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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
13265 | 1 | null | null | 2 | 289 | We know that due to LD, we can get significant p-values for markers near a causal marker (or a marker closes to the causal region) in [GWAS](https://secure.wikimedia.org/wikipedia/en/wiki/Genome-wide_association_study) studies. I've seen [attempts](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2276357/?tool=pubmed) looking at LD to do multiple testing correction, by counting the effective number of tests, but as far as I know, they are simply trying to fix the correct global p-value cutoff for a certain error rate. My question is regarding the actual problem of selecting causal markers. What attempts have their been to remove these false positives that occur due to them being in LD with the causal markers? It seems regression and penalized regression will remove some correlated markers, but are their any techniques other than penalized regression to remove these false positives? Regression is obviously also intractable when we have a large number of markers.
| Countering false positives in GWAS due to linkage disequilibrium | CC BY-SA 3.0 | null | 2011-07-19T19:00:40.587 | 2011-07-20T20:12:36.737 | 2011-07-20T20:12:36.737 | 2728 | 2728 | [
"regression",
"correlation",
"genetics"
] |
13266 | 1 | 13269 | null | 21 | 60099 | I have run a simple linear regression of the natural log of 2 variables to determine if they correlate. My output is this:
```
R^2 = 0.0893
slope = 0.851
p < 0.001
```
I am confused. Looking at the $R^2$ value, I would say that the two variables are not correlated, since it is so close to $0$. However, the slope of the regression line is almost $1$ (despite looking as though it's almost horizontal in the plot), and the p-value indicates that the regression is highly significant.
Does this mean that the two variables are highly correlated? If so, what does the $R^2$ value indicate?
I should add that the Durbin-Watson statistic was tested in my software, and did not reject the null hypothesis (it equalled $1.357$). I thought that this tested for independence between the $2$ variables. In this case, I would expect the variables to be dependent, since they are $2$ measurements of an individual bird. I'm doing this regression as part of a published method to determine body condition of an individual, so I assumed that using a regression in this way made sense. However, given these outputs, I'm thinking that maybe for these birds, this method isn't suitable. Does this seem a reasonable conclusion?
| Simple linear regression output interpretation | CC BY-SA 3.0 | null | 2011-07-19T19:09:06.103 | 2017-05-08T17:37:15.853 | 2012-06-05T11:14:32.473 | 4856 | 4238 | [
"regression",
"r-squared"
] |
13267 | 1 | 13271 | null | 10 | 10793 | If I have two variables following two different distributions and having different standard deviations... How do I need to transform two variables so that when I sum the two result is not "driven" by more volatile one.
For example... Variable A is less volatile than variable B (ranges from 0 to 3000) and variable B goes fro. 300 to 350.
If simply add the two variables together the result will obviously be driven by A.
| How to sum two variables that are on different scales? | CC BY-SA 3.0 | null | 2011-07-19T19:14:43.033 | 2012-05-21T14:50:04.840 | 2012-05-21T14:50:04.840 | 4856 | 333 | [
"distributions"
] |
13268 | 2 | null | 13264 | 8 | null | AFAIK, there is no closed form for the distribution. Using R, the naive implementation of getting the exact distribution works for me up to group sizes of at least 12 - that takes less than 1 minute on a Core i5 using Windows7 64bit and current R. For R's own more clever algorithm in C that's used in `pwilcox()`, you can check the source file src/nmath/wilcox.c
```
n1 <- 12 # size group 1
n2 <- 12 # size group 2
N <- n1 + n2 # total number of subjects
```
Now generate all possible cases for the ranks within group 1. These are all ${N \choose n_{1}}$ different samples from the numbers $1, \ldots, N$ of size $n_{1}$. Then calculate the rank sum (= test statistic) for each of these cases. Tabulate these rank sums to get the probability density function from the relative frequencies, the cumulative sum of these relative frequencies is the cumulative distribution function.
```
rankMat <- combn(1:N, n1) # all possible ranks within group 1
LnPl <- colSums(rankMat) # all possible rank sums for group 1
dWRS <- table(LnPl) / choose(N, n1) # relative frequencies of rank sums: pdf
pWRS <- cumsum(dWRS) # cumulative sums: cdf
```
Compare the exact distribution against the asymptotically correct normal distribution.
```
muLnPl <- (n1 * (N+1)) / 2 # expected value
varLnPl <- (n1*n2 * (N+1)) / 12 # variance
plot(names(pWRS), pWRS, main="Wilcoxon RS, N=(12, 12): exact vs. asymptotic",
type="n", xlab="ln+", ylab="P(Ln+ <= ln+)", cex.lab=1.4)
curve(pnorm(x, mean=muLnPl, sd=sqrt(varLnPl)), lwd=4, n=200, add=TRUE)
points(names(pWRS), pWRS, pch=16, col="red", cex=0.7)
abline(h=0.95, col="blue")
legend(x="bottomright", legend=c("exact", "asymptotic"),
pch=c(16, NA), col=c("red", "black"), lty=c(NA, 1), lwd=c(NA, 2))
```
| null | CC BY-SA 3.0 | null | 2011-07-19T19:22:42.417 | 2011-07-19T20:07:06.050 | 2011-07-19T20:07:06.050 | 1909 | 1909 | null |
13269 | 2 | null | 13266 | 25 | null | The estimated value of the slope does not, by itself, tell you the strength of the relationship. The strength of the relationship depends on the size of the error variance, and the range of the predictor. Also, a significant $p$-value doesn't tell you necessarily that there is a strong relationship; the $p$-value is simply testing whether the slope is exactly 0. For a sufficiently large sample size, even small departures from that hypothesis (e.g. ones not of practical importance) will yield a significant $p$-value.
Of the three quantities you presented, $R^2$, the coefficient of determination, gives the greatest indication of the strength of the relationship. In your case, $R^{2} = .089$, means that $8.9\%$ of the variation in your response variable can be explained a linear relationship with the predictor. What constitutes a "large" $R^2$ is discipline dependent. For example, in social sciences $R^2 = .2$ might be "large" but in controlled environments like a factory setting, $R^2 > .9$ may be required to say there is a "strong" relationship. In most situations $.089$ is a very small $R^2$, so your conclusion that there is a weak linear relationship is probably reasonable.
| null | CC BY-SA 3.0 | null | 2011-07-19T19:28:53.157 | 2011-07-19T21:37:28.857 | 2011-07-19T21:37:28.857 | 4856 | 4856 | null |
13270 | 2 | null | 13266 | 9 | null | The $R^2$ value tells you how much variation in the data is explained by the fitted model.
The low $R^2$ value in your study suggests that your data is probably spread widely around the regression line, meaning that the regression model can only explain (very little) 8.9% of the variation in the data.
Have you checked to see whether a linear model is appropriate? Have a look at the distribution of your residuals, as you can use this to assess the fit of the model to your data. Ideally, your residuals should not show a relation with your $x$ values, and if it does, you may want to think of rescaling your variables in a suitable way, or fitting a more appropriate model.
| null | CC BY-SA 3.0 | null | 2011-07-19T19:32:57.373 | 2012-03-13T22:36:51.647 | 2012-03-13T22:36:51.647 | 4856 | 5473 | null |
13271 | 2 | null | 13267 | 16 | null | A common practice is to [standardize](http://en.wikipedia.org/wiki/Standard_score#Calculation_from_raw_score) the two variables, $A,B$, to place them on the same scale by subtracting the sample mean and dividing by the sample standard deviation. Once you've done this, both variables will be on the same scale in the sense that they each have a sample mean of 0 and sample standard deviation of 1. Thus, they can be added without one variable having an undue influence due simply to scale.
That is, calculate
$$ \frac{ A - \overline{A} }{ {\rm SD}(A) }, \ \ \frac{ B - \overline{B} }{ {\rm SD}(B) } $$
where $\overline{A}, {\rm SD}(A)$ denotes the sample mean and standard deviation of $A$, and similarly for B. The standardized versions of the variables are interpreted as the number of standard deviations above/below the mean a particular observation is.
| null | CC BY-SA 3.0 | null | 2011-07-19T19:38:35.613 | 2012-05-21T14:49:33.623 | 2012-05-21T14:49:33.623 | 4856 | 4856 | null |
13272 | 1 | 13274 | null | 31 | 28050 | Consider the following experiment: a group of people is given a list of cities, and asked to mark the corresponding locations on an (otherwise unlabeled) map of the world. For each city, you will get a scattering of points roughly centered at the respective city. Some cities, say Istanbul, will exhibit less scattering than others, say Moscow.
Let's assume that for a given city, we get a set of 2D samples $\{(x_i, y_i)\}$, representing the $(x, y)$ position of the city (e.g. in a local coordinate system) on the map assigned by test subject $i$. I would like to express the amount of "dispersion" of the points in this set as a single number in the appropriate units (km).
For a 1D problem, I would choose the standard deviation, but is there a 2D analog that could reasonably be chosen for the situation as described above?
| 2D analog of standard deviation? | CC BY-SA 3.0 | null | 2011-07-19T19:42:51.087 | 2019-10-08T03:58:40.813 | 2011-07-20T18:35:11.860 | 1036 | 5474 | [
"standard-deviation",
"spatial"
] |
13274 | 2 | null | 13272 | 20 | null | One thing you could use is a distance measure from a central point, ${\bf c}=(c_{1},c_{2})$, such as the sample mean of the points $(\overline{x}, \overline{y})$, or perhaps the centroid of the observed points. Then a measure of dispersion would be the average distance from that central point:
$$ \frac{1}{n} \sum_{i=1}^{n} || {\bf z}_{i} - {\bf c} || $$
where ${\bf z}_{i} = \{ x_{i}, y_{i} \}$. There are many potential choices for a distance measure but the $L_{2}$ norm (e.g. euclidean distance) may be a reasonable choice:
$$ || {\bf z}_{i} - {\bf c} || = \sqrt{ (x_{i}-c_{1})^{2} + (y_{i}-c_{2})^{2} } $$
There are lots of other potential choices, though. See [http://en.wikipedia.org/wiki/Norm_%28mathematics%29](http://en.wikipedia.org/wiki/Norm_%28mathematics%29)
| null | CC BY-SA 3.0 | null | 2011-07-19T19:50:02.047 | 2011-07-19T19:50:02.047 | null | null | 4856 | null |
13275 | 1 | 13281 | null | 3 | 15026 | Running a crude model using Bayesian inference, I get some results > 1 (ie, more than 100% "certain") for some combinations of "evidence". For instance, for one bit of evidence the conditional probability of the null hypothesis is 0.85 while the marginal probability is 0.77. If the prior probability is 0.9, the computed posterior probability is 1.008. Which maybe could be ascribed to rounding error, except that the next bit of evidence raises the posterior probability to 1.34.
It stands to reason that, for a problem with two hypotheses, one would have a conditional probability less than the marginal probability and the other would be greater. So the resulting P(E|H) / P(E) multiplier would be > 1. So it's hard to see how such > 1 results can be avoided in the general case.
Is this just the way Bayesian inference works, or do I likely have an error somewhere in my calcs?
## Data
```
One "evidence":
Total 6134 samples
Total actual positives 2845
Total actual negatives 3289
Test was true 1623 times
Test was true 465 times when the "gold standard" was positive
Test was true 1158 times when the "gold standard" was negative
conditional probability of positive hypothesis = 465/2845 = 0.1634
conditional probability of negative hypothesis = 1158/3289 = 0.3521
marginal probability of test being true = 1623/6134 = 0.2646
Bayes multiplier for the negative hypothesis = 0.3521 / 0.2646 = 1.3307
```
As can be seen, the multiplier is significantly > 1, and with several such tests back-to-back it seems hard to avoid probabilities > 1. (Of course, I suppose one can argue that the tests aren't truly independent, and that puts the fly in the ointment.)
## Fudging
So does anyone have any suggestions as to how to "fudge" non-independent measurements to improve an estimate? For the two most egregious cases I can come up with a fair estimate of how connected the measurements are, but I don't have a feel for how to factor that knowledge in.
| Bayesian probability > 1 -- is it possible? | CC BY-SA 3.0 | null | 2011-07-19T21:28:44.167 | 2011-09-03T13:01:54.787 | 2020-06-11T14:32:37.003 | -1 | 5392 | [
"bayesian",
"conditional-probability"
] |
13278 | 2 | null | 13272 | 2 | null | I think you should use 'Mahalanobis Distance' rather than Euclidean distance norms, as it takes into account the correlation of the data set and is 'scale-invariant'. Here is the link:
[http://en.wikipedia.org/wiki/Mahalanobis_distance](http://en.wikipedia.org/wiki/Mahalanobis_distance)
You could also use 'Half-Space Depth'. It is a bit more complicated but shares many attractive properties. The Half space Depth (also known as Location depth) of a given point a relative to a data set P is the minimum number of points of P lying in any closed halfplane determined by a line through a.
Here are the links:
[http://www.cs.unb.ca/~bremner/research/talks/depth-survey.pdf](http://www.cs.unb.ca/~bremner/research/talks/depth-survey.pdf)
http://depth.johnhugg.com/DepthExplorerALENEXslides.pdf
| null | CC BY-SA 3.0 | null | 2011-07-19T21:44:56.047 | 2011-07-19T21:44:56.047 | null | null | 1831 | null |
13279 | 2 | null | 13260 | 1 | null | If your looking for basic statistics I recommend Khan Academy, they have an excellent series on the theory of [regression lines](http://www.khanacademy.org/video/squared-error-of-regression-line?playlist=Statistics). You can alson check out [this thread](https://stats.stackexchange.com/questions/170/free-statistical-textbooks) about free statistical textbooks.
| null | CC BY-SA 3.0 | null | 2011-07-19T21:47:02.553 | 2011-07-19T21:47:02.553 | 2017-04-13T12:44:33.977 | -1 | 5429 | null |
13280 | 2 | null | 13260 | 2 | null | A classic reference for the theory of general linear models is Searle: Linear Models
@article{searle1971linear,
title={Linear models},
author={Searle, SR},
year={1971},
publisher={Wiley Series in Probability and Mathematical Statistics}
}
| null | CC BY-SA 3.0 | null | 2011-07-19T22:50:35.123 | 2011-07-19T22:50:35.123 | null | null | 5020 | null |
13281 | 2 | null | 13275 | 11 | null | You're making a mistake somewhere.
Yes, $P(E|H)$ can be greater than $P(E)$, but $P(E|H)P(H)$ should never be greater than $P(E)$, since
$$ P(E) = P(E|H)P(H) + P(E|\ {\rm not} \ H)P( \ {\rm not} \ H). $$
EDIT: Presumably the "evidence" is the uncertain test, and the "hypothesis" is the result of the "gold standard test". As the other listed probabilities appear to be right (though they're misleading named; I would instead list them as "conditional probability of positive test result given (positive) negative hypothesis), the marginal hypothesis probabilities (which you don't list) must be wrong, and indeed the Bayes multiplier is right, and greater than one. (But it's a bad test where a positive result increases your belief of a negative hypothesis.)
Initially $P(H = t)$ is just the fraction of gold-standard tests that are true, 2845/6134. Similarly, $P(H = f) = 3289 / 6134$.
$$
\begin{align*}
P(H = f | E = t ) &= P(E = f | H = t) P(H = f) / P(E = t)
\\ &= (1158/3289)*(3289/6134)/(1623/6134)
\\ &= 1158/1623
\end{align*}
$$
As should be expected, when you know the test returns true, you get out the fraction of true tests that are revealed by the gold standard test to be negative.
You'll have to be careful when trying to iterate Bayes rule. The next "$P(H | E)$" needs to be "$P(H | E_1, E_2)$", because you're interested in the chances after collecting both bits of data. It's not that hard to setup the formulas, but you have to be extremely careful about what your symbols mean. Jaynes recommended always writing probabilities as probabilities conditioned on "background information". Adding this conditioning to the standard formula may help you see how the formula must be used when iterating.
| null | CC BY-SA 3.0 | null | 2011-07-19T23:17:39.157 | 2011-07-20T03:30:07.573 | 2011-07-20T03:30:07.573 | 4925 | 4925 | null |
13282 | 2 | null | 13061 | 1 | null | The 2010 UCSD data mining contest has a training data set with transaction data from 130K customers. There are several hundred transactions (amount in whole dollars only, no other data) per customer. While the accompanying information is very sparse, it is fairly clear from the dataset where the transactions are recorded.
[http://mill.ucsd.edu/](http://mill.ucsd.edu/)
| null | CC BY-SA 3.0 | null | 2011-07-19T23:30:08.100 | 2011-07-19T23:30:08.100 | null | null | 4062 | null |
13284 | 2 | null | 13266 | 7 | null | For a linear regression, the fitted slope is going to be the correlation (which, when squared, gives the coefficient of determination, the $R^2$) times the empirical standard deviation of the regressand (the $y$) divided by the empirical standard deviation of the regressor (the $x$). Depending on the scaling of the $x$ and $y$, you can have a fit slope equal to one but an arbitrarily small $R^2$ value.
In short, the slope is not a good indicator of model 'fit' unless you are certain that the scales of the dependent and independent variables must be equal to each other.
| null | CC BY-SA 3.0 | null | 2011-07-19T23:39:53.670 | 2011-07-19T23:39:53.670 | null | null | 795 | null |
13285 | 2 | null | 13252 | 4 | null | I think dot-chart is the best while visualizing categorical data vs numeric observations on these categorical data. You would keep the categorical data on the y-axis and the x-axis has the numeric observations. Ordering of the categories is recommended to let the patterns be clearer. Therefore, I don't think you'd get much out of dot-plot for time-series data. You can definitely draw it, but there are better ways to represent time-series data. However, I must point out that your hunch was right in replacing a bar chart with a dot-chart. In fact, dot-chart can replace every bar-chart and pie chart in a much better way! (unfortunately, I don't see people doing it though).
Time-series data come with the natural uni-directional time axis and you'd lose key patterns if they weren't on the x-axis. Check out the excellent STL decomposition [here](http://stat.ethz.ch/R-manual/R-devel/library/stats/html/stl.html) in stats package of R and in latticeExtra package [here](http://latticeextra.r-forge.r-project.org/#xyplot.stl). They are time-series-specific visualization methods. They are also emphasized in Cleveland's Visualizing Data book in Chapter 4.
As Andy W points out, line plots are the best! Like Gelman and Fung, I also think line plots are under-rated.
| null | CC BY-SA 3.0 | null | 2011-07-19T23:40:59.223 | 2011-07-20T00:03:31.463 | 2011-07-20T00:03:31.463 | 1307 | 1307 | null |
13286 | 2 | null | 13259 | 4 | null | I answer this: "Arbitrarily group the draws into n groups with m values in each group. Look at the minimum value in each group. Take the group that has the greatest of these minima. Now, what is the distribution that defines the maximum value in that group?"
Let $X_{i,j}$ the i-th random variable in group j and $f(x_{i,j})$ ($F(x_{i,j})$) its density (cdf) function.
Let $X_{\max,j}, X_{\min,j}$ the maximum and minimum in group $j$. Let $X_{final}$ the variable that results at the end of all process.
We want to calculate $P(X_{final}<x)$ which is
$$P(X_{\max,j_0}<x \hbox{ and } X_{\min,j_0}=\max_j{X_{\min,j}} \hbox { and } 1\leq j_0\leq n)$$
$$=nP(X_{max,1}<x \hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})$$
$$=nmP(X_{1,1}<x\hbox{ and } X_{1,1}=\max_i(X_{i,1})\hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})$$
$$=nmP(X_{1,1}<x, X_{1,1}>X_{2,1}>\max_{j=2\ldots n} X_{min,j},\ldots,X_{1,1}>X_{m,1}>\max_{j=2\ldots n} X_{min,j})$$
Now, let $Y=\max_{j=2\ldots n} X_{min,j}$ and $W=X_{1,1}$.
A reminder: if $X_1,\ldots X_n$ are iid with pdf (cdf) $h$ ($H$), then $X_{\min}$ has pdf $h_{\min}=nh(1-H)^{n-1}$ and $X_{\max}$ has pdf $h_{max}=nhH^{n-1}$.
Using this, we get the pdf of $Y$ is
$$g(y)=(n-1)mf(1-F)^{m-1}[\int_0^y mf(z)(1-F(z))^{m-1} dz]^{n-2},n\geq 2$$
Note that $Y$ is a statistics that is independent of group 1 so its joint density with any variable in the group 1 is the product of densities.
Now the above probability becomes
$$nm\int_0^x f(w)[\int_0^w \int_y^w f(x_{2,1})dx_{2,1}\ldots\int_y^w f(x_{m,1})dx_{m,1}g(y)dy]dw$$
$$=nm\int_0^x f(w)[\int_0^w (F(w)-F(y))^{m-1}g(y)dy]dw$$
By taking derivative of this integral wrt $x$ and using binomial formula we obtain the pdf of $X_{final}$.
Example: $X$ is uniform, $n=4$, $m=3$. Then
$$g(y)=9(1-y)^2(3y+y^3-3y^2)^2,$$
$$P(X_{final}<x)=(1/55)x^{12}-(12/55)x^{11}$$
$$+ (6/5)x^{10}-(27/7)x^9+(54/7)x^8-(324/35)x^7+(27/5)x^6. $$
Mean of $X_{final}$ is $374/455=0.822$ and its s.d. is $0.145$ .
| null | CC BY-SA 3.0 | null | 2011-07-19T23:49:13.103 | 2011-07-28T07:06:56.833 | 2011-07-28T07:06:56.833 | 2116 | 3454 | null |
13287 | 1 | null | null | 0 | 1058 | Please excuse me if my question is dumb but I am not a Statistics student.
I have a project where I was exploring options for time series analysis where I found some papers on cointegration.
I have 4 variables. One of them response variable. I wanted to to the forecasting using the cointegration technique. Please tell me if I am wrong whether the technique to use is a Johansen test. However I understand that the test needs a requirement that the variables are unit root (right?).
So I took the software R and did the ADF test in R on each variable (and the first difference ) . But some variables dont pass the adf test. My question is:
- Is my thinking above right?
- Then, what do i do for the variables which dont pass the ADF test
for unit root ? Can I still take those in the johansen test? If so,
what is the purpose of the ADF test at all?
@mpiktas Here I performed the ADF test and here are the results I got:
- yt.adf Null Hypothesis not rejected (Non Stationary)
- dyt.adf Null Hypothesis not rejected (Non Stationary)
- X1t.adf Null hypothesis rejected (Stationary)
- dX1t.adf Null hypothesis rejected (Stationary)
- X3t.adf Null hypothesis rejected (Stationary)
- dX3t.adf Null Hypothesis not rejected (Non Stationary)
- X4t.adf Null Hypothesis not rejected (Non Stationary)
- dX4t.adf Null Hypothesis not rejected (Non Stationary)
yt.adf is the response variable while dyt.adf is the firs difference. Similarly other variables.
As seen here, I suppose the inference is that none (yt, X1t,X2t,X2t,X4t) are unit root?
So then I cant apply johansen test for cointegration,can I?
| Help with cointegration problem | CC BY-SA 3.0 | null | 2011-07-19T23:52:08.123 | 2011-07-21T02:12:55.510 | 2011-07-21T02:12:55.510 | 5479 | 5479 | [
"cointegration"
] |
13288 | 2 | null | 10564 | 2 | null | The more appropriate comparator is the proportional odds model, which contains the Wilcoxon test as a special case. And note that 'multivariate' refers to the simultaneous analysis of more than one dependent variable. I think you meant to say 'multivariable'.
| null | CC BY-SA 3.0 | null | 2011-07-20T00:58:55.880 | 2011-07-20T00:58:55.880 | null | null | 4253 | null |
13289 | 2 | null | 13230 | 1 | null | Welcome to the wonderful world of Statistics. Despite a Masters degree in Medical Statistics, it never ceases to amaze me that the one thing I am learning over and over again is how little I actually understand about this fascinating modality!!
For what it is worth from a non-practising and not very experienced statistician, you need to look at a time-series data analysis with a time varying covariate. The Cox Proportional Hazards model can be used for this data with relative ease.
Check out this paper on the CRAN website:
[Cox Proportional Hazards Regression (J Fox, 2002)](http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-cox-regression.pdf)
My two cents - hope it's useful.
| null | CC BY-SA 3.0 | null | 2011-07-20T01:19:15.733 | 2011-07-20T01:19:15.733 | null | null | 5317 | null |
13290 | 2 | null | 12873 | 12 | null | You can use the `Amelia` package to impute the data (full disclosure: I am one of the authors of `Amelia`). The [package vignette](http://cran.r-project.org/web/packages/Amelia/vignettes/amelia.pdf) has an extended example of how to use it to impute missing data.
It seems as though you have units which are district-gender-ageGroup observed at the monthly level. First you create a factor variable for each type of unit (that is, one level for each district-gender-ageGroup). Let's call this `group`. Then, you would need a variable for time, which is probably the number of months since January 2003. Thus, this variable would be 13 in January of 2004. Call this variable `time`. Amelia will allow you to impute based on the time trends with the following commands:
```
library(Amelia)
a.out <- amelia(my.data, ts = "time", cs = "group", splinetime = 2, intercs = TRUE)
```
The `ts` and `cs` arguments simply denote the time and unit variables. The `splinetime` argument sets how flexible should time be used to impute the missing data. Here, a 2 means that the imputation will use a quadratic function of time, but higher values will be more flexible. The `intercs` argument here tells Amelia to use a separate time trend for each district-gender-ageGroup. This adds many parameters to the model, so if you run into trouble, you can set this to `FALSE` to try to debug.
In any event, this will get you imputations using the time information in your data. Since the missing data is bounded at zero, you can use the `bounds` argument to force imputations into those logical bounds.
EDIT: How to create group/time variables
The time variable might be the easiest to create, because you just need to count from 2002 (assuming that is the lowest year in your data):
```
my.data$time <- my.data$Month + 12 * (my.data$Year - 2002)
```
The group variable is slightly harder but a quick way to do it is using the paste command:
```
my.data$group <- with(my.data,
as.factor(paste(District, Gender, AgeGroup, sep = ".")))
```
With these variables created, you want to remove the original variables from the imputation. To do that you can use the `idvars` argument:
```
a.out <- amelia(my.data, ts = "time", cs = "group", splinetime = 2, intercs = TRUE,
idvars = c("District", "Gender", "Month", "Year", "AgeGroup"))
```
| null | CC BY-SA 3.0 | null | 2011-07-20T03:24:29.877 | 2011-07-22T17:05:28.327 | 2011-07-22T17:05:28.327 | 4160 | 4160 | null |
13291 | 2 | null | 13118 | 9 | null | I worked out the details omitted from JMS's answer.
Let $P_0 = \Sigma_0^{-1}$ and $P_1 = \Sigma_1^{-1}$
We have
$$KL(N_0||N_1) = \int N_0(x) \frac{1}{2}[\ln |P_0| - \ln|P_1| + (x-\mu_0)^T P_0 (x-\mu_0) - (x-\mu_1)^T P_1 (x-\mu_1)] dx$$
$$
= \frac{1}{2}\ln|P_0\Sigma_1| - \frac{1}{2} \mathbb{E}[(x-\mu_0)^T P_0 (x-\mu_0) - (x-\mu_1)^T P_1 (x-\mu_1)]
$$
$$
= \frac{1}{2}\ln|P_0\Sigma_1| + \frac{1}{2}(\mu_0-\mu_1)^T P_1 (\mu_0-\mu_1)- \frac{1}{2} \mathbb{E}[x^T P_0 x - x^T P_1 x - \mu_0^T (P_0 - P_1) \mu_0]
$$
$$
= \frac{1}{2}\ln|P_0\Sigma_1| + \frac{1}{2}(\mu_0-\mu_1)^T P_1 (\mu_0-\mu_1)- \frac{1}{2} \mathbb{E}[(x-\mu_0)^T (P_0-P_1) (x-\mu_0)]
$$
To simplify notation, assume $\mu_0 = 0$. It remains to show that
$$\mathbb{E}[x^T P_0x-x^T P_1 x] = k - tr(P_1\Sigma_0)$$
But $x^T P_0 x = x^T \Sigma_0 x$ has a Chi-squared distribution and therefore has expectation $k$. Meanwhile,
$$\mathbb{E}[x^T P_1 x] = \mathbb{E}[tr(x^T P_1 x)] = \mathbb{E}[tr(P_1 xx^T )]= tr(\mathbb{E}(P_1 xx^T)) = tr(P_1 \Sigma_0)$$
where the second equality can be obtained by treating $x$ as a square matrix padded by zeroes.
| null | CC BY-SA 3.0 | null | 2011-07-20T03:25:45.493 | 2011-07-20T03:40:33.020 | 2011-07-20T03:40:33.020 | 3567 | 3567 | null |
13292 | 2 | null | 13272 | 7 | null | A good reference on metrics for the spatial distribution of point patterns is the [CrimeStat manual](http://www.icpsr.umich.edu/CrimeStat/download.html) (in particular for this question, [Chapter 4](http://www.icpsr.umich.edu/CrimeStat/files/CrimeStatChapter.4.pdf) will be of interest). Similar to the metric Macro suggested, the Standard Distance Deviation is similar to a 2D standard deviation (the only difference is that you would divide by "n-2" not "n" in the first formula Macro gave).
Your example experiment actually reminds me a bit of how studies evaluate [Geographic Offender Profiling](http://www.nij.gov/maps/gp.htm), and hence the metrics used in those works may be of interest. In particular the terms precision and accuracy are used quite a bit and would be pertinent to the study. Guesses could have a small standard deviation (i.e. precise) but still have a very low accuracy.
| null | CC BY-SA 3.0 | null | 2011-07-20T04:11:06.267 | 2011-07-20T04:11:06.267 | null | null | 1036 | null |
13293 | 1 | null | null | 2 | 371 | So as I understand it SOM is primarily a visualization tool and clustering is a logical next step after you construct a SOM from data. Typically, the clustering is subjective in that after looking at your SOM you can 'see' N clusters and then from there you would go on to cluster (k means, hierarchical etc.) the SOM nodes with this N as your parameter. Is there a nonsubjective (maybe nonparametric?) way to cluster the SOM nodes (i.e. without first looking at the SOM to determine the number of distinct clusters)? Thanks
| SOM automated/objective clustering | CC BY-SA 3.0 | null | 2011-07-20T05:50:54.657 | 2016-09-30T07:34:28.910 | null | null | 5464 | [
"clustering",
"unsupervised-learning",
"self-organizing-maps"
] |
13294 | 2 | null | 13287 | 3 | null | Johansen test is for testing whether the variables are cointegrated. If they are, the appropriate model is vector error correction model (VECM) which is a special case of vector autoregression models (VAR).
Cointegration is defined for unit root (integrated) time-series. To be more precise a set of variables are said to be cointegrated if their linear combination is stationary. Since any linear combination of stationary time series is stationary the cointegration is specifically defined for non-stationary, i.e. unit root time-series. So yes there is a requirement to test whether the time series in question are unit roots (which by the way is always a good idea if you are using time series in regression).
So to sum up, the answer to your first question is yes, more or less :) It would be good to know how exactly you have determined that ADF test was not passed. It has a lot of variations, and it is easy to make mistakes.
If some of the variables are stationary (did not pass ADF test in your terminology) it is still possible to use VECM. For this however more information is needed about your desired model. If you are only interested in forecasting response variable, simply differencing unit root time series might be sufficient. Then a lot depends whether the response variable is unit root or not. If it is not, then using VECM might not be appropriate.
| null | CC BY-SA 3.0 | null | 2011-07-20T09:04:55.940 | 2011-07-20T09:04:55.940 | null | null | 2116 | null |
13295 | 2 | null | 13252 | 0 | null | I wouldn't use a bar graph, regardless of the sample size, I'd use a line plot.
But I'm not sure WHY you would want a Cleveland plot of time series data. Line plots seem like the right tool; if you have more time points, you can look at smoothed curves of various types, and seasonality and all sorts of things, but I would start with a line plot
| null | CC BY-SA 3.0 | null | 2011-07-20T10:15:43.813 | 2011-07-20T10:15:43.813 | null | null | 686 | null |
13296 | 1 | 13297 | null | 4 | 4115 | I have a text corpus, from which I have computed the unigram probabilities of every word.
Now, let $p_{i}$ & $p_{j}$ be the unigram probabilities of $i$ and $j$ respectively, what will be the probability with which a pair of words $i$ and $j$ will co-occur(within a maximum gap of 1 word) when independence between $i$ and $j$ is assumed?
EDIT:
Here $p_{i}$ is the probability of any given word being $i$.
By co-occurrence probability, I mean the chance that word $i$ & $j$ co-occur in a random sequence of three words drawn from the corpus.
By co-occurring I mean that $i$ and $j$ occurring anywhere in this 3 word random sequence within a maximum word distance of 1. (For example, $i$*$j$ and $ij$ are both valid co-occurrences of $i$ & $j$ but $i**j$ is not a valid co-occurrence since the distance between $i$ & $j$ is more than 1.)
And finally, $ijij$ counts as 3 co-occurrences.
| How to compute co-occurrence probability assuming independence from unigram probabilities? | CC BY-SA 3.0 | null | 2011-07-20T10:56:57.087 | 2011-07-20T13:47:18.050 | 2011-07-20T12:25:50.727 | 4966 | 4966 | [
"probability",
"conditional-probability"
] |
13297 | 2 | null | 13296 | 3 | null | By the [Principle of Inclusion-Exclusion](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), the probability of no co-occurrences of i and j in three words equals the probability that i fails to appear plus the probability that j fails to appear minus the probability that both i and j fail to appear. Subtracting this from 1 will yield the answer. The independence assumption lets us compute all three probabilities by multiplying the relevant "unigram" probabilities, whence the probability of a co-occurrence of i and j equals
$$1 - (1-p_i)^3 - (1-p_j)^3 + (1 - p_i - p_j)^3$$
$$= 3 p_i p_j (2 - p_i - p_j).$$
For example, when $p_i=p_j=1/3$, this formula gives $4/9$. The 12 possible patterns of co-occurrence can be symbolized as iij, iji, ijj, ij*, i*j, jii, jij, ji*, jji, j*i, *ij, *ji, each with equal probability of $(1/3)^3$. The chance therefore is $12/3^3=4/9$, agreeing with the formula.
| null | CC BY-SA 3.0 | null | 2011-07-20T12:59:34.110 | 2011-07-20T13:47:18.050 | 2011-07-20T13:47:18.050 | 919 | 919 | null |
13298 | 2 | null | 13275 | 6 | null | The mistake is in thinking that $P(E)$, the marginal probability, is some sort of "absolute" probability, which never changes. It should really be written $P(E|I)$ to denote that it is still conditional on something. Here's why
If we denote two pieces of evidence $E_1$ and $E_2$, then we have:
$$P(H|E_1 E_2 I)=\frac{P(H|I)P(E_1 E_2|HI)}{P(E_1 E_2|I)}$$
Now we can use the product rule to decompose $P(E_1 E_2|I)=P(E_1|I)P(E_2|E_1 I)$, and similarly, $P(E_1 E_2|HI)=P(E_1|HI)P(E_2|E_1HI)$. We now get
$$P(H|E_1 E_2 I)=\frac{P(H|I)P(E_1|HI)}{P(E_1|I)}\times\frac{P(E_2 | E_1HI)}{P(E_2 | E_1I)}$$
But the first term is simply the posterior after observing $E_1$, denoted by $P(H|E_1I)$. We now have:
$$P(H|E_1 E_2 I)=P(H|E_1I)\frac{P(E_2 | E_1HI)}{P(E_2 | E_1I)}$$
If I was to write $X=E_1I$, then we have $P(H|E_2 X)=P(H|X)\frac{P(E_2 | HX)}{P(E_2 | X)}$. This gives you the usual "posterior to prior" relationship in sequentially digesting data. The posterior after observing $E_1$ is equal to the prior before observing $E_2$
But we also have that the marginal probability (and the condtional probability) now depends on $X=E_1I$, and not just on $I$. The mistake comes from assuming that they are equal - this is an easy thing to do if your notation does not explicitly recognise prior information somewhere.
The "Bayes multiplier", while not less than $1$, is bounded. To see this, expand $P(E|I)$ using the sum rule as follows:
$$P(E|I)=P(EH|I)+P(E\overline{H}|I)\geq P(EH|I)=P(H|I)P(E|HI)$$
$$\implies \frac{1}{P(H|I)}\geq\frac{P(E|HI)}{P(E|I)}$$
And note that this upper bound is just enough to ensure the posterior probability is less than or equal to 1
| null | CC BY-SA 3.0 | null | 2011-07-20T14:03:31.127 | 2011-07-20T14:03:31.127 | null | null | 2392 | null |
13300 | 1 | 13311 | null | 5 | 41347 | I'm a novice to data mining and started to read about it. What's the exact difference between experimental data and observation data? Both are obviously data; and many say observation data can lead to errors. But I guess it's not possible to do an experiment for all data sets. I'm really confused, explain me what is experimental data and observation data and say when these should be used?
Thanks in advance.
| Difference between experimental data and observational data? | CC BY-SA 3.0 | null | 2011-07-20T15:26:49.323 | 2015-03-06T18:19:43.007 | 2011-07-20T19:39:55.823 | 919 | 5351 | [
"dataset",
"data-mining",
"causality"
] |
13301 | 1 | null | null | 1 | 621 | does it make sense to assume $u\sim N(0,\sigma^2)$ when I know from a histogram that $y$ is highly skewed. Because from the assumption $u\sim N(0,\sigma^2)$ it follows that $y\sim N(x\beta,\sigma^2)$ and I'm absoluteley not sure if the assumption $u\sim N(0,\sigma^2)$ makes sense in a case where I know that the distribution of $y$ is not bell shaped. The alternative would be just to make OLS without any assumption about the error term, but in this case I can't analyze outliers and leverages (what I'd really like to do, because otherwise I can't present much more than a line which minimizes the sqaured sum of the residuals). [addendum: I can't make an outlier analysis because I can not define "outlier" in a context where I don't assume a normal distribution, because there is no outlying without a distribution] Besides your answers I'd really like to have a recommendation for a good book, where I can find some thoughts about what assumptions should we make when y is obviously not normal distributed.
Thanks for your thoughts, they'll be helpful to me!
Regards
| Assuming $u\sim N(0,\sigma^2)$ when y is highly skewed | CC BY-SA 3.0 | 0 | 2011-07-20T15:29:31.560 | 2011-07-20T16:50:47.897 | 2011-07-20T16:36:47.853 | null | 4496 | [
"regression",
"normal-distribution",
"residuals",
"linear-model"
] |
13302 | 2 | null | 12002 | 0 | null | I think the CI for grand mean is too wide [17,62] even for the range of original data.
This experiments are VERY common in chemistry. For example, in certification of reference materials you have to pick up some bottles from whole lot in a random way, and you have to carry out replicate analysis on each bottles. How do you calculate the reference value and its uncertainty? There are a lot of way to do it, but the most sofisticated (and correct, I think) is applying meta-analysis or ML (Dersimonian-Laird, Vangel-Rukhin, etc)
What about bootstrap estimates?
| null | CC BY-SA 3.0 | null | 2011-07-20T15:42:33.267 | 2011-07-20T15:42:33.267 | null | null | 5486 | null |
13303 | 1 | 13304 | null | 6 | 371 | Suppose I am building some predictive models and then creating a report detailing how "good" those models were in various ways. Is there a generic (maybe even non-technical) term for the various measures of correctness (e.g. precision, recall, etc.)? A layman might use "accuracy," but accuracy actually has a very specific meaning and is a subset of possible measures to be included in such a report. In describing the contents of such a report, should I just use "accuracy" even though there will be other measures of correctness included in such a report?
| Is there a generic term for measures of correctness like "precision" and "recall"? | CC BY-SA 3.0 | null | 2011-07-20T16:06:30.230 | 2011-08-01T21:20:57.973 | 2011-07-20T16:28:54.067 | null | 2485 | [
"machine-learning",
"predictive-models",
"terminology"
] |
13304 | 2 | null | 13303 | 4 | null | I don't know if there is a generally accepted generic term, but I think you might say "classifier performance metrics/measures" (like in the `R` package `ROCR`), or "measures of predictive/classification performance".
The widely cited paper by [Fawcett](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.98.4088&rep=rep1&type=pdf), for example, talks about "common performance metrics" and lists true positive rate (tpr), fpr, sensitivity, speficitiy, precision, and recall.
| null | CC BY-SA 3.0 | null | 2011-07-20T16:24:06.213 | 2011-07-20T16:24:06.213 | null | null | 5020 | null |
13305 | 2 | null | 13301 | 6 | null | If u is the residual from a regression of y on some other variable x, then I think this is a variant of an [earlier question](https://stats.stackexchange.com/questions/12262/what-if-residuals-are-normally-distributed-but-y-is-not/12266#12266). The residuals of u can be Gaussian even if the distribution of y is highly skewed as it may simply be that the distribution of x is highly skewed.
Consider an example of estimating temperature (y) as a function of lattitude (x); here u represents the measurment error of the thermometer (and is Gaussian). The distribution of y values in our sample will depend on where we choose to site out weather stations. If we place them all either at the poles or the equator, then we will have a bimodal distribution. If we place them on a regular equal area grid, we will get a unimodal distribution of y values, even though the physics of climate is the same for both samples and the measurement uncertainty u is normal.
| null | CC BY-SA 3.0 | null | 2011-07-20T16:50:47.897 | 2011-07-20T16:50:47.897 | 2017-04-13T12:44:24.677 | -1 | 887 | null |
13306 | 1 | 13313 | null | 10 | 10793 | I am trying to understand how to use Bayes' theorem to calculate a posterior but am getting stuck with the computational approach, e.g., in the following case it is not clear to me how to take the product of the prior and likelihood and then calculate the posterior:
For this example, I am interested in calculating the posterior probability of $\mu$ and I use a standard normal prior on $\mu$ $p(\mu)\sim N(\mu = 0, \sigma = 1)$, but I want to know how to calculate the posterior from a prior on $\mu$ that is represented by an MCMC chain, so I will use 1000 samples as my starting point.
- sample 1000 from the prior.
set.seed(0)
prior.mu <- 0
prior.sigma <- 1
prior.samples <- sort(rnorm(1000, prior.mu, prior.sigma))
- make some observations:
observations <- c(0.4, 0.5, 0.8, 0.1)
- and calculate the likelihood, e.g. $p(y | \mu, \sigma)$:
likelihood <- prod(dnorm(observations, mean(prior.samplse), sd(prior.samples)))
what I don't quite understand is:
- when / how to multiply the prior by the likelihood?
- when / how to normalize the posterior density?
please note: I am interested in the general computational solution that could be generalizable problems with no analytical solution
| How can I compute a posterior density estimate from a prior and likelihood? | CC BY-SA 3.0 | null | 2011-07-20T17:02:38.780 | 2017-07-11T02:33:13.140 | 2011-07-20T20:36:31.880 | 2750 | 2750 | [
"bayesian",
"simulation",
"computational-statistics"
] |
13307 | 1 | 13308 | null | 12 | 4242 | I have a series of functions, each one supposedly representing the density of a random variable across agents. Each function also has a domain, which describes what values of the random variable are valid.
Now, if I remember my stats classes correctly, if I take the integral of one of the functions across the values described by the function's domain, I should get a value of 1.0. This does not happen however.
Is there a normalization technique that can turn a function into a true probability density, yet maintains the shape of the function?
All the functions are of the form $\frac{a}{bx}+c$, where $x$ is the random variable, and $a,b,c$ are varying constants.
| How to turn a function into a probability density whilst maintaining the shape of the function? | CC BY-SA 3.0 | null | 2011-07-20T17:45:39.233 | 2012-06-19T11:32:14.357 | 2012-06-19T11:32:14.357 | 183 | 5487 | [
"distributions",
"probability"
] |
13308 | 2 | null | 13307 | 16 | null | If you have a non-negative integrable function $f$ with domain $D$ such that
$$ k = \int_{D} f(x) dx < \infty $$
Then $f(x)/k$ is a probability density on $D$. The value $k$ is known as the normalizing constant.
Edit: In your example you said that $f(x) = \frac{a}{bx} + c$ for known constants $a,b,c$. In that case, the indefinite integral is simple to compute and the normalizing constant would be
$$ k = \left[ \frac{a \log(x) }{b} + cx \right]_{D}$$
if $D$ is an interval $(A,B)$ then this simplifies to
$$ k = \frac{a}{b} \cdot \log \left( \frac{B}{A} \right) + c(B-A) $$ Therefore $$ g(x) = \frac{\frac{a}{bx} + c}{\frac{a}{b} \cdot \log \left( \frac{B}{A} \right) + c(B-A)}$$ is a probability density on $(A,B)$.
| null | CC BY-SA 3.0 | null | 2011-07-20T17:50:24.377 | 2012-06-18T22:13:36.307 | 2012-06-18T22:13:36.307 | 4856 | 4856 | null |
13310 | 2 | null | 10075 | 4 | null | In portfolio theory, variance is additive. In other words, just as the return of a portfolio is the weighted average of the returns of its members, so to is the portfolio variance the weighted average of the securities' variances. However, this property does not hold true for standard deviation.
| null | CC BY-SA 3.0 | null | 2011-07-20T19:01:34.797 | 2011-07-20T19:01:34.797 | null | null | 8101 | null |
13311 | 2 | null | 13300 | 11 | null | wow, that's a tough one :-)
That question is far more widely relevant than just in data mining. It comes up in medicine and in the social sciences including psychology all the time.
The distinction is necessary when it comes to drawing conclusions about causality, that is, when you want to know if something (e.g. a medical treatment) causes another thing (e.g. recovery of a patient). Hordes of scientists and philosophers debate whether you can draw conclusions about causality from observational studies or not. You might want to look at the question [statistics and causal inference?](https://stats.stackexchange.com/questions/2245/statistics-and-causal-inference).
So what is an experiment? Concisely, an experiment is often defined as random assignment of observational units to different conditions, and conditions differ by the treatment of observational units. Treatment is a generic term, which translates most easily in medical applications (e.g. patients are treated differently under different conditions), but it also applies to other areas. There are variations of experiments --- you might want to start by reading the wikipedia entries for [Experiment](http://en.wikipedia.org/wiki/Experiment) and [randomized experiment](http://en.wikipedia.org/wiki/Randomized_experiment) --- but the one crucial point is random assignment of subjects to conditions.
With that in mind, it is definitely not possible to do an experiment for all kinds of hypotheses you want to test. For example, you sometimes can't do experiments for ethical reasons, e.g. you don't want people to suffer because of a treatment. In other cases, it might be physically impossible to conduct an experiment.
So whereas experimentation (controlled randomized assignment to treatment conditions) is the primary way to draw conclusions about causality --- and for some, it is the only way --- people still want to do something empirical in those cases where experiments are not possible. That's when you want to do an observational study.
To define an observational study, I draw on Paul Rosenbaums [entry](http://www-stat.wharton.upenn.edu/~rosenbap/BehStatObs.pdf) in the encyclopedia of statistics in behavioral science: An observational study is "an empiric comparison of treated and control groups in which the objective is to elucidate cause-and-effect relationships
[. . . in which it] is not feasible to use controlled experimentation, in the sense of being able to impose the procedures or treatments whose effects it is desired to discover, or to assign subjects at random to different procedures." In an observational study, you try to measure as many variables as possible, and you want to test hypotheses about what changes in a set of those variables are associated with changes in other sets of variables, often with the goal of drawing conclusions about causality in these associations (see [Under what conditions does correlation imply causation](https://stats.stackexchange.com/questions/534/under-what-conditions-does-correlation-imply-causation)
In what ways can observational studies lead to errors? Primarily if you want to draw conclusions about causality. The issue that arises is that there might always be the chance that some variables you did not observe are the "real" causes (often called "unmeasured confounding"), so you might falsely assume that one of your measured variables is causing something, whereas "in truth" it is one of the unmeasured confounders. In experiments, the general assumption is that by random assignment potential confounders will get canceled out.
If you want to know more, start by going through the links provided, and look at publications from people like [Paul Rosenbaum](http://www-stat.wharton.upenn.edu/~rosenbap/index.html) or the book-link provided by [iopsych](https://stats.stackexchange.com/users/1334/iopsych): [Experimental and Quasi-Experimental Designs for Generalized Causal Inference (Shadish, Cook, and Campbell, (2002)](http://rads.stackoverflow.com/amzn/click/0395615569)
| null | CC BY-SA 3.0 | null | 2011-07-20T19:12:30.563 | 2011-07-20T19:12:30.563 | 2017-04-13T12:44:49.953 | -1 | 5020 | null |
13312 | 2 | null | 13300 | 2 | null | Very much in a nutshell: only data for which you have all covariates under control, and have either randomization over possible confounders or enough information on them to properly account for them, can be truly called experimental. This could e.g. be the case in plant research where genetically identical and similarly grown plants are feasible: you can then make sure that only your variable of interest differs between groups of interest.
The place where (in statistically correct research) this matters most, is in trying to find a causal relation. A classical example is people taking aspirin, and the effect it has on heart disease: if you pick 100 people who take aspirin and 100 people who don't, and then somehow measure their heart condition, then even if the aspirin takers are at a lower risk frmo this research, you cannot conclude that people should all take aspirin: perhaps the aspirin taking and heart 'improvement' are both consequences of 'better living' or similar.
So, basically (since in reality we almost always want to show that A is a consequence f B): if it is available/attainable: prefer experimental data.
| null | CC BY-SA 3.0 | null | 2011-07-20T19:18:46.843 | 2011-07-20T19:18:46.843 | null | null | 4257 | null |
13313 | 2 | null | 13306 | 10 | null | You have several things mixed up. The theory talks about multiplying the prior distribution and the likelihood, not samples from the prior distribution. Also it is not clear what you have the prior of, is this a prior on the mean of something? or something else?
Then you have things reversed in the likelihood, your observations should be x with either prior draws or known fixed constants as the mean and standard deviation. And even then it would really be the product of 4 calls to dnorm with each of your observations as x and the same mean and standard deviation.
What is really no clear is what you are trying to do. What is your question? which parameters are you interested in? what prior(s) do you have on those parameters? are there other parameters? do you have priors or fixed values for those?
Trying to go about things the way you currently are will only confuse you more until you work out exactly what your question is and work from there.
Below is added due after the editing of the original question.
You are still missing some pieces, and probably not understanding everything, but we can start from where you are at.
I think you are confusing a few concepts. There is the likelihood that shows the relationship between the data and the parameters, you are using the normal which has 2 parameters, the mean and the standard deviation (or variance, or precision). Then there is the prior distributions on the parameters, you have specified a normal prior with mean 0 and sd 1, but that mean and standard deviation are completely different from the mean and standard deviation of the likelihood. To be complete you need to either know the likelihood SD or place a prior on the likelihood SD, for simplicity (but less real) I will assume we know the likelihood SD is $\frac12$ (no good reason other than it works and is different from 1).
So we can start similar to what you did and generate from the prior:
```
> obs <- c(0.4, 0.5, 0.8, 0.1)
> pri <- rnorm(10000, 0, 1)
```
Now we need to compute the likelihoods, this is based on the prior draws of the mean, the likelihood with the data, and the known value of the SD. The dnorm function will give us the likelihood of a single point, but we need to multiply together the values for each of the observations, here is a function to do that:
```
> likfun <- function(theta) {
+ sapply( theta, function(t) prod( dnorm(obs, t, 0.5) ) )
+ }
```
Now we can compute the likelihood for each draw from the prior for the mean
```
> tmp <- likfun(pri)
```
Now to get the posterior we need to do a new type of draw, one approach that is similar to rejection sampling is to sample from the prior mean draws proportional to the likelihood for each prior draw (this is the closest to the multiplication step you were asking about):
```
> post <- sample( pri, 100000, replace=TRUE, prob=tmp )
```
Now we can look at the results of the posterior draws:
```
> mean(post)
[1] 0.4205842
> sd(post)
[1] 0.2421079
>
> hist(post)
> abline(v=mean(post), col='green')
```
and compare the above results to the closed form values from the theory
```
> (1/1^2*mean(pri) + length(obs)/0.5^2 * mean(obs))/( 1/1^2 + length(obs)/0.5^2 )
[1] 0.4233263
> sqrt(1/(1+4*4))
[1] 0.2425356
```
Not a bad approximation, but it would probably work better to use a built-in McMC tool to draw from the posterior. Most of these tools sample one point at a time not in batches like above.
More realistically we would not know the SD of the likelihood and would need a prior for that as well (often the prior on the variance is a $\chi^2$ or gamma), but then it is more complicated to compute (McMC comes in handy) and there is no closed form to compare with.
The general solution is to use existing tools for doing the McMC calculations such as WinBugs or OpenBugs (BRugs in R gives an interface between R and Bugs) or packages such LearnBayes in R.
| null | CC BY-SA 3.0 | null | 2011-07-20T20:24:27.690 | 2011-07-21T15:47:25.103 | 2011-07-21T15:47:25.103 | 4505 | 4505 | null |
13314 | 1 | 13317 | null | 282 | 44949 | I was skimming through [some lecture notes](http://www.stat.cmu.edu/~cshalizi/402/) by Cosma Shalizi (in particular, section 2.1.1 of the [second lecture](http://www.stat.cmu.edu/~cshalizi/402/lectures/02-truth-about-linear-regression/lecture-02.pdf)), and was reminded that you can get very low $R^2$ even when you have a completely linear model.
To paraphrase Shalizi's example: suppose you have a model $Y = aX + \epsilon$, where $a$ is known. Then $\newcommand{\Var}{\mathrm{Var}}\Var[Y] = a^2 \Var[x] + \Var[\epsilon]$ and the amount of explained variance is $a^2 \Var[X]$, so $R^2 = \frac{a^2 \Var[x]}{a^2 \Var[X] + \Var[\epsilon]}$. This goes to 0 as $\Var[X] \rightarrow 0$ and to 1 as $\Var[X] \rightarrow \infty$.
Conversely, you can get high $R^2$ even when your model is noticeably non-linear. (Anyone have a good example offhand?)
So when is $R^2$ a useful statistic, and when should it be ignored?
| Is $R^2$ useful or dangerous? | CC BY-SA 3.0 | null | 2011-07-20T20:32:55.510 | 2020-12-18T15:02:32.140 | 2013-01-10T12:45:33.370 | 17230 | 1106 | [
"regression",
"r-squared"
] |
13315 | 2 | null | 13314 | 17 | null | One situation you would want to avoid $R^2$ is multiple regression, where adding irrelevant predictor variables to the model can in some cases increase $R^2$. This can be addressed by using the [adjusted $R^2$](http://en.wikipedia.org/wiki/Coefficient_of_determination#Adjusted_R2) value instead, calculated as
$\bar{R}^2 = 1 - (1-R^2)\frac{n-1}{n-p-1}$ where $n$ is the number of data samples, and $p$ is the number of regressors not counting the constant term.
| null | CC BY-SA 3.0 | null | 2011-07-20T20:43:32.047 | 2012-01-09T18:55:47.337 | 2012-01-09T18:55:47.337 | 919 | 5473 | null |
13316 | 2 | null | 13314 | 18 | null | When you have a single predictor $R^{2}$ is exactly interpreted as the proportion of variation in $Y$ that can be explained by the linear relationship with $X$. This interpretation must be kept in mind when looking at the value of $R^2$.
You can get a large $R^2$ from a non-linear relationship only when the relationship is close to linear. For example, suppose $Y = e^{X} + \varepsilon$ where $X \sim {\rm Uniform}(2,3)$ and $\varepsilon \sim N(0,1)$. If you do the calculation of
$$ R^{2} = {\rm cor}(X, e^{X} + \varepsilon)^{2} $$
you will find it to be around $.914$ (I only approximated this by simulation) despite that the relationship is clearly not linear. The reason is that $e^{X}$ looks an awful lot like a linear function over the interval $(2,3)$.
| null | CC BY-SA 3.0 | null | 2011-07-20T20:44:40.037 | 2011-07-20T20:50:41.637 | 2011-07-20T20:50:41.637 | 4856 | 4856 | null |
13317 | 2 | null | 13314 | 320 | null | To address the first question, consider the model
$$Y = X + \sin(X) + \varepsilon$$
with iid $\varepsilon$ of mean zero and finite variance. As the range of $X$ (thought of as fixed or random) increases, $R^2$ goes to 1. Nevertheless, if the variance of $\varepsilon$ is small (around 1 or less), the data are "noticeably non-linear." In the plots, $var(\varepsilon)=1$.
Incidentally, an easy way to get a small $R^2$ is to slice the independent variables into narrow ranges. The regression (using exactly the same model) within each range will have a low $R^2$ even when the full regression based on all the data has a high $R^2$. Contemplating this situation is an informative exercise and good preparation for the second question.
Both the following plots use the same data. The $R^2$ for the full regression is 0.86. The $R^2$ for the slices (of width 1/2 from -5/2 to 5/2) are .16, .18, .07, .14, .08, .17, .20, .12, .01, .00, reading left to right. If anything, the fits get better in the sliced situation because the 10 separate lines can more closely conform to the data within their narrow ranges. Although the $R^2$ for all the slices are far below the full $R^2$, neither the strength of the relationship, the linearity, nor indeed any aspect of the data (except the range of $X$ used for the regression) has changed.
(One might object that this slicing procedure changes the distribution of $X$. That is true, but it nevertheless corresponds with the most common use of $R^2$ in fixed-effects modeling and reveals the degree to which $R^2$ is telling us about the variance of $X$ in the random-effects situation. In particular, when $X$ is constrained to vary within a smaller interval of its natural range, $R^2$ will usually drop.)
The basic problem with $R^2$ is that it depends on too many things (even when adjusted in multiple regression), but most especially on the variance of the independent variables and the variance of the residuals. Normally it tells us nothing about "linearity" or "strength of relationship" or even "goodness of fit" for comparing a sequence of models.
Most of the time you can find a better statistic than $R^2$. For model selection you can look to AIC and BIC; for expressing the adequacy of a model, look at the variance of the residuals.
This brings us finally to the second question. One situation in which $R^2$ might have some use is when the independent variables are set to standard values, essentially controlling for the effect of their variance. Then $1 - R^2$ is really a proxy for the variance of the residuals, suitably standardized.
| null | CC BY-SA 3.0 | null | 2011-07-20T21:35:25.190 | 2011-07-20T21:35:25.190 | null | null | 919 | null |
13318 | 1 | null | null | 4 | 2576 | I need to calculate effect size for a mixed ANOVA design. I need it because I'm using GPower3 to calculate a rough sample size for a full study. I have pilot data. Based on this page:
[http://www.xlstat.com/en/features/statistical-power-analysis-of-variance-anova.htm](http://www.xlstat.com/en/features/statistical-power-analysis-of-variance-anova.htm)
and the GPower3 software, it looks like there are 3 ways to calculate an effect size:
- eta²
- use variances
- use group means
Is there any rational for picking one over another? Since I have the pilot data, I guess I was just going to use the variances to calculate the effect size. The formula I have for that is:
f (effect size) = sqrt(Variance Explained / Variance Error)
Based on the summation rule:
Total Variance = Variance Explained + Variance Error
Is this correct?
If I have unequal group sizes, could I use the pooled variance formula for the Variance Error term?
I also noticed that this effect size formula using the variances is quite similar to the F statistic:
F = Variance Explained / Variance Error
Is this correct and/or is there a relationship between them?
Thanks in advance.
| Effect size calculations in mixed ANOVA design for power analysis | CC BY-SA 3.0 | null | 2011-07-20T22:05:49.523 | 2011-07-21T14:00:25.303 | 2011-07-21T01:14:50.207 | 183 | 5491 | [
"anova",
"variance",
"statistical-power",
"effect-size"
] |
13319 | 1 | null | null | 2 | 438 | Are there any packages that supports weighted average semi-parametric regressions in R? An example of such a regression is in the links below.
I see that there is package [GAM](http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBoQFjAA&url=http://eprints.lse.ac.uk/24504/1/dp599.pdf&ei=FOYoTqo0zdOAB8qn8Z8L&usg=AFQjCNGx4pn6837FopcKifg8qwK4d2Setg) in R for Generalized Additive Models. However, it is not clear to me whether this procedure is the same as weighted average semi-parametric regression. There is also a package called [SPM](http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-nonparametric-regression.pdf)
Some background on semi-parametric methods as used by Connor is [here](http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBoQFjAA&url=http://eprints.lse.ac.uk/24504/1/dp599.pdf&ei=FOYoTqo0zdOAB8qn8Z8L&usg=AFQjCNGx4pn6837FopcKifg8qwK4d2Setg). This paper refers extensively to improvements in the methodology vs. a prior paper by Connor [here](http://sticerd.lse.ac.uk/dps/em/Em506.pdf)
Also, attached is a paper that discusses [non-parametric regressions in R](http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-nonparametric-regression.pdf)
Thank you
| Weighted average semi-parametric regression in R | CC BY-SA 3.0 | null | 2011-07-20T22:10:49.667 | 2011-07-29T20:23:35.010 | 2011-07-29T20:23:35.010 | 8101 | 8101 | [
"r",
"regression",
"algorithms",
"nonparametric"
] |
13320 | 1 | 13330 | null | 5 | 540 | After doing some reading on stochastic processes for work I've found that a proof that the specific process exists is often one of the first things presented.
Could someone please explain, in layman's terms, the purpose/necessity of this proof?
| Meaning of the "existence" proof | CC BY-SA 3.0 | null | 2011-07-21T00:37:56.327 | 2011-07-22T22:56:08.763 | 2011-07-21T00:51:16.313 | 601 | 1913 | [
"stochastic-processes"
] |
13321 | 1 | null | null | 2 | 115 | The basic problem I have is very similar to a classic Bayesian estimation problem: there is a real valued parameters $p\in[0,1]$, that I would like to estimate via observations $o_i$.
It is also the case that
$$
p=\lim_{N\to\infty}{1\over N}\sum_{i=1}^N o_i,
$$
so this problem is very closely analogous to estimating a probability.
Here is the catch: $o_i\in[-1,1]$
Due to the similarity with estimating a probability, I would like to do something like a Bayesian update about my knowledge of $p$, where the prior would be a Beta distributed RV. However, since $p\in[0,1]$ it would seem like the prior would only have support in $[0,1]$, but that would mean that any $o_i\in[-1,0]$ would have not lead to any useful information about $p$, which is simply incorrect.
On the other hand, if I have the prior have support over the whole $[-1,1]$ interval and perform the usual Bayesian update, the posterior distribution will indicate $\Pr(p\in[-1,0])\not=0$, which is also incorrect.
What is the appropriate prior and update rule in this case? Has anyone seen problems similar to this one?
| Bayesian estimation of a mean with tighter constraints than the observations | CC BY-SA 3.0 | null | 2011-07-21T01:15:33.293 | 2016-09-12T19:30:49.557 | 2016-09-12T19:30:49.557 | 28666 | 5493 | [
"bayesian",
"mean"
] |
13322 | 1 | 13324 | null | 4 | 3247 | I'm reading a paper by Stephane Adjémian on DSGE modeling with a zero lower bound for the nominal interest rate, and he's using what he describes as the simulated method of moments / extended path. Has anyone worked with these techniques? What would you say is the next step toward gaining familiarity with them for someone who has a bit of a background in GMM estimation.
I know that the main paper is McFadden (1989), but does anyone know of a textbook treatment of this material? I'd like to avoid having to worry about probability theory if possible.
| What's a good introduction to simulated method of moments and the extended path technique? | CC BY-SA 3.0 | null | 2011-07-21T02:33:09.417 | 2018-02-26T18:06:02.137 | 2018-02-26T18:06:02.137 | 53690 | 2251 | [
"references",
"simulation",
"method-of-moments"
] |
13323 | 1 | 13325 | null | 2 | 431 | I'm trying to finish a proof for a review exercise and I'm asked to show that
$E\left[(y-E(y|x))(E(y|x)-f(x))\right]=0$
where $y$ is the dependent variable and $f(x)$ is a linear predictor of $y$.
I'm almost finished, but I just want to check whether or not
$E\left[ E \left[ y E(y|x)\,|\,x \right] \right]=E\left[ E(y|x)E \left[ E(y|x)\,|\,x \right] \right]$
Basically, can I take $y$ out of the conditional expectation as $E(y|x)$, or worded differently, are conditional expectations multiplacative in this way?
Ordinarily you can back a function out of the expectation if it is a function of the variable being conditioned on - i.e., $E\left[ f(x)y|x \right]=f(x)E(y|x)$, but I think what I'm trying to do above is different.
Also, a second, related question – if $E\left[E(y|x)\right]=E(y)$ by the Law of Total Expectations, then does $E\left[E(y|x)E(y|x)\right]=E(y^2)$ by the same token?
| How to show that $E\left[ y E(y|x) \right] = E\left[ E(y|x)E(y|x) \right]$ - linearity of conditional expectations? | CC BY-SA 3.0 | null | 2011-07-21T02:58:29.200 | 2013-03-22T18:39:55.880 | 2011-07-21T06:53:51.567 | 2251 | 2251 | [
"econometrics",
"expected-value",
"conditional-expectation"
] |
13324 | 2 | null | 13322 | 7 | null | Chapter 5: [http://press.princeton.edu/titles/8434.html](http://press.princeton.edu/titles/8434.html)
Chapter 12: (Cameron&Trivedi textbook)
Chapter 15: (Greene's 7th edition)
This one discusses the general issues and is freely available: [http://elsa.berkeley.edu/books/choice2.html](http://elsa.berkeley.edu/books/choice2.html)
And I also recall that the handbook of econometrics chapter on simulation (40) is particularly readable.
Edited to remove extra links as only 2 are allowed.
| null | CC BY-SA 3.0 | null | 2011-07-21T03:39:50.040 | 2011-07-21T03:39:50.040 | null | null | 5494 | null |
13325 | 2 | null | 13323 | 7 | null | You answer your own question. There is nothing different than backing out function of $x$ from the conditional expectation conditioned on $x$.
Introduce definition
$$g(x)=E(y|x)$$
Then
$$Eyg(x)=E[E[yg(x)|x]]=E[g(x)E[y|x]]=E[(E[y|x])^2]$$
And we get your result. However your last statement is false. $E(y|X)$ is a random variable which is different from $y$, so its second moment should not be in general the same as $y$. To see that condition let $x$ be the random variable $1_A(w)$, where set $A\subset \Omega$ and $1_A$ is the set indicator function. Then
$$E(y|x)=\frac{Ey1_A}{P(A)}1_A+\frac{Ey1_{A^c}}{1-P(A)}1_{A^c}$$
where $A^c=\Omega\backslash A$. Then it is clear that $EE(y|x)=Ey$. However
$$[E(y|x)]^2=\left(\frac{Ey1_A}{P(A)}\right)^21_A+\left(\frac{Ey1_{A^c}}{1-P(A)}\right)^21_{A^c}$$
and taking the expectation we see that it is not equal to $Ey^2$. In fact $Ey^2$ can even be undefined, yet $E(E(y|x))^2$ exists.
| null | CC BY-SA 3.0 | null | 2011-07-21T06:51:34.367 | 2011-07-21T07:16:56.460 | 2011-07-21T07:16:56.460 | 2116 | 2116 | null |
13326 | 1 | 13328 | null | 23 | 17079 | Is it OK to use the Kolmogorov-Smirnov goodness-of-fit test to compare two empirical distributions to determine whether they appear to have come from the same underlying distribution, rather than to compare one empirical distribution to a pre-specified reference distribution?
Let me try asking this another way. I collect N samples from some distribution at one location. I collect M samples at another location. The data is continuous (each sample is a real number between 0 and 10, say) but not normally distributed. I want to test whether these N+M samples all come from the same underlying distribution. Is it reasonable to use the Kolmogorov-Smirnov test for this purpose?
In particular, I could compute the empirical distribution $F_0$ from the $N$ samples, and the empirical distribution $F_1$ from the $M$ samples. Then, I could compute the Kolmogorov-Smirnov test statistic to measure the distance between $F_0$ and $F_1$: i.e., compute $D = \sup_x |F_0(x) - F_1(x)|$, and use $D$ as my test statistic as in the Kolmogorov-Smirnov test for goodness of fit. Is this a reasonable approach?
(I read elsewhere that the Kolmogorov-Smirnov test for goodness of fit [is not valid for discrete distributions](https://stats.stackexchange.com/questions/1047/is-kolmogorov-smirnov-test-valid-with-discrete-distributions), but I admit I don't understand what this means or why it might be true. Does that means my proposed approach is a bad one?)
Or, do you recommend something else instead?
| Can I use Kolmogorov-Smirnov to compare two empirical distributions? | CC BY-SA 3.0 | null | 2011-07-21T07:19:58.793 | 2017-10-07T18:05:48.463 | 2017-10-07T18:05:48.463 | 128677 | 2921 | [
"hypothesis-testing",
"distributions",
"kolmogorov-smirnov-test"
] |
13327 | 1 | 13338 | null | 6 | 1927 | I have a hypothetical set of diet composition, showing how much of each fruit does a particular group of people eat something like:
- How can I visualize in one page, all the compositions for groups A to G and more importantly, their change over the past year?
- I know having more than 4 colors in design is a no-no, but what can I do here?
| How do I visualize changes in proportions compared to another period? | CC BY-SA 3.0 | null | 2011-07-21T07:22:45.950 | 2011-07-22T13:49:11.357 | 2011-07-21T21:04:09.540 | 919 | 5495 | [
"data-visualization"
] |
13328 | 2 | null | 13326 | 28 | null | That is OK, and quite reasonable. It is referred to as the [two-sample Kolmogorov-Smirnov test](http://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Smirnov_test#Two-sample_Kolmogorov.E2.80.93Smirnov_test). Measuring the difference between two distribution functions by the supnorm is always sensible, but to do a formal test you want to know the distribution under the hypothesis that the two samples are independent and each i.i.d. from the same underlying distribution. To rely on the usual asymptotic theory you will need continuity of the underlying common distribution (not of the empirical distributions). See the Wikipedia page linked to above for more details.
In R, you can use the [ks.test](http://www.stat.psu.edu/~dhunter/R/html/stats/html/ks.test.html), which computes exact $p$-values for small sample sizes.
| null | CC BY-SA 3.0 | null | 2011-07-21T08:25:01.413 | 2011-07-21T17:48:04.780 | 2011-07-21T17:48:04.780 | 4376 | 4376 | null |
13329 | 1 | 15417 | null | 15 | 5474 | Does anyone know how to compute the prediction interval for ridge regression? and what's its relation with prediction interval of OLS regression?
| Calculate prediction interval for ridge regression? | CC BY-SA 3.0 | null | 2011-07-21T08:39:45.820 | 2022-06-06T13:58:14.863 | 2011-07-21T09:20:20.373 | null | 5496 | [
"ridge-regression",
"prediction-interval"
] |
13330 | 2 | null | 13320 | 9 | null | Existence proofs are notoriously difficult to justify and appreciate. One reason is that they don't seem to have any consequences. Whether or not you read and understand an existence proof does not change your subsequent work with a given stochastic process because that work relies on properties of the process and not the fact that it exists.
The purpose of existence proofs is to justify the mathematical foundation of working with a given process. Without these proofs it is like in Star Trek. They can do all sorts of cool things, but in reality the materials and technology do not exist. Hence it is all fiction. If your stochastic process does not exists, that is, if there is no mathematical object with a given set of properties, your subsequent work is all fiction. So it is a fundamental step in science to justify that we are doing science and not science fiction.
Edit: In response to @Nick's comment. When it comes to stochastic processes it is, indeed, of central importance to be able to produce an infinite dimensional measure from a consistent family of finite dimensional distributions. This is what Kolmogorov's consistency theorem is about. This theorem, or a variation of it, often lurks in the background even if it is not used explicitly. For instance, a stochastic process could be a solution to a stochastic differential equation (SDE) involving Brownian motion. Then the existence of the process is a question about whether the SDE has a solution - based on the existence of Brownian motion. The existence of Brownian motion can be proved using Kolmogorov's consistency theorem.
There are, however, alternative ways to obtain existence of processes. My favorite alternative is via compactness arguments.
I kept my reply above non-specific because it goes for all other mathematical topics as well. The existence of, for instance, the uniform distribution on $[0,1]$, which is absolutely non-trivial. And the existence of the real numbers for that matter. I wonder how many on this site have proved that the real numbers exist, or do we just take for granted that we can "fill out the gaps" in the rational numbers to get completeness?
N.B. Just to clarify, by the first line in my original reply it was not my intention to imply that an existence proof can not be justified or appreciated, but to a newcomer it may look like a soccer match where the players are given an object and the referee says it's a ball, but then the players use the entire first half to clarify that, indeed, it is a ball before they start playing.
| null | CC BY-SA 3.0 | null | 2011-07-21T08:49:27.393 | 2011-07-22T22:56:08.763 | 2011-07-22T22:56:08.763 | 4376 | 4376 | null |
13331 | 1 | null | null | 8 | 6066 | Assume that I'm going to estimate a linear regression where I assume $u\sim N(0,\sigma^2)$. What is the benefit of OLS against ML estimation? I know that we need to know a distribution of $u$ when we use ML Methods, but since I assume $u\sim N(0,\sigma^2)$ whether I use ML or OLS this point seems to be irrelevant. Thus the only advantage of OLS should be in the asymptotic features of the $\beta$ estimators. Or do we have other advantages of the OLS method?
| Estimating linear regression with OLS vs. ML | CC BY-SA 3.0 | null | 2011-07-21T10:06:49.620 | 2012-12-09T14:10:09.690 | 2011-07-21T10:58:25.503 | 2116 | 4496 | [
"regression",
"least-squares",
"linear-model"
] |
13332 | 2 | null | 13331 | 14 | null | Using the usual notations, the log-likelihood of the ML method is
$l(\beta_0, \beta_1 ; y_1, \ldots, y_n) = \sum_{i=1}^n \left\{ -\frac{1}{2} \log (2\pi\sigma^2) - \frac{(y_{i} - (\beta_0 + \beta_1 x_{i}))^{2}}{2 \sigma^2} \right\}$.
It has to be maximised with respect to $\beta_0$ and $\beta_1$.
But, it is easy to see that this is equivalent to minimising
$\sum_{i=1}^{n} (y_{i} - (\beta_0 + \beta_1 x_{i}))^{2} $.
Hence, both ML and OLS lead to the same solution.
More details are provided in these [nice lecture notes](http://data.princeton.edu/wws509/notes/c2s2.html).
| null | CC BY-SA 3.0 | null | 2011-07-21T10:23:57.200 | 2011-07-21T10:36:47.723 | 2011-07-21T10:36:47.723 | 3019 | 3019 | null |
13334 | 1 | 13350 | null | 4 | 4007 | I am using SAS to estimate some logistic models. Usually, I work with either MDs or social scientists, and odds ratios are the preferred metric. But I am now working with a client in economics/law and she wants the marginal effects and their standard errors, and she wants them at the means of the other variables.
This isn't easy in SAS, but, with help from tech support, I found that you can do this with PROC NLMIXED, I believe you then need the out = der option. Something like this
```
proc nlmixed data=olivia.small;
p=1/(1+exp(-(Intercept+ba*log_fund_age + bb*log_fund_size + bc*yield + bd*loaded + be*log_assets)));
model vote_code_num ~ binomial(1,p);
parms intercept 36.43 ba -14.55 bb -0.98 bc -0.37 bd 2.2 be -0.07;
predict p*(1-p)*ba out=a der;
predict p*(1-p)*bb out=b der;
predict p*(1-p)*bc out=c der;
predict p*(1-p)*bd out=d der;
predict p*(1-p)*be out=e der;
where year = 2003;
run;
```
but then the output data sets a, b, c, d and e have the derivatives and their standard deviations for each observation in the data set, not for the mean of the other variables. It's easy to find the mean of all those derivatives, but 1) Is that the same as the marginal at the mean of the other variables? and 2) How then to get the standard errors?
Peter
| Marginal effects of a logistic model, and their standard errors | CC BY-SA 3.0 | null | 2011-07-21T11:11:33.160 | 2014-10-30T15:20:14.220 | null | null | 686 | [
"logistic",
"sas",
"marginal-distribution"
] |
13335 | 1 | null | null | 7 | 2056 | I'm currently trying to predict the probability for low probability events (~1%).
I have large DB with ~200,000 vectors (~2000 plus examples) with ~200 features.
I'm trying to find the the best features for my problem. What are the recommended method? (preferred in Python or R, but not necessarily)
| Feature selection for low probability event prediction | CC BY-SA 4.0 | null | 2011-07-21T12:41:02.757 | 2019-01-29T11:40:00.723 | 2019-01-29T11:40:00.723 | 128677 | 5497 | [
"machine-learning",
"feature-selection",
"unbalanced-classes"
] |
13336 | 2 | null | 13334 | 6 | null | Some people would find such 'marginal effects' difficult to interpret and non-unique. There are other ways to get 'marginal effects' in binary logistic regression. Because of non-collapsibility of the odds ratio, marginal estimates are not well defined in general, and they can represent quantities that are not weighted averages over the factors you are unconditioning on. Mitch Gail has an example where the partial odds ratio for an exposure x2 is 9 for both x1=0 and x1=1 but is 5.44 when not holding x1 constant.
```
@ARTICLE{gai84bia,
author = {Gail, M.H. and Wieand, S. and Piantadosi, S.},
year = 1984,
title = {Biased estimates of treatment effect in randomized experiments with
nonlinear regressions and omitted covariates},
journal = Biometrika,
volume = 71,
pages = {431-444},
annote = {covariable adjustment;bias if omitted covariables and model is
nonlinear}
}
```
I wonder also whether you meant 'marginal effect' or 'effect on the original scale'. That would involve two different considerations. Effects on the log odds scale are easier to deal with, and you can relate odds ratios to absolute risk changes (as a function of starting risk) using a simple chart.
| null | CC BY-SA 3.0 | null | 2011-07-21T12:41:20.020 | 2014-10-30T15:20:14.220 | 2014-10-30T15:20:14.220 | 25138 | 4253 | null |
13337 | 2 | null | 13335 | 7 | null | My first advice would be that unless identifying the informative features is a goal of the analysis, don't bother with feature selection and just use a regularised model, such a penalised logistic regression, ridge regression or SVM, and let the regularisation handle the over-fitting. It is often said that feature selection improves classifier performance, but it isn't always true.
To deal with the class imbalance problem, give different weights to the patterns from each class in calculating the loss function used to fit the model. Choose the ratio of weights by cross-validation (for a probabilistic classifier you can work out the asymptically optimal weights, but it generally won't give optimal results on a finite sample). If you are using a classifier that can't give different weights to each class, then sub-sample the majority class instead, where again the ratio of positive and negative patterns is determined by cross-validation (make sure the test partition in each fold of the cross-validation procedure has the same relative class frequencies you expect to see in operation).
Lastly, it is often the case in practical application with a class imbalance that false-positives and false-negatives are not of equal seriousness, so incorporate this into the construction of the classifier.
| null | CC BY-SA 3.0 | null | 2011-07-21T13:17:33.747 | 2011-07-21T13:17:33.747 | null | null | 887 | null |
13338 | 2 | null | 13327 | 5 | null | What is more important for you - between group comparison or the intra-group composition? For the former, a parallel coordinates plot seems to be a natural choice: [http://charliepark.org/slopegraphs/](http://charliepark.org/slopegraphs/)
For the latter, a time series of percent stacked charts might look fine - you do not have to use 7 colors, just alternate them to emphasize the pattern.
| null | CC BY-SA 3.0 | null | 2011-07-21T13:34:54.837 | 2011-07-21T13:34:54.837 | null | null | 5494 | null |
13339 | 2 | null | 13318 | 2 | null | I am not familiar with Gpower3 so this does not exactly address your question, but here is another approach that you may want to consider.
For power and sample size studies more complicated than simple 2 sample tests I prefer to use simulation rather than prepackaged routines. The basic steps are:
- Think through what your data will look like (structure, means, variances, distributions, ...)
- Think through how you will analyse the data
- Generate data based on current assumptions and analyze it
- Repeate step 3 many times and compute the percentage of times that the null is rejected (this is the power)
- Repeate steps 3 and 4 for different sets of assumptions
One of the advantages of this approach is that you know exactly what assumptions are being made and what the nature of the data and analysis is. The main disadvantage is that it takes a bit more work and time than just plugging in a few numbers to a canned routine.
| null | CC BY-SA 3.0 | null | 2011-07-21T14:00:25.303 | 2011-07-21T14:00:25.303 | null | null | 4505 | null |
13340 | 1 | 13343 | null | 23 | 11503 | I am interested in testing a simple mediation model with one IV, one DV, and one mediator. The indirect effect is significant as tested by the Preacher and Hayes SPSS macro, which suggests the mediator does serve to statistically mediate the relationship.
When reading about mediation I have read things such as "Note that a mediational model is a causal model." - [David Kenny](http://davidakenny.net/cm/mediate.htm). I can certainly appreciate the use of mediation models as causal models, and indeed, if a model is theoretically sound, I can see this as very useful.
In my model, however, the mediator (a trait considered to be a diathesis for anxiety disorders) is not caused by the independent variable (symptoms of an anxiety disorder). Rather, the mediator and independent variables are related, and I believe the association between the independent variable and the dependent variable can be explained largely by variance between the IV-mediator-DV. In essence I am trying to demonstrate that previous reports of the IV-DV relationship can be explained by a related mediator that is not caused by the IV.
Mediation is useful in this case because it explains how the IV-DV relationship can be statistically explained by the IV-Mediator-DV relationship. My problem is the question of causation. Could a review come back and tell us that the mediation is not appropriate because the IV does not in fact cause the mediator (which I would have never argued in the first place)?
Does this make sense? Any feedback on this matter would be greatly appreciated!
Edit: What I mean to say is that X is correlated with Y not because it causes Y, but because Z causes Y (partially) and because X and Z are highly correlated. A bit confusing, but that is it. The causal relationships in this instance are not really in question and this manuscript is not so much about causation. I simply seek to demonstrate that variance between X and Y can be explained by variance between Z and Y. So basically, that X is correlated indirectly to Y through Z (the "mediator" in this case).
| Are mediation analyses inherently causal? | CC BY-SA 3.0 | null | 2011-07-21T15:00:49.527 | 2019-02-05T20:12:24.137 | 2011-07-22T00:38:18.910 | 183 | 3262 | [
"causality",
"mediation"
] |
13342 | 1 | 13418 | null | 11 | 7941 | Say I have a 1-dimensional continuous random variable $X$, with PDF $f(X)$, CDF $F(X)$ and inverse CDF $F^{-1}$. What is the best way to discretize $X$?
To keep things clear, let $Y$ denote the discretized version of $X$, and let $g(Y)$, $G(Y)$ and $G^{-1}$ refer to the PMF, CDF and inverse CDF of $Y$.
Ideally, I would like for $E[X] = E[Y]$ and $Var(X) = Var(Y)$. However, I understand if these properties cannot hold for finite $N$ due to discretization error.
Some thoughts:
- Should I just discretize $X$ into $N$ equally spaced points? Or should I discretize more at regions of high probability, by first discretizing a Uniform $[0,1]$ into equidistant points $u_1,u_2...u_N$ and then generating discrete values $Y_1 = F^{-1}(u_1), Y_2 = F^{-1}(u_2)$ and $Y_N = F^{-1}(u_N)$
- Say I have discretized values $Y_1, Y_2... Y_N$. What is the best way to create $g(Y)$, $G(Y)$ and $G^{-1}$ from these values? My thoughts here are that each $Y_k$ corresponds to an interval of $X$ with width $w_k$, and that the $Y_k$ should be the "midpoints" of these intervals. That is, $g(Y_k) = p(Y_k - w_k < X < Y_k + w_k) = F(Y_k + w_k) - F(Y_k - w_k)$.
| What is the best way to discretize a 1D continuous random variable? | CC BY-SA 3.0 | null | 2011-07-21T16:04:06.377 | 2012-03-24T01:27:55.750 | 2011-07-22T19:36:05.167 | 3572 | 3572 | [
"continuous-data",
"density-function",
"discrete-data",
"cumulative-distribution-function"
] |
13343 | 2 | null | 13340 | 20 | null | A. "Mediation" conceptually means causation (as Kenny quote indicates). Path models that treat a variable as a mediator thus mean to convey that some treatment is influencing an outcome variable through its effect on the mediator, variance in which in turn causes the outcome to vary. But modeling something as a "mediator" doesn't mean it really is a mediator--this is the causation issue. Your post & comment in response to Macro suggest that you have in mind a path analysis in which a variable is modeled as a mediator but isn't viewed as "causal"; I'm not quite seeing why, though. Are you positing that the relationship is spurious--that there is some 3rd variable that is causing both the "independent variable" and the "mediator"? And maybe that both the "independent variable" & the "mediator" in your analysis are in fact mediators of the 3rd variable's influence on the outcome variable? If so, then a reviewer (or any thoughtful person) will want to know what the 3rd variable is & what evidence you have that it is responsible for spurious relationships between what are in fact mediators. This will get you into issues posed by Macro's answer.
B. To extend Macro's post, this is a notorious thicket, overgrown with dogma and scholasticism. But here are some highlights:
- Some people think that you can only "prove" mediation if you experimentally manipulate the mediator as well as the influence that is hypothesized to exert the causal effect. Accordingly, if you did an experiment that manipulated only the causal influence & observed that its impact on the outcome variable was mirrored by changes in the mediator, they'd so "nope! Not good enough!" Basically, though, they just don't think observational methods ever support causal inferences & unmanipulated mediators in experiments are just a special case for them.
- Other people, who don't exclude causal inferences from observational studies out of hand, nevertheless believe that if you use really really really complicated statistical methods (including but not limited to structural equation models that compare the covariance matrix for the posited mediating relationship with those for various alternatives), you can effectively silence the critics I just mentioned. Basically this is Baron & Kenny, but on steroids. Empirically speaking, they haven't silenced them; logically, I don't see how they could.
- Still others, most notably, Judea Pearl, say that the soundness of causal inferences in either experimental or observational studies can never be proven w/ statistics; the strength of the inference inheres in the validity of the design. Statistics only confirm the effect causal inference contemplates or depends on.
Some readings (all of which are good, not dogmatic or scholastic):
- Green, D.P., Ha, S.E. & Bullock, J.G. Enough Already about “Black
Box” Experiments: Studying Mediation Is More Difficult than Most
Scholars Suppose. The ANNALS of the American Academy of Political and
Social Science 628, 200-208 (2010).
- Sobel, M.E. Identification of Causal Parameters in Randomized Studies
With Mediating Variables. Journal of Educational and Behavioral
Statistics 33, 230-251 (2008).
- Pearl, J. An Introduction to Causal Inference. The International
Journal of Biostatistics 6, Article 7 (2010).
Last but by no means least, part of a cool exchange between Gelman & Pearl on causal inference in which mediation was central focus: [http://andrewgelman.com/2007/07/identification/](http://andrewgelman.com/2007/07/identification/)
| null | CC BY-SA 3.0 | null | 2011-07-21T16:53:29.940 | 2011-07-21T17:15:36.953 | 2011-07-21T17:15:36.953 | 11954 | 11954 | null |
13344 | 2 | null | 13321 | 1 | null | What if you define $q=\frac{p+1}{2}$ and use a beta prior on $q$? Then $p$ can be in the $-1$ to $1$ range, but $q$ conforms to the support of the beta distribution.
| null | CC BY-SA 3.0 | null | 2011-07-21T17:03:29.233 | 2011-07-21T17:03:29.233 | null | null | 4505 | null |
13345 | 2 | null | 13331 | 4 | null | You are focusing on the wrong part of the concept in your question. The beauty of least squares is that it gives a nice easy answer regaurdless of the distribution, and if the true distribution happens to be normal, then it is the maiximum likelihood answer as well (I think this is the Gauss-Markov thereom). When you have a distribution other than the normal then ML and OLS will give different answers (but if the true distribution is close to normal then the answers will be similar).
| null | CC BY-SA 3.0 | null | 2011-07-21T17:35:23.680 | 2012-07-17T17:16:39.467 | 2012-07-17T17:16:39.467 | 4505 | 4505 | null |
13346 | 1 | 13349 | null | 19 | 4353 | A [Skellam Distribution](http://en.wikipedia.org/wiki/Skellam_distribution) describes the difference between two variables that have Poisson distributions. Is there a similar distribution that describes the difference between variables that follow negative binomial distributions?
My data is produced by a Poisson process, but includes a fair amount of noise, leading to overdispersion in the distribution. Thus, modeling the data with a negative binomial (NB) distribution works well. If I want to model the difference between two of these NB data sets, what are my options? If it helps, assume similar means and variance for the two sets.
| Distribution that describes the difference between negative binomial distributed variables? | CC BY-SA 3.0 | null | 2011-07-21T19:08:22.733 | 2016-11-18T14:55:44.030 | 2016-06-29T13:54:53.473 | 1352 | 54 | [
"distributions",
"modeling",
"poisson-distribution",
"negative-binomial-distribution",
"skellam-distribution"
] |
13347 | 1 | 13373 | null | 3 | 367 | I have a series of continuous probability distributions and the intervals for each distribution of possible values. Now I want to use the distributions to distribute values of a random variable among agents. I have thought of one way to do this, and I'm curious if this is an acceptable method. This is how I would do it:
```
Let P(A<X<B) be a given pdf, where A,B are min and max values for random
variable X.
Divide the spread between A and B into an arbitrarily large amount of 'chunks'
or ranges.
For each chunk, represented as a range, say between a and b, make a table
where P(a<X<b) relates to the range a,b.
For each agent receiving a value of random variable X, use a random number
generator, look up the result in the table, give the agent a value in the
range (a,b).
```
Assuming that this is an acceptable technique, I'm still curious as to what an acceptable amount of granularity for the 'chunks' would be. I hope this wan't too confusing and any feedback would be much appreciated.
edit: All of my functions take the form of f(x) = a/(b*x)+c over some interval. They are all easily integrable.
| Distributing values according to a given continuous probability distribution | CC BY-SA 3.0 | null | 2011-07-21T19:23:00.187 | 2012-06-04T14:40:38.060 | 2012-06-04T14:27:37.157 | 4856 | 5487 | [
"distributions",
"probability",
"algorithms"
] |
13349 | 2 | null | 13346 | 22 | null | I don't know the name of this distribution but you can just derive it from the law of total probability. Suppose $X, Y$ each have negative binomial distributions with parameters $(r_{1}, p_{1})$ and $(r_{2}, p_{2})$, respectively. I'm using the parameterization where $X,Y$ represent the number of successes before the $r_{1}$'th, and $r_{2}$'th failures, respectively. Then,
$$ P(X - Y = k) = E_{Y} \Big( P(X-Y = k) \Big) = E_{Y} \Big( P(X = k+Y) \Big) =
\sum_{y=0}^{\infty} P(Y=y)P(X = k+y) $$
We know
$$ P(X = k + y) = {k+y+r_{1}-1 \choose k+y} (1-p_{1})^{r_{1}} p_{1}^{k+y} $$
and
$$ P(Y = y) = {y+r_{2}-1 \choose y} (1-p_{2})^{r_{2}} p_{2}^{y} $$
so
$$ P(X-Y=k) = \sum_{y=0}^{\infty} {y+r_{2}-1 \choose y} (1-p_{2})^{r_{2}} p_{2}^{y} \cdot
{k+y+r_{1}-1 \choose k+y} (1-p_{1})^{r_{1}} p_{1}^{k+y} $$
That's not pretty (yikes!). The only simplification I see right off is
$$ p_{1}^{k} (1-p_{1})^{r_{1}} (1-p_{2})^{r_{2}}
\sum_{y=0}^{\infty} (p_{1}p_{2})^{y} {y+r_{2}-1 \choose y}
{k+y+r_{1}-1 \choose k+y} $$
which is still pretty ugly. I'm not sure if this is helpful but this can also be re-written as
$$ \frac{ p_{1}^{k} (1-p_{1})^{r_{1}} (1-p_{2})^{r_{2}} }{ (r_{1}-1)! (r_{2}-1)! }
\sum_{y=0}^{\infty}
(p_{1}p_{2})^{y}
\frac{ (y+r_{2}-1)! (k+y+r_{1}-1)! }{y! (k+y)! } $$
I'm not sure if there is a simplified expression for this sum but it could be approximated numerically if you only need it to calculate $p$-values
I verified with simulation that the above calculation is correct. Here is a crude R function to calculate this mass function and carry out a few simulations
```
f = function(k,r1,r2,p1,p2,UB)
{
S=0
const = (p1^k) * ((1-p1)^r1) * ((1-p2)^r2)
const = const/( factorial(r1-1) * factorial(r2-1) )
for(y in 0:UB)
{
iy = ((p1*p2)^y) * factorial(y+r2-1)*factorial(k+y+r1-1)
iy = iy/( factorial(y)*factorial(y+k) )
S = S + iy
}
return(S*const)
}
### Sims
r1 = 6; r2 = 4;
p1 = .7; p2 = .53;
X = rnbinom(1e5,r1,p1)
Y = rnbinom(1e5,r2,p2)
mean( (X-Y) == 2 )
[1] 0.08508
f(2,r1,r2,1-p1,1-p2,20)
[1] 0.08509068
mean( (X-Y) == 1 )
[1] 0.11581
f(1,r1,r2,1-p1,1-p2,20)
[1] 0.1162279
mean( (X-Y) == 0 )
[1] 0.13888
f(0,r1,r2,1-p1,1-p2,20)
[1] 0.1363209
```
I've found the sum converges very quickly for all of the values I tried, so setting UB higher than 10 or so
is not necessary. Note that R's built in rnbinom function parameterizes the negative binomial in terms of
the number of failures before the $r$'th success, in which case you'd need to replace all of the $p_{1}, p_{2}$'s
in the above formulas with $1-p_{1}, 1-p_{2}$ for compatibility.
| null | CC BY-SA 3.0 | null | 2011-07-21T20:12:29.753 | 2014-01-09T16:51:28.137 | 2014-01-09T16:51:28.137 | 19999 | 4856 | null |
13350 | 2 | null | 13334 | 2 | null | SAS Tech support solved this with the (now obvious) idea of adding a variable to the data set that has the average of all the variables.
Peter
| null | CC BY-SA 3.0 | null | 2011-07-21T21:54:04.930 | 2011-07-21T21:54:04.930 | null | null | 686 | null |
13351 | 1 | 13354 | null | 3 | 221 | In my application I'm receiving a continuous stream of 50 changing numbers, mostly these samples are contained with a nice Gaussian distribution, but now and then there are two+ distributions within the samples. The samples are always positive and within the range of 0 to 1. If there is more than one distribution present, the distribution I'm always interested has the lowest mean.
So ideally I want to identify the lower bound and roughly the mean of the main distribution. The lower bound is straight-forward, its always the minimum value of the samples, but I'm stumped as to how find roughly the mean of the main distribution when there is more than one distribution present.
The really tricky thing is that its a real time system and I have very little time to process the samples before the next set arrives, so other than finding the min, max, mean and at a push median I don't have time to carry out much more processing.
Anyone have any suggestions for how I could achieve my goals?! Many thanks and sorry for the stats-newbie nature of my question
| Distinguishing two distributions | CC BY-SA 3.0 | null | 2011-07-21T21:58:14.380 | 2011-07-21T22:36:47.140 | 2011-07-21T22:36:47.140 | 5500 | 5500 | [
"distributions",
"confidence-interval",
"normal-distribution"
] |
13352 | 1 | 13363 | null | 3 | 446 | Suppose I have a questionnaire and I ask respondents how often they eat at McDonalds:
- Never
- Less than once a month
- At least once a month but less than once a week
- 1-3 times a week
- More than 3 times a week
I then correlate these answers with whether the respondents are wearing brown shoes.
- Brown 65 -- not brown 38
- Brown 32 -- not brown 62
- Brown 17 -- not brown 53
- Brown 10 -- not brown 48
- Brown 9 -- not brown 6
The thing I can't get my head around is this: If a respondent picks #5, he (statistically) has a higher probability of wearing brown shoes than not. But, in a sense, his response subsumes responses 2-4, and if you accumulate their statistics (ie, "eats at McDonalds sometimes") he has a higher probability of not wearing brown shoes.
Now I realize that there are a bunch of caveats here -- sampling error in the stats, etc. But is there ever a valid argument for "rolling up" the stats (so that the values used in inference for #5 eg, would consist of the sums of the 2-5 values, or some other scheme), or is this concept just a product of my twisted mind?
(Note that I'm not talking about "collapsing" the stats into fewer observations, which I assume would be perfectly valid, but rather adjusting the probabilities that are used in inference, based on the knowledge that the various possible mutually-exclusive observations are on a sliding scale.)
| How do you deal with a "multiple choice" observation in Bayesian inference, when the choices are on a scale? | CC BY-SA 3.0 | null | 2011-07-21T22:02:49.060 | 2011-07-22T11:48:04.797 | null | null | 5392 | [
"bayesian",
"inference"
] |
13353 | 1 | null | null | 9 | 19623 | I will give my examples with R calls. First a simple example of a linear regression with a dependent variable 'lifespan', and two continuous explanatory variables.
```
data.frame(height=runif(4000,160,200))->human.life
human.life$weight=runif(4000,50,120)
human.life$lifespan=sample(45:90,4000,replace=TRUE)
summary(lm(lifespan~1+height+weight,data=human.life))
Call:
lm(formula = lifespan ~ 1 + height + weight, data = human.life)
Residuals:
Min 1Q Median 3Q Max
-23.0257 -11.9124 -0.0565 11.3755 23.8591
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 63.635709 3.486426 18.252 <2e-16 ***
height 0.007485 0.018665 0.401 0.6884
weight 0.024544 0.010428 2.354 0.0186 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.41 on 3997 degrees of freedom
Multiple R-squared: 0.001425, Adjusted R-squared: 0.0009257
F-statistic: 2.853 on 2 and 3997 DF, p-value: 0.05781
```
In order to find the estimate of 'lifespan' when the value of 'weight' is 1, I add (Intercept)+height=63.64319
Now what if I have a similar data frame, but one where one of the explanatory variables is categorical?
```
data.frame(animal=rep(c("dog","fox","pig","wolf"),1000))->animal.life
animal.life$weight=runif(4000,8,50)
animal.life$lifespan=sample(1:10,replace=TRUE)
summary(lm(lifespan~1+animal+weight,data=animal.life))
Call:
lm(formula = lifespan ~ 1 + animal + weight, data = animal.life)
Residuals:
Min 1Q Median 3Q Max
-4.7677 -2.7796 -0.1025 3.1972 4.3691
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.565556 0.145851 38.159 < 2e-16 ***
animalfox 0.806634 0.131198 6.148 8.6e-10 ***
animalpig 0.010635 0.131259 0.081 0.9354
animalwolf 0.806650 0.131198 6.148 8.6e-10 ***
weight 0.007946 0.003815 2.083 0.0373 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.933 on 3995 degrees of freedom
Multiple R-squared: 0.01933, Adjusted R-squared: 0.01835
F-statistic: 19.69 on 4 and 3995 DF, p-value: 4.625e-16
```
In this case, to find the estimate of 'lifespan' when the value of 'weight' is 1, should I add each of the coefficients for 'animal' to the intercept: (Intercept)+animalfox+animalpig+animalwolf? Or what is the proper way to do this?
Thanks
Sverre
| Interpreting coefficient in a linear regression model with categorical variables | CC BY-SA 3.0 | null | 2011-07-21T22:12:17.597 | 2013-11-08T01:30:52.750 | 2012-07-14T13:00:27.373 | 4856 | 5501 | [
"r",
"multiple-regression",
"interpretation",
"regression-coefficients"
] |
13354 | 2 | null | 13351 | 6 | null | When you have a sample where different points potentially come from different distributions, the data generating distribution is often envisaged as a finite mixture. That is, the data comes from a density
$$ f(x) = \sum_{j=1}^{M} p_{j} f_{j}(x) $$
such that $\sum p_{j} = 1$ where each $f_{j}$ is a density. Typically in finite mixture modeling the purpose is to estimate the number of mixture components, $M$, under some assumptions about the constituent densities, $f_{j}$. Also, posterior probabilities that particular observations came from component $j$ are usually of interest, giving a means for classifying the observations.
In the case you described you know a priori there are only two groups, so one could imagine observed data comes from a density $f$ such that
$$ f(x) = p f_{1}(x) + (1-p) f_{0}(x) $$
That is, $p \in (0,1)$ proportion of the samples come from a population with density $f_{1}$ while the rest come from a density $f_{0}$. It may suffice to model your observed data as a mixture of Gaussians with different means (despite them being bounded between 0 and 1). That and more complex procedures related to mixture models can be found in the FlexMix package in R. Here is some documentation:
[http://cran.r-project.org/web/packages/flexmix/vignettes/flexmix-intro.pdf](http://cran.r-project.org/web/packages/flexmix/vignettes/flexmix-intro.pdf)
The package mixtools also appears to be pretty popular as well:
[http://www.stat.psu.edu/~dhunter/papers/mixtools.pdf](http://www.stat.psu.edu/~dhunter/papers/mixtools.pdf)
although I've never used it
| null | CC BY-SA 3.0 | null | 2011-07-21T22:18:04.363 | 2011-07-21T22:23:35.187 | 2011-07-21T22:23:35.187 | 4856 | 4856 | null |
13355 | 2 | null | 13353 | 12 | null | No, you shouldn't add all of the coefficients together. You essentially have the model
$$ {\rm lifespan} = \beta_{0} + \beta_{1} \cdot {\rm fox} + \beta_{2} \cdot {\rm pig} + \beta_{3} \cdot {\rm wolf} + \beta_{4} \cdot {\rm weight} + \varepsilon $$
where, for example, ${\rm pig} = 1$ if the animal was a pig and 0 otherwise. So, to calculate $\beta_{0} + \beta_{1} + \beta_{2} + \beta_{3} + \beta_{4}$ as you've suggested for getting the overall average when ${\rm weight}=1$ is like saying "if you were a pig, a wolf, and a fox, and your weight was 1, what is your expected lifespan?". Clearly since each animal is only one of those things, that doesn't make much sense.
You will have to do this separately for each animal. For example, $\beta_{0} + \beta_{2} + \beta_{4}$ is the expected lifespan for a pig when its weight is 1.
| null | CC BY-SA 3.0 | null | 2011-07-21T22:30:40.173 | 2011-07-22T03:48:00.533 | 2011-07-22T03:48:00.533 | 5020 | 4856 | null |
13356 | 1 | null | null | -1 | 1310 | I'm trying to extend this graph via the provided trendline formula. I'm inserting the formula into the next row and trying to assign X to be the date value, but it doesn't appear to be working. Do I need to apply a function to the date value in order for it to work? Basically I'm trying to predict what June 17th's downloads will be based on the formula.
Anyone know how do to this?
| What does X represent in this linear trendline? | CC BY-SA 3.0 | null | 2011-07-21T22:33:07.437 | 2011-07-22T11:53:49.067 | null | null | 3626 | [
"regression",
"excel",
"linear-model"
] |
13358 | 2 | null | 13356 | 2 | null | $X$ might represent some time ticks starting from $1$ or whatever integer value the Excel assigns to it. Since your downloads are nonnegative, it is not the best way to predict the trend without prior transformation of the response variable, probably some logistic curve would fit it better.
In any case if you still want to do a linear regression model in Excel, you may transform you time points to integers by letting them to start from $1, 2, \dots, T$ and to do the regression using build in formulas (or the [analysis tool pack add-in](http://www.jeremymiles.co.uk/regressionbook/extras/appendix2/excel/)). Probably this [link](http://phoenix.phys.clemson.edu/tutorials/excel/regression.html) or the MS [link](http://office.microsoft.com/en-us/excel-help/perform-a-regression-analysis-HA001111963.aspx) will guide you. I personally as many others on this site would recommend to use either [gretl](http://gretl.sourceforge.net/) or [R](http://www.r-project.org/).
| null | CC BY-SA 3.0 | null | 2011-07-21T23:40:05.443 | 2011-07-21T23:40:05.443 | null | null | 2645 | null |
13359 | 2 | null | 13335 | 2 | null | The problem of estimating probabilities falls under the category of "regression," since the probability is a conditional mean. Classical methods for feature selection (AKA "subset selection" or "model selection") methods for regression include best-k, forward- and backward- stepwise, and forward stagewise, all described in Chapter 3 of [Elements of Statistical Learning](http://www-stat.stanford.edu/~tibs/ElemStatLearn/). However, such methods are generally costly, and given the number of features in your dataset, my choice would be to use [glmpath](http://cran.r-project.org/web/packages/glmpath/glmpath.pdf), which implements L1-regularized regression using a modification of the fantastically efficient LARS algorithm.
EDIT: More details on L1 regularization.
The LARS algorithm produces the entire "Lasso" path for $\lambda$ (the regularization constant) ranging from 0 to $\infty$. At $\lambda=0$, all features are used; at $\lambda=\infty$, none of the features have nonzero coefficients. In between there are values of $\lambda$ for which anywhere from 1 to 199 features are used.
Using the results from LARS one can select the values of $\lambda$ with the best performance (according to whatever criteria). Then, using only the features with nonzero coefficients for a particular $\lambda$, one can then fit an unregularized logistic regression model for the final prediction.
| null | CC BY-SA 3.0 | null | 2011-07-22T00:04:47.163 | 2011-07-22T12:52:48.370 | 2011-07-22T12:52:48.370 | 3567 | 3567 | null |
13360 | 2 | null | 13340 | 14 | null |
### Causality and Mediation
- A mediation model makes theoretical claims about causality.
The model proposes that the IV causes the DV and that this effect is totally or partially explained by a chain of causality whereby the IV causes the MEDIATOR which in turn causes the DV.
- Support for a mediational model does not prove the proposed causal pathway.
Statistical tests of mediation are typically based on observational studies. The range of alternative causal interpretations is large (e.g., third variables, alternative directions, reciprocity, etc.)
I am typically not persuaded by arguments (if any) presented by researchers who propose causal claims implied in mediation models.
- Support for a mediational model may provide evidence to supplement other sources of evidence when building an argument for a causal claim.
In summary, correlation does not prove causation, but it can provide supplementary evidence.
- Despite the limitations of tests of mediation in observational studies, (a) mediation models are good for getting researchers thinking about causal pathways, and (b) there are better and worse ways to write up mediation models, where better ways acknowledge nuances in interpretation and provide thorough theoretical discussion of the evidence both for the proposed causal pathway and alternative causal pathways (see this page of tips that I prepared).
- @dmk38 has provided some excellent references and additional discussion.
### Showing that a variable explains the prediction of another variable
- Based on your description, mediation does NOT appear to be aligned with your research question. As such I would avoid using the language of mediation in your analyses.
- As I understand it, your research question is concerned with whether the prediction of one variable (lets call it X1 instead of IV) on the DV is explained by a second variable (lets call it X2 instead of MEDIATOR).
You may also be making causal claims like X2 causes DV but X1 is only correlated with X2 and does not cause DV.
- There are several statistical tests that might be suitable for testing this research question:
Compare zero-order (X1 with DV) with semi-partial correlations (X1 partialling out X2 with DV). I imagine the interesting element would be the degree of reduction and not so much the statistical significance (although of course you would want to get some confidence intervals on that reduction).
Or similarly, compare incremental R-square of a hierarchical regression where you add X2 in block 1 and X1 in block 2 with the R-square of a model with just X1 predicting DV.
I imagine you could also draw a path diagram that aligned with your causal assumptions (e.g., double headed arrows between X1 and X2 and a single headed arrow between X2 and DV.
| null | CC BY-SA 3.0 | null | 2011-07-22T01:15:41.637 | 2014-05-23T06:01:55.143 | 2014-05-23T06:01:55.143 | 183 | 183 | null |
13361 | 1 | null | null | 1 | 2577 | (cross-posted from the R-sig ME list - maybe this is too basic for it)
Hello
I would like to fit piecewise linear growth curves to a number of
subjects (each with a number of observations at different abscissa),
where all the slopes for a common interval (e.g, 2 <= x < 3)
presumable arise from the same normal distribution, and therefore
should be "shrunk" towards each other. What is the best way to about
doing this in lme4? I have a toy, cleaned up (although I am not very
good with R) example of the type of data I am looking at.
Thank you for your time
```
set.seed(1)
# our growth functions are piecewise linear with these breaks
breaks <- seq(0,5)
mean.slopes <- seq(5,2,-.6)
# generate the true slope parameters for 20 subjects (variance isn't
really constant or known, just simplifying)...
true.slopes <- sapply(mean.slopes,function(x) rnorm(n=20,sd=3,mean=x))
# ...and the intercepts - they really don't come from any particular
distribution
true.intercepts <- runif(n=20,min=30,max=50)
# generate sample data - we don't necessarily have the same number of
observations per subject
mat <- matrix(nr=20*10,nc=3)
for (i in 1:20)
{
xs <- runif(n=10,min=0,max=6)
slopes <- true.slopes[i,]
ys <- true.intercepts[i] + rnorm(n=10,mean=0,sd=0.9) +
sapply(xs, function(x) sum(slopes[1:floor(x)]) + (x - floor(x)) *
slopes[ceiling(x)])
id <- rep(i,10)
mat[(1+10*(i-1)):(10*i),] <- cbind(xs,ys,id)
}
mat <- as.data.frame(mat)
names(mat) <- c("x","y","id")
# qplot(x=x,y=y,colour=as.factor(id),data=mat)
```
Eventually I might like to shrink the slopes for each individual
towards each other, in the sense that if for a given subject the slope
in the interval [2,3) is very small compared to the slopes of other
subjects in the same interval, then the same would hold for the slope
in [1,2), but I would like to get the basics first.
| Fitting piecewise linear curves with lmer | CC BY-SA 3.0 | null | 2011-07-22T05:39:37.907 | 2011-07-22T20:00:26.247 | 2011-07-22T06:04:21.537 | 2116 | 5438 | [
"r",
"mixed-model",
"lme4-nlme"
] |
13362 | 1 | null | null | 4 | 121 | I am comparing the error between wind turbine production and predictions from 24 hours ahead. The graphs below are using the normal distribution.
But can I do better?
| Suggestions for improving the fit of probability density | CC BY-SA 3.0 | null | 2011-07-22T09:31:33.130 | 2011-07-22T15:34:13.693 | 2011-07-22T09:40:56.477 | 2116 | 3579 | [
"distributions"
] |
13363 | 2 | null | 13352 | 2 | null | Are you being confused by the correlation / regression distinction perhaps? When you say 'I correlate these answers' with shoe colour, you could mean either
- 'I compute the conditional probability of shoe colour given eating habits', or
- 'I compute a measure of linear relatedness underlying the joint distribution of eating habits and shoe colour choices'.
Number 1 would be realised as a regression which, unlike a correlation, would completely ignore the dominance structure in the eating habits answers. Hence your mystification and interest in an data aggregation method that will embody this structure.
Number 2 is a job for a correlation coefficient. There are special, structure-respecting correlation coefficients that might be better for this sort of dominance-structured data (see the excellent [tetrachoric and polychoric correlations page](http://www.john-uebersax.com/stat/tetra.htm) by John Uebersax, for an overview.)
Alternatively there are more explicit models in the latent trait / IRT family that will provide the raw materials for the various marginal and conditional distributional inferences that are, I think, the 'thing you can't get your head around'.
| null | CC BY-SA 3.0 | null | 2011-07-22T09:39:49.177 | 2011-07-22T09:39:49.177 | null | null | 1739 | null |
13364 | 2 | null | 13353 | 2 | null | If you want the average lifespan when weight is 1 then you can just take out "animal" in this call:
```
lm(formula = lifespan ~ 1 + animal + weight, data = animal.life)
```
| null | CC BY-SA 3.0 | null | 2011-07-22T10:55:54.383 | 2011-07-22T10:55:54.383 | null | null | 686 | null |
13365 | 2 | null | 13352 | 1 | null | I don't think it's at all correct that the "5" response "subsumes responses 2-4." The answer options are, as you acknowledge at the end, all mutually exclusive. For example, "more than 3" is definitely not inclusive of "1 to 3." So all that's left, that I can make out at least, is the question as to whether to collapse categories. That's more of a domain-knowledge issue than a statistical one.
| null | CC BY-SA 3.0 | null | 2011-07-22T11:48:04.797 | 2011-07-22T11:48:04.797 | null | null | 2669 | null |
13368 | 1 | 13372 | null | 7 | 2591 | Consider the problem of randomly generating a $p \times p$ covariance matrix $\Sigma$ with the diagonal entries constrained to be 1, and the off-diagonal entries $\Sigma_{ij}=\Sigma_{ji} \sim Unif[-a,a]$. What is the maximum value of $a$, as a function of $p$, such that $\Sigma$ is positive definite with probability 1?
| Off-diagonal range for guaranteed positive definiteness? | CC BY-SA 3.0 | null | 2011-07-22T13:06:14.103 | 2011-07-22T13:39:17.583 | 2011-07-22T13:39:17.583 | 919 | 3567 | [
"correlation",
"matrix"
] |
13369 | 2 | null | 9724 | 2 | null | First of all, I'm not sure whether the replicates are really what a statistician would call replicates (statisticians please comment). Mice that are grown together (same supplier, same cage, or even same litter) are often more similar to each other than mice from different cages, suppliers, litters. So the "cage" is a confounding variable that you otherwise neglect to measure.
In consequence, I woudn't say that you have 3 separate result sets and would go for Prof. Harrell's grand model.
Obviously whether you want to experiment 3 x on 5 mice or 5 x on 3 mice depends on the between cage and between mice variance (plus the fact that 3 x 5 is cheaper than 5 x 3).
| null | CC BY-SA 3.0 | null | 2011-07-22T13:09:00.340 | 2011-07-22T13:09:00.340 | null | null | 4598 | null |
13371 | 1 | 20278 | null | 10 | 1917 | I'm not that familiar with this literature, so please forgive me if this is an obvious question.
Since AIC and BIC depend on maximizing the likelihood, it seems that they can only be used to make relative comparisons between a set of models attempting to fit a given data-set. According to my understanding, it wouldn't make sense to calculate the AIC for Model A on data-set 1, calculate the AIC for Model B on data-set 2, and then compare the two AIC values and judge that (for example) Model A fits data-set 1 better than Model B fits data-set 2. Or perhaps I am mistaken and that is a reasonable thing to do. Please let me know.
My question is this: does there exist a model fit statistic that can be used for absolute instead of just relative comparisons? For linear models, something like $R^2$ would work; it has a defined range and discipline specific ideas about what is a "good" value. I'm looking for something more general and thought that I could start by pinging the experts here. I'm sure that someone has thought of this kind of thing before, but I don't quite know the right terms to make a productive search on Google Scholar.
Any help would be appreciated.
| Does there exist a model fit statistic (like AIC or BIC) that can be used for absolute instead of just relative comparisons? | CC BY-SA 3.0 | null | 2011-07-22T13:32:33.640 | 2015-06-02T18:27:47.967 | 2011-07-22T14:10:24.293 | 930 | 5507 | [
"model-selection",
"aic",
"bic"
] |
13372 | 2 | null | 13368 | 4 | null | The maximum is $1/(p-1)$.
To see this, note first that the eigenvalues of the matrix with all off-diagonal entries equal to a constant $x$ are $1-x$ (with multiplicity $p-1$) and $1+(p-1)x$. When $x \lt -1/(p-1)$, the smallest eigenvalue will therefore be negative implying the matrix is not positive definite. Because the smallest eigenvalue is a continuous function of the entries, we can find a positive $\epsilon$ such that when all off-diagonal entries are in the interval $[x, x+\epsilon]$ (but no longer all equal to each other), the smallest eigenvalue remains negative.
Now suppose $a \gt 1/(p-1)$. Setting $x=-a$, choose an $\epsilon$ as just described and if necessary make it even smaller, but still positive, to assure that $a - \epsilon \gt 1/(p-1)$. Assuming the off-diagonal entries are independently generated, the probability that all entries lie in the interval $[-a, -a+\epsilon]$ equals $(\epsilon / (2a))^{p(p-1)/2} \gt 0$, showing that the matrix has a positive probability of not being positive definite.
This has established $1/(p-1)$ as an upper bound for $a$. We need to show that it suffices. Consider an arbitrary symmetric $p$ by $p$ matrix $(a_{ij})$ with unit diagonal and all entries in size less than $1/p$. By a suitable induction on $p$, and by virtue of [Sylvester's Criterion](http://en.wikipedia.org/wiki/Positive-definite_matrix#Characterizations), it suffices to show this matrix has positive determinant. Row-reduction using the first row reduces this question to considering the sign of a $p-1$ by $p-1$ determinant with entries $(a_{ij} / (1 + a_{1i})$. Because $-1/p \lt a_{1i} \lt 1/p$, these clearly are less than $1/(p-1)$ in absolute value, so we are done by induction. (The base case $p=2$ is trivial.)
| null | CC BY-SA 3.0 | null | 2011-07-22T13:36:32.047 | 2011-07-22T13:36:32.047 | null | null | 919 | null |
13373 | 2 | null | 13347 | 3 | null | In your problem it sounds like the CDF will have a simple closed form that is easily inverted. In that case I'd suggest what is called the "inversion method" for generating from the distribution:
Let $F(x)=P(X≤x)$. Then set $y=F(x)$ and solve for $x$. The general solution to that inversion problem is the inverse CDF, $F^{-1}$. If you generate Uniform(0,1) random numbers and plug them into $F^{-1}$ you have random numbers with CDF $F$.
For example, $F(x)=1−e^{-x}$ for $x>0$ is the exponential CDF. The general solution to $y=1−e^{−x}$ in terms of $x$ is
$$ x=F^{-1}(y)=−\log(1−y).$$
Then, it follows that if $u$ is a ${\rm Uniform}(0,1)$ random variable, then $-\log(1−u)$ has an exponential distribution.
Edit: From your problem statement, the density is proportional to $f(x) = \frac{a}{bx} + c$ on some interval $(A,B)$. Therefore, the CDF is
$$ F(x) = \frac{1}{k} \int_{A}^{x} \frac{a}{by} + c \ dy = \frac{1}{k} \cdot \left( \frac{a}{b} \cdot \log(x/A) + c(x - A) \right) $$
where $k$ is the normalizing constant, which is calculated [here](https://stats.stackexchange.com/questions/13307/how-to-turn-a-function-into-a-probability-density/13308#13308). This closed form of the inverse of this CDF involves the [Lambert W Function](http://mathworld.wolfram.com/LambertW-Function.html) (thanks, @whuber), so I'm going to be lazy and solve for it numerically. Here is a very crude method of doing that in `R` using [bisection](http://en.wikipedia.org/wiki/Bisection_method), implemented as the `uniroot` function:
```
# Find the point at which the CDF above equals p, for parameters values
# a,b,c and predefined domain (A,B)
F_inv = function(p,a,b,c,A,B)
{
k = (a * log(B/A))/b + c(B-A)
br = c(A,B)
G = function(x) (a * log(x/A)/b + c(x-A))/k - p
return( uniroot(G,br)$root )
}
# x is a sample from this distribution with a=2, b=5, c=.8, A=10, B=20.
x = rep(0,1000)
for(i in 1:1000) x[i] = F_inv(runif(1),2,5,.8,10,20)
```
| null | CC BY-SA 3.0 | null | 2011-07-22T13:38:15.477 | 2012-06-04T14:40:38.060 | 2017-04-13T12:44:46.680 | -1 | 4856 | null |
13374 | 2 | null | 13327 | 2 | null | To me the slope graph looks really messy and I think I'd have trouble looking at it, especially across eight time series.
I am not an expert in graph design, so this may also be a no-go, but have you considered four colors with three types of plot type?
Though, I think there is an even better approach. I know you say that 4 colors is a no-go, I'm about to ignore that. It is probably canonically true... but you are describing fruits. These have canonical colors as well as shapes associated with them. If you use those colors and shapes I think it would be hard to go wrong.
Using colors alone there might be some confusion, e.g. green apple vs honeydew, red apple vs watermelon, etc. But using colors poses an additional problem, color blind individuals. You can test for the extent to which this would be a problem by creating an image of your different colors and looking here: [http://www.vischeck.com/vischeck/](http://www.vischeck.com/vischeck/). Protanopia and Deuteranopia, forms of red-green color blindness, are by far the most common (occurring almost always in males). Even so, color blindness is a misnomer and if you select your pallet carefully the differences in shades may be sufficient.
In conjunction with a color approach, you want to use fruit shaped points. These are unlikely to be a default in a plotting program and you may have to spend some time in photoshop to make it look right. Even if you can't take that time, differing geometric plot points AND color should make things reasonable.
Moreover, you could use the approach I suggest with a slope graph.
As a side note, if you have a technically/mathematically astute audience, a Y-axis in log odds might mean more to them that percentages alone.
| null | CC BY-SA 3.0 | null | 2011-07-22T13:40:32.653 | 2011-07-22T13:49:11.357 | 2011-07-22T13:49:11.357 | 196 | 196 | null |
13375 | 2 | null | 13362 | 2 | null | Maybe use the laplace distribution, which has density
$$ \frac{1}{2s} {\rm exp} \left( -\frac{|x-\mu|}{s} \right ) $$
where $\mu \in \mathbb{R}, s > 0$ are parameters. As Wolfgang pointed out, your distribution appears to be symmetric but has higher kurtosis than the normal distribution; the laplace distribution has these properties and the QQ plot looks pretty similar to what you're observed from your data:
```
library(VGAM)
x=rlaplace(1000)
qqnorm(x)
qqline(x)
```
(you will need to download the package VGAM). The logistic distribution could also work but the QQ plots don't agree with what you've observed as well.
| null | CC BY-SA 3.0 | null | 2011-07-22T14:34:07.277 | 2011-07-22T14:34:07.277 | null | null | 4856 | null |
13376 | 1 | 13380 | null | 4 | 8530 | I would like to run a simulation where the sum of 20 random variables are generated 1000 times and plotted in a histogram.
I generate the random numbers using runif(20,0,1) then sum these random numbers using the sum function, but I was wondering if there is an easier method to do this 1000 then to write a for loop.
| How to run and plot simulation in R of sum of 20 random variables? | CC BY-SA 3.0 | null | 2011-07-22T15:08:35.917 | 2013-06-10T23:02:16.450 | 2011-07-24T08:42:11.767 | 183 | null | [
"r",
"simulation"
] |
13377 | 2 | null | 10159 | 130 | null |
# Usage of the density function is not necessary
## Integrate 1 minus the CDF
When you have a random variable $X$ that has a support that is non-negative (that is, the variable has nonzero density/probability for only positive values), you can use the following property:
$$
E(X) = \int_0^\infty \left( 1 - F_X(x) \right) \,\mathrm{d}x
$$
A similar property applies in the case of a discrete random variable.
## Proof
Since $1 - F_X(x) = P(X\geq x) = \int_x^\infty f_X(t) \,\mathrm{d}t$,
$$
\int_0^\infty \left( 1 - F_X(x) \right) \,\mathrm{d}x = \int_0^\infty P(X\geq x) \,\mathrm{d}x = \int_0^\infty \int_x^\infty f_X(t) \,\mathrm{d}t \mathrm{d}x
$$
Then change the order of integration:
$$
= \int_0^\infty \int_0^t f_X(t) \,\mathrm{d}x \mathrm{d}t = \int_0^\infty \left[xf_X(t)\right]_0^t \,\mathrm{d}t = \int_0^\infty t f_X(t) \,\mathrm{d}t
$$
Recognizing that $t$ is a dummy variable, or taking the simple substitution $t=x$ and $\mathrm{d}t = \mathrm{d}x$,
$$
= \int_0^\infty x f_X(x) \,\mathrm{d}x = \mathrm{E}(X)
$$
## Attribution
I used the [Formulas for special cases](http://en.wikipedia.org/wiki/Expected_value#Formulas_for_special_cases) section of the [Expected value](http://en.wikipedia.org/wiki/Expected_value) article on [Wikipedia](http://en.wikipedia.org) to refresh my memory on the proof. That section also contains proofs for the discrete random variable case and also for the case that no density function exists.
| null | CC BY-SA 3.0 | null | 2011-07-22T15:30:25.007 | 2011-07-22T15:30:25.007 | null | null | 1583 | null |
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