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 |
|---|---|---|---|---|---|---|
44,901 | Intuition for the standard error of the difference of sample means | You seem to be thinking that $\sqrt{\text{Var}(\bar X-\bar Y)} = \sqrt{\text{Var}(\bar X)} + \sqrt{\text{Var}(\bar Y)}$.
This is not the case for independent variables.
For $X,Y$ independent, $\text{Var}(\bar X-\bar Y) = \text{Var}(\bar X) + \text{Var}(\bar Y)$
Further,
$\text{Var}(\bar X) = \text{Var}(\frac{1}{n}\su... | Intuition for the standard error of the difference of sample means | You seem to be thinking that $\sqrt{\text{Var}(\bar X-\bar Y)} = \sqrt{\text{Var}(\bar X)} + \sqrt{\text{Var}(\bar Y)}$.
This is not the case for independent variables.
For $X,Y$ independent, $\text{ | Intuition for the standard error of the difference of sample means
You seem to be thinking that $\sqrt{\text{Var}(\bar X-\bar Y)} = \sqrt{\text{Var}(\bar X)} + \sqrt{\text{Var}(\bar Y)}$.
This is not the case for independent variables.
For $X,Y$ independent, $\text{Var}(\bar X-\bar Y) = \text{Var}(\bar X) + \text{Var}... | Intuition for the standard error of the difference of sample means
You seem to be thinking that $\sqrt{\text{Var}(\bar X-\bar Y)} = \sqrt{\text{Var}(\bar X)} + \sqrt{\text{Var}(\bar Y)}$.
This is not the case for independent variables.
For $X,Y$ independent, $\text{ |
44,902 | Intuition for the standard error of the difference of sample means | Algebraic intuition
The standard error of the mean for $n$ independent observations is $\frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation.
So if we have two independent samples we have the standard errors for the means of group 1 and group 2.
$$\sigma_{\mu_1}=\frac{\sigma_1}{\sqrt{n_1}}$$
$$\sigma_{\mu_... | Intuition for the standard error of the difference of sample means | Algebraic intuition
The standard error of the mean for $n$ independent observations is $\frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation.
So if we have two independent samples we have | Intuition for the standard error of the difference of sample means
Algebraic intuition
The standard error of the mean for $n$ independent observations is $\frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation.
So if we have two independent samples we have the standard errors for the means of group 1 and gro... | Intuition for the standard error of the difference of sample means
Algebraic intuition
The standard error of the mean for $n$ independent observations is $\frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation.
So if we have two independent samples we have |
44,903 | Intuition for the standard error of the difference of sample means | You don't square the sum of the variances, you take the square root of the sum of the variances. You do this for the same reason that the standard deviation is the square root of the variance: It make the units the same as the original ones, rather than squared units.
Although we often lose sight of it while doing sta... | Intuition for the standard error of the difference of sample means | You don't square the sum of the variances, you take the square root of the sum of the variances. You do this for the same reason that the standard deviation is the square root of the variance: It make | Intuition for the standard error of the difference of sample means
You don't square the sum of the variances, you take the square root of the sum of the variances. You do this for the same reason that the standard deviation is the square root of the variance: It make the units the same as the original ones, rather than... | Intuition for the standard error of the difference of sample means
You don't square the sum of the variances, you take the square root of the sum of the variances. You do this for the same reason that the standard deviation is the square root of the variance: It make |
44,904 | does rstandard standardize in z? | rstandard() produces standardised residuals via normalisation to unit variance using the overall error variance of the residuals/model.
rstudent() produces Studentized residuals in the same way, but it uses a leave-one-out estimate of the error variance.
The key line in rstandard() is
res <- infl$wt.res/(sd * sqrt(1 - ... | does rstandard standardize in z? | rstandard() produces standardised residuals via normalisation to unit variance using the overall error variance of the residuals/model.
rstudent() produces Studentized residuals in the same way, but i | does rstandard standardize in z?
rstandard() produces standardised residuals via normalisation to unit variance using the overall error variance of the residuals/model.
rstudent() produces Studentized residuals in the same way, but it uses a leave-one-out estimate of the error variance.
The key line in rstandard() is
r... | does rstandard standardize in z?
rstandard() produces standardised residuals via normalisation to unit variance using the overall error variance of the residuals/model.
rstudent() produces Studentized residuals in the same way, but i |
44,905 | does rstandard standardize in z? | The two functions do different things, as I understand it.
residuals(model) gives the response minus the fitted values. (from help for lm)
rstandard(model) is part of leave one out influence diagnostics (from help for rstandard), which says:
This suite of functions can be used to compute some of the regression
(leav... | does rstandard standardize in z? | The two functions do different things, as I understand it.
residuals(model) gives the response minus the fitted values. (from help for lm)
rstandard(model) is part of leave one out influence diagnosti | does rstandard standardize in z?
The two functions do different things, as I understand it.
residuals(model) gives the response minus the fitted values. (from help for lm)
rstandard(model) is part of leave one out influence diagnostics (from help for rstandard), which says:
This suite of functions can be used to compu... | does rstandard standardize in z?
The two functions do different things, as I understand it.
residuals(model) gives the response minus the fitted values. (from help for lm)
rstandard(model) is part of leave one out influence diagnosti |
44,906 | What is the current 'standard' for modern statistical computing hardware? | It's a little tough to provide specific recommendations, particularly without knowing too much about your budget and goals. However:
A lot of data analysis can now be done on...nearly anything. If you plan on doing a lot of $t$-tests, ANOVAs, or regression modeling, I think you would be hard-pressed to find a system t... | What is the current 'standard' for modern statistical computing hardware? | It's a little tough to provide specific recommendations, particularly without knowing too much about your budget and goals. However:
A lot of data analysis can now be done on...nearly anything. If yo | What is the current 'standard' for modern statistical computing hardware?
It's a little tough to provide specific recommendations, particularly without knowing too much about your budget and goals. However:
A lot of data analysis can now be done on...nearly anything. If you plan on doing a lot of $t$-tests, ANOVAs, or... | What is the current 'standard' for modern statistical computing hardware?
It's a little tough to provide specific recommendations, particularly without knowing too much about your budget and goals. However:
A lot of data analysis can now be done on...nearly anything. If yo |
44,907 | Multiple t-tests vs. one-way ANOVA | If your goal is to see which methods are better than the baseline, then method 1 is correct. If your goal is to see which methods are better than each other, then method 2 is correct. Method 2 with Dunnett's test could also be used, this makes corrections for the multiple comparisons involved. However, the question of ... | Multiple t-tests vs. one-way ANOVA | If your goal is to see which methods are better than the baseline, then method 1 is correct. If your goal is to see which methods are better than each other, then method 2 is correct. Method 2 with Du | Multiple t-tests vs. one-way ANOVA
If your goal is to see which methods are better than the baseline, then method 1 is correct. If your goal is to see which methods are better than each other, then method 2 is correct. Method 2 with Dunnett's test could also be used, this makes corrections for the multiple comparisons ... | Multiple t-tests vs. one-way ANOVA
If your goal is to see which methods are better than the baseline, then method 1 is correct. If your goal is to see which methods are better than each other, then method 2 is correct. Method 2 with Du |
44,908 | Multiple t-tests vs. one-way ANOVA | There are two problems with your first approach: Multiple testing and the interpretation of the difference between significant and non-significant.
First, when you perform several tests, you increase the overall error rate. If you set $\alpha = .05$, you will still reject the null hypothesis 5% of the time when it is a... | Multiple t-tests vs. one-way ANOVA | There are two problems with your first approach: Multiple testing and the interpretation of the difference between significant and non-significant.
First, when you perform several tests, you increase | Multiple t-tests vs. one-way ANOVA
There are two problems with your first approach: Multiple testing and the interpretation of the difference between significant and non-significant.
First, when you perform several tests, you increase the overall error rate. If you set $\alpha = .05$, you will still reject the null hyp... | Multiple t-tests vs. one-way ANOVA
There are two problems with your first approach: Multiple testing and the interpretation of the difference between significant and non-significant.
First, when you perform several tests, you increase |
44,909 | Another p-value fallacy | A test procedure goes like this:
(1) Define the sample space: 1024 outcomes of tossing a coin 10 times
(2) State the null hypothesis: A fair coin; i.e. $\mathsf{H}$ & $\mathsf{T}$ equiprobable, tosses independent
(3) Define a test statistic: You can use the sum of heads, or the number of runs, or whatever you like
(4)... | Another p-value fallacy | A test procedure goes like this:
(1) Define the sample space: 1024 outcomes of tossing a coin 10 times
(2) State the null hypothesis: A fair coin; i.e. $\mathsf{H}$ & $\mathsf{T}$ equiprobable, tosse | Another p-value fallacy
A test procedure goes like this:
(1) Define the sample space: 1024 outcomes of tossing a coin 10 times
(2) State the null hypothesis: A fair coin; i.e. $\mathsf{H}$ & $\mathsf{T}$ equiprobable, tosses independent
(3) Define a test statistic: You can use the sum of heads, or the number of runs, ... | Another p-value fallacy
A test procedure goes like this:
(1) Define the sample space: 1024 outcomes of tossing a coin 10 times
(2) State the null hypothesis: A fair coin; i.e. $\mathsf{H}$ & $\mathsf{T}$ equiprobable, tosse |
44,910 | Another p-value fallacy | I don't think this is anything to do with P-values. In any case you nowhere specify what test you have in mind.
The usual definition of a fair coin I take to be that heads and tails are equally probable, but nothing rules out "fairness" being a vague concept that can be made precise in several ways. In practice one sh... | Another p-value fallacy | I don't think this is anything to do with P-values. In any case you nowhere specify what test you have in mind.
The usual definition of a fair coin I take to be that heads and tails are equally proba | Another p-value fallacy
I don't think this is anything to do with P-values. In any case you nowhere specify what test you have in mind.
The usual definition of a fair coin I take to be that heads and tails are equally probable, but nothing rules out "fairness" being a vague concept that can be made precise in several ... | Another p-value fallacy
I don't think this is anything to do with P-values. In any case you nowhere specify what test you have in mind.
The usual definition of a fair coin I take to be that heads and tails are equally proba |
44,911 | Another p-value fallacy | What's the implicit model?
The probability p = 1/1024 is derived from the fair coin model of $\Pr(H)=\Pr(T)=0.5$ with independent throws, i.e. $Cov(n_i,n_{i-1})=1$. Note that this model is invariant to the sequence of throws, i.e. $\Pr(HHT)=\Pr(HTH)=\Pr(THH)$.
Under this model, a sequence such as HTHTHTHT is suspect, b... | Another p-value fallacy | What's the implicit model?
The probability p = 1/1024 is derived from the fair coin model of $\Pr(H)=\Pr(T)=0.5$ with independent throws, i.e. $Cov(n_i,n_{i-1})=1$. Note that this model is invariant t | Another p-value fallacy
What's the implicit model?
The probability p = 1/1024 is derived from the fair coin model of $\Pr(H)=\Pr(T)=0.5$ with independent throws, i.e. $Cov(n_i,n_{i-1})=1$. Note that this model is invariant to the sequence of throws, i.e. $\Pr(HHT)=\Pr(HTH)=\Pr(THH)$.
Under this model, a sequence such a... | Another p-value fallacy
What's the implicit model?
The probability p = 1/1024 is derived from the fair coin model of $\Pr(H)=\Pr(T)=0.5$ with independent throws, i.e. $Cov(n_i,n_{i-1})=1$. Note that this model is invariant t |
44,912 | Another p-value fallacy | On the chance of misunderstanding you, I will guess about what you might mean.
I guess your hypothesis is that the coin is fair, right?
And your problem is how to test this if, in fact, all sequences by themselves have the same probability to occur. However when we are talking about p-values, this usually doesn't matt... | Another p-value fallacy | On the chance of misunderstanding you, I will guess about what you might mean.
I guess your hypothesis is that the coin is fair, right?
And your problem is how to test this if, in fact, all sequences | Another p-value fallacy
On the chance of misunderstanding you, I will guess about what you might mean.
I guess your hypothesis is that the coin is fair, right?
And your problem is how to test this if, in fact, all sequences by themselves have the same probability to occur. However when we are talking about p-values, t... | Another p-value fallacy
On the chance of misunderstanding you, I will guess about what you might mean.
I guess your hypothesis is that the coin is fair, right?
And your problem is how to test this if, in fact, all sequences |
44,913 | Estimating the covariance of the means from two samples? | \begin{eqnarray}
\text{cov}(\bar X_n, \bar Y_n)
&=& \text{cov}(1/n \sum X_i, 1/n \sum Y_j)\\
&=& 1/n^2 \cdot \text{cov}( \sum X_i, \sum Y_j)\\
&=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j)
\end{eqnarray}
To go further, we need to specify something about the covariances. If the samples are iid random samples whe... | Estimating the covariance of the means from two samples? | \begin{eqnarray}
\text{cov}(\bar X_n, \bar Y_n)
&=& \text{cov}(1/n \sum X_i, 1/n \sum Y_j)\\
&=& 1/n^2 \cdot \text{cov}( \sum X_i, \sum Y_j)\\
&=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j)
\en | Estimating the covariance of the means from two samples?
\begin{eqnarray}
\text{cov}(\bar X_n, \bar Y_n)
&=& \text{cov}(1/n \sum X_i, 1/n \sum Y_j)\\
&=& 1/n^2 \cdot \text{cov}( \sum X_i, \sum Y_j)\\
&=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j)
\end{eqnarray}
To go further, we need to specify something about t... | Estimating the covariance of the means from two samples?
\begin{eqnarray}
\text{cov}(\bar X_n, \bar Y_n)
&=& \text{cov}(1/n \sum X_i, 1/n \sum Y_j)\\
&=& 1/n^2 \cdot \text{cov}( \sum X_i, \sum Y_j)\\
&=& 1/n^2 \cdot \sum_i \sum_j \text{cov}( X_i, Y_j)
\en |
44,914 | Estimating the covariance of the means from two samples? | Here is an answer derived using the theory of 'moments of moments', using power sum notation, and leaving the grunt work to mathStatica. In particular, in power sum notation, let:
$$s_{a,b}=\sum _{i=1}^n X_i^a Y_i^b$$
Then, $\operatorname{cov}(\bar X_n, \bar Y_n)$ = $\operatorname{cov}(\frac{s_{1,0}}{n}$, $\frac{s_{0,1... | Estimating the covariance of the means from two samples? | Here is an answer derived using the theory of 'moments of moments', using power sum notation, and leaving the grunt work to mathStatica. In particular, in power sum notation, let:
$$s_{a,b}=\sum _{i=1 | Estimating the covariance of the means from two samples?
Here is an answer derived using the theory of 'moments of moments', using power sum notation, and leaving the grunt work to mathStatica. In particular, in power sum notation, let:
$$s_{a,b}=\sum _{i=1}^n X_i^a Y_i^b$$
Then, $\operatorname{cov}(\bar X_n, \bar Y_n)... | Estimating the covariance of the means from two samples?
Here is an answer derived using the theory of 'moments of moments', using power sum notation, and leaving the grunt work to mathStatica. In particular, in power sum notation, let:
$$s_{a,b}=\sum _{i=1 |
44,915 | When is it appropriate to use a paired Wilcoxon test? | (Sorry, but I will have to use the word "paired")
There are two aspects to this question:
1) When to use paired tests
2) When to use Wilcoxon
Paired tests are appropriate when the data are not independent and when the dependency results in a 1 to 1 match. For example suppose I want to study heights of men and women. I... | When is it appropriate to use a paired Wilcoxon test? | (Sorry, but I will have to use the word "paired")
There are two aspects to this question:
1) When to use paired tests
2) When to use Wilcoxon
Paired tests are appropriate when the data are not indepe | When is it appropriate to use a paired Wilcoxon test?
(Sorry, but I will have to use the word "paired")
There are two aspects to this question:
1) When to use paired tests
2) When to use Wilcoxon
Paired tests are appropriate when the data are not independent and when the dependency results in a 1 to 1 match. For examp... | When is it appropriate to use a paired Wilcoxon test?
(Sorry, but I will have to use the word "paired")
There are two aspects to this question:
1) When to use paired tests
2) When to use Wilcoxon
Paired tests are appropriate when the data are not indepe |
44,916 | Machine learning book with code examples | Machine Learning, Stephen Marsland.
One of the best practical, Python based, texts I've come across. | Machine learning book with code examples | Machine Learning, Stephen Marsland.
One of the best practical, Python based, texts I've come across. | Machine learning book with code examples
Machine Learning, Stephen Marsland.
One of the best practical, Python based, texts I've come across. | Machine learning book with code examples
Machine Learning, Stephen Marsland.
One of the best practical, Python based, texts I've come across. |
44,917 | Machine learning book with code examples | Bayesian Reasoning and Machine Learning by Barber. It is freely available. In addition, there is Matlab toolbox of the book. | Machine learning book with code examples | Bayesian Reasoning and Machine Learning by Barber. It is freely available. In addition, there is Matlab toolbox of the book. | Machine learning book with code examples
Bayesian Reasoning and Machine Learning by Barber. It is freely available. In addition, there is Matlab toolbox of the book. | Machine learning book with code examples
Bayesian Reasoning and Machine Learning by Barber. It is freely available. In addition, there is Matlab toolbox of the book. |
44,918 | Machine learning book with code examples | The recently published
Machine learning: a probabilistic perspective, By Kevin Murphy
comes with an excellent and very extensive Matlab toolkit for Machine learning, which includes code for examples discussed in the book. | Machine learning book with code examples | The recently published
Machine learning: a probabilistic perspective, By Kevin Murphy
comes with an excellent and very extensive Matlab toolkit for Machine learning, which includes code for example | Machine learning book with code examples
The recently published
Machine learning: a probabilistic perspective, By Kevin Murphy
comes with an excellent and very extensive Matlab toolkit for Machine learning, which includes code for examples discussed in the book. | Machine learning book with code examples
The recently published
Machine learning: a probabilistic perspective, By Kevin Murphy
comes with an excellent and very extensive Matlab toolkit for Machine learning, which includes code for example |
44,919 | Machine learning book with code examples | WEKA is fully implemented in Java.
http://www.cs.waikato.ac.nz/ml/weka/
It has its own manual and book. You can find the manual in the WEKA directory if you install it or straight away in the zip file if you don't download the self-extracting. | Machine learning book with code examples | WEKA is fully implemented in Java.
http://www.cs.waikato.ac.nz/ml/weka/
It has its own manual and book. You can find the manual in the WEKA directory if you install it or straight away in the zip fil | Machine learning book with code examples
WEKA is fully implemented in Java.
http://www.cs.waikato.ac.nz/ml/weka/
It has its own manual and book. You can find the manual in the WEKA directory if you install it or straight away in the zip file if you don't download the self-extracting. | Machine learning book with code examples
WEKA is fully implemented in Java.
http://www.cs.waikato.ac.nz/ml/weka/
It has its own manual and book. You can find the manual in the WEKA directory if you install it or straight away in the zip fil |
44,920 | Problems with mixed model simulation | As ?lmeControl says, the default optimizer function is nlminb rather than optim. ?nlminb, however, says that optim is preferred and, indeed, the following works.
lme(y~1+x1, data=d1, random=~1+x1|j, control = lmeControl(opt = "optim"))
Linear mixed-effects model fit by REML
Data: d1
Log-restricted-likelihood: 3208... | Problems with mixed model simulation | As ?lmeControl says, the default optimizer function is nlminb rather than optim. ?nlminb, however, says that optim is preferred and, indeed, the following works.
lme(y~1+x1, data=d1, random=~1+x1|j, c | Problems with mixed model simulation
As ?lmeControl says, the default optimizer function is nlminb rather than optim. ?nlminb, however, says that optim is preferred and, indeed, the following works.
lme(y~1+x1, data=d1, random=~1+x1|j, control = lmeControl(opt = "optim"))
Linear mixed-effects model fit by REML
Data: ... | Problems with mixed model simulation
As ?lmeControl says, the default optimizer function is nlminb rather than optim. ?nlminb, however, says that optim is preferred and, indeed, the following works.
lme(y~1+x1, data=d1, random=~1+x1|j, c |
44,921 | Problems with mixed model simulation | d1 <- data.frame(j, y=g00+u0j+(g10+u1j)*x1+e, x1) # You need to add error terms
My intuition is that when we simulatated a mixed model data set without error terms, sometimes it may be difficult to converge. This is likely to be a zero residual problem. | Problems with mixed model simulation | d1 <- data.frame(j, y=g00+u0j+(g10+u1j)*x1+e, x1) # You need to add error terms
My intuition is that when we simulatated a mixed model data set without error terms, sometimes it may be difficult to co | Problems with mixed model simulation
d1 <- data.frame(j, y=g00+u0j+(g10+u1j)*x1+e, x1) # You need to add error terms
My intuition is that when we simulatated a mixed model data set without error terms, sometimes it may be difficult to converge. This is likely to be a zero residual problem. | Problems with mixed model simulation
d1 <- data.frame(j, y=g00+u0j+(g10+u1j)*x1+e, x1) # You need to add error terms
My intuition is that when we simulatated a mixed model data set without error terms, sometimes it may be difficult to co |
44,922 | Problems with mixed model simulation | I might be missing a simpler answer (i.e., mis-specification in your model), but /if/ I was sure that the mixed model should converge, I would move on to using openbugs, which will let you drag out a much longer mcmc sample. | Problems with mixed model simulation | I might be missing a simpler answer (i.e., mis-specification in your model), but /if/ I was sure that the mixed model should converge, I would move on to using openbugs, which will let you drag out a | Problems with mixed model simulation
I might be missing a simpler answer (i.e., mis-specification in your model), but /if/ I was sure that the mixed model should converge, I would move on to using openbugs, which will let you drag out a much longer mcmc sample. | Problems with mixed model simulation
I might be missing a simpler answer (i.e., mis-specification in your model), but /if/ I was sure that the mixed model should converge, I would move on to using openbugs, which will let you drag out a |
44,923 | Problems with mixed model simulation | This works for me. Instead of using nlme or lmer, I use mgcv::gamm and specify a gaussian link as follows:
GAMM(y~1+x1, data=d1, random=~1+x1|j,family = gaussian)
The estimation results are close to nlme/lmer mixed models and is applicable for most of the cases. In fact, the HLM/mixed model is a specific case of GAMM. | Problems with mixed model simulation | This works for me. Instead of using nlme or lmer, I use mgcv::gamm and specify a gaussian link as follows:
GAMM(y~1+x1, data=d1, random=~1+x1|j,family = gaussian)
The estimation results are close to | Problems with mixed model simulation
This works for me. Instead of using nlme or lmer, I use mgcv::gamm and specify a gaussian link as follows:
GAMM(y~1+x1, data=d1, random=~1+x1|j,family = gaussian)
The estimation results are close to nlme/lmer mixed models and is applicable for most of the cases. In fact, the HLM/mi... | Problems with mixed model simulation
This works for me. Instead of using nlme or lmer, I use mgcv::gamm and specify a gaussian link as follows:
GAMM(y~1+x1, data=d1, random=~1+x1|j,family = gaussian)
The estimation results are close to |
44,924 | Could logistic regression be used to detect large errors in least squares regression? | I think this would only work if the logistic regression model had access to relevant covariates that are missing from the OLS model. In such a situation (i.e., model misspecification), there could be regions in which the observed response values diverge strongly from the predicted values and the logistic regression mo... | Could logistic regression be used to detect large errors in least squares regression? | I think this would only work if the logistic regression model had access to relevant covariates that are missing from the OLS model. In such a situation (i.e., model misspecification), there could be | Could logistic regression be used to detect large errors in least squares regression?
I think this would only work if the logistic regression model had access to relevant covariates that are missing from the OLS model. In such a situation (i.e., model misspecification), there could be regions in which the observed res... | Could logistic regression be used to detect large errors in least squares regression?
I think this would only work if the logistic regression model had access to relevant covariates that are missing from the OLS model. In such a situation (i.e., model misspecification), there could be |
44,925 | Could logistic regression be used to detect large errors in least squares regression? | You are trying to use logistic regression to find a possible structure to your residuals. Residuals should be unstructured. If logistic regression picks up something the model is misspecified. Great.
Important: logistic regression looks for a very specific structure. Your method works only if residuals are broken in a ... | Could logistic regression be used to detect large errors in least squares regression? | You are trying to use logistic regression to find a possible structure to your residuals. Residuals should be unstructured. If logistic regression picks up something the model is misspecified. Great.
| Could logistic regression be used to detect large errors in least squares regression?
You are trying to use logistic regression to find a possible structure to your residuals. Residuals should be unstructured. If logistic regression picks up something the model is misspecified. Great.
Important: logistic regression loo... | Could logistic regression be used to detect large errors in least squares regression?
You are trying to use logistic regression to find a possible structure to your residuals. Residuals should be unstructured. If logistic regression picks up something the model is misspecified. Great.
|
44,926 | Could logistic regression be used to detect large errors in least squares regression? | It would be possible in the case where the original model was heteroskedastic and the heteroskedasticity was related to the covariates. For example,
$y_i \sim \text{N}(x_i^T\beta, \sigma^2x_{i,1}^2)$
where the variance of the $i^{th}$ observation is proportional to the square of the first covariate.
One can imagine, ... | Could logistic regression be used to detect large errors in least squares regression? | It would be possible in the case where the original model was heteroskedastic and the heteroskedasticity was related to the covariates. For example,
$y_i \sim \text{N}(x_i^T\beta, \sigma^2x_{i,1}^2) | Could logistic regression be used to detect large errors in least squares regression?
It would be possible in the case where the original model was heteroskedastic and the heteroskedasticity was related to the covariates. For example,
$y_i \sim \text{N}(x_i^T\beta, \sigma^2x_{i,1}^2)$
where the variance of the $i^{th... | Could logistic regression be used to detect large errors in least squares regression?
It would be possible in the case where the original model was heteroskedastic and the heteroskedasticity was related to the covariates. For example,
$y_i \sim \text{N}(x_i^T\beta, \sigma^2x_{i,1}^2) |
44,927 | Could logistic regression be used to detect large errors in least squares regression? | There are circumstances where such a thing might be made to work - assuming you can supply suitable predictors to your logistic regression, but
(i) dichotomizing the spread may be less useful than leaving it continuous
(ii) I think formal hypothesis testing is a bad idea for assessing model assumptions, since it doesn'... | Could logistic regression be used to detect large errors in least squares regression? | There are circumstances where such a thing might be made to work - assuming you can supply suitable predictors to your logistic regression, but
(i) dichotomizing the spread may be less useful than lea | Could logistic regression be used to detect large errors in least squares regression?
There are circumstances where such a thing might be made to work - assuming you can supply suitable predictors to your logistic regression, but
(i) dichotomizing the spread may be less useful than leaving it continuous
(ii) I think fo... | Could logistic regression be used to detect large errors in least squares regression?
There are circumstances where such a thing might be made to work - assuming you can supply suitable predictors to your logistic regression, but
(i) dichotomizing the spread may be less useful than lea |
44,928 | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | I am not an expert, but I think it is common sense to not merge two studies if they have strongly opposite outcomes. For the sake of the example, say they measure a variable $X$ (glucose in the blood etc.) in identical conditions.
Since the outcome is different, you can imagine that there is "something" unknown, so tha... | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | I am not an expert, but I think it is common sense to not merge two studies if they have strongly opposite outcomes. For the sake of the example, say they measure a variable $X$ (glucose in the blood | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
I am not an expert, but I think it is common sense to not merge two studies if they have strongly opposite outcomes. For the sake of the example, say they measure a variable $X$ (glucose in the blood etc.) in identical con... | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
I am not an expert, but I think it is common sense to not merge two studies if they have strongly opposite outcomes. For the sake of the example, say they measure a variable $X$ (glucose in the blood |
44,929 | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | try to find the confounding variable.. see if you get any interactions or mabe run another test with different methodology (and more variables that may give you a clearer picture. | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | try to find the confounding variable.. see if you get any interactions or mabe run another test with different methodology (and more variables that may give you a clearer picture. | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
try to find the confounding variable.. see if you get any interactions or mabe run another test with different methodology (and more variables that may give you a clearer picture. | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
try to find the confounding variable.. see if you get any interactions or mabe run another test with different methodology (and more variables that may give you a clearer picture. |
44,930 | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | Generally, there can not be bimodal findings. The theoretical underpinnings must be considered. Moreover, two sets of outcomes can not be combined in this way. If you are keen to apply meta-analysis, the outcomes showing excellent results and outcomes showing bad results may be meta-anlysed separately. Be sure that num... | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model? | Generally, there can not be bimodal findings. The theoretical underpinnings must be considered. Moreover, two sets of outcomes can not be combined in this way. If you are keen to apply meta-analysis, | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
Generally, there can not be bimodal findings. The theoretical underpinnings must be considered. Moreover, two sets of outcomes can not be combined in this way. If you are keen to apply meta-analysis, the outcomes showing e... | Two studies, opposite results: Am I allowed to compute an overall mean using random-effect model?
Generally, there can not be bimodal findings. The theoretical underpinnings must be considered. Moreover, two sets of outcomes can not be combined in this way. If you are keen to apply meta-analysis, |
44,931 | How to judge if a datapoint deviates substantially from the norm | Outlier detection in time series encompasses a large body of literature. First you would want to have a time series model that fit well to the data when there were no suspect observations. If for example an ARMA model works you might assume that the noise distribution is Gaussian. There are at least two types of out... | How to judge if a datapoint deviates substantially from the norm | Outlier detection in time series encompasses a large body of literature. First you would want to have a time series model that fit well to the data when there were no suspect observations. If for ex | How to judge if a datapoint deviates substantially from the norm
Outlier detection in time series encompasses a large body of literature. First you would want to have a time series model that fit well to the data when there were no suspect observations. If for example an ARMA model works you might assume that the noi... | How to judge if a datapoint deviates substantially from the norm
Outlier detection in time series encompasses a large body of literature. First you would want to have a time series model that fit well to the data when there were no suspect observations. If for ex |
44,932 | How to judge if a datapoint deviates substantially from the norm | I'm going to say something that's a bit down at the simpler end, in case you end up drowning in ARMA models (although that of course it the correct way to go).
A graph with a fitted line and some measure of variability on could improve your eyeballing. I drew this graph yesterday:
Using this code with ggplot2 in R:
gg... | How to judge if a datapoint deviates substantially from the norm | I'm going to say something that's a bit down at the simpler end, in case you end up drowning in ARMA models (although that of course it the correct way to go).
A graph with a fitted line and some meas | How to judge if a datapoint deviates substantially from the norm
I'm going to say something that's a bit down at the simpler end, in case you end up drowning in ARMA models (although that of course it the correct way to go).
A graph with a fitted line and some measure of variability on could improve your eyeballing. I ... | How to judge if a datapoint deviates substantially from the norm
I'm going to say something that's a bit down at the simpler end, in case you end up drowning in ARMA models (although that of course it the correct way to go).
A graph with a fitted line and some meas |
44,933 | Calculating % unsampled in sampling with replacement | Let $Z_{ij}$ be the binary indicator that subject $i = 1, ..., N$ was selected as the $j=1,...,k$'th sampled unit. Since the sampling (assumed to be simple random sampling) is with replacement, each of the $Z_{ij}$ are independent bernoulli trials with success probability $1/N$. Therefore the number of times subject $i... | Calculating % unsampled in sampling with replacement | Let $Z_{ij}$ be the binary indicator that subject $i = 1, ..., N$ was selected as the $j=1,...,k$'th sampled unit. Since the sampling (assumed to be simple random sampling) is with replacement, each o | Calculating % unsampled in sampling with replacement
Let $Z_{ij}$ be the binary indicator that subject $i = 1, ..., N$ was selected as the $j=1,...,k$'th sampled unit. Since the sampling (assumed to be simple random sampling) is with replacement, each of the $Z_{ij}$ are independent bernoulli trials with success probab... | Calculating % unsampled in sampling with replacement
Let $Z_{ij}$ be the binary indicator that subject $i = 1, ..., N$ was selected as the $j=1,...,k$'th sampled unit. Since the sampling (assumed to be simple random sampling) is with replacement, each o |
44,934 | Calculating % unsampled in sampling with replacement | If $N$ is large, the distribution of per item sampling frequencies is approximately Poisson distributed, with mean $k/N$. So you can estimate the unsampled proportion as $e^{-k/N}$. An exact solution for small $N$ is a bit of a chore to derive. | Calculating % unsampled in sampling with replacement | If $N$ is large, the distribution of per item sampling frequencies is approximately Poisson distributed, with mean $k/N$. So you can estimate the unsampled proportion as $e^{-k/N}$. An exact solutio | Calculating % unsampled in sampling with replacement
If $N$ is large, the distribution of per item sampling frequencies is approximately Poisson distributed, with mean $k/N$. So you can estimate the unsampled proportion as $e^{-k/N}$. An exact solution for small $N$ is a bit of a chore to derive. | Calculating % unsampled in sampling with replacement
If $N$ is large, the distribution of per item sampling frequencies is approximately Poisson distributed, with mean $k/N$. So you can estimate the unsampled proportion as $e^{-k/N}$. An exact solutio |
44,935 | How to plot a learning curve based on a sequence of date stamped successes and failures? | If your dates are in string or factor form, convert them to Dates with as.Date(DateVariable, format="%Y-%m-%d") (potentially with an as.character around DateVariable if it is a factor.
Sort the data.frame with the date and outcome by surgery date. To get the y variable you describe, the easiest way is to take the cumu... | How to plot a learning curve based on a sequence of date stamped successes and failures? | If your dates are in string or factor form, convert them to Dates with as.Date(DateVariable, format="%Y-%m-%d") (potentially with an as.character around DateVariable if it is a factor.
Sort the data.f | How to plot a learning curve based on a sequence of date stamped successes and failures?
If your dates are in string or factor form, convert them to Dates with as.Date(DateVariable, format="%Y-%m-%d") (potentially with an as.character around DateVariable if it is a factor.
Sort the data.frame with the date and outcome ... | How to plot a learning curve based on a sequence of date stamped successes and failures?
If your dates are in string or factor form, convert them to Dates with as.Date(DateVariable, format="%Y-%m-%d") (potentially with an as.character around DateVariable if it is a factor.
Sort the data.f |
44,936 | How to plot a learning curve based on a sequence of date stamped successes and failures? | An informative "learning curve" will indicate the current performance of success. A cumulative average, however, doesn't do that, because it includes old results along with the new. When learning truly improves, the cumulative average will be biased low.
A good solution is to use a Lowess smooth of the response plott... | How to plot a learning curve based on a sequence of date stamped successes and failures? | An informative "learning curve" will indicate the current performance of success. A cumulative average, however, doesn't do that, because it includes old results along with the new. When learning tr | How to plot a learning curve based on a sequence of date stamped successes and failures?
An informative "learning curve" will indicate the current performance of success. A cumulative average, however, doesn't do that, because it includes old results along with the new. When learning truly improves, the cumulative av... | How to plot a learning curve based on a sequence of date stamped successes and failures?
An informative "learning curve" will indicate the current performance of success. A cumulative average, however, doesn't do that, because it includes old results along with the new. When learning tr |
44,937 | How to plot a learning curve based on a sequence of date stamped successes and failures? | Use as.POSIXct to convert to dates, which are stored as seconds from the epoch (and are thus easily plotted) but can be easily displayed in human-readable formats. The help file documents how to import from arbitrary character representations.
Use geom_line for plotting. You can use aggregate or by to bin the observati... | How to plot a learning curve based on a sequence of date stamped successes and failures? | Use as.POSIXct to convert to dates, which are stored as seconds from the epoch (and are thus easily plotted) but can be easily displayed in human-readable formats. The help file documents how to impor | How to plot a learning curve based on a sequence of date stamped successes and failures?
Use as.POSIXct to convert to dates, which are stored as seconds from the epoch (and are thus easily plotted) but can be easily displayed in human-readable formats. The help file documents how to import from arbitrary character repr... | How to plot a learning curve based on a sequence of date stamped successes and failures?
Use as.POSIXct to convert to dates, which are stored as seconds from the epoch (and are thus easily plotted) but can be easily displayed in human-readable formats. The help file documents how to impor |
44,938 | Intra- and inter-rater reliability on the same data | I assume that A through D are different symptoms, say, and 1 and 2 are the two raters. As you tagged this in Stata, I will build a Stata example. Let us first simulate some data: we have a bunch of subjects with two uncorrelated traits, and a battery of questions, tapping upon these traits. The two raters have differen... | Intra- and inter-rater reliability on the same data | I assume that A through D are different symptoms, say, and 1 and 2 are the two raters. As you tagged this in Stata, I will build a Stata example. Let us first simulate some data: we have a bunch of su | Intra- and inter-rater reliability on the same data
I assume that A through D are different symptoms, say, and 1 and 2 are the two raters. As you tagged this in Stata, I will build a Stata example. Let us first simulate some data: we have a bunch of subjects with two uncorrelated traits, and a battery of questions, tap... | Intra- and inter-rater reliability on the same data
I assume that A through D are different symptoms, say, and 1 and 2 are the two raters. As you tagged this in Stata, I will build a Stata example. Let us first simulate some data: we have a bunch of su |
44,939 | Combining 2 sets of coefficients, weighting one of the sets | You are retaining $p$ (=3 in this case) values for each regression: the estimated coefficients. If you are willing to retain $p(p+1)$ (=12) values per regression, you can weight your results in a way that is equivalent to having all the data and performing a weighted least squares regression with them en masse.
The an... | Combining 2 sets of coefficients, weighting one of the sets | You are retaining $p$ (=3 in this case) values for each regression: the estimated coefficients. If you are willing to retain $p(p+1)$ (=12) values per regression, you can weight your results in a way | Combining 2 sets of coefficients, weighting one of the sets
You are retaining $p$ (=3 in this case) values for each regression: the estimated coefficients. If you are willing to retain $p(p+1)$ (=12) values per regression, you can weight your results in a way that is equivalent to having all the data and performing a ... | Combining 2 sets of coefficients, weighting one of the sets
You are retaining $p$ (=3 in this case) values for each regression: the estimated coefficients. If you are willing to retain $p(p+1)$ (=12) values per regression, you can weight your results in a way |
44,940 | Combining 2 sets of coefficients, weighting one of the sets | This is "taylor made" almost for a Bayesian regression. First of all, there is nothing "fundamentally wrong" with what you suggest. You result may not be optimal by some mathematical standard, but it will almost certainly be optimal time wise. Most other methods will involve much more time than a straight multiplica... | Combining 2 sets of coefficients, weighting one of the sets | This is "taylor made" almost for a Bayesian regression. First of all, there is nothing "fundamentally wrong" with what you suggest. You result may not be optimal by some mathematical standard, but i | Combining 2 sets of coefficients, weighting one of the sets
This is "taylor made" almost for a Bayesian regression. First of all, there is nothing "fundamentally wrong" with what you suggest. You result may not be optimal by some mathematical standard, but it will almost certainly be optimal time wise. Most other me... | Combining 2 sets of coefficients, weighting one of the sets
This is "taylor made" almost for a Bayesian regression. First of all, there is nothing "fundamentally wrong" with what you suggest. You result may not be optimal by some mathematical standard, but i |
44,941 | Combining 2 sets of coefficients, weighting one of the sets | There is no reason accounting for the use of convex linear combinations of coefficients in order to "average" two models.
At best, you could consider the three coefficients for each dataset are realizations of the same three random variables, and you would be interested in the distribution of each random variable.
What... | Combining 2 sets of coefficients, weighting one of the sets | There is no reason accounting for the use of convex linear combinations of coefficients in order to "average" two models.
At best, you could consider the three coefficients for each dataset are realiz | Combining 2 sets of coefficients, weighting one of the sets
There is no reason accounting for the use of convex linear combinations of coefficients in order to "average" two models.
At best, you could consider the three coefficients for each dataset are realizations of the same three random variables, and you would be ... | Combining 2 sets of coefficients, weighting one of the sets
There is no reason accounting for the use of convex linear combinations of coefficients in order to "average" two models.
At best, you could consider the three coefficients for each dataset are realiz |
44,942 | Combining 2 sets of coefficients, weighting one of the sets | Maybe you should look into "stacking". Or even "feature-weighed stacking".
The former is using a cross validation method to determine the weights you should use to linearly stack them. The latter is using "meta-parameters" to give even more insight on how to weight the parameters depending on what is being predicted. T... | Combining 2 sets of coefficients, weighting one of the sets | Maybe you should look into "stacking". Or even "feature-weighed stacking".
The former is using a cross validation method to determine the weights you should use to linearly stack them. The latter is u | Combining 2 sets of coefficients, weighting one of the sets
Maybe you should look into "stacking". Or even "feature-weighed stacking".
The former is using a cross validation method to determine the weights you should use to linearly stack them. The latter is using "meta-parameters" to give even more insight on how to w... | Combining 2 sets of coefficients, weighting one of the sets
Maybe you should look into "stacking". Or even "feature-weighed stacking".
The former is using a cross validation method to determine the weights you should use to linearly stack them. The latter is u |
44,943 | Random effect slopes in linear mixed models | First, you should compare models from lmer after fitting with ML (maximum likelihood) since the default is REML. So something like:
fit.nc <- update(NoCor, REML=FALSE)
fit.wc <- udpate(WithCor, REML=FALSE)
anova(fit.nc, fit.wc)
It would help to see the output of the random effects variation from your fits. For examp... | Random effect slopes in linear mixed models | First, you should compare models from lmer after fitting with ML (maximum likelihood) since the default is REML. So something like:
fit.nc <- update(NoCor, REML=FALSE)
fit.wc <- udpate(WithCor, REML= | Random effect slopes in linear mixed models
First, you should compare models from lmer after fitting with ML (maximum likelihood) since the default is REML. So something like:
fit.nc <- update(NoCor, REML=FALSE)
fit.wc <- udpate(WithCor, REML=FALSE)
anova(fit.nc, fit.wc)
It would help to see the output of the random ... | Random effect slopes in linear mixed models
First, you should compare models from lmer after fitting with ML (maximum likelihood) since the default is REML. So something like:
fit.nc <- update(NoCor, REML=FALSE)
fit.wc <- udpate(WithCor, REML= |
44,944 | Random effect slopes in linear mixed models | In this case it would not be expected to find substantive differences between the 'NoCor' model and the 'WithCor' model that you have specified. This is because 'Size' is a factor, not a numeric covariate, and what changes between the two models is that instead of the random effects being referenced to the intercept (b... | Random effect slopes in linear mixed models | In this case it would not be expected to find substantive differences between the 'NoCor' model and the 'WithCor' model that you have specified. This is because 'Size' is a factor, not a numeric covar | Random effect slopes in linear mixed models
In this case it would not be expected to find substantive differences between the 'NoCor' model and the 'WithCor' model that you have specified. This is because 'Size' is a factor, not a numeric covariate, and what changes between the two models is that instead of the random ... | Random effect slopes in linear mixed models
In this case it would not be expected to find substantive differences between the 'NoCor' model and the 'WithCor' model that you have specified. This is because 'Size' is a factor, not a numeric covar |
44,945 | What is the normality test for binary data? | There is no such thing as normality of categorical variable. Normal distribution is a continuous distribution so in assumption don't cover categorical output. | What is the normality test for binary data? | There is no such thing as normality of categorical variable. Normal distribution is a continuous distribution so in assumption don't cover categorical output. | What is the normality test for binary data?
There is no such thing as normality of categorical variable. Normal distribution is a continuous distribution so in assumption don't cover categorical output. | What is the normality test for binary data?
There is no such thing as normality of categorical variable. Normal distribution is a continuous distribution so in assumption don't cover categorical output. |
44,946 | The "Risk" game dice problem | You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).
It's very difficult to to do a proper calculation of post hoc probability, because what other events would have been deemed at least as surprising depends on what the ... | The "Risk" game dice problem | You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).
It's very difficult to to do a proper calculati | The "Risk" game dice problem
You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).
It's very difficult to to do a proper calculation of post hoc probability, because what other events would have been deemed at least as su... | The "Risk" game dice problem
You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).
It's very difficult to to do a proper calculati |
44,947 | Bayesian Analysis in the Absence of Prior Information? | If you have no information, you can use “uninformative” priors. Those priors aim to bring as little information as possible, but as you already learned from the What is an "uninformative prior"? Can we ever have one with truly no information? thread, the name is a little bit misleading because even such priors bring s... | Bayesian Analysis in the Absence of Prior Information? | If you have no information, you can use “uninformative” priors. Those priors aim to bring as little information as possible, but as you already learned from the What is an "uninformative prior"? Can | Bayesian Analysis in the Absence of Prior Information?
If you have no information, you can use “uninformative” priors. Those priors aim to bring as little information as possible, but as you already learned from the What is an "uninformative prior"? Can we ever have one with truly no information? thread, the name is a... | Bayesian Analysis in the Absence of Prior Information?
If you have no information, you can use “uninformative” priors. Those priors aim to bring as little information as possible, but as you already learned from the What is an "uninformative prior"? Can |
44,948 | Bayesian Analysis in the Absence of Prior Information? | I think that the posterior inherits meaning from the prior, which implies that if the prior is meaningless, so is the posterior.
Now there are different varieties of Bayesians when it comes to interpreting probabilities, i.e., assigning meaning to priors (and posteriors). Most (but not all) Bayesian interpretation of p... | Bayesian Analysis in the Absence of Prior Information? | I think that the posterior inherits meaning from the prior, which implies that if the prior is meaningless, so is the posterior.
Now there are different varieties of Bayesians when it comes to interpr | Bayesian Analysis in the Absence of Prior Information?
I think that the posterior inherits meaning from the prior, which implies that if the prior is meaningless, so is the posterior.
Now there are different varieties of Bayesians when it comes to interpreting probabilities, i.e., assigning meaning to priors (and poste... | Bayesian Analysis in the Absence of Prior Information?
I think that the posterior inherits meaning from the prior, which implies that if the prior is meaningless, so is the posterior.
Now there are different varieties of Bayesians when it comes to interpr |
44,949 | Bayesian Analysis in the Absence of Prior Information? | When we don't have a lot of confidence on information that can be used
to construct Bayesian Priors - is it generally better to stick to
Frequentist Approaches instead of "guessing" which Priors will better
fit the data and the choice of model?
In my opinion, the full power of a Bayesian analysis is unleashed when a p... | Bayesian Analysis in the Absence of Prior Information? | When we don't have a lot of confidence on information that can be used
to construct Bayesian Priors - is it generally better to stick to
Frequentist Approaches instead of "guessing" which Priors will | Bayesian Analysis in the Absence of Prior Information?
When we don't have a lot of confidence on information that can be used
to construct Bayesian Priors - is it generally better to stick to
Frequentist Approaches instead of "guessing" which Priors will better
fit the data and the choice of model?
In my opinion, the ... | Bayesian Analysis in the Absence of Prior Information?
When we don't have a lot of confidence on information that can be used
to construct Bayesian Priors - is it generally better to stick to
Frequentist Approaches instead of "guessing" which Priors will |
44,950 | Is there an equivalent to an ECDF with a "<" sign? | Remember how the CDF is a right-continuous function? That comes from using the $\le$.
If you use $<$, then you change the function to be left-continuous.
Regarding your exact question about if such a function exists, the answer is that it does, and you wrote out exactly what it is. That function isn’t the classical CDF... | Is there an equivalent to an ECDF with a "<" sign? | Remember how the CDF is a right-continuous function? That comes from using the $\le$.
If you use $<$, then you change the function to be left-continuous.
Regarding your exact question about if such a | Is there an equivalent to an ECDF with a "<" sign?
Remember how the CDF is a right-continuous function? That comes from using the $\le$.
If you use $<$, then you change the function to be left-continuous.
Regarding your exact question about if such a function exists, the answer is that it does, and you wrote out exactl... | Is there an equivalent to an ECDF with a "<" sign?
Remember how the CDF is a right-continuous function? That comes from using the $\le$.
If you use $<$, then you change the function to be left-continuous.
Regarding your exact question about if such a |
44,951 | Do Statistical Binning Algorithms Exist? | The most rational and elegant solution, and best performing in terms of mean squared error of estimates, is to use a method that borrows information across groups: either penalized maximum likelihood estimation, mixed effects model, or a Bayesian model that connects the groups through the use of random effects. But a... | Do Statistical Binning Algorithms Exist? | The most rational and elegant solution, and best performing in terms of mean squared error of estimates, is to use a method that borrows information across groups: either penalized maximum likelihood | Do Statistical Binning Algorithms Exist?
The most rational and elegant solution, and best performing in terms of mean squared error of estimates, is to use a method that borrows information across groups: either penalized maximum likelihood estimation, mixed effects model, or a Bayesian model that connects the groups ... | Do Statistical Binning Algorithms Exist?
The most rational and elegant solution, and best performing in terms of mean squared error of estimates, is to use a method that borrows information across groups: either penalized maximum likelihood |
44,952 | What is the derivation for "Partial Expectation"? | One way is to think about the conditional density. The density of $X|X>k$ is zero when $X\leq k$, so it's proportional to $f_X(x)I(X>k)$. The constant of proportionality is given by the fact that a density has to integrate to 1, so the conditional density is
$$g_X(x)=\frac{f_X(x)I(X>k)}{P(X>k)}$$
So,
$$E[X|X>k]=\int_... | What is the derivation for "Partial Expectation"? | One way is to think about the conditional density. The density of $X|X>k$ is zero when $X\leq k$, so it's proportional to $f_X(x)I(X>k)$. The constant of proportionality is given by the fact that a | What is the derivation for "Partial Expectation"?
One way is to think about the conditional density. The density of $X|X>k$ is zero when $X\leq k$, so it's proportional to $f_X(x)I(X>k)$. The constant of proportionality is given by the fact that a density has to integrate to 1, so the conditional density is
$$g_X(x)=... | What is the derivation for "Partial Expectation"?
One way is to think about the conditional density. The density of $X|X>k$ is zero when $X\leq k$, so it's proportional to $f_X(x)I(X>k)$. The constant of proportionality is given by the fact that a |
44,953 | What is the derivation for "Partial Expectation"? | The first equation can be derived using two fundamental formulas:
The formula for the expectation of a random variable $X$ given event $A$:
$$E(X\mid A)=\frac{E(XI(A))}{P(A)}\tag1$$
The formula for the expectation of a function $h$ of a random variable $X$ (often described as the law of the unconscious statistician):... | What is the derivation for "Partial Expectation"? | The first equation can be derived using two fundamental formulas:
The formula for the expectation of a random variable $X$ given event $A$:
$$E(X\mid A)=\frac{E(XI(A))}{P(A)}\tag1$$
The formula for | What is the derivation for "Partial Expectation"?
The first equation can be derived using two fundamental formulas:
The formula for the expectation of a random variable $X$ given event $A$:
$$E(X\mid A)=\frac{E(XI(A))}{P(A)}\tag1$$
The formula for the expectation of a function $h$ of a random variable $X$ (often desc... | What is the derivation for "Partial Expectation"?
The first equation can be derived using two fundamental formulas:
The formula for the expectation of a random variable $X$ given event $A$:
$$E(X\mid A)=\frac{E(XI(A))}{P(A)}\tag1$$
The formula for |
44,954 | Paired, discrete hypothesis testing | Lazy version: permutation testing.
The null hypothesis is that their interests would be unchanged. Operationally, that means that for each student, you might as well flip a coin to pick their "before" rating and "after" rating. So do that, a large number of times. If the result is more extreme than the one you have oft... | Paired, discrete hypothesis testing | Lazy version: permutation testing.
The null hypothesis is that their interests would be unchanged. Operationally, that means that for each student, you might as well flip a coin to pick their "before" | Paired, discrete hypothesis testing
Lazy version: permutation testing.
The null hypothesis is that their interests would be unchanged. Operationally, that means that for each student, you might as well flip a coin to pick their "before" rating and "after" rating. So do that, a large number of times. If the result is mo... | Paired, discrete hypothesis testing
Lazy version: permutation testing.
The null hypothesis is that their interests would be unchanged. Operationally, that means that for each student, you might as well flip a coin to pick their "before" |
44,955 | Paired, discrete hypothesis testing | The sign test or the Wilcoxon sign-rank test present alternatives to a permutation test.
The former has really low power, but a reasonably comprehensible null and alternative hypothesis:
$\text{H}_{0}\text{: }P(X_{\text{before}} > X_{\text{after}}) = 0.5$ , with
$\text{H}_{\text{A}}\text{: }P(X_{\text{before}} > X_{\te... | Paired, discrete hypothesis testing | The sign test or the Wilcoxon sign-rank test present alternatives to a permutation test.
The former has really low power, but a reasonably comprehensible null and alternative hypothesis:
$\text{H}_{0} | Paired, discrete hypothesis testing
The sign test or the Wilcoxon sign-rank test present alternatives to a permutation test.
The former has really low power, but a reasonably comprehensible null and alternative hypothesis:
$\text{H}_{0}\text{: }P(X_{\text{before}} > X_{\text{after}}) = 0.5$ , with
$\text{H}_{\text{A}}\... | Paired, discrete hypothesis testing
The sign test or the Wilcoxon sign-rank test present alternatives to a permutation test.
The former has really low power, but a reasonably comprehensible null and alternative hypothesis:
$\text{H}_{0} |
44,956 | Can any Models be "Bagged"? | You ask "Or is 'bagging' only used for very specific models and instances (e.g. Random Forest)?"
If you truly want to, you can bag any model. The reason that bagging is almost synonymous with random forest is that bagging is more effective for high-variance models like decision trees. Again, Elements of Statistical Le... | Can any Models be "Bagged"? | You ask "Or is 'bagging' only used for very specific models and instances (e.g. Random Forest)?"
If you truly want to, you can bag any model. The reason that bagging is almost synonymous with random f | Can any Models be "Bagged"?
You ask "Or is 'bagging' only used for very specific models and instances (e.g. Random Forest)?"
If you truly want to, you can bag any model. The reason that bagging is almost synonymous with random forest is that bagging is more effective for high-variance models like decision trees. Again... | Can any Models be "Bagged"?
You ask "Or is 'bagging' only used for very specific models and instances (e.g. Random Forest)?"
If you truly want to, you can bag any model. The reason that bagging is almost synonymous with random f |
44,957 | Can any Models be "Bagged"? | Bagging is usually used for model so called the "weak learner" like Decision Tree. The reason behind is those weak learner usually overfit and lack of generalization power (High Variance).
With the application of Bagging , researcher found that it can reduce the variance of model (prevent overfitting). Althought some s... | Can any Models be "Bagged"? | Bagging is usually used for model so called the "weak learner" like Decision Tree. The reason behind is those weak learner usually overfit and lack of generalization power (High Variance).
With the ap | Can any Models be "Bagged"?
Bagging is usually used for model so called the "weak learner" like Decision Tree. The reason behind is those weak learner usually overfit and lack of generalization power (High Variance).
With the application of Bagging , researcher found that it can reduce the variance of model (prevent ov... | Can any Models be "Bagged"?
Bagging is usually used for model so called the "weak learner" like Decision Tree. The reason behind is those weak learner usually overfit and lack of generalization power (High Variance).
With the ap |
44,958 | Non-linearity of neural network classification | A neural network with a single sigmoid neuron is also a linear classifier when the output is thresholded for classification. But, more than one layer produces a non-linear one because the decision rule can't be written in the following form:
$$\sum w_i x_i+b>\tau$$
where $\tau$ is threshold, $w_i$ are learnable weights... | Non-linearity of neural network classification | A neural network with a single sigmoid neuron is also a linear classifier when the output is thresholded for classification. But, more than one layer produces a non-linear one because the decision rul | Non-linearity of neural network classification
A neural network with a single sigmoid neuron is also a linear classifier when the output is thresholded for classification. But, more than one layer produces a non-linear one because the decision rule can't be written in the following form:
$$\sum w_i x_i+b>\tau$$
where $... | Non-linearity of neural network classification
A neural network with a single sigmoid neuron is also a linear classifier when the output is thresholded for classification. But, more than one layer produces a non-linear one because the decision rul |
44,959 | How is time series analysis a different problem than forecasting? | Forecasting tries to answer questions like "can we predict the distribution of values of a time series variable at some point in the future?" Consider Sugihara's simplex projection (a kind of state space reconstruction method), which can make reasonable short term forecasts on a time series variable, even when that var... | How is time series analysis a different problem than forecasting? | Forecasting tries to answer questions like "can we predict the distribution of values of a time series variable at some point in the future?" Consider Sugihara's simplex projection (a kind of state sp | How is time series analysis a different problem than forecasting?
Forecasting tries to answer questions like "can we predict the distribution of values of a time series variable at some point in the future?" Consider Sugihara's simplex projection (a kind of state space reconstruction method), which can make reasonable ... | How is time series analysis a different problem than forecasting?
Forecasting tries to answer questions like "can we predict the distribution of values of a time series variable at some point in the future?" Consider Sugihara's simplex projection (a kind of state sp |
44,960 | Is "Permutation Test" sufficient for a/b testing? | Ultimately permutation tests are a type of statistical significance test; we could have use bootstrapping if we want another frequentist non-parametric approach (e.g. see "How do bootstrap and permutation tests work?" (2003) by Janssen & Pauls for a relevant comparison). Whether or not they should be used instead of st... | Is "Permutation Test" sufficient for a/b testing? | Ultimately permutation tests are a type of statistical significance test; we could have use bootstrapping if we want another frequentist non-parametric approach (e.g. see "How do bootstrap and permuta | Is "Permutation Test" sufficient for a/b testing?
Ultimately permutation tests are a type of statistical significance test; we could have use bootstrapping if we want another frequentist non-parametric approach (e.g. see "How do bootstrap and permutation tests work?" (2003) by Janssen & Pauls for a relevant comparison)... | Is "Permutation Test" sufficient for a/b testing?
Ultimately permutation tests are a type of statistical significance test; we could have use bootstrapping if we want another frequentist non-parametric approach (e.g. see "How do bootstrap and permuta |
44,961 | Is "Permutation Test" sufficient for a/b testing? | Consider the two fictitious normal samples below, in which
sample sizes, sample means, and sample variances all differ.
set.seed(2021)
x1 = rnorm( 70, 50, 5)
x2 = rnorm(100, 53.2, 9)
Someone who did not notice the different variances, might
do a pooled 2-sample t test, find no difference in means
at the 5% level, an... | Is "Permutation Test" sufficient for a/b testing? | Consider the two fictitious normal samples below, in which
sample sizes, sample means, and sample variances all differ.
set.seed(2021)
x1 = rnorm( 70, 50, 5)
x2 = rnorm(100, 53.2, 9)
Someone who di | Is "Permutation Test" sufficient for a/b testing?
Consider the two fictitious normal samples below, in which
sample sizes, sample means, and sample variances all differ.
set.seed(2021)
x1 = rnorm( 70, 50, 5)
x2 = rnorm(100, 53.2, 9)
Someone who did not notice the different variances, might
do a pooled 2-sample t tes... | Is "Permutation Test" sufficient for a/b testing?
Consider the two fictitious normal samples below, in which
sample sizes, sample means, and sample variances all differ.
set.seed(2021)
x1 = rnorm( 70, 50, 5)
x2 = rnorm(100, 53.2, 9)
Someone who di |
44,962 | Is "Permutation Test" sufficient for a/b testing? | It depends. That’s a very broad question.
Permutation testing is one way, of many, for hypothesis testing. In contrast to the more common tests based on sampling distribution, permutation tests tend to be more conservative to accept new evidence (lower power) and less sensitive to assumptions about the population. A do... | Is "Permutation Test" sufficient for a/b testing? | It depends. That’s a very broad question.
Permutation testing is one way, of many, for hypothesis testing. In contrast to the more common tests based on sampling distribution, permutation tests tend t | Is "Permutation Test" sufficient for a/b testing?
It depends. That’s a very broad question.
Permutation testing is one way, of many, for hypothesis testing. In contrast to the more common tests based on sampling distribution, permutation tests tend to be more conservative to accept new evidence (lower power) and less s... | Is "Permutation Test" sufficient for a/b testing?
It depends. That’s a very broad question.
Permutation testing is one way, of many, for hypothesis testing. In contrast to the more common tests based on sampling distribution, permutation tests tend t |
44,963 | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test? | Another thing to bring up in addition to the answers that already exist: AIC, BIC etc. can be really good (i.e. cheap to evaluate, use all the data, let you do things like AIC-model averaging etc.) in the specific circumstances when you can define them and they are valid. What do I mean by this limitation? E.g. for som... | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test? | Another thing to bring up in addition to the answers that already exist: AIC, BIC etc. can be really good (i.e. cheap to evaluate, use all the data, let you do things like AIC-model averaging etc.) in | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test?
Another thing to bring up in addition to the answers that already exist: AIC, BIC etc. can be really good (i.e. cheap to evaluate, use all the data, let you do things like AIC-model averaging etc.) in the specific circumstanc... | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test?
Another thing to bring up in addition to the answers that already exist: AIC, BIC etc. can be really good (i.e. cheap to evaluate, use all the data, let you do things like AIC-model averaging etc.) in |
44,964 | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test? | @RichardHardy already gave a partial answer
1) AIC does have an interpretation as twice the negative expected log-likelihood (as mentioned e.g. here). Hence, it is not only a relative measure. Moreover, it is in a sense a measure of error. 2) Cross validation used to be computationally infeasible for some complex mode... | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test? | @RichardHardy already gave a partial answer
1) AIC does have an interpretation as twice the negative expected log-likelihood (as mentioned e.g. here). Hence, it is not only a relative measure. Moreov | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test?
@RichardHardy already gave a partial answer
1) AIC does have an interpretation as twice the negative expected log-likelihood (as mentioned e.g. here). Hence, it is not only a relative measure. Moreover, it is in a sense a me... | Is the Cross Validation Error more "Informative" compared to AIC, BIC and the Likelihood Test?
@RichardHardy already gave a partial answer
1) AIC does have an interpretation as twice the negative expected log-likelihood (as mentioned e.g. here). Hence, it is not only a relative measure. Moreov |
44,965 | T-tests provide info about the margin of mean difference? | if I can use the t-test to claim that one dataset provides significantly better results than the other
I wanted to point out that the expression "statistically significant" has been discouraged by the American Statistical Association. (But personally, I don't feel too strongly against it)
use confidence intervals (or... | T-tests provide info about the margin of mean difference? | if I can use the t-test to claim that one dataset provides significantly better results than the other
I wanted to point out that the expression "statistically significant" has been discouraged by th | T-tests provide info about the margin of mean difference?
if I can use the t-test to claim that one dataset provides significantly better results than the other
I wanted to point out that the expression "statistically significant" has been discouraged by the American Statistical Association. (But personally, I don't f... | T-tests provide info about the margin of mean difference?
if I can use the t-test to claim that one dataset provides significantly better results than the other
I wanted to point out that the expression "statistically significant" has been discouraged by th |
44,966 | T-tests provide info about the margin of mean difference? | Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)
mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863
The sample means show $\bar X_1 < \bar X_2$ suggesting that $\mu_1$ may be 'significantly' smaller than $\mu_2.$ Welch two-sample t test in R shows that... | T-tests provide info about the margin of mean difference? | Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)
mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863
The sample means show $\bar X_1 | T-tests provide info about the margin of mean difference?
Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)
mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863
The sample means show $\bar X_1 < \bar X_2$ suggesting that $\mu_1$ may be 'significantly' sma... | T-tests provide info about the margin of mean difference?
Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)
mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863
The sample means show $\bar X_1 |
44,967 | T-tests provide info about the margin of mean difference? | If the null hypothesis is $t=0$, then this is testing if the two means are the same or not. But you can also test $t > a$, then this tests if the margin is larger than some number. | T-tests provide info about the margin of mean difference? | If the null hypothesis is $t=0$, then this is testing if the two means are the same or not. But you can also test $t > a$, then this tests if the margin is larger than some number. | T-tests provide info about the margin of mean difference?
If the null hypothesis is $t=0$, then this is testing if the two means are the same or not. But you can also test $t > a$, then this tests if the margin is larger than some number. | T-tests provide info about the margin of mean difference?
If the null hypothesis is $t=0$, then this is testing if the two means are the same or not. But you can also test $t > a$, then this tests if the margin is larger than some number. |
44,968 | Paired t-test with multiple observations per pair | Averaging the data will result in a loss information and statistical power, so it is best avoided.
Since you have repeated measures for websites, you can account the differences between websites (or equivalently, the non-independence of observations within each website, since observations on one website are more likely... | Paired t-test with multiple observations per pair | Averaging the data will result in a loss information and statistical power, so it is best avoided.
Since you have repeated measures for websites, you can account the differences between websites (or e | Paired t-test with multiple observations per pair
Averaging the data will result in a loss information and statistical power, so it is best avoided.
Since you have repeated measures for websites, you can account the differences between websites (or equivalently, the non-independence of observations within each website,... | Paired t-test with multiple observations per pair
Averaging the data will result in a loss information and statistical power, so it is best avoided.
Since you have repeated measures for websites, you can account the differences between websites (or e |
44,969 | Paired t-test with multiple observations per pair | If each group has the same number of members, then the mean of the group means is the same as the mean over all the observations. The sd can be more complicated, depending on what you're taking as your null hypothesis. One of the simplest null hypotheses is that each observation is equal to $\mu(W)+\epsilon$, where $\m... | Paired t-test with multiple observations per pair | If each group has the same number of members, then the mean of the group means is the same as the mean over all the observations. The sd can be more complicated, depending on what you're taking as you | Paired t-test with multiple observations per pair
If each group has the same number of members, then the mean of the group means is the same as the mean over all the observations. The sd can be more complicated, depending on what you're taking as your null hypothesis. One of the simplest null hypotheses is that each ob... | Paired t-test with multiple observations per pair
If each group has the same number of members, then the mean of the group means is the same as the mean over all the observations. The sd can be more complicated, depending on what you're taking as you |
44,970 | Lowercase $x$ vs Uppercase $X$ in statistics | In this context, capitalisation is usually used to distinguish random variables from fixed values, so you are correct to interpret $X: \Omega \rightarrow \mathbb{R}$ as a random variable. The post specifies that it is considering a sequence of random variables $X_n : \Omega \rightarrow \mathbb{R}$, so there are an inf... | Lowercase $x$ vs Uppercase $X$ in statistics | In this context, capitalisation is usually used to distinguish random variables from fixed values, so you are correct to interpret $X: \Omega \rightarrow \mathbb{R}$ as a random variable. The post sp | Lowercase $x$ vs Uppercase $X$ in statistics
In this context, capitalisation is usually used to distinguish random variables from fixed values, so you are correct to interpret $X: \Omega \rightarrow \mathbb{R}$ as a random variable. The post specifies that it is considering a sequence of random variables $X_n : \Omega... | Lowercase $x$ vs Uppercase $X$ in statistics
In this context, capitalisation is usually used to distinguish random variables from fixed values, so you are correct to interpret $X: \Omega \rightarrow \mathbb{R}$ as a random variable. The post sp |
44,971 | Lowercase $x$ vs Uppercase $X$ in statistics | You are correct that $X$ is a Borel function $X : \Omega\to\mathbb R$. A realization of $X$ is a particular real number $X(\omega)$, and such a number may be referred to as $x$. This means, for example, the event $\{X = x\}$ is more formally written as $\{\omega\in\Omega : X(\omega) = x\}$.
We can define as many functi... | Lowercase $x$ vs Uppercase $X$ in statistics | You are correct that $X$ is a Borel function $X : \Omega\to\mathbb R$. A realization of $X$ is a particular real number $X(\omega)$, and such a number may be referred to as $x$. This means, for exampl | Lowercase $x$ vs Uppercase $X$ in statistics
You are correct that $X$ is a Borel function $X : \Omega\to\mathbb R$. A realization of $X$ is a particular real number $X(\omega)$, and such a number may be referred to as $x$. This means, for example, the event $\{X = x\}$ is more formally written as $\{\omega\in\Omega : X... | Lowercase $x$ vs Uppercase $X$ in statistics
You are correct that $X$ is a Borel function $X : \Omega\to\mathbb R$. A realization of $X$ is a particular real number $X(\omega)$, and such a number may be referred to as $x$. This means, for exampl |
44,972 | Comparing two, or more, independent paired t-tests | Using the difference between t2 and t1 as feature should work; one can use ANOVA to test the equality of the means.
Note that there are post-hoc tests that allow you to get more insight into the differences, if any, found. This allows you to identify which means deviate. These post-hoc tests take into account that mult... | Comparing two, or more, independent paired t-tests | Using the difference between t2 and t1 as feature should work; one can use ANOVA to test the equality of the means.
Note that there are post-hoc tests that allow you to get more insight into the diffe | Comparing two, or more, independent paired t-tests
Using the difference between t2 and t1 as feature should work; one can use ANOVA to test the equality of the means.
Note that there are post-hoc tests that allow you to get more insight into the differences, if any, found. This allows you to identify which means deviat... | Comparing two, or more, independent paired t-tests
Using the difference between t2 and t1 as feature should work; one can use ANOVA to test the equality of the means.
Note that there are post-hoc tests that allow you to get more insight into the diffe |
44,973 | Comparing two, or more, independent paired t-tests | Dunnett's test is the canonical test for a one-way ANOVA of various treatments versus control. It has slightly more statistical power compared to multiple independent t-tests with Bonferoni corrections under the baseline assumptions.
See the Wikipedia entry on Dunnett's test: https://en.wikipedia.org/wiki/Dunnett%27s... | Comparing two, or more, independent paired t-tests | Dunnett's test is the canonical test for a one-way ANOVA of various treatments versus control. It has slightly more statistical power compared to multiple independent t-tests with Bonferoni correctio | Comparing two, or more, independent paired t-tests
Dunnett's test is the canonical test for a one-way ANOVA of various treatments versus control. It has slightly more statistical power compared to multiple independent t-tests with Bonferoni corrections under the baseline assumptions.
See the Wikipedia entry on Dunnett... | Comparing two, or more, independent paired t-tests
Dunnett's test is the canonical test for a one-way ANOVA of various treatments versus control. It has slightly more statistical power compared to multiple independent t-tests with Bonferoni correctio |
44,974 | Comparing two, or more, independent paired t-tests | There is no reason you cannot just apply ANOVA with a standard regression model in this case. Taking three seperate T-tests and combining them into a single inference is inferior to conducting a single ANOVA with a model that includes data from all the groups.
In regard to your concerns about ANOVA, it is important to... | Comparing two, or more, independent paired t-tests | There is no reason you cannot just apply ANOVA with a standard regression model in this case. Taking three seperate T-tests and combining them into a single inference is inferior to conducting a sing | Comparing two, or more, independent paired t-tests
There is no reason you cannot just apply ANOVA with a standard regression model in this case. Taking three seperate T-tests and combining them into a single inference is inferior to conducting a single ANOVA with a model that includes data from all the groups.
In rega... | Comparing two, or more, independent paired t-tests
There is no reason you cannot just apply ANOVA with a standard regression model in this case. Taking three seperate T-tests and combining them into a single inference is inferior to conducting a sing |
44,975 | Comparing two, or more, independent paired t-tests | Yes, you can compare multiple independent t-tests, but note that by doing so you increase the probability of making a Type-I error. Assuming an $\alpha$ of .05, the probability of a Type-I error is 14.3% if comparing between three groups.
Instead of comparing the start and end weights as a pair, compare the difference ... | Comparing two, or more, independent paired t-tests | Yes, you can compare multiple independent t-tests, but note that by doing so you increase the probability of making a Type-I error. Assuming an $\alpha$ of .05, the probability of a Type-I error is 14 | Comparing two, or more, independent paired t-tests
Yes, you can compare multiple independent t-tests, but note that by doing so you increase the probability of making a Type-I error. Assuming an $\alpha$ of .05, the probability of a Type-I error is 14.3% if comparing between three groups.
Instead of comparing the start... | Comparing two, or more, independent paired t-tests
Yes, you can compare multiple independent t-tests, but note that by doing so you increase the probability of making a Type-I error. Assuming an $\alpha$ of .05, the probability of a Type-I error is 14 |
44,976 | Comparing two, or more, independent paired t-tests | If you compare multiple independent t-test(s) you should counteract the multiple comparison problem.
If you want to go down this path, one way to tackle the multi comparison problem is to use theBonferroni correction but others are available. you can find a bunch of them here. | Comparing two, or more, independent paired t-tests | If you compare multiple independent t-test(s) you should counteract the multiple comparison problem.
If you want to go down this path, one way to tackle the multi comparison problem is to use theBonfe | Comparing two, or more, independent paired t-tests
If you compare multiple independent t-test(s) you should counteract the multiple comparison problem.
If you want to go down this path, one way to tackle the multi comparison problem is to use theBonferroni correction but others are available. you can find a bunch of th... | Comparing two, or more, independent paired t-tests
If you compare multiple independent t-test(s) you should counteract the multiple comparison problem.
If you want to go down this path, one way to tackle the multi comparison problem is to use theBonfe |
44,977 | Comparing two, or more, independent paired t-tests | Instead of doing ANOVA, I want to make three independent paired t-tests, i.e, one paired t-test for each one of the groups (measuring weight before and weight after).
...
Can I compare the outputs from the three paired t-tests?
You can do this with Tukey's range method
It is actually relatively similar and there are s... | Comparing two, or more, independent paired t-tests | Instead of doing ANOVA, I want to make three independent paired t-tests, i.e, one paired t-test for each one of the groups (measuring weight before and weight after).
...
Can I compare the outputs fro | Comparing two, or more, independent paired t-tests
Instead of doing ANOVA, I want to make three independent paired t-tests, i.e, one paired t-test for each one of the groups (measuring weight before and weight after).
...
Can I compare the outputs from the three paired t-tests?
You can do this with Tukey's range metho... | Comparing two, or more, independent paired t-tests
Instead of doing ANOVA, I want to make three independent paired t-tests, i.e, one paired t-test for each one of the groups (measuring weight before and weight after).
...
Can I compare the outputs fro |
44,978 | How many must you sample with no negatives to conclude there is no negatives in the population? | Suppose we want to expect that there are no defective widgets at the 95% confidence level. What this means is that we must test enough widgets that if there were a defective widget, then we would have a 95% chance of finding the defective one in our tests.
Of course, if we want to have a 95% chance of finding the defec... | How many must you sample with no negatives to conclude there is no negatives in the population? | Suppose we want to expect that there are no defective widgets at the 95% confidence level. What this means is that we must test enough widgets that if there were a defective widget, then we would have | How many must you sample with no negatives to conclude there is no negatives in the population?
Suppose we want to expect that there are no defective widgets at the 95% confidence level. What this means is that we must test enough widgets that if there were a defective widget, then we would have a 95% chance of finding... | How many must you sample with no negatives to conclude there is no negatives in the population?
Suppose we want to expect that there are no defective widgets at the 95% confidence level. What this means is that we must test enough widgets that if there were a defective widget, then we would have |
44,979 | How many must you sample with no negatives to conclude there is no negatives in the population? | As explained in the Wikipedia link, the "Rule of Three" is for (binomial)
sampling with replacement or sampling from a theoetical infinite distribution. I am skeptical that this rule applies to your question.
Of course, to be 100% sure not even one of your 1000 widgets is defective, you will
have to look at all of them... | How many must you sample with no negatives to conclude there is no negatives in the population? | As explained in the Wikipedia link, the "Rule of Three" is for (binomial)
sampling with replacement or sampling from a theoetical infinite distribution. I am skeptical that this rule applies to your q | How many must you sample with no negatives to conclude there is no negatives in the population?
As explained in the Wikipedia link, the "Rule of Three" is for (binomial)
sampling with replacement or sampling from a theoetical infinite distribution. I am skeptical that this rule applies to your question.
Of course, to b... | How many must you sample with no negatives to conclude there is no negatives in the population?
As explained in the Wikipedia link, the "Rule of Three" is for (binomial)
sampling with replacement or sampling from a theoetical infinite distribution. I am skeptical that this rule applies to your q |
44,980 | How many must you sample with no negatives to conclude there is no negatives in the population? | I looked at a version of this problem back in April, to see what we could say about Covid elimination based on reasonable numbers of tests (spoiler: not a lot).
As @keith's answer shows, if you want to be sure there isn't even one failure in the population, you need sample a large fraction of the population. That's obv... | How many must you sample with no negatives to conclude there is no negatives in the population? | I looked at a version of this problem back in April, to see what we could say about Covid elimination based on reasonable numbers of tests (spoiler: not a lot).
As @keith's answer shows, if you want t | How many must you sample with no negatives to conclude there is no negatives in the population?
I looked at a version of this problem back in April, to see what we could say about Covid elimination based on reasonable numbers of tests (spoiler: not a lot).
As @keith's answer shows, if you want to be sure there isn't ev... | How many must you sample with no negatives to conclude there is no negatives in the population?
I looked at a version of this problem back in April, to see what we could say about Covid elimination based on reasonable numbers of tests (spoiler: not a lot).
As @keith's answer shows, if you want t |
44,981 | How many must you sample with no negatives to conclude there is no negatives in the population? | The Rule of three states that if I sample N then I know I have a rate less than 3/N at 95% CL
My expectation, E, is my rate times the number of total events, T.
E = T*3/N
The total events remaining is just the number unsampled T=1000-N
E = (1000-N)*3/N
I want to expect at most 1 so E=1
Solving for N gives
N = 10003/(1... | How many must you sample with no negatives to conclude there is no negatives in the population? | The Rule of three states that if I sample N then I know I have a rate less than 3/N at 95% CL
My expectation, E, is my rate times the number of total events, T.
E = T*3/N
The total events remaining i | How many must you sample with no negatives to conclude there is no negatives in the population?
The Rule of three states that if I sample N then I know I have a rate less than 3/N at 95% CL
My expectation, E, is my rate times the number of total events, T.
E = T*3/N
The total events remaining is just the number unsamp... | How many must you sample with no negatives to conclude there is no negatives in the population?
The Rule of three states that if I sample N then I know I have a rate less than 3/N at 95% CL
My expectation, E, is my rate times the number of total events, T.
E = T*3/N
The total events remaining i |
44,982 | Distribution of $\frac{1}{1+X}$ if $X$ is Lognormal | I will answer a simplified version, so leave the generalization as an exercise. Let $Z$ be a standard normal random variable so $X=e^Z$ is standard lognormal. Since $X>0 $ we have $Y=\frac1{1+X}$ is in the unit interval. Let $\phi, \Phi$ be the density and cdf (cumulative distribution ) functions of the standard norma... | Distribution of $\frac{1}{1+X}$ if $X$ is Lognormal | I will answer a simplified version, so leave the generalization as an exercise. Let $Z$ be a standard normal random variable so $X=e^Z$ is standard lognormal. Since $X>0 $ we have $Y=\frac1{1+X}$ is | Distribution of $\frac{1}{1+X}$ if $X$ is Lognormal
I will answer a simplified version, so leave the generalization as an exercise. Let $Z$ be a standard normal random variable so $X=e^Z$ is standard lognormal. Since $X>0 $ we have $Y=\frac1{1+X}$ is in the unit interval. Let $\phi, \Phi$ be the density and cdf (cumul... | Distribution of $\frac{1}{1+X}$ if $X$ is Lognormal
I will answer a simplified version, so leave the generalization as an exercise. Let $Z$ be a standard normal random variable so $X=e^Z$ is standard lognormal. Since $X>0 $ we have $Y=\frac1{1+X}$ is |
44,983 | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book | That's because the author is considering in that part that the activation function of a neuron $j$ only depends on the value of $z_j^L$. This happens for example with the sigmoid or ReLU activation functions.
However, when using activation functions like Softmax (described in Chapter 3.1.4), we have that:
$$a_j^L = \fr... | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book | That's because the author is considering in that part that the activation function of a neuron $j$ only depends on the value of $z_j^L$. This happens for example with the sigmoid or ReLU activation fu | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book
That's because the author is considering in that part that the activation function of a neuron $j$ only depends on the value of $z_j^L$. This happens for example with the sigmoid or ReLU activation functions.
However, when... | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book
That's because the author is considering in that part that the activation function of a neuron $j$ only depends on the value of $z_j^L$. This happens for example with the sigmoid or ReLU activation fu |
44,984 | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book | Yes, in this case, it's obvious. The author is simply writing a general equation for taking derivatives in a multivariate setting. If $f(x,y)$ is a function of $x,y$ and we look for the derivative of $f$ wrt $t$, it's customary to write the following:
$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\... | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book | Yes, in this case, it's obvious. The author is simply writing a general equation for taking derivatives in a multivariate setting. If $f(x,y)$ is a function of $x,y$ and we look for the derivative of | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book
Yes, in this case, it's obvious. The author is simply writing a general equation for taking derivatives in a multivariate setting. If $f(x,y)$ is a function of $x,y$ and we look for the derivative of $f$ wrt $t$, it's cust... | Can't understand the proof of the first backpropagation equation in Nielsen's neural network book
Yes, in this case, it's obvious. The author is simply writing a general equation for taking derivatives in a multivariate setting. If $f(x,y)$ is a function of $x,y$ and we look for the derivative of |
44,985 | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ of the variables, can we recover the $n$th value? | Since $$\text{Var}(X_1 + \cdots + X_n) = n + n(n-1) \left(-\frac{1}{n-1}\right) = 0$$
the sum of the $n$ random variables is a constant. If you know this constant, then yes knowing the realizations of $n-1$ of the random variables will give you the last one.
However, if we don't know this constant, without further info... | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ | Since $$\text{Var}(X_1 + \cdots + X_n) = n + n(n-1) \left(-\frac{1}{n-1}\right) = 0$$
the sum of the $n$ random variables is a constant. If you know this constant, then yes knowing the realizations of | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ of the variables, can we recover the $n$th value?
Since $$\text{Var}(X_1 + \cdots + X_n) = n + n(n-1) \left(-\frac{1}{n-1}\right) = 0$$
the sum of the $n$ random variables is a constant. If you know this constant, then ... | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$
Since $$\text{Var}(X_1 + \cdots + X_n) = n + n(n-1) \left(-\frac{1}{n-1}\right) = 0$$
the sum of the $n$ random variables is a constant. If you know this constant, then yes knowing the realizations of |
44,986 | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ of the variables, can we recover the $n$th value? | As a concrete example of @angryavian's point, with $n=2$
If you take $X_1\sim N(0,1)$, you could have $X_2=42-X_1$ or $X_2=69-X_1$ or $X_2=17-X_1$ and these would all have unit variance and correlation $-1$. In general, $X_n = A-\sum_{i=1}^{n-1} X_i$, and $A$ can be chosen freely.
If you also required $E[X_i]=0$ ther... | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ | As a concrete example of @angryavian's point, with $n=2$
If you take $X_1\sim N(0,1)$, you could have $X_2=42-X_1$ or $X_2=69-X_1$ or $X_2=17-X_1$ and these would all have unit variance and correlati | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ of the variables, can we recover the $n$th value?
As a concrete example of @angryavian's point, with $n=2$
If you take $X_1\sim N(0,1)$, you could have $X_2=42-X_1$ or $X_2=69-X_1$ or $X_2=17-X_1$ and these would all h... | Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$
As a concrete example of @angryavian's point, with $n=2$
If you take $X_1\sim N(0,1)$, you could have $X_2=42-X_1$ or $X_2=69-X_1$ or $X_2=17-X_1$ and these would all have unit variance and correlati |
44,987 | least square estimator of regression x onto y [duplicate] | To prove that the reverse regression is not a good estimator for $1/\beta$, recall that OLS is generally consistent (when regressing $y$ on $x$) for $cov(x,y)/var(x)$. Correspondingly, it is consistent for $cov(x,y)/var(y)$ when regressing $x$ on $y$.
When the relationship between error and regressor is (essentially, p... | least square estimator of regression x onto y [duplicate] | To prove that the reverse regression is not a good estimator for $1/\beta$, recall that OLS is generally consistent (when regressing $y$ on $x$) for $cov(x,y)/var(x)$. Correspondingly, it is consisten | least square estimator of regression x onto y [duplicate]
To prove that the reverse regression is not a good estimator for $1/\beta$, recall that OLS is generally consistent (when regressing $y$ on $x$) for $cov(x,y)/var(x)$. Correspondingly, it is consistent for $cov(x,y)/var(y)$ when regressing $x$ on $y$.
When the r... | least square estimator of regression x onto y [duplicate]
To prove that the reverse regression is not a good estimator for $1/\beta$, recall that OLS is generally consistent (when regressing $y$ on $x$) for $cov(x,y)/var(x)$. Correspondingly, it is consisten |
44,988 | least square estimator of regression x onto y [duplicate] | No, in general you will obtain a different line wih ordinary least squares if you interchange x and y. You can easily check this with your formula by interchanging x and y and comparing it to $1/\beta$. The reaaon for this descrepancy is that ordinary least squares is not about fitting a line through points, but about ... | least square estimator of regression x onto y [duplicate] | No, in general you will obtain a different line wih ordinary least squares if you interchange x and y. You can easily check this with your formula by interchanging x and y and comparing it to $1/\beta | least square estimator of regression x onto y [duplicate]
No, in general you will obtain a different line wih ordinary least squares if you interchange x and y. You can easily check this with your formula by interchanging x and y and comparing it to $1/\beta$. The reaaon for this descrepancy is that ordinary least squa... | least square estimator of regression x onto y [duplicate]
No, in general you will obtain a different line wih ordinary least squares if you interchange x and y. You can easily check this with your formula by interchanging x and y and comparing it to $1/\beta |
44,989 | Relationship between KL divergence and entropy | I will reformulate using my answer at Intuition on the Kullback-Leibler (KL) Divergence. Since I am a statistician, I am more comfortable with likelihoods than with entropies, but I also think that gives more intuition here. Now entropy $ \DeclareMathOperator{\E}{\mathbb{E}} H(X) =-\sum_x p(x) \log p(x) =-\E_p \log p$.... | Relationship between KL divergence and entropy | I will reformulate using my answer at Intuition on the Kullback-Leibler (KL) Divergence. Since I am a statistician, I am more comfortable with likelihoods than with entropies, but I also think that gi | Relationship between KL divergence and entropy
I will reformulate using my answer at Intuition on the Kullback-Leibler (KL) Divergence. Since I am a statistician, I am more comfortable with likelihoods than with entropies, but I also think that gives more intuition here. Now entropy $ \DeclareMathOperator{\E}{\mathbb{E... | Relationship between KL divergence and entropy
I will reformulate using my answer at Intuition on the Kullback-Leibler (KL) Divergence. Since I am a statistician, I am more comfortable with likelihoods than with entropies, but I also think that gi |
44,990 | Relationship between KL divergence and entropy | Firstly, the KL divergence $D_{KL}(p||q)$ is synonymous with the relative entropy, relative entropy of $p$ with respect to $q$.
Entropy then becomes the self‐information of a random variable. Mutual information is a special case of a more general quantity called relative entropy, which is a measure of the distance be... | Relationship between KL divergence and entropy | Firstly, the KL divergence $D_{KL}(p||q)$ is synonymous with the relative entropy, relative entropy of $p$ with respect to $q$.
Entropy then becomes the self‐information of a random variable. Mutual | Relationship between KL divergence and entropy
Firstly, the KL divergence $D_{KL}(p||q)$ is synonymous with the relative entropy, relative entropy of $p$ with respect to $q$.
Entropy then becomes the self‐information of a random variable. Mutual information is a special case of a more general quantity called relative... | Relationship between KL divergence and entropy
Firstly, the KL divergence $D_{KL}(p||q)$ is synonymous with the relative entropy, relative entropy of $p$ with respect to $q$.
Entropy then becomes the self‐information of a random variable. Mutual |
44,991 | Why glm() can't recover the true parameters? | Poisson regression model is
$$
\log (\operatorname{E}(Y\mid\mathbf{x}))=\alpha + \mathbf{\beta}' \mathbf{x}
$$
So you are fitting different function to your data, then the one that generated it, so it would have different parameters. Poisson regression uses by default the log as a link function. For your simulation to ... | Why glm() can't recover the true parameters? | Poisson regression model is
$$
\log (\operatorname{E}(Y\mid\mathbf{x}))=\alpha + \mathbf{\beta}' \mathbf{x}
$$
So you are fitting different function to your data, then the one that generated it, so it | Why glm() can't recover the true parameters?
Poisson regression model is
$$
\log (\operatorname{E}(Y\mid\mathbf{x}))=\alpha + \mathbf{\beta}' \mathbf{x}
$$
So you are fitting different function to your data, then the one that generated it, so it would have different parameters. Poisson regression uses by default the lo... | Why glm() can't recover the true parameters?
Poisson regression model is
$$
\log (\operatorname{E}(Y\mid\mathbf{x}))=\alpha + \mathbf{\beta}' \mathbf{x}
$$
So you are fitting different function to your data, then the one that generated it, so it |
44,992 | When is a data set a time series? | This raises an under-appreciated point about the scope of "time-series analysis"
I think you have hit on an under-appreciated point in time-series analysis here, which is that mathematically these models can also legitimately model data indexed by some other variable that does not represent "time" in the real world. I... | When is a data set a time series? | This raises an under-appreciated point about the scope of "time-series analysis"
I think you have hit on an under-appreciated point in time-series analysis here, which is that mathematically these mod | When is a data set a time series?
This raises an under-appreciated point about the scope of "time-series analysis"
I think you have hit on an under-appreciated point in time-series analysis here, which is that mathematically these models can also legitimately model data indexed by some other variable that does not repr... | When is a data set a time series?
This raises an under-appreciated point about the scope of "time-series analysis"
I think you have hit on an under-appreciated point in time-series analysis here, which is that mathematically these mod |
44,993 | When is a data set a time series? | I would say the simplest answer is that a time-series is any dataset where event time is a feature. It doesn't need to be correlated with anything in any way,
but you also can't assume it's not.
Event time here is defined as a unique numeric index $T$ corresponding to a partial ordering of the set of events such that ... | When is a data set a time series? | I would say the simplest answer is that a time-series is any dataset where event time is a feature. It doesn't need to be correlated with anything in any way,
but you also can't assume it's not.
Even | When is a data set a time series?
I would say the simplest answer is that a time-series is any dataset where event time is a feature. It doesn't need to be correlated with anything in any way,
but you also can't assume it's not.
Event time here is defined as a unique numeric index $T$ corresponding to a partial orderi... | When is a data set a time series?
I would say the simplest answer is that a time-series is any dataset where event time is a feature. It doesn't need to be correlated with anything in any way,
but you also can't assume it's not.
Even |
44,994 | When is a data set a time series? | A time-series data set is just data collected through time. For calling it a time series, it doesn't matter if futures values are a function of past values or not.
So in your moon brightness example, if you measure the brightness and record the time of measurement, it's time-series data. Even the number of road acciden... | When is a data set a time series? | A time-series data set is just data collected through time. For calling it a time series, it doesn't matter if futures values are a function of past values or not.
So in your moon brightness example, | When is a data set a time series?
A time-series data set is just data collected through time. For calling it a time series, it doesn't matter if futures values are a function of past values or not.
So in your moon brightness example, if you measure the brightness and record the time of measurement, it's time-series dat... | When is a data set a time series?
A time-series data set is just data collected through time. For calling it a time series, it doesn't matter if futures values are a function of past values or not.
So in your moon brightness example, |
44,995 | When is a data set a time series? | I think that, in practice, a time-series is a method, not a data type, one of those developed such as ARIMA to address problems when data is causally ordered and thus X can predict Y differently at different points in time, and autocorrelation is an issue. But that is another way of saying I guess that if autocorrelati... | When is a data set a time series? | I think that, in practice, a time-series is a method, not a data type, one of those developed such as ARIMA to address problems when data is causally ordered and thus X can predict Y differently at di | When is a data set a time series?
I think that, in practice, a time-series is a method, not a data type, one of those developed such as ARIMA to address problems when data is causally ordered and thus X can predict Y differently at different points in time, and autocorrelation is an issue. But that is another way of sa... | When is a data set a time series?
I think that, in practice, a time-series is a method, not a data type, one of those developed such as ARIMA to address problems when data is causally ordered and thus X can predict Y differently at di |
44,996 | Linear regression with aggregated data or not? | I would suggest taking advantage of all the data by fitting a linear mixed effects model. This is a model that includes fixed effects and random effects. If you aggregate the data then you lose information, while if you fail to account for the probability that costs in one lane are more similar to costs in that lane th... | Linear regression with aggregated data or not? | I would suggest taking advantage of all the data by fitting a linear mixed effects model. This is a model that includes fixed effects and random effects. If you aggregate the data then you lose inform | Linear regression with aggregated data or not?
I would suggest taking advantage of all the data by fitting a linear mixed effects model. This is a model that includes fixed effects and random effects. If you aggregate the data then you lose information, while if you fail to account for the probability that costs in one... | Linear regression with aggregated data or not?
I would suggest taking advantage of all the data by fitting a linear mixed effects model. This is a model that includes fixed effects and random effects. If you aggregate the data then you lose inform |
44,997 | Linear regression with aggregated data or not? | First, I'd try to answer the questions. Are the cost/distance relationships between each lane statistically different? If yes, are the differences large enough to matter for our application / problem? If no, use the aggregated data. Otherwise, we need to do some simple thought experiments to see what using aggregated /... | Linear regression with aggregated data or not? | First, I'd try to answer the questions. Are the cost/distance relationships between each lane statistically different? If yes, are the differences large enough to matter for our application / problem? | Linear regression with aggregated data or not?
First, I'd try to answer the questions. Are the cost/distance relationships between each lane statistically different? If yes, are the differences large enough to matter for our application / problem? If no, use the aggregated data. Otherwise, we need to do some simple tho... | Linear regression with aggregated data or not?
First, I'd try to answer the questions. Are the cost/distance relationships between each lane statistically different? If yes, are the differences large enough to matter for our application / problem? |
44,998 | Suppose $X$ follows a normal mixture. In which cases is $X$ itself normally distributed? | When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2,$ $F$ cannot be Normal.
A simple way to see why this assertion is true uses the characteristic function. Let the Normal components have variances $\sigma_1^2 \... | Suppose $X$ follows a normal mixture. In which cases is $X$ itself normally distributed? | When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2,$ $F$ cannot be Normal.
A simple way to | Suppose $X$ follows a normal mixture. In which cases is $X$ itself normally distributed?
When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2,$ $F$ cannot be Normal.
A simple way to see why this assertion is true ... | Suppose $X$ follows a normal mixture. In which cases is $X$ itself normally distributed?
When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2,$ $F$ cannot be Normal.
A simple way to |
44,999 | Does order of events matter in Bayesian update? | AFAIK, you cannot say that $p > \frac{1}{2}$ or even $\mathbb E[p]>\frac{1}{2}$ is an "observation" or "event", but rather a constraint on your model parameter(s). The term "observation" is usually reserved for specific realizations of a random variable (i.e. draws from a distribution). In your model, one could plausi... | Does order of events matter in Bayesian update? | AFAIK, you cannot say that $p > \frac{1}{2}$ or even $\mathbb E[p]>\frac{1}{2}$ is an "observation" or "event", but rather a constraint on your model parameter(s). The term "observation" is usually re | Does order of events matter in Bayesian update?
AFAIK, you cannot say that $p > \frac{1}{2}$ or even $\mathbb E[p]>\frac{1}{2}$ is an "observation" or "event", but rather a constraint on your model parameter(s). The term "observation" is usually reserved for specific realizations of a random variable (i.e. draws from a... | Does order of events matter in Bayesian update?
AFAIK, you cannot say that $p > \frac{1}{2}$ or even $\mathbb E[p]>\frac{1}{2}$ is an "observation" or "event", but rather a constraint on your model parameter(s). The term "observation" is usually re |
45,000 | Does order of events matter in Bayesian update? | In Bayesian inference, terms like "observation" and "event" are just conveniences; there is no fundamental importance to them, so don't get hung up on them.
In particular, there is no physical causality or time's arrow -- no "events". Whether you can carry out some calculations in more than one order depends solely on... | Does order of events matter in Bayesian update? | In Bayesian inference, terms like "observation" and "event" are just conveniences; there is no fundamental importance to them, so don't get hung up on them.
In particular, there is no physical causal | Does order of events matter in Bayesian update?
In Bayesian inference, terms like "observation" and "event" are just conveniences; there is no fundamental importance to them, so don't get hung up on them.
In particular, there is no physical causality or time's arrow -- no "events". Whether you can carry out some calcu... | Does order of events matter in Bayesian update?
In Bayesian inference, terms like "observation" and "event" are just conveniences; there is no fundamental importance to them, so don't get hung up on them.
In particular, there is no physical causal |
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