idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
|---|---|---|---|---|---|---|
48,201 | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lifetime cumulative incidence? | Point prevalence: The proportion of individuals who manifest a disorder at a given point in time.
Period prevalence: The proportion of individuals who manifest a disorder over a specific period of time (e.g., over one year).
Lifetime prevalence: The proportion of individuals in the population who have ever manifested a... | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lif | Point prevalence: The proportion of individuals who manifest a disorder at a given point in time.
Period prevalence: The proportion of individuals who manifest a disorder over a specific period of tim | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lifetime cumulative incidence?
Point prevalence: The proportion of individuals who manifest a disorder at a given point in time.
Period prevalence: The proportion of individuals who manifest a disorder over a specific perio... | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lif
Point prevalence: The proportion of individuals who manifest a disorder at a given point in time.
Period prevalence: The proportion of individuals who manifest a disorder over a specific period of tim |
48,202 | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lifetime cumulative incidence? | From Kessler et al, 2012:
In studies of lifetime disorders, furthermore, some researchers report estimates of lifetime prevalence (i.e. the proportion of people who have had the disorder at sometime in their life), while others report estimates of lifetime morbid risk (LMR) (i.e. the proportion of people who will eve... | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lif | From Kessler et al, 2012:
In studies of lifetime disorders, furthermore, some researchers report estimates of lifetime prevalence (i.e. the proportion of people who have had the disorder at sometime | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lifetime cumulative incidence?
From Kessler et al, 2012:
In studies of lifetime disorders, furthermore, some researchers report estimates of lifetime prevalence (i.e. the proportion of people who have had the disorder at ... | What is the difference between lifetime risk, lifetime morbid risk, and lifetime prevalence, and lif
From Kessler et al, 2012:
In studies of lifetime disorders, furthermore, some researchers report estimates of lifetime prevalence (i.e. the proportion of people who have had the disorder at sometime |
48,203 | Sophisticated models for classifying short pieces of texts | Is your SVM implementation parallel? One simple idea would be to split your 10-fold CV across 10 machines (or cores). This should reduce the algorithm's runtime to almost 1/10 of its current running time.
What do you mean by "info gain?" Have you tried applying LibLINEAR on a dataset with no features removed? | Sophisticated models for classifying short pieces of texts | Is your SVM implementation parallel? One simple idea would be to split your 10-fold CV across 10 machines (or cores). This should reduce the algorithm's runtime to almost 1/10 of its current running | Sophisticated models for classifying short pieces of texts
Is your SVM implementation parallel? One simple idea would be to split your 10-fold CV across 10 machines (or cores). This should reduce the algorithm's runtime to almost 1/10 of its current running time.
What do you mean by "info gain?" Have you tried apply... | Sophisticated models for classifying short pieces of texts
Is your SVM implementation parallel? One simple idea would be to split your 10-fold CV across 10 machines (or cores). This should reduce the algorithm's runtime to almost 1/10 of its current running |
48,204 | Sophisticated models for classifying short pieces of texts | The first thing you need to do is to figure out is the source of your generalization error. Is it "bias" or is it "variance"? (or perhaps something else?) If it is variance, your training set might be small for what you are trying to accomplish, and you might need more training data. If it is bias, then changing to a... | Sophisticated models for classifying short pieces of texts | The first thing you need to do is to figure out is the source of your generalization error. Is it "bias" or is it "variance"? (or perhaps something else?) If it is variance, your training set might b | Sophisticated models for classifying short pieces of texts
The first thing you need to do is to figure out is the source of your generalization error. Is it "bias" or is it "variance"? (or perhaps something else?) If it is variance, your training set might be small for what you are trying to accomplish, and you might ... | Sophisticated models for classifying short pieces of texts
The first thing you need to do is to figure out is the source of your generalization error. Is it "bias" or is it "variance"? (or perhaps something else?) If it is variance, your training set might b |
48,205 | Sample size for a variable number of answers | You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).
Let $p_k$ be the true proportion of people in the population w... | Sample size for a variable number of answers | You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ ( | Sample size for a variable number of answers
You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).
Let $p_k$ be the ... | Sample size for a variable number of answers
You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ ( |
48,206 | Sample size for a variable number of answers | How is this different than asking independent questions on a survey such as who will you vote for - red, yellow, etc. If I've read the question correct, your answer is 364 observations. This will provide +/-5% at the 95% level of confidence given a normal distribution.
Dr. Doug | Sample size for a variable number of answers | How is this different than asking independent questions on a survey such as who will you vote for - red, yellow, etc. If I've read the question correct, your answer is 364 observations. This will pr | Sample size for a variable number of answers
How is this different than asking independent questions on a survey such as who will you vote for - red, yellow, etc. If I've read the question correct, your answer is 364 observations. This will provide +/-5% at the 95% level of confidence given a normal distribution.
Dr.... | Sample size for a variable number of answers
How is this different than asking independent questions on a survey such as who will you vote for - red, yellow, etc. If I've read the question correct, your answer is 364 observations. This will pr |
48,207 | Data sets and problems for learning hypothesis testing | You are supposed to form your hypotheses before seeing any of the actual data. These hypotheses come from some sort of conceptual frame work. Your best bet may be to form a hypothesis on a topic of interest to you and then try to find a data set to test that hypothesis. | Data sets and problems for learning hypothesis testing | You are supposed to form your hypotheses before seeing any of the actual data. These hypotheses come from some sort of conceptual frame work. Your best bet may be to form a hypothesis on a topic of | Data sets and problems for learning hypothesis testing
You are supposed to form your hypotheses before seeing any of the actual data. These hypotheses come from some sort of conceptual frame work. Your best bet may be to form a hypothesis on a topic of interest to you and then try to find a data set to test that hypo... | Data sets and problems for learning hypothesis testing
You are supposed to form your hypotheses before seeing any of the actual data. These hypotheses come from some sort of conceptual frame work. Your best bet may be to form a hypothesis on a topic of |
48,208 | Data sets and problems for learning hypothesis testing | R has a whole bunch of datasets built in - they're used by all the examples for each function. Some are quite detailed, and you could do some really interesting stuff with them. The Quick-R site uses these datasets (the cars set especially) for most of their examples. | Data sets and problems for learning hypothesis testing | R has a whole bunch of datasets built in - they're used by all the examples for each function. Some are quite detailed, and you could do some really interesting stuff with them. The Quick-R site uses | Data sets and problems for learning hypothesis testing
R has a whole bunch of datasets built in - they're used by all the examples for each function. Some are quite detailed, and you could do some really interesting stuff with them. The Quick-R site uses these datasets (the cars set especially) for most of their exampl... | Data sets and problems for learning hypothesis testing
R has a whole bunch of datasets built in - they're used by all the examples for each function. Some are quite detailed, and you could do some really interesting stuff with them. The Quick-R site uses |
48,209 | How to sample randomly from a population? | It's a fine method in theory. To see why, we need to check two things.
Are all individuals selected with equal chances? Yes, because the distributions of floats assigned to each individual are identical (they are uniform in $[0,1)$).
Are the selections independent? Yes, because the floats assigned to the individual... | How to sample randomly from a population? | It's a fine method in theory. To see why, we need to check two things.
Are all individuals selected with equal chances? Yes, because the distributions of floats assigned to each individual are iden | How to sample randomly from a population?
It's a fine method in theory. To see why, we need to check two things.
Are all individuals selected with equal chances? Yes, because the distributions of floats assigned to each individual are identical (they are uniform in $[0,1)$).
Are the selections independent? Yes, bec... | How to sample randomly from a population?
It's a fine method in theory. To see why, we need to check two things.
Are all individuals selected with equal chances? Yes, because the distributions of floats assigned to each individual are iden |
48,210 | How to find P(X > F) in ANOVA F-test? | The answer can not be retrieved without the aid of computation.
If you look at the F distribution table for F(3,19), you'll see that (for F(3,20):
...
.050 | 3.10
.025 | 3.86
...
Which means 0.025 < p < 0.05.
I'm guessing that they "cheated" with MATLAB or something. | How to find P(X > F) in ANOVA F-test? | The answer can not be retrieved without the aid of computation.
If you look at the F distribution table for F(3,19), you'll see that (for F(3,20):
...
.050 | 3.10
.025 | 3.86
...
Which means 0.025 < p | How to find P(X > F) in ANOVA F-test?
The answer can not be retrieved without the aid of computation.
If you look at the F distribution table for F(3,19), you'll see that (for F(3,20):
...
.050 | 3.10
.025 | 3.86
...
Which means 0.025 < p < 0.05.
I'm guessing that they "cheated" with MATLAB or something. | How to find P(X > F) in ANOVA F-test?
The answer can not be retrieved without the aid of computation.
If you look at the F distribution table for F(3,19), you'll see that (for F(3,20):
...
.050 | 3.10
.025 | 3.86
...
Which means 0.025 < p |
48,211 | How to find P(X > F) in ANOVA F-test? | Since the test statistics follows F-ratio distribution, the $\mathbb{P}(X>z)$ is known in closed form.
Let $X \sim F(n,m)$. Then
$$
\mathbb{P}(X > z) = \tilde{B}_{\frac{m}{m+ n z}}\left( \frac{m}{2}, \frac{n}{2} \right)
$$
For $n=3$ and $m=19$ and $z=3.77$, Wolfram Alpha produces the answer of $0.0280768$. | How to find P(X > F) in ANOVA F-test? | Since the test statistics follows F-ratio distribution, the $\mathbb{P}(X>z)$ is known in closed form.
Let $X \sim F(n,m)$. Then
$$
\mathbb{P}(X > z) = \tilde{B}_{\frac{m}{m+ n z}}\left( \frac{m}{ | How to find P(X > F) in ANOVA F-test?
Since the test statistics follows F-ratio distribution, the $\mathbb{P}(X>z)$ is known in closed form.
Let $X \sim F(n,m)$. Then
$$
\mathbb{P}(X > z) = \tilde{B}_{\frac{m}{m+ n z}}\left( \frac{m}{2}, \frac{n}{2} \right)
$$
For $n=3$ and $m=19$ and $z=3.77$, Wolfram Alpha produ... | How to find P(X > F) in ANOVA F-test?
Since the test statistics follows F-ratio distribution, the $\mathbb{P}(X>z)$ is known in closed form.
Let $X \sim F(n,m)$. Then
$$
\mathbb{P}(X > z) = \tilde{B}_{\frac{m}{m+ n z}}\left( \frac{m}{ |
48,212 | What are the sampling distributions of higher moments of the normal distribution? | I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$?
$\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the moment-generating function $\exp(\sigma^2t^2/2 + \mu t)$ of the $N(\mu, \sigma^2)$ random variable $X_i$ (or from the characteris... | What are the sampling distributions of higher moments of the normal distribution? | I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$?
$\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the moment-ge | What are the sampling distributions of higher moments of the normal distribution?
I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$?
$\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the moment-generating function $\exp(\sigma^2t^2/2 ... | What are the sampling distributions of higher moments of the normal distribution?
I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$?
$\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the moment-ge |
48,213 | Parallel solving Ax=b? | You can decompose this operation into a set of smaller operations that are easy to parallelize.
Suppose you wish to solve $\mathbf{m}\mathbf{v}=\mathbf{u}$ for an $N$ by $N$ matrix $\mathbf{m}$ and $N$-vector $\mathbf{u}$. Writing $N=n+m$ (intending $n\approx m$), decompose $\mathbf{m}$ into four blocks $\mathbf{a}_{n... | Parallel solving Ax=b? | You can decompose this operation into a set of smaller operations that are easy to parallelize.
Suppose you wish to solve $\mathbf{m}\mathbf{v}=\mathbf{u}$ for an $N$ by $N$ matrix $\mathbf{m}$ and $N | Parallel solving Ax=b?
You can decompose this operation into a set of smaller operations that are easy to parallelize.
Suppose you wish to solve $\mathbf{m}\mathbf{v}=\mathbf{u}$ for an $N$ by $N$ matrix $\mathbf{m}$ and $N$-vector $\mathbf{u}$. Writing $N=n+m$ (intending $n\approx m$), decompose $\mathbf{m}$ into fou... | Parallel solving Ax=b?
You can decompose this operation into a set of smaller operations that are easy to parallelize.
Suppose you wish to solve $\mathbf{m}\mathbf{v}=\mathbf{u}$ for an $N$ by $N$ matrix $\mathbf{m}$ and $N |
48,214 | Parallel solving Ax=b? | Have you tried QR decomposition? See Theorem 3 here for solving $Ax=b$.
Finding the inverse of a matrix (even a small one) is a slow process. Methods such as QR or Cholesky decomposition are used in practice when 'inverting' is needed (at least in my experience in statistical programming). | Parallel solving Ax=b? | Have you tried QR decomposition? See Theorem 3 here for solving $Ax=b$.
Finding the inverse of a matrix (even a small one) is a slow process. Methods such as QR or Cholesky decomposition are used in | Parallel solving Ax=b?
Have you tried QR decomposition? See Theorem 3 here for solving $Ax=b$.
Finding the inverse of a matrix (even a small one) is a slow process. Methods such as QR or Cholesky decomposition are used in practice when 'inverting' is needed (at least in my experience in statistical programming). | Parallel solving Ax=b?
Have you tried QR decomposition? See Theorem 3 here for solving $Ax=b$.
Finding the inverse of a matrix (even a small one) is a slow process. Methods such as QR or Cholesky decomposition are used in |
48,215 | Odd problem with a histogram in R with a relative frequency axis | One explanation is that the standard deviation of your data is much less than one, and the histogram is giving something like the probability density.
For example, see how the density on the histogram changes when I divide a uniform random variable with range (0, 1) by 1000:
set.seed(4444)
x <- runif(100)
y <- x / 1000... | Odd problem with a histogram in R with a relative frequency axis | One explanation is that the standard deviation of your data is much less than one, and the histogram is giving something like the probability density.
For example, see how the density on the histogram | Odd problem with a histogram in R with a relative frequency axis
One explanation is that the standard deviation of your data is much less than one, and the histogram is giving something like the probability density.
For example, see how the density on the histogram changes when I divide a uniform random variable with r... | Odd problem with a histogram in R with a relative frequency axis
One explanation is that the standard deviation of your data is much less than one, and the histogram is giving something like the probability density.
For example, see how the density on the histogram |
48,216 | Odd problem with a histogram in R with a relative frequency axis | As other have noted, frequency=FALSE only makes the integral over the histogram equal to 1, not the sum over all values. (The parameter probability=TRUE is only there for S compatibility, by the way, and probably therefore a misnomer. Probability density would be better.)
Here is some code to relabel the y axis to plo... | Odd problem with a histogram in R with a relative frequency axis | As other have noted, frequency=FALSE only makes the integral over the histogram equal to 1, not the sum over all values. (The parameter probability=TRUE is only there for S compatibility, by the way, | Odd problem with a histogram in R with a relative frequency axis
As other have noted, frequency=FALSE only makes the integral over the histogram equal to 1, not the sum over all values. (The parameter probability=TRUE is only there for S compatibility, by the way, and probably therefore a misnomer. Probability density... | Odd problem with a histogram in R with a relative frequency axis
As other have noted, frequency=FALSE only makes the integral over the histogram equal to 1, not the sum over all values. (The parameter probability=TRUE is only there for S compatibility, by the way, |
48,217 | Odd problem with a histogram in R with a relative frequency axis | If you pass probability=TRUE (or frequency=FALSE) you should indeed see densities on the plot.
Note that this does not make it impossible for them to be above 1 if your number of breaks is relatively low and your bin widths are small (way below 1). Looking at the code of hist.default you can see that the densities are ... | Odd problem with a histogram in R with a relative frequency axis | If you pass probability=TRUE (or frequency=FALSE) you should indeed see densities on the plot.
Note that this does not make it impossible for them to be above 1 if your number of breaks is relatively | Odd problem with a histogram in R with a relative frequency axis
If you pass probability=TRUE (or frequency=FALSE) you should indeed see densities on the plot.
Note that this does not make it impossible for them to be above 1 if your number of breaks is relatively low and your bin widths are small (way below 1). Lookin... | Odd problem with a histogram in R with a relative frequency axis
If you pass probability=TRUE (or frequency=FALSE) you should indeed see densities on the plot.
Note that this does not make it impossible for them to be above 1 if your number of breaks is relatively |
48,218 | How to report general precision in estimating correlations within a context of justifying sample size? | I do think constructing confidence intervals around parameter estimates (such as correlations, among other kinds) is a good thing to do. I strongly recommend it. Moreover, I don't think it should matter if the observed value is 0 or any other focal value. If someone claims that they did a study and found that the co... | How to report general precision in estimating correlations within a context of justifying sample siz | I do think constructing confidence intervals around parameter estimates (such as correlations, among other kinds) is a good thing to do. I strongly recommend it. Moreover, I don't think it should ma | How to report general precision in estimating correlations within a context of justifying sample size?
I do think constructing confidence intervals around parameter estimates (such as correlations, among other kinds) is a good thing to do. I strongly recommend it. Moreover, I don't think it should matter if the obser... | How to report general precision in estimating correlations within a context of justifying sample siz
I do think constructing confidence intervals around parameter estimates (such as correlations, among other kinds) is a good thing to do. I strongly recommend it. Moreover, I don't think it should ma |
48,219 | How to report general precision in estimating correlations within a context of justifying sample size? | You can construct a confidence interval for the correlation; see here and here.
I don't understand what you mean by "... 95% confidence intervals on correlations when the correlation is zero, or perhaps some other focal value." | How to report general precision in estimating correlations within a context of justifying sample siz | You can construct a confidence interval for the correlation; see here and here.
I don't understand what you mean by "... 95% confidence intervals on correlations when the correlation is zero, or perh | How to report general precision in estimating correlations within a context of justifying sample size?
You can construct a confidence interval for the correlation; see here and here.
I don't understand what you mean by "... 95% confidence intervals on correlations when the correlation is zero, or perhaps some other fo... | How to report general precision in estimating correlations within a context of justifying sample siz
You can construct a confidence interval for the correlation; see here and here.
I don't understand what you mean by "... 95% confidence intervals on correlations when the correlation is zero, or perh |
48,220 | Maximum likelihood estimation procedures for state-space linear models | If you consider gaussian state-space models, maximum likelihood is conceptually straightforward, and several packages in R (including dlm, kfas, or FKF, to name a few)
include routines for doing so: you can find a short review here (disclaimer: I am the author) and a monographic issue of the Journal of Statistical Soft... | Maximum likelihood estimation procedures for state-space linear models | If you consider gaussian state-space models, maximum likelihood is conceptually straightforward, and several packages in R (including dlm, kfas, or FKF, to name a few)
include routines for doing so: y | Maximum likelihood estimation procedures for state-space linear models
If you consider gaussian state-space models, maximum likelihood is conceptually straightforward, and several packages in R (including dlm, kfas, or FKF, to name a few)
include routines for doing so: you can find a short review here (disclaimer: I am... | Maximum likelihood estimation procedures for state-space linear models
If you consider gaussian state-space models, maximum likelihood is conceptually straightforward, and several packages in R (including dlm, kfas, or FKF, to name a few)
include routines for doing so: y |
48,221 | Calculating standard error and attaching an error bar on ggplot2 bar chart | Here's an example from the ggplot2 homepage: https://ggplot2.tidyverse.org/reference/geom_errorbarh.html as others have mentioned in the comments, you have to calculate SE on your own and append this information to the data.frame
df <- data.frame(
trt = factor(c(1, 1, 2, 2)),
resp = c(1, 5, 3, 4),
group = fact... | Calculating standard error and attaching an error bar on ggplot2 bar chart | Here's an example from the ggplot2 homepage: https://ggplot2.tidyverse.org/reference/geom_errorbarh.html as others have mentioned in the comments, you have to calculate SE on your own and append this | Calculating standard error and attaching an error bar on ggplot2 bar chart
Here's an example from the ggplot2 homepage: https://ggplot2.tidyverse.org/reference/geom_errorbarh.html as others have mentioned in the comments, you have to calculate SE on your own and append this information to the data.frame
df <- data.fram... | Calculating standard error and attaching an error bar on ggplot2 bar chart
Here's an example from the ggplot2 homepage: https://ggplot2.tidyverse.org/reference/geom_errorbarh.html as others have mentioned in the comments, you have to calculate SE on your own and append this |
48,222 | Trend analysis with three time points in a repeated measures design | If you have a constant term in your model, the midpoint is not being ignored, even though it is given a value of zero in this coding. The value equal to the mean of the two other values indicates to the software that the time spacings are equal. You can code your time as 1, 2 and 3, but then it will be collinear with t... | Trend analysis with three time points in a repeated measures design | If you have a constant term in your model, the midpoint is not being ignored, even though it is given a value of zero in this coding. The value equal to the mean of the two other values indicates to t | Trend analysis with three time points in a repeated measures design
If you have a constant term in your model, the midpoint is not being ignored, even though it is given a value of zero in this coding. The value equal to the mean of the two other values indicates to the software that the time spacings are equal. You ca... | Trend analysis with three time points in a repeated measures design
If you have a constant term in your model, the midpoint is not being ignored, even though it is given a value of zero in this coding. The value equal to the mean of the two other values indicates to t |
48,223 | Trend analysis with three time points in a repeated measures design | Sorry to disappoint, but I would NOT try to categorize this as trend analysis, at least not in the usual time series sense.
That said, you can still do productive modeling. You look at testing the hypothesis that the difference between (t1,t2) is non-zero (i.e. test against the null hypothesis). You could do the same... | Trend analysis with three time points in a repeated measures design | Sorry to disappoint, but I would NOT try to categorize this as trend analysis, at least not in the usual time series sense.
That said, you can still do productive modeling. You look at testing the hy | Trend analysis with three time points in a repeated measures design
Sorry to disappoint, but I would NOT try to categorize this as trend analysis, at least not in the usual time series sense.
That said, you can still do productive modeling. You look at testing the hypothesis that the difference between (t1,t2) is non-... | Trend analysis with three time points in a repeated measures design
Sorry to disappoint, but I would NOT try to categorize this as trend analysis, at least not in the usual time series sense.
That said, you can still do productive modeling. You look at testing the hy |
48,224 | Estimating standard deviation in Poisson regression | Poisson regression finds a value $\hat{\beta}$ maximizing the likelihood of the data. For any value of $x$, you would then suppose $Y$ has a Poisson($\exp(x \hat{\beta})$) distribution. The standard deviation of that distribution equals $\exp(x \hat{\beta}/2)$. This appears to be what you mean by $\sqrt{\widehat{\mu... | Estimating standard deviation in Poisson regression | Poisson regression finds a value $\hat{\beta}$ maximizing the likelihood of the data. For any value of $x$, you would then suppose $Y$ has a Poisson($\exp(x \hat{\beta})$) distribution. The standard | Estimating standard deviation in Poisson regression
Poisson regression finds a value $\hat{\beta}$ maximizing the likelihood of the data. For any value of $x$, you would then suppose $Y$ has a Poisson($\exp(x \hat{\beta})$) distribution. The standard deviation of that distribution equals $\exp(x \hat{\beta}/2)$. Thi... | Estimating standard deviation in Poisson regression
Poisson regression finds a value $\hat{\beta}$ maximizing the likelihood of the data. For any value of $x$, you would then suppose $Y$ has a Poisson($\exp(x \hat{\beta})$) distribution. The standard |
48,225 | Estimating standard deviation in Poisson regression | You are thinking too much in terms of "normally distributed" here. For a normal distribution, you have two parameters then mean $\mu$ and the variance, $\sigma^2$. So you require two pieces of information to characterize the probability distribution for the normal case.
However, in the Poisson distributed case, there... | Estimating standard deviation in Poisson regression | You are thinking too much in terms of "normally distributed" here. For a normal distribution, you have two parameters then mean $\mu$ and the variance, $\sigma^2$. So you require two pieces of infor | Estimating standard deviation in Poisson regression
You are thinking too much in terms of "normally distributed" here. For a normal distribution, you have two parameters then mean $\mu$ and the variance, $\sigma^2$. So you require two pieces of information to characterize the probability distribution for the normal c... | Estimating standard deviation in Poisson regression
You are thinking too much in terms of "normally distributed" here. For a normal distribution, you have two parameters then mean $\mu$ and the variance, $\sigma^2$. So you require two pieces of infor |
48,226 | Testing the importance of an item among a finite set of items | The naive approach would be to compute the marginal distribution of rankings (e.g., mean score for each item), but it would throw away a lot of information as it does not account for the within-person relationship between ranks.
As an extension to paired preference model (e.g., the Bradley-Terry model, described in Agr... | Testing the importance of an item among a finite set of items | The naive approach would be to compute the marginal distribution of rankings (e.g., mean score for each item), but it would throw away a lot of information as it does not account for the within-person | Testing the importance of an item among a finite set of items
The naive approach would be to compute the marginal distribution of rankings (e.g., mean score for each item), but it would throw away a lot of information as it does not account for the within-person relationship between ranks.
As an extension to paired pre... | Testing the importance of an item among a finite set of items
The naive approach would be to compute the marginal distribution of rankings (e.g., mean score for each item), but it would throw away a lot of information as it does not account for the within-person |
48,227 | Two poisson random variables and likelihood ratio test | The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:
$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\lambda_{2}
\end{array}
$$
Now neither hypothesis says what the parameters are, so the actual values are nuisance paramete... | Two poisson random variables and likelihood ratio test | The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:
$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\l | Two poisson random variables and likelihood ratio test
The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:
$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\lambda_{2}
\end{array}
$$
Now neither hypothesis says what the par... | Two poisson random variables and likelihood ratio test
The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:
$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\l |
48,228 | Two poisson random variables and likelihood ratio test | Did your reading suggest that the likelihood ratio test statistic had problems? or that the chi-squared approximation was not very good?
I expect that most of the problems are more likely the latter, the test statistic is fine, but we don't know the distribution of it under the null hypothesis. With modern computers w... | Two poisson random variables and likelihood ratio test | Did your reading suggest that the likelihood ratio test statistic had problems? or that the chi-squared approximation was not very good?
I expect that most of the problems are more likely the latter, | Two poisson random variables and likelihood ratio test
Did your reading suggest that the likelihood ratio test statistic had problems? or that the chi-squared approximation was not very good?
I expect that most of the problems are more likely the latter, the test statistic is fine, but we don't know the distribution of... | Two poisson random variables and likelihood ratio test
Did your reading suggest that the likelihood ratio test statistic had problems? or that the chi-squared approximation was not very good?
I expect that most of the problems are more likely the latter, |
48,229 | What do "real values" refer to in supervised classification? | Regarding the "Real Values"
The "Real Values" are better called "confidences" or (from my pov the most common term) "scores".
Such scores are often normalized so that they sum up to 1 for all classes. They represent a measure how, well, confident the model is that the presented example belongs to a certain class. They... | What do "real values" refer to in supervised classification? | Regarding the "Real Values"
The "Real Values" are better called "confidences" or (from my pov the most common term) "scores".
Such scores are often normalized so that they sum up to 1 for all classes | What do "real values" refer to in supervised classification?
Regarding the "Real Values"
The "Real Values" are better called "confidences" or (from my pov the most common term) "scores".
Such scores are often normalized so that they sum up to 1 for all classes. They represent a measure how, well, confident the model i... | What do "real values" refer to in supervised classification?
Regarding the "Real Values"
The "Real Values" are better called "confidences" or (from my pov the most common term) "scores".
Such scores are often normalized so that they sum up to 1 for all classes |
48,230 | Making a heatmap with a precomputed distance matrix and data matrix in R | Ok, so you can just look at the code by typing the name of the function at the R prompt, or use edit(pheatmap) to see it in your default editor.
Around line 14 and 23, you'll see that another function is called for computing the distance matrices (for rows and columns), given a distance function (R dist) and a method (... | Making a heatmap with a precomputed distance matrix and data matrix in R | Ok, so you can just look at the code by typing the name of the function at the R prompt, or use edit(pheatmap) to see it in your default editor.
Around line 14 and 23, you'll see that another function | Making a heatmap with a precomputed distance matrix and data matrix in R
Ok, so you can just look at the code by typing the name of the function at the R prompt, or use edit(pheatmap) to see it in your default editor.
Around line 14 and 23, you'll see that another function is called for computing the distance matrices ... | Making a heatmap with a precomputed distance matrix and data matrix in R
Ok, so you can just look at the code by typing the name of the function at the R prompt, or use edit(pheatmap) to see it in your default editor.
Around line 14 and 23, you'll see that another function |
48,231 | Making a heatmap with a precomputed distance matrix and data matrix in R | As of version 1.0.10, the input parameter clustering_distance_rows and clustering_distance_cols can take a dist object to impose the precomputed distance for clustering. | Making a heatmap with a precomputed distance matrix and data matrix in R | As of version 1.0.10, the input parameter clustering_distance_rows and clustering_distance_cols can take a dist object to impose the precomputed distance for clustering. | Making a heatmap with a precomputed distance matrix and data matrix in R
As of version 1.0.10, the input parameter clustering_distance_rows and clustering_distance_cols can take a dist object to impose the precomputed distance for clustering. | Making a heatmap with a precomputed distance matrix and data matrix in R
As of version 1.0.10, the input parameter clustering_distance_rows and clustering_distance_cols can take a dist object to impose the precomputed distance for clustering. |
48,232 | Learning a univariate transform (kernel?) for novelty detection | Your setting is pretty hard. I have no solution, but a couple of points.
Energy based models can give you a scalar corresponding to a "grade of belief" that an input is generated by the distribution of your data. It comes down to chosing a model and a good loss function. Check out Yann Lecun's tutorial on energy based... | Learning a univariate transform (kernel?) for novelty detection | Your setting is pretty hard. I have no solution, but a couple of points.
Energy based models can give you a scalar corresponding to a "grade of belief" that an input is generated by the distribution | Learning a univariate transform (kernel?) for novelty detection
Your setting is pretty hard. I have no solution, but a couple of points.
Energy based models can give you a scalar corresponding to a "grade of belief" that an input is generated by the distribution of your data. It comes down to chosing a model and a goo... | Learning a univariate transform (kernel?) for novelty detection
Your setting is pretty hard. I have no solution, but a couple of points.
Energy based models can give you a scalar corresponding to a "grade of belief" that an input is generated by the distribution |
48,233 | Learning a univariate transform (kernel?) for novelty detection | Would the nearest-neighbor distance distribution help ?
For each of the 150 observations, you have distances
$d_1\ d_2$ ... to its nearest, 2nd nearest ... neighbors,
and an averaged distance distribution, call it DD.
A query point gives you the distribution $d_1 .. d_{150}$:
compare that to DD.
The metric between poin... | Learning a univariate transform (kernel?) for novelty detection | Would the nearest-neighbor distance distribution help ?
For each of the 150 observations, you have distances
$d_1\ d_2$ ... to its nearest, 2nd nearest ... neighbors,
and an averaged distance distribu | Learning a univariate transform (kernel?) for novelty detection
Would the nearest-neighbor distance distribution help ?
For each of the 150 observations, you have distances
$d_1\ d_2$ ... to its nearest, 2nd nearest ... neighbors,
and an averaged distance distribution, call it DD.
A query point gives you the distributi... | Learning a univariate transform (kernel?) for novelty detection
Would the nearest-neighbor distance distribution help ?
For each of the 150 observations, you have distances
$d_1\ d_2$ ... to its nearest, 2nd nearest ... neighbors,
and an averaged distance distribu |
48,234 | Correct specification of longitudinal model in lme4 | I'll imagine a concrete example, with more context, to make things easy. Assume you measure the score on test of 3k students of 200 schools and you measured each student at 4 time points (say, at each quarter). You have a covariate at student level that doesn't vary by time (like sex), that you called pred1.obs and a c... | Correct specification of longitudinal model in lme4 | I'll imagine a concrete example, with more context, to make things easy. Assume you measure the score on test of 3k students of 200 schools and you measured each student at 4 time points (say, at each | Correct specification of longitudinal model in lme4
I'll imagine a concrete example, with more context, to make things easy. Assume you measure the score on test of 3k students of 200 schools and you measured each student at 4 time points (say, at each quarter). You have a covariate at student level that doesn't vary b... | Correct specification of longitudinal model in lme4
I'll imagine a concrete example, with more context, to make things easy. Assume you measure the score on test of 3k students of 200 schools and you measured each student at 4 time points (say, at each |
48,235 | Correct specification of longitudinal model in lme4 | You have a large number of groups, so I speculate that (depending on the setting) you may think about group as a random effect, so your (…|grp) terms are probably justified. It may also be reasonable to associate random effects with individuals ((…|id) terms). However you have time as a covariate in all your models, so... | Correct specification of longitudinal model in lme4 | You have a large number of groups, so I speculate that (depending on the setting) you may think about group as a random effect, so your (…|grp) terms are probably justified. It may also be reasonable | Correct specification of longitudinal model in lme4
You have a large number of groups, so I speculate that (depending on the setting) you may think about group as a random effect, so your (…|grp) terms are probably justified. It may also be reasonable to associate random effects with individuals ((…|id) terms). However... | Correct specification of longitudinal model in lme4
You have a large number of groups, so I speculate that (depending on the setting) you may think about group as a random effect, so your (…|grp) terms are probably justified. It may also be reasonable |
48,236 | Review of case control matching algorithms? | For an overview of some matching algorithms as well as clear examples of applications in everything from education to medical experiments, I would suggest:
Paul R. Rosenbaum (2010). Design of Observational Studies. Springer.
Rosenbaum's earlier book provides a more technical review, though because matching is such a ... | Review of case control matching algorithms? | For an overview of some matching algorithms as well as clear examples of applications in everything from education to medical experiments, I would suggest:
Paul R. Rosenbaum (2010). Design of Observa | Review of case control matching algorithms?
For an overview of some matching algorithms as well as clear examples of applications in everything from education to medical experiments, I would suggest:
Paul R. Rosenbaum (2010). Design of Observational Studies. Springer.
Rosenbaum's earlier book provides a more technica... | Review of case control matching algorithms?
For an overview of some matching algorithms as well as clear examples of applications in everything from education to medical experiments, I would suggest:
Paul R. Rosenbaum (2010). Design of Observa |
48,237 | Review of case control matching algorithms? | Try the What's New in Econometrics section by Imbens and Wooldridge here:
http://www.nber.org/WNE/lect_1_match_fig.pdf
There is a section on matching and it is aimed at the informed practitioner, not suitable for the complete novice. | Review of case control matching algorithms? | Try the What's New in Econometrics section by Imbens and Wooldridge here:
http://www.nber.org/WNE/lect_1_match_fig.pdf
There is a section on matching and it is aimed at the informed practitioner, not | Review of case control matching algorithms?
Try the What's New in Econometrics section by Imbens and Wooldridge here:
http://www.nber.org/WNE/lect_1_match_fig.pdf
There is a section on matching and it is aimed at the informed practitioner, not suitable for the complete novice. | Review of case control matching algorithms?
Try the What's New in Econometrics section by Imbens and Wooldridge here:
http://www.nber.org/WNE/lect_1_match_fig.pdf
There is a section on matching and it is aimed at the informed practitioner, not |
48,238 | Modelling the effect of a 2 by 4 mixed design on a three-level nominal dependent variable | A log-linear model or any model that fails to model the dependence of responses could underestimate (or overestimate) standard errors because they do not take into account potential subject-level association of responses. For example, if some subjects are likely to have responses patterns like (A,A,A,A) and others like... | Modelling the effect of a 2 by 4 mixed design on a three-level nominal dependent variable | A log-linear model or any model that fails to model the dependence of responses could underestimate (or overestimate) standard errors because they do not take into account potential subject-level asso | Modelling the effect of a 2 by 4 mixed design on a three-level nominal dependent variable
A log-linear model or any model that fails to model the dependence of responses could underestimate (or overestimate) standard errors because they do not take into account potential subject-level association of responses. For exam... | Modelling the effect of a 2 by 4 mixed design on a three-level nominal dependent variable
A log-linear model or any model that fails to model the dependence of responses could underestimate (or overestimate) standard errors because they do not take into account potential subject-level asso |
48,239 | Variance of a distribution's product with itself | $Var(X^i)$ = $\mathbb{E}[(X^i - \mathbb{E}[X^i])^2]$ = $\mathbb{E}[X^{2i}] - (\mathbb{E}[X^i])^2$ by definition. This expresses the variance of $X^i$ in terms of moments of $X$.
The generalization is false, because $\mathbb{E}[(\lambda X)^{2i}]$ = $|\lambda|^{2i}\mathbb{E}[X^{2i}]$ implies that the scale parameter of ... | Variance of a distribution's product with itself | $Var(X^i)$ = $\mathbb{E}[(X^i - \mathbb{E}[X^i])^2]$ = $\mathbb{E}[X^{2i}] - (\mathbb{E}[X^i])^2$ by definition. This expresses the variance of $X^i$ in terms of moments of $X$.
The generalization is | Variance of a distribution's product with itself
$Var(X^i)$ = $\mathbb{E}[(X^i - \mathbb{E}[X^i])^2]$ = $\mathbb{E}[X^{2i}] - (\mathbb{E}[X^i])^2$ by definition. This expresses the variance of $X^i$ in terms of moments of $X$.
The generalization is false, because $\mathbb{E}[(\lambda X)^{2i}]$ = $|\lambda|^{2i}\mathbb... | Variance of a distribution's product with itself
$Var(X^i)$ = $\mathbb{E}[(X^i - \mathbb{E}[X^i])^2]$ = $\mathbb{E}[X^{2i}] - (\mathbb{E}[X^i])^2$ by definition. This expresses the variance of $X^i$ in terms of moments of $X$.
The generalization is |
48,240 | Variance of a distribution's product with itself | Quick note, you may find useful discussion of why the formula for estimating the standard deviation for a sample uses $n-1$ as opposed to $n$.
Paul Savory Why divide by (n-1) for sample standard deviation
Graphpad Why use n-1 when calculating a standard deviation?
Andrew Hardwick Why there is a Minus One in Standard D... | Variance of a distribution's product with itself | Quick note, you may find useful discussion of why the formula for estimating the standard deviation for a sample uses $n-1$ as opposed to $n$.
Paul Savory Why divide by (n-1) for sample standard devi | Variance of a distribution's product with itself
Quick note, you may find useful discussion of why the formula for estimating the standard deviation for a sample uses $n-1$ as opposed to $n$.
Paul Savory Why divide by (n-1) for sample standard deviation
Graphpad Why use n-1 when calculating a standard deviation?
Andre... | Variance of a distribution's product with itself
Quick note, you may find useful discussion of why the formula for estimating the standard deviation for a sample uses $n-1$ as opposed to $n$.
Paul Savory Why divide by (n-1) for sample standard devi |
48,241 | Overfit by removing misclassified objects? | The following is not restricted to NB + LogRes
Overfitting = Loss of generalization.
When you train a model on dataset you generally assume that the data you use for training has a similar structure than the data the model is applied to later (the general assumption of predicting the future from the past). So if you re... | Overfit by removing misclassified objects? | The following is not restricted to NB + LogRes
Overfitting = Loss of generalization.
When you train a model on dataset you generally assume that the data you use for training has a similar structure t | Overfit by removing misclassified objects?
The following is not restricted to NB + LogRes
Overfitting = Loss of generalization.
When you train a model on dataset you generally assume that the data you use for training has a similar structure than the data the model is applied to later (the general assumption of predict... | Overfit by removing misclassified objects?
The following is not restricted to NB + LogRes
Overfitting = Loss of generalization.
When you train a model on dataset you generally assume that the data you use for training has a similar structure t |
48,242 | Overfit by removing misclassified objects? | Naive Bayes and Logistic Regression (Classification) are both linear classifiers. If you remove all misclassified instances, then you will allow an infinite number of separators to have 0 training error. In the case of the logistic regression, this translate to your information matrix being singular (The information ma... | Overfit by removing misclassified objects? | Naive Bayes and Logistic Regression (Classification) are both linear classifiers. If you remove all misclassified instances, then you will allow an infinite number of separators to have 0 training err | Overfit by removing misclassified objects?
Naive Bayes and Logistic Regression (Classification) are both linear classifiers. If you remove all misclassified instances, then you will allow an infinite number of separators to have 0 training error. In the case of the logistic regression, this translate to your informatio... | Overfit by removing misclassified objects?
Naive Bayes and Logistic Regression (Classification) are both linear classifiers. If you remove all misclassified instances, then you will allow an infinite number of separators to have 0 training err |
48,243 | Efficient way to classify with SVM | I would do two things. First, to address your issue with accuracy due to imbalanced data, you need to set the cost of misclassifying positive and negative examples separately. A reasonable rule of thumb in your case would be to set the cost to 5 for the larger class and to 95 for the smaller class. This way misclassify... | Efficient way to classify with SVM | I would do two things. First, to address your issue with accuracy due to imbalanced data, you need to set the cost of misclassifying positive and negative examples separately. A reasonable rule of thu | Efficient way to classify with SVM
I would do two things. First, to address your issue with accuracy due to imbalanced data, you need to set the cost of misclassifying positive and negative examples separately. A reasonable rule of thumb in your case would be to set the cost to 5 for the larger class and to 95 for the ... | Efficient way to classify with SVM
I would do two things. First, to address your issue with accuracy due to imbalanced data, you need to set the cost of misclassifying positive and negative examples separately. A reasonable rule of thu |
48,244 | Efficient way to classify with SVM | By Default libSVM find the optimal hyper-parameters, for the SVM model using cross validation methods and by using Accuracy (for classification), or Mean Square Error (for regression) as a measure for evaluation.
Weka has several other evaluation metric to find the optimal parameters (using the gridSearch)
If the metri... | Efficient way to classify with SVM | By Default libSVM find the optimal hyper-parameters, for the SVM model using cross validation methods and by using Accuracy (for classification), or Mean Square Error (for regression) as a measure for | Efficient way to classify with SVM
By Default libSVM find the optimal hyper-parameters, for the SVM model using cross validation methods and by using Accuracy (for classification), or Mean Square Error (for regression) as a measure for evaluation.
Weka has several other evaluation metric to find the optimal parameters ... | Efficient way to classify with SVM
By Default libSVM find the optimal hyper-parameters, for the SVM model using cross validation methods and by using Accuracy (for classification), or Mean Square Error (for regression) as a measure for |
48,245 | Analyzing a 2x3 repeated measures design using a logit mixed model | I would definitely use all your data and add direction of intended bias induction as a variable. Since you already have variables in the model describing the difference between the tasks, I don't believe that adding task as a random effect is necessary. The model would be:
my_model = lmer(
correct ~ (1|subject) + c... | Analyzing a 2x3 repeated measures design using a logit mixed model | I would definitely use all your data and add direction of intended bias induction as a variable. Since you already have variables in the model describing the difference between the tasks, I don't beli | Analyzing a 2x3 repeated measures design using a logit mixed model
I would definitely use all your data and add direction of intended bias induction as a variable. Since you already have variables in the model describing the difference between the tasks, I don't believe that adding task as a random effect is necessary.... | Analyzing a 2x3 repeated measures design using a logit mixed model
I would definitely use all your data and add direction of intended bias induction as a variable. Since you already have variables in the model describing the difference between the tasks, I don't beli |
48,246 | Analyzing a 2x3 repeated measures design using a logit mixed model | I disagree with the other responder, you shouldn't just fit varying intercept, fixed slope by default.
y~1+condition*distractor*direction+(1+condition*distractor*direction|subject) should at least be fit for model testing. Otherwise, you are just assuming the subject effect is only on the intercept and you are assuming... | Analyzing a 2x3 repeated measures design using a logit mixed model | I disagree with the other responder, you shouldn't just fit varying intercept, fixed slope by default.
y~1+condition*distractor*direction+(1+condition*distractor*direction|subject) should at least be | Analyzing a 2x3 repeated measures design using a logit mixed model
I disagree with the other responder, you shouldn't just fit varying intercept, fixed slope by default.
y~1+condition*distractor*direction+(1+condition*distractor*direction|subject) should at least be fit for model testing. Otherwise, you are just assumi... | Analyzing a 2x3 repeated measures design using a logit mixed model
I disagree with the other responder, you shouldn't just fit varying intercept, fixed slope by default.
y~1+condition*distractor*direction+(1+condition*distractor*direction|subject) should at least be |
48,247 | Alternative ways for interpretation of odds | The answer is near the bottom of p166. It's using a linear approximation (what social scientists would call a 'marginal effect'). A small change $\delta x$ in $x$ gives a change in probability of:
$$\delta\pi \approx \frac{\partial \pi(x)}{\partial x} \delta x.$$
With $\operatorname{logit}(\pi(x)) = \alpha + \beta x$, ... | Alternative ways for interpretation of odds | The answer is near the bottom of p166. It's using a linear approximation (what social scientists would call a 'marginal effect'). A small change $\delta x$ in $x$ gives a change in probability of:
$$\ | Alternative ways for interpretation of odds
The answer is near the bottom of p166. It's using a linear approximation (what social scientists would call a 'marginal effect'). A small change $\delta x$ in $x$ gives a change in probability of:
$$\delta\pi \approx \frac{\partial \pi(x)}{\partial x} \delta x.$$
With $\opera... | Alternative ways for interpretation of odds
The answer is near the bottom of p166. It's using a linear approximation (what social scientists would call a 'marginal effect'). A small change $\delta x$ in $x$ gives a change in probability of:
$$\ |
48,248 | What is the conjugate of the noncentral hypergeometric distribution? | Perhaps a conjugate prior does not exist for the noncentral hypergeometric distribution.
If someone wants to confirm this, it is worth noting that the conjugate to a univariate hypergeometric distribution is the beta-binomial; the conjugate to a multivariate hypergeometric distribution is the Dirichlet-Multinomial (fro... | What is the conjugate of the noncentral hypergeometric distribution? | Perhaps a conjugate prior does not exist for the noncentral hypergeometric distribution.
If someone wants to confirm this, it is worth noting that the conjugate to a univariate hypergeometric distribu | What is the conjugate of the noncentral hypergeometric distribution?
Perhaps a conjugate prior does not exist for the noncentral hypergeometric distribution.
If someone wants to confirm this, it is worth noting that the conjugate to a univariate hypergeometric distribution is the beta-binomial; the conjugate to a multi... | What is the conjugate of the noncentral hypergeometric distribution?
Perhaps a conjugate prior does not exist for the noncentral hypergeometric distribution.
If someone wants to confirm this, it is worth noting that the conjugate to a univariate hypergeometric distribu |
48,249 | What is the conjugate of the noncentral hypergeometric distribution? | From the statement of the question, it seems as though you don't require conjugacy per se, rather you would like an analytical solution to your integration. From the form of the distribution, it would appear at first glance that the analytics of most solutions would be rather messy and difficult to interpret. The "an... | What is the conjugate of the noncentral hypergeometric distribution? | From the statement of the question, it seems as though you don't require conjugacy per se, rather you would like an analytical solution to your integration. From the form of the distribution, it woul | What is the conjugate of the noncentral hypergeometric distribution?
From the statement of the question, it seems as though you don't require conjugacy per se, rather you would like an analytical solution to your integration. From the form of the distribution, it would appear at first glance that the analytics of most... | What is the conjugate of the noncentral hypergeometric distribution?
From the statement of the question, it seems as though you don't require conjugacy per se, rather you would like an analytical solution to your integration. From the form of the distribution, it woul |
48,250 | Find out pseudo R square value for a Logistic Regression analysis [closed] | Take a look at the lrm() function from the Design package. It features everything you need for fitting GLM. The Hosmer and Lemeshow test has limited power and depends on arbitrary discretization; it is discussed in Harrell, Regression Modeling Strategies (p. 231) and on the R-help mailing-list. There is also a comparis... | Find out pseudo R square value for a Logistic Regression analysis [closed] | Take a look at the lrm() function from the Design package. It features everything you need for fitting GLM. The Hosmer and Lemeshow test has limited power and depends on arbitrary discretization; it i | Find out pseudo R square value for a Logistic Regression analysis [closed]
Take a look at the lrm() function from the Design package. It features everything you need for fitting GLM. The Hosmer and Lemeshow test has limited power and depends on arbitrary discretization; it is discussed in Harrell, Regression Modeling S... | Find out pseudo R square value for a Logistic Regression analysis [closed]
Take a look at the lrm() function from the Design package. It features everything you need for fitting GLM. The Hosmer and Lemeshow test has limited power and depends on arbitrary discretization; it i |
48,251 | Find out pseudo R square value for a Logistic Regression analysis [closed] | Pseudo R Square is very easy to calculate manually. You just need to look up the -2LL value for the baseline based on the average probability of occurrence of the binomial event. And, you need the -2LL value for the actual Logistic regression.
Let's say the -2LL value for the baseline is 10 and for the Logistic reg... | Find out pseudo R square value for a Logistic Regression analysis [closed] | Pseudo R Square is very easy to calculate manually. You just need to look up the -2LL value for the baseline based on the average probability of occurrence of the binomial event. And, you need the - | Find out pseudo R square value for a Logistic Regression analysis [closed]
Pseudo R Square is very easy to calculate manually. You just need to look up the -2LL value for the baseline based on the average probability of occurrence of the binomial event. And, you need the -2LL value for the actual Logistic regression.... | Find out pseudo R square value for a Logistic Regression analysis [closed]
Pseudo R Square is very easy to calculate manually. You just need to look up the -2LL value for the baseline based on the average probability of occurrence of the binomial event. And, you need the - |
48,252 | How can the IID assumption be checked in a given dataset? | You don't frame the two problems the right way.
Given a random dataset, ie a collection of observations $x_{ij}$ lying in general position you can always make the $n$ $x_{i}\in\mathbb{R}^p$ independent from one another by randomly shuffling the $n$ indexes. The real question is whether you will lose information doing ... | How can the IID assumption be checked in a given dataset? | You don't frame the two problems the right way.
Given a random dataset, ie a collection of observations $x_{ij}$ lying in general position you can always make the $n$ $x_{i}\in\mathbb{R}^p$ independe | How can the IID assumption be checked in a given dataset?
You don't frame the two problems the right way.
Given a random dataset, ie a collection of observations $x_{ij}$ lying in general position you can always make the $n$ $x_{i}\in\mathbb{R}^p$ independent from one another by randomly shuffling the $n$ indexes. The... | How can the IID assumption be checked in a given dataset?
You don't frame the two problems the right way.
Given a random dataset, ie a collection of observations $x_{ij}$ lying in general position you can always make the $n$ $x_{i}\in\mathbb{R}^p$ independe |
48,253 | How can the IID assumption be checked in a given dataset? | Whether a set of observations are iid or not is a decision that is typically taken after a consideration of the underlying data generating process. In your case, the underlying data generating process seems to be the measurements of the speed of a river. I would not consider these observations to be independent. If a p... | How can the IID assumption be checked in a given dataset? | Whether a set of observations are iid or not is a decision that is typically taken after a consideration of the underlying data generating process. In your case, the underlying data generating process | How can the IID assumption be checked in a given dataset?
Whether a set of observations are iid or not is a decision that is typically taken after a consideration of the underlying data generating process. In your case, the underlying data generating process seems to be the measurements of the speed of a river. I would... | How can the IID assumption be checked in a given dataset?
Whether a set of observations are iid or not is a decision that is typically taken after a consideration of the underlying data generating process. In your case, the underlying data generating process |
48,254 | Checking assumptions for random effects in nested mixed-effects models in R / S-Plus | It seems you are using the nlme package. Maybe it would be worth trying R and the lme4 instead, although it is not fully comparable wrt. syntax or function call.
In your case, I would suggest to specify the level when you called ranef(), see ?ranef.lme:
level: an optional vector of positive integers giving the lev... | Checking assumptions for random effects in nested mixed-effects models in R / S-Plus | It seems you are using the nlme package. Maybe it would be worth trying R and the lme4 instead, although it is not fully comparable wrt. syntax or function call.
In your case, I would suggest to spec | Checking assumptions for random effects in nested mixed-effects models in R / S-Plus
It seems you are using the nlme package. Maybe it would be worth trying R and the lme4 instead, although it is not fully comparable wrt. syntax or function call.
In your case, I would suggest to specify the level when you called ranef... | Checking assumptions for random effects in nested mixed-effects models in R / S-Plus
It seems you are using the nlme package. Maybe it would be worth trying R and the lme4 instead, although it is not fully comparable wrt. syntax or function call.
In your case, I would suggest to spec |
48,255 | How to draw a probable outcome from a distribution? | I also think that it's not clear what you want. But if you want a set of deterministically chosen points, so that they preserve the moments of the initial distribution, you can use the sigma point selection method that applies to the unscented Kalman filter.
Say that you want to select $2L+1$ points that fulfill thos... | How to draw a probable outcome from a distribution? | I also think that it's not clear what you want. But if you want a set of deterministically chosen points, so that they preserve the moments of the initial distribution, you can use the sigma point sel | How to draw a probable outcome from a distribution?
I also think that it's not clear what you want. But if you want a set of deterministically chosen points, so that they preserve the moments of the initial distribution, you can use the sigma point selection method that applies to the unscented Kalman filter.
Say tha... | How to draw a probable outcome from a distribution?
I also think that it's not clear what you want. But if you want a set of deterministically chosen points, so that they preserve the moments of the initial distribution, you can use the sigma point sel |
48,256 | How to draw a probable outcome from a distribution? | To summarise (please correct me if I'm wrong):
You have a set of points for a number of parameters/states.
The points provide a joint distribution of the parameters states
You want to simulate from a model using some typical states.
The problem you have is that you can't write down a nice closed form density.
To tack... | How to draw a probable outcome from a distribution? | To summarise (please correct me if I'm wrong):
You have a set of points for a number of parameters/states.
The points provide a joint distribution of the parameters states
You want to simulate from a | How to draw a probable outcome from a distribution?
To summarise (please correct me if I'm wrong):
You have a set of points for a number of parameters/states.
The points provide a joint distribution of the parameters states
You want to simulate from a model using some typical states.
The problem you have is that you ... | How to draw a probable outcome from a distribution?
To summarise (please correct me if I'm wrong):
You have a set of points for a number of parameters/states.
The points provide a joint distribution of the parameters states
You want to simulate from a |
48,257 | How to draw a probable outcome from a distribution? | One thing that you could do is to plot the position of all your experiments in the 2D plane, one point for each object, maybe colored by experiment (if you have a lot of experiments you may just plot a random subset of them).
If there is a pattern in the position of the objects it should emerge when doing this.
Also, d... | How to draw a probable outcome from a distribution? | One thing that you could do is to plot the position of all your experiments in the 2D plane, one point for each object, maybe colored by experiment (if you have a lot of experiments you may just plot | How to draw a probable outcome from a distribution?
One thing that you could do is to plot the position of all your experiments in the 2D plane, one point for each object, maybe colored by experiment (if you have a lot of experiments you may just plot a random subset of them).
If there is a pattern in the position of t... | How to draw a probable outcome from a distribution?
One thing that you could do is to plot the position of all your experiments in the 2D plane, one point for each object, maybe colored by experiment (if you have a lot of experiments you may just plot |
48,258 | How to draw a probable outcome from a distribution? | Maybe you could use a smoothed scatterplot? It is an analogy to kernel density approximation, but in 2D. | How to draw a probable outcome from a distribution? | Maybe you could use a smoothed scatterplot? It is an analogy to kernel density approximation, but in 2D. | How to draw a probable outcome from a distribution?
Maybe you could use a smoothed scatterplot? It is an analogy to kernel density approximation, but in 2D. | How to draw a probable outcome from a distribution?
Maybe you could use a smoothed scatterplot? It is an analogy to kernel density approximation, but in 2D. |
48,259 | How can one speed up this correlation calculation in R without multicore? | While making a call to diag you throw out a lot of information, so you can save time by simply not calculating it. You code is equivalent to:
sapply(1:100,function(i) cor(x[i,],y[i,]))
Extended to reflect comments: This code will be slower for small matrices since it does not use the full "vectorization power" of cor.... | How can one speed up this correlation calculation in R without multicore? | While making a call to diag you throw out a lot of information, so you can save time by simply not calculating it. You code is equivalent to:
sapply(1:100,function(i) cor(x[i,],y[i,]))
Extended to re | How can one speed up this correlation calculation in R without multicore?
While making a call to diag you throw out a lot of information, so you can save time by simply not calculating it. You code is equivalent to:
sapply(1:100,function(i) cor(x[i,],y[i,]))
Extended to reflect comments: This code will be slower for s... | How can one speed up this correlation calculation in R without multicore?
While making a call to diag you throw out a lot of information, so you can save time by simply not calculating it. You code is equivalent to:
sapply(1:100,function(i) cor(x[i,],y[i,]))
Extended to re |
48,260 | How can one speed up this correlation calculation in R without multicore? | This really depends on the relative numbers of "scores" and "subjects". The method you use calculates lots of cross-correlations which are not required. However, if there are relatively few "subjects" relative to "scores", then this probably doesn't matter too much, and the method you suggest is probably as good as any... | How can one speed up this correlation calculation in R without multicore? | This really depends on the relative numbers of "scores" and "subjects". The method you use calculates lots of cross-correlations which are not required. However, if there are relatively few "subjects" | How can one speed up this correlation calculation in R without multicore?
This really depends on the relative numbers of "scores" and "subjects". The method you use calculates lots of cross-correlations which are not required. However, if there are relatively few "subjects" relative to "scores", then this probably does... | How can one speed up this correlation calculation in R without multicore?
This really depends on the relative numbers of "scores" and "subjects". The method you use calculates lots of cross-correlations which are not required. However, if there are relatively few "subjects" |
48,261 | How can one speed up this correlation calculation in R without multicore? | It might be one of those cases where using a different BLAS engine would help. But I am not sure of it - it needs testing (and depends on your machine) | How can one speed up this correlation calculation in R without multicore? | It might be one of those cases where using a different BLAS engine would help. But I am not sure of it - it needs testing (and depends on your machine) | How can one speed up this correlation calculation in R without multicore?
It might be one of those cases where using a different BLAS engine would help. But I am not sure of it - it needs testing (and depends on your machine) | How can one speed up this correlation calculation in R without multicore?
It might be one of those cases where using a different BLAS engine would help. But I am not sure of it - it needs testing (and depends on your machine) |
48,262 | How could XGBoost beat perfect logistic regression? | I have upvoted Sycorax's answer as useful. Nevertheless, I think there is a serious issue with the question: sklearn.linear_model.LogisticRegression.score returns the mean accuracy, not AUC-ROC. If we used LR.predict_proba(df_1)[:,1] to get the predicted probabilistic estimates AUC-ROC values both in the training and... | How could XGBoost beat perfect logistic regression? | I have upvoted Sycorax's answer as useful. Nevertheless, I think there is a serious issue with the question: sklearn.linear_model.LogisticRegression.score returns the mean accuracy, not AUC-ROC. If w | How could XGBoost beat perfect logistic regression?
I have upvoted Sycorax's answer as useful. Nevertheless, I think there is a serious issue with the question: sklearn.linear_model.LogisticRegression.score returns the mean accuracy, not AUC-ROC. If we used LR.predict_proba(df_1)[:,1] to get the predicted probabilist... | How could XGBoost beat perfect logistic regression?
I have upvoted Sycorax's answer as useful. Nevertheless, I think there is a serious issue with the question: sklearn.linear_model.LogisticRegression.score returns the mean accuracy, not AUC-ROC. If w |
48,263 | Average marginal means with marginaleffects | Yes, they are actually not called by marginal means in marginaleffects but rather adjusted predictions, and they can be estimated using the predictions() function. To compare adjusted predictions, you can uset he hypothesis argument of predicitions() or the comparisons() function, which is specifically designed for g-c... | Average marginal means with marginaleffects | Yes, they are actually not called by marginal means in marginaleffects but rather adjusted predictions, and they can be estimated using the predictions() function. To compare adjusted predictions, you | Average marginal means with marginaleffects
Yes, they are actually not called by marginal means in marginaleffects but rather adjusted predictions, and they can be estimated using the predictions() function. To compare adjusted predictions, you can uset he hypothesis argument of predicitions() or the comparisons() func... | Average marginal means with marginaleffects
Yes, they are actually not called by marginal means in marginaleffects but rather adjusted predictions, and they can be estimated using the predictions() function. To compare adjusted predictions, you |
48,264 | Average marginal means with marginaleffects | This truly is a different answer...
You won't believe this, but this can be done via a new counterfactuals argument that I added to ref_grid():
> emmeans(mod, "species", counterfact = "species")
species prob SE df asymp.LCL asymp.UCL
Adelie 0.5821 0.0367 Inf 0.510 0.6540
Chinstrap 0.0602 0.0190 In... | Average marginal means with marginaleffects | This truly is a different answer...
You won't believe this, but this can be done via a new counterfactuals argument that I added to ref_grid():
> emmeans(mod, "species", counterfact = "species")
spec | Average marginal means with marginaleffects
This truly is a different answer...
You won't believe this, but this can be done via a new counterfactuals argument that I added to ref_grid():
> emmeans(mod, "species", counterfact = "species")
species prob SE df asymp.LCL asymp.UCL
Adelie 0.5821 0.0367 Inf ... | Average marginal means with marginaleffects
This truly is a different answer...
You won't believe this, but this can be done via a new counterfactuals argument that I added to ref_grid():
> emmeans(mod, "species", counterfact = "species")
spec |
48,265 | Gaussian fourth-moment formulas? | I have never seen an closed form expression for this. Probably because it is quite ugly. I have worked with a similar expression before, and I'd be happy to see if my expression is stands up to yours. So here is one way to compute the matrix expectation.
I will work with $\frac{X^TX}{m}$ rather than $X^TX$, since that ... | Gaussian fourth-moment formulas? | I have never seen an closed form expression for this. Probably because it is quite ugly. I have worked with a similar expression before, and I'd be happy to see if my expression is stands up to yours. | Gaussian fourth-moment formulas?
I have never seen an closed form expression for this. Probably because it is quite ugly. I have worked with a similar expression before, and I'd be happy to see if my expression is stands up to yours. So here is one way to compute the matrix expectation.
I will work with $\frac{X^TX}{m}... | Gaussian fourth-moment formulas?
I have never seen an closed form expression for this. Probably because it is quite ugly. I have worked with a similar expression before, and I'd be happy to see if my expression is stands up to yours. |
48,266 | How to prove that $\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B)$? | The answer is a one-liner, but that's disingenuous because (as you can see) it's going to take me some time to get to it!
Before setting out, let's establish a general yet accurate way to visualize what's going on.
Seeing things
This figure schematically draws $\Omega$ as the interior of a rectangle.
When $X:\Omega\to... | How to prove that $\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B)$? | The answer is a one-liner, but that's disingenuous because (as you can see) it's going to take me some time to get to it!
Before setting out, let's establish a general yet accurate way to visualize wh | How to prove that $\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B)$?
The answer is a one-liner, but that's disingenuous because (as you can see) it's going to take me some time to get to it!
Before setting out, let's establish a general yet accurate way to visualize what's going on.
Seeing ... | How to prove that $\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B)$?
The answer is a one-liner, but that's disingenuous because (as you can see) it's going to take me some time to get to it!
Before setting out, let's establish a general yet accurate way to visualize wh |
48,267 | When are cluster robust standard errors a valid "alternative" to mixed models? | The answer from @Eoin gets to the heart of the matter. One models marginal, population-level values while the other models conditional, subject-level values.
Here are a few additional suggestions, drawing from the McNeish et al. paper recommended by @Noah in a comment, "On the unnecessary ubiquity of hierarchical linea... | When are cluster robust standard errors a valid "alternative" to mixed models? | The answer from @Eoin gets to the heart of the matter. One models marginal, population-level values while the other models conditional, subject-level values.
Here are a few additional suggestions, dra | When are cluster robust standard errors a valid "alternative" to mixed models?
The answer from @Eoin gets to the heart of the matter. One models marginal, population-level values while the other models conditional, subject-level values.
Here are a few additional suggestions, drawing from the McNeish et al. paper recomm... | When are cluster robust standard errors a valid "alternative" to mixed models?
The answer from @Eoin gets to the heart of the matter. One models marginal, population-level values while the other models conditional, subject-level values.
Here are a few additional suggestions, dra |
48,268 | When are cluster robust standard errors a valid "alternative" to mixed models? | Good question!
I think the short answer is that the working independence linear model and the random effects model by design estimate different quantities, and in your example these quantities are not the same. In the simplest example...
m = lm(y ~ 1, data = data)
mixed_m = lmer(y ~ 1 + (1|id), data = data)
$$
\begin{... | When are cluster robust standard errors a valid "alternative" to mixed models? | Good question!
I think the short answer is that the working independence linear model and the random effects model by design estimate different quantities, and in your example these quantities are not | When are cluster robust standard errors a valid "alternative" to mixed models?
Good question!
I think the short answer is that the working independence linear model and the random effects model by design estimate different quantities, and in your example these quantities are not the same. In the simplest example...
m =... | When are cluster robust standard errors a valid "alternative" to mixed models?
Good question!
I think the short answer is that the working independence linear model and the random effects model by design estimate different quantities, and in your example these quantities are not |
48,269 | Expected Fisher information isn't positive definite for truncated normal with heteroskedasticity | The expected Fisher information is positive-definite by definition, so there must be some mistake in your code.
Here is exactly the same calculation using MATLAB's symbolic math toolbox, which produces a positive-definite matrix with the same inputs:
(it's enough to check the case $k=1$, since a sum of positive definit... | Expected Fisher information isn't positive definite for truncated normal with heteroskedasticity | The expected Fisher information is positive-definite by definition, so there must be some mistake in your code.
Here is exactly the same calculation using MATLAB's symbolic math toolbox, which produce | Expected Fisher information isn't positive definite for truncated normal with heteroskedasticity
The expected Fisher information is positive-definite by definition, so there must be some mistake in your code.
Here is exactly the same calculation using MATLAB's symbolic math toolbox, which produces a positive-definite m... | Expected Fisher information isn't positive definite for truncated normal with heteroskedasticity
The expected Fisher information is positive-definite by definition, so there must be some mistake in your code.
Here is exactly the same calculation using MATLAB's symbolic math toolbox, which produce |
48,270 | 3-D random walk: average distance after N steps | Let's solve this in all dimensions $d=1,2,3,\ldots.$
The (vector) increments of the walk are $\mathbf{X}_i = (x_{1i}, x_{2i}, \ldots, x_{di}).$ After $n$ such independent steps the walk has reached the point $\mathbf{S}_n = \mathbf{X}_1 + \mathbf{X}_2 + \cdots + \mathbf{X}_n$ with corresponding components $s_{1n}, \ld... | 3-D random walk: average distance after N steps | Let's solve this in all dimensions $d=1,2,3,\ldots.$
The (vector) increments of the walk are $\mathbf{X}_i = (x_{1i}, x_{2i}, \ldots, x_{di}).$ After $n$ such independent steps the walk has reached t | 3-D random walk: average distance after N steps
Let's solve this in all dimensions $d=1,2,3,\ldots.$
The (vector) increments of the walk are $\mathbf{X}_i = (x_{1i}, x_{2i}, \ldots, x_{di}).$ After $n$ such independent steps the walk has reached the point $\mathbf{S}_n = \mathbf{X}_1 + \mathbf{X}_2 + \cdots + \mathbf{... | 3-D random walk: average distance after N steps
Let's solve this in all dimensions $d=1,2,3,\ldots.$
The (vector) increments of the walk are $\mathbf{X}_i = (x_{1i}, x_{2i}, \ldots, x_{di}).$ After $n$ such independent steps the walk has reached t |
48,271 | Measure of similarity between two distributions - with variable starting point (wrap-around) | This problem is more challenging than it might look, for many reasons--some of which will become apparent in discussing one set of solutions. I was moved to post this discussion because of the emergence of several surprising results, illustrated at the end.
In the interests of space, I will focus on presenting one sol... | Measure of similarity between two distributions - with variable starting point (wrap-around) | This problem is more challenging than it might look, for many reasons--some of which will become apparent in discussing one set of solutions. I was moved to post this discussion because of the emerge | Measure of similarity between two distributions - with variable starting point (wrap-around)
This problem is more challenging than it might look, for many reasons--some of which will become apparent in discussing one set of solutions. I was moved to post this discussion because of the emergence of several surprising r... | Measure of similarity between two distributions - with variable starting point (wrap-around)
This problem is more challenging than it might look, for many reasons--some of which will become apparent in discussing one set of solutions. I was moved to post this discussion because of the emerge |
48,272 | Measure of similarity between two distributions - with variable starting point (wrap-around) | Since there are still no other answers to this question and several comments by @whuber highlighted flaws in using Pearson's Chi-Square test, here is a second answer that uses the Mann-Whitney U test. Before answering, I'll point out that the question itself includes 3 possible solutions that the author seemed interest... | Measure of similarity between two distributions - with variable starting point (wrap-around) | Since there are still no other answers to this question and several comments by @whuber highlighted flaws in using Pearson's Chi-Square test, here is a second answer that uses the Mann-Whitney U test. | Measure of similarity between two distributions - with variable starting point (wrap-around)
Since there are still no other answers to this question and several comments by @whuber highlighted flaws in using Pearson's Chi-Square test, here is a second answer that uses the Mann-Whitney U test. Before answering, I'll poi... | Measure of similarity between two distributions - with variable starting point (wrap-around)
Since there are still no other answers to this question and several comments by @whuber highlighted flaws in using Pearson's Chi-Square test, here is a second answer that uses the Mann-Whitney U test. |
48,273 | Measure of similarity between two distributions - with variable starting point (wrap-around) | Here is what I would do in your situation. Whether it works for you depends on the format and quantity of your data, but I do think that it will yield the answer you're looking for if the data allows this method. The method is a regular Pearson's chi-squared test. The reason this choice makes sense is that music is arr... | Measure of similarity between two distributions - with variable starting point (wrap-around) | Here is what I would do in your situation. Whether it works for you depends on the format and quantity of your data, but I do think that it will yield the answer you're looking for if the data allows | Measure of similarity between two distributions - with variable starting point (wrap-around)
Here is what I would do in your situation. Whether it works for you depends on the format and quantity of your data, but I do think that it will yield the answer you're looking for if the data allows this method. The method is ... | Measure of similarity between two distributions - with variable starting point (wrap-around)
Here is what I would do in your situation. Whether it works for you depends on the format and quantity of your data, but I do think that it will yield the answer you're looking for if the data allows |
48,274 | How do Bayesians interpret unobservable model parameters? | You can find a good primer on the Bayesian interpretation of these types of models in Bernardo and Smith (1994). In that work they take an "operational" approach where model parameters are interpreted as limiting quantities that are functions of the observable sequence. You can also find a more detailed discussion of... | How do Bayesians interpret unobservable model parameters? | You can find a good primer on the Bayesian interpretation of these types of models in Bernardo and Smith (1994). In that work they take an "operational" approach where model parameters are interprete | How do Bayesians interpret unobservable model parameters?
You can find a good primer on the Bayesian interpretation of these types of models in Bernardo and Smith (1994). In that work they take an "operational" approach where model parameters are interpreted as limiting quantities that are functions of the observable ... | How do Bayesians interpret unobservable model parameters?
You can find a good primer on the Bayesian interpretation of these types of models in Bernardo and Smith (1994). In that work they take an "operational" approach where model parameters are interprete |
48,275 | Why is the size of the feature bagging sample typically the square root of the total predictor set size? | It's just a recommended default value. Leo Breiman observed that this value tends to work well on the classification problems that he worked on, but I'm not aware of any rigorous demonstration that this value must work best on all problems; indeed, the discovery that a different value works better on a specific problem... | Why is the size of the feature bagging sample typically the square root of the total predictor set s | It's just a recommended default value. Leo Breiman observed that this value tends to work well on the classification problems that he worked on, but I'm not aware of any rigorous demonstration that th | Why is the size of the feature bagging sample typically the square root of the total predictor set size?
It's just a recommended default value. Leo Breiman observed that this value tends to work well on the classification problems that he worked on, but I'm not aware of any rigorous demonstration that this value must w... | Why is the size of the feature bagging sample typically the square root of the total predictor set s
It's just a recommended default value. Leo Breiman observed that this value tends to work well on the classification problems that he worked on, but I'm not aware of any rigorous demonstration that th |
48,276 | Independent copula vs Student-$t$ copula with zero correlation matrix? | The uncorrelated $t$ copula is not the same as the independence copula. It is based on the multivariate $t$-distribution, which is an elliptical family, and the only elliptical distribution for which zero correlation implies independence is the normal. The difference can be quite large.
Below we will illustrate this us... | Independent copula vs Student-$t$ copula with zero correlation matrix? | The uncorrelated $t$ copula is not the same as the independence copula. It is based on the multivariate $t$-distribution, which is an elliptical family, and the only elliptical distribution for which | Independent copula vs Student-$t$ copula with zero correlation matrix?
The uncorrelated $t$ copula is not the same as the independence copula. It is based on the multivariate $t$-distribution, which is an elliptical family, and the only elliptical distribution for which zero correlation implies independence is the norm... | Independent copula vs Student-$t$ copula with zero correlation matrix?
The uncorrelated $t$ copula is not the same as the independence copula. It is based on the multivariate $t$-distribution, which is an elliptical family, and the only elliptical distribution for which |
48,277 | Is there any reason that mean squared log error uses the natural log? | By the change of base property of logarithms,
$$\log_a b = \frac{\log_c b}{\log_c a} = \frac{\ln b}{\ln a}, $$
so changing from any fixed base $a$ to any other base $c$ (or $e$) simply means dividing by a fixed value $\log_c a$, so a change of base is just a scaling. All logarithms are equivalent.
The natural logarithm... | Is there any reason that mean squared log error uses the natural log? | By the change of base property of logarithms,
$$\log_a b = \frac{\log_c b}{\log_c a} = \frac{\ln b}{\ln a}, $$
so changing from any fixed base $a$ to any other base $c$ (or $e$) simply means dividing | Is there any reason that mean squared log error uses the natural log?
By the change of base property of logarithms,
$$\log_a b = \frac{\log_c b}{\log_c a} = \frac{\ln b}{\ln a}, $$
so changing from any fixed base $a$ to any other base $c$ (or $e$) simply means dividing by a fixed value $\log_c a$, so a change of base i... | Is there any reason that mean squared log error uses the natural log?
By the change of base property of logarithms,
$$\log_a b = \frac{\log_c b}{\log_c a} = \frac{\ln b}{\ln a}, $$
so changing from any fixed base $a$ to any other base $c$ (or $e$) simply means dividing |
48,278 | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for a multivariate distribution? | Approach:
First, we approach through a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $||x_n||\to\infty$ as $n \to \infty.$ We fix an $x_n$ first.
We take an $L_\epsilon$ such that $P(||X||>L_\epsilon) <\epsilon$, i.e., we first take the n-ball where 'most of the $X$' lies.
Then, we will show that for all such 'good' v... | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\i | Approach:
First, we approach through a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $||x_n||\to\infty$ as $n \to \infty.$ We fix an $x_n$ first.
We take an $L_\epsilon$ such that $P(||X||>L_\epsilon | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for a multivariate distribution?
Approach:
First, we approach through a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $||x_n||\to\infty$ as $n \to \infty.$ We fix an $x_n$ first.
We take an $L_\epsilon$ such tha... | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\i
Approach:
First, we approach through a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $||x_n||\to\infty$ as $n \to \infty.$ We fix an $x_n$ first.
We take an $L_\epsilon$ such that $P(||X||>L_\epsilon |
48,279 | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for a multivariate distribution? | $X$ is the random variable. The quantity you are taking the expectation of can be viewed in two ways: as written, it looks like the unit vector pointing from $X$ to $x.$ One route to a simple solution is to negate this, so that it is the unit vector pointing from $x$ to $X.$ This negation changes nothing, because it ... | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\i | $X$ is the random variable. The quantity you are taking the expectation of can be viewed in two ways: as written, it looks like the unit vector pointing from $X$ to $x.$ One route to a simple solutio | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for a multivariate distribution?
$X$ is the random variable. The quantity you are taking the expectation of can be viewed in two ways: as written, it looks like the unit vector pointing from $X$ to $x.$ One rout... | How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\i
$X$ is the random variable. The quantity you are taking the expectation of can be viewed in two ways: as written, it looks like the unit vector pointing from $X$ to $x.$ One route to a simple solutio |
48,280 | Why is the ROC curve two-dimensional instead of three-dimensional? | The ROC curve is two-dimensional because it is defined as the plot of TPR and FPR. Yes, you might be interested in the threshold used to generate the TPR and FPR, and you might want to plot that on a third axis, but then it is not a ROC curve but something else (arguably more informative, arguably harder to draw, argua... | Why is the ROC curve two-dimensional instead of three-dimensional? | The ROC curve is two-dimensional because it is defined as the plot of TPR and FPR. Yes, you might be interested in the threshold used to generate the TPR and FPR, and you might want to plot that on a | Why is the ROC curve two-dimensional instead of three-dimensional?
The ROC curve is two-dimensional because it is defined as the plot of TPR and FPR. Yes, you might be interested in the threshold used to generate the TPR and FPR, and you might want to plot that on a third axis, but then it is not a ROC curve but someth... | Why is the ROC curve two-dimensional instead of three-dimensional?
The ROC curve is two-dimensional because it is defined as the plot of TPR and FPR. Yes, you might be interested in the threshold used to generate the TPR and FPR, and you might want to plot that on a |
48,281 | Basic properties of the kernel density estimator | You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k(u) u^mdu = o(h^m)$$
by the properties of the kernel function given in the assignment. Hence,
$$\mathbb{E}[f_n(t)] = f(t... | Basic properties of the kernel density estimator | You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k | Basic properties of the kernel density estimator
You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k(u) u^mdu = o(h^m)$$
by the properties of the kernel function given in ... | Basic properties of the kernel density estimator
You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k |
48,282 | AdaBoost - why decision stumps instead of trees? | The reason for using 'stumps' in boosting but full-height trees in random forests is to do with how the aggregation and fitting is done.
In random forests, the trees in the ensemble are fitted independently to independent bootstrap samples, so any error caused by growing the trees too far is independent for each tree a... | AdaBoost - why decision stumps instead of trees? | The reason for using 'stumps' in boosting but full-height trees in random forests is to do with how the aggregation and fitting is done.
In random forests, the trees in the ensemble are fitted indepen | AdaBoost - why decision stumps instead of trees?
The reason for using 'stumps' in boosting but full-height trees in random forests is to do with how the aggregation and fitting is done.
In random forests, the trees in the ensemble are fitted independently to independent bootstrap samples, so any error caused by growing... | AdaBoost - why decision stumps instead of trees?
The reason for using 'stumps' in boosting but full-height trees in random forests is to do with how the aggregation and fitting is done.
In random forests, the trees in the ensemble are fitted indepen |
48,283 | AdaBoost - why decision stumps instead of trees? | If you think about the gradient-boosted CART (aka an atom of gbm), the model is "boxes"(1).
This (data):
Is represented by a (CART fit) as this:
Each leaf-tip is a mean. Each split is perpendicular to an axis. It is trying to adjust the box bounds and the mean height to minimize error in representation.
From the ab... | AdaBoost - why decision stumps instead of trees? | If you think about the gradient-boosted CART (aka an atom of gbm), the model is "boxes"(1).
This (data):
Is represented by a (CART fit) as this:
Each leaf-tip is a mean. Each split is perpendicular | AdaBoost - why decision stumps instead of trees?
If you think about the gradient-boosted CART (aka an atom of gbm), the model is "boxes"(1).
This (data):
Is represented by a (CART fit) as this:
Each leaf-tip is a mean. Each split is perpendicular to an axis. It is trying to adjust the box bounds and the mean height... | AdaBoost - why decision stumps instead of trees?
If you think about the gradient-boosted CART (aka an atom of gbm), the model is "boxes"(1).
This (data):
Is represented by a (CART fit) as this:
Each leaf-tip is a mean. Each split is perpendicular |
48,284 | Does a Binomial converge to Poisson or Normal? | The difficulty disappears when you are careful in formulating the limits. In the first case, $p$ is not constant, so it would be more precise to write it as $p_n$, as $p$ varies with $n$. We can write $n \cdot p_n \to \lambda>0$ another way as $p_n \sim \lambda/n$, where $\sim$ means that the quotient between the two s... | Does a Binomial converge to Poisson or Normal? | The difficulty disappears when you are careful in formulating the limits. In the first case, $p$ is not constant, so it would be more precise to write it as $p_n$, as $p$ varies with $n$. We can write | Does a Binomial converge to Poisson or Normal?
The difficulty disappears when you are careful in formulating the limits. In the first case, $p$ is not constant, so it would be more precise to write it as $p_n$, as $p$ varies with $n$. We can write $n \cdot p_n \to \lambda>0$ another way as $p_n \sim \lambda/n$, where $... | Does a Binomial converge to Poisson or Normal?
The difficulty disappears when you are careful in formulating the limits. In the first case, $p$ is not constant, so it would be more precise to write it as $p_n$, as $p$ varies with $n$. We can write |
48,285 | Does a Binomial converge to Poisson or Normal? | One can write the CLT as:
$$
\frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1)
$$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identically distributed, recalling that the mean and variance of a Bernoulli random variable are $p$ and $p(1-p)$ respectively, ... | Does a Binomial converge to Poisson or Normal? | One can write the CLT as:
$$
\frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1)
$$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identica | Does a Binomial converge to Poisson or Normal?
One can write the CLT as:
$$
\frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1)
$$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identically distributed, recalling that the mean and variance of a Bernoulli rand... | Does a Binomial converge to Poisson or Normal?
One can write the CLT as:
$$
\frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1)
$$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identica |
48,286 | What is MBConv that EfficientNetv2 is using? | The bottleneck_block used as the basic building block of MobileNetv2 is the MBConv (building block of EfficientNets).
Please refer bottle_neck method in https://towardsdatascience.com/mobilenetv2-inverted-residuals-and-linear-bottlenecks-8a4362f4ffd5
MBConvs seek inverted Res nature, which is the exact contrary to the ... | What is MBConv that EfficientNetv2 is using? | The bottleneck_block used as the basic building block of MobileNetv2 is the MBConv (building block of EfficientNets).
Please refer bottle_neck method in https://towardsdatascience.com/mobilenetv2-inve | What is MBConv that EfficientNetv2 is using?
The bottleneck_block used as the basic building block of MobileNetv2 is the MBConv (building block of EfficientNets).
Please refer bottle_neck method in https://towardsdatascience.com/mobilenetv2-inverted-residuals-and-linear-bottlenecks-8a4362f4ffd5
MBConvs seek inverted Re... | What is MBConv that EfficientNetv2 is using?
The bottleneck_block used as the basic building block of MobileNetv2 is the MBConv (building block of EfficientNets).
Please refer bottle_neck method in https://towardsdatascience.com/mobilenetv2-inve |
48,287 | Variance involving two independent variables | The inequality should follow from the Law of Total Variance. Also, I'm assuming $\alpha \in [0,1]$, as otherwise the construction doesn't make sense.
We need a slightly more formal definition of your compound variable. Let $Z$ be an independent, binary variable with $P(Z=1)=\alpha$, $P(Z=0)=1-\alpha$, and thus we can d... | Variance involving two independent variables | The inequality should follow from the Law of Total Variance. Also, I'm assuming $\alpha \in [0,1]$, as otherwise the construction doesn't make sense.
We need a slightly more formal definition of your | Variance involving two independent variables
The inequality should follow from the Law of Total Variance. Also, I'm assuming $\alpha \in [0,1]$, as otherwise the construction doesn't make sense.
We need a slightly more formal definition of your compound variable. Let $Z$ be an independent, binary variable with $P(Z=1)=... | Variance involving two independent variables
The inequality should follow from the Law of Total Variance. Also, I'm assuming $\alpha \in [0,1]$, as otherwise the construction doesn't make sense.
We need a slightly more formal definition of your |
48,288 | Why is the explanatory variable non-stochastic or fixed in repeated samples? | The main reason is simply for teaching purposes: Assuming fixed explanatory variables ensures that the error term is independent of the (deterministic) variables, $E(u|X) = 0$ holds by definition, see also here. Sometimes you started with fixed / deterministic explanatory variables in a first undergraduate course to ex... | Why is the explanatory variable non-stochastic or fixed in repeated samples? | The main reason is simply for teaching purposes: Assuming fixed explanatory variables ensures that the error term is independent of the (deterministic) variables, $E(u|X) = 0$ holds by definition, see | Why is the explanatory variable non-stochastic or fixed in repeated samples?
The main reason is simply for teaching purposes: Assuming fixed explanatory variables ensures that the error term is independent of the (deterministic) variables, $E(u|X) = 0$ holds by definition, see also here. Sometimes you started with fixe... | Why is the explanatory variable non-stochastic or fixed in repeated samples?
The main reason is simply for teaching purposes: Assuming fixed explanatory variables ensures that the error term is independent of the (deterministic) variables, $E(u|X) = 0$ holds by definition, see |
48,289 | Why is the explanatory variable non-stochastic or fixed in repeated samples? | I assume this is a regression model. The main reason is inferential, the explanatory variable(s) are treated as fixed for purposes of inference. The regression model $Y= X\beta + \epsilon$ is a model for the conditional expectation of $Y$ given $X$: $\DeclareMathOperator{\E}{\mathbb{E}} \E\left\{ Y | X=x\right\} = x^T ... | Why is the explanatory variable non-stochastic or fixed in repeated samples? | I assume this is a regression model. The main reason is inferential, the explanatory variable(s) are treated as fixed for purposes of inference. The regression model $Y= X\beta + \epsilon$ is a model | Why is the explanatory variable non-stochastic or fixed in repeated samples?
I assume this is a regression model. The main reason is inferential, the explanatory variable(s) are treated as fixed for purposes of inference. The regression model $Y= X\beta + \epsilon$ is a model for the conditional expectation of $Y$ give... | Why is the explanatory variable non-stochastic or fixed in repeated samples?
I assume this is a regression model. The main reason is inferential, the explanatory variable(s) are treated as fixed for purposes of inference. The regression model $Y= X\beta + \epsilon$ is a model |
48,290 | Why do we need Gumbel distribution? | Given a sample of iid $\{ X_1, ..., X_n \}$, we can compute its maximum $Y = \max \{ X_1, ..., X_n \}$. The question is, if the values of the sample follow a distribution $P(X)$, what is the distribution for the maximum value of the sample, $P(Y)$?
It can be shown that depending on the form of $P(X)$ in the limit $X\to... | Why do we need Gumbel distribution? | Given a sample of iid $\{ X_1, ..., X_n \}$, we can compute its maximum $Y = \max \{ X_1, ..., X_n \}$. The question is, if the values of the sample follow a distribution $P(X)$, what is the distribut | Why do we need Gumbel distribution?
Given a sample of iid $\{ X_1, ..., X_n \}$, we can compute its maximum $Y = \max \{ X_1, ..., X_n \}$. The question is, if the values of the sample follow a distribution $P(X)$, what is the distribution for the maximum value of the sample, $P(Y)$?
It can be shown that depending on t... | Why do we need Gumbel distribution?
Given a sample of iid $\{ X_1, ..., X_n \}$, we can compute its maximum $Y = \max \{ X_1, ..., X_n \}$. The question is, if the values of the sample follow a distribution $P(X)$, what is the distribut |
48,291 | Multivariate Normal Distribution: Divide each random variable by its standard deviation | So, let's say $X=[X_1,...,X_n]$, and your new vector is $Y=[X_1/\sigma_1, ...,X_n/\sigma_n]$. For each pair $i,j$, the new covariance entries will be
$$(\Sigma_Y)_{ij}=\operatorname{cov}(X_i/\sigma_i, X_j/\sigma_j)=\frac{\operatorname{cov}(X_i,X_j)}{\sigma_i\sigma_j}=\frac{(\Sigma_X)_{ij}}{\sigma_i\sigma_j}$$
Note that... | Multivariate Normal Distribution: Divide each random variable by its standard deviation | So, let's say $X=[X_1,...,X_n]$, and your new vector is $Y=[X_1/\sigma_1, ...,X_n/\sigma_n]$. For each pair $i,j$, the new covariance entries will be
$$(\Sigma_Y)_{ij}=\operatorname{cov}(X_i/\sigma_i, | Multivariate Normal Distribution: Divide each random variable by its standard deviation
So, let's say $X=[X_1,...,X_n]$, and your new vector is $Y=[X_1/\sigma_1, ...,X_n/\sigma_n]$. For each pair $i,j$, the new covariance entries will be
$$(\Sigma_Y)_{ij}=\operatorname{cov}(X_i/\sigma_i, X_j/\sigma_j)=\frac{\operatorna... | Multivariate Normal Distribution: Divide each random variable by its standard deviation
So, let's say $X=[X_1,...,X_n]$, and your new vector is $Y=[X_1/\sigma_1, ...,X_n/\sigma_n]$. For each pair $i,j$, the new covariance entries will be
$$(\Sigma_Y)_{ij}=\operatorname{cov}(X_i/\sigma_i, |
48,292 | Help finding MGF of mixture distribution | You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
The information given in the problem statement gives you
$$\mathbb{E}[e^{tX}\mid N]=(1-p+e^tp)^N. \qquad\qquad \text{(why?)}$$
Defining $e^u:=1-p+e^tp$, and using t... | Help finding MGF of mixture distribution | You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
The information given in the problem stateme | Help finding MGF of mixture distribution
You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
The information given in the problem statement gives you
$$\mathbb{E}[e^{tX}\mid N]=(1-p+e^tp)^N. \qquad\qquad \text{(why?)... | Help finding MGF of mixture distribution
You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
The information given in the problem stateme |
48,293 | Non-mathematical explanation of how to interpret and evaluate scoring rules in R | Scoring rules are cost functions that evaluate how far probabilistic predictions deviate from observations. Proper scoring rules have minimum cost when true probabilities are reported; strictly proper scoring rules have minimum cost only when true probabilities are reported.
The choice of scoring rule is closely connec... | Non-mathematical explanation of how to interpret and evaluate scoring rules in R | Scoring rules are cost functions that evaluate how far probabilistic predictions deviate from observations. Proper scoring rules have minimum cost when true probabilities are reported; strictly proper | Non-mathematical explanation of how to interpret and evaluate scoring rules in R
Scoring rules are cost functions that evaluate how far probabilistic predictions deviate from observations. Proper scoring rules have minimum cost when true probabilities are reported; strictly proper scoring rules have minimum cost only w... | Non-mathematical explanation of how to interpret and evaluate scoring rules in R
Scoring rules are cost functions that evaluate how far probabilistic predictions deviate from observations. Proper scoring rules have minimum cost when true probabilities are reported; strictly proper |
48,294 | Asymptotic equivalence of Likelihood Ratio Statistic and Wald Statistic? | Either the number of binomial trials or the number of observations will do; usually we think of this result as applying more generally than binomial data and so think of the number of observations as the $n\to\infty$.
It's also important to note that the asymptotic equivalence is local. Suppose 0 is the null value of ... | Asymptotic equivalence of Likelihood Ratio Statistic and Wald Statistic? | Either the number of binomial trials or the number of observations will do; usually we think of this result as applying more generally than binomial data and so think of the number of observations as | Asymptotic equivalence of Likelihood Ratio Statistic and Wald Statistic?
Either the number of binomial trials or the number of observations will do; usually we think of this result as applying more generally than binomial data and so think of the number of observations as the $n\to\infty$.
It's also important to note t... | Asymptotic equivalence of Likelihood Ratio Statistic and Wald Statistic?
Either the number of binomial trials or the number of observations will do; usually we think of this result as applying more generally than binomial data and so think of the number of observations as |
48,295 | Whether the minimal sufficient statistic is complete for a translated exponential distribution | Lemma The minimal sufficient statistic $\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$ is not complete.
Proof. The joint distribution of $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$$
is the product of an Exponential $\mathcal E(n/\theta^2)$ translated by $\theta$ and of a $\mathcal Ga(n-1,1/\theta... | Whether the minimal sufficient statistic is complete for a translated exponential distribution | Lemma The minimal sufficient statistic $\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$ is not complete.
Proof. The joint distribution of $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\righ | Whether the minimal sufficient statistic is complete for a translated exponential distribution
Lemma The minimal sufficient statistic $\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$ is not complete.
Proof. The joint distribution of $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$$
is the product of an... | Whether the minimal sufficient statistic is complete for a translated exponential distribution
Lemma The minimal sufficient statistic $\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$ is not complete.
Proof. The joint distribution of $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\righ |
48,296 | Can we ALWAYS assume normal distribution if n >30? | My strategy is to look at the distribution of the data before statistical tests between two groups via histograms, q-q plots, and Shapiro Wilk test. If the data is approximately normal I use an appropriate test (t-test, ANOVA, Linear Regression etc). If not I use an appropriate non-parametric method (Mann-Whitney Test,... | Can we ALWAYS assume normal distribution if n >30? | My strategy is to look at the distribution of the data before statistical tests between two groups via histograms, q-q plots, and Shapiro Wilk test. If the data is approximately normal I use an approp | Can we ALWAYS assume normal distribution if n >30?
My strategy is to look at the distribution of the data before statistical tests between two groups via histograms, q-q plots, and Shapiro Wilk test. If the data is approximately normal I use an appropriate test (t-test, ANOVA, Linear Regression etc). If not I use an ap... | Can we ALWAYS assume normal distribution if n >30?
My strategy is to look at the distribution of the data before statistical tests between two groups via histograms, q-q plots, and Shapiro Wilk test. If the data is approximately normal I use an approp |
48,297 | Can we ALWAYS assume normal distribution if n >30? | The data does NOT get closer to being normally distributed as the sample size grows.
Rather, the thing that gets closer to being normally distributed is the sample mean or the sample sum.
And if the population distribution is very skewed, then you may need far more than $30,$ and it it isn't then maybe $10$ would be en... | Can we ALWAYS assume normal distribution if n >30? | The data does NOT get closer to being normally distributed as the sample size grows.
Rather, the thing that gets closer to being normally distributed is the sample mean or the sample sum.
And if the p | Can we ALWAYS assume normal distribution if n >30?
The data does NOT get closer to being normally distributed as the sample size grows.
Rather, the thing that gets closer to being normally distributed is the sample mean or the sample sum.
And if the population distribution is very skewed, then you may need far more tha... | Can we ALWAYS assume normal distribution if n >30?
The data does NOT get closer to being normally distributed as the sample size grows.
Rather, the thing that gets closer to being normally distributed is the sample mean or the sample sum.
And if the p |
48,298 | Forward algorithm vs Forward–backward algorithm | What you are referring to is called fixed-lag smoothing (see the Wikipedia article on the topic). The idea of fixed-lag smoothing is to use $k$ observations to infer the hidden state at time $k-N$. In other words, you wait for a time $N$ (which is called the lag) to gather more information about the state at time $k-N$... | Forward algorithm vs Forward–backward algorithm | What you are referring to is called fixed-lag smoothing (see the Wikipedia article on the topic). The idea of fixed-lag smoothing is to use $k$ observations to infer the hidden state at time $k-N$. In | Forward algorithm vs Forward–backward algorithm
What you are referring to is called fixed-lag smoothing (see the Wikipedia article on the topic). The idea of fixed-lag smoothing is to use $k$ observations to infer the hidden state at time $k-N$. In other words, you wait for a time $N$ (which is called the lag) to gathe... | Forward algorithm vs Forward–backward algorithm
What you are referring to is called fixed-lag smoothing (see the Wikipedia article on the topic). The idea of fixed-lag smoothing is to use $k$ observations to infer the hidden state at time $k-N$. In |
48,299 | What is the role of non-anchor items when equating scales across studies using IRT? | It is entirely possible to equate the test forms if the items common across all three forms contain no response bias (i.e., DIF). All that would be required is to estimate a multiple-group IRT model with equality constraints for the respective anchor items across all test versions (there may be larger blocks of missing... | What is the role of non-anchor items when equating scales across studies using IRT? | It is entirely possible to equate the test forms if the items common across all three forms contain no response bias (i.e., DIF). All that would be required is to estimate a multiple-group IRT model w | What is the role of non-anchor items when equating scales across studies using IRT?
It is entirely possible to equate the test forms if the items common across all three forms contain no response bias (i.e., DIF). All that would be required is to estimate a multiple-group IRT model with equality constraints for the res... | What is the role of non-anchor items when equating scales across studies using IRT?
It is entirely possible to equate the test forms if the items common across all three forms contain no response bias (i.e., DIF). All that would be required is to estimate a multiple-group IRT model w |
48,300 | Inverse Gaussian chi square connection | The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :
First, define $Y=\min\{X,\mu^2/X\}$ which belongs to $(0,\mu)$. The density of $Y$ can be derived from $(y<\mu)$
$$\mathbb P(Y\le y) = \mathbb P(X\le y)+\mathbb P... | Inverse Gaussian chi square connection | The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :
First, define $Y=\min\{X,\mu^2/X\}$ which | Inverse Gaussian chi square connection
The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :
First, define $Y=\min\{X,\mu^2/X\}$ which belongs to $(0,\mu)$. The density of $Y$ can be derived from $(y<\mu)$
$$\mathbb ... | Inverse Gaussian chi square connection
The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :
First, define $Y=\min\{X,\mu^2/X\}$ which |
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