, and the test statistic is based on a distance from the richer set of parameters under the alternative to the set with restrictions under the null; or, in nonparametric rank/order statistics world, the distribution under the null can be derived by the complete enumeration of all possible samples and outcomes (often approximated by something normal in large samples though).
Taking the null as something different (e.g., that the means of the two groups differ by at most 0.01, with the alternative differ by more than 0.01) requires a more complicated set of derivations, e.g., looking at the worst possible situations, which in the above case would still boil down to the point null against a one-sided alternative. The worst case on the right is $H_0: \mu_2 = \mu_1 + 0.01$ vs. $H_1: \mu_2 > \mu_1 +1$, and on the left it is $H_0: \mu_2 = \mu_1 - 0.01$ vs. $H_1: \mu_2 < \mu_1 - 1$.
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Why is the null hypothesis often sought to be rejected?
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It is not taken for granted that the null is always to be rejected. In model fit testing, the null is usually that the model fits well, and that is something desirable that we would hate to reject. It
|
Why is the null hypothesis often sought to be rejected?
It is not taken for granted that the null is always to be rejected. In model fit testing, the null is usually that the model fits well, and that is something desirable that we would hate to reject. It is, however, usually true that the sampling distribution of the test statistic is easier to derive under the null, which is usually far more restrictive than the alternative. The null that the mean difference between two groups is zero leads to a $t$-test; the null that the two distributions are the same leading to Kolmogorov-Smirnov test; the null that the linear regression model does not need nonlinear terms via Ramsey RESET test; the null that a latent variable model describes the observed covariance matrix adequately leads to a parametric space of lower dimension than an unrestricted alternative, and an asymptotic chi-square test of the distance to the smooth parametric surface in the space of $p(p+1)/2$ covariances defined by the model. So my take on this is that, as @whuber put it in the comment below, the null is usually a crucial albeit convenient technical assumption. The null is either a point (potentially multivariate) in the parametric space, so that the sampling distribution is fully specified; or a restricted parametric space, with the alternative that can be formulated to be complementary in that space, and the test statistic is based on a distance from the richer set of parameters under the alternative to the set with restrictions under the null; or, in nonparametric rank/order statistics world, the distribution under the null can be derived by the complete enumeration of all possible samples and outcomes (often approximated by something normal in large samples though).
Taking the null as something different (e.g., that the means of the two groups differ by at most 0.01, with the alternative differ by more than 0.01) requires a more complicated set of derivations, e.g., looking at the worst possible situations, which in the above case would still boil down to the point null against a one-sided alternative. The worst case on the right is $H_0: \mu_2 = \mu_1 + 0.01$ vs. $H_1: \mu_2 > \mu_1 +1$, and on the left it is $H_0: \mu_2 = \mu_1 - 0.01$ vs. $H_1: \mu_2 < \mu_1 - 1$.
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Why is the null hypothesis often sought to be rejected?
It is not taken for granted that the null is always to be rejected. In model fit testing, the null is usually that the model fits well, and that is something desirable that we would hate to reject. It
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6,756
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Why is the null hypothesis often sought to be rejected?
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This is a fair and good question. @Tim already gave you all you need to answer your question in a formal way, however if you are not familiar with statistical hypothesis testing you could conceptualize the null hypothesis by thinking about it in a more familiar setting.
Suppose you are being accused of having conducted a crime. Until proven guilty, you are innocent (null hypothesis). The attorney provides evidence that you are guilty (alternative hypothesis), your lawyers try to invalidate this evidence during the trial (the experiment) and in the end the judge rules whether you are innocent given the facts provided by the attorney and the lawyers. If the facts against you are overwhelming, i.e. the probability that you are innocent is very low, the judge (or jury) will conclude that your are guilty given the evidence.
Now with this in mind, you could also conceptualize features of statistical hypothesis testing, for instance why independent measurements (or evidence) are important, since after all your deserve a fair trial.
However, this is example has its limitations and eventually you have to formally understand the concept of the null hypothesis.
So to answer your questions:
Yes there is a reason for the null hypothesis (as described above).
No it's not just a convention, the null hypothesis is the core or statistical hypothesis testing or else it wouldn't work they way it is intended to.
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Why is the null hypothesis often sought to be rejected?
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This is a fair and good question. @Tim already gave you all you need to answer your question in a formal way, however if you are not familiar with statistical hypothesis testing you could conceptuali
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Why is the null hypothesis often sought to be rejected?
This is a fair and good question. @Tim already gave you all you need to answer your question in a formal way, however if you are not familiar with statistical hypothesis testing you could conceptualize the null hypothesis by thinking about it in a more familiar setting.
Suppose you are being accused of having conducted a crime. Until proven guilty, you are innocent (null hypothesis). The attorney provides evidence that you are guilty (alternative hypothesis), your lawyers try to invalidate this evidence during the trial (the experiment) and in the end the judge rules whether you are innocent given the facts provided by the attorney and the lawyers. If the facts against you are overwhelming, i.e. the probability that you are innocent is very low, the judge (or jury) will conclude that your are guilty given the evidence.
Now with this in mind, you could also conceptualize features of statistical hypothesis testing, for instance why independent measurements (or evidence) are important, since after all your deserve a fair trial.
However, this is example has its limitations and eventually you have to formally understand the concept of the null hypothesis.
So to answer your questions:
Yes there is a reason for the null hypothesis (as described above).
No it's not just a convention, the null hypothesis is the core or statistical hypothesis testing or else it wouldn't work they way it is intended to.
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Why is the null hypothesis often sought to be rejected?
This is a fair and good question. @Tim already gave you all you need to answer your question in a formal way, however if you are not familiar with statistical hypothesis testing you could conceptuali
|
6,757
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Why is the null hypothesis often sought to be rejected?
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The law of parsimony (also known as Occam's razor) is a general principle of science. Under that principle, we assume a simple world until it can be shown that the world is more complicated. So, we assume the simpler world of the null hypothesis until it can be falsified. For example:
We assume treatment A and treatment B work the same until we show differently. We assume the weather is the same in San Diego as in Halifax until we show differently, we assume men and women are paid the same until we show differently, etc.
For more, see https://en.wikipedia.org/wiki/Occam%27s_razor
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Why is the null hypothesis often sought to be rejected?
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The law of parsimony (also known as Occam's razor) is a general principle of science. Under that principle, we assume a simple world until it can be shown that the world is more complicated. So, we
|
Why is the null hypothesis often sought to be rejected?
The law of parsimony (also known as Occam's razor) is a general principle of science. Under that principle, we assume a simple world until it can be shown that the world is more complicated. So, we assume the simpler world of the null hypothesis until it can be falsified. For example:
We assume treatment A and treatment B work the same until we show differently. We assume the weather is the same in San Diego as in Halifax until we show differently, we assume men and women are paid the same until we show differently, etc.
For more, see https://en.wikipedia.org/wiki/Occam%27s_razor
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Why is the null hypothesis often sought to be rejected?
The law of parsimony (also known as Occam's razor) is a general principle of science. Under that principle, we assume a simple world until it can be shown that the world is more complicated. So, we
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6,758
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Why is the null hypothesis often sought to be rejected?
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If I can draw an analogy to logic, a general way to prove something is to assume the opposite and see if that leads to a contradiction. Here the null hypothesis is like the opposite, and rejecting it (i.e. showing that it is very unlikely) is like deriving the contradiction.
You do it that way because it's a way to make an unambiguous statement.
Like in my field, it's much easier to say "The statement 'this drug has no benefit' has 5% chance of being right" than it is to say "The statement 'this drug has benefit' has 90% chance of being right".
Of course, people want to know how much benefit is being claimed,
but first they want to know it isn't zero.
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Why is the null hypothesis often sought to be rejected?
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If I can draw an analogy to logic, a general way to prove something is to assume the opposite and see if that leads to a contradiction. Here the null hypothesis is like the opposite, and rejecting it
|
Why is the null hypothesis often sought to be rejected?
If I can draw an analogy to logic, a general way to prove something is to assume the opposite and see if that leads to a contradiction. Here the null hypothesis is like the opposite, and rejecting it (i.e. showing that it is very unlikely) is like deriving the contradiction.
You do it that way because it's a way to make an unambiguous statement.
Like in my field, it's much easier to say "The statement 'this drug has no benefit' has 5% chance of being right" than it is to say "The statement 'this drug has benefit' has 90% chance of being right".
Of course, people want to know how much benefit is being claimed,
but first they want to know it isn't zero.
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Why is the null hypothesis often sought to be rejected?
If I can draw an analogy to logic, a general way to prove something is to assume the opposite and see if that leads to a contradiction. Here the null hypothesis is like the opposite, and rejecting it
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6,759
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Why is the null hypothesis often sought to be rejected?
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The null hypothesis is always formed with the intention to reject it that is the basic idea of hypothesis testing. When you are trying to show that something is likely to be true (e.g. a treatment improves or worsens a disease), then the null hypothesis is the default position (e.g. the treatment does not make a difference to the disease). You generate evidence for your desired claim by accumulating data that is (hopefully) so far away from what should have happened under the null hypothesis (in the example patients that are randomized to receive the treatment or a placebo having the same expected outcome) that one concludes that is very unlikely to have arisen under the null hypothesis so that you can reject the null hypothesis. In contrast failure to reject the null hypothesis does not necessarily make it very likely that the null hypothesis is true (just because a clinical trial failed to show a drug works does not mean that the drug really does nothing).
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Why is the null hypothesis often sought to be rejected?
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The null hypothesis is always formed with the intention to reject it that is the basic idea of hypothesis testing. When you are trying to show that something is likely to be true (e.g. a treatment imp
|
Why is the null hypothesis often sought to be rejected?
The null hypothesis is always formed with the intention to reject it that is the basic idea of hypothesis testing. When you are trying to show that something is likely to be true (e.g. a treatment improves or worsens a disease), then the null hypothesis is the default position (e.g. the treatment does not make a difference to the disease). You generate evidence for your desired claim by accumulating data that is (hopefully) so far away from what should have happened under the null hypothesis (in the example patients that are randomized to receive the treatment or a placebo having the same expected outcome) that one concludes that is very unlikely to have arisen under the null hypothesis so that you can reject the null hypothesis. In contrast failure to reject the null hypothesis does not necessarily make it very likely that the null hypothesis is true (just because a clinical trial failed to show a drug works does not mean that the drug really does nothing).
|
Why is the null hypothesis often sought to be rejected?
The null hypothesis is always formed with the intention to reject it that is the basic idea of hypothesis testing. When you are trying to show that something is likely to be true (e.g. a treatment imp
|
6,760
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
|
Causal theory offers another explanation for how two variables could be unconditionally independent yet conditionally dependent. I am not an expert on causal theory and am grateful for any criticism that will correct any misguidance below.
To illustrate, I will use directed acyclic graphs (DAG). In these graphs, edges ($-$) between variables represent direct causal relationships. Arrowheads ($\leftarrow$ or $\rightarrow$) indicate the direction of causal relationships. Thus $A \rightarrow B$ infers that $A$ directly causes $B$, and $A \leftarrow B$ infers that $A$ is directly caused by $B$. $A \rightarrow B \rightarrow C$ is a causal path that infers that $A$ indirectly causes $C$ through $B$. For simplicity, assume all causal relationships are linear.
First, consider a simple example of confounder bias:
Here, a simple bivariable regression will suggest a dependence between $X$ and $Y$. However, there is no direct causal relationship between $X$ and $Y$. Instead, both are directly caused by $Z$, and in the simple bivariable regression, observing $Z$ induces a dependency between $X$ and $Y$, resulting in bias by confounding. However, a multivariable regression conditioning on $Z$ will remove the bias and suggest no dependence between $X$ and $Y$.
Second, consider an example of collider bias (also known as Berkson's bias or berksonian bias, of which selection bias is a special type):
Here, a simple bivariable regression will suggest no dependence between $X$ and $Y$. This agrees with the DAG, which infers no direct causal relationship between $X$ and $Y$. However, a multivariable regression conditioning on $Z$ will induce a dependence between $X$ and $Y$, suggesting that a direct causal relationship between the two variables may exist when in fact, none exist. The inclusion of $Z$ in the multivariable regression results in collider bias.
Third, consider an example of incidental cancellation:
Let us assume that $\alpha$, $\beta$, and $\gamma$ are path coefficients and that $\beta = -\alpha\gamma$. A simple bivariable regression will suggest no dependence between $X$ and $Y$. Although $X$ is, in fact, a direct cause of $Y$, the confounding effect of $Z$ on $X$ and $Y$ incidentally cancels out the effect of $X$ on $Y$. A multivariable regression conditioning on $Z$ will remove the confounding effect of $Z$ on $X$ and $Y$, allowing for the estimation of the direct effect of $X$ on $Y$, assuming the DAG of the causal model is correct.
To summarize:
Confounder example: $X$ and $Y$ are dependent in bivariable regression and independent in multivariable regression conditioning on confounder $Z$.
Collider example: $X$ and $Y$ are independent in bivariable regression and dependent in multivariable regression conditioning on collider $Z$.
Incidental cancellation example: $X$ and $Y$ are independent in bivariable regression and dependent in multivariable regression conditioning on confounder $Z$.
Discussion:
The results of your analysis are not compatible with the confounder example but are compatible with both the collider example and the incidental cancellation example. Thus, a potential explanation is that you have incorrectly conditioned on a collider variable in your multivariable regression and have induced an association between $X$ and $Y$ even though $X$ is not a cause of $Y$ and $Y$ is not a cause of $X$. Alternatively, you might have correctly conditioned on a confounder in your multivariable regression that was incidentally cancelling out the true effect of $X$ on $Y$ in your bivariable regression.
I find using background knowledge to construct causal models helpful when considering which variables to include in statistical models. For example, if previous high-quality randomized studies concluded that $X$ causes $Z$ and $Y$ causes $Z$, I could make a strong assumption that $Z$ is a collider of $X$ and $Y$ and not condition upon it in a statistical model. However, if I merely had an intuition that $X$ causes $Z$, and $Y$ causes $Z$, but no strong scientific evidence to support my intuition, I could only make a weak assumption that $Z$ is a collider of $X$ and $Y$, as human intuition has a history of being misguided. Subsequently, I would be skeptical of inferring causal relationships between $X$ and $Y$ without further investigating their causal relationships with $Z$. In lieu of or in addition to background knowledge, there are also algorithms designed to infer causal models from the data using a series of tests of association (e.g. PC algorithm and FCI algorithm, see TETRAD for Java implementation, PCalg for R implementation). These algorithms are very interesting, but I would not recommend relying on them without a strong understanding of the power and limitations of causal calculus and causal models in causal theory.
Conclusion:
Contemplation of causal models does not excuse the investigator from addressing the statistical considerations discussed in other answers here. However, I feel that causal models can nevertheless provide a helpful framework when thinking of potential explanations for observed statistical dependence and independence in statistical models, especially when visualizing potential confounders and colliders.
Further reading:
Gelman, Andrew. 2011. "Causality and Statistical Learning." Am. J. Sociology 117 (3) (November): 955–966.
Greenland, S, J Pearl, and J M Robins. 1999. “Causal Diagrams for Epidemiologic Research.” Epidemiology (Cambridge, Mass.) 10 (1) (January): 37–48.
Greenland, Sander. 2003. “Quantifying Biases in Causal Models: Classical Confounding Vs Collider-Stratification Bias.” Epidemiology 14 (3) (May 1): 300–306.
Pearl, Judea. 1998. Why There Is No Statistical Test For Confounding, Why Many Think There Is, And Why They Are Almost Right.
Pearl, Judea. 2009. Causality: Models, Reasoning and Inference. 2nd ed. Cambridge University Press.
Spirtes, Peter, Clark Glymour, and Richard Scheines. 2001. Causation, Prediction, and Search, Second Edition. A Bradford Book.
Update: Judea Pearl discusses the theory of causal inference and the need to incorporate causal inference into introductory statistics courses in the November 2012 edition of Amstat News. His Turing Award Lecture, entitled "The mechanization of causal inference: A 'mini' Turing Test and beyond" is also of interest.
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
|
Causal theory offers another explanation for how two variables could be unconditionally independent yet conditionally dependent. I am not an expert on causal theory and am grateful for any criticism t
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
Causal theory offers another explanation for how two variables could be unconditionally independent yet conditionally dependent. I am not an expert on causal theory and am grateful for any criticism that will correct any misguidance below.
To illustrate, I will use directed acyclic graphs (DAG). In these graphs, edges ($-$) between variables represent direct causal relationships. Arrowheads ($\leftarrow$ or $\rightarrow$) indicate the direction of causal relationships. Thus $A \rightarrow B$ infers that $A$ directly causes $B$, and $A \leftarrow B$ infers that $A$ is directly caused by $B$. $A \rightarrow B \rightarrow C$ is a causal path that infers that $A$ indirectly causes $C$ through $B$. For simplicity, assume all causal relationships are linear.
First, consider a simple example of confounder bias:
Here, a simple bivariable regression will suggest a dependence between $X$ and $Y$. However, there is no direct causal relationship between $X$ and $Y$. Instead, both are directly caused by $Z$, and in the simple bivariable regression, observing $Z$ induces a dependency between $X$ and $Y$, resulting in bias by confounding. However, a multivariable regression conditioning on $Z$ will remove the bias and suggest no dependence between $X$ and $Y$.
Second, consider an example of collider bias (also known as Berkson's bias or berksonian bias, of which selection bias is a special type):
Here, a simple bivariable regression will suggest no dependence between $X$ and $Y$. This agrees with the DAG, which infers no direct causal relationship between $X$ and $Y$. However, a multivariable regression conditioning on $Z$ will induce a dependence between $X$ and $Y$, suggesting that a direct causal relationship between the two variables may exist when in fact, none exist. The inclusion of $Z$ in the multivariable regression results in collider bias.
Third, consider an example of incidental cancellation:
Let us assume that $\alpha$, $\beta$, and $\gamma$ are path coefficients and that $\beta = -\alpha\gamma$. A simple bivariable regression will suggest no dependence between $X$ and $Y$. Although $X$ is, in fact, a direct cause of $Y$, the confounding effect of $Z$ on $X$ and $Y$ incidentally cancels out the effect of $X$ on $Y$. A multivariable regression conditioning on $Z$ will remove the confounding effect of $Z$ on $X$ and $Y$, allowing for the estimation of the direct effect of $X$ on $Y$, assuming the DAG of the causal model is correct.
To summarize:
Confounder example: $X$ and $Y$ are dependent in bivariable regression and independent in multivariable regression conditioning on confounder $Z$.
Collider example: $X$ and $Y$ are independent in bivariable regression and dependent in multivariable regression conditioning on collider $Z$.
Incidental cancellation example: $X$ and $Y$ are independent in bivariable regression and dependent in multivariable regression conditioning on confounder $Z$.
Discussion:
The results of your analysis are not compatible with the confounder example but are compatible with both the collider example and the incidental cancellation example. Thus, a potential explanation is that you have incorrectly conditioned on a collider variable in your multivariable regression and have induced an association between $X$ and $Y$ even though $X$ is not a cause of $Y$ and $Y$ is not a cause of $X$. Alternatively, you might have correctly conditioned on a confounder in your multivariable regression that was incidentally cancelling out the true effect of $X$ on $Y$ in your bivariable regression.
I find using background knowledge to construct causal models helpful when considering which variables to include in statistical models. For example, if previous high-quality randomized studies concluded that $X$ causes $Z$ and $Y$ causes $Z$, I could make a strong assumption that $Z$ is a collider of $X$ and $Y$ and not condition upon it in a statistical model. However, if I merely had an intuition that $X$ causes $Z$, and $Y$ causes $Z$, but no strong scientific evidence to support my intuition, I could only make a weak assumption that $Z$ is a collider of $X$ and $Y$, as human intuition has a history of being misguided. Subsequently, I would be skeptical of inferring causal relationships between $X$ and $Y$ without further investigating their causal relationships with $Z$. In lieu of or in addition to background knowledge, there are also algorithms designed to infer causal models from the data using a series of tests of association (e.g. PC algorithm and FCI algorithm, see TETRAD for Java implementation, PCalg for R implementation). These algorithms are very interesting, but I would not recommend relying on them without a strong understanding of the power and limitations of causal calculus and causal models in causal theory.
Conclusion:
Contemplation of causal models does not excuse the investigator from addressing the statistical considerations discussed in other answers here. However, I feel that causal models can nevertheless provide a helpful framework when thinking of potential explanations for observed statistical dependence and independence in statistical models, especially when visualizing potential confounders and colliders.
Further reading:
Gelman, Andrew. 2011. "Causality and Statistical Learning." Am. J. Sociology 117 (3) (November): 955–966.
Greenland, S, J Pearl, and J M Robins. 1999. “Causal Diagrams for Epidemiologic Research.” Epidemiology (Cambridge, Mass.) 10 (1) (January): 37–48.
Greenland, Sander. 2003. “Quantifying Biases in Causal Models: Classical Confounding Vs Collider-Stratification Bias.” Epidemiology 14 (3) (May 1): 300–306.
Pearl, Judea. 1998. Why There Is No Statistical Test For Confounding, Why Many Think There Is, And Why They Are Almost Right.
Pearl, Judea. 2009. Causality: Models, Reasoning and Inference. 2nd ed. Cambridge University Press.
Spirtes, Peter, Clark Glymour, and Richard Scheines. 2001. Causation, Prediction, and Search, Second Edition. A Bradford Book.
Update: Judea Pearl discusses the theory of causal inference and the need to incorporate causal inference into introductory statistics courses in the November 2012 edition of Amstat News. His Turing Award Lecture, entitled "The mechanization of causal inference: A 'mini' Turing Test and beyond" is also of interest.
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
Causal theory offers another explanation for how two variables could be unconditionally independent yet conditionally dependent. I am not an expert on causal theory and am grateful for any criticism t
|
6,761
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
|
I think @jthetzel's approach is the right one (+1). In order to interpret these results you will have to think about / have some theory of why the relationships manifest as they do. That is, you will need to think about the pattern of causal relationships that underlies your data. You need to recognize that, as @jthetzel points out, your results are consistent with several different data generating processes. I don't think that any amount of additional statistical tests on the same dataset will allow you to distinguish amongst those possibilities (although further experiments certainly could). So thinking hard about what's known about the topic is vital here.
I want to point out another possible underlying situation that could generate results like yours: Suppression. This is more difficult to illustrate using the arrow diagrams, but if I can augment them slightly, we could think of it like this:
What's important about this situation is that the $\text{Other Variable}$ is made up of two parts, an unrelated ($\text{U}$) part, and a related ($\text{R}$) part. The $\text{Suppressor}$ will be uncorrelated with $\text{Y}$, but may very well be 'significant' in a multiple regression model. Furthermore, the $\text{Other Variable}$ may or may not be 'significantly' correlated with the $\text{Suppressor}$ or $\text{Y}$ on its own. Moreover, your variable X could be playing the role of either the $\text{Suppressor}$ or the $\text{Other Variable}$ in this situation (and thus, again, you need to think about what the underlying pattern might be based on your knowledge of the area).
I don't know if you can read R code, but here's an example I worked up. (This particular example fits better with X playing the role of the $\text{Suppressor}$, but both are not 'significantly' correlated with $\text{Y}$; it should be possible to get the correlation between the $\text{Other Variable}$ and $\text{Y}$ close to 0 and match the other descriptives with just the right settings.)
set.seed(888) # for reproducibility
S = rnorm(60, mean=0, sd=1.0) # the Suppressor is normally distributed
U = 1.1*S + rnorm(60, mean=0, sd=0.1) # U (unrelated) is Suppressor plus error
R = rnorm(60, mean=0, sd=1.0) # related part; normally distributed
OV = U + R # the Other Variable is U plus R
Y = R + rnorm(60, mean=0, sd=2) # Y is R plus error
cor.test(S, Y) # Suppressor uncorrelated w/ Y
# t = 0.0283, df = 58, p-value = 0.9775
# cor 0.003721616
cor.test(S, OV) # Suppressor correlated w/ Other Variable
# t = 8.655, df = 58, p-value = 4.939e-12
# cor 0.7507423
cor.test(OV,Y) # Other Var not significantly cor w/ Y
# t = 1.954, df = 58, p-value = 0.05553
# cor 0.2485251
summary(lm(Y~OV+S)) # both Suppressor & Other Var sig in mult reg
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.2752 0.2396 1.148 0.25557
# OV 0.7232 0.2390 3.026 0.00372 **
# S -0.7690 0.3415 -2.251 0.02823 *
My point here isn't that this situation is the one that underlies your data. I don't know if this is more or less likely than the options @jthetzel suggests. I only offer this as more food for thought. To interpret your current results, you need to think about these possibilities and decide what makes the most sense. To confirm your choice, careful experimentation will be needed.
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
|
I think @jthetzel's approach is the right one (+1). In order to interpret these results you will have to think about / have some theory of why the relationships manifest as they do. That is, you wil
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
I think @jthetzel's approach is the right one (+1). In order to interpret these results you will have to think about / have some theory of why the relationships manifest as they do. That is, you will need to think about the pattern of causal relationships that underlies your data. You need to recognize that, as @jthetzel points out, your results are consistent with several different data generating processes. I don't think that any amount of additional statistical tests on the same dataset will allow you to distinguish amongst those possibilities (although further experiments certainly could). So thinking hard about what's known about the topic is vital here.
I want to point out another possible underlying situation that could generate results like yours: Suppression. This is more difficult to illustrate using the arrow diagrams, but if I can augment them slightly, we could think of it like this:
What's important about this situation is that the $\text{Other Variable}$ is made up of two parts, an unrelated ($\text{U}$) part, and a related ($\text{R}$) part. The $\text{Suppressor}$ will be uncorrelated with $\text{Y}$, but may very well be 'significant' in a multiple regression model. Furthermore, the $\text{Other Variable}$ may or may not be 'significantly' correlated with the $\text{Suppressor}$ or $\text{Y}$ on its own. Moreover, your variable X could be playing the role of either the $\text{Suppressor}$ or the $\text{Other Variable}$ in this situation (and thus, again, you need to think about what the underlying pattern might be based on your knowledge of the area).
I don't know if you can read R code, but here's an example I worked up. (This particular example fits better with X playing the role of the $\text{Suppressor}$, but both are not 'significantly' correlated with $\text{Y}$; it should be possible to get the correlation between the $\text{Other Variable}$ and $\text{Y}$ close to 0 and match the other descriptives with just the right settings.)
set.seed(888) # for reproducibility
S = rnorm(60, mean=0, sd=1.0) # the Suppressor is normally distributed
U = 1.1*S + rnorm(60, mean=0, sd=0.1) # U (unrelated) is Suppressor plus error
R = rnorm(60, mean=0, sd=1.0) # related part; normally distributed
OV = U + R # the Other Variable is U plus R
Y = R + rnorm(60, mean=0, sd=2) # Y is R plus error
cor.test(S, Y) # Suppressor uncorrelated w/ Y
# t = 0.0283, df = 58, p-value = 0.9775
# cor 0.003721616
cor.test(S, OV) # Suppressor correlated w/ Other Variable
# t = 8.655, df = 58, p-value = 4.939e-12
# cor 0.7507423
cor.test(OV,Y) # Other Var not significantly cor w/ Y
# t = 1.954, df = 58, p-value = 0.05553
# cor 0.2485251
summary(lm(Y~OV+S)) # both Suppressor & Other Var sig in mult reg
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.2752 0.2396 1.148 0.25557
# OV 0.7232 0.2390 3.026 0.00372 **
# S -0.7690 0.3415 -2.251 0.02823 *
My point here isn't that this situation is the one that underlies your data. I don't know if this is more or less likely than the options @jthetzel suggests. I only offer this as more food for thought. To interpret your current results, you need to think about these possibilities and decide what makes the most sense. To confirm your choice, careful experimentation will be needed.
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
I think @jthetzel's approach is the right one (+1). In order to interpret these results you will have to think about / have some theory of why the relationships manifest as they do. That is, you wil
|
6,762
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
|
Just some visualization that it is possible.
On picture (a) "normal" or "intuitive" regressional situation is shown. This pic is the same as for example found (and explained) here or here.
The variables are drawn as vectors. Angles between them (their cosines) are the variables' correlations. $Y'$ here designates the variable of predicted values (more often notated as $\hat Y$). Skew coordinate of its edge onto a predictor vector (skew projection, parallel to the other predictor) - notch $b$ - is proportional to the regression coefficient of that predictor.
On pic (a), all three variables correlate positively, and both $b_1$ and $b_2$ are also positive regression coefficients. $X_1$ and $X_2$ "compete" in the regression, with the regression coefficients being their score in that contest.
On picture (b) shown is situation where predictor $X_1$ correlates with $Y$ positively, still it's regression coefficient is zero: the endpoint of the prediction $Y'$ projects at the origin of vector $X_1$. Note that this fact coincides with that $Y'$ and $X_2$ superimpose, which means that the predicted values absolutely correlate with that other predictor.
On picture (c) is the situation where $X_1$ does not correlate with $Y$ (their vectors are orthogonal), yet the regression coefficient of the predictor is not zero: it is negative (the projection falls behind $X_1$ vector).
Data and analysis approximately corresponding to pic (b):
y x1 x2
1.644540 1.063845 .351188
1.785204 1.203146 .200000
-1.36357 -.466514 -.961069
.314549 1.175054 .800000
.317955 .100612 .858597
.970097 2.438904 1.000000
.664388 1.204048 .292670
-.870252 -.993857 -1.89018
1.962192 .587540 -.275352
1.036381 -.110834 -.246448
.007415 -.069234 1.447422
1.634353 .965370 .467095
.219813 .553268 .348095
-.285774 .358621 .166708
1.498758 -2.87971 -1.13757
1.671538 -.310708 .396034
1.462036 .057677 1.401522
-.563266 .904716 -.744522
.297874 .561898 -.929709
-1.54898 -.898084 -.838295
Data and analysis approximately corresponding to pic (c):
y x1 x2
1.644540 1.063845 .351188
1.785204 -1.20315 .200000
-1.36357 -.466514 -.961069
.314549 1.175054 .800000
.317955 -.100612 .858597
.970097 1.438904 1.000000
.664388 1.204048 .292670
-.870252 -.993857 -1.89018
1.962192 -.587540 -.275352
1.036381 -.110834 -.246448
.007415 -.069234 1.447422
1.634353 .965370 .467095
.219813 .553268 .348095
-.285774 .358621 .166708
1.498758 -2.87971 -1.13757
1.671538 -.810708 .396034
1.462036 -.057677 1.401522
-.563266 .904716 -.744522
.297874 .561898 -.929709
-1.54898 -1.26108 -.838295
Observe that $X_1$ in the last example served as suppressor. Its zero-order correlation with $Y$ is practically zero but its part correlation is much larger by magnitude, $-.224$. It strengthened to some extent the predictive force of $X_2$ (from $.419$, a would-be beta in simple regression with it, to beta $.538$ in the multiple regression).
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
|
Just some visualization that it is possible.
On picture (a) "normal" or "intuitive" regressional situation is shown. This pic is the same as for example found (and explained) here or here.
The variabl
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
Just some visualization that it is possible.
On picture (a) "normal" or "intuitive" regressional situation is shown. This pic is the same as for example found (and explained) here or here.
The variables are drawn as vectors. Angles between them (their cosines) are the variables' correlations. $Y'$ here designates the variable of predicted values (more often notated as $\hat Y$). Skew coordinate of its edge onto a predictor vector (skew projection, parallel to the other predictor) - notch $b$ - is proportional to the regression coefficient of that predictor.
On pic (a), all three variables correlate positively, and both $b_1$ and $b_2$ are also positive regression coefficients. $X_1$ and $X_2$ "compete" in the regression, with the regression coefficients being their score in that contest.
On picture (b) shown is situation where predictor $X_1$ correlates with $Y$ positively, still it's regression coefficient is zero: the endpoint of the prediction $Y'$ projects at the origin of vector $X_1$. Note that this fact coincides with that $Y'$ and $X_2$ superimpose, which means that the predicted values absolutely correlate with that other predictor.
On picture (c) is the situation where $X_1$ does not correlate with $Y$ (their vectors are orthogonal), yet the regression coefficient of the predictor is not zero: it is negative (the projection falls behind $X_1$ vector).
Data and analysis approximately corresponding to pic (b):
y x1 x2
1.644540 1.063845 .351188
1.785204 1.203146 .200000
-1.36357 -.466514 -.961069
.314549 1.175054 .800000
.317955 .100612 .858597
.970097 2.438904 1.000000
.664388 1.204048 .292670
-.870252 -.993857 -1.89018
1.962192 .587540 -.275352
1.036381 -.110834 -.246448
.007415 -.069234 1.447422
1.634353 .965370 .467095
.219813 .553268 .348095
-.285774 .358621 .166708
1.498758 -2.87971 -1.13757
1.671538 -.310708 .396034
1.462036 .057677 1.401522
-.563266 .904716 -.744522
.297874 .561898 -.929709
-1.54898 -.898084 -.838295
Data and analysis approximately corresponding to pic (c):
y x1 x2
1.644540 1.063845 .351188
1.785204 -1.20315 .200000
-1.36357 -.466514 -.961069
.314549 1.175054 .800000
.317955 -.100612 .858597
.970097 1.438904 1.000000
.664388 1.204048 .292670
-.870252 -.993857 -1.89018
1.962192 -.587540 -.275352
1.036381 -.110834 -.246448
.007415 -.069234 1.447422
1.634353 .965370 .467095
.219813 .553268 .348095
-.285774 .358621 .166708
1.498758 -2.87971 -1.13757
1.671538 -.810708 .396034
1.462036 -.057677 1.401522
-.563266 .904716 -.744522
.297874 .561898 -.929709
-1.54898 -1.26108 -.838295
Observe that $X_1$ in the last example served as suppressor. Its zero-order correlation with $Y$ is practically zero but its part correlation is much larger by magnitude, $-.224$. It strengthened to some extent the predictive force of $X_2$ (from $.419$, a would-be beta in simple regression with it, to beta $.538$ in the multiple regression).
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
Just some visualization that it is possible.
On picture (a) "normal" or "intuitive" regressional situation is shown. This pic is the same as for example found (and explained) here or here.
The variabl
|
6,763
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
|
I agree with the previous answer but hope I can contribute by giving more details.
The correlation coefficient is just measuring the linear dependence between $X$ and $Y$ and it's not controlling for the fact that other variables might be involved in the relationship as well. In fact the correlation coefficient equals the slope parameter of the following regression scaled by $x$ and $y$ standard deviations :
$Y = a + \beta x + u$
where $\hat \rho_{yx} = \hat \beta \hat\sigma_x/\hat\sigma_y$
But what happens if $Y$ is generated by other variables as well, thus the real model is something like:
$Y = a + \beta x + \sum_j\alpha_jz_j + u$
Under this real model, it becomes obvious that estimating the first one (only with x) will yield a biased $\beta$ estimate as that model is omitting the $z_j$ regressors(this implies that $\rho$ is also biased !). So your results are in line with the fact that the omitted variables are relevant. To deal with this issue , theory on correlation analysis provides the partial correlation coefficient (I'm sure you will find references on this) which basically calculates $\rho_{xy|z}$ from the latter estimating equation that controls for $z_j$.
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
|
I agree with the previous answer but hope I can contribute by giving more details.
The correlation coefficient is just measuring the linear dependence between $X$ and $Y$ and it's not controlling for
|
X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it mean?
I agree with the previous answer but hope I can contribute by giving more details.
The correlation coefficient is just measuring the linear dependence between $X$ and $Y$ and it's not controlling for the fact that other variables might be involved in the relationship as well. In fact the correlation coefficient equals the slope parameter of the following regression scaled by $x$ and $y$ standard deviations :
$Y = a + \beta x + u$
where $\hat \rho_{yx} = \hat \beta \hat\sigma_x/\hat\sigma_y$
But what happens if $Y$ is generated by other variables as well, thus the real model is something like:
$Y = a + \beta x + \sum_j\alpha_jz_j + u$
Under this real model, it becomes obvious that estimating the first one (only with x) will yield a biased $\beta$ estimate as that model is omitting the $z_j$ regressors(this implies that $\rho$ is also biased !). So your results are in line with the fact that the omitted variables are relevant. To deal with this issue , theory on correlation analysis provides the partial correlation coefficient (I'm sure you will find references on this) which basically calculates $\rho_{xy|z}$ from the latter estimating equation that controls for $z_j$.
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X and Y are not correlated, but X is significant predictor of Y in multiple regression. What does it
I agree with the previous answer but hope I can contribute by giving more details.
The correlation coefficient is just measuring the linear dependence between $X$ and $Y$ and it's not controlling for
|
6,764
|
Difference between feedback RNN and LSTM/GRU
|
All RNNs have feedback loops in the recurrent layer. This lets them maintain information in 'memory' over time. But, it can be difficult to train standard RNNs to solve problems that require learning long-term temporal dependencies. This is because the gradient of the loss function decays exponentially with time (called the vanishing gradient problem). LSTM networks are a type of RNN that uses special units in addition to standard units. LSTM units include a 'memory cell' that can maintain information in memory for long periods of time. A set of gates is used to control when information enters the memory, when it's output, and when it's forgotten. This architecture lets them learn longer-term dependencies. GRUs are similar to LSTMs, but use a simplified structure. They also use a set of gates to control the flow of information, but they don't use separate memory cells, and they use fewer gates.
This paper gives a good overview:
Chung et al. (2014). Empirical Evaluation of Gated Recurrent Neural Networks on Sequence Modeling.
|
Difference between feedback RNN and LSTM/GRU
|
All RNNs have feedback loops in the recurrent layer. This lets them maintain information in 'memory' over time. But, it can be difficult to train standard RNNs to solve problems that require learning
|
Difference between feedback RNN and LSTM/GRU
All RNNs have feedback loops in the recurrent layer. This lets them maintain information in 'memory' over time. But, it can be difficult to train standard RNNs to solve problems that require learning long-term temporal dependencies. This is because the gradient of the loss function decays exponentially with time (called the vanishing gradient problem). LSTM networks are a type of RNN that uses special units in addition to standard units. LSTM units include a 'memory cell' that can maintain information in memory for long periods of time. A set of gates is used to control when information enters the memory, when it's output, and when it's forgotten. This architecture lets them learn longer-term dependencies. GRUs are similar to LSTMs, but use a simplified structure. They also use a set of gates to control the flow of information, but they don't use separate memory cells, and they use fewer gates.
This paper gives a good overview:
Chung et al. (2014). Empirical Evaluation of Gated Recurrent Neural Networks on Sequence Modeling.
|
Difference between feedback RNN and LSTM/GRU
All RNNs have feedback loops in the recurrent layer. This lets them maintain information in 'memory' over time. But, it can be difficult to train standard RNNs to solve problems that require learning
|
6,765
|
Difference between feedback RNN and LSTM/GRU
|
Standard RNNs (Recurrent Neural Networks) suffer from vanishing and exploding gradient problems. LSTMs (Long Short Term Memory) deal with these problems by introducing new gates, such as input and forget gates, which allow for a better control over the gradient flow and enable better preservation of “long-range dependencies”. The long range dependency in RNN is resolved by increasing the number of repeating layer in LSTM.
RNN
LSTM
For Further Details :Understanding LSTM
|
Difference between feedback RNN and LSTM/GRU
|
Standard RNNs (Recurrent Neural Networks) suffer from vanishing and exploding gradient problems. LSTMs (Long Short Term Memory) deal with these problems by introducing new gates, such as input and fo
|
Difference between feedback RNN and LSTM/GRU
Standard RNNs (Recurrent Neural Networks) suffer from vanishing and exploding gradient problems. LSTMs (Long Short Term Memory) deal with these problems by introducing new gates, such as input and forget gates, which allow for a better control over the gradient flow and enable better preservation of “long-range dependencies”. The long range dependency in RNN is resolved by increasing the number of repeating layer in LSTM.
RNN
LSTM
For Further Details :Understanding LSTM
|
Difference between feedback RNN and LSTM/GRU
Standard RNNs (Recurrent Neural Networks) suffer from vanishing and exploding gradient problems. LSTMs (Long Short Term Memory) deal with these problems by introducing new gates, such as input and fo
|
6,766
|
Difference between feedback RNN and LSTM/GRU
|
LSTMs are often referred to as fancy RNNs. Vanilla RNNs do not have a cell state. They only have hidden states and those hidden states serve as the memory for RNNs.
Meanwhile, LSTM has both cell states and a hidden states. The cell state has the ability to remove or add information to the cell, regulated by "gates". And because of this "cell", in theory, LSTM should be able to handle the long-term dependency (in practice, it's difficult to do so.)
|
Difference between feedback RNN and LSTM/GRU
|
LSTMs are often referred to as fancy RNNs. Vanilla RNNs do not have a cell state. They only have hidden states and those hidden states serve as the memory for RNNs.
Meanwhile, LSTM has both cell stat
|
Difference between feedback RNN and LSTM/GRU
LSTMs are often referred to as fancy RNNs. Vanilla RNNs do not have a cell state. They only have hidden states and those hidden states serve as the memory for RNNs.
Meanwhile, LSTM has both cell states and a hidden states. The cell state has the ability to remove or add information to the cell, regulated by "gates". And because of this "cell", in theory, LSTM should be able to handle the long-term dependency (in practice, it's difficult to do so.)
|
Difference between feedback RNN and LSTM/GRU
LSTMs are often referred to as fancy RNNs. Vanilla RNNs do not have a cell state. They only have hidden states and those hidden states serve as the memory for RNNs.
Meanwhile, LSTM has both cell stat
|
6,767
|
Difference between feedback RNN and LSTM/GRU
|
TL;DR
We can say that, when we move from RNN to LSTM (Long Short-Term Memory), we are introducing
more & more controlling knobs, which control the flow and mixing of
Inputs as per trained Weights. And thus, bringing in more flexibility
in controlling the outputs. So, LSTM gives us the most Control-ability
and thus, Better Results. But also comes with more Complexity and
Operating Cost.
Reference
[NOTE]:
GRU is better than LSTM as it is easy to modify and doesn't need memory units, therefore, faster to train than LSTM and give as per performance. Actually, the key difference comes out to be more than that: Long-short term (LSTM) perceptrons are made up using the momentum and gradient descent algorithms.
This image demonstrates the difference between them:
|
Difference between feedback RNN and LSTM/GRU
|
TL;DR
We can say that, when we move from RNN to LSTM (Long Short-Term Memory), we are introducing
more & more controlling knobs, which control the flow and mixing of
Inputs as per trained Weights. An
|
Difference between feedback RNN and LSTM/GRU
TL;DR
We can say that, when we move from RNN to LSTM (Long Short-Term Memory), we are introducing
more & more controlling knobs, which control the flow and mixing of
Inputs as per trained Weights. And thus, bringing in more flexibility
in controlling the outputs. So, LSTM gives us the most Control-ability
and thus, Better Results. But also comes with more Complexity and
Operating Cost.
Reference
[NOTE]:
GRU is better than LSTM as it is easy to modify and doesn't need memory units, therefore, faster to train than LSTM and give as per performance. Actually, the key difference comes out to be more than that: Long-short term (LSTM) perceptrons are made up using the momentum and gradient descent algorithms.
This image demonstrates the difference between them:
|
Difference between feedback RNN and LSTM/GRU
TL;DR
We can say that, when we move from RNN to LSTM (Long Short-Term Memory), we are introducing
more & more controlling knobs, which control the flow and mixing of
Inputs as per trained Weights. An
|
6,768
|
Difference between feedback RNN and LSTM/GRU
|
I think the difference between regular RNNs and the so-called "gated RNNs" is well explained in the existing answers to this question. However, I would like to add my two cents by pointing out the exact differences and similarities between LSTM and GRU.
The original definitions mostly come in the following form (I omitted bias terms, as well as time indices where not needed, in order to reduce some noise in the formulas):
$$\begin{align}
& \text{GRU} & & \text{LSTM} \\ \\
\boldsymbol{z} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) &
\boldsymbol{i} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
\boldsymbol{r} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) &
\boldsymbol{o} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
&&
\boldsymbol{f} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
\hat{\boldsymbol{h}}[t] &= \phi(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{r} \odot \boldsymbol{h}[t-1])) &
\boldsymbol{c}[t] &= \boldsymbol{f} \odot \boldsymbol{c}[t-1] + \boldsymbol{i} \odot \phi_1(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} \boldsymbol{h}[t-1]) \\
\boldsymbol{h}[t] &= (\boldsymbol{1} - \boldsymbol{z}) \odot \boldsymbol{h}[t-1] + \boldsymbol{z} \odot \hat{\boldsymbol{h}}[t] &
\boldsymbol{h}[t] &= \boldsymbol{o} \odot \phi_2(\boldsymbol{c}[t]).
\end{align}$$
Note that the GRU has only 2 gates, whereas the LSTM has 3. Also, the LSTM has two activation functions, $\phi_1$ and $\phi_2$, whereas the GRU has only 1, $\phi$. This immediately gives the idea that GRU is slightly less complex than the LSTM.
Now, if we rewrite the recurrences of both models so that they fit on a single line (instead of spreading over two lines), we would end up with the following equations:
\begin{align*}
\boldsymbol{h}[t] &= (\boldsymbol{1} - \boldsymbol{z}) \odot \boldsymbol{h}[t-1] + \boldsymbol{z} \odot \phi(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{r} \odot \boldsymbol{h}[t-1])) \tag{GRU} \\
\boldsymbol{c}[t] &= \boldsymbol{f} \odot \boldsymbol{c}[t-1] + \boldsymbol{i} \odot \phi_1(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{o}\odot \phi_2(\boldsymbol{c}[t-1]))) \tag{LSTM}
\end{align*}
By putting the recurrences next to each other like this, it should become clear that if all of the follwing equalities hold
$$\begin{align}
\boldsymbol{f} & = \boldsymbol{1} - \boldsymbol{z} &
\boldsymbol{i} & = \boldsymbol{z} &
\boldsymbol{o} & = \boldsymbol{r} &
\phi_2(\boldsymbol{x}) & = \boldsymbol{x},
\end{align}$$
the GRU and LSTM models implement the same recurrence. The only difference that remains, is that, in the LSTM, the gates are computed from $\boldsymbol{c}[t-1]$, whereas the GRU directly uses the result of the recurrence, $\boldsymbol{h}[t-1]$.
In this sense, the GRU is a strictly simplified version of the LSTM.
|
Difference between feedback RNN and LSTM/GRU
|
I think the difference between regular RNNs and the so-called "gated RNNs" is well explained in the existing answers to this question. However, I would like to add my two cents by pointing out the exa
|
Difference between feedback RNN and LSTM/GRU
I think the difference between regular RNNs and the so-called "gated RNNs" is well explained in the existing answers to this question. However, I would like to add my two cents by pointing out the exact differences and similarities between LSTM and GRU.
The original definitions mostly come in the following form (I omitted bias terms, as well as time indices where not needed, in order to reduce some noise in the formulas):
$$\begin{align}
& \text{GRU} & & \text{LSTM} \\ \\
\boldsymbol{z} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) &
\boldsymbol{i} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
\boldsymbol{r} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) &
\boldsymbol{o} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
&&
\boldsymbol{f} &= \mathrm{gate}(\boldsymbol{x}, \boldsymbol{h}) \\
\hat{\boldsymbol{h}}[t] &= \phi(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{r} \odot \boldsymbol{h}[t-1])) &
\boldsymbol{c}[t] &= \boldsymbol{f} \odot \boldsymbol{c}[t-1] + \boldsymbol{i} \odot \phi_1(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} \boldsymbol{h}[t-1]) \\
\boldsymbol{h}[t] &= (\boldsymbol{1} - \boldsymbol{z}) \odot \boldsymbol{h}[t-1] + \boldsymbol{z} \odot \hat{\boldsymbol{h}}[t] &
\boldsymbol{h}[t] &= \boldsymbol{o} \odot \phi_2(\boldsymbol{c}[t]).
\end{align}$$
Note that the GRU has only 2 gates, whereas the LSTM has 3. Also, the LSTM has two activation functions, $\phi_1$ and $\phi_2$, whereas the GRU has only 1, $\phi$. This immediately gives the idea that GRU is slightly less complex than the LSTM.
Now, if we rewrite the recurrences of both models so that they fit on a single line (instead of spreading over two lines), we would end up with the following equations:
\begin{align*}
\boldsymbol{h}[t] &= (\boldsymbol{1} - \boldsymbol{z}) \odot \boldsymbol{h}[t-1] + \boldsymbol{z} \odot \phi(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{r} \odot \boldsymbol{h}[t-1])) \tag{GRU} \\
\boldsymbol{c}[t] &= \boldsymbol{f} \odot \boldsymbol{c}[t-1] + \boldsymbol{i} \odot \phi_1(\boldsymbol{W} \boldsymbol{x} + \boldsymbol{U} (\boldsymbol{o}\odot \phi_2(\boldsymbol{c}[t-1]))) \tag{LSTM}
\end{align*}
By putting the recurrences next to each other like this, it should become clear that if all of the follwing equalities hold
$$\begin{align}
\boldsymbol{f} & = \boldsymbol{1} - \boldsymbol{z} &
\boldsymbol{i} & = \boldsymbol{z} &
\boldsymbol{o} & = \boldsymbol{r} &
\phi_2(\boldsymbol{x}) & = \boldsymbol{x},
\end{align}$$
the GRU and LSTM models implement the same recurrence. The only difference that remains, is that, in the LSTM, the gates are computed from $\boldsymbol{c}[t-1]$, whereas the GRU directly uses the result of the recurrence, $\boldsymbol{h}[t-1]$.
In this sense, the GRU is a strictly simplified version of the LSTM.
|
Difference between feedback RNN and LSTM/GRU
I think the difference between regular RNNs and the so-called "gated RNNs" is well explained in the existing answers to this question. However, I would like to add my two cents by pointing out the exa
|
6,769
|
How to interpret the coefficient of variation?
|
In examples like yours when data differ just additively, i.e. we add some constant $k$ to everything, then as you point out the standard deviation is unchanged, the mean is changed by exactly that constant, and so the coefficient of variation changes from $\sigma / \mu$ to $\sigma / (\mu + k)$, which is neither interesting nor useful.
It's multiplicative change that's interesting and where the coefficient of variation has some use. For multiplying everything by some constant $k$ implies that the coefficient of variation becomes $k \sigma/k \mu$, i.e. remains the same as before. Changing of units of measurement is a case in point, as in the answers of @Aksalal and @Macond.
As the coefficient of variation is unit-free, so also it is dimension-free, as whatever units or dimensions are possessed by the underlying variable are washed out by the division. That makes the coefficient of variation a measure of relative variability, so the relative variability of lengths may be compared with that of weights, and so forth. One field where the coefficient of variation has found some descriptive use is the morphometrics of organism size in biology.
In principle and practice the coefficient of variation is only defined fully and at all useful for variables that are entirely positive. Hence in detail your first sample with a value of $0$ is not an appropriate example. Another way of seeing this is to note that were the mean ever zero the coefficient would be indeterminate and were the mean ever negative the coefficient would be negative, assuming in the latter case that the standard deviation is positive. Either case would make the measure useless as a measure of relative variability, or indeed for any other purpose.
An equivalent statement is that the coefficient of variation is interesting and useful only if logarithms are defined in the usual way for all values, and indeed using coefficients of variation is equivalent to looking at variability of logarithms.
Although it should seem incredible to readers here, I have seen climatological and geographical publications in which the coefficients of variation of Celsius temperatures have puzzled naive scientists who note that coefficients can explode as mean temperatures get close to $0^\circ$C and become negative for mean temperatures below freezing. Even more bizarrely, I have seen suggestions that the problem is solved by using Fahrenheit instead. Conversely, the coefficient of variation is often mentioned correctly as a summary measure defined if and only if measurement scales qualify as ratio scale. As it happens, the coefficient of variation is not especially useful even for temperatures measured in kelvin, but for physical reasons rather than mathematical or statistical.
As in the case of the bizarre examples from climatology, which I leave unreferenced as the authors deserve neither the credit nor the shame, the coefficient of variation has been over-used in some fields. There is occasionally a tendency to regard it as a kind of magic summary measure that encapsulates both mean and standard deviation. This is naturally primitive thinking, as even when the ratio makes sense, the mean and standard deviation cannot be recovered from it.
In statistics the coefficient of variation is a fairly natural parameter if variation follows either the gamma or the lognormal, as may be seen by looking at the form of the coefficient of variation for those distributions.
Although the coefficient of variation can be of some use, in cases where it applies the more useful step is to work on logarithmic scale, either by logarithmic transformation or by using a logarithmic link function in a generalized linear model.
EDIT: If all values are negative, then we can regard the sign as just a convention that can be ignored. Equivalently in that case, $\sigma / |\mu|$ is effectively an identical twin of coefficient of variation.
EDIT 25 May 2020: Good detailed discussion in Simpson, G.G., Roe, A. and Lewontin, R.C. 1960. Quantitative Zoology. New York: Harcourt, Brace, pp.89-94. This text is inevitably dated in several respects, but includes many lucid explanations and pugnacious comments and criticisms.
See also Lewontin, R.C. 1966. On the measurement of relative variability.
Systematic Biology 15: 141–142. https://doi.org/10.2307/sysbio/15.2.141
|
How to interpret the coefficient of variation?
|
In examples like yours when data differ just additively, i.e. we add some constant $k$ to everything, then as you point out the standard deviation is unchanged, the mean is changed by exactly that con
|
How to interpret the coefficient of variation?
In examples like yours when data differ just additively, i.e. we add some constant $k$ to everything, then as you point out the standard deviation is unchanged, the mean is changed by exactly that constant, and so the coefficient of variation changes from $\sigma / \mu$ to $\sigma / (\mu + k)$, which is neither interesting nor useful.
It's multiplicative change that's interesting and where the coefficient of variation has some use. For multiplying everything by some constant $k$ implies that the coefficient of variation becomes $k \sigma/k \mu$, i.e. remains the same as before. Changing of units of measurement is a case in point, as in the answers of @Aksalal and @Macond.
As the coefficient of variation is unit-free, so also it is dimension-free, as whatever units or dimensions are possessed by the underlying variable are washed out by the division. That makes the coefficient of variation a measure of relative variability, so the relative variability of lengths may be compared with that of weights, and so forth. One field where the coefficient of variation has found some descriptive use is the morphometrics of organism size in biology.
In principle and practice the coefficient of variation is only defined fully and at all useful for variables that are entirely positive. Hence in detail your first sample with a value of $0$ is not an appropriate example. Another way of seeing this is to note that were the mean ever zero the coefficient would be indeterminate and were the mean ever negative the coefficient would be negative, assuming in the latter case that the standard deviation is positive. Either case would make the measure useless as a measure of relative variability, or indeed for any other purpose.
An equivalent statement is that the coefficient of variation is interesting and useful only if logarithms are defined in the usual way for all values, and indeed using coefficients of variation is equivalent to looking at variability of logarithms.
Although it should seem incredible to readers here, I have seen climatological and geographical publications in which the coefficients of variation of Celsius temperatures have puzzled naive scientists who note that coefficients can explode as mean temperatures get close to $0^\circ$C and become negative for mean temperatures below freezing. Even more bizarrely, I have seen suggestions that the problem is solved by using Fahrenheit instead. Conversely, the coefficient of variation is often mentioned correctly as a summary measure defined if and only if measurement scales qualify as ratio scale. As it happens, the coefficient of variation is not especially useful even for temperatures measured in kelvin, but for physical reasons rather than mathematical or statistical.
As in the case of the bizarre examples from climatology, which I leave unreferenced as the authors deserve neither the credit nor the shame, the coefficient of variation has been over-used in some fields. There is occasionally a tendency to regard it as a kind of magic summary measure that encapsulates both mean and standard deviation. This is naturally primitive thinking, as even when the ratio makes sense, the mean and standard deviation cannot be recovered from it.
In statistics the coefficient of variation is a fairly natural parameter if variation follows either the gamma or the lognormal, as may be seen by looking at the form of the coefficient of variation for those distributions.
Although the coefficient of variation can be of some use, in cases where it applies the more useful step is to work on logarithmic scale, either by logarithmic transformation or by using a logarithmic link function in a generalized linear model.
EDIT: If all values are negative, then we can regard the sign as just a convention that can be ignored. Equivalently in that case, $\sigma / |\mu|$ is effectively an identical twin of coefficient of variation.
EDIT 25 May 2020: Good detailed discussion in Simpson, G.G., Roe, A. and Lewontin, R.C. 1960. Quantitative Zoology. New York: Harcourt, Brace, pp.89-94. This text is inevitably dated in several respects, but includes many lucid explanations and pugnacious comments and criticisms.
See also Lewontin, R.C. 1966. On the measurement of relative variability.
Systematic Biology 15: 141–142. https://doi.org/10.2307/sysbio/15.2.141
|
How to interpret the coefficient of variation?
In examples like yours when data differ just additively, i.e. we add some constant $k$ to everything, then as you point out the standard deviation is unchanged, the mean is changed by exactly that con
|
6,770
|
How to interpret the coefficient of variation?
|
Imagine I said "There are 1,625,330 people in this town. Plus or minus five." You'd be impressed by my accurate demographic knowledge.
But if I said "There are five people in this house. Plus or minus five." You'd think I had no clue how many people were in the house.
Same standard deviation, much different CV's.
|
How to interpret the coefficient of variation?
|
Imagine I said "There are 1,625,330 people in this town. Plus or minus five." You'd be impressed by my accurate demographic knowledge.
But if I said "There are five people in this house. Plus or minu
|
How to interpret the coefficient of variation?
Imagine I said "There are 1,625,330 people in this town. Plus or minus five." You'd be impressed by my accurate demographic knowledge.
But if I said "There are five people in this house. Plus or minus five." You'd think I had no clue how many people were in the house.
Same standard deviation, much different CV's.
|
How to interpret the coefficient of variation?
Imagine I said "There are 1,625,330 people in this town. Plus or minus five." You'd be impressed by my accurate demographic knowledge.
But if I said "There are five people in this house. Plus or minu
|
6,771
|
How to interpret the coefficient of variation?
|
Normally, you use coefficient of variation for variable of different units of measure or very different scales. You can think of it as noise/signal ratio. For instance, you may want to compare variability of the weight and height of students; variability of GDP of USA and Monaco.
In your case, coefficient of variation may not make much sense at all, since the values are not much different.
|
How to interpret the coefficient of variation?
|
Normally, you use coefficient of variation for variable of different units of measure or very different scales. You can think of it as noise/signal ratio. For instance, you may want to compare variabi
|
How to interpret the coefficient of variation?
Normally, you use coefficient of variation for variable of different units of measure or very different scales. You can think of it as noise/signal ratio. For instance, you may want to compare variability of the weight and height of students; variability of GDP of USA and Monaco.
In your case, coefficient of variation may not make much sense at all, since the values are not much different.
|
How to interpret the coefficient of variation?
Normally, you use coefficient of variation for variable of different units of measure or very different scales. You can think of it as noise/signal ratio. For instance, you may want to compare variabi
|
6,772
|
How to interpret the coefficient of variation?
|
Sample with higher values has less variation relative to its mean, as the definition ($s / \bar{x} $) suggests. It is actually pretty straight-forward. Coefficient of variation is useful when comparing variation between samples (or populations) of different scales. Consider you are dealing with wages among countries. Comparing variation in wages in US and Japan is less informative if you use variance instead of coefficient of variation as your statistic, because 1 USD ~= 100 JPY and a 1 unit difference in wages doesn't mean same thing in both samples. Well, in this example you can convert everything to USD and then do the calculations, but it is not always obvious how to convert between different scales. When comparing variation in body weights of different species for instance.
|
How to interpret the coefficient of variation?
|
Sample with higher values has less variation relative to its mean, as the definition ($s / \bar{x} $) suggests. It is actually pretty straight-forward. Coefficient of variation is useful when comparin
|
How to interpret the coefficient of variation?
Sample with higher values has less variation relative to its mean, as the definition ($s / \bar{x} $) suggests. It is actually pretty straight-forward. Coefficient of variation is useful when comparing variation between samples (or populations) of different scales. Consider you are dealing with wages among countries. Comparing variation in wages in US and Japan is less informative if you use variance instead of coefficient of variation as your statistic, because 1 USD ~= 100 JPY and a 1 unit difference in wages doesn't mean same thing in both samples. Well, in this example you can convert everything to USD and then do the calculations, but it is not always obvious how to convert between different scales. When comparing variation in body weights of different species for instance.
|
How to interpret the coefficient of variation?
Sample with higher values has less variation relative to its mean, as the definition ($s / \bar{x} $) suggests. It is actually pretty straight-forward. Coefficient of variation is useful when comparin
|
6,773
|
How to interpret the coefficient of variation?
|
In actuality, both statistics can be misleading if you do not know or understand your hypothesis and experiment. Consider this gruesome example... Walking across two high rise buildings on a tightrope as opposed to walking on a plank. Let's say that the tightrope has a 1 inch diameter, whereas the plank is 12 inches wide. 5 people were asked to walk the rope and 5 were asked to walk the plank. We found the following results:
The average distance of each step from the edge (or side) of the rope (inches):
0.5, 0.2, 0.3, 0.6, 0.1
The average distance of each step from the edge (or side) of the plank (inches):
5.5, 5.2, 5.3, 5.6, 5.1
Just as in your example, this example will results in equal standard deviations as the values for the plank are simply a +5 difference to those for the tightrope. However, if I told you that the standard deviation for each experiment was 0.2074 you might say well then the two experiments were equivalent. However, if I told you that the CV for the tightrope experiment was almost 61% compared to under 4% for the plank, you might be inclined to ask me how many people fell off of the rope.
|
How to interpret the coefficient of variation?
|
In actuality, both statistics can be misleading if you do not know or understand your hypothesis and experiment. Consider this gruesome example... Walking across two high rise buildings on a tightrope
|
How to interpret the coefficient of variation?
In actuality, both statistics can be misleading if you do not know or understand your hypothesis and experiment. Consider this gruesome example... Walking across two high rise buildings on a tightrope as opposed to walking on a plank. Let's say that the tightrope has a 1 inch diameter, whereas the plank is 12 inches wide. 5 people were asked to walk the rope and 5 were asked to walk the plank. We found the following results:
The average distance of each step from the edge (or side) of the rope (inches):
0.5, 0.2, 0.3, 0.6, 0.1
The average distance of each step from the edge (or side) of the plank (inches):
5.5, 5.2, 5.3, 5.6, 5.1
Just as in your example, this example will results in equal standard deviations as the values for the plank are simply a +5 difference to those for the tightrope. However, if I told you that the standard deviation for each experiment was 0.2074 you might say well then the two experiments were equivalent. However, if I told you that the CV for the tightrope experiment was almost 61% compared to under 4% for the plank, you might be inclined to ask me how many people fell off of the rope.
|
How to interpret the coefficient of variation?
In actuality, both statistics can be misleading if you do not know or understand your hypothesis and experiment. Consider this gruesome example... Walking across two high rise buildings on a tightrope
|
6,774
|
How to interpret the coefficient of variation?
|
CV is a relative variability that is used to compare the variability of different sample dataset.
For a you example, the same standard deviation/variance with smaller mean will generate a smaller CV. it indicates that smaller CV dataset has smaller relative variability.
Assume You earn 10000 monthly, and I earn 100.(different mean)
we all probably loss 100 monthly (vriation), I will be hurt far more than you since I get a bigger CV(cv=1 compared to yours 0.01), relative greater variation.
|
How to interpret the coefficient of variation?
|
CV is a relative variability that is used to compare the variability of different sample dataset.
For a you example, the same standard deviation/variance with smaller mean will generate a smaller CV.
|
How to interpret the coefficient of variation?
CV is a relative variability that is used to compare the variability of different sample dataset.
For a you example, the same standard deviation/variance with smaller mean will generate a smaller CV. it indicates that smaller CV dataset has smaller relative variability.
Assume You earn 10000 monthly, and I earn 100.(different mean)
we all probably loss 100 monthly (vriation), I will be hurt far more than you since I get a bigger CV(cv=1 compared to yours 0.01), relative greater variation.
|
How to interpret the coefficient of variation?
CV is a relative variability that is used to compare the variability of different sample dataset.
For a you example, the same standard deviation/variance with smaller mean will generate a smaller CV.
|
6,775
|
How to interpret the coefficient of variation?
|
in this case, cv is not the right statistical tool to explain the result.
depending on the nature of the research carried out hence the objective, researcher has a specific hypothesis or point to proof. He or she must design, execute experiment and analyse data using the best and appropriate statistical tool i.e. if the experiment is to compare growth of group 1 and group 2, although cv of both are the same, but using T-test or paired T-test or Anova (bigger experiment) it could easily prove the different between the two group.
The key here is to apply the appropriate statistical tool to give a meaningful explanation about the result.
Remember cv is just one of the choices in Descriptive statistic.
my 2 cents
|
How to interpret the coefficient of variation?
|
in this case, cv is not the right statistical tool to explain the result.
depending on the nature of the research carried out hence the objective, researcher has a specific hypothesis or point to proo
|
How to interpret the coefficient of variation?
in this case, cv is not the right statistical tool to explain the result.
depending on the nature of the research carried out hence the objective, researcher has a specific hypothesis or point to proof. He or she must design, execute experiment and analyse data using the best and appropriate statistical tool i.e. if the experiment is to compare growth of group 1 and group 2, although cv of both are the same, but using T-test or paired T-test or Anova (bigger experiment) it could easily prove the different between the two group.
The key here is to apply the appropriate statistical tool to give a meaningful explanation about the result.
Remember cv is just one of the choices in Descriptive statistic.
my 2 cents
|
How to interpret the coefficient of variation?
in this case, cv is not the right statistical tool to explain the result.
depending on the nature of the research carried out hence the objective, researcher has a specific hypothesis or point to proo
|
6,776
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row?
|
Below we compute the probability in four ways:
Computation with Markov Chain 0.473981098314993
Computation with generating function 0.473981098314988
Estimation false method 0.536438013618686
Estimation correct method 0.473304632462677
The first two are exact methods and differ only a little (probably some round of errors), the third method is a naive estimation that does not give the correct number, the fourth method is better and gives a result that is very close to the exact method.
Computationally:
Markov Chain
You can model this computationally with a transition matrix
Say the column vector $X_{k,j} = \lbrace x_1,x_2,x_3,x_4,x_5 \rbrace_{j}$ is the probability to have $k$ of the same numbers in a row in the $j$-th dice roll. Then (when assuming a 6-sided dice)
$$X_{k,j} = M \cdot X_{k,j-1}$$ with
$$M = \begin{bmatrix} \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & 0 \\
\frac{1}{6} & 0& 0 & 0 & 0 \\
0& \frac{1}{6} & 0& 0 & 0 \\
0 & 0& \frac{1}{6} & 0& 0 \\
0&0 & 0& \frac{1}{6} & 1 \\
\end{bmatrix}$$
where this last entry $M_{5,5} = 1$ relates to 5 of the same in a row being an absorbing state where we 'stop' the experiment.
After the first roll you will be certainly in state 1 (there's certainly only 1 of the same number in a row).
$$X_{k,1} = \lbrace 1,0,0,0,0 \rbrace$$
After the $j$-th roll this will be multiplied with $M$ a $j-1$ times
$$X_{k,j} = M^{j-1} \lbrace 1,0,0,0,0 \rbrace$$
R-Code:
library(matrixcalc) ### allows us to use matrix.power
M <- matrix(c(5/6, 5/6, 5/6, 5/6, 0,
1/6, 0 , 0 , 0 , 0,
0, 1/6, 0 , 0 , 0,
0, 0 , 1/6, 0 , 0,
0, 0 , 0 , 1/6, 1),
5, byrow = TRUE)
start <- c(1,0,0,0,0)
matrix.power(M,999) %*% start
The result is $$X_{k,1000} = \begin{bmatrix}
0.438631855\\
0.073152468\\
0.012199943\\
0.002034635\\
\color{red}{0.473981098}\end{bmatrix}$$
and this last entry 0.473981098 is the probability to roll the same number 5 times in a row in 1000 rolls.
generating function
Our question is:
How to calculate the probability of rolling any number at least $k$ times in a row, out of $n$ tries?
This is equivalent to the question
How to calculate the probability of rolling the number 6 at least $k-1$ times in a row, out of $n-1$ tries?
You can see it as tracking whether the dice roll $m$ is the same number as the number of the dice roll $m-1$ (which has 1/6-th probabilty). And this needs to happen $k-1$ times in a row (in our case 4 times).
In this Q&A the alternative question is solved as a combinatorial problem: How many ways can we roll the dice $n$ times without the number '6' occuring $k$ or more times in a row.
This is found by finding all possible combinations of ways that we can combine the strings 'x', 'x6', 'x66', 'x666' (where 'x' is any number 1,2,3,4,5) into a string of length $n+1$ ($n+1$ instead of $n$ because in this way of constructing strings the first letter is always $x$ here). In this way we counted all possibilities to make a string of length $n$ but with only 1, 2, or 3 times a 6 in a row (and not 4 or more times).
Those combinations can be found by using an equivalent polynomial. This is very similar to the binomial coefficients which relate to the coefficients when we expand the power $(x+y)^n$, but it also relates to a combination.
The polynomial is
$$\begin{array}{rcl}
P(x) &=& \sum_{k=0}^\infty (5x+5x^2+5x^3+5x^4)^k\\
&=& \frac{1}{1-(5x+5x^2+5x^3+5x^4)} \\
&=& \frac{1}{1-5\frac{x-x^5}{1-x}}\\
&=& \frac{1-x}{1-6x+5x^5}
\end{array}$$
The coefficient of the $x^n$ relates to the number of ways to arrange the numbers 1,2,3,4,5,6 in a string of length $n-1$ without 4 or more 6's in a row. This coefficient can be found by a recursive relation. $$P(x) (1-6x+5x^5) = 1-x$$ which implies that the coefficients follow the relation
$$a_n - 6a_{n-1} + 5 a_{n-5} = 0$$
and the first coefficients can be computed manually
$$a_1,a_2,a_3,a_4,a_5,a_6,a_7 = 5,30,180,1080,6475,38825,232800$$
With this, you can compute $a_{1000}$ and $1-a_{1000}/6^{999}$ will be the probability to roll the same number 5 times in a row 5.
In the R-code below we compute this (and we include a division by 6 inside the recursion because the numbers $a_{1000}$ and $6^{999}$ are too large to compute directly). The result is $0.473981098314988$, the same as the computation with the Markov Chain.
x <- 6/5*c(5/6,30/6^2,180/6^3,1080/6^4,6475/6^5,38825/6^6,232800/6^7)
for (i in 1:1000) {
t <- tail(x,5)
x <- c(x,(6/6*t[5]-5/6^5*t[1])) ### this adds a new number to the back of the vector x
}
1-x[1000]
Analytic/Estimate
Method 1: wrong
You might think, the probability to have in any set of 5 neighboring dices, 5 of the same numbers, is $\frac{1}{6^4} = \frac{1}{1296}$, and since there are 996 sets of 5 neighboring dices the probability to have in at least one of these sets 5 of the same dices is:
$$ 1-(1-\frac{1}{6^4})^{996} \approx 0.536$$
But this is wrong. The reason is that the 996 sets are overlapping and not independent.
Method 2: correct
A better way is to approximate the Markov chain that we computed above. After some time you will get that the occupation of the states, with 1,2,3,4 of the same number in a row, are more or less stable and the ratio's will be roughly $1/6,1/6^2,1/6^3,1/6^4$ (*). Thus the fraction of the time that we have 4 in a row is:
$$\text{frequency 4 in a row} = \frac{1/6^4}{1/6+1/6^2+1/6^3+1/6^4}$$
If we have these 4 in a row then we have a 1/6-th probability to finish the game. So the frequency of finishing the game is
$$\text{finish-rate} = \frac{1}{6} \text{frequency 4 in a row} = \frac{1}{1554}$$
and the probability to be finished after $k$ steps is approximately
$$P_k \approx 1-(1-\frac{1}{1554})^{k-4} \underbrace{\approx 0.47330}_{\text{if $k=1000$}}$$
much closer to the exact computation.
(*) The occupation in state $k$ during roll $j$ will relate to the occupation in state $k-1$ during roll $j-1$. We will have $x_{k,j} = \frac{1}{6} x_{k-1,j-1} \approx \frac{1}{6} x_{k-1,j}$. Note that this requires that you have $x_{k-1,j} \approx x_{k-1,j-1}$, which occurs when the finish-rate is small. If this is not the case, then you could apply a factor to compensate, but the assumption of relatively steady ratio's will be wrong as well.
Related problems
Limit distribution associated with counts (non-trivial combinatoric problem)
Checking if a coin is fair based on how often a subsequence occurs
What is the probability of rolling all faces of a die after n number of rolls
Probability of a similar sub-sequence of length X in two sequences of length Y and Z
This latter related problem gives a different approximation based on expectation values and estimates the distribution as an overdispersed Poisson distribution. Giving an approximation $1- \exp \left(-(1000-5+1)\left(\frac{1}{6^4}\right) /1.2 \right)\approx 0.4729354$ which isn't bad either.
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a ro
|
Below we compute the probability in four ways:
Computation with Markov Chain 0.473981098314993
Computation with generating function 0.473981098314988
Estimation false method
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row?
Below we compute the probability in four ways:
Computation with Markov Chain 0.473981098314993
Computation with generating function 0.473981098314988
Estimation false method 0.536438013618686
Estimation correct method 0.473304632462677
The first two are exact methods and differ only a little (probably some round of errors), the third method is a naive estimation that does not give the correct number, the fourth method is better and gives a result that is very close to the exact method.
Computationally:
Markov Chain
You can model this computationally with a transition matrix
Say the column vector $X_{k,j} = \lbrace x_1,x_2,x_3,x_4,x_5 \rbrace_{j}$ is the probability to have $k$ of the same numbers in a row in the $j$-th dice roll. Then (when assuming a 6-sided dice)
$$X_{k,j} = M \cdot X_{k,j-1}$$ with
$$M = \begin{bmatrix} \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & 0 \\
\frac{1}{6} & 0& 0 & 0 & 0 \\
0& \frac{1}{6} & 0& 0 & 0 \\
0 & 0& \frac{1}{6} & 0& 0 \\
0&0 & 0& \frac{1}{6} & 1 \\
\end{bmatrix}$$
where this last entry $M_{5,5} = 1$ relates to 5 of the same in a row being an absorbing state where we 'stop' the experiment.
After the first roll you will be certainly in state 1 (there's certainly only 1 of the same number in a row).
$$X_{k,1} = \lbrace 1,0,0,0,0 \rbrace$$
After the $j$-th roll this will be multiplied with $M$ a $j-1$ times
$$X_{k,j} = M^{j-1} \lbrace 1,0,0,0,0 \rbrace$$
R-Code:
library(matrixcalc) ### allows us to use matrix.power
M <- matrix(c(5/6, 5/6, 5/6, 5/6, 0,
1/6, 0 , 0 , 0 , 0,
0, 1/6, 0 , 0 , 0,
0, 0 , 1/6, 0 , 0,
0, 0 , 0 , 1/6, 1),
5, byrow = TRUE)
start <- c(1,0,0,0,0)
matrix.power(M,999) %*% start
The result is $$X_{k,1000} = \begin{bmatrix}
0.438631855\\
0.073152468\\
0.012199943\\
0.002034635\\
\color{red}{0.473981098}\end{bmatrix}$$
and this last entry 0.473981098 is the probability to roll the same number 5 times in a row in 1000 rolls.
generating function
Our question is:
How to calculate the probability of rolling any number at least $k$ times in a row, out of $n$ tries?
This is equivalent to the question
How to calculate the probability of rolling the number 6 at least $k-1$ times in a row, out of $n-1$ tries?
You can see it as tracking whether the dice roll $m$ is the same number as the number of the dice roll $m-1$ (which has 1/6-th probabilty). And this needs to happen $k-1$ times in a row (in our case 4 times).
In this Q&A the alternative question is solved as a combinatorial problem: How many ways can we roll the dice $n$ times without the number '6' occuring $k$ or more times in a row.
This is found by finding all possible combinations of ways that we can combine the strings 'x', 'x6', 'x66', 'x666' (where 'x' is any number 1,2,3,4,5) into a string of length $n+1$ ($n+1$ instead of $n$ because in this way of constructing strings the first letter is always $x$ here). In this way we counted all possibilities to make a string of length $n$ but with only 1, 2, or 3 times a 6 in a row (and not 4 or more times).
Those combinations can be found by using an equivalent polynomial. This is very similar to the binomial coefficients which relate to the coefficients when we expand the power $(x+y)^n$, but it also relates to a combination.
The polynomial is
$$\begin{array}{rcl}
P(x) &=& \sum_{k=0}^\infty (5x+5x^2+5x^3+5x^4)^k\\
&=& \frac{1}{1-(5x+5x^2+5x^3+5x^4)} \\
&=& \frac{1}{1-5\frac{x-x^5}{1-x}}\\
&=& \frac{1-x}{1-6x+5x^5}
\end{array}$$
The coefficient of the $x^n$ relates to the number of ways to arrange the numbers 1,2,3,4,5,6 in a string of length $n-1$ without 4 or more 6's in a row. This coefficient can be found by a recursive relation. $$P(x) (1-6x+5x^5) = 1-x$$ which implies that the coefficients follow the relation
$$a_n - 6a_{n-1} + 5 a_{n-5} = 0$$
and the first coefficients can be computed manually
$$a_1,a_2,a_3,a_4,a_5,a_6,a_7 = 5,30,180,1080,6475,38825,232800$$
With this, you can compute $a_{1000}$ and $1-a_{1000}/6^{999}$ will be the probability to roll the same number 5 times in a row 5.
In the R-code below we compute this (and we include a division by 6 inside the recursion because the numbers $a_{1000}$ and $6^{999}$ are too large to compute directly). The result is $0.473981098314988$, the same as the computation with the Markov Chain.
x <- 6/5*c(5/6,30/6^2,180/6^3,1080/6^4,6475/6^5,38825/6^6,232800/6^7)
for (i in 1:1000) {
t <- tail(x,5)
x <- c(x,(6/6*t[5]-5/6^5*t[1])) ### this adds a new number to the back of the vector x
}
1-x[1000]
Analytic/Estimate
Method 1: wrong
You might think, the probability to have in any set of 5 neighboring dices, 5 of the same numbers, is $\frac{1}{6^4} = \frac{1}{1296}$, and since there are 996 sets of 5 neighboring dices the probability to have in at least one of these sets 5 of the same dices is:
$$ 1-(1-\frac{1}{6^4})^{996} \approx 0.536$$
But this is wrong. The reason is that the 996 sets are overlapping and not independent.
Method 2: correct
A better way is to approximate the Markov chain that we computed above. After some time you will get that the occupation of the states, with 1,2,3,4 of the same number in a row, are more or less stable and the ratio's will be roughly $1/6,1/6^2,1/6^3,1/6^4$ (*). Thus the fraction of the time that we have 4 in a row is:
$$\text{frequency 4 in a row} = \frac{1/6^4}{1/6+1/6^2+1/6^3+1/6^4}$$
If we have these 4 in a row then we have a 1/6-th probability to finish the game. So the frequency of finishing the game is
$$\text{finish-rate} = \frac{1}{6} \text{frequency 4 in a row} = \frac{1}{1554}$$
and the probability to be finished after $k$ steps is approximately
$$P_k \approx 1-(1-\frac{1}{1554})^{k-4} \underbrace{\approx 0.47330}_{\text{if $k=1000$}}$$
much closer to the exact computation.
(*) The occupation in state $k$ during roll $j$ will relate to the occupation in state $k-1$ during roll $j-1$. We will have $x_{k,j} = \frac{1}{6} x_{k-1,j-1} \approx \frac{1}{6} x_{k-1,j}$. Note that this requires that you have $x_{k-1,j} \approx x_{k-1,j-1}$, which occurs when the finish-rate is small. If this is not the case, then you could apply a factor to compensate, but the assumption of relatively steady ratio's will be wrong as well.
Related problems
Limit distribution associated with counts (non-trivial combinatoric problem)
Checking if a coin is fair based on how often a subsequence occurs
What is the probability of rolling all faces of a die after n number of rolls
Probability of a similar sub-sequence of length X in two sequences of length Y and Z
This latter related problem gives a different approximation based on expectation values and estimates the distribution as an overdispersed Poisson distribution. Giving an approximation $1- \exp \left(-(1000-5+1)\left(\frac{1}{6^4}\right) /1.2 \right)\approx 0.4729354$ which isn't bad either.
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a ro
Below we compute the probability in four ways:
Computation with Markov Chain 0.473981098314993
Computation with generating function 0.473981098314988
Estimation false method
|
6,777
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row?
|
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll matches the results of the previous 4 rolls, a flag is set to TRUE. The mean of this flag column and the mean of the runs is then reported. I get ~0.07% as the probability of seeing 5 rolls in a row of the same number.
In R,
tibble(
run = rep(seq(1:1000), each = 1000),
roll = rep(seq(1:1000), 1000),
x = sample(1:6, 1000000, replace = T)
) %>%
group_by(run) %>%
mutate(
same_five = x == lag(x, 1) & x == lag(x, 2) & x == lag(x, 3) & x == lag(x, 4)
) %>%
summarize(
p_same_five = mean(same_five, na.rm = TRUE), .groups = "drop"
) %>%
summarize(mean(p_same_five)) * 100
mean(p_same_five)
1 0.07208702
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a ro
|
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll mat
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row?
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll matches the results of the previous 4 rolls, a flag is set to TRUE. The mean of this flag column and the mean of the runs is then reported. I get ~0.07% as the probability of seeing 5 rolls in a row of the same number.
In R,
tibble(
run = rep(seq(1:1000), each = 1000),
roll = rep(seq(1:1000), 1000),
x = sample(1:6, 1000000, replace = T)
) %>%
group_by(run) %>%
mutate(
same_five = x == lag(x, 1) & x == lag(x, 2) & x == lag(x, 3) & x == lag(x, 4)
) %>%
summarize(
p_same_five = mean(same_five, na.rm = TRUE), .groups = "drop"
) %>%
summarize(mean(p_same_five)) * 100
mean(p_same_five)
1 0.07208702
|
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a ro
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll mat
|
6,778
|
Is cosine similarity identical to l2-normalized euclidean distance?
|
For $\ell^2$-normalized vectors $\mathbf{x}, \mathbf{y}$,
$$||\mathbf{x}||_2 = ||\mathbf{y}||_2 = 1,$$
we have that the squared Euclidean distance is proportional to the cosine distance,
\begin{align}
||\mathbf{x} - \mathbf{y}||_2^2
&= (\mathbf{x} - \mathbf{y})^\top (\mathbf{x} - \mathbf{y}) \\
&= \mathbf{x}^\top \mathbf{x} - 2 \mathbf{x}^\top \mathbf{y} + \mathbf{y}^\top \mathbf{y} \\
&= 2 - 2\mathbf{x}^\top \mathbf{y} \\
&= 2 - 2 \cos\angle(\mathbf{x}, \mathbf{y})
\end{align}
That is, even if you normalized your data and your algorithm was invariant to scaling of the distances, you would still expect differences because of the squaring.
|
Is cosine similarity identical to l2-normalized euclidean distance?
|
For $\ell^2$-normalized vectors $\mathbf{x}, \mathbf{y}$,
$$||\mathbf{x}||_2 = ||\mathbf{y}||_2 = 1,$$
we have that the squared Euclidean distance is proportional to the cosine distance,
\begin{align}
|
Is cosine similarity identical to l2-normalized euclidean distance?
For $\ell^2$-normalized vectors $\mathbf{x}, \mathbf{y}$,
$$||\mathbf{x}||_2 = ||\mathbf{y}||_2 = 1,$$
we have that the squared Euclidean distance is proportional to the cosine distance,
\begin{align}
||\mathbf{x} - \mathbf{y}||_2^2
&= (\mathbf{x} - \mathbf{y})^\top (\mathbf{x} - \mathbf{y}) \\
&= \mathbf{x}^\top \mathbf{x} - 2 \mathbf{x}^\top \mathbf{y} + \mathbf{y}^\top \mathbf{y} \\
&= 2 - 2\mathbf{x}^\top \mathbf{y} \\
&= 2 - 2 \cos\angle(\mathbf{x}, \mathbf{y})
\end{align}
That is, even if you normalized your data and your algorithm was invariant to scaling of the distances, you would still expect differences because of the squaring.
|
Is cosine similarity identical to l2-normalized euclidean distance?
For $\ell^2$-normalized vectors $\mathbf{x}, \mathbf{y}$,
$$||\mathbf{x}||_2 = ||\mathbf{y}||_2 = 1,$$
we have that the squared Euclidean distance is proportional to the cosine distance,
\begin{align}
|
6,779
|
Is cosine similarity identical to l2-normalized euclidean distance?
|
Standard cosine similarity is defined as follows in a Euclidian space, assuming column vectors $\mathbf{u}$ and $\mathbf{v}$:
$$
\cos(\mathbf{u}, \mathbf{v}) = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \cdot \|\mathbf{v}\|} = \frac{\mathbf{u}^T\mathbf{v}}{\|\mathbf{u}\| \cdot \|\mathbf{v}\|} \in [-1, 1].
$$
This reduces to the standard inner product if your vectors are normalized to unit norm (in l2). In text mining this kind of normalization is not unheard of, but I wouldn't consider that the standard.
|
Is cosine similarity identical to l2-normalized euclidean distance?
|
Standard cosine similarity is defined as follows in a Euclidian space, assuming column vectors $\mathbf{u}$ and $\mathbf{v}$:
$$
\cos(\mathbf{u}, \mathbf{v}) = \frac{\langle \mathbf{u}, \mathbf{v} \ra
|
Is cosine similarity identical to l2-normalized euclidean distance?
Standard cosine similarity is defined as follows in a Euclidian space, assuming column vectors $\mathbf{u}$ and $\mathbf{v}$:
$$
\cos(\mathbf{u}, \mathbf{v}) = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \cdot \|\mathbf{v}\|} = \frac{\mathbf{u}^T\mathbf{v}}{\|\mathbf{u}\| \cdot \|\mathbf{v}\|} \in [-1, 1].
$$
This reduces to the standard inner product if your vectors are normalized to unit norm (in l2). In text mining this kind of normalization is not unheard of, but I wouldn't consider that the standard.
|
Is cosine similarity identical to l2-normalized euclidean distance?
Standard cosine similarity is defined as follows in a Euclidian space, assuming column vectors $\mathbf{u}$ and $\mathbf{v}$:
$$
\cos(\mathbf{u}, \mathbf{v}) = \frac{\langle \mathbf{u}, \mathbf{v} \ra
|
6,780
|
What are the most common biases humans make when collecting or interpreting data?
|
I think in academia, p-values are very commonly misinterpreted. People tend to forget that the p-value expresses a conditional probability. Even if an experiment has been perfectly conducted and all requisites of the chosen statistical test are met, the false discovery rate is typically much higher than the significance level alpha. The false discovery rate increases with a decrease in statistical power and prevalence of true positives (Colquhoun, 2014; Nuzzo, 2014).
In addition people tend to regard their estimates as the truth and the parameter they estimate as random (Haller & Kraus, 2002). For example when they say that in “95% of the cases this identified confidence interval covers the parameter”...
Confusion of correlation and causation is probably also a very common error in data interpretation.
In terms of data collection, I think a common error is to take the most easily accessible rather than the most representative sample.
Colquhoun, D. (2014). An investigation of the false discovery rate and the misinterpretation of P values. Royal Society Open Science, 1–15.
Nuzzo, R. (2014). Statistical errors: P values, the “gold standard” of statistical validity are not as reliable as many scientists assume. Nature, 506, 150–152.
Haller, H. & Kraus, S. (2002): Misinterpretations of Significance: A Problem Students Share with Their Teachers? Methods of Psychological Research Online, Vol.7, No.1
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What are the most common biases humans make when collecting or interpreting data?
|
I think in academia, p-values are very commonly misinterpreted. People tend to forget that the p-value expresses a conditional probability. Even if an experiment has been perfectly conducted and all r
|
What are the most common biases humans make when collecting or interpreting data?
I think in academia, p-values are very commonly misinterpreted. People tend to forget that the p-value expresses a conditional probability. Even if an experiment has been perfectly conducted and all requisites of the chosen statistical test are met, the false discovery rate is typically much higher than the significance level alpha. The false discovery rate increases with a decrease in statistical power and prevalence of true positives (Colquhoun, 2014; Nuzzo, 2014).
In addition people tend to regard their estimates as the truth and the parameter they estimate as random (Haller & Kraus, 2002). For example when they say that in “95% of the cases this identified confidence interval covers the parameter”...
Confusion of correlation and causation is probably also a very common error in data interpretation.
In terms of data collection, I think a common error is to take the most easily accessible rather than the most representative sample.
Colquhoun, D. (2014). An investigation of the false discovery rate and the misinterpretation of P values. Royal Society Open Science, 1–15.
Nuzzo, R. (2014). Statistical errors: P values, the “gold standard” of statistical validity are not as reliable as many scientists assume. Nature, 506, 150–152.
Haller, H. & Kraus, S. (2002): Misinterpretations of Significance: A Problem Students Share with Their Teachers? Methods of Psychological Research Online, Vol.7, No.1
|
What are the most common biases humans make when collecting or interpreting data?
I think in academia, p-values are very commonly misinterpreted. People tend to forget that the p-value expresses a conditional probability. Even if an experiment has been perfectly conducted and all r
|
6,781
|
What are the most common biases humans make when collecting or interpreting data?
|
I would say a general inability to appreciate what true randomness looks like. People seem to expect too few spurious patterns than actually occur in sequences of random events. This also shows up when we try to simulate randomness on our own.
Another fairly common one is not understanding independence, as in the gambler's fallacy. We sometimes think that prior events can affect future ones even when it's clearly impossible, like the previous deal of a shuffled deck of cards impacting a future one.
|
What are the most common biases humans make when collecting or interpreting data?
|
I would say a general inability to appreciate what true randomness looks like. People seem to expect too few spurious patterns than actually occur in sequences of random events. This also shows up w
|
What are the most common biases humans make when collecting or interpreting data?
I would say a general inability to appreciate what true randomness looks like. People seem to expect too few spurious patterns than actually occur in sequences of random events. This also shows up when we try to simulate randomness on our own.
Another fairly common one is not understanding independence, as in the gambler's fallacy. We sometimes think that prior events can affect future ones even when it's clearly impossible, like the previous deal of a shuffled deck of cards impacting a future one.
|
What are the most common biases humans make when collecting or interpreting data?
I would say a general inability to appreciate what true randomness looks like. People seem to expect too few spurious patterns than actually occur in sequences of random events. This also shows up w
|
6,782
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What are the most common biases humans make when collecting or interpreting data?
|
It has already been pointed out that many of the behaviors and thought processes labeled "irrational" or "biased" by (behavioral) economists are actually highly adaptive and efficient in the real world. Nonetheless, OP's question is interesting. I think, however, that it may be profitable to refer to more fundamental, descriptive knowledge about our cognitive processes, rather than go looking for specific "biases" that correspond to those discussed in the economic literature (e.g., loss aversion, endowment effect, baserate neglect etc.).
For instance, evaluability is certainly an issue in data analysis. Evaluability theory states that we overweight information that we find easy to interpret or evaluate.
Consider the case of a regression coefficient. Evaluating the "real-world" consequences of a coefficient can be hard work. We need to consider the units of the independent and the dependent variable as well was the distributions of our independent and dependent variable to understand whether a coefficient has practical relevance. Evaluating the significance of a coefficient, on the other hand, is easy: I merely compare its p-value to my alpha level. Given the greater evaluability of the p-value compared the coefficient itself, it is scarcely surprising that so much is made of p-values.
(Standardizing increases the evaluability of a coefficient, but it may increase ambiguity: the sense that relevant information is unavailable or withheld, because the "original" form of the data we are processing is not available to us.)
A related cognitive "bias" is the concreteness principle, the tendency to overweight information that is "right there" in a decision context, and does not require retrieval from memory. (The concreteness principle also states that we are likely to use information in the format in which it is given and tend to avoid performing transformations.) Interpreting a p-value can be done by merely looking at the regression output; it does not require me to retrieve any substantive knowledge about the thing that I am modeling.
I expect that many biases in the interpretation of statistical data can be traced to the general understanding that we are likely to take the easy route when solving a problem or forming a judgment (see "cognitive miser", "bounded rationality" and so on). Relatedly, doing something "with ease" usually increases the confidence with which we hold the resulting beliefs (fluency theory). (One might also consider the possibility that data which are easier to articulate - to ourselves or to others - are overweighted in our analyses.) I think this becomes particularly interesting when we consider possible exceptions. Some psychological research suggests, for instance, that if we believe that a problem should be difficult to solve, then we may favor approaches and solutions which are less concrete and more difficult, e.g., choose a more arcane method over a simple one.
|
What are the most common biases humans make when collecting or interpreting data?
|
It has already been pointed out that many of the behaviors and thought processes labeled "irrational" or "biased" by (behavioral) economists are actually highly adaptive and efficient in the real worl
|
What are the most common biases humans make when collecting or interpreting data?
It has already been pointed out that many of the behaviors and thought processes labeled "irrational" or "biased" by (behavioral) economists are actually highly adaptive and efficient in the real world. Nonetheless, OP's question is interesting. I think, however, that it may be profitable to refer to more fundamental, descriptive knowledge about our cognitive processes, rather than go looking for specific "biases" that correspond to those discussed in the economic literature (e.g., loss aversion, endowment effect, baserate neglect etc.).
For instance, evaluability is certainly an issue in data analysis. Evaluability theory states that we overweight information that we find easy to interpret or evaluate.
Consider the case of a regression coefficient. Evaluating the "real-world" consequences of a coefficient can be hard work. We need to consider the units of the independent and the dependent variable as well was the distributions of our independent and dependent variable to understand whether a coefficient has practical relevance. Evaluating the significance of a coefficient, on the other hand, is easy: I merely compare its p-value to my alpha level. Given the greater evaluability of the p-value compared the coefficient itself, it is scarcely surprising that so much is made of p-values.
(Standardizing increases the evaluability of a coefficient, but it may increase ambiguity: the sense that relevant information is unavailable or withheld, because the "original" form of the data we are processing is not available to us.)
A related cognitive "bias" is the concreteness principle, the tendency to overweight information that is "right there" in a decision context, and does not require retrieval from memory. (The concreteness principle also states that we are likely to use information in the format in which it is given and tend to avoid performing transformations.) Interpreting a p-value can be done by merely looking at the regression output; it does not require me to retrieve any substantive knowledge about the thing that I am modeling.
I expect that many biases in the interpretation of statistical data can be traced to the general understanding that we are likely to take the easy route when solving a problem or forming a judgment (see "cognitive miser", "bounded rationality" and so on). Relatedly, doing something "with ease" usually increases the confidence with which we hold the resulting beliefs (fluency theory). (One might also consider the possibility that data which are easier to articulate - to ourselves or to others - are overweighted in our analyses.) I think this becomes particularly interesting when we consider possible exceptions. Some psychological research suggests, for instance, that if we believe that a problem should be difficult to solve, then we may favor approaches and solutions which are less concrete and more difficult, e.g., choose a more arcane method over a simple one.
|
What are the most common biases humans make when collecting or interpreting data?
It has already been pointed out that many of the behaviors and thought processes labeled "irrational" or "biased" by (behavioral) economists are actually highly adaptive and efficient in the real worl
|
6,783
|
What are the most common biases humans make when collecting or interpreting data?
|
The biggest single factor I can think of is broadly known as "confirmation bias". Having settled upon what I think my study will show, I uncritically accept data that lead to that conclusion, while making excuses for all data points that appear to refute it. I may unconsciously reject as "obvious instrument error" (or some equivalent) any data points that don't fit my conclusion. In some cases, it won't be quite as blatant; rather than throwing out those data points entirely, I'll concoct some formula to remove the "error", which will conveniently steer the results toward confirming my preordained conclusion.
There's nothing particularly nefarious about this; it's just how our brains work. It takes a great deal of effort to filter out such bias, and it's one of the reasons why scientists like to concoct double-blind studies, such that the person performing the measurements does not know what the experiment is expected to prove. It then requires enormous discipline not to adjust away what he has faithfully measured.
|
What are the most common biases humans make when collecting or interpreting data?
|
The biggest single factor I can think of is broadly known as "confirmation bias". Having settled upon what I think my study will show, I uncritically accept data that lead to that conclusion, while m
|
What are the most common biases humans make when collecting or interpreting data?
The biggest single factor I can think of is broadly known as "confirmation bias". Having settled upon what I think my study will show, I uncritically accept data that lead to that conclusion, while making excuses for all data points that appear to refute it. I may unconsciously reject as "obvious instrument error" (or some equivalent) any data points that don't fit my conclusion. In some cases, it won't be quite as blatant; rather than throwing out those data points entirely, I'll concoct some formula to remove the "error", which will conveniently steer the results toward confirming my preordained conclusion.
There's nothing particularly nefarious about this; it's just how our brains work. It takes a great deal of effort to filter out such bias, and it's one of the reasons why scientists like to concoct double-blind studies, such that the person performing the measurements does not know what the experiment is expected to prove. It then requires enormous discipline not to adjust away what he has faithfully measured.
|
What are the most common biases humans make when collecting or interpreting data?
The biggest single factor I can think of is broadly known as "confirmation bias". Having settled upon what I think my study will show, I uncritically accept data that lead to that conclusion, while m
|
6,784
|
What are the most common biases humans make when collecting or interpreting data?
|
Linearity.
I think a common bias during data interpretation/analysis is that people usually are quick to assume linear relations. Mathematically, a regression model assumes that its deterministic component is a linear function of the predictors; unfortunately that is not always true. I recently went to an undergraduate poster conference and the amount of bluntly quadratic or non-linear trends I saw being fitted with a linear model was worrying to say the least.
(This is in addition to the mentions of gambler's fallacy, $p$-value misinterpretation and true randomness; +1 to all relevant posts.)
|
What are the most common biases humans make when collecting or interpreting data?
|
Linearity.
I think a common bias during data interpretation/analysis is that people usually are quick to assume linear relations. Mathematically, a regression model assumes that its deterministic com
|
What are the most common biases humans make when collecting or interpreting data?
Linearity.
I think a common bias during data interpretation/analysis is that people usually are quick to assume linear relations. Mathematically, a regression model assumes that its deterministic component is a linear function of the predictors; unfortunately that is not always true. I recently went to an undergraduate poster conference and the amount of bluntly quadratic or non-linear trends I saw being fitted with a linear model was worrying to say the least.
(This is in addition to the mentions of gambler's fallacy, $p$-value misinterpretation and true randomness; +1 to all relevant posts.)
|
What are the most common biases humans make when collecting or interpreting data?
Linearity.
I think a common bias during data interpretation/analysis is that people usually are quick to assume linear relations. Mathematically, a regression model assumes that its deterministic com
|
6,785
|
What are the most common biases humans make when collecting or interpreting data?
|
An intersting case is the discussions of the Gamblers Fallacy.
Should the existing data be included or exluded? If I am already ahead with 6 sixes, are these to be included in my run of a dozen tries? Be clear about prior data.
When should I change from absolute numbers to ratio's? It takes a long time for the advantage gained during an intial winning streak to return to zero (a random walk).
0.1% of a million dollars may not be much to a big company, but to loose $1000 could be life and death to a sole trader (which is why investors want 'driven' people to invest in). Being able to shift to percentages can be a bias.
Even statisticians have biases.
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What are the most common biases humans make when collecting or interpreting data?
|
An intersting case is the discussions of the Gamblers Fallacy.
Should the existing data be included or exluded? If I am already ahead with 6 sixes, are these to be included in my run of a dozen tries
|
What are the most common biases humans make when collecting or interpreting data?
An intersting case is the discussions of the Gamblers Fallacy.
Should the existing data be included or exluded? If I am already ahead with 6 sixes, are these to be included in my run of a dozen tries? Be clear about prior data.
When should I change from absolute numbers to ratio's? It takes a long time for the advantage gained during an intial winning streak to return to zero (a random walk).
0.1% of a million dollars may not be much to a big company, but to loose $1000 could be life and death to a sole trader (which is why investors want 'driven' people to invest in). Being able to shift to percentages can be a bias.
Even statisticians have biases.
|
What are the most common biases humans make when collecting or interpreting data?
An intersting case is the discussions of the Gamblers Fallacy.
Should the existing data be included or exluded? If I am already ahead with 6 sixes, are these to be included in my run of a dozen tries
|
6,786
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What are the most common biases humans make when collecting or interpreting data?
|
I would recommend "Thinking, Fast and Slow" by Daniel Kahneman, which explains many cognitive biases in lucid language.
You may also refer to "http://www.burns-stat.com/review-thinking-fast-slow-daniel-kahneman/" which summarizes some of the biases in the above book.
For more detailed chapter wise summary you may want to read "https://erikreads.files.wordpress.com/2014/04/thinking-fast-and-slow-book-summary.pdf".
|
What are the most common biases humans make when collecting or interpreting data?
|
I would recommend "Thinking, Fast and Slow" by Daniel Kahneman, which explains many cognitive biases in lucid language.
You may also refer to "http://www.burns-stat.com/review-thinking-fast-slow-danie
|
What are the most common biases humans make when collecting or interpreting data?
I would recommend "Thinking, Fast and Slow" by Daniel Kahneman, which explains many cognitive biases in lucid language.
You may also refer to "http://www.burns-stat.com/review-thinking-fast-slow-daniel-kahneman/" which summarizes some of the biases in the above book.
For more detailed chapter wise summary you may want to read "https://erikreads.files.wordpress.com/2014/04/thinking-fast-and-slow-book-summary.pdf".
|
What are the most common biases humans make when collecting or interpreting data?
I would recommend "Thinking, Fast and Slow" by Daniel Kahneman, which explains many cognitive biases in lucid language.
You may also refer to "http://www.burns-stat.com/review-thinking-fast-slow-danie
|
6,787
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Simple way to algorithmically identify a spike in recorded errors
|
It has been 5 months since you asked this question, and hopefully you figured something out. I'm going to make a few different suggestions here, hoping that you find some use for them in other scenarios.
For your use-case I don't think you need to look at spike-detection algorithms.
So here goes:
Let's start with a picture of the errors occurring on a timeline:
What you want is a numerical indicator, a "measure" of how fast the errors are coming. And this measure should be amenable to thresholding - your sysadmins should be able to set limits which control with what sensitivity errors turn into warnings.
Measure 1
You mentioned "spikes", the easiest way to get a spike is to draw a histogram over every 20-minute interval:
Your sysadmins would set the sensitivity based on the heights of the bars i.e. the most errors tolerable in a 20-minute interval.
(At this point you may be wondering if that 20-minute window length can't be adjusted. It can, and you can think of the window length as defining the word together in the phrase errors appearing together.)
What's the problem with this method for your particular scenario? Well, your variable is an integer, probably less than 3. You wouldn't set your threshold to 1, since that just means "every error is a warning" which doesn't require an algorithm. So your choices for the threshold are going to be 2 and 3. This doesn't give your sysadmins a whole lot of fine-grained control.
Measure 2
Instead of counting errors in a time window, keep track of the number of minutes between the current and last errors. When this value gets too small, it means your errors are getting too frequent and you need to raise a warning.
Your sysadmins will probably set the limit at 10 (i.e. if errors are happening less than 10 minutes apart, it's a problem) or 20 minutes. Maybe 30 minutes for a less mission-critical system.
This measure provides more flexibility. Unlike Measure 1, for which there was a small set of values you could work with, now you have a measure which provides a good 20-30 values. Your sysadmins will therefore have more scope for fine-tuning.
Friendly Advice
There is another way to approach this problem. Rather than looking at the error frequencies, it may be possible to predict the errors before they occur.
You mentioned that this behavior was occurring on a single server, which is known to have performance issues. You could monitor certain Key Performance Indicators on that machine, and have them tell you when an error is going to happen. Specifically, you would look at CPU usage, Memory usage, and KPIs relating to Disk I/O. If your CPU usage crosses 80%, the system's going to slow down.
(I know you said you didn't want to install any software, and it's true that you could do this using PerfMon. But there are free tools out there which will do this for you, like Nagios and Zenoss.)
And for people who came here hoping to find something about spike detection in a time-series:
Spike Detection in a Time-Series
The simplest thing you should start by doing is to compute a moving average of your input values. If your series is $x_1, x_2,...$, then you would compute a moving average after each observation as:
$M_k = (1 - \alpha) M_{k-1} + \alpha x_k$
where the $\alpha$ would determine how much weight give the latest value of $x_k$.
If your new value has moved too far away from the moving average, for example
$\frac{x_k - M_k}{M_k} > 20\%$
then you raise a warning.
Moving averages are nice when working with real-time data. But suppose you already have a bunch of data in a table, and you just want to run SQL queries against it to find the spikes.
I would suggest:
Compute the mean value of your time-series
Compute the standard deviation $\sigma$
Isolate those values which are more than $2\sigma$ above the mean (you may need to adjust that factor of "2")
More fun stuff about time series
Many real-world time-series exhibit cyclic behavior. There is a model called ARIMA which helps you extract these cycles from your time-series.
Moving averages which take into account cyclic behavior: Holt and Winters
|
Simple way to algorithmically identify a spike in recorded errors
|
It has been 5 months since you asked this question, and hopefully you figured something out. I'm going to make a few different suggestions here, hoping that you find some use for them in other scenari
|
Simple way to algorithmically identify a spike in recorded errors
It has been 5 months since you asked this question, and hopefully you figured something out. I'm going to make a few different suggestions here, hoping that you find some use for them in other scenarios.
For your use-case I don't think you need to look at spike-detection algorithms.
So here goes:
Let's start with a picture of the errors occurring on a timeline:
What you want is a numerical indicator, a "measure" of how fast the errors are coming. And this measure should be amenable to thresholding - your sysadmins should be able to set limits which control with what sensitivity errors turn into warnings.
Measure 1
You mentioned "spikes", the easiest way to get a spike is to draw a histogram over every 20-minute interval:
Your sysadmins would set the sensitivity based on the heights of the bars i.e. the most errors tolerable in a 20-minute interval.
(At this point you may be wondering if that 20-minute window length can't be adjusted. It can, and you can think of the window length as defining the word together in the phrase errors appearing together.)
What's the problem with this method for your particular scenario? Well, your variable is an integer, probably less than 3. You wouldn't set your threshold to 1, since that just means "every error is a warning" which doesn't require an algorithm. So your choices for the threshold are going to be 2 and 3. This doesn't give your sysadmins a whole lot of fine-grained control.
Measure 2
Instead of counting errors in a time window, keep track of the number of minutes between the current and last errors. When this value gets too small, it means your errors are getting too frequent and you need to raise a warning.
Your sysadmins will probably set the limit at 10 (i.e. if errors are happening less than 10 minutes apart, it's a problem) or 20 minutes. Maybe 30 minutes for a less mission-critical system.
This measure provides more flexibility. Unlike Measure 1, for which there was a small set of values you could work with, now you have a measure which provides a good 20-30 values. Your sysadmins will therefore have more scope for fine-tuning.
Friendly Advice
There is another way to approach this problem. Rather than looking at the error frequencies, it may be possible to predict the errors before they occur.
You mentioned that this behavior was occurring on a single server, which is known to have performance issues. You could monitor certain Key Performance Indicators on that machine, and have them tell you when an error is going to happen. Specifically, you would look at CPU usage, Memory usage, and KPIs relating to Disk I/O. If your CPU usage crosses 80%, the system's going to slow down.
(I know you said you didn't want to install any software, and it's true that you could do this using PerfMon. But there are free tools out there which will do this for you, like Nagios and Zenoss.)
And for people who came here hoping to find something about spike detection in a time-series:
Spike Detection in a Time-Series
The simplest thing you should start by doing is to compute a moving average of your input values. If your series is $x_1, x_2,...$, then you would compute a moving average after each observation as:
$M_k = (1 - \alpha) M_{k-1} + \alpha x_k$
where the $\alpha$ would determine how much weight give the latest value of $x_k$.
If your new value has moved too far away from the moving average, for example
$\frac{x_k - M_k}{M_k} > 20\%$
then you raise a warning.
Moving averages are nice when working with real-time data. But suppose you already have a bunch of data in a table, and you just want to run SQL queries against it to find the spikes.
I would suggest:
Compute the mean value of your time-series
Compute the standard deviation $\sigma$
Isolate those values which are more than $2\sigma$ above the mean (you may need to adjust that factor of "2")
More fun stuff about time series
Many real-world time-series exhibit cyclic behavior. There is a model called ARIMA which helps you extract these cycles from your time-series.
Moving averages which take into account cyclic behavior: Holt and Winters
|
Simple way to algorithmically identify a spike in recorded errors
It has been 5 months since you asked this question, and hopefully you figured something out. I'm going to make a few different suggestions here, hoping that you find some use for them in other scenari
|
6,788
|
Simple way to algorithmically identify a spike in recorded errors
|
+1 for Statistical process control, there's some useful information here on Step Detection.
For SPC it's not too hard to write an implementation of either the Western Electric Rules or the Nelson Rules.
Just make a USP in SQL server that will iterate through a data set and ping each point against the rules using its neighbouring points. Maybe sum up the number of errors by hour (depending on your needs).
This kind of relates to a question I posted on Stack Overflow a while back (have just penned a quick answer if it helps): Statistical Process Control Charts in SQL Server 2008 R2
|
Simple way to algorithmically identify a spike in recorded errors
|
+1 for Statistical process control, there's some useful information here on Step Detection.
For SPC it's not too hard to write an implementation of either the Western Electric Rules or the Nelson Rule
|
Simple way to algorithmically identify a spike in recorded errors
+1 for Statistical process control, there's some useful information here on Step Detection.
For SPC it's not too hard to write an implementation of either the Western Electric Rules or the Nelson Rules.
Just make a USP in SQL server that will iterate through a data set and ping each point against the rules using its neighbouring points. Maybe sum up the number of errors by hour (depending on your needs).
This kind of relates to a question I posted on Stack Overflow a while back (have just penned a quick answer if it helps): Statistical Process Control Charts in SQL Server 2008 R2
|
Simple way to algorithmically identify a spike in recorded errors
+1 for Statistical process control, there's some useful information here on Step Detection.
For SPC it's not too hard to write an implementation of either the Western Electric Rules or the Nelson Rule
|
6,789
|
Simple way to algorithmically identify a spike in recorded errors
|
A search for Online detection algorithms would be a start.
More information located on stackoverflow: Peak Dection of measured signal
A python implementation of a naive peak detection routine is to be found at github
|
Simple way to algorithmically identify a spike in recorded errors
|
A search for Online detection algorithms would be a start.
More information located on stackoverflow: Peak Dection of measured signal
A python implementation of a naive peak detection routine is to be
|
Simple way to algorithmically identify a spike in recorded errors
A search for Online detection algorithms would be a start.
More information located on stackoverflow: Peak Dection of measured signal
A python implementation of a naive peak detection routine is to be found at github
|
Simple way to algorithmically identify a spike in recorded errors
A search for Online detection algorithms would be a start.
More information located on stackoverflow: Peak Dection of measured signal
A python implementation of a naive peak detection routine is to be
|
6,790
|
Simple way to algorithmically identify a spike in recorded errors
|
You may want to look at statistical process control. Or time series monitoring. There are tons of work in this direction, and the optimal answer probably depends a lot on what exactly you are doing (do you need to filter out yearly or weekly seasonalities in load before detecting anomalies etc.).
|
Simple way to algorithmically identify a spike in recorded errors
|
You may want to look at statistical process control. Or time series monitoring. There are tons of work in this direction, and the optimal answer probably depends a lot on what exactly you are doing (d
|
Simple way to algorithmically identify a spike in recorded errors
You may want to look at statistical process control. Or time series monitoring. There are tons of work in this direction, and the optimal answer probably depends a lot on what exactly you are doing (do you need to filter out yearly or weekly seasonalities in load before detecting anomalies etc.).
|
Simple way to algorithmically identify a spike in recorded errors
You may want to look at statistical process control. Or time series monitoring. There are tons of work in this direction, and the optimal answer probably depends a lot on what exactly you are doing (d
|
6,791
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If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
Many frequentist confidence intervals (CIs) are based on the likelihood function.
If the prior distribution is truly non-informative, then the a Bayesian
posterior has essentially the same information as the likelihood function.
Consequently, in practice, a Bayesian probability interval (or credible interval)
may be very similar numerically to a frequentist confidence interval.
[Of course, even if numerically similar, there are philosophical differences
in interpretation between frequentist and Bayesian interval estimates.]
Here is a simple example, estimating binomial success probability $\theta.$
Suppose we have $n = 100$ observations (trials) with $X = 73$ successes.
Frequentist: The traditional Wald interval uses the point estimate
$\hat \theta = X/n = 73/100 = 0.73.$ And the 95% CI is of the form
$$\hat \theta \pm 1.96\sqrt{\frac{\hat \theta(1-\hat \theta)} {n}},$$
which computes to $(0.643,\,0.817).$
n = 100; x = 73; th.w = x/n; pm = c(-1,1)
ci.w = th.w + pm*1.96*sqrt(th.w*(1-th.w)/n); ci.w
[1] 0.6429839 0.8170161
This form of CI assumes that relevant
binomial distributions can be approximated by normal ones and that the
margin of error $\sqrt{\theta(1-\theta)/n}$ is well approximated by
$\sqrt{\hat\theta(1-\hat\theta)/n}.$ Particularly for small $n,$ these
assumptions need not be true. [The cases where $X = 0$ or $X = n$ are
especially problematic.]
The Agresti-Coull CI has been shown to have more accurate coverage probability.
This interval 'adds two Success and two Failures' as a trick to get
a coverage probability nearer to 95%. It begins with the point estimate
$\tilde \theta = (X+2)/\tilde n,$ where $\tilde n + 4.$ Then a 95% CI is of
the form
$$\tilde \theta \pm 1.96\sqrt{\frac{\tilde \theta(1-\tilde \theta)} {\tilde n}},$$
which computes to $(0.612, 0.792).$ For $n > 100$ and $0.3 < \tilde \theta < 0.7,$ the difference between
these two styles of confidence intervals is nearly negligible.
ci.a = th.a + pm*1.96*sqrt(th.a*(1-th.a)/n); ci.a
[1] 0.6122700 0.7915761
Bayesian:
One popular noninformative prior in this situation is $\mathsf{Beta}(1,1) \equiv \mathsf{Unif}(0,1).$ The likelihood function is proportional to
$\theta^x(1-\theta)^{n-x}.$ Multiplying the kernels of the prior and
likelihood we have the kernel of the posterior distribution
$\mathsf{Beta}(x+1,\, n-x+1).$
Then a 95% Bayesian interval estimate uses
quantiles 0.025 and 0.975 of the posterior distribution to get $(0.635, 0.807).$
When the prior distribution is 'flat' or 'noninformative' the numerical difference between the Bayesian probability interval and the Agresti-Coull
confidence interval is slight.
qbeta(c(.025, .975), 74, 28)
[1] 0.6353758 0.8072313
Notes: (a) In this situation, some Bayesians prefer the noninformative prior $\mathsf{Beta}(.5, .5).$ (b) For confidence levels other than 95%, the Agresti-Coull CI uses a slightly different point estimate. (c) For data other than binomial, there may be no available 'flat' prior,
but one can choose a prior with a huge variance (small precision) that
carries very little information. (d) For more discussion of Agresti-Coull CIs, graphs of coverage probabilities, and some references, perhaps also see this Q & A.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
Many frequentist confidence intervals (CIs) are based on the likelihood function.
If the prior distribution is truly non-informative, then the a Bayesian
posterior has essentially the same information
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
Many frequentist confidence intervals (CIs) are based on the likelihood function.
If the prior distribution is truly non-informative, then the a Bayesian
posterior has essentially the same information as the likelihood function.
Consequently, in practice, a Bayesian probability interval (or credible interval)
may be very similar numerically to a frequentist confidence interval.
[Of course, even if numerically similar, there are philosophical differences
in interpretation between frequentist and Bayesian interval estimates.]
Here is a simple example, estimating binomial success probability $\theta.$
Suppose we have $n = 100$ observations (trials) with $X = 73$ successes.
Frequentist: The traditional Wald interval uses the point estimate
$\hat \theta = X/n = 73/100 = 0.73.$ And the 95% CI is of the form
$$\hat \theta \pm 1.96\sqrt{\frac{\hat \theta(1-\hat \theta)} {n}},$$
which computes to $(0.643,\,0.817).$
n = 100; x = 73; th.w = x/n; pm = c(-1,1)
ci.w = th.w + pm*1.96*sqrt(th.w*(1-th.w)/n); ci.w
[1] 0.6429839 0.8170161
This form of CI assumes that relevant
binomial distributions can be approximated by normal ones and that the
margin of error $\sqrt{\theta(1-\theta)/n}$ is well approximated by
$\sqrt{\hat\theta(1-\hat\theta)/n}.$ Particularly for small $n,$ these
assumptions need not be true. [The cases where $X = 0$ or $X = n$ are
especially problematic.]
The Agresti-Coull CI has been shown to have more accurate coverage probability.
This interval 'adds two Success and two Failures' as a trick to get
a coverage probability nearer to 95%. It begins with the point estimate
$\tilde \theta = (X+2)/\tilde n,$ where $\tilde n + 4.$ Then a 95% CI is of
the form
$$\tilde \theta \pm 1.96\sqrt{\frac{\tilde \theta(1-\tilde \theta)} {\tilde n}},$$
which computes to $(0.612, 0.792).$ For $n > 100$ and $0.3 < \tilde \theta < 0.7,$ the difference between
these two styles of confidence intervals is nearly negligible.
ci.a = th.a + pm*1.96*sqrt(th.a*(1-th.a)/n); ci.a
[1] 0.6122700 0.7915761
Bayesian:
One popular noninformative prior in this situation is $\mathsf{Beta}(1,1) \equiv \mathsf{Unif}(0,1).$ The likelihood function is proportional to
$\theta^x(1-\theta)^{n-x}.$ Multiplying the kernels of the prior and
likelihood we have the kernel of the posterior distribution
$\mathsf{Beta}(x+1,\, n-x+1).$
Then a 95% Bayesian interval estimate uses
quantiles 0.025 and 0.975 of the posterior distribution to get $(0.635, 0.807).$
When the prior distribution is 'flat' or 'noninformative' the numerical difference between the Bayesian probability interval and the Agresti-Coull
confidence interval is slight.
qbeta(c(.025, .975), 74, 28)
[1] 0.6353758 0.8072313
Notes: (a) In this situation, some Bayesians prefer the noninformative prior $\mathsf{Beta}(.5, .5).$ (b) For confidence levels other than 95%, the Agresti-Coull CI uses a slightly different point estimate. (c) For data other than binomial, there may be no available 'flat' prior,
but one can choose a prior with a huge variance (small precision) that
carries very little information. (d) For more discussion of Agresti-Coull CIs, graphs of coverage probabilities, and some references, perhaps also see this Q & A.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
Many frequentist confidence intervals (CIs) are based on the likelihood function.
If the prior distribution is truly non-informative, then the a Bayesian
posterior has essentially the same information
|
6,792
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
BruceET's answer is excellent but pretty long, so here's a quick practical summary:
if the prior is flat, likelihood and posterior have the same shape
the intervals, however, are not necessarily the same, because they are constructed in different ways. A standard Bayesian 90% CI covers the central 90% of the posterior. A frequentist CI is usually defined by a point-wise comparison (see BruceET's answer). For an unbounded location parameter (e.g. estimating the mean of a normal distribution), difference are usually small, but if you estimate a bounded parameter (e.g. binomial mean) close to the boundaries (0/1), differences can be substantial.
of course, the interpretation is different too, but I interpret the question mainly as "when will the values be the same?"
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
BruceET's answer is excellent but pretty long, so here's a quick practical summary:
if the prior is flat, likelihood and posterior have the same shape
the intervals, however, are not necessarily the
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
BruceET's answer is excellent but pretty long, so here's a quick practical summary:
if the prior is flat, likelihood and posterior have the same shape
the intervals, however, are not necessarily the same, because they are constructed in different ways. A standard Bayesian 90% CI covers the central 90% of the posterior. A frequentist CI is usually defined by a point-wise comparison (see BruceET's answer). For an unbounded location parameter (e.g. estimating the mean of a normal distribution), difference are usually small, but if you estimate a bounded parameter (e.g. binomial mean) close to the boundaries (0/1), differences can be substantial.
of course, the interpretation is different too, but I interpret the question mainly as "when will the values be the same?"
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
BruceET's answer is excellent but pretty long, so here's a quick practical summary:
if the prior is flat, likelihood and posterior have the same shape
the intervals, however, are not necessarily the
|
6,793
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
While one can solve for a prior that yields a credible interval that equals the frequentist confidence interval, it is important to realize how narrow the scope of application is. The entire discussion is assuming that the sample size was fixed and is not a random variable. It assumes that there was only one look at the data, and that sequential inference was not done. It assumes there was only one dependent variable and no other parameters were of interest. Where there are multiplicities, the Bayesian and frequentist intervals diverge (Bayesian posterior probabilities are in forward-time predictive mode and don't need to consider "how we got here", thus have no way or need to adjust for multiple looks). In addition, in the frequentist world the interpretation of confidence intervals is extremely strange and has confused many a student and caused some frequentist statisticians to become Bayesian.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
While one can solve for a prior that yields a credible interval that equals the frequentist confidence interval, it is important to realize how narrow the scope of application is. The entire discussi
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
While one can solve for a prior that yields a credible interval that equals the frequentist confidence interval, it is important to realize how narrow the scope of application is. The entire discussion is assuming that the sample size was fixed and is not a random variable. It assumes that there was only one look at the data, and that sequential inference was not done. It assumes there was only one dependent variable and no other parameters were of interest. Where there are multiplicities, the Bayesian and frequentist intervals diverge (Bayesian posterior probabilities are in forward-time predictive mode and don't need to consider "how we got here", thus have no way or need to adjust for multiple looks). In addition, in the frequentist world the interpretation of confidence intervals is extremely strange and has confused many a student and caused some frequentist statisticians to become Bayesian.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
While one can solve for a prior that yields a credible interval that equals the frequentist confidence interval, it is important to realize how narrow the scope of application is. The entire discussi
|
6,794
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
Likelihood $\neq$ Bayesian with flat prior
The likelihood function, and associated the confidence interval, are not the same (concept) as a Bayesian posterior probability constructed with a prior that specifies a uniform distribution.
In part 1 and 2 of this answer it is argued why likelihood should not be viewed as a Bayesian posterior probability based on a flat prior.
In part 3 an example is given where the confidence interval and credible interval are widely varying. Also it is pointed out how this discrepancy arises.
1 Different behavior when variable is transformed
Probabilities transform in a particular way. If we know the probability distribution distribution $f_x(x)$ then we also know the distribution of $f_\xi(\xi)$ for the variable $\xi$ defined by any function $x=\chi(\xi)$, according to the transformation rule:
$$f_\xi(\xi) = f_x(\chi(\xi)) \frac{d\chi}{d\xi} d\xi $$
If you transform a variable then the mean and the mode may vary due to this change of the distribution function. That means $\bar{x} \neq \chi(\bar{\xi})$ and $x_{\max f(x)} \neq \chi(\xi_{\max f(\xi)})$.
The likelihood function does not transform in this way. This is the contrasts between the likelihood function and the posterior probability. The (maximum of the) likelihood function remains the same when you transform the variable.
$$\mathcal{L}_\xi(\xi) = \mathcal{L}_x(\chi(\xi)) $$
Related:
The flat prior is ambiguous. It depends on the form of the particular statistic.
For instance, if $X$ is uniform distributed (e.g. $\mathcal{U}(0,1))$, then $X^2$ is not a uniform distributed variable.
There is no single flat prior that you can relate the Likelihood function to. It is different when you define the flat prior for $X$ or some transformed variable like $X^2$. For the likelihood this dependency does not exist.
The boundaries of probabilities (credibility intervals) will be different when you transform the variable, (for likelihood functions this is not the case). E.g for some parameter $a$ and a monotonic transformation $f(a)$ (e.g. logarithm) you get the equivalent likelihood intervals
$$\begin{array}{ccccc}
a_{\min} &<& a &<& a_{\max}\\
f(a_{\min}) &<& f(a) &<& f(a_{\max})
\end{array}$$
2 Different concept: confidence intervals are independent from the prior
Suppose you sample a variable $X$ from a population with (unknown) parameter $\theta$ which itself (the population with parameter $\theta$) is sampled from a super-population (with possibly varying values for $\theta$).
One can make an inverse statement trying to infer what the original $\theta$ may have been based on observing some values $x_i$ for the variable $X$.
Bayesian methods do this by supposing a prior distribution for the distribution of possible $\theta$
This contrasts with the likelihood function and confidence interval, which are independent from the prior distribution.
The confidence interval does not use information of a prior like the credible interval does (confidence is not a probability).
Regardless of the prior distribution (uniform or not) the x%-confidence interval will contain the true parameter in $x%$ of the cases (confidence intervals refer to the success rate, type I error, of the method, not of a particular case).
In the case of the credible interval this concept ($%$ of time that the interval contains the true parameter) is not even applicable, but we may interpret it in a frequentist sense and then we observe that the credible interval will contain the true parameter only $x%$ of the time when the (uniform) prior is correctly describing the super-population of parameters that we may encounter. The interval may effectively be performing higher or lower than the x% (not that this matters since the Bayesian approach answers different questions, but it is just to note the difference).
3 Difference between confidence and credible intervals
In the example below we examine the likelihood function for the exponential distribution as function of the rate parameter $\lambda$, the sample mean $\bar{x}$, and sample size $n$:
$$\mathcal{L}(\lambda,\bar{x},n) = \frac{n^n}{(n-1)!} x^{n-1} \lambda^n e^{-\lambda n \bar{x}}$$
this functions expresses the probability to observe (for a given $n$ and $\lambda$) a sample mean between $\bar{x}$ and $\bar{x}+dx$.
note: the rate parameter $\lambda$ goes from $0$ to $\infty$ (unlike the OP 'request' from $0$ to $1$). The prior in this case will be an improper prior. The principles however does not change. I am using this perspective for easier illustration. Distributions with parameters between $0$ and $1$ are often discrete distributions (difficult to drawing continuous lines) or a beta distribution (difficult to calculate)
The image below illustrates this likelihood function (the blue colored map), for sample size $n=4$, and also draws the boundaries for the 95% intervals (both confidence and credible).
The boundaries are created obtaining the (one-dimensional) cumulative distribution function. But, this integration/cumulation can be done in two directions.
The difference between the intervals occurs because the 5% area's are made in different ways.
The 95% confidence interval contains values $\lambda$ for which the observed value $\bar{x}$ would occur at least in 95% of the cases. In this way. whatever the value $\lambda$, we would only make a wrong judgement in 95% of the cases.
For any $\lambda$ you have north and south of the boundaries (changing $\bar{x}$) 2.5% of the weight of the likelihood function.
The 95% credible interval contains values $\lambda$ which are most likely to cause the observed value $\bar{x}$ (given a flat prior).
Even when the observed result $\bar{x}$ is less than 5% likely for a given $\lambda$, the particular $\lambda$ may be inside the credible interval. In the particular example higher values of $\lambda$ are 'preferred' for the credible interval.
For any $\bar{x}$ you have west and east of the boundaries (changing $\lambda$) 2.5% of the weight of the likelihood function.
A case where confidence interval and credible interval (based on improper prior) coincide is for estimating the mean of a Gaussian distributed variable (the distribution is illustrated here: https://stats.stackexchange.com/a/351333/164061 ).
An obvious case where confidence interval and credible interval do not coincide is illustrated here (https://stats.stackexchange.com/a/369909/164061). The confidence interval for this case may have one or even both of the (upper/lower) bounds at infinity.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
Likelihood $\neq$ Bayesian with flat prior
The likelihood function, and associated the confidence interval, are not the same (concept) as a Bayesian posterior probability constructed with a prior that
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
Likelihood $\neq$ Bayesian with flat prior
The likelihood function, and associated the confidence interval, are not the same (concept) as a Bayesian posterior probability constructed with a prior that specifies a uniform distribution.
In part 1 and 2 of this answer it is argued why likelihood should not be viewed as a Bayesian posterior probability based on a flat prior.
In part 3 an example is given where the confidence interval and credible interval are widely varying. Also it is pointed out how this discrepancy arises.
1 Different behavior when variable is transformed
Probabilities transform in a particular way. If we know the probability distribution distribution $f_x(x)$ then we also know the distribution of $f_\xi(\xi)$ for the variable $\xi$ defined by any function $x=\chi(\xi)$, according to the transformation rule:
$$f_\xi(\xi) = f_x(\chi(\xi)) \frac{d\chi}{d\xi} d\xi $$
If you transform a variable then the mean and the mode may vary due to this change of the distribution function. That means $\bar{x} \neq \chi(\bar{\xi})$ and $x_{\max f(x)} \neq \chi(\xi_{\max f(\xi)})$.
The likelihood function does not transform in this way. This is the contrasts between the likelihood function and the posterior probability. The (maximum of the) likelihood function remains the same when you transform the variable.
$$\mathcal{L}_\xi(\xi) = \mathcal{L}_x(\chi(\xi)) $$
Related:
The flat prior is ambiguous. It depends on the form of the particular statistic.
For instance, if $X$ is uniform distributed (e.g. $\mathcal{U}(0,1))$, then $X^2$ is not a uniform distributed variable.
There is no single flat prior that you can relate the Likelihood function to. It is different when you define the flat prior for $X$ or some transformed variable like $X^2$. For the likelihood this dependency does not exist.
The boundaries of probabilities (credibility intervals) will be different when you transform the variable, (for likelihood functions this is not the case). E.g for some parameter $a$ and a monotonic transformation $f(a)$ (e.g. logarithm) you get the equivalent likelihood intervals
$$\begin{array}{ccccc}
a_{\min} &<& a &<& a_{\max}\\
f(a_{\min}) &<& f(a) &<& f(a_{\max})
\end{array}$$
2 Different concept: confidence intervals are independent from the prior
Suppose you sample a variable $X$ from a population with (unknown) parameter $\theta$ which itself (the population with parameter $\theta$) is sampled from a super-population (with possibly varying values for $\theta$).
One can make an inverse statement trying to infer what the original $\theta$ may have been based on observing some values $x_i$ for the variable $X$.
Bayesian methods do this by supposing a prior distribution for the distribution of possible $\theta$
This contrasts with the likelihood function and confidence interval, which are independent from the prior distribution.
The confidence interval does not use information of a prior like the credible interval does (confidence is not a probability).
Regardless of the prior distribution (uniform or not) the x%-confidence interval will contain the true parameter in $x%$ of the cases (confidence intervals refer to the success rate, type I error, of the method, not of a particular case).
In the case of the credible interval this concept ($%$ of time that the interval contains the true parameter) is not even applicable, but we may interpret it in a frequentist sense and then we observe that the credible interval will contain the true parameter only $x%$ of the time when the (uniform) prior is correctly describing the super-population of parameters that we may encounter. The interval may effectively be performing higher or lower than the x% (not that this matters since the Bayesian approach answers different questions, but it is just to note the difference).
3 Difference between confidence and credible intervals
In the example below we examine the likelihood function for the exponential distribution as function of the rate parameter $\lambda$, the sample mean $\bar{x}$, and sample size $n$:
$$\mathcal{L}(\lambda,\bar{x},n) = \frac{n^n}{(n-1)!} x^{n-1} \lambda^n e^{-\lambda n \bar{x}}$$
this functions expresses the probability to observe (for a given $n$ and $\lambda$) a sample mean between $\bar{x}$ and $\bar{x}+dx$.
note: the rate parameter $\lambda$ goes from $0$ to $\infty$ (unlike the OP 'request' from $0$ to $1$). The prior in this case will be an improper prior. The principles however does not change. I am using this perspective for easier illustration. Distributions with parameters between $0$ and $1$ are often discrete distributions (difficult to drawing continuous lines) or a beta distribution (difficult to calculate)
The image below illustrates this likelihood function (the blue colored map), for sample size $n=4$, and also draws the boundaries for the 95% intervals (both confidence and credible).
The boundaries are created obtaining the (one-dimensional) cumulative distribution function. But, this integration/cumulation can be done in two directions.
The difference between the intervals occurs because the 5% area's are made in different ways.
The 95% confidence interval contains values $\lambda$ for which the observed value $\bar{x}$ would occur at least in 95% of the cases. In this way. whatever the value $\lambda$, we would only make a wrong judgement in 95% of the cases.
For any $\lambda$ you have north and south of the boundaries (changing $\bar{x}$) 2.5% of the weight of the likelihood function.
The 95% credible interval contains values $\lambda$ which are most likely to cause the observed value $\bar{x}$ (given a flat prior).
Even when the observed result $\bar{x}$ is less than 5% likely for a given $\lambda$, the particular $\lambda$ may be inside the credible interval. In the particular example higher values of $\lambda$ are 'preferred' for the credible interval.
For any $\bar{x}$ you have west and east of the boundaries (changing $\lambda$) 2.5% of the weight of the likelihood function.
A case where confidence interval and credible interval (based on improper prior) coincide is for estimating the mean of a Gaussian distributed variable (the distribution is illustrated here: https://stats.stackexchange.com/a/351333/164061 ).
An obvious case where confidence interval and credible interval do not coincide is illustrated here (https://stats.stackexchange.com/a/369909/164061). The confidence interval for this case may have one or even both of the (upper/lower) bounds at infinity.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
Likelihood $\neq$ Bayesian with flat prior
The likelihood function, and associated the confidence interval, are not the same (concept) as a Bayesian posterior probability constructed with a prior that
|
6,795
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
This is not generally true, but it may seem so because of the most frequently considered special cases.
Consider $X,Y\sim\operatorname{i.i.d}\sim\operatorname{Uniform}[\theta-1/2,\, \theta+1/2].$ The interval $\big(\min\{X,Y\},\max\{X,Y\}\big)$ is a $50\%$ confidence interval for $\theta,$ albeit not one that anyone with any common sense would use. It does not coincide with a $50\%$ credible interval from the posterior from a flat prior.
Fisher's technique of conditioning on an ancillary statistic does in this case yield a confidence interval that coincides with that credible interval.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
This is not generally true, but it may seem so because of the most frequently considered special cases.
Consider $X,Y\sim\operatorname{i.i.d}\sim\operatorname{Uniform}[\theta-1/2,\, \theta+1/2].$ The
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
This is not generally true, but it may seem so because of the most frequently considered special cases.
Consider $X,Y\sim\operatorname{i.i.d}\sim\operatorname{Uniform}[\theta-1/2,\, \theta+1/2].$ The interval $\big(\min\{X,Y\},\max\{X,Y\}\big)$ is a $50\%$ confidence interval for $\theta,$ albeit not one that anyone with any common sense would use. It does not coincide with a $50\%$ credible interval from the posterior from a flat prior.
Fisher's technique of conditioning on an ancillary statistic does in this case yield a confidence interval that coincides with that credible interval.
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
This is not generally true, but it may seem so because of the most frequently considered special cases.
Consider $X,Y\sim\operatorname{i.i.d}\sim\operatorname{Uniform}[\theta-1/2,\, \theta+1/2].$ The
|
6,796
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
|
From my reading, I thought this statement is true asymptotically, i.e. for large sample size, and if one uses an uninformative prior.
A simple numerical example would seem to confirm this - the 90% profile maximum likelihood intervals and 90% credible intervals of a ML binomial GLM and Bayesian binomial GLM are indeed virtually identical for n=1000, though the discrepancy would become larger for small n :
# simulate some data
set.seed(123)
n = 1000 # sample size
x1 = rnorm(n) # two continuous covariates
x2 = rnorm(n)
z = 0.1 + 2*x1 + 3*x2 # predicted values on logit scale
y = rbinom(n,1,plogis(z)) # bernoulli response variable
d = data.frame(y=y, x1=x1, x2=x2)
# fit a regular GLM and calculate 90% confidence intervals
glmfit = glm(y ~ x1 + x2, family = "binomial", data = d)
library(MASS)
# coefficients and 90% profile confidence intervals :
round(cbind(coef(glmfit), confint(glmfit, level=0.9)), 2)
# 5 % 95 %
# (Intercept) 0.00 -0.18 0.17
# x1 2.04 1.77 2.34
# x2 3.42 3.05 3.81
# fit a Bayesian GLM using rstanarm
library(rstanarm)
t_prior = student_t(df = 3, location = 0, scale = 100) # we set scale to large value to specify an uninformative prior
bfit1 = stan_glm(y ~ x1 + x2, data = d,
family = binomial(link = "logit"),
prior = t_prior, prior_intercept = t_prior,
chains = 1, cores = 4, seed = 123, iter = 10000)
# coefficients and 90% credible intervals :
round(cbind(coef(bfit1), posterior_interval(bfit1, prob = 0.9)), 2)
# 5% 95%
# (Intercept) -0.01 -0.18 0.17
# x1 2.06 1.79 2.37
# x2 3.45 3.07 3.85
# fit a Bayesian GLM using brms
library(brms)
priors = c(
prior(student_t(3, 0, 100), class = "Intercept"),
prior(student_t(3, 0, 100), class = "b")
)
bfit2 = brm(
y ~ x1 + x2,
data = d,
prior = priors,
family = "bernoulli",
seed = 123
)
# coefficients and 90% credible intervals :
summary(bfit2, prob=0.9)
# Population-Level Effects:
# Estimate Est.Error l-90% CI u-90% CI Eff.Sample Rhat
# Intercept -0.01 0.11 -0.18 0.18 2595 1.00
# x1 2.06 0.17 1.79 2.35 2492 1.00
# x2 3.45 0.23 3.07 3.83 2594 1.00
# fit a Bayesian GLM using arm
library(arm)
# we set prior.scale to Inf to specify an uninformative prior
bfit3 = bayesglm(y ~ x1 + x2, family = "binomial", data = d, prior.scale = Inf)
sims = coef(sim(bfit3, n.sims=1000000))
# coefficients and 90% credible intervals :
round(cbind(coef(bfit3), t(apply(sims, 2, function (col) quantile(col,c(.05, .95))))),2)
# 5% 95%
# (Intercept) 0.00 -0.18 0.17
# x1 2.04 1.76 2.33
# x2 3.42 3.03 3.80
As you can see, in the example above, for n=1000, the 90% profile confidence intervals of a binomial GLM are virtually identical to the 90% credible intervals of a Bayesian binomial GLM (the difference is also within the bounds of using different seeds and different nrs of iterations in the bayesian fits, and an exact equivalence can also not be obtained since specifying a 100% uninformative prior is also not possible with rstanarm or brms).
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
|
From my reading, I thought this statement is true asymptotically, i.e. for large sample size, and if one uses an uninformative prior.
A simple numerical example would seem to confirm this - the 90% pr
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval?
From my reading, I thought this statement is true asymptotically, i.e. for large sample size, and if one uses an uninformative prior.
A simple numerical example would seem to confirm this - the 90% profile maximum likelihood intervals and 90% credible intervals of a ML binomial GLM and Bayesian binomial GLM are indeed virtually identical for n=1000, though the discrepancy would become larger for small n :
# simulate some data
set.seed(123)
n = 1000 # sample size
x1 = rnorm(n) # two continuous covariates
x2 = rnorm(n)
z = 0.1 + 2*x1 + 3*x2 # predicted values on logit scale
y = rbinom(n,1,plogis(z)) # bernoulli response variable
d = data.frame(y=y, x1=x1, x2=x2)
# fit a regular GLM and calculate 90% confidence intervals
glmfit = glm(y ~ x1 + x2, family = "binomial", data = d)
library(MASS)
# coefficients and 90% profile confidence intervals :
round(cbind(coef(glmfit), confint(glmfit, level=0.9)), 2)
# 5 % 95 %
# (Intercept) 0.00 -0.18 0.17
# x1 2.04 1.77 2.34
# x2 3.42 3.05 3.81
# fit a Bayesian GLM using rstanarm
library(rstanarm)
t_prior = student_t(df = 3, location = 0, scale = 100) # we set scale to large value to specify an uninformative prior
bfit1 = stan_glm(y ~ x1 + x2, data = d,
family = binomial(link = "logit"),
prior = t_prior, prior_intercept = t_prior,
chains = 1, cores = 4, seed = 123, iter = 10000)
# coefficients and 90% credible intervals :
round(cbind(coef(bfit1), posterior_interval(bfit1, prob = 0.9)), 2)
# 5% 95%
# (Intercept) -0.01 -0.18 0.17
# x1 2.06 1.79 2.37
# x2 3.45 3.07 3.85
# fit a Bayesian GLM using brms
library(brms)
priors = c(
prior(student_t(3, 0, 100), class = "Intercept"),
prior(student_t(3, 0, 100), class = "b")
)
bfit2 = brm(
y ~ x1 + x2,
data = d,
prior = priors,
family = "bernoulli",
seed = 123
)
# coefficients and 90% credible intervals :
summary(bfit2, prob=0.9)
# Population-Level Effects:
# Estimate Est.Error l-90% CI u-90% CI Eff.Sample Rhat
# Intercept -0.01 0.11 -0.18 0.18 2595 1.00
# x1 2.06 0.17 1.79 2.35 2492 1.00
# x2 3.45 0.23 3.07 3.83 2594 1.00
# fit a Bayesian GLM using arm
library(arm)
# we set prior.scale to Inf to specify an uninformative prior
bfit3 = bayesglm(y ~ x1 + x2, family = "binomial", data = d, prior.scale = Inf)
sims = coef(sim(bfit3, n.sims=1000000))
# coefficients and 90% credible intervals :
round(cbind(coef(bfit3), t(apply(sims, 2, function (col) quantile(col,c(.05, .95))))),2)
# 5% 95%
# (Intercept) 0.00 -0.18 0.17
# x1 2.04 1.76 2.33
# x2 3.42 3.03 3.80
As you can see, in the example above, for n=1000, the 90% profile confidence intervals of a binomial GLM are virtually identical to the 90% credible intervals of a Bayesian binomial GLM (the difference is also within the bounds of using different seeds and different nrs of iterations in the bayesian fits, and an exact equivalence can also not be obtained since specifying a 100% uninformative prior is also not possible with rstanarm or brms).
|
If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interv
From my reading, I thought this statement is true asymptotically, i.e. for large sample size, and if one uses an uninformative prior.
A simple numerical example would seem to confirm this - the 90% pr
|
6,797
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
From your question and in particular your comments to other answers, it seems to me that you are mainly confused about the "big picture" here: namely, what does "positive dependency" refer to in this context at all -- as opposed to what is the technical meaning of the PRDS condition. So I will talk about the big picture.
The big picture
Imagine that you are testing $N$ null hypotheses, and imagine that all of them are true. Each of the $N$ $p$-values is a random variable; repeating the experiment over and over again would yield a different $p$-value each time, so one can talk about a distribution of $p$-values (under the null). It is well-known that for any test, a distribution of $p$-values under the null must be uniform; so, in the case of multiple testing, all $N$ marginal distributions of $p$-values will be uniform.
If all the data and all $N$ tests are independent from each other, then the joint $N$-dimensional distribution of $p$-values will also be uniform. This will be true e.g. in a classic "jelly-bean" situation when a bunch of independent things are being tested:
However, it does not have to be like that. Any pair of $p$-values can in principle be correlated, either positively or negatively, or be dependent in some more complicated way. Consider testing all pairwise differences in means between four groups; this is $N=4\cdot 3/2=6$ tests. Each of the six $p$-values alone is uniformly distributed. But they are all positively correlated: if (on a given attempt) group A by chance has particularly low mean, then A-vs-B comparison might yield a low $p$-value (this would be a false positive). But in this situation it is likely that A-vs-C, as well as A-vs-D, will also yield low $p$-values. So the $p$-values are obviously non-independent and moreover they are positively correlated between each other.
This is, informally, what "positive dependency" refers to.
This seems to be a common situation in multiple testing. Another example would be testing for differences in several variables that are correlated between each other. Obtaining a significant difference in one of them increases the chances of obtaining a significant difference in another.
It is tricky to come up with a natural example where $p$-values would be "negatively dependent". @user43849 remarked in the comments above that for one-sided tests it is easy:
Imagine I am testing whether A = 0 and also whether B = 0 against one-tailed alternatives (A > 0 and B > 0). Further imagine that B depends on A. For example, imagine I want to know if a population contains more women than men, and also if the population contains more ovaries than testes. Clearly knowing the p-value of the first question changes our expectation of the p-value for the second. Both p-values change in the same direction, and this is PRD. But if I instead test the second hypothesis that population 2 has more testes than ovaries, our expectation for the second p-value decreases as the first p-value increases. This is not PRD.
But I have so far been unable to come up with a natural example with point nulls.
Now, the exact mathematical formulation of "positive dependency" that guarantees the validity of Benjamini-Hochberg procedure is rather tricky. As mentioned in other answers, the main reference is Benjamini & Yekutieli 2001; they show that PRDS property ("positive regression dependency on each one
from a subset") entails Benjamini-Hochberg procedure. It is a relaxed form of the PRD ("positive regression dependency") property, meaning that PRD implies PRDS and hence also entails Benjamini-Hochberg procedure.
For the definitions of PRD/PRDS see @user43849's answer (+1) and Benjamini & Yekutieli paper. The definitions are rather technical and I do not have a good intuitive understanding of them. In fact, B&Y mention several other related concepts as well: multivariate total positivity of order two (MTP2) and positive association. According to B&Y, they are related as follows (the diagram is mine):
$\hskip{10em}$
MTP2 implies PRD that implies PRDS that guarantees correctness of B-H procedure. PRD also implies PA, but PA$\ne$PRDS.
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
From your question and in particular your comments to other answers, it seems to me that you are mainly confused about the "big picture" here: namely, what does "positive dependency" refer to in this
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
From your question and in particular your comments to other answers, it seems to me that you are mainly confused about the "big picture" here: namely, what does "positive dependency" refer to in this context at all -- as opposed to what is the technical meaning of the PRDS condition. So I will talk about the big picture.
The big picture
Imagine that you are testing $N$ null hypotheses, and imagine that all of them are true. Each of the $N$ $p$-values is a random variable; repeating the experiment over and over again would yield a different $p$-value each time, so one can talk about a distribution of $p$-values (under the null). It is well-known that for any test, a distribution of $p$-values under the null must be uniform; so, in the case of multiple testing, all $N$ marginal distributions of $p$-values will be uniform.
If all the data and all $N$ tests are independent from each other, then the joint $N$-dimensional distribution of $p$-values will also be uniform. This will be true e.g. in a classic "jelly-bean" situation when a bunch of independent things are being tested:
However, it does not have to be like that. Any pair of $p$-values can in principle be correlated, either positively or negatively, or be dependent in some more complicated way. Consider testing all pairwise differences in means between four groups; this is $N=4\cdot 3/2=6$ tests. Each of the six $p$-values alone is uniformly distributed. But they are all positively correlated: if (on a given attempt) group A by chance has particularly low mean, then A-vs-B comparison might yield a low $p$-value (this would be a false positive). But in this situation it is likely that A-vs-C, as well as A-vs-D, will also yield low $p$-values. So the $p$-values are obviously non-independent and moreover they are positively correlated between each other.
This is, informally, what "positive dependency" refers to.
This seems to be a common situation in multiple testing. Another example would be testing for differences in several variables that are correlated between each other. Obtaining a significant difference in one of them increases the chances of obtaining a significant difference in another.
It is tricky to come up with a natural example where $p$-values would be "negatively dependent". @user43849 remarked in the comments above that for one-sided tests it is easy:
Imagine I am testing whether A = 0 and also whether B = 0 against one-tailed alternatives (A > 0 and B > 0). Further imagine that B depends on A. For example, imagine I want to know if a population contains more women than men, and also if the population contains more ovaries than testes. Clearly knowing the p-value of the first question changes our expectation of the p-value for the second. Both p-values change in the same direction, and this is PRD. But if I instead test the second hypothesis that population 2 has more testes than ovaries, our expectation for the second p-value decreases as the first p-value increases. This is not PRD.
But I have so far been unable to come up with a natural example with point nulls.
Now, the exact mathematical formulation of "positive dependency" that guarantees the validity of Benjamini-Hochberg procedure is rather tricky. As mentioned in other answers, the main reference is Benjamini & Yekutieli 2001; they show that PRDS property ("positive regression dependency on each one
from a subset") entails Benjamini-Hochberg procedure. It is a relaxed form of the PRD ("positive regression dependency") property, meaning that PRD implies PRDS and hence also entails Benjamini-Hochberg procedure.
For the definitions of PRD/PRDS see @user43849's answer (+1) and Benjamini & Yekutieli paper. The definitions are rather technical and I do not have a good intuitive understanding of them. In fact, B&Y mention several other related concepts as well: multivariate total positivity of order two (MTP2) and positive association. According to B&Y, they are related as follows (the diagram is mine):
$\hskip{10em}$
MTP2 implies PRD that implies PRDS that guarantees correctness of B-H procedure. PRD also implies PA, but PA$\ne$PRDS.
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
From your question and in particular your comments to other answers, it seems to me that you are mainly confused about the "big picture" here: namely, what does "positive dependency" refer to in this
|
6,798
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
Great question! Let's step back and understand what Bonferroni did, and why it was necessary for Benjamini and Hochberg to develop an alternative.
It has become necessary and compulsory in recent years to perform a procedure called multiple testing correction. This is due to the increasing numbers of tests being performed simultaneously with high throughput sciences, especially in genetics with the advent of whole genome association studies (GWAS). Excuse my reference to genetics, as it is my area of work. If we are performing 1,000,000 tests simultaneously at $P = 0.05$, we would expect $50,000$ false positives. This is ludicrously large, and thus we must control the level at which significance is assessed. The bonferroni correction, that is, dividing the acceptance threshold (0.05) by the number of independent tests $(0.05/M)$ corrects for the family wise error rate ($FWER$).
This is true because the FWER is related to test-wise error rate ($TWER$) by the equation $FWER = 1 - (1 - TWER)^M$. That is, 100 percent minus 1 subtract the test wise error rate raised to the power of the number of independent tests performed. Making the assumption that $(1- 0.05)^{1/M} = 1-\frac{0.05}{M}$ gives $TWER \approx \frac{0.05}{M}$, which is the acceptance P value adjusted for M completely independent tests.
The problem that we encounter now, as did Benjamini and Hochberg, is that not all tests are completely independant. Thus, the Bonferroni correction, though robust and flexible, is an overcorrection. Consider the case in genetics where two genes are linked in a case called linkage disequilibrium; that is, when one gene has a mutation, another is more likely to be expressed. These are obviously not independent tests, though in the bonferroni correction they are assumed to be. It is here where we start to see that dividing the P value by M is creating a threshold that is artificially low because of assumed independent tests which really influence each other, ergo creating an M that is too large for our real situation, where things aren't independent.
The procedure suggested by Benjamini and Hochberg, and augmented by Yekutieli (and many others) is more liberal than Bonferroni, and in fact Bonferroni correction is only used in the very largest of studies now. This is because, in FDR, we assume some interdependence on the part of the tests and thus an M which is too large and unrealistic and getting rid of results that we, in reality, care about. Therefore in the case of 1000 tests which are not independent, the true M would not be 1000, but something smaller because of dependencies. Thus when we divide 0.05 by 1000, the threshold is too strict and avoids some tests which may be of interest.
I'm not sure if you care about the mechanics behind the controlling for dependency, though if you do I have linked the Yekutieli paper for your reference. I'll also attach a few other things for your information and curiosity.
Hope this has helped in some way, if I have misrepresented anything please do let me know.
~ ~ ~
References
Yekutieli paper on positive dependencies -- http://www.math.tau.ac.il/~ybenja/MyPapers/benjamini_yekutieli_ANNSTAT2001.pdf
( see 1.3 -- The Problem. )
Explaination of Bonferroni and other things of interest -- Nature Genetics reviews. Statistical Power and significance testing in large-scale genetic studies -- Pak C Sham and Shaun M Purcell
( see box 3. )
http://en.wikipedia.org/wiki/Familywise_error_rate
EDIT:
In my previous answer I did not directly define positive dependency, which was what was asked. In the Yekutieli paper, section 2.2 is entitled Positive dependence, and I suggest this as it is very detailed. However, I believe that we can make it a little bit more succinct.
The paper at first begins by talking about positive dependancy, using it as a vague term that is interpretable but not specific. If you read the proofs, the thing that is mentioned as positive dependency is called PRSD, defined earlier as "Positive regression dependancy on each one from a subset $I_0$". $I_0$ is the subset of tests that correctly support the null hypothesis (0). PRDS is then defined as the following.
$X$ is our whole set of test statistics, and $I_0$ is our set of test statistics which correctly support the null. Thus, for $X$ to be PRDS (positively dependent) on $I_0$, the probability of $X$ being an element of $I_0$ (nulls) increases in non decreasing set of test statistics $x$ (elements of $X$).
Interpreting this, as we order our $P$-values from lowest to highest, the probability of being part of the null set of test statistics is the lowest at the smallest P value, and increases from there. The FDR sets a boundary on this list of test statistics such that the probability of being part of the null set is 0.05. This is what we are doing when controlling for FDR.
In summation, the property of positive dependency is really the property of positive regression dependency of our whole set of test statistics upon our set of true null test statistics, and we control for an FDR of 0.05; thus as P values go from the bottom up (the step up procedure), they increase in probability of being part of the null set.
My former answer in the comments about the covariance matrix was not incorrect, just a little bit vague. I hope this helps a little bit more.
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
Great question! Let's step back and understand what Bonferroni did, and why it was necessary for Benjamini and Hochberg to develop an alternative.
It has become necessary and compulsory in recent ye
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
Great question! Let's step back and understand what Bonferroni did, and why it was necessary for Benjamini and Hochberg to develop an alternative.
It has become necessary and compulsory in recent years to perform a procedure called multiple testing correction. This is due to the increasing numbers of tests being performed simultaneously with high throughput sciences, especially in genetics with the advent of whole genome association studies (GWAS). Excuse my reference to genetics, as it is my area of work. If we are performing 1,000,000 tests simultaneously at $P = 0.05$, we would expect $50,000$ false positives. This is ludicrously large, and thus we must control the level at which significance is assessed. The bonferroni correction, that is, dividing the acceptance threshold (0.05) by the number of independent tests $(0.05/M)$ corrects for the family wise error rate ($FWER$).
This is true because the FWER is related to test-wise error rate ($TWER$) by the equation $FWER = 1 - (1 - TWER)^M$. That is, 100 percent minus 1 subtract the test wise error rate raised to the power of the number of independent tests performed. Making the assumption that $(1- 0.05)^{1/M} = 1-\frac{0.05}{M}$ gives $TWER \approx \frac{0.05}{M}$, which is the acceptance P value adjusted for M completely independent tests.
The problem that we encounter now, as did Benjamini and Hochberg, is that not all tests are completely independant. Thus, the Bonferroni correction, though robust and flexible, is an overcorrection. Consider the case in genetics where two genes are linked in a case called linkage disequilibrium; that is, when one gene has a mutation, another is more likely to be expressed. These are obviously not independent tests, though in the bonferroni correction they are assumed to be. It is here where we start to see that dividing the P value by M is creating a threshold that is artificially low because of assumed independent tests which really influence each other, ergo creating an M that is too large for our real situation, where things aren't independent.
The procedure suggested by Benjamini and Hochberg, and augmented by Yekutieli (and many others) is more liberal than Bonferroni, and in fact Bonferroni correction is only used in the very largest of studies now. This is because, in FDR, we assume some interdependence on the part of the tests and thus an M which is too large and unrealistic and getting rid of results that we, in reality, care about. Therefore in the case of 1000 tests which are not independent, the true M would not be 1000, but something smaller because of dependencies. Thus when we divide 0.05 by 1000, the threshold is too strict and avoids some tests which may be of interest.
I'm not sure if you care about the mechanics behind the controlling for dependency, though if you do I have linked the Yekutieli paper for your reference. I'll also attach a few other things for your information and curiosity.
Hope this has helped in some way, if I have misrepresented anything please do let me know.
~ ~ ~
References
Yekutieli paper on positive dependencies -- http://www.math.tau.ac.il/~ybenja/MyPapers/benjamini_yekutieli_ANNSTAT2001.pdf
( see 1.3 -- The Problem. )
Explaination of Bonferroni and other things of interest -- Nature Genetics reviews. Statistical Power and significance testing in large-scale genetic studies -- Pak C Sham and Shaun M Purcell
( see box 3. )
http://en.wikipedia.org/wiki/Familywise_error_rate
EDIT:
In my previous answer I did not directly define positive dependency, which was what was asked. In the Yekutieli paper, section 2.2 is entitled Positive dependence, and I suggest this as it is very detailed. However, I believe that we can make it a little bit more succinct.
The paper at first begins by talking about positive dependancy, using it as a vague term that is interpretable but not specific. If you read the proofs, the thing that is mentioned as positive dependency is called PRSD, defined earlier as "Positive regression dependancy on each one from a subset $I_0$". $I_0$ is the subset of tests that correctly support the null hypothesis (0). PRDS is then defined as the following.
$X$ is our whole set of test statistics, and $I_0$ is our set of test statistics which correctly support the null. Thus, for $X$ to be PRDS (positively dependent) on $I_0$, the probability of $X$ being an element of $I_0$ (nulls) increases in non decreasing set of test statistics $x$ (elements of $X$).
Interpreting this, as we order our $P$-values from lowest to highest, the probability of being part of the null set of test statistics is the lowest at the smallest P value, and increases from there. The FDR sets a boundary on this list of test statistics such that the probability of being part of the null set is 0.05. This is what we are doing when controlling for FDR.
In summation, the property of positive dependency is really the property of positive regression dependency of our whole set of test statistics upon our set of true null test statistics, and we control for an FDR of 0.05; thus as P values go from the bottom up (the step up procedure), they increase in probability of being part of the null set.
My former answer in the comments about the covariance matrix was not incorrect, just a little bit vague. I hope this helps a little bit more.
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
Great question! Let's step back and understand what Bonferroni did, and why it was necessary for Benjamini and Hochberg to develop an alternative.
It has become necessary and compulsory in recent ye
|
6,799
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
I found this pre-print helpful in understanding the meaning. It should be said that I offer this answer not as an expert in the topic, but as an attempt at understanding to be vetted and validated by the community.
Thanks to Amoeba for very helpful observations about the difference between PRD and PRDS, see comments
Positive regression dependency (PRD) means the following:
Consider the subset of p-values (or equivalently, test statistics) that correspond to true null hypotheses. Call the vector of these p-values $p$. Let $C$ be a set of vectors with length equal to the length of $p$ and let $C$ have the following property:
If some vector $q$ is in $C$, and
We construct some vector $r$ of the same length as $q$ so that all elements of $r$ are less than the corresponding elements of $q$ ($r_i < q_i$ for all $i$), then
$r$ is also in $C$
(This means that $C$ is a "decreasing set".)
Assume we know something about the values of some of the elements of $p$. Namely, $p_1 ... p_{n} < B_1 ... B_n$. PRD means that the probability that $p$ is in $C$ never increases as $B_1 ... B_n$ increases.
In plain language, notice that we can formulate an expectation for any element $p_i$. Since $p_i$ corresponds to a true null, it's unconditional expectation should be a uniform distribution from 0 to 1. But if the p-values are not independent, then our conditional expectation for $p_i$ given some other elements of $p_1 ... p_n$ might not be uniform. PRD means that raising increasing the value $p_1 ... p_n$ can never increase the probability that another element $p_i$ has lower value.
Benjamini and Yekutieli (2001) show that the Benjamini and Hochberg procedure for controlling FDR requires a condition they term positive regression dependence on a subset (PRDS). PRDS is similar to, and implied by, PRD. However, it is a weaker condition because it only conditions on one of $p_1 ... p_n$ at a time.
To rephrase in plain language: again consider the set of p-values that correspond to true null hypotheses. For any one of these p-values (call it $p_n$), imagine that we know $p_n < B$, where $B$ is some constant. Then we can formulate a conditional expectation for the remaining p-values, given that $p_n < B$. If the p-values are independent, then our expectation for the remaining p-values is the uniform distribution from 0 to 1. But if the p-values are not independent, then knowing $p_n < B$ might change our expectation for the remaining p-values. PRDS says that increasing the value of $B$ must not decrease our expectation for any of the remaining p-values corresponding to the true null hypotheses.
Edited to add:
Here's a putative example of a system that is not PRDS (R code below). The logic is that when samples a and b are very similar, it is more likely that their product will be atypical. I suspect that this effect (and not the non-uniformity of p-values under the null for the (a*b), (c*d) comparison) is driving the negative correlation in the p-values, but I cannot be sure. The same effect appears if we do a t-test for the second comparison (rather than a Wilcoxon), but the distribution of p-values still isn't uniform, presumably due to violations of the normality assumption.
ab <- rep(NA, 100000) # We'll repeat the comparison many times to assess the relationships among p-values.
abcd <- rep(NA, 100000)
for(i in 1:100000){
a <- rnorm(10) # Draw 4 samples from identical populations.
b <- rnorm(10)
c <- rnorm(10)
d <- rnorm(10)
ab[i] <- t.test(a,b)$p.value # We perform 2 comparisons and extract p-values
abcd[i] <- wilcox.test((a*b),(c*d))$p.value
}
summary(lm(abcd ~ ab)) # The p-values are negatively correlated
ks.test(ab, punif) # The p-values are uniform for the first test
ks.test(abcd, punif) # but non-uniform for the second test.
hist(abcd)
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
I found this pre-print helpful in understanding the meaning. It should be said that I offer this answer not as an expert in the topic, but as an attempt at understanding to be vetted and validated by
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
I found this pre-print helpful in understanding the meaning. It should be said that I offer this answer not as an expert in the topic, but as an attempt at understanding to be vetted and validated by the community.
Thanks to Amoeba for very helpful observations about the difference between PRD and PRDS, see comments
Positive regression dependency (PRD) means the following:
Consider the subset of p-values (or equivalently, test statistics) that correspond to true null hypotheses. Call the vector of these p-values $p$. Let $C$ be a set of vectors with length equal to the length of $p$ and let $C$ have the following property:
If some vector $q$ is in $C$, and
We construct some vector $r$ of the same length as $q$ so that all elements of $r$ are less than the corresponding elements of $q$ ($r_i < q_i$ for all $i$), then
$r$ is also in $C$
(This means that $C$ is a "decreasing set".)
Assume we know something about the values of some of the elements of $p$. Namely, $p_1 ... p_{n} < B_1 ... B_n$. PRD means that the probability that $p$ is in $C$ never increases as $B_1 ... B_n$ increases.
In plain language, notice that we can formulate an expectation for any element $p_i$. Since $p_i$ corresponds to a true null, it's unconditional expectation should be a uniform distribution from 0 to 1. But if the p-values are not independent, then our conditional expectation for $p_i$ given some other elements of $p_1 ... p_n$ might not be uniform. PRD means that raising increasing the value $p_1 ... p_n$ can never increase the probability that another element $p_i$ has lower value.
Benjamini and Yekutieli (2001) show that the Benjamini and Hochberg procedure for controlling FDR requires a condition they term positive regression dependence on a subset (PRDS). PRDS is similar to, and implied by, PRD. However, it is a weaker condition because it only conditions on one of $p_1 ... p_n$ at a time.
To rephrase in plain language: again consider the set of p-values that correspond to true null hypotheses. For any one of these p-values (call it $p_n$), imagine that we know $p_n < B$, where $B$ is some constant. Then we can formulate a conditional expectation for the remaining p-values, given that $p_n < B$. If the p-values are independent, then our expectation for the remaining p-values is the uniform distribution from 0 to 1. But if the p-values are not independent, then knowing $p_n < B$ might change our expectation for the remaining p-values. PRDS says that increasing the value of $B$ must not decrease our expectation for any of the remaining p-values corresponding to the true null hypotheses.
Edited to add:
Here's a putative example of a system that is not PRDS (R code below). The logic is that when samples a and b are very similar, it is more likely that their product will be atypical. I suspect that this effect (and not the non-uniformity of p-values under the null for the (a*b), (c*d) comparison) is driving the negative correlation in the p-values, but I cannot be sure. The same effect appears if we do a t-test for the second comparison (rather than a Wilcoxon), but the distribution of p-values still isn't uniform, presumably due to violations of the normality assumption.
ab <- rep(NA, 100000) # We'll repeat the comparison many times to assess the relationships among p-values.
abcd <- rep(NA, 100000)
for(i in 1:100000){
a <- rnorm(10) # Draw 4 samples from identical populations.
b <- rnorm(10)
c <- rnorm(10)
d <- rnorm(10)
ab[i] <- t.test(a,b)$p.value # We perform 2 comparisons and extract p-values
abcd[i] <- wilcox.test((a*b),(c*d))$p.value
}
summary(lm(abcd ~ ab)) # The p-values are negatively correlated
ks.test(ab, punif) # The p-values are uniform for the first test
ks.test(abcd, punif) # but non-uniform for the second test.
hist(abcd)
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
I found this pre-print helpful in understanding the meaning. It should be said that I offer this answer not as an expert in the topic, but as an attempt at understanding to be vetted and validated by
|
6,800
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
In their paper, Benjamini and Yekutieli provide some examples of how positive regression dependence (PRD) is different from just being positively associated. The FDR control procedure relies on a weaker form of PRD which they call PRDS (i.e. PRD on each one from a subset of variables).
Positive dependency was originally proposed in the bivariate setting by Lehmann, but the multivariate version of this concept, known as positive regression dependency is what is relevant to multiple testing.
Here is a relevant excerpt from pg.6
Nevertheless, PRDS and positive association do not imply one another, and
the difference is of some importance. For example, a multivariate normal distribution
is positively associated iff all correlations are nonnegative. Not all
correlations need be nonnegative for the PRDS property to hold (see Section
3.1, Case 1 below). On the other hand, a bivariate distribution may be positively
associated, yet not positive regression dependent [Lehmann (1966)], and
therefore also not PRDS on any subset. A stricter notion of positive association,
Rosenbaum’s (1984) conditional (positive) association, is enough to imply
PRDS: $\mathbf{X}$ is conditionally associated, if for any partition $(\mathbf{X}_1, \mathbf{X}_2)$ of $\mathbf{X}$, and any
function $h(\mathbf{X}_1)$, $\mathbf{X}_2$ given $h(\mathbf{X}_1)$ is positively associated.
It is important to note that all of the above properties, including PRDS,
remain invariant to taking comonotone transformations in each of the coordinates
[Eaton (1986)]. $$\ldots$$
Background on these concepts is clearly presented
in Eaton (1986), supplemented by Holland and Rosenbaum (1986).
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
|
In their paper, Benjamini and Yekutieli provide some examples of how positive regression dependence (PRD) is different from just being positively associated. The FDR control procedure relies on a weak
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
In their paper, Benjamini and Yekutieli provide some examples of how positive regression dependence (PRD) is different from just being positively associated. The FDR control procedure relies on a weaker form of PRD which they call PRDS (i.e. PRD on each one from a subset of variables).
Positive dependency was originally proposed in the bivariate setting by Lehmann, but the multivariate version of this concept, known as positive regression dependency is what is relevant to multiple testing.
Here is a relevant excerpt from pg.6
Nevertheless, PRDS and positive association do not imply one another, and
the difference is of some importance. For example, a multivariate normal distribution
is positively associated iff all correlations are nonnegative. Not all
correlations need be nonnegative for the PRDS property to hold (see Section
3.1, Case 1 below). On the other hand, a bivariate distribution may be positively
associated, yet not positive regression dependent [Lehmann (1966)], and
therefore also not PRDS on any subset. A stricter notion of positive association,
Rosenbaum’s (1984) conditional (positive) association, is enough to imply
PRDS: $\mathbf{X}$ is conditionally associated, if for any partition $(\mathbf{X}_1, \mathbf{X}_2)$ of $\mathbf{X}$, and any
function $h(\mathbf{X}_1)$, $\mathbf{X}_2$ given $h(\mathbf{X}_1)$ is positively associated.
It is important to note that all of the above properties, including PRDS,
remain invariant to taking comonotone transformations in each of the coordinates
[Eaton (1986)]. $$\ldots$$
Background on these concepts is clearly presented
in Eaton (1986), supplemented by Holland and Rosenbaum (1986).
|
The meaning of "positive dependency" as a condition to use the usual method for FDR control
In their paper, Benjamini and Yekutieli provide some examples of how positive regression dependence (PRD) is different from just being positively associated. The FDR control procedure relies on a weak
|
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