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 |
|---|---|---|---|---|---|---|
50,501 | How does co-adaptation occur in deep neural nets | Since the weights are not initialized properly and groups of neurons end up in the same local minima, according to their (similar) initialization.
To overcome this, you could use dropout / drop connect to break symmetry.
Hinton, G. E., Srivastava, N., Krizhevsky, A., Sutskever, I., & Salakhutdinov, R. R. (2012). Improv... | How does co-adaptation occur in deep neural nets | Since the weights are not initialized properly and groups of neurons end up in the same local minima, according to their (similar) initialization.
To overcome this, you could use dropout / drop connec | How does co-adaptation occur in deep neural nets
Since the weights are not initialized properly and groups of neurons end up in the same local minima, according to their (similar) initialization.
To overcome this, you could use dropout / drop connect to break symmetry.
Hinton, G. E., Srivastava, N., Krizhevsky, A., Sut... | How does co-adaptation occur in deep neural nets
Since the weights are not initialized properly and groups of neurons end up in the same local minima, according to their (similar) initialization.
To overcome this, you could use dropout / drop connec |
50,502 | Gelman & Hill ARM textbook, Question 3.2, R-squared | With these kinds of questions it is usually best to eschew computer coding until you are at least able to write down the various algebraic equations you are using. The key for these questions is to be able to interpret the written information to obtain corresponding algebraic equations from your model. Once you have ... | Gelman & Hill ARM textbook, Question 3.2, R-squared | With these kinds of questions it is usually best to eschew computer coding until you are at least able to write down the various algebraic equations you are using. The key for these questions is to b | Gelman & Hill ARM textbook, Question 3.2, R-squared
With these kinds of questions it is usually best to eschew computer coding until you are at least able to write down the various algebraic equations you are using. The key for these questions is to be able to interpret the written information to obtain corresponding ... | Gelman & Hill ARM textbook, Question 3.2, R-squared
With these kinds of questions it is usually best to eschew computer coding until you are at least able to write down the various algebraic equations you are using. The key for these questions is to b |
50,503 | Gelman & Hill ARM textbook, Question 3.2, R-squared | log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log earnings (Gelman has heights in the book, but I think this is an error) at 5(%), R2 = 1 - 1.78 /5 = .64 | Gelman & Hill ARM textbook, Question 3.2, R-squared | log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log earnings (Gelman has heights in the book, but I think this is an error) | Gelman & Hill ARM textbook, Question 3.2, R-squared
log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log earnings (Gelman has heights in the book, but I think this is an error) at 5(%), R2 = 1 - 1.78 /5 = .64 | Gelman & Hill ARM textbook, Question 3.2, R-squared
log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log earnings (Gelman has heights in the book, but I think this is an error) |
50,504 | Gelman & Hill ARM textbook, Question 3.2, R-squared | I had a different interpretation of (b). we have $\log(y) = a + b\log(x)$ from the regression. Hence,
$$sd(\log(y)) = |b|sd(\log(x)) = (0.8)\times(0.05)=0.04,$$
since $sd(\log(x))$ is given. Squaring this we get the regression sum of squares $SSR = 0.0016$. From (a) we have the residual standard deviation ($0.049$). S... | Gelman & Hill ARM textbook, Question 3.2, R-squared | I had a different interpretation of (b). we have $\log(y) = a + b\log(x)$ from the regression. Hence,
$$sd(\log(y)) = |b|sd(\log(x)) = (0.8)\times(0.05)=0.04,$$
since $sd(\log(x))$ is given. Squaring | Gelman & Hill ARM textbook, Question 3.2, R-squared
I had a different interpretation of (b). we have $\log(y) = a + b\log(x)$ from the regression. Hence,
$$sd(\log(y)) = |b|sd(\log(x)) = (0.8)\times(0.05)=0.04,$$
since $sd(\log(x))$ is given. Squaring this we get the regression sum of squares $SSR = 0.0016$. From (a) ... | Gelman & Hill ARM textbook, Question 3.2, R-squared
I had a different interpretation of (b). we have $\log(y) = a + b\log(x)$ from the regression. Hence,
$$sd(\log(y)) = |b|sd(\log(x)) = (0.8)\times(0.05)=0.04,$$
since $sd(\log(x))$ is given. Squaring |
50,505 | Comparison of medians in samples with unequal variance, size and shape | I would suggest that you consider relative summary effects/relative treatment effects methods of Akritas, Arnold, Brunner, etc. The best book on the subject, generally, is Nonparametric Analysis of Longitudinal Data in Factorial Experiments by Brunner, Domhof, and Langer which is very well-written and clear.
These met... | Comparison of medians in samples with unequal variance, size and shape | I would suggest that you consider relative summary effects/relative treatment effects methods of Akritas, Arnold, Brunner, etc. The best book on the subject, generally, is Nonparametric Analysis of L | Comparison of medians in samples with unequal variance, size and shape
I would suggest that you consider relative summary effects/relative treatment effects methods of Akritas, Arnold, Brunner, etc. The best book on the subject, generally, is Nonparametric Analysis of Longitudinal Data in Factorial Experiments by Brun... | Comparison of medians in samples with unequal variance, size and shape
I would suggest that you consider relative summary effects/relative treatment effects methods of Akritas, Arnold, Brunner, etc. The best book on the subject, generally, is Nonparametric Analysis of L |
50,506 | Comparison of medians in samples with unequal variance, size and shape | One (very conservative/low power) option would be to use Mood's median test. It does not assume normality or homogeneity of variance (or at least quite robust to unequal variance).
The test is done by making a cross tabulation of the number of observations less than or equal to the overall median vs. greater than the o... | Comparison of medians in samples with unequal variance, size and shape | One (very conservative/low power) option would be to use Mood's median test. It does not assume normality or homogeneity of variance (or at least quite robust to unequal variance).
The test is done by | Comparison of medians in samples with unequal variance, size and shape
One (very conservative/low power) option would be to use Mood's median test. It does not assume normality or homogeneity of variance (or at least quite robust to unequal variance).
The test is done by making a cross tabulation of the number of obser... | Comparison of medians in samples with unequal variance, size and shape
One (very conservative/low power) option would be to use Mood's median test. It does not assume normality or homogeneity of variance (or at least quite robust to unequal variance).
The test is done by |
50,507 | Comparison of medians in samples with unequal variance, size and shape | My personal opinion is that you have the order of your tests backwards. If the Kolmogorov–Smirnov test shows that the samples come from a similar distribution, then use the t-test, otherwise use the Mann-Whitney U test.
For unequal variances, there is a procedure by Fligner and Policello (1981) who describe a methodo... | Comparison of medians in samples with unequal variance, size and shape | My personal opinion is that you have the order of your tests backwards. If the Kolmogorov–Smirnov test shows that the samples come from a similar distribution, then use the t-test, otherwise use the | Comparison of medians in samples with unequal variance, size and shape
My personal opinion is that you have the order of your tests backwards. If the Kolmogorov–Smirnov test shows that the samples come from a similar distribution, then use the t-test, otherwise use the Mann-Whitney U test.
For unequal variances, ther... | Comparison of medians in samples with unequal variance, size and shape
My personal opinion is that you have the order of your tests backwards. If the Kolmogorov–Smirnov test shows that the samples come from a similar distribution, then use the t-test, otherwise use the |
50,508 | creating random variable with certain auto-correlation in R | In this case, each value returned by filter is a weighted sum of $m$ out of $n$ normal random variates $\{x_1,...,x_n\}$ with the $m$ weights, $w$, set equal to acf.target in the example provided by the OP. The $i^\text{th}$ term is:
$$y_i=\sum_{j=1}^m{w_j{}x_{\text{mod}(i-j+\lfloor(m-1)/2\rfloor,n)+1}}$$
The $i^{th}$ ... | creating random variable with certain auto-correlation in R | In this case, each value returned by filter is a weighted sum of $m$ out of $n$ normal random variates $\{x_1,...,x_n\}$ with the $m$ weights, $w$, set equal to acf.target in the example provided by t | creating random variable with certain auto-correlation in R
In this case, each value returned by filter is a weighted sum of $m$ out of $n$ normal random variates $\{x_1,...,x_n\}$ with the $m$ weights, $w$, set equal to acf.target in the example provided by the OP. The $i^\text{th}$ term is:
$$y_i=\sum_{j=1}^m{w_j{}x_... | creating random variable with certain auto-correlation in R
In this case, each value returned by filter is a weighted sum of $m$ out of $n$ normal random variates $\{x_1,...,x_n\}$ with the $m$ weights, $w$, set equal to acf.target in the example provided by t |
50,509 | Approximating P(A,B,C) using P(A,B), P(A,C), P(B,C), and P(A), P(B), P(C) | Bounding
$$
P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)
$$
$\implies$
$$
P(A \cap B \cap C) = P(A \cup B \cup C) - \big(P(A) + P(B) + P(C)\big) + \big(P(A \cap B) + P(B \cap C) + P(C \cap A)\big)
$$
and
$$ 0 \leq P(A \cup B \cup C) \leq 1$$
$\implies$
$$
- \... | Approximating P(A,B,C) using P(A,B), P(A,C), P(B,C), and P(A), P(B), P(C) | Bounding
$$
P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)
$$
$\implies$
$$
P(A \cap B \cap C) = P(A \cup B \cup C) - \big(P(A) + P(B) + P(C)\ | Approximating P(A,B,C) using P(A,B), P(A,C), P(B,C), and P(A), P(B), P(C)
Bounding
$$
P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)
$$
$\implies$
$$
P(A \cap B \cap C) = P(A \cup B \cup C) - \big(P(A) + P(B) + P(C)\big) + \big(P(A \cap B) + P(B \cap C) + P(C \c... | Approximating P(A,B,C) using P(A,B), P(A,C), P(B,C), and P(A), P(B), P(C)
Bounding
$$
P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)
$$
$\implies$
$$
P(A \cap B \cap C) = P(A \cup B \cup C) - \big(P(A) + P(B) + P(C)\ |
50,510 | Adjusting time-series before a sudden increase(the reason & time of the increase are known) | Detecting and incorporating Level Shifts (Step Shifts) is an integral part or both non-causal and causal modelling. The term Intervention Detection is often used. Some software implementations restrict identification to non-causal models while others do not. See http://www.unc.edu/~jbhill/tsay.pdf for a seminal article... | Adjusting time-series before a sudden increase(the reason & time of the increase are known) | Detecting and incorporating Level Shifts (Step Shifts) is an integral part or both non-causal and causal modelling. The term Intervention Detection is often used. Some software implementations restric | Adjusting time-series before a sudden increase(the reason & time of the increase are known)
Detecting and incorporating Level Shifts (Step Shifts) is an integral part or both non-causal and causal modelling. The term Intervention Detection is often used. Some software implementations restrict identification to non-caus... | Adjusting time-series before a sudden increase(the reason & time of the increase are known)
Detecting and incorporating Level Shifts (Step Shifts) is an integral part or both non-causal and causal modelling. The term Intervention Detection is often used. Some software implementations restric |
50,511 | Basu's Theorem Proof | The first questions were already answered in the comments. You could add that the inner expectation in
$E_{\theta}E_{\theta}[I_{V \in A}|T]$
is taken "with a fixed T = t", giving a function $g(t)=E_{\theta}[I_{V \in A}|T=t]$, as Xi'an said. The second $E_{\theta}$ then takes the expectation of g(t) (so you vary t now)... | Basu's Theorem Proof | The first questions were already answered in the comments. You could add that the inner expectation in
$E_{\theta}E_{\theta}[I_{V \in A}|T]$
is taken "with a fixed T = t", giving a function $g(t)=E_{ | Basu's Theorem Proof
The first questions were already answered in the comments. You could add that the inner expectation in
$E_{\theta}E_{\theta}[I_{V \in A}|T]$
is taken "with a fixed T = t", giving a function $g(t)=E_{\theta}[I_{V \in A}|T=t]$, as Xi'an said. The second $E_{\theta}$ then takes the expectation of g(t... | Basu's Theorem Proof
The first questions were already answered in the comments. You could add that the inner expectation in
$E_{\theta}E_{\theta}[I_{V \in A}|T]$
is taken "with a fixed T = t", giving a function $g(t)=E_{ |
50,512 | Are There Evaluation Criteria For Instrumental Variables? | Suppose you have several valid instrumental variables. Then you can think of the following when choosing among them:
If they capture the same local average treatment effects (i.e. they generate the same set of compliers), you can perhaps choose both and conduct an overidentifying restrictions test, or choose the one t... | Are There Evaluation Criteria For Instrumental Variables? | Suppose you have several valid instrumental variables. Then you can think of the following when choosing among them:
If they capture the same local average treatment effects (i.e. they generate the s | Are There Evaluation Criteria For Instrumental Variables?
Suppose you have several valid instrumental variables. Then you can think of the following when choosing among them:
If they capture the same local average treatment effects (i.e. they generate the same set of compliers), you can perhaps choose both and conduct... | Are There Evaluation Criteria For Instrumental Variables?
Suppose you have several valid instrumental variables. Then you can think of the following when choosing among them:
If they capture the same local average treatment effects (i.e. they generate the s |
50,513 | Is it possible to construct a hypothesis test for the existence of a mean of a symmetric distribution? | Not really. See here for example: Test for finite variance? | Is it possible to construct a hypothesis test for the existence of a mean of a symmetric distributio | Not really. See here for example: Test for finite variance? | Is it possible to construct a hypothesis test for the existence of a mean of a symmetric distribution?
Not really. See here for example: Test for finite variance? | Is it possible to construct a hypothesis test for the existence of a mean of a symmetric distributio
Not really. See here for example: Test for finite variance? |
50,514 | Plotting a tree timeline (evolution history) | So, as the question is perfectly on-topic here, as it deals with "Data Visualization", I would reproduce the comment as an answer, so that future viewers can be benefitted.
What is this type of graph called?
This graph is called as a "Tree diagram" and sometimes also called a dendrogram
Are there existing tools to pro... | Plotting a tree timeline (evolution history) | So, as the question is perfectly on-topic here, as it deals with "Data Visualization", I would reproduce the comment as an answer, so that future viewers can be benefitted.
What is this type of graph | Plotting a tree timeline (evolution history)
So, as the question is perfectly on-topic here, as it deals with "Data Visualization", I would reproduce the comment as an answer, so that future viewers can be benefitted.
What is this type of graph called?
This graph is called as a "Tree diagram" and sometimes also called ... | Plotting a tree timeline (evolution history)
So, as the question is perfectly on-topic here, as it deals with "Data Visualization", I would reproduce the comment as an answer, so that future viewers can be benefitted.
What is this type of graph |
50,515 | Comparing CV Predictions across Folds for Random Forest | For each fold, you are building a classifier that makes predictions for the observations. The classifiers within each fold have slightly different training sets and different weights, but they are all attempting to estimate the same underlying model. So yes, you can combine the predictions. If you have multiple predict... | Comparing CV Predictions across Folds for Random Forest | For each fold, you are building a classifier that makes predictions for the observations. The classifiers within each fold have slightly different training sets and different weights, but they are all | Comparing CV Predictions across Folds for Random Forest
For each fold, you are building a classifier that makes predictions for the observations. The classifiers within each fold have slightly different training sets and different weights, but they are all attempting to estimate the same underlying model. So yes, you c... | Comparing CV Predictions across Folds for Random Forest
For each fold, you are building a classifier that makes predictions for the observations. The classifiers within each fold have slightly different training sets and different weights, but they are all |
50,516 | Comparing CV Predictions across Folds for Random Forest | After speaking with a few other folks about this problem, I think that technially you can't directly compare probabilities predicted for different folds, but practically, in most cases, you can.
The time when you would not be able to is if you have a small, potentially diverse positive set. Then when you divide the pos... | Comparing CV Predictions across Folds for Random Forest | After speaking with a few other folks about this problem, I think that technially you can't directly compare probabilities predicted for different folds, but practically, in most cases, you can.
The t | Comparing CV Predictions across Folds for Random Forest
After speaking with a few other folks about this problem, I think that technially you can't directly compare probabilities predicted for different folds, but practically, in most cases, you can.
The time when you would not be able to is if you have a small, potent... | Comparing CV Predictions across Folds for Random Forest
After speaking with a few other folks about this problem, I think that technially you can't directly compare probabilities predicted for different folds, but practically, in most cases, you can.
The t |
50,517 | Comparing CV Predictions across Folds for Random Forest | I am not sure you can combine all the predictions of the k-folds.
However, you could stratified your K folds so that you have similar number of positives in each fold and ROC performance would not vary due to imbalanced dataset.
In python, there is this package from scikit learn that works very well: http://scikit-lea... | Comparing CV Predictions across Folds for Random Forest | I am not sure you can combine all the predictions of the k-folds.
However, you could stratified your K folds so that you have similar number of positives in each fold and ROC performance would not va | Comparing CV Predictions across Folds for Random Forest
I am not sure you can combine all the predictions of the k-folds.
However, you could stratified your K folds so that you have similar number of positives in each fold and ROC performance would not vary due to imbalanced dataset.
In python, there is this package f... | Comparing CV Predictions across Folds for Random Forest
I am not sure you can combine all the predictions of the k-folds.
However, you could stratified your K folds so that you have similar number of positives in each fold and ROC performance would not va |
50,518 | One class SVM with caret in R using cross validation [closed] | Simple option is not to use caret and just use the tune function from E1071.
svm_model <- tune(svm(training,y=NULL, type='one-classification', nu=0.01, gamma=0.002, scale=TRUE, kernel="radial", tunecontrol = tune.control(nrepeat = 3))
The default setting from tune is 10 fold CV. Using tune.control you can adjust this... | One class SVM with caret in R using cross validation [closed] | Simple option is not to use caret and just use the tune function from E1071.
svm_model <- tune(svm(training,y=NULL, type='one-classification', nu=0.01, gamma=0.002, scale=TRUE, kernel="radial", tunec | One class SVM with caret in R using cross validation [closed]
Simple option is not to use caret and just use the tune function from E1071.
svm_model <- tune(svm(training,y=NULL, type='one-classification', nu=0.01, gamma=0.002, scale=TRUE, kernel="radial", tunecontrol = tune.control(nrepeat = 3))
The default setting f... | One class SVM with caret in R using cross validation [closed]
Simple option is not to use caret and just use the tune function from E1071.
svm_model <- tune(svm(training,y=NULL, type='one-classification', nu=0.01, gamma=0.002, scale=TRUE, kernel="radial", tunec |
50,519 | Controlling FWER using minP/maxT methods | if you feel comfortable with the Bonferroni correction, you should not have a problem with minP method.
Suppose that you have K p-values: $P_1,P_2,...,P_K$. Let's consider the Bonferroni correction, each p-value is adjusted by the number of tests, $P_i^{Bon}= K*P_i$, then you compare your p-values with $\alpha$ to dete... | Controlling FWER using minP/maxT methods | if you feel comfortable with the Bonferroni correction, you should not have a problem with minP method.
Suppose that you have K p-values: $P_1,P_2,...,P_K$. Let's consider the Bonferroni correction, e | Controlling FWER using minP/maxT methods
if you feel comfortable with the Bonferroni correction, you should not have a problem with minP method.
Suppose that you have K p-values: $P_1,P_2,...,P_K$. Let's consider the Bonferroni correction, each p-value is adjusted by the number of tests, $P_i^{Bon}= K*P_i$, then you co... | Controlling FWER using minP/maxT methods
if you feel comfortable with the Bonferroni correction, you should not have a problem with minP method.
Suppose that you have K p-values: $P_1,P_2,...,P_K$. Let's consider the Bonferroni correction, e |
50,520 | Expectation Maximization clarification questions | Just to elaborate a little further: In terms of $\theta$ we have
\begin{align*}
\sum_{i=1}^M \sum_{z^{(i)}=1}^K & Q(z^{(i)}) \log\left(\frac{P(x^{(i)},z^{(i)};\theta)}{Q(z^{(i)})}\right) \\
&= \sum_{i=1}^M \sum_{z^{(i)}=1}^K Q(z^{(i)}) \log P(x^{(i)},z^{(i)};\theta) - \sum_{i=1}^M \sum_{z^{(i)}=1}^K Q(z^{(i)}) \log Q... | Expectation Maximization clarification questions | Just to elaborate a little further: In terms of $\theta$ we have
\begin{align*}
\sum_{i=1}^M \sum_{z^{(i)}=1}^K & Q(z^{(i)}) \log\left(\frac{P(x^{(i)},z^{(i)};\theta)}{Q(z^{(i)})}\right) \\
&= \sum_ | Expectation Maximization clarification questions
Just to elaborate a little further: In terms of $\theta$ we have
\begin{align*}
\sum_{i=1}^M \sum_{z^{(i)}=1}^K & Q(z^{(i)}) \log\left(\frac{P(x^{(i)},z^{(i)};\theta)}{Q(z^{(i)})}\right) \\
&= \sum_{i=1}^M \sum_{z^{(i)}=1}^K Q(z^{(i)}) \log P(x^{(i)},z^{(i)};\theta) - ... | Expectation Maximization clarification questions
Just to elaborate a little further: In terms of $\theta$ we have
\begin{align*}
\sum_{i=1}^M \sum_{z^{(i)}=1}^K & Q(z^{(i)}) \log\left(\frac{P(x^{(i)},z^{(i)};\theta)}{Q(z^{(i)})}\right) \\
&= \sum_ |
50,521 | Find the MLE of the proportion of employees falling in $[I_1,I_2]$ | Unless I've made an error, you're very close to the right answer.
I don't see how $X_{(n)}$ comes into the MLE. It looks to me like you can work out the MLE of $c$ and $A$ (neither of which involve $X_{(n)}$) and substitute them into the relevant places to get the MLE of the probability. After removing reference to $X... | Find the MLE of the proportion of employees falling in $[I_1,I_2]$ | Unless I've made an error, you're very close to the right answer.
I don't see how $X_{(n)}$ comes into the MLE. It looks to me like you can work out the MLE of $c$ and $A$ (neither of which involve $ | Find the MLE of the proportion of employees falling in $[I_1,I_2]$
Unless I've made an error, you're very close to the right answer.
I don't see how $X_{(n)}$ comes into the MLE. It looks to me like you can work out the MLE of $c$ and $A$ (neither of which involve $X_{(n)}$) and substitute them into the relevant place... | Find the MLE of the proportion of employees falling in $[I_1,I_2]$
Unless I've made an error, you're very close to the right answer.
I don't see how $X_{(n)}$ comes into the MLE. It looks to me like you can work out the MLE of $c$ and $A$ (neither of which involve $ |
50,522 | Linear OLS v Mixed-Effects Model with Correlated Regressors | Consistent with the conversation in the comments, it may actually be quite obvious: in mixed-effects, the overall increasing values of $y$ with increasing $x1$ as shown in:
with a positive correlation of cor(x1,y) [1] 0.7924759, is nicely captured in the plane formed by lm(y ~ x1 + x2) (depicted in the OP plots) or in... | Linear OLS v Mixed-Effects Model with Correlated Regressors | Consistent with the conversation in the comments, it may actually be quite obvious: in mixed-effects, the overall increasing values of $y$ with increasing $x1$ as shown in:
with a positive correlatio | Linear OLS v Mixed-Effects Model with Correlated Regressors
Consistent with the conversation in the comments, it may actually be quite obvious: in mixed-effects, the overall increasing values of $y$ with increasing $x1$ as shown in:
with a positive correlation of cor(x1,y) [1] 0.7924759, is nicely captured in the plan... | Linear OLS v Mixed-Effects Model with Correlated Regressors
Consistent with the conversation in the comments, it may actually be quite obvious: in mixed-effects, the overall increasing values of $y$ with increasing $x1$ as shown in:
with a positive correlatio |
50,523 | Division with lag operator | First of all it is important to notice that $L$ is an operator that works on the following random variable. Then a short answer is No you cannot. Let me give you an example,
Assume you have a time series of the form of
$y_t- y_{t-1}=e_t - e_{t-1}$. then using backward shift operator we have, $(1-L)y_t=(1-L)e_t$. Itis ... | Division with lag operator | First of all it is important to notice that $L$ is an operator that works on the following random variable. Then a short answer is No you cannot. Let me give you an example,
Assume you have a time ser | Division with lag operator
First of all it is important to notice that $L$ is an operator that works on the following random variable. Then a short answer is No you cannot. Let me give you an example,
Assume you have a time series of the form of
$y_t- y_{t-1}=e_t - e_{t-1}$. then using backward shift operator we have,... | Division with lag operator
First of all it is important to notice that $L$ is an operator that works on the following random variable. Then a short answer is No you cannot. Let me give you an example,
Assume you have a time ser |
50,524 | Bayesian Updating | This isn't a typical Bayesian update setup - what is the sequence $B_i$? Usually these are the observed variables, while $A_i$ is a sequence of latent variables, the ones we wish to estimate. In that case,
we predict, based on the $B_1,...,B_{n-1}$, using
$$
P(A_n|\{B_1,...,B_{n-1}\}) = \int P(A_n|A_{n-1})P(A_{n-1}|\{... | Bayesian Updating | This isn't a typical Bayesian update setup - what is the sequence $B_i$? Usually these are the observed variables, while $A_i$ is a sequence of latent variables, the ones we wish to estimate. In that | Bayesian Updating
This isn't a typical Bayesian update setup - what is the sequence $B_i$? Usually these are the observed variables, while $A_i$ is a sequence of latent variables, the ones we wish to estimate. In that case,
we predict, based on the $B_1,...,B_{n-1}$, using
$$
P(A_n|\{B_1,...,B_{n-1}\}) = \int P(A_n|A_... | Bayesian Updating
This isn't a typical Bayesian update setup - what is the sequence $B_i$? Usually these are the observed variables, while $A_i$ is a sequence of latent variables, the ones we wish to estimate. In that |
50,525 | Do I need to adjust the degrees of freedom returned by pool.compare() in MICE? | You posed two questions, so I will simply comment on them in order:
Question 1: High degrees of freedom:
Such high degrees of freedoms are normal with pool.compare. The function implements the procedure by Meng & Rubin (1992), in which the denominator degrees of freedom for the test statistic $D_m$ are derived under th... | Do I need to adjust the degrees of freedom returned by pool.compare() in MICE? | You posed two questions, so I will simply comment on them in order:
Question 1: High degrees of freedom:
Such high degrees of freedoms are normal with pool.compare. The function implements the procedu | Do I need to adjust the degrees of freedom returned by pool.compare() in MICE?
You posed two questions, so I will simply comment on them in order:
Question 1: High degrees of freedom:
Such high degrees of freedoms are normal with pool.compare. The function implements the procedure by Meng & Rubin (1992), in which the d... | Do I need to adjust the degrees of freedom returned by pool.compare() in MICE?
You posed two questions, so I will simply comment on them in order:
Question 1: High degrees of freedom:
Such high degrees of freedoms are normal with pool.compare. The function implements the procedu |
50,526 | In learning theory, why can't we bound like $P[|E_{in}(g)-E_{out}(g)|>\epsilon] \leq 2e^{-2\epsilon^{2}N}$?($g$ is our learned hypothesis) | I believe the error is in the following equality
$$\sum_{h \in H} P[|E_{in}(g)-E_{out}(g)|>\epsilon \;\lvert \;g=h]\;P[g=h]
= \sum_{h \in H} P[|E_{in}(h)-E_{out}(h)|>\epsilon]\;P[g=h]$$
but its hidden by the notation.
Consider your equality
$$ P [|E_{in}(g)-E_{out}(g)|>\epsilon \;\lvert \;g=h] = P [|E_{in}(h)-E_{out}(h... | In learning theory, why can't we bound like $P[|E_{in}(g)-E_{out}(g)|>\epsilon] \leq 2e^{-2\epsilon^ | I believe the error is in the following equality
$$\sum_{h \in H} P[|E_{in}(g)-E_{out}(g)|>\epsilon \;\lvert \;g=h]\;P[g=h]
= \sum_{h \in H} P[|E_{in}(h)-E_{out}(h)|>\epsilon]\;P[g=h]$$
but its hidden | In learning theory, why can't we bound like $P[|E_{in}(g)-E_{out}(g)|>\epsilon] \leq 2e^{-2\epsilon^{2}N}$?($g$ is our learned hypothesis)
I believe the error is in the following equality
$$\sum_{h \in H} P[|E_{in}(g)-E_{out}(g)|>\epsilon \;\lvert \;g=h]\;P[g=h]
= \sum_{h \in H} P[|E_{in}(h)-E_{out}(h)|>\epsilon]\;P[g=... | In learning theory, why can't we bound like $P[|E_{in}(g)-E_{out}(g)|>\epsilon] \leq 2e^{-2\epsilon^
I believe the error is in the following equality
$$\sum_{h \in H} P[|E_{in}(g)-E_{out}(g)|>\epsilon \;\lvert \;g=h]\;P[g=h]
= \sum_{h \in H} P[|E_{in}(h)-E_{out}(h)|>\epsilon]\;P[g=h]$$
but its hidden |
50,527 | Prior distribution on/of a parameter | Being that the first two results of a Google search (in English) for "prior distribution on" yield papers by Andrew Gelman, I'm sufficiently convinced that "on" is common enough not to cause confusion. Actually, there are contexts in which using "of" would strike some native English speakers as odd. For instance, the p... | Prior distribution on/of a parameter | Being that the first two results of a Google search (in English) for "prior distribution on" yield papers by Andrew Gelman, I'm sufficiently convinced that "on" is common enough not to cause confusion | Prior distribution on/of a parameter
Being that the first two results of a Google search (in English) for "prior distribution on" yield papers by Andrew Gelman, I'm sufficiently convinced that "on" is common enough not to cause confusion. Actually, there are contexts in which using "of" would strike some native English... | Prior distribution on/of a parameter
Being that the first two results of a Google search (in English) for "prior distribution on" yield papers by Andrew Gelman, I'm sufficiently convinced that "on" is common enough not to cause confusion |
50,528 | Prior distribution on/of a parameter | Why do you want to use on instead of of?
I think that of is more correct because it refers to the fact that we are talking about the distribution of $\mu$ before (prior -> pre -> before) --- the observation of data.
Similarly, I think it is correct "posterior distribution of $\mu$", because it refers to the distributio... | Prior distribution on/of a parameter | Why do you want to use on instead of of?
I think that of is more correct because it refers to the fact that we are talking about the distribution of $\mu$ before (prior -> pre -> before) --- the obser | Prior distribution on/of a parameter
Why do you want to use on instead of of?
I think that of is more correct because it refers to the fact that we are talking about the distribution of $\mu$ before (prior -> pre -> before) --- the observation of data.
Similarly, I think it is correct "posterior distribution of $\mu$",... | Prior distribution on/of a parameter
Why do you want to use on instead of of?
I think that of is more correct because it refers to the fact that we are talking about the distribution of $\mu$ before (prior -> pre -> before) --- the obser |
50,529 | Autoencoder doesn't work (can't learn features) | (Primary author of theanets here.) As hinted in the comments on your question, this is actually a difficult learning problem! The network is, as indicated by the optimized loss value during training, learning the optimal filters for representing this set of input data as well as it can.
The important thing to think abo... | Autoencoder doesn't work (can't learn features) | (Primary author of theanets here.) As hinted in the comments on your question, this is actually a difficult learning problem! The network is, as indicated by the optimized loss value during training, | Autoencoder doesn't work (can't learn features)
(Primary author of theanets here.) As hinted in the comments on your question, this is actually a difficult learning problem! The network is, as indicated by the optimized loss value during training, learning the optimal filters for representing this set of input data as ... | Autoencoder doesn't work (can't learn features)
(Primary author of theanets here.) As hinted in the comments on your question, this is actually a difficult learning problem! The network is, as indicated by the optimized loss value during training, |
50,530 | Why non-negative regression? | This is adding prior knowledge to your model, in a somewhat Bayesian sense. Per the comments, not everyone would call this "regularization". I would, because it constrains the parameters to not vary wildly all over the place, just in a different way than (say) L1 or L2 regularization.
When does this make sense? Anytime... | Why non-negative regression? | This is adding prior knowledge to your model, in a somewhat Bayesian sense. Per the comments, not everyone would call this "regularization". I would, because it constrains the parameters to not vary w | Why non-negative regression?
This is adding prior knowledge to your model, in a somewhat Bayesian sense. Per the comments, not everyone would call this "regularization". I would, because it constrains the parameters to not vary wildly all over the place, just in a different way than (say) L1 or L2 regularization.
When ... | Why non-negative regression?
This is adding prior knowledge to your model, in a somewhat Bayesian sense. Per the comments, not everyone would call this "regularization". I would, because it constrains the parameters to not vary w |
50,531 | How to deal with missing coefficients while bootstrapping regressions | One method that can be used (with caution!!) is a stratified bootstrap. That is, suppose we have 20 subjects in group 1 and 20 in group 2. Then we can resample our data, conditional on these sample sizes (i.e. we resample 20 from group 1 and 20 from group 2). Because of this, we are now insured that the estimator of th... | How to deal with missing coefficients while bootstrapping regressions | One method that can be used (with caution!!) is a stratified bootstrap. That is, suppose we have 20 subjects in group 1 and 20 in group 2. Then we can resample our data, conditional on these sample si | How to deal with missing coefficients while bootstrapping regressions
One method that can be used (with caution!!) is a stratified bootstrap. That is, suppose we have 20 subjects in group 1 and 20 in group 2. Then we can resample our data, conditional on these sample sizes (i.e. we resample 20 from group 1 and 20 from ... | How to deal with missing coefficients while bootstrapping regressions
One method that can be used (with caution!!) is a stratified bootstrap. That is, suppose we have 20 subjects in group 1 and 20 in group 2. Then we can resample our data, conditional on these sample si |
50,532 | How are subjects with only one observation used in fixed effect models? | In a frequentist approach this fixed effect model has an unindentifiability problem (https://en.wikipedia.org/wiki/Identifiability) for $\beta_1$ and $\alpha_i$ for those subjects that have only one measurement. These subjects do not have a slope and do not contribute to the estimation of $\beta_2$.
In a Bayesian appro... | How are subjects with only one observation used in fixed effect models? | In a frequentist approach this fixed effect model has an unindentifiability problem (https://en.wikipedia.org/wiki/Identifiability) for $\beta_1$ and $\alpha_i$ for those subjects that have only one m | How are subjects with only one observation used in fixed effect models?
In a frequentist approach this fixed effect model has an unindentifiability problem (https://en.wikipedia.org/wiki/Identifiability) for $\beta_1$ and $\alpha_i$ for those subjects that have only one measurement. These subjects do not have a slope a... | How are subjects with only one observation used in fixed effect models?
In a frequentist approach this fixed effect model has an unindentifiability problem (https://en.wikipedia.org/wiki/Identifiability) for $\beta_1$ and $\alpha_i$ for those subjects that have only one m |
50,533 | Post propensity score matching analysis | Stuart et al. mention the R package Zelig, which seemlessly works for post-matching analysis after matching with MatchIt. It is mentioned quite often that you should NOT simply compare the means after matching, although this is quite common practice. You can make optimal use of the matching process by using regression ... | Post propensity score matching analysis | Stuart et al. mention the R package Zelig, which seemlessly works for post-matching analysis after matching with MatchIt. It is mentioned quite often that you should NOT simply compare the means after | Post propensity score matching analysis
Stuart et al. mention the R package Zelig, which seemlessly works for post-matching analysis after matching with MatchIt. It is mentioned quite often that you should NOT simply compare the means after matching, although this is quite common practice. You can make optimal use of t... | Post propensity score matching analysis
Stuart et al. mention the R package Zelig, which seemlessly works for post-matching analysis after matching with MatchIt. It is mentioned quite often that you should NOT simply compare the means after |
50,534 | Difference between Bag of words and Vector space model | I find existing answer very misleading.
The Word vector (aka word embedding) is concept coming from probabilistic language models (see [1]). It describes contextual similarity between the words in the language model and came into existence several decades after the VSM was proposed and successfully applied for text cat... | Difference between Bag of words and Vector space model | I find existing answer very misleading.
The Word vector (aka word embedding) is concept coming from probabilistic language models (see [1]). It describes contextual similarity between the words in the | Difference between Bag of words and Vector space model
I find existing answer very misleading.
The Word vector (aka word embedding) is concept coming from probabilistic language models (see [1]). It describes contextual similarity between the words in the language model and came into existence several decades after the... | Difference between Bag of words and Vector space model
I find existing answer very misleading.
The Word vector (aka word embedding) is concept coming from probabilistic language models (see [1]). It describes contextual similarity between the words in the |
50,535 | Difference between Bag of words and Vector space model | Note that the word "bag" means multiset, i.e., it allows multiple instances for each word. Thus:
Bag of words indicates the count of each word in the document. This simple model is used, for example, in naive Bayes
Word vector generalizes the idea of bag of words assigning a ranking to each word in the document. Often... | Difference between Bag of words and Vector space model | Note that the word "bag" means multiset, i.e., it allows multiple instances for each word. Thus:
Bag of words indicates the count of each word in the document. This simple model is used, for example, | Difference between Bag of words and Vector space model
Note that the word "bag" means multiset, i.e., it allows multiple instances for each word. Thus:
Bag of words indicates the count of each word in the document. This simple model is used, for example, in naive Bayes
Word vector generalizes the idea of bag of words ... | Difference between Bag of words and Vector space model
Note that the word "bag" means multiset, i.e., it allows multiple instances for each word. Thus:
Bag of words indicates the count of each word in the document. This simple model is used, for example, |
50,536 | Simple way for histograms classification | You can use nearest neighbor classification, with an appropriate distance metric. For example histogram intersection distance, $\chi^2$ distance, F-divergence, jensen-shannon divergence, or any other of the divergence measures you like. | Simple way for histograms classification | You can use nearest neighbor classification, with an appropriate distance metric. For example histogram intersection distance, $\chi^2$ distance, F-divergence, jensen-shannon divergence, or any other | Simple way for histograms classification
You can use nearest neighbor classification, with an appropriate distance metric. For example histogram intersection distance, $\chi^2$ distance, F-divergence, jensen-shannon divergence, or any other of the divergence measures you like. | Simple way for histograms classification
You can use nearest neighbor classification, with an appropriate distance metric. For example histogram intersection distance, $\chi^2$ distance, F-divergence, jensen-shannon divergence, or any other |
50,537 | Denoising Autoencoders weights at test time | Yes, assuming the noise has the same nature (ie. setting each input to 0 with some probability p), the same reasoning applies and you should increase the weights at test time. Conversely, you could reduce them during training, which is not exactly equivalent because of the non-linearities, but in practice seems to beha... | Denoising Autoencoders weights at test time | Yes, assuming the noise has the same nature (ie. setting each input to 0 with some probability p), the same reasoning applies and you should increase the weights at test time. Conversely, you could re | Denoising Autoencoders weights at test time
Yes, assuming the noise has the same nature (ie. setting each input to 0 with some probability p), the same reasoning applies and you should increase the weights at test time. Conversely, you could reduce them during training, which is not exactly equivalent because of the no... | Denoising Autoencoders weights at test time
Yes, assuming the noise has the same nature (ie. setting each input to 0 with some probability p), the same reasoning applies and you should increase the weights at test time. Conversely, you could re |
50,538 | What will be the estimator for these parameters | I need to find an estimator for the parameters $(y_0,σ^2_η)$ from the
observations $y(t)$,in order to get an estimator for the parameter
$z_d$.
In your formulation, $$y(t) = y_0 + \eta(t)$$with $\eta(t)\stackrel{\text{iid}}{\sim}N(0,\sigma^2_\eta)$, i.e.$$y(t) \stackrel{\text{iid}}{\sim}N(y_0,\sigma^2_\eta)\qquad ... | What will be the estimator for these parameters | I need to find an estimator for the parameters $(y_0,σ^2_η)$ from the
observations $y(t)$,in order to get an estimator for the parameter
$z_d$.
In your formulation, $$y(t) = y_0 + \eta(t)$$with $ | What will be the estimator for these parameters
I need to find an estimator for the parameters $(y_0,σ^2_η)$ from the
observations $y(t)$,in order to get an estimator for the parameter
$z_d$.
In your formulation, $$y(t) = y_0 + \eta(t)$$with $\eta(t)\stackrel{\text{iid}}{\sim}N(0,\sigma^2_\eta)$, i.e.$$y(t) \stack... | What will be the estimator for these parameters
I need to find an estimator for the parameters $(y_0,σ^2_η)$ from the
observations $y(t)$,in order to get an estimator for the parameter
$z_d$.
In your formulation, $$y(t) = y_0 + \eta(t)$$with $ |
50,539 | Estimating robust standard errors in panel data regressions | 1) Given that you have specified "id" in the regression (I guess individuals or some other unit you follow over time), the cluster="group" standard errors are clustered at the individual level. This makes sense given that a person's error today may be correlated with her error of yesterday. For more information see pag... | Estimating robust standard errors in panel data regressions | 1) Given that you have specified "id" in the regression (I guess individuals or some other unit you follow over time), the cluster="group" standard errors are clustered at the individual level. This m | Estimating robust standard errors in panel data regressions
1) Given that you have specified "id" in the regression (I guess individuals or some other unit you follow over time), the cluster="group" standard errors are clustered at the individual level. This makes sense given that a person's error today may be correlat... | Estimating robust standard errors in panel data regressions
1) Given that you have specified "id" in the regression (I guess individuals or some other unit you follow over time), the cluster="group" standard errors are clustered at the individual level. This m |
50,540 | Is a logit model with a pseudo-R^2 of less than 0.5 a worse model than a coin toss? | (Percent correct)/(Total count) is usually termed the Correct Classification Rate. This is not one of the pseudo-R-squared indicators, and it's generally considered an inferior way of assessing model fit because it simplifies so much; it doesn't take into account the differences in predicted probability from observati... | Is a logit model with a pseudo-R^2 of less than 0.5 a worse model than a coin toss? | (Percent correct)/(Total count) is usually termed the Correct Classification Rate. This is not one of the pseudo-R-squared indicators, and it's generally considered an inferior way of assessing model | Is a logit model with a pseudo-R^2 of less than 0.5 a worse model than a coin toss?
(Percent correct)/(Total count) is usually termed the Correct Classification Rate. This is not one of the pseudo-R-squared indicators, and it's generally considered an inferior way of assessing model fit because it simplifies so much; ... | Is a logit model with a pseudo-R^2 of less than 0.5 a worse model than a coin toss?
(Percent correct)/(Total count) is usually termed the Correct Classification Rate. This is not one of the pseudo-R-squared indicators, and it's generally considered an inferior way of assessing model |
50,541 | Elastic net: dealing with wide data with outliers | Couple of things I just thought I will mention. It is hard to mention specifics without actually looking at the results but hope this is helpful. Most of these things I am sure you already know but in case you missed something
1) The 10% v/s 90% check if you used glmnet or cv.glmnet. You should be using cv.glmnet. The ... | Elastic net: dealing with wide data with outliers | Couple of things I just thought I will mention. It is hard to mention specifics without actually looking at the results but hope this is helpful. Most of these things I am sure you already know but in | Elastic net: dealing with wide data with outliers
Couple of things I just thought I will mention. It is hard to mention specifics without actually looking at the results but hope this is helpful. Most of these things I am sure you already know but in case you missed something
1) The 10% v/s 90% check if you used glmnet... | Elastic net: dealing with wide data with outliers
Couple of things I just thought I will mention. It is hard to mention specifics without actually looking at the results but hope this is helpful. Most of these things I am sure you already know but in |
50,542 | Comparing different methods of discrete-time survival analysis | A Cox proportional hazards model with "exact" tie resolution, a.k.a a conditional logistic regression ...
A standard logistic regression with one data point per subject-month, with time represented as a categorical variable
A conditional logistic regression model and a standard binary regression model with the logisti... | Comparing different methods of discrete-time survival analysis | A Cox proportional hazards model with "exact" tie resolution, a.k.a a conditional logistic regression ...
A standard logistic regression with one data point per subject-month, with time represented as | Comparing different methods of discrete-time survival analysis
A Cox proportional hazards model with "exact" tie resolution, a.k.a a conditional logistic regression ...
A standard logistic regression with one data point per subject-month, with time represented as a categorical variable
A conditional logistic regressio... | Comparing different methods of discrete-time survival analysis
A Cox proportional hazards model with "exact" tie resolution, a.k.a a conditional logistic regression ...
A standard logistic regression with one data point per subject-month, with time represented as |
50,543 | Categorical variable coding to compare all levels to all levels | You want all possible pairwise comparisons of levels, but there are much more pairs than there are degrees of freedom in the factor. Say the factor has five levels, then you need 4 parameters to code it, but there are $\binom{5}{2}$ pairs, that is, 10 pairs. So it is impossible to find a coding with one parameter for ... | Categorical variable coding to compare all levels to all levels | You want all possible pairwise comparisons of levels, but there are much more pairs than there are degrees of freedom in the factor. Say the factor has five levels, then you need 4 parameters to code | Categorical variable coding to compare all levels to all levels
You want all possible pairwise comparisons of levels, but there are much more pairs than there are degrees of freedom in the factor. Say the factor has five levels, then you need 4 parameters to code it, but there are $\binom{5}{2}$ pairs, that is, 10 pair... | Categorical variable coding to compare all levels to all levels
You want all possible pairwise comparisons of levels, but there are much more pairs than there are degrees of freedom in the factor. Say the factor has five levels, then you need 4 parameters to code |
50,544 | Categorical variable coding to compare all levels to all levels | In the case of ANOVA regression, when you have all the categorical variables, usually one of those will be represented by the intercept (the one you choose). You can verify this: the intercept should be the mean of the variable that was "left out" from your model.
the logic of the variable that is represented in the r... | Categorical variable coding to compare all levels to all levels | In the case of ANOVA regression, when you have all the categorical variables, usually one of those will be represented by the intercept (the one you choose). You can verify this: the intercept should | Categorical variable coding to compare all levels to all levels
In the case of ANOVA regression, when you have all the categorical variables, usually one of those will be represented by the intercept (the one you choose). You can verify this: the intercept should be the mean of the variable that was "left out" from you... | Categorical variable coding to compare all levels to all levels
In the case of ANOVA regression, when you have all the categorical variables, usually one of those will be represented by the intercept (the one you choose). You can verify this: the intercept should |
50,545 | Sensitivity analysis of machine learning techniques | You can use partial dependence plots, which will give you an estimate of the sensitivity of the predicted output with regards to each of the independent variables.
See chapter 10.13.2 of the Elements of Statistical Learning by Hastie, Tibshirani, Friedman. | Sensitivity analysis of machine learning techniques | You can use partial dependence plots, which will give you an estimate of the sensitivity of the predicted output with regards to each of the independent variables.
See chapter 10.13.2 of the Elements | Sensitivity analysis of machine learning techniques
You can use partial dependence plots, which will give you an estimate of the sensitivity of the predicted output with regards to each of the independent variables.
See chapter 10.13.2 of the Elements of Statistical Learning by Hastie, Tibshirani, Friedman. | Sensitivity analysis of machine learning techniques
You can use partial dependence plots, which will give you an estimate of the sensitivity of the predicted output with regards to each of the independent variables.
See chapter 10.13.2 of the Elements |
50,546 | Modelling flight delays with negative values | First, I agree that this is not count data.
If there are many flights that are canceled, then you might think of it as time to event data and look into survival analysis methods. This might depend on where and when you are: More flights are cancelled from Chicago in winter than from Phoenix in May.
Other than that, y... | Modelling flight delays with negative values | First, I agree that this is not count data.
If there are many flights that are canceled, then you might think of it as time to event data and look into survival analysis methods. This might depend on | Modelling flight delays with negative values
First, I agree that this is not count data.
If there are many flights that are canceled, then you might think of it as time to event data and look into survival analysis methods. This might depend on where and when you are: More flights are cancelled from Chicago in winter ... | Modelling flight delays with negative values
First, I agree that this is not count data.
If there are many flights that are canceled, then you might think of it as time to event data and look into survival analysis methods. This might depend on |
50,547 | Model Selection and RFE using caret | You are fitting a @#$^ ton of models, even with the adaptive resampling. You can do it, but the tuning will take a lot of time regardless.
You would probably be better off fitting a model with built-in feature selection instead of using a feature selection wrapper.
sbf could work, and it probably wouldn't be hard to ... | Model Selection and RFE using caret | You are fitting a @#$^ ton of models, even with the adaptive resampling. You can do it, but the tuning will take a lot of time regardless.
You would probably be better off fitting a model with built- | Model Selection and RFE using caret
You are fitting a @#$^ ton of models, even with the adaptive resampling. You can do it, but the tuning will take a lot of time regardless.
You would probably be better off fitting a model with built-in feature selection instead of using a feature selection wrapper.
sbf could work, ... | Model Selection and RFE using caret
You are fitting a @#$^ ton of models, even with the adaptive resampling. You can do it, but the tuning will take a lot of time regardless.
You would probably be better off fitting a model with built- |
50,548 | non-classical measurement error in a binary outcome model | Citing from the survey article by Chen et al. (2011) "Nonlinear Models of Measurement Errors", Journal of Economic Literature:
The approximate bias depends on the derivatives of the regression
function with respect to the mismeasured regressor and the curvature
of the distribution functions of the true regressor a... | non-classical measurement error in a binary outcome model | Citing from the survey article by Chen et al. (2011) "Nonlinear Models of Measurement Errors", Journal of Economic Literature:
The approximate bias depends on the derivatives of the regression
func | non-classical measurement error in a binary outcome model
Citing from the survey article by Chen et al. (2011) "Nonlinear Models of Measurement Errors", Journal of Economic Literature:
The approximate bias depends on the derivatives of the regression
function with respect to the mismeasured regressor and the curvatu... | non-classical measurement error in a binary outcome model
Citing from the survey article by Chen et al. (2011) "Nonlinear Models of Measurement Errors", Journal of Economic Literature:
The approximate bias depends on the derivatives of the regression
func |
50,549 | "Only if" and "if" direction in Kolmogorov's Existence Theorem | I contacted the author: it was a typo, and he's added it to http://probability.ca/jeff/ftpdir/errata2.pdf. | "Only if" and "if" direction in Kolmogorov's Existence Theorem | I contacted the author: it was a typo, and he's added it to http://probability.ca/jeff/ftpdir/errata2.pdf. | "Only if" and "if" direction in Kolmogorov's Existence Theorem
I contacted the author: it was a typo, and he's added it to http://probability.ca/jeff/ftpdir/errata2.pdf. | "Only if" and "if" direction in Kolmogorov's Existence Theorem
I contacted the author: it was a typo, and he's added it to http://probability.ca/jeff/ftpdir/errata2.pdf. |
50,550 | Least-square fit with uneven distribution of data | Assuming that some data is redundant simply because they are similar in value (on the x-axis) then your approach is correct (if ignoring issues of outliers). The technique you are looking for is called Kernel density estimation. The kernel bandwidth should be chosen based on the context of the data. If x-values with... | Least-square fit with uneven distribution of data | Assuming that some data is redundant simply because they are similar in value (on the x-axis) then your approach is correct (if ignoring issues of outliers). The technique you are looking for is call | Least-square fit with uneven distribution of data
Assuming that some data is redundant simply because they are similar in value (on the x-axis) then your approach is correct (if ignoring issues of outliers). The technique you are looking for is called Kernel density estimation. The kernel bandwidth should be chosen b... | Least-square fit with uneven distribution of data
Assuming that some data is redundant simply because they are similar in value (on the x-axis) then your approach is correct (if ignoring issues of outliers). The technique you are looking for is call |
50,551 | Least-square fit with uneven distribution of data | For this problem, I found this article to be of great interest.
For what I understood, there is usually 2 ways to deal with such non well distributed dataset :
either we resample the dataset, usually by deleting some data to transform the dataset into a well distributed one,
or we weight the data points, for example h... | Least-square fit with uneven distribution of data | For this problem, I found this article to be of great interest.
For what I understood, there is usually 2 ways to deal with such non well distributed dataset :
either we resample the dataset, usually | Least-square fit with uneven distribution of data
For this problem, I found this article to be of great interest.
For what I understood, there is usually 2 ways to deal with such non well distributed dataset :
either we resample the dataset, usually by deleting some data to transform the dataset into a well distribute... | Least-square fit with uneven distribution of data
For this problem, I found this article to be of great interest.
For what I understood, there is usually 2 ways to deal with such non well distributed dataset :
either we resample the dataset, usually |
50,552 | Least-square fit with uneven distribution of data | You can use cubic splines. One advantage of splines over standard polynomial regression is that data points influence is more local. This is due to irregularity at knots. | Least-square fit with uneven distribution of data | You can use cubic splines. One advantage of splines over standard polynomial regression is that data points influence is more local. This is due to irregularity at knots. | Least-square fit with uneven distribution of data
You can use cubic splines. One advantage of splines over standard polynomial regression is that data points influence is more local. This is due to irregularity at knots. | Least-square fit with uneven distribution of data
You can use cubic splines. One advantage of splines over standard polynomial regression is that data points influence is more local. This is due to irregularity at knots. |
50,553 | Distribution of "normalised" Gaussian random variables | Not intended as an answer ... but more a comment that is too long for the comment box ...
Updated for OP's change of sample variance to sample standard deviation
To get an idea of the difficulty of the problem ... consider the simplest possible form this question can take, namely:
a sample of size $n = 2$, where ...
$... | Distribution of "normalised" Gaussian random variables | Not intended as an answer ... but more a comment that is too long for the comment box ...
Updated for OP's change of sample variance to sample standard deviation
To get an idea of the difficulty of th | Distribution of "normalised" Gaussian random variables
Not intended as an answer ... but more a comment that is too long for the comment box ...
Updated for OP's change of sample variance to sample standard deviation
To get an idea of the difficulty of the problem ... consider the simplest possible form this question c... | Distribution of "normalised" Gaussian random variables
Not intended as an answer ... but more a comment that is too long for the comment box ...
Updated for OP's change of sample variance to sample standard deviation
To get an idea of the difficulty of th |
50,554 | Robust option in Stata: why are the p values computed using a Student distribution? | Robust variance estimators require large samples to be valid. In small samples, they are biased downward, and the normal-distribution-based confidence intervals may have coverage way below nominal coverage rates.
Using a $t_{n-k}$-distribution approximations to be conservative is one possible solution: you hope that th... | Robust option in Stata: why are the p values computed using a Student distribution? | Robust variance estimators require large samples to be valid. In small samples, they are biased downward, and the normal-distribution-based confidence intervals may have coverage way below nominal cov | Robust option in Stata: why are the p values computed using a Student distribution?
Robust variance estimators require large samples to be valid. In small samples, they are biased downward, and the normal-distribution-based confidence intervals may have coverage way below nominal coverage rates.
Using a $t_{n-k}$-distr... | Robust option in Stata: why are the p values computed using a Student distribution?
Robust variance estimators require large samples to be valid. In small samples, they are biased downward, and the normal-distribution-based confidence intervals may have coverage way below nominal cov |
50,555 | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is small and potentially non-Gaussian? | In terms of the p-value, the answer can be found in an earlier post. Basically, use the permutation test for n<20. A generally normalizing transformation, such as rankit, will work for larger n's and will be more powerful (Bishara & Hittner, 2012). Of course, if you transform, you're no longer looking at the linear r... | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is s | In terms of the p-value, the answer can be found in an earlier post. Basically, use the permutation test for n<20. A generally normalizing transformation, such as rankit, will work for larger n's and | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is small and potentially non-Gaussian?
In terms of the p-value, the answer can be found in an earlier post. Basically, use the permutation test for n<20. A generally normalizing transformation, such as rankit, will work for... | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is s
In terms of the p-value, the answer can be found in an earlier post. Basically, use the permutation test for n<20. A generally normalizing transformation, such as rankit, will work for larger n's and |
50,556 | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is small and potentially non-Gaussian? | You could certainly perform a permutation test (of the null that the two are uncorrelated) in the manner you suggest, but you wouldn't normally "use that distribution to get a confidence interval" for the correlation.
You would instead use that distribution to get a p-value, or an acceptance (/rejection) region.
You co... | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is s | You could certainly perform a permutation test (of the null that the two are uncorrelated) in the manner you suggest, but you wouldn't normally "use that distribution to get a confidence interval" for | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is small and potentially non-Gaussian?
You could certainly perform a permutation test (of the null that the two are uncorrelated) in the manner you suggest, but you wouldn't normally "use that distribution to get a confidenc... | How to get p-values or confidence intervals for pearson correlation coefficient when the sample is s
You could certainly perform a permutation test (of the null that the two are uncorrelated) in the manner you suggest, but you wouldn't normally "use that distribution to get a confidence interval" for |
50,557 | Violation of Gauss-Markov assumptions | There've been a couple answers and none of em have touched on what I thought were the most interesting questions asked, the bias and consistency of misspecified linear models. Since it seems pretty clear from the residuals that the model is misspecified with a quadratic term, let's take a look at what happens to our es... | Violation of Gauss-Markov assumptions | There've been a couple answers and none of em have touched on what I thought were the most interesting questions asked, the bias and consistency of misspecified linear models. Since it seems pretty cl | Violation of Gauss-Markov assumptions
There've been a couple answers and none of em have touched on what I thought were the most interesting questions asked, the bias and consistency of misspecified linear models. Since it seems pretty clear from the residuals that the model is misspecified with a quadratic term, let's... | Violation of Gauss-Markov assumptions
There've been a couple answers and none of em have touched on what I thought were the most interesting questions asked, the bias and consistency of misspecified linear models. Since it seems pretty cl |
50,558 | Violation of Gauss-Markov assumptions | It seems that your data does not follow a linear model. | Violation of Gauss-Markov assumptions | It seems that your data does not follow a linear model. | Violation of Gauss-Markov assumptions
It seems that your data does not follow a linear model. | Violation of Gauss-Markov assumptions
It seems that your data does not follow a linear model. |
50,559 | Violation of Gauss-Markov assumptions | First note that the data look to have a quadratic as opposed to linear relationship. This casts doubt on the linearity assumption:
$$ E[Y\,|\,einkommen] = \beta_{0}+\beta_{1}einkommen$$
My hint to you is this: assume to the contrary that the model is linear, perform the regression (either hypothetically or literally... | Violation of Gauss-Markov assumptions | First note that the data look to have a quadratic as opposed to linear relationship. This casts doubt on the linearity assumption:
$$ E[Y\,|\,einkommen] = \beta_{0}+\beta_{1}einkommen$$
My hint to | Violation of Gauss-Markov assumptions
First note that the data look to have a quadratic as opposed to linear relationship. This casts doubt on the linearity assumption:
$$ E[Y\,|\,einkommen] = \beta_{0}+\beta_{1}einkommen$$
My hint to you is this: assume to the contrary that the model is linear, perform the regressi... | Violation of Gauss-Markov assumptions
First note that the data look to have a quadratic as opposed to linear relationship. This casts doubt on the linearity assumption:
$$ E[Y\,|\,einkommen] = \beta_{0}+\beta_{1}einkommen$$
My hint to |
50,560 | Violation of Gauss-Markov assumptions | The main issue is with assumption #5, i.e. homoscedasticity. Your error variance seems to change with income. It's higher in the middle than at the ends. | Violation of Gauss-Markov assumptions | The main issue is with assumption #5, i.e. homoscedasticity. Your error variance seems to change with income. It's higher in the middle than at the ends. | Violation of Gauss-Markov assumptions
The main issue is with assumption #5, i.e. homoscedasticity. Your error variance seems to change with income. It's higher in the middle than at the ends. | Violation of Gauss-Markov assumptions
The main issue is with assumption #5, i.e. homoscedasticity. Your error variance seems to change with income. It's higher in the middle than at the ends. |
50,561 | How to take advantage of multiples series with the same behaviour for forecasting? | You could use generalized regression model for producing hierarchical forecasts from the individual forecasts.
Here is a link:
https://www.otexts.org/fpp/9/4 | How to take advantage of multiples series with the same behaviour for forecasting? | You could use generalized regression model for producing hierarchical forecasts from the individual forecasts.
Here is a link:
https://www.otexts.org/fpp/9/4 | How to take advantage of multiples series with the same behaviour for forecasting?
You could use generalized regression model for producing hierarchical forecasts from the individual forecasts.
Here is a link:
https://www.otexts.org/fpp/9/4 | How to take advantage of multiples series with the same behaviour for forecasting?
You could use generalized regression model for producing hierarchical forecasts from the individual forecasts.
Here is a link:
https://www.otexts.org/fpp/9/4 |
50,562 | How to take advantage of multiples series with the same behaviour for forecasting? | There is an article on the inside-R website which uses signal decomposition for training and testing and then forecasting with a neural net and which might be a useful alternative to generalized regression models. R code is supplied.
link here | How to take advantage of multiples series with the same behaviour for forecasting? | There is an article on the inside-R website which uses signal decomposition for training and testing and then forecasting with a neural net and which might be a useful alternative to generalized regre | How to take advantage of multiples series with the same behaviour for forecasting?
There is an article on the inside-R website which uses signal decomposition for training and testing and then forecasting with a neural net and which might be a useful alternative to generalized regression models. R code is supplied.
li... | How to take advantage of multiples series with the same behaviour for forecasting?
There is an article on the inside-R website which uses signal decomposition for training and testing and then forecasting with a neural net and which might be a useful alternative to generalized regre |
50,563 | How to take advantage of multiples series with the same behaviour for forecasting? | Try finding the common factors, and model these factors. For instance, you could run PCA, and see if there's a few factors that explain the variance of your 300 series. It is possible that you may find a handful of principal components explain a huge chunk of the variance of all 300 series. In this case you'll model th... | How to take advantage of multiples series with the same behaviour for forecasting? | Try finding the common factors, and model these factors. For instance, you could run PCA, and see if there's a few factors that explain the variance of your 300 series. It is possible that you may fin | How to take advantage of multiples series with the same behaviour for forecasting?
Try finding the common factors, and model these factors. For instance, you could run PCA, and see if there's a few factors that explain the variance of your 300 series. It is possible that you may find a handful of principal components e... | How to take advantage of multiples series with the same behaviour for forecasting?
Try finding the common factors, and model these factors. For instance, you could run PCA, and see if there's a few factors that explain the variance of your 300 series. It is possible that you may fin |
50,564 | Labeling a pool of unlabelled samples iteratively | To answer my own question, the optimal way to pick an initial sample according to information criteria such as entropy is a notorious problem called maximal entropy sampling. This turns out to be NP-hard, so I will probably select a small uniform sample of the data and then try to apply maximal entropy sampling afterwa... | Labeling a pool of unlabelled samples iteratively | To answer my own question, the optimal way to pick an initial sample according to information criteria such as entropy is a notorious problem called maximal entropy sampling. This turns out to be NP-h | Labeling a pool of unlabelled samples iteratively
To answer my own question, the optimal way to pick an initial sample according to information criteria such as entropy is a notorious problem called maximal entropy sampling. This turns out to be NP-hard, so I will probably select a small uniform sample of the data and ... | Labeling a pool of unlabelled samples iteratively
To answer my own question, the optimal way to pick an initial sample according to information criteria such as entropy is a notorious problem called maximal entropy sampling. This turns out to be NP-h |
50,565 | How to compute confidence interval from a confidence distribution | In general you'd do it numerically. In some cases you could do it algebraically, but often there will be no explicit closed form algebraic solution.
I would like to figure out how to find confidence intervals for confidence distributions such as Betas or Gammas and I find it difficult to reach a closed-form solution f... | How to compute confidence interval from a confidence distribution | In general you'd do it numerically. In some cases you could do it algebraically, but often there will be no explicit closed form algebraic solution.
I would like to figure out how to find confidence | How to compute confidence interval from a confidence distribution
In general you'd do it numerically. In some cases you could do it algebraically, but often there will be no explicit closed form algebraic solution.
I would like to figure out how to find confidence intervals for confidence distributions such as Betas o... | How to compute confidence interval from a confidence distribution
In general you'd do it numerically. In some cases you could do it algebraically, but often there will be no explicit closed form algebraic solution.
I would like to figure out how to find confidence |
50,566 | Intuition for the "information matrix equality" result? | I have always found this result to be counter-intuitive as well. In my case, this is due to a tendency to confuse the score function and the maximum likelihood estimator. (Oh, the shame!). In fact, they sort of pull in opposite directions.
Consider the one parameter case. In large samples, the log likelihood tends towa... | Intuition for the "information matrix equality" result? | I have always found this result to be counter-intuitive as well. In my case, this is due to a tendency to confuse the score function and the maximum likelihood estimator. (Oh, the shame!). In fact, th | Intuition for the "information matrix equality" result?
I have always found this result to be counter-intuitive as well. In my case, this is due to a tendency to confuse the score function and the maximum likelihood estimator. (Oh, the shame!). In fact, they sort of pull in opposite directions.
Consider the one paramet... | Intuition for the "information matrix equality" result?
I have always found this result to be counter-intuitive as well. In my case, this is due to a tendency to confuse the score function and the maximum likelihood estimator. (Oh, the shame!). In fact, th |
50,567 | Intuition for the "information matrix equality" result? | I have to say that I am not sure what you are confused about.
"it seems like they would be zero or the same as the main diagonal?"
Not really - expected value of the score is zero but the cross products aren't (the different components of the score vector are not independent) - it is a matrix multiplication. | Intuition for the "information matrix equality" result? | I have to say that I am not sure what you are confused about.
"it seems like they would be zero or the same as the main diagonal?"
Not really - expected value of the score is zero but the cross produc | Intuition for the "information matrix equality" result?
I have to say that I am not sure what you are confused about.
"it seems like they would be zero or the same as the main diagonal?"
Not really - expected value of the score is zero but the cross products aren't (the different components of the score vector are not ... | Intuition for the "information matrix equality" result?
I have to say that I am not sure what you are confused about.
"it seems like they would be zero or the same as the main diagonal?"
Not really - expected value of the score is zero but the cross produc |
50,568 | Paired or not paired? Comparing groups after propensity score matching | I personally find results are very similar when you use paired and unpaired tests. Yet, my recommendation, built upon studying quite extensively the topic, and following authoritative sources, such as this one from Austin, is now to use tests that recognize the clustering features of the dataset.
Thus, if I am using pr... | Paired or not paired? Comparing groups after propensity score matching | I personally find results are very similar when you use paired and unpaired tests. Yet, my recommendation, built upon studying quite extensively the topic, and following authoritative sources, such as | Paired or not paired? Comparing groups after propensity score matching
I personally find results are very similar when you use paired and unpaired tests. Yet, my recommendation, built upon studying quite extensively the topic, and following authoritative sources, such as this one from Austin, is now to use tests that r... | Paired or not paired? Comparing groups after propensity score matching
I personally find results are very similar when you use paired and unpaired tests. Yet, my recommendation, built upon studying quite extensively the topic, and following authoritative sources, such as |
50,569 | Weibull Mixture question | The Weibull survival function with shape parameter $k$ and scale parameter $\lambda$ (both positive) has the form
$$S(x; \lambda, k) = \exp\left(-(x/\lambda)^k\right)$$
for $x \gt 0.$ A finite mixture of $n$ such distributions is determined by positive mixture weights $p_i$ (necessarily summing to unity) and correspon... | Weibull Mixture question | The Weibull survival function with shape parameter $k$ and scale parameter $\lambda$ (both positive) has the form
$$S(x; \lambda, k) = \exp\left(-(x/\lambda)^k\right)$$
for $x \gt 0.$ A finite mixtur | Weibull Mixture question
The Weibull survival function with shape parameter $k$ and scale parameter $\lambda$ (both positive) has the form
$$S(x; \lambda, k) = \exp\left(-(x/\lambda)^k\right)$$
for $x \gt 0.$ A finite mixture of $n$ such distributions is determined by positive mixture weights $p_i$ (necessarily summin... | Weibull Mixture question
The Weibull survival function with shape parameter $k$ and scale parameter $\lambda$ (both positive) has the form
$$S(x; \lambda, k) = \exp\left(-(x/\lambda)^k\right)$$
for $x \gt 0.$ A finite mixtur |
50,570 | How many levels in multilevel modeling is too many? | This is hard to answer without much context. But in general, parameters of additional levels will be harder to estimate. For each additional level you will need much more data, specially for the variance-covariance parameters of the higher levels. See here for a related discussion. | How many levels in multilevel modeling is too many? | This is hard to answer without much context. But in general, parameters of additional levels will be harder to estimate. For each additional level you will need much more data, specially for the varia | How many levels in multilevel modeling is too many?
This is hard to answer without much context. But in general, parameters of additional levels will be harder to estimate. For each additional level you will need much more data, specially for the variance-covariance parameters of the higher levels. See here for a relat... | How many levels in multilevel modeling is too many?
This is hard to answer without much context. But in general, parameters of additional levels will be harder to estimate. For each additional level you will need much more data, specially for the varia |
50,571 | What is the "pdm" stat in the "rms" R package? | I don't know yet about the background of the statistic but below is an illustration how it is computed.
$$pdm = \frac{1}{n} \sum_{k=1}^n \left| \hat{P}(Y \geq median|X_k) - 0.5 \right| $$
It is an indication of how much the conditional predicted probability varies around the point of the marginal median.
library(rms)
... | What is the "pdm" stat in the "rms" R package? | I don't know yet about the background of the statistic but below is an illustration how it is computed.
$$pdm = \frac{1}{n} \sum_{k=1}^n \left| \hat{P}(Y \geq median|X_k) - 0.5 \right| $$
It is an ind | What is the "pdm" stat in the "rms" R package?
I don't know yet about the background of the statistic but below is an illustration how it is computed.
$$pdm = \frac{1}{n} \sum_{k=1}^n \left| \hat{P}(Y \geq median|X_k) - 0.5 \right| $$
It is an indication of how much the conditional predicted probability varies around t... | What is the "pdm" stat in the "rms" R package?
I don't know yet about the background of the statistic but below is an illustration how it is computed.
$$pdm = \frac{1}{n} \sum_{k=1}^n \left| \hat{P}(Y \geq median|X_k) - 0.5 \right| $$
It is an ind |
50,572 | What is the "pdm" stat in the "rms" R package? | This link: https://www.rdocumentation.org/packages/rms/versions/5.1-0/topics/validate.lrm
describes pdm as a "new" metric, which would imply that is has not been used before. | What is the "pdm" stat in the "rms" R package? | This link: https://www.rdocumentation.org/packages/rms/versions/5.1-0/topics/validate.lrm
describes pdm as a "new" metric, which would imply that is has not been used before. | What is the "pdm" stat in the "rms" R package?
This link: https://www.rdocumentation.org/packages/rms/versions/5.1-0/topics/validate.lrm
describes pdm as a "new" metric, which would imply that is has not been used before. | What is the "pdm" stat in the "rms" R package?
This link: https://www.rdocumentation.org/packages/rms/versions/5.1-0/topics/validate.lrm
describes pdm as a "new" metric, which would imply that is has not been used before. |
50,573 | Maximum likelihood of multivariate t-distributed variable with scaled covariance | The EM algorithm is typically used to find MLEs of the parameters from an iid multivariate t sample. Instead of writting a long answer as to how to implement the algorithm I refer you to McLachlan and Krishnan "The EM algorithm and Extensions", second edition. They show the MLE procedure for both known and unknown deg... | Maximum likelihood of multivariate t-distributed variable with scaled covariance | The EM algorithm is typically used to find MLEs of the parameters from an iid multivariate t sample. Instead of writting a long answer as to how to implement the algorithm I refer you to McLachlan an | Maximum likelihood of multivariate t-distributed variable with scaled covariance
The EM algorithm is typically used to find MLEs of the parameters from an iid multivariate t sample. Instead of writting a long answer as to how to implement the algorithm I refer you to McLachlan and Krishnan "The EM algorithm and Extens... | Maximum likelihood of multivariate t-distributed variable with scaled covariance
The EM algorithm is typically used to find MLEs of the parameters from an iid multivariate t sample. Instead of writting a long answer as to how to implement the algorithm I refer you to McLachlan an |
50,574 | Unrealistically high significance when marginalizing over large number of parameters | As Cyan pointed out, this behaviour arises from the fact that the model does not put enough emphasis on the case $\mathbf{a}=0$. You say that you don't want to impose a 'tighter' prior on $\mathbf{a}$. I'm not sure what you mean by 'tighter', but increasing the prior probability of $\mathbf{a}=0$ is the only solution... | Unrealistically high significance when marginalizing over large number of parameters | As Cyan pointed out, this behaviour arises from the fact that the model does not put enough emphasis on the case $\mathbf{a}=0$. You say that you don't want to impose a 'tighter' prior on $\mathbf{a} | Unrealistically high significance when marginalizing over large number of parameters
As Cyan pointed out, this behaviour arises from the fact that the model does not put enough emphasis on the case $\mathbf{a}=0$. You say that you don't want to impose a 'tighter' prior on $\mathbf{a}$. I'm not sure what you mean by '... | Unrealistically high significance when marginalizing over large number of parameters
As Cyan pointed out, this behaviour arises from the fact that the model does not put enough emphasis on the case $\mathbf{a}=0$. You say that you don't want to impose a 'tighter' prior on $\mathbf{a} |
50,575 | What is the limiting distribution of the sample mean? | You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d $ a constant. A different way to frame a similar question is to say, what is an approximating distribution of $\bar{X}_... | What is the limiting distribution of the sample mean? | You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d | What is the limiting distribution of the sample mean?
You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d $ a constant. A different way to frame a similar question is to s... | What is the limiting distribution of the sample mean?
You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d |
50,576 | Gamma vs tweedie distribution for large productivity dataset | The question you need to ask yourself is if your response variable takes 0 values (not if it takes very small values). Normally if you have 0s on your data you should'nt be able to fit a gamma distribution.
I would suggest then trying lognormal, gamma and inverse normal, which are the most common positive distributio... | Gamma vs tweedie distribution for large productivity dataset | The question you need to ask yourself is if your response variable takes 0 values (not if it takes very small values). Normally if you have 0s on your data you should'nt be able to fit a gamma distrib | Gamma vs tweedie distribution for large productivity dataset
The question you need to ask yourself is if your response variable takes 0 values (not if it takes very small values). Normally if you have 0s on your data you should'nt be able to fit a gamma distribution.
I would suggest then trying lognormal, gamma and i... | Gamma vs tweedie distribution for large productivity dataset
The question you need to ask yourself is if your response variable takes 0 values (not if it takes very small values). Normally if you have 0s on your data you should'nt be able to fit a gamma distrib |
50,577 | Random variables with some properties (conditional expectation) | By keeping things as simple as possible we can construct a rather pretty solution.
Step 0 We have to begin somewhere. Since the variables are supposed to have strictly positive values, take the simplest positive number, $1$, and since $X$ appears first alphabetically, suppose $X=1$. In order to obtain $\mathbb{E}(Y|... | Random variables with some properties (conditional expectation) | By keeping things as simple as possible we can construct a rather pretty solution.
Step 0 We have to begin somewhere. Since the variables are supposed to have strictly positive values, take the simp | Random variables with some properties (conditional expectation)
By keeping things as simple as possible we can construct a rather pretty solution.
Step 0 We have to begin somewhere. Since the variables are supposed to have strictly positive values, take the simplest positive number, $1$, and since $X$ appears first a... | Random variables with some properties (conditional expectation)
By keeping things as simple as possible we can construct a rather pretty solution.
Step 0 We have to begin somewhere. Since the variables are supposed to have strictly positive values, take the simp |
50,578 | Fitting a Gaussian to a histogram when the bin size is significant | If you know that $y_i \in [x_j, x_{j+1})$, where $x_j$'s are cut points from a bin, then you can treat this as interval censored data. In other words, for your case, you can define your likelihood function as
$\displaystyle \prod_{i = 1}^n (\Phi(r_i|\mu, \sigma) - \Phi(l_i|\mu, \sigma) )$
Where $l_i$ and $r_i$ are the... | Fitting a Gaussian to a histogram when the bin size is significant | If you know that $y_i \in [x_j, x_{j+1})$, where $x_j$'s are cut points from a bin, then you can treat this as interval censored data. In other words, for your case, you can define your likelihood fun | Fitting a Gaussian to a histogram when the bin size is significant
If you know that $y_i \in [x_j, x_{j+1})$, where $x_j$'s are cut points from a bin, then you can treat this as interval censored data. In other words, for your case, you can define your likelihood function as
$\displaystyle \prod_{i = 1}^n (\Phi(r_i|\m... | Fitting a Gaussian to a histogram when the bin size is significant
If you know that $y_i \in [x_j, x_{j+1})$, where $x_j$'s are cut points from a bin, then you can treat this as interval censored data. In other words, for your case, you can define your likelihood fun |
50,579 | Fitting a Gaussian to a histogram when the bin size is significant | You should treat each bin as if it were generating random points uniformly within its bounds. Therefore calculate a weighted average for each bin $(x_l, x_h]$ of $E(x) = \frac{x_h + x_l}{2}$ and $E(x^2) = \frac{x_h^2 + x_lx_h + x_l^2}{3}$. This weighted average determines a Gaussian.
You can incorporate a prior by tre... | Fitting a Gaussian to a histogram when the bin size is significant | You should treat each bin as if it were generating random points uniformly within its bounds. Therefore calculate a weighted average for each bin $(x_l, x_h]$ of $E(x) = \frac{x_h + x_l}{2}$ and $E(x^ | Fitting a Gaussian to a histogram when the bin size is significant
You should treat each bin as if it were generating random points uniformly within its bounds. Therefore calculate a weighted average for each bin $(x_l, x_h]$ of $E(x) = \frac{x_h + x_l}{2}$ and $E(x^2) = \frac{x_h^2 + x_lx_h + x_l^2}{3}$. This weighte... | Fitting a Gaussian to a histogram when the bin size is significant
You should treat each bin as if it were generating random points uniformly within its bounds. Therefore calculate a weighted average for each bin $(x_l, x_h]$ of $E(x) = \frac{x_h + x_l}{2}$ and $E(x^ |
50,580 | Fitting a Gaussian to a histogram when the bin size is significant | Given whuber's comment on my last answer, I suggest you use that answer to find a mean and variance $\mu, \sigma^2$ as a starting point. Then, calculate the log-likelihood of having observed the bin counts you got $\ell$. Finally, optimize the mean and variance by gradient descent. It should be easy to calculate the... | Fitting a Gaussian to a histogram when the bin size is significant | Given whuber's comment on my last answer, I suggest you use that answer to find a mean and variance $\mu, \sigma^2$ as a starting point. Then, calculate the log-likelihood of having observed the bin | Fitting a Gaussian to a histogram when the bin size is significant
Given whuber's comment on my last answer, I suggest you use that answer to find a mean and variance $\mu, \sigma^2$ as a starting point. Then, calculate the log-likelihood of having observed the bin counts you got $\ell$. Finally, optimize the mean an... | Fitting a Gaussian to a histogram when the bin size is significant
Given whuber's comment on my last answer, I suggest you use that answer to find a mean and variance $\mu, \sigma^2$ as a starting point. Then, calculate the log-likelihood of having observed the bin |
50,581 | Methods of fitting a dynamic linear model | The Kalman filter is the "forward filtering" part of FFBS, while the "backward sampling" part provides a draw from the joint distribution for the $\Theta_t$ for all $t$.
All the other ways of performing statistical parameter estimation, e.g. maximum likelihood, can be used for DLMs. Yes a particle filter could be used... | Methods of fitting a dynamic linear model | The Kalman filter is the "forward filtering" part of FFBS, while the "backward sampling" part provides a draw from the joint distribution for the $\Theta_t$ for all $t$.
All the other ways of perform | Methods of fitting a dynamic linear model
The Kalman filter is the "forward filtering" part of FFBS, while the "backward sampling" part provides a draw from the joint distribution for the $\Theta_t$ for all $t$.
All the other ways of performing statistical parameter estimation, e.g. maximum likelihood, can be used for... | Methods of fitting a dynamic linear model
The Kalman filter is the "forward filtering" part of FFBS, while the "backward sampling" part provides a draw from the joint distribution for the $\Theta_t$ for all $t$.
All the other ways of perform |
50,582 | What are desirable characteristics of a test statistic? | For a test-statistic to be a statistical test you need to know the sampling distribution of that statistic if the null hypothesis is true. For some statistics it is easier to derive (asymptotically) what that distribution would be, and these statistics have been given names like t-statistic, F-statistic, etc. There exi... | What are desirable characteristics of a test statistic? | For a test-statistic to be a statistical test you need to know the sampling distribution of that statistic if the null hypothesis is true. For some statistics it is easier to derive (asymptotically) w | What are desirable characteristics of a test statistic?
For a test-statistic to be a statistical test you need to know the sampling distribution of that statistic if the null hypothesis is true. For some statistics it is easier to derive (asymptotically) what that distribution would be, and these statistics have been g... | What are desirable characteristics of a test statistic?
For a test-statistic to be a statistical test you need to know the sampling distribution of that statistic if the null hypothesis is true. For some statistics it is easier to derive (asymptotically) w |
50,583 | What are desirable characteristics of a test statistic? | tl;dr: your test needs to have statistical power, the concept of statistical power invalidates the whole raison d'etre of null hypothesis testing
Suppose you have run an experiment in a lab and you'd like to test is an effect. Anathema! Don't you know you can only reject the absence of an effect? Haven't you read Poppe... | What are desirable characteristics of a test statistic? | tl;dr: your test needs to have statistical power, the concept of statistical power invalidates the whole raison d'etre of null hypothesis testing
Suppose you have run an experiment in a lab and you'd | What are desirable characteristics of a test statistic?
tl;dr: your test needs to have statistical power, the concept of statistical power invalidates the whole raison d'etre of null hypothesis testing
Suppose you have run an experiment in a lab and you'd like to test is an effect. Anathema! Don't you know you can only... | What are desirable characteristics of a test statistic?
tl;dr: your test needs to have statistical power, the concept of statistical power invalidates the whole raison d'etre of null hypothesis testing
Suppose you have run an experiment in a lab and you'd |
50,584 | What are desirable characteristics of a test statistic? | From more of an intro stat perspective, some tests are useful with some data and others are not. For example, if your data is normally distributed (or if you're looking at the means of samples greater than 30 in size) you can use a Z-test. They're easy to compute.
If you're forced to use a small sample, a better distri... | What are desirable characteristics of a test statistic? | From more of an intro stat perspective, some tests are useful with some data and others are not. For example, if your data is normally distributed (or if you're looking at the means of samples greater | What are desirable characteristics of a test statistic?
From more of an intro stat perspective, some tests are useful with some data and others are not. For example, if your data is normally distributed (or if you're looking at the means of samples greater than 30 in size) you can use a Z-test. They're easy to compute.... | What are desirable characteristics of a test statistic?
From more of an intro stat perspective, some tests are useful with some data and others are not. For example, if your data is normally distributed (or if you're looking at the means of samples greater |
50,585 | Find conditional expectation given a discrete random variable whose range is N | Since
$$
X_N=\sum_{n\geqslant 1} X_n\mathbf{1}_{N=n}
$$
holds pointwise, we have
$$
{\rm E}[X_N]=\sum_{n\geqslant 1}\mu_np_n
$$
agreeing with your expression. Similarly,
$$
{\rm E}[X_N^2]=\sum_{n\geqslant 1}{\rm E}[X_n^2]p_n=\sum_{n\geqslant 1}(\sigma_n^2+\mu_n^2)p_n
$$
and hence
$$
{\rm Var}(X_N)=\sum_{n\geqslant 1} ... | Find conditional expectation given a discrete random variable whose range is N | Since
$$
X_N=\sum_{n\geqslant 1} X_n\mathbf{1}_{N=n}
$$
holds pointwise, we have
$$
{\rm E}[X_N]=\sum_{n\geqslant 1}\mu_np_n
$$
agreeing with your expression. Similarly,
$$
{\rm E}[X_N^2]=\sum_{n\geq | Find conditional expectation given a discrete random variable whose range is N
Since
$$
X_N=\sum_{n\geqslant 1} X_n\mathbf{1}_{N=n}
$$
holds pointwise, we have
$$
{\rm E}[X_N]=\sum_{n\geqslant 1}\mu_np_n
$$
agreeing with your expression. Similarly,
$$
{\rm E}[X_N^2]=\sum_{n\geqslant 1}{\rm E}[X_n^2]p_n=\sum_{n\geqslan... | Find conditional expectation given a discrete random variable whose range is N
Since
$$
X_N=\sum_{n\geqslant 1} X_n\mathbf{1}_{N=n}
$$
holds pointwise, we have
$$
{\rm E}[X_N]=\sum_{n\geqslant 1}\mu_np_n
$$
agreeing with your expression. Similarly,
$$
{\rm E}[X_N^2]=\sum_{n\geq |
50,586 | Find conditional expectation given a discrete random variable whose range is N | This question is another clear case of applying identity: $E[f(X,Y)|Y=y]=E[f(X,y)|Y=y]$.
$$E[X_N|N=n]=E[X_n|N=n]=E[X_n]=\mu _n$$.
In the same way, for the variance we have:
$$Var[X_N|N=n]=E[(X_N-\mu_N)^2|N=n]=E[(X_n-\mu_n)^2|N=n]=E[(X_n-\mu_n)^2|=\sigma_n^2$$ | Find conditional expectation given a discrete random variable whose range is N | This question is another clear case of applying identity: $E[f(X,Y)|Y=y]=E[f(X,y)|Y=y]$.
$$E[X_N|N=n]=E[X_n|N=n]=E[X_n]=\mu _n$$.
In the same way, for the variance we have:
$$Var[X_N|N=n]=E[(X_N-\mu_N | Find conditional expectation given a discrete random variable whose range is N
This question is another clear case of applying identity: $E[f(X,Y)|Y=y]=E[f(X,y)|Y=y]$.
$$E[X_N|N=n]=E[X_n|N=n]=E[X_n]=\mu _n$$.
In the same way, for the variance we have:
$$Var[X_N|N=n]=E[(X_N-\mu_N)^2|N=n]=E[(X_n-\mu_n)^2|N=n]=E[(X_n-\mu_... | Find conditional expectation given a discrete random variable whose range is N
This question is another clear case of applying identity: $E[f(X,Y)|Y=y]=E[f(X,y)|Y=y]$.
$$E[X_N|N=n]=E[X_n|N=n]=E[X_n]=\mu _n$$.
In the same way, for the variance we have:
$$Var[X_N|N=n]=E[(X_N-\mu_N |
50,587 | What are some interesting examples of wrong or crazy inferences being drawn from Big Data? | One example could be Google's failure to predict flu trends. See for instance this Guardian article. | What are some interesting examples of wrong or crazy inferences being drawn from Big Data? | One example could be Google's failure to predict flu trends. See for instance this Guardian article. | What are some interesting examples of wrong or crazy inferences being drawn from Big Data?
One example could be Google's failure to predict flu trends. See for instance this Guardian article. | What are some interesting examples of wrong or crazy inferences being drawn from Big Data?
One example could be Google's failure to predict flu trends. See for instance this Guardian article. |
50,588 | Extracting city name from free text? | This task is typically referred as named entity normalization. Fuzzy string matching can be a good baseline if words are not too close (in terms of Levenshtein distance) in your dictionary. I have used the Python package fuzzywuzzy in the past for that purpose. | Extracting city name from free text? | This task is typically referred as named entity normalization. Fuzzy string matching can be a good baseline if words are not too close (in terms of Levenshtein distance) in your dictionary. I have us | Extracting city name from free text?
This task is typically referred as named entity normalization. Fuzzy string matching can be a good baseline if words are not too close (in terms of Levenshtein distance) in your dictionary. I have used the Python package fuzzywuzzy in the past for that purpose. | Extracting city name from free text?
This task is typically referred as named entity normalization. Fuzzy string matching can be a good baseline if words are not too close (in terms of Levenshtein distance) in your dictionary. I have us |
50,589 | Extracting city name from free text? | What I currently do to normalize places name is resorting to geocoding using api such as, e.g. navitia or google map, which already deal with the normalization process.
Once the lat/lng in hand, I reverse-geocode them, of course always using the same api so as to get normalized outputs.
Furthermore, these api return m... | Extracting city name from free text? | What I currently do to normalize places name is resorting to geocoding using api such as, e.g. navitia or google map, which already deal with the normalization process.
Once the lat/lng in hand, I re | Extracting city name from free text?
What I currently do to normalize places name is resorting to geocoding using api such as, e.g. navitia or google map, which already deal with the normalization process.
Once the lat/lng in hand, I reverse-geocode them, of course always using the same api so as to get normalized out... | Extracting city name from free text?
What I currently do to normalize places name is resorting to geocoding using api such as, e.g. navitia or google map, which already deal with the normalization process.
Once the lat/lng in hand, I re |
50,590 | Propensity score matching: using alternative methods to create a distance measure | Logistic regression is mostly likely used because of historic convenience, well studied convergence properties, relative data-frugality in comparison with other ML learners as well as being readily available pretty much everywhere. Also the resulting probabilities are usually "well-calibrated" out-of-the-box and this i... | Propensity score matching: using alternative methods to create a distance measure | Logistic regression is mostly likely used because of historic convenience, well studied convergence properties, relative data-frugality in comparison with other ML learners as well as being readily av | Propensity score matching: using alternative methods to create a distance measure
Logistic regression is mostly likely used because of historic convenience, well studied convergence properties, relative data-frugality in comparison with other ML learners as well as being readily available pretty much everywhere. Also t... | Propensity score matching: using alternative methods to create a distance measure
Logistic regression is mostly likely used because of historic convenience, well studied convergence properties, relative data-frugality in comparison with other ML learners as well as being readily av |
50,591 | When do (and don't) confidence intervals and credible intervals coincide? | I'm not sure you can consider this a complete answer so you can double check yourself, however, here goes.
By the definition of confidence intervals that
there's an $X\%$ chance that when computing the $X\%$ confidence intervals (CI) the true value $y$ will fall within computed CI,
then you can synthesize an experim... | When do (and don't) confidence intervals and credible intervals coincide? | I'm not sure you can consider this a complete answer so you can double check yourself, however, here goes.
By the definition of confidence intervals that
there's an $X\%$ chance that when computing | When do (and don't) confidence intervals and credible intervals coincide?
I'm not sure you can consider this a complete answer so you can double check yourself, however, here goes.
By the definition of confidence intervals that
there's an $X\%$ chance that when computing the $X\%$ confidence intervals (CI) the true v... | When do (and don't) confidence intervals and credible intervals coincide?
I'm not sure you can consider this a complete answer so you can double check yourself, however, here goes.
By the definition of confidence intervals that
there's an $X\%$ chance that when computing |
50,592 | How to statistically compare groups for multiple density plots? | The problem with a chi-square is it ignores the ordering, leading to a loss of power.
One possibility: there is a k-group version of a Kolmogorov-Smirnov test$^{[1]}$.
Another is a k-sample Anderson-Darling test. E.g. see Wikipedia.
A third possibility might be to look at an orthogonal-polynomial decomposition of a ch... | How to statistically compare groups for multiple density plots? | The problem with a chi-square is it ignores the ordering, leading to a loss of power.
One possibility: there is a k-group version of a Kolmogorov-Smirnov test$^{[1]}$.
Another is a k-sample Anderson- | How to statistically compare groups for multiple density plots?
The problem with a chi-square is it ignores the ordering, leading to a loss of power.
One possibility: there is a k-group version of a Kolmogorov-Smirnov test$^{[1]}$.
Another is a k-sample Anderson-Darling test. E.g. see Wikipedia.
A third possibility mi... | How to statistically compare groups for multiple density plots?
The problem with a chi-square is it ignores the ordering, leading to a loss of power.
One possibility: there is a k-group version of a Kolmogorov-Smirnov test$^{[1]}$.
Another is a k-sample Anderson- |
50,593 | Significant output in Levene's test for equality of variances in MANOVA; what to do? | First, make sure you look at the boxplots of the residuals instead of just using Levene's tests. The significant result could be due to outliers, a bimodal distribution, or skewness that you may need to address.
To get equal variances, try log-transforming (ln) your dependent variables before you run MANOVA.
Also, t... | Significant output in Levene's test for equality of variances in MANOVA; what to do? | First, make sure you look at the boxplots of the residuals instead of just using Levene's tests. The significant result could be due to outliers, a bimodal distribution, or skewness that you may need | Significant output in Levene's test for equality of variances in MANOVA; what to do?
First, make sure you look at the boxplots of the residuals instead of just using Levene's tests. The significant result could be due to outliers, a bimodal distribution, or skewness that you may need to address.
To get equal variance... | Significant output in Levene's test for equality of variances in MANOVA; what to do?
First, make sure you look at the boxplots of the residuals instead of just using Levene's tests. The significant result could be due to outliers, a bimodal distribution, or skewness that you may need |
50,594 | Question about Dynkin Lehmann Scheffe Theorem | The following claim makes no sense:
$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient
Lehmann and Scheffé proved:
If $T(x)=T(y)$ $\iff$ $\theta \mapsto \dfrac{p(x,\theta)}{p(y,\theta)}$ is a constant function, then $T$ is minimal sufficient.
Call $(*)$ = "$\theta \mapsto \frac{p(x,\theta)}{p(y,\theta)}$ is a con... | Question about Dynkin Lehmann Scheffe Theorem | The following claim makes no sense:
$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient
Lehmann and Scheffé proved:
If $T(x)=T(y)$ $\iff$ $\theta \mapsto \dfrac{p(x,\theta)}{p(y,\theta)}$ is a con | Question about Dynkin Lehmann Scheffe Theorem
The following claim makes no sense:
$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient
Lehmann and Scheffé proved:
If $T(x)=T(y)$ $\iff$ $\theta \mapsto \dfrac{p(x,\theta)}{p(y,\theta)}$ is a constant function, then $T$ is minimal sufficient.
Call $(*)$ = "$\theta \ma... | Question about Dynkin Lehmann Scheffe Theorem
The following claim makes no sense:
$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient
Lehmann and Scheffé proved:
If $T(x)=T(y)$ $\iff$ $\theta \mapsto \dfrac{p(x,\theta)}{p(y,\theta)}$ is a con |
50,595 | Drawing numbered balls from an urn | I would suggest the following reason:
You have a random variable X - the sum of balls in a round. Which is an independent identically distributed variable ( by assumption?) ( whereas $\mu*$ is not)
You are interested in $Z=\frac{1}{N_x}\sum_{j=1}^{10^6}\sum_{j=1}^{n_j} x_{ij}$.
Well the central limit theorem tells you... | Drawing numbered balls from an urn | I would suggest the following reason:
You have a random variable X - the sum of balls in a round. Which is an independent identically distributed variable ( by assumption?) ( whereas $\mu*$ is not)
Y | Drawing numbered balls from an urn
I would suggest the following reason:
You have a random variable X - the sum of balls in a round. Which is an independent identically distributed variable ( by assumption?) ( whereas $\mu*$ is not)
You are interested in $Z=\frac{1}{N_x}\sum_{j=1}^{10^6}\sum_{j=1}^{n_j} x_{ij}$.
Well ... | Drawing numbered balls from an urn
I would suggest the following reason:
You have a random variable X - the sum of balls in a round. Which is an independent identically distributed variable ( by assumption?) ( whereas $\mu*$ is not)
Y |
50,596 | Drawing numbered balls from an urn | Is the mean really the best estimator and why?
If you're looking for the highest reward (or highest sum of ball numbers), then yes, the mean is the best estimator. It's also consistent with common sense to pick the robot which had the highest average in earlier rounds. Especially because the robots don't draw the bal... | Drawing numbered balls from an urn | Is the mean really the best estimator and why?
If you're looking for the highest reward (or highest sum of ball numbers), then yes, the mean is the best estimator. It's also consistent with common se | Drawing numbered balls from an urn
Is the mean really the best estimator and why?
If you're looking for the highest reward (or highest sum of ball numbers), then yes, the mean is the best estimator. It's also consistent with common sense to pick the robot which had the highest average in earlier rounds. Especially be... | Drawing numbered balls from an urn
Is the mean really the best estimator and why?
If you're looking for the highest reward (or highest sum of ball numbers), then yes, the mean is the best estimator. It's also consistent with common se |
50,597 | Weighted least squares | For weighted regression through the origin, I presume you either know or can show that $b_1=\frac{\sum_i w_ix_iy_i}{\sum_i w_ix_i^2} = \frac{\sum_i (w_ix_i)y_i}{\sum_i (w_ix_i)x_i}$
Since in this case $b_1=\frac{\sum_i y_i}{\sum x_i}$, you can see by inspection that the weights must be such that $w_ix_i$ is a constant.... | Weighted least squares | For weighted regression through the origin, I presume you either know or can show that $b_1=\frac{\sum_i w_ix_iy_i}{\sum_i w_ix_i^2} = \frac{\sum_i (w_ix_i)y_i}{\sum_i (w_ix_i)x_i}$
Since in this case | Weighted least squares
For weighted regression through the origin, I presume you either know or can show that $b_1=\frac{\sum_i w_ix_iy_i}{\sum_i w_ix_i^2} = \frac{\sum_i (w_ix_i)y_i}{\sum_i (w_ix_i)x_i}$
Since in this case $b_1=\frac{\sum_i y_i}{\sum x_i}$, you can see by inspection that the weights must be such that ... | Weighted least squares
For weighted regression through the origin, I presume you either know or can show that $b_1=\frac{\sum_i w_ix_iy_i}{\sum_i w_ix_i^2} = \frac{\sum_i (w_ix_i)y_i}{\sum_i (w_ix_i)x_i}$
Since in this case |
50,598 | Cumulative Hazard Function where "status" is dependent on "time" | I'm not going to give you a full answer on how to address this at the moment, but I do want to alert you a very serious statistical issue in your dataset.
Standard survival analysis methods (such as the standard KM curves, Cox PH models and aft models) all assume that the censoring is independent of event time. But th... | Cumulative Hazard Function where "status" is dependent on "time" | I'm not going to give you a full answer on how to address this at the moment, but I do want to alert you a very serious statistical issue in your dataset.
Standard survival analysis methods (such as | Cumulative Hazard Function where "status" is dependent on "time"
I'm not going to give you a full answer on how to address this at the moment, but I do want to alert you a very serious statistical issue in your dataset.
Standard survival analysis methods (such as the standard KM curves, Cox PH models and aft models) a... | Cumulative Hazard Function where "status" is dependent on "time"
I'm not going to give you a full answer on how to address this at the moment, but I do want to alert you a very serious statistical issue in your dataset.
Standard survival analysis methods (such as |
50,599 | Cumulative Hazard Function where "status" is dependent on "time" | I don´t know if I understood well your question but here's my point of view about this... In my experience with real data ,even though with nothing related with health data, I usually prefer not to deal with cox models because I never met a case with proportional hazards. I've been always in front of data where hazard... | Cumulative Hazard Function where "status" is dependent on "time" | I don´t know if I understood well your question but here's my point of view about this... In my experience with real data ,even though with nothing related with health data, I usually prefer not to de | Cumulative Hazard Function where "status" is dependent on "time"
I don´t know if I understood well your question but here's my point of view about this... In my experience with real data ,even though with nothing related with health data, I usually prefer not to deal with cox models because I never met a case with prop... | Cumulative Hazard Function where "status" is dependent on "time"
I don´t know if I understood well your question but here's my point of view about this... In my experience with real data ,even though with nothing related with health data, I usually prefer not to de |
50,600 | Estimating Poisson process intensity using GLM | Can I assume the sojourn times to be exponentially i.i.d. given that the Poisson process is inhomogeneous?
In general the inter-event intervals will not be exponentially distributed. This paper by Yakovlev et al. (2008) derives an expression for the inter-event distribution for one-dimensional non-homogeneous Poisson ... | Estimating Poisson process intensity using GLM | Can I assume the sojourn times to be exponentially i.i.d. given that the Poisson process is inhomogeneous?
In general the inter-event intervals will not be exponentially distributed. This paper by Ya | Estimating Poisson process intensity using GLM
Can I assume the sojourn times to be exponentially i.i.d. given that the Poisson process is inhomogeneous?
In general the inter-event intervals will not be exponentially distributed. This paper by Yakovlev et al. (2008) derives an expression for the inter-event distributi... | Estimating Poisson process intensity using GLM
Can I assume the sojourn times to be exponentially i.i.d. given that the Poisson process is inhomogeneous?
In general the inter-event intervals will not be exponentially distributed. This paper by Ya |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.