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 |
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48,901 | Robust regression or ANOVA for non-normal dependent variable | First a comment: "robust" usually refers to approaches guarding against outliers and violations of distributional assumptions. In your case, the problem is obviously a violation of distributional assumption, but it seems to depend on your DV (sorry for the pun).
What method to use depends on whether 100 is "truely" the... | Robust regression or ANOVA for non-normal dependent variable | First a comment: "robust" usually refers to approaches guarding against outliers and violations of distributional assumptions. In your case, the problem is obviously a violation of distributional assu | Robust regression or ANOVA for non-normal dependent variable
First a comment: "robust" usually refers to approaches guarding against outliers and violations of distributional assumptions. In your case, the problem is obviously a violation of distributional assumption, but it seems to depend on your DV (sorry for the pu... | Robust regression or ANOVA for non-normal dependent variable
First a comment: "robust" usually refers to approaches guarding against outliers and violations of distributional assumptions. In your case, the problem is obviously a violation of distributional assu |
48,902 | Robust regression or ANOVA for non-normal dependent variable | Rank based tests work by transforming the data to a uniform distribution then relying on the central limit theorem to justify approximate normality (the clt kicks in for the uniform around n=5 or 6), this helps counter the effects of skewness or outliers. Your data has the opposite problem and the rank transform is un... | Robust regression or ANOVA for non-normal dependent variable | Rank based tests work by transforming the data to a uniform distribution then relying on the central limit theorem to justify approximate normality (the clt kicks in for the uniform around n=5 or 6), | Robust regression or ANOVA for non-normal dependent variable
Rank based tests work by transforming the data to a uniform distribution then relying on the central limit theorem to justify approximate normality (the clt kicks in for the uniform around n=5 or 6), this helps counter the effects of skewness or outliers. Yo... | Robust regression or ANOVA for non-normal dependent variable
Rank based tests work by transforming the data to a uniform distribution then relying on the central limit theorem to justify approximate normality (the clt kicks in for the uniform around n=5 or 6), |
48,903 | Robust regression or ANOVA for non-normal dependent variable | Without knowing that the data are really about it's hard to say. One potential very general solution is to consider that 100 isn't really 100 (sometimes). What to do with that is what you need to work out. You need to come up with a model about what other values 100 is. Would some people have wanted to pick 1000? 1... | Robust regression or ANOVA for non-normal dependent variable | Without knowing that the data are really about it's hard to say. One potential very general solution is to consider that 100 isn't really 100 (sometimes). What to do with that is what you need to wo | Robust regression or ANOVA for non-normal dependent variable
Without knowing that the data are really about it's hard to say. One potential very general solution is to consider that 100 isn't really 100 (sometimes). What to do with that is what you need to work out. You need to come up with a model about what other ... | Robust regression or ANOVA for non-normal dependent variable
Without knowing that the data are really about it's hard to say. One potential very general solution is to consider that 100 isn't really 100 (sometimes). What to do with that is what you need to wo |
48,904 | Incorporating random effects in the logistic regression formula in R | Short answer is you can't - well, not without recoding a version of stepAIC() that knows how to handle S4 objects. stepAIC() knows nothing about lmer() and glmer() models, and there is no equivalent code in lme4 that will allow you to do this sort of stepping.
I also think your whole process needs carefully rethinking ... | Incorporating random effects in the logistic regression formula in R | Short answer is you can't - well, not without recoding a version of stepAIC() that knows how to handle S4 objects. stepAIC() knows nothing about lmer() and glmer() models, and there is no equivalent c | Incorporating random effects in the logistic regression formula in R
Short answer is you can't - well, not without recoding a version of stepAIC() that knows how to handle S4 objects. stepAIC() knows nothing about lmer() and glmer() models, and there is no equivalent code in lme4 that will allow you to do this sort of ... | Incorporating random effects in the logistic regression formula in R
Short answer is you can't - well, not without recoding a version of stepAIC() that knows how to handle S4 objects. stepAIC() knows nothing about lmer() and glmer() models, and there is no equivalent c |
48,905 | Heckman sample selection | The answer is yes, you do not need to use the parameters of inverse Mills ratios. But you must include them in the regression nevertheless, or your other parameters will be biased.
According to the article yes. Although if different variables are statistically significant in different regression there is no problem. Ju... | Heckman sample selection | The answer is yes, you do not need to use the parameters of inverse Mills ratios. But you must include them in the regression nevertheless, or your other parameters will be biased.
According to the ar | Heckman sample selection
The answer is yes, you do not need to use the parameters of inverse Mills ratios. But you must include them in the regression nevertheless, or your other parameters will be biased.
According to the article yes. Although if different variables are statistically significant in different regressio... | Heckman sample selection
The answer is yes, you do not need to use the parameters of inverse Mills ratios. But you must include them in the regression nevertheless, or your other parameters will be biased.
According to the ar |
48,906 | One-inflated negative binomial? | Sure. You can write
$Pr(X=x) = p*f(x) + (1-p)1(x=1)$
where $f$ is the NB pmf, $1()$ is just an indicator function and $p$ is some probability. Of course in general the $x=1$ could be $x=k$ for any integer $k$, and $f$ could be any pmf.
In principle it shouldn't be harder to fit than a ZI model, but an off-the-shelf mo... | One-inflated negative binomial? | Sure. You can write
$Pr(X=x) = p*f(x) + (1-p)1(x=1)$
where $f$ is the NB pmf, $1()$ is just an indicator function and $p$ is some probability. Of course in general the $x=1$ could be $x=k$ for any int | One-inflated negative binomial?
Sure. You can write
$Pr(X=x) = p*f(x) + (1-p)1(x=1)$
where $f$ is the NB pmf, $1()$ is just an indicator function and $p$ is some probability. Of course in general the $x=1$ could be $x=k$ for any integer $k$, and $f$ could be any pmf.
In principle it shouldn't be harder to fit than a Z... | One-inflated negative binomial?
Sure. You can write
$Pr(X=x) = p*f(x) + (1-p)1(x=1)$
where $f$ is the NB pmf, $1()$ is just an indicator function and $p$ is some probability. Of course in general the $x=1$ could be $x=k$ for any int |
48,907 | Given two responses for two groups, how to decide what to test on response or profit? | The reason why you are conducting this test is to determine which policy is more valuable, and if value is measured in profitability, then it makes no sense to do statistical testing on any other variable. A properly conducted test on profitability gives you all the information needed for your companies' decision: onc... | Given two responses for two groups, how to decide what to test on response or profit? | The reason why you are conducting this test is to determine which policy is more valuable, and if value is measured in profitability, then it makes no sense to do statistical testing on any other vari | Given two responses for two groups, how to decide what to test on response or profit?
The reason why you are conducting this test is to determine which policy is more valuable, and if value is measured in profitability, then it makes no sense to do statistical testing on any other variable. A properly conducted test o... | Given two responses for two groups, how to decide what to test on response or profit?
The reason why you are conducting this test is to determine which policy is more valuable, and if value is measured in profitability, then it makes no sense to do statistical testing on any other vari |
48,908 | Calculation of natural cubic splines in R | Wikipedia has a nice explanation of spline interpolation
I posted the code to create cubic Bezier splines on Rosettacode a while ago.
Also, you can have a look at this discussion on SO about spline extrapolation. | Calculation of natural cubic splines in R | Wikipedia has a nice explanation of spline interpolation
I posted the code to create cubic Bezier splines on Rosettacode a while ago.
Also, you can have a look at this discussion on SO about spline ex | Calculation of natural cubic splines in R
Wikipedia has a nice explanation of spline interpolation
I posted the code to create cubic Bezier splines on Rosettacode a while ago.
Also, you can have a look at this discussion on SO about spline extrapolation. | Calculation of natural cubic splines in R
Wikipedia has a nice explanation of spline interpolation
I posted the code to create cubic Bezier splines on Rosettacode a while ago.
Also, you can have a look at this discussion on SO about spline ex |
48,909 | Calculation of natural cubic splines in R | I learnt about the use of splines in regression from the book "Regression Modeling Strategies" by Frank Harrell. Harrell's R package rms allows you to easily fit regression models in which some predictor variables are represented as splines. | Calculation of natural cubic splines in R | I learnt about the use of splines in regression from the book "Regression Modeling Strategies" by Frank Harrell. Harrell's R package rms allows you to easily fit regression models in which some predic | Calculation of natural cubic splines in R
I learnt about the use of splines in regression from the book "Regression Modeling Strategies" by Frank Harrell. Harrell's R package rms allows you to easily fit regression models in which some predictor variables are represented as splines. | Calculation of natural cubic splines in R
I learnt about the use of splines in regression from the book "Regression Modeling Strategies" by Frank Harrell. Harrell's R package rms allows you to easily fit regression models in which some predic |
48,910 | Pivotal quantities, test statistics and hypothesis tests | The first thing you should do is challenge your lecturer to explain these things clearly. If anything whatsoever seems counter-intuitive or backwards, them demand that he/she explains why it is intuitive. Statistics always makes sense if you think about it in the "right" way.
Calculating pivotal quantities is a very ... | Pivotal quantities, test statistics and hypothesis tests | The first thing you should do is challenge your lecturer to explain these things clearly. If anything whatsoever seems counter-intuitive or backwards, them demand that he/she explains why it is intui | Pivotal quantities, test statistics and hypothesis tests
The first thing you should do is challenge your lecturer to explain these things clearly. If anything whatsoever seems counter-intuitive or backwards, them demand that he/she explains why it is intuitive. Statistics always makes sense if you think about it in t... | Pivotal quantities, test statistics and hypothesis tests
The first thing you should do is challenge your lecturer to explain these things clearly. If anything whatsoever seems counter-intuitive or backwards, them demand that he/she explains why it is intui |
48,911 | Calculating predicted values from categorical predictors in logistic regression | My initial thought would have been to display the probability of of acceptance as a function of relative GPA for each of your four schools, using some kind of trellis displays. In this case, facetting should do the job well as the number of schools is not so large. This is very easy to do with lattice (y ~ gpa | school... | Calculating predicted values from categorical predictors in logistic regression | My initial thought would have been to display the probability of of acceptance as a function of relative GPA for each of your four schools, using some kind of trellis displays. In this case, facetting | Calculating predicted values from categorical predictors in logistic regression
My initial thought would have been to display the probability of of acceptance as a function of relative GPA for each of your four schools, using some kind of trellis displays. In this case, facetting should do the job well as the number of... | Calculating predicted values from categorical predictors in logistic regression
My initial thought would have been to display the probability of of acceptance as a function of relative GPA for each of your four schools, using some kind of trellis displays. In this case, facetting |
48,912 | Interpreting significance of predictor vs significance of predictor coeffs in multinomial logistic regression | The p-value itself cannot tell you how strong the relationship is, because the p-value is so influenced by sample size, among other things. But assuming your N is something on the order of 100-150, I'd say there's a reasonably strong effect involving Size whereby as Size increases, the log of the odds of Y being 1 is ... | Interpreting significance of predictor vs significance of predictor coeffs in multinomial logistic r | The p-value itself cannot tell you how strong the relationship is, because the p-value is so influenced by sample size, among other things. But assuming your N is something on the order of 100-150, I | Interpreting significance of predictor vs significance of predictor coeffs in multinomial logistic regression
The p-value itself cannot tell you how strong the relationship is, because the p-value is so influenced by sample size, among other things. But assuming your N is something on the order of 100-150, I'd say the... | Interpreting significance of predictor vs significance of predictor coeffs in multinomial logistic r
The p-value itself cannot tell you how strong the relationship is, because the p-value is so influenced by sample size, among other things. But assuming your N is something on the order of 100-150, I |
48,913 | Hellwig's method of selection of variables | After spending too long on web research, I'm pretty sure the source of 'Hellwig's method' is:
Hellwig, Zdzisław. On the optimal choice of predictors. Study VI in Z. Gostkowski (ed.): Toward a system of quantitative indicators of components of human resources development; Paris: UNESCO, 1968; 23 pages. [pdf]
Google Scho... | Hellwig's method of selection of variables | After spending too long on web research, I'm pretty sure the source of 'Hellwig's method' is:
Hellwig, Zdzisław. On the optimal choice of predictors. Study VI in Z. Gostkowski (ed.): Toward a system o | Hellwig's method of selection of variables
After spending too long on web research, I'm pretty sure the source of 'Hellwig's method' is:
Hellwig, Zdzisław. On the optimal choice of predictors. Study VI in Z. Gostkowski (ed.): Toward a system of quantitative indicators of components of human resources development; Paris... | Hellwig's method of selection of variables
After spending too long on web research, I'm pretty sure the source of 'Hellwig's method' is:
Hellwig, Zdzisław. On the optimal choice of predictors. Study VI in Z. Gostkowski (ed.): Toward a system o |
48,914 | Any good reference books/material to help me build a txn level fraud detection model? | Fraud detection is a rare class problem. Chapter Six of Charles Elkan's Notes for his Graduate Course in Data Mining and Predictive Analytics at UCSD walks you through the prediction of a rare class, and the pitfalls and proper ways to evaluate the success of such a model. The methods he specifically uses are Isotonic ... | Any good reference books/material to help me build a txn level fraud detection model? | Fraud detection is a rare class problem. Chapter Six of Charles Elkan's Notes for his Graduate Course in Data Mining and Predictive Analytics at UCSD walks you through the prediction of a rare class, | Any good reference books/material to help me build a txn level fraud detection model?
Fraud detection is a rare class problem. Chapter Six of Charles Elkan's Notes for his Graduate Course in Data Mining and Predictive Analytics at UCSD walks you through the prediction of a rare class, and the pitfalls and proper ways t... | Any good reference books/material to help me build a txn level fraud detection model?
Fraud detection is a rare class problem. Chapter Six of Charles Elkan's Notes for his Graduate Course in Data Mining and Predictive Analytics at UCSD walks you through the prediction of a rare class, |
48,915 | Splitting one variable according to bins from another variable | > A <- round(rnorm(100, 100, 15), 2) # generate some data
> age <- sample(18:65, 100, replace=TRUE)
> sex <- factor(sample(0:1, 100, replace=TRUE), labels=c("f", "m"))
# 1) bin age into 4 groups of similar size
> ageFac <- cut(age, breaks=quantile(age, probs=seq(from=0, to=1, by=0.25)),
+ include... | Splitting one variable according to bins from another variable | > A <- round(rnorm(100, 100, 15), 2) # generate some data
> age <- sample(18:65, 100, replace=TRUE)
> sex <- factor(sample(0:1, 100, replace=TRUE), labels=c("f", "m"))
# 1) bin age into 4 gro | Splitting one variable according to bins from another variable
> A <- round(rnorm(100, 100, 15), 2) # generate some data
> age <- sample(18:65, 100, replace=TRUE)
> sex <- factor(sample(0:1, 100, replace=TRUE), labels=c("f", "m"))
# 1) bin age into 4 groups of similar size
> ageFac <- cut(age, breaks=quantile(... | Splitting one variable according to bins from another variable
> A <- round(rnorm(100, 100, 15), 2) # generate some data
> age <- sample(18:65, 100, replace=TRUE)
> sex <- factor(sample(0:1, 100, replace=TRUE), labels=c("f", "m"))
# 1) bin age into 4 gro |
48,916 | Correct way to calibrate means | There is something called "small error propagation", and it says that the error of a function $f$ of variables $x_1,x_2,\cdots,x_n$ with errors $\Delta x_1,\Delta x_2,\cdots,\Delta x_n$ equals
$$\Delta f=\sqrt{\sum_i\left(\frac{\partial f}{\partial x_i}\Delta x_i\right)^2},$$
so for $f(a,b):=a-b$ the error is $\Delta f... | Correct way to calibrate means | There is something called "small error propagation", and it says that the error of a function $f$ of variables $x_1,x_2,\cdots,x_n$ with errors $\Delta x_1,\Delta x_2,\cdots,\Delta x_n$ equals
$$\Delt | Correct way to calibrate means
There is something called "small error propagation", and it says that the error of a function $f$ of variables $x_1,x_2,\cdots,x_n$ with errors $\Delta x_1,\Delta x_2,\cdots,\Delta x_n$ equals
$$\Delta f=\sqrt{\sum_i\left(\frac{\partial f}{\partial x_i}\Delta x_i\right)^2},$$
so for $f(a,... | Correct way to calibrate means
There is something called "small error propagation", and it says that the error of a function $f$ of variables $x_1,x_2,\cdots,x_n$ with errors $\Delta x_1,\Delta x_2,\cdots,\Delta x_n$ equals
$$\Delt |
48,917 | Saturation in ARIMA (et al) models? | If Y is customer demand, than you are observing X=min(Y,1000) due to resource constraints. The actual Y could be larger, but you never observe it. So if you fit a time series model to X, you can set the forecasts to min(F,1000) where F is the forecast from the time series model. I don't think there is a need to do anyt... | Saturation in ARIMA (et al) models? | If Y is customer demand, than you are observing X=min(Y,1000) due to resource constraints. The actual Y could be larger, but you never observe it. So if you fit a time series model to X, you can set t | Saturation in ARIMA (et al) models?
If Y is customer demand, than you are observing X=min(Y,1000) due to resource constraints. The actual Y could be larger, but you never observe it. So if you fit a time series model to X, you can set the forecasts to min(F,1000) where F is the forecast from the time series model. I do... | Saturation in ARIMA (et al) models?
If Y is customer demand, than you are observing X=min(Y,1000) due to resource constraints. The actual Y could be larger, but you never observe it. So if you fit a time series model to X, you can set t |
48,918 | Saturation in ARIMA (et al) models? | Perhaps your ARIMA model needs to be tempered with identifiable deterministic structure such as "changes in intercept" or changes in trend. These models would then be classified as robust ARIMA models or Transfer FunctionModels. If there is a True Limiting Value then the data might suggest that as it grows towards that... | Saturation in ARIMA (et al) models? | Perhaps your ARIMA model needs to be tempered with identifiable deterministic structure such as "changes in intercept" or changes in trend. These models would then be classified as robust ARIMA models | Saturation in ARIMA (et al) models?
Perhaps your ARIMA model needs to be tempered with identifiable deterministic structure such as "changes in intercept" or changes in trend. These models would then be classified as robust ARIMA models or Transfer FunctionModels. If there is a True Limiting Value then the data might s... | Saturation in ARIMA (et al) models?
Perhaps your ARIMA model needs to be tempered with identifiable deterministic structure such as "changes in intercept" or changes in trend. These models would then be classified as robust ARIMA models |
48,919 | Data mining algorithm suggestion | Your first task is to find a reasonable model relating an outcome $Y$ to the sequence of workouts that preceded it. One might start by supposing that the outcome depends quite generally on a linear combination of time-weighted workout efforts $X$, but such a model would be unidentifiable (from having more parameters t... | Data mining algorithm suggestion | Your first task is to find a reasonable model relating an outcome $Y$ to the sequence of workouts that preceded it. One might start by supposing that the outcome depends quite generally on a linear c | Data mining algorithm suggestion
Your first task is to find a reasonable model relating an outcome $Y$ to the sequence of workouts that preceded it. One might start by supposing that the outcome depends quite generally on a linear combination of time-weighted workout efforts $X$, but such a model would be unidentifiab... | Data mining algorithm suggestion
Your first task is to find a reasonable model relating an outcome $Y$ to the sequence of workouts that preceded it. One might start by supposing that the outcome depends quite generally on a linear c |
48,920 | Coefficient / model averaging to control for exogenous circumstances in prediction | I am not sure you need any special tricks as long as the relevant factors are captured by the model. To keep things simple I will discuss the issue in the context of linear regression. The same intuition carries over to the time series setting.
Suppose, that you want to predict the monthly sales of cell phones for bran... | Coefficient / model averaging to control for exogenous circumstances in prediction | I am not sure you need any special tricks as long as the relevant factors are captured by the model. To keep things simple I will discuss the issue in the context of linear regression. The same intuit | Coefficient / model averaging to control for exogenous circumstances in prediction
I am not sure you need any special tricks as long as the relevant factors are captured by the model. To keep things simple I will discuss the issue in the context of linear regression. The same intuition carries over to the time series s... | Coefficient / model averaging to control for exogenous circumstances in prediction
I am not sure you need any special tricks as long as the relevant factors are captured by the model. To keep things simple I will discuss the issue in the context of linear regression. The same intuit |
48,921 | Interpreting 2D correspondence analysis plots (Part II) | I'm an ecologist, so I apologise in advance is this sounds a bit strange :-)
I like to think of these plots in terms of weighted averages. The region points are at the weighted averages of the smoking status classes and vice versa.
The problem with the above figure is the axis scaling and the fact that you can't displa... | Interpreting 2D correspondence analysis plots (Part II) | I'm an ecologist, so I apologise in advance is this sounds a bit strange :-)
I like to think of these plots in terms of weighted averages. The region points are at the weighted averages of the smoking | Interpreting 2D correspondence analysis plots (Part II)
I'm an ecologist, so I apologise in advance is this sounds a bit strange :-)
I like to think of these plots in terms of weighted averages. The region points are at the weighted averages of the smoking status classes and vice versa.
The problem with the above figur... | Interpreting 2D correspondence analysis plots (Part II)
I'm an ecologist, so I apologise in advance is this sounds a bit strange :-)
I like to think of these plots in terms of weighted averages. The region points are at the weighted averages of the smoking |
48,922 | How to compute the standard error of an L-estimator? | As you know, from
$$Var[q] = Var[\sum_i w_i x_{(i)}] = \sum_i\sum_j w_i w_j Cov[x_{(i)}, x_{(j)}]$$
it follows you need only compute the variances and covariances of the order statistics. To do this, diagonalize the covariance matrix! Although this cannot be done in general, M. A. Stephens has obtained (heuristically... | How to compute the standard error of an L-estimator? | As you know, from
$$Var[q] = Var[\sum_i w_i x_{(i)}] = \sum_i\sum_j w_i w_j Cov[x_{(i)}, x_{(j)}]$$
it follows you need only compute the variances and covariances of the order statistics. To do this, | How to compute the standard error of an L-estimator?
As you know, from
$$Var[q] = Var[\sum_i w_i x_{(i)}] = \sum_i\sum_j w_i w_j Cov[x_{(i)}, x_{(j)}]$$
it follows you need only compute the variances and covariances of the order statistics. To do this, diagonalize the covariance matrix! Although this cannot be done i... | How to compute the standard error of an L-estimator?
As you know, from
$$Var[q] = Var[\sum_i w_i x_{(i)}] = \sum_i\sum_j w_i w_j Cov[x_{(i)}, x_{(j)}]$$
it follows you need only compute the variances and covariances of the order statistics. To do this, |
48,923 | How to compute the standard error of an L-estimator? | It looks like I am probably stuck with a bootstrap. One interesting possibility here is to compute the 'exact bootstrap covariance', as outlined by Hutson & Ernst. Presumably the bootstrap covariance gives a good estimate of the standard error, asymptotically. However, the approach of Hutson & Ernst requires computatio... | How to compute the standard error of an L-estimator? | It looks like I am probably stuck with a bootstrap. One interesting possibility here is to compute the 'exact bootstrap covariance', as outlined by Hutson & Ernst. Presumably the bootstrap covariance | How to compute the standard error of an L-estimator?
It looks like I am probably stuck with a bootstrap. One interesting possibility here is to compute the 'exact bootstrap covariance', as outlined by Hutson & Ernst. Presumably the bootstrap covariance gives a good estimate of the standard error, asymptotically. Howeve... | How to compute the standard error of an L-estimator?
It looks like I am probably stuck with a bootstrap. One interesting possibility here is to compute the 'exact bootstrap covariance', as outlined by Hutson & Ernst. Presumably the bootstrap covariance |
48,924 | Few machine learning problems | For (1), as ebony1 suggests, there are several incremental or on-line SVM algorithms you could try, the only thing I would mention is that the hyper-parameters (regularisation and kernel parameters) may also need tuning as you go along as well, and there are fewer algorithmic tricks to help with that. The regularisati... | Few machine learning problems | For (1), as ebony1 suggests, there are several incremental or on-line SVM algorithms you could try, the only thing I would mention is that the hyper-parameters (regularisation and kernel parameters) m | Few machine learning problems
For (1), as ebony1 suggests, there are several incremental or on-line SVM algorithms you could try, the only thing I would mention is that the hyper-parameters (regularisation and kernel parameters) may also need tuning as you go along as well, and there are fewer algorithmic tricks to hel... | Few machine learning problems
For (1), as ebony1 suggests, there are several incremental or on-line SVM algorithms you could try, the only thing I would mention is that the hyper-parameters (regularisation and kernel parameters) m |
48,925 | Few machine learning problems | I'm not really sure if an implementation exists to address all your needs.
For (1), you can use any of the online implementations of SVM such as Pegasos or LASVM. If you want something simpler, you may use Perceptron or Kernel Perceptron. Basically, in all these algorithms, given an already learned weight vector (say w... | Few machine learning problems | I'm not really sure if an implementation exists to address all your needs.
For (1), you can use any of the online implementations of SVM such as Pegasos or LASVM. If you want something simpler, you ma | Few machine learning problems
I'm not really sure if an implementation exists to address all your needs.
For (1), you can use any of the online implementations of SVM such as Pegasos or LASVM. If you want something simpler, you may use Perceptron or Kernel Perceptron. Basically, in all these algorithms, given an alread... | Few machine learning problems
I'm not really sure if an implementation exists to address all your needs.
For (1), you can use any of the online implementations of SVM such as Pegasos or LASVM. If you want something simpler, you ma |
48,926 | Best way to show these or similar count data are not independent? | You could just plot the ACF and check if the first coefficient is inside the critical values. The critical values are ok for non-Gaussian time series (at least asymptotically).
Alternatively, fit a simple count time series model such as the INAR(1) and see if the coefficient is significantly different from zero. | Best way to show these or similar count data are not independent? | You could just plot the ACF and check if the first coefficient is inside the critical values. The critical values are ok for non-Gaussian time series (at least asymptotically).
Alternatively, fit a si | Best way to show these or similar count data are not independent?
You could just plot the ACF and check if the first coefficient is inside the critical values. The critical values are ok for non-Gaussian time series (at least asymptotically).
Alternatively, fit a simple count time series model such as the INAR(1) and s... | Best way to show these or similar count data are not independent?
You could just plot the ACF and check if the first coefficient is inside the critical values. The critical values are ok for non-Gaussian time series (at least asymptotically).
Alternatively, fit a si |
48,927 | What are the indications that one should be using interaction variables in their linear regression model? | The short, but perhaps unsatisfying answer is: when you have a prior reason to think that the effect of one variable might depend on what's going on with another variable.
For example, let's say I'm trying to model student scores on a math test as a function of math test scores in the previous year and a binary variabl... | What are the indications that one should be using interaction variables in their linear regression m | The short, but perhaps unsatisfying answer is: when you have a prior reason to think that the effect of one variable might depend on what's going on with another variable.
For example, let's say I'm t | What are the indications that one should be using interaction variables in their linear regression model?
The short, but perhaps unsatisfying answer is: when you have a prior reason to think that the effect of one variable might depend on what's going on with another variable.
For example, let's say I'm trying to model... | What are the indications that one should be using interaction variables in their linear regression m
The short, but perhaps unsatisfying answer is: when you have a prior reason to think that the effect of one variable might depend on what's going on with another variable.
For example, let's say I'm t |
48,928 | What are the indications that one should be using interaction variables in their linear regression model? | The prior for any interaction is that it is likely very, perhaps vanishingly, small (see Tosh 2021). As such, you should not explore or add-in interaction terms unless you expect them, a priori, to be large or informative. Doing so creates a massive risk for false-positives. Generally, one should only add in interactio... | What are the indications that one should be using interaction variables in their linear regression m | The prior for any interaction is that it is likely very, perhaps vanishingly, small (see Tosh 2021). As such, you should not explore or add-in interaction terms unless you expect them, a priori, to be | What are the indications that one should be using interaction variables in their linear regression model?
The prior for any interaction is that it is likely very, perhaps vanishingly, small (see Tosh 2021). As such, you should not explore or add-in interaction terms unless you expect them, a priori, to be large or info... | What are the indications that one should be using interaction variables in their linear regression m
The prior for any interaction is that it is likely very, perhaps vanishingly, small (see Tosh 2021). As such, you should not explore or add-in interaction terms unless you expect them, a priori, to be |
48,929 | What does "predictive discrimination" mean and how is it different from classification? | Frank Harrell's comment [emphasis is mine]:
Predictive discrimination is the degree to which predictive signals can separate those with good outcomes from those with worse outcomes. The most popular measures of discrimination are R2 and c-index (concordance probability; equal to AUROC when Y is binary). Rank correlati... | What does "predictive discrimination" mean and how is it different from classification? | Frank Harrell's comment [emphasis is mine]:
Predictive discrimination is the degree to which predictive signals can separate those with good outcomes from those with worse outcomes. The most popular | What does "predictive discrimination" mean and how is it different from classification?
Frank Harrell's comment [emphasis is mine]:
Predictive discrimination is the degree to which predictive signals can separate those with good outcomes from those with worse outcomes. The most popular measures of discrimination are R... | What does "predictive discrimination" mean and how is it different from classification?
Frank Harrell's comment [emphasis is mine]:
Predictive discrimination is the degree to which predictive signals can separate those with good outcomes from those with worse outcomes. The most popular |
48,930 | What does "predictive discrimination" mean and how is it different from classification? | Predictive discrimination is the ability of a model to produce (distributions of) predicted values that are separated when the observed values are distinct, and to have the correct order, and I think this is totally consistent with the quote in the answer by paperskilltrees. This is not the same as making quality predi... | What does "predictive discrimination" mean and how is it different from classification? | Predictive discrimination is the ability of a model to produce (distributions of) predicted values that are separated when the observed values are distinct, and to have the correct order, and I think | What does "predictive discrimination" mean and how is it different from classification?
Predictive discrimination is the ability of a model to produce (distributions of) predicted values that are separated when the observed values are distinct, and to have the correct order, and I think this is totally consistent with ... | What does "predictive discrimination" mean and how is it different from classification?
Predictive discrimination is the ability of a model to produce (distributions of) predicted values that are separated when the observed values are distinct, and to have the correct order, and I think |
48,931 | When should one use a Tweedie GLM over a Zero-Inflated GLM? | Tweedie GLMs are true GLMs and enjoy the usual properties of GLMs.
ZI GLMs are more complex models that assume a GLM plus an extra zero-inflation process, so they are obviously more flexible but at the cost of extra parameters. If the simpler Tweedie GLM fits the data adequately then it is the preferable model. It is d... | When should one use a Tweedie GLM over a Zero-Inflated GLM? | Tweedie GLMs are true GLMs and enjoy the usual properties of GLMs.
ZI GLMs are more complex models that assume a GLM plus an extra zero-inflation process, so they are obviously more flexible but at th | When should one use a Tweedie GLM over a Zero-Inflated GLM?
Tweedie GLMs are true GLMs and enjoy the usual properties of GLMs.
ZI GLMs are more complex models that assume a GLM plus an extra zero-inflation process, so they are obviously more flexible but at the cost of extra parameters. If the simpler Tweedie GLM fits ... | When should one use a Tweedie GLM over a Zero-Inflated GLM?
Tweedie GLMs are true GLMs and enjoy the usual properties of GLMs.
ZI GLMs are more complex models that assume a GLM plus an extra zero-inflation process, so they are obviously more flexible but at th |
48,932 | does convolution of a probability distribution with itself converge to its mean | $Y_0=\lambda X + (1-\lambda) X'$ so
$\text{var}(Y_0) = (\lambda^2 + (1-\lambda)^2)\text{var}(X)$
define $v = (\lambda^2 + (1-\lambda)^2)$
note $0.5< v < 1$ for $0< \lambda<1$
$Y_{1}=\lambda Y_0 + (1-\lambda) Y_0'$ and we have the general pattern
$Y_{n+1}=\lambda Y_n + (1-\lambda) Y_n'$
since $Y_n$ and $Y_n'$ are indepe... | does convolution of a probability distribution with itself converge to its mean | $Y_0=\lambda X + (1-\lambda) X'$ so
$\text{var}(Y_0) = (\lambda^2 + (1-\lambda)^2)\text{var}(X)$
define $v = (\lambda^2 + (1-\lambda)^2)$
note $0.5< v < 1$ for $0< \lambda<1$
$Y_{1}=\lambda Y_0 + (1-\ | does convolution of a probability distribution with itself converge to its mean
$Y_0=\lambda X + (1-\lambda) X'$ so
$\text{var}(Y_0) = (\lambda^2 + (1-\lambda)^2)\text{var}(X)$
define $v = (\lambda^2 + (1-\lambda)^2)$
note $0.5< v < 1$ for $0< \lambda<1$
$Y_{1}=\lambda Y_0 + (1-\lambda) Y_0'$ and we have the general pa... | does convolution of a probability distribution with itself converge to its mean
$Y_0=\lambda X + (1-\lambda) X'$ so
$\text{var}(Y_0) = (\lambda^2 + (1-\lambda)^2)\text{var}(X)$
define $v = (\lambda^2 + (1-\lambda)^2)$
note $0.5< v < 1$ for $0< \lambda<1$
$Y_{1}=\lambda Y_0 + (1-\ |
48,933 | does convolution of a probability distribution with itself converge to its mean | This is just an example that your statement seems to be wrong, at least in some scenarios.
Suppose there are two same but independent distributions:
$Y_0$: 50% we get 1, 50% we get 0.
$Y_0'$: 50% we get 1, 50% we get 0.
Let $Y_1=1/2Y_0+1/2Y_0'$
Then $Y_1$ is: 25% we get 1, 50% we get 0.5, 25% we get 0.
The 25% comes fr... | does convolution of a probability distribution with itself converge to its mean | This is just an example that your statement seems to be wrong, at least in some scenarios.
Suppose there are two same but independent distributions:
$Y_0$: 50% we get 1, 50% we get 0.
$Y_0'$: 50% we g | does convolution of a probability distribution with itself converge to its mean
This is just an example that your statement seems to be wrong, at least in some scenarios.
Suppose there are two same but independent distributions:
$Y_0$: 50% we get 1, 50% we get 0.
$Y_0'$: 50% we get 1, 50% we get 0.
Let $Y_1=1/2Y_0+1/2Y... | does convolution of a probability distribution with itself converge to its mean
This is just an example that your statement seems to be wrong, at least in some scenarios.
Suppose there are two same but independent distributions:
$Y_0$: 50% we get 1, 50% we get 0.
$Y_0'$: 50% we g |
48,934 | Why is it that if you undersample or oversample you have to calibrate your output probabilities? | I'm taking a lot of my answer from Agresti's "Categorical Data Analysis". For some context, let's say we want to predict if an email is spam or ham using a single binary variable, namely if the subject line contains the word "Viagra". Let's call this predictor $x$. This is a simplification of our more general problem... | Why is it that if you undersample or oversample you have to calibrate your output probabilities? | I'm taking a lot of my answer from Agresti's "Categorical Data Analysis". For some context, let's say we want to predict if an email is spam or ham using a single binary variable, namely if the subje | Why is it that if you undersample or oversample you have to calibrate your output probabilities?
I'm taking a lot of my answer from Agresti's "Categorical Data Analysis". For some context, let's say we want to predict if an email is spam or ham using a single binary variable, namely if the subject line contains the wo... | Why is it that if you undersample or oversample you have to calibrate your output probabilities?
I'm taking a lot of my answer from Agresti's "Categorical Data Analysis". For some context, let's say we want to predict if an email is spam or ham using a single binary variable, namely if the subje |
48,935 | Linear regression to predict both mean and SD of dependent variable | As you are interested in modeling percentiles, you should have a look at quantile regression methods. Instead of modeling conditional means (as in linear regression), quantile regression allows you to model (conditional) quantiles.
As mentioned in the comments, a good introduction to quantile regression is the vignette... | Linear regression to predict both mean and SD of dependent variable | As you are interested in modeling percentiles, you should have a look at quantile regression methods. Instead of modeling conditional means (as in linear regression), quantile regression allows you to | Linear regression to predict both mean and SD of dependent variable
As you are interested in modeling percentiles, you should have a look at quantile regression methods. Instead of modeling conditional means (as in linear regression), quantile regression allows you to model (conditional) quantiles.
As mentioned in the ... | Linear regression to predict both mean and SD of dependent variable
As you are interested in modeling percentiles, you should have a look at quantile regression methods. Instead of modeling conditional means (as in linear regression), quantile regression allows you to |
48,936 | Linear regression to predict both mean and SD of dependent variable | In this case, it would make sense to try to predict the log of daily expenditure - this will probably be closer to linear and so easier to predict. | Linear regression to predict both mean and SD of dependent variable | In this case, it would make sense to try to predict the log of daily expenditure - this will probably be closer to linear and so easier to predict. | Linear regression to predict both mean and SD of dependent variable
In this case, it would make sense to try to predict the log of daily expenditure - this will probably be closer to linear and so easier to predict. | Linear regression to predict both mean and SD of dependent variable
In this case, it would make sense to try to predict the log of daily expenditure - this will probably be closer to linear and so easier to predict. |
48,937 | What is "natural" about the natural parameterization of an exponential family and the natural parameter space? | "Natural" is the qualification chosen for this particular class of exponential families, so one can accept it as a definition without seeking further reasons. (The alternative qualification "canonical" is also sometimes employed.)
If one looks at the generic production of exponential families, one starts$ ^\dagger$ wit... | What is "natural" about the natural parameterization of an exponential family and the natural parame | "Natural" is the qualification chosen for this particular class of exponential families, so one can accept it as a definition without seeking further reasons. (The alternative qualification "canonical | What is "natural" about the natural parameterization of an exponential family and the natural parameter space?
"Natural" is the qualification chosen for this particular class of exponential families, so one can accept it as a definition without seeking further reasons. (The alternative qualification "canonical" is also... | What is "natural" about the natural parameterization of an exponential family and the natural parame
"Natural" is the qualification chosen for this particular class of exponential families, so one can accept it as a definition without seeking further reasons. (The alternative qualification "canonical |
48,938 | What is the difference between a non-zero nugget and a noise term in Kriging/GPR? | Random noise and nugget effect are indeed quite similar to some extent. The difference between the two appears
when there are repeated observations (i.e., several observations at the same location), and
when you compute the predicted value at an observation point.
The random noise model assumes that observations are... | What is the difference between a non-zero nugget and a noise term in Kriging/GPR? | Random noise and nugget effect are indeed quite similar to some extent. The difference between the two appears
when there are repeated observations (i.e., several observations at the same location), | What is the difference between a non-zero nugget and a noise term in Kriging/GPR?
Random noise and nugget effect are indeed quite similar to some extent. The difference between the two appears
when there are repeated observations (i.e., several observations at the same location), and
when you compute the predicted va... | What is the difference between a non-zero nugget and a noise term in Kriging/GPR?
Random noise and nugget effect are indeed quite similar to some extent. The difference between the two appears
when there are repeated observations (i.e., several observations at the same location), |
48,939 | What is the frequentist interpretation of uncertainty vs. variability? | Short answer: Any uncertainty (confidence) we have surrounding an unknown population quantity is due to the sampling variability of the estimation or testing procedure that uses a limited sample size.
Full answer: To the frequentist, population-level quantities (typically denoted by greek characters) are fixed and unkn... | What is the frequentist interpretation of uncertainty vs. variability? | Short answer: Any uncertainty (confidence) we have surrounding an unknown population quantity is due to the sampling variability of the estimation or testing procedure that uses a limited sample size. | What is the frequentist interpretation of uncertainty vs. variability?
Short answer: Any uncertainty (confidence) we have surrounding an unknown population quantity is due to the sampling variability of the estimation or testing procedure that uses a limited sample size.
Full answer: To the frequentist, population-leve... | What is the frequentist interpretation of uncertainty vs. variability?
Short answer: Any uncertainty (confidence) we have surrounding an unknown population quantity is due to the sampling variability of the estimation or testing procedure that uses a limited sample size. |
48,940 | What is the frequentist interpretation of uncertainty vs. variability? | Uncertainty means we do not know the value (or outcome) of some quantity, eg the average porosity of a specific reservoir (or the porosity of a core-sized piece of rock at some point within the reservoir). Variability refers to the multiple values a quantity has at different locations, times or instances – eg the avera... | What is the frequentist interpretation of uncertainty vs. variability? | Uncertainty means we do not know the value (or outcome) of some quantity, eg the average porosity of a specific reservoir (or the porosity of a core-sized piece of rock at some point within the reserv | What is the frequentist interpretation of uncertainty vs. variability?
Uncertainty means we do not know the value (or outcome) of some quantity, eg the average porosity of a specific reservoir (or the porosity of a core-sized piece of rock at some point within the reservoir). Variability refers to the multiple values a... | What is the frequentist interpretation of uncertainty vs. variability?
Uncertainty means we do not know the value (or outcome) of some quantity, eg the average porosity of a specific reservoir (or the porosity of a core-sized piece of rock at some point within the reserv |
48,941 | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected loss? | I think the answer is no as the expected loss depends on $P(Y = 1)$, and this information isn't given by a ROC curve.
Let's say you have a binary random variable $Y$ with $p = P(Y = 1)$, and note $\hat{Y}_t$ a classifier depending on a treshold (or more generally a parameter) $t$.
The expected loss of classifier $\hat{... | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected | I think the answer is no as the expected loss depends on $P(Y = 1)$, and this information isn't given by a ROC curve.
Let's say you have a binary random variable $Y$ with $p = P(Y = 1)$, and note $\ha | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected loss?
I think the answer is no as the expected loss depends on $P(Y = 1)$, and this information isn't given by a ROC curve.
Let's say you have a binary random variable $Y$ with $p = P(Y = 1)$, and note $\hat{Y}_t$ a clas... | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected
I think the answer is no as the expected loss depends on $P(Y = 1)$, and this information isn't given by a ROC curve.
Let's say you have a binary random variable $Y$ with $p = P(Y = 1)$, and note $\ha |
48,942 | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected loss? | Given a threshold* $t$, model 1 has lower estimated expected loss than model 2 if the corresponding ROC point of model 1 dominates** the ROC point of model 2. Here is why.
Let the confusion matrix corresponding to a particular threshold $t$ be
$$
\text{Conf}_t=\begin{pmatrix} j_t & k_t\\ l_t & m_t \end{pmatrix}
$$
with... | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected | Given a threshold* $t$, model 1 has lower estimated expected loss than model 2 if the corresponding ROC point of model 1 dominates** the ROC point of model 2. Here is why.
Let the confusion matrix cor | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected loss?
Given a threshold* $t$, model 1 has lower estimated expected loss than model 2 if the corresponding ROC point of model 1 dominates** the ROC point of model 2. Here is why.
Let the confusion matrix corresponding to ... | Is a pair of threshold-specific points on two ROC curves sufficient to rank classifiers by expected
Given a threshold* $t$, model 1 has lower estimated expected loss than model 2 if the corresponding ROC point of model 1 dominates** the ROC point of model 2. Here is why.
Let the confusion matrix cor |
48,943 | Is there an intuition to the mean of a Gumbel distribution being the Euler constant vis-à-vis the modeling of extreme events? | The Euler-Mascheroni constant $\gamma$ is like many other constants ($e$, $\pi$, and , $\varphi$) used in a simple pattern. That is why it shows up in many places and also in different fields that do not seem to be connected at first glance.
Below we give two 'reasons' behind the occurrence of $\gamma$ in the expressio... | Is there an intuition to the mean of a Gumbel distribution being the Euler constant vis-à-vis the mo | The Euler-Mascheroni constant $\gamma$ is like many other constants ($e$, $\pi$, and , $\varphi$) used in a simple pattern. That is why it shows up in many places and also in different fields that do | Is there an intuition to the mean of a Gumbel distribution being the Euler constant vis-à-vis the modeling of extreme events?
The Euler-Mascheroni constant $\gamma$ is like many other constants ($e$, $\pi$, and , $\varphi$) used in a simple pattern. That is why it shows up in many places and also in different fields th... | Is there an intuition to the mean of a Gumbel distribution being the Euler constant vis-à-vis the mo
The Euler-Mascheroni constant $\gamma$ is like many other constants ($e$, $\pi$, and , $\varphi$) used in a simple pattern. That is why it shows up in many places and also in different fields that do |
48,944 | When to use dot-product as a similarity metric | Euclidean distance (norm of difference) and dot predict are proportional to each other, while they are not equal, but roughly the same.
After normalizing $a$ and $b$ such that$\|a\| = 1$ and $\|b\| = 1$,
these three measures are related as:
Euclidean distance = $\| a - b \| = \sqrt{\| a \|^2 + \|b\|^2 - 2
a^Tb} =\sqr... | When to use dot-product as a similarity metric | Euclidean distance (norm of difference) and dot predict are proportional to each other, while they are not equal, but roughly the same.
After normalizing $a$ and $b$ such that$\|a\| = 1$ and $\|b\| = | When to use dot-product as a similarity metric
Euclidean distance (norm of difference) and dot predict are proportional to each other, while they are not equal, but roughly the same.
After normalizing $a$ and $b$ such that$\|a\| = 1$ and $\|b\| = 1$,
these three measures are related as:
Euclidean distance = $\| a - b ... | When to use dot-product as a similarity metric
Euclidean distance (norm of difference) and dot predict are proportional to each other, while they are not equal, but roughly the same.
After normalizing $a$ and $b$ such that$\|a\| = 1$ and $\|b\| = |
48,945 | Loss function in Supervised Learning vs Statistical Decision Theory | I would say this is more a difference in the form of the decision than the loss. The loss function in both cases is Loss(true state of nature, your decision), but it simplifies differently depending on the form of the decision
In point prediction settings (such as a lot of ML), the decision is a potential value of the ... | Loss function in Supervised Learning vs Statistical Decision Theory | I would say this is more a difference in the form of the decision than the loss. The loss function in both cases is Loss(true state of nature, your decision), but it simplifies differently depending o | Loss function in Supervised Learning vs Statistical Decision Theory
I would say this is more a difference in the form of the decision than the loss. The loss function in both cases is Loss(true state of nature, your decision), but it simplifies differently depending on the form of the decision
In point prediction setti... | Loss function in Supervised Learning vs Statistical Decision Theory
I would say this is more a difference in the form of the decision than the loss. The loss function in both cases is Loss(true state of nature, your decision), but it simplifies differently depending o |
48,946 | Estimating growth rate from noisy data? | It makes sense to smooth and then calculate growth rate.
I assume you are interested in the growth rate of the latent measurement (i.e. the noise free value). Smoothing is a method of estimating this latent quantity, and so calculating the growth rate on the smooth is closer to what you want. | Estimating growth rate from noisy data? | It makes sense to smooth and then calculate growth rate.
I assume you are interested in the growth rate of the latent measurement (i.e. the noise free value). Smoothing is a method of estimating this | Estimating growth rate from noisy data?
It makes sense to smooth and then calculate growth rate.
I assume you are interested in the growth rate of the latent measurement (i.e. the noise free value). Smoothing is a method of estimating this latent quantity, and so calculating the growth rate on the smooth is closer to ... | Estimating growth rate from noisy data?
It makes sense to smooth and then calculate growth rate.
I assume you are interested in the growth rate of the latent measurement (i.e. the noise free value). Smoothing is a method of estimating this |
48,947 | Estimating growth rate from noisy data? | I think that the smooth-then-derive method is OK. However there are ways to avoid the 2 steps, I think. For example, one could model the derivative of the trend directly. The idea is the following:
We observe a series of data $\{x_{i}, y_{i}\}_{i=1}^{n}$ with $y_{i} \sim \mathcal{N}(\mu_{i}, \sigma^{2})$
$\mu_{i}$ is ... | Estimating growth rate from noisy data? | I think that the smooth-then-derive method is OK. However there are ways to avoid the 2 steps, I think. For example, one could model the derivative of the trend directly. The idea is the following:
W | Estimating growth rate from noisy data?
I think that the smooth-then-derive method is OK. However there are ways to avoid the 2 steps, I think. For example, one could model the derivative of the trend directly. The idea is the following:
We observe a series of data $\{x_{i}, y_{i}\}_{i=1}^{n}$ with $y_{i} \sim \mathca... | Estimating growth rate from noisy data?
I think that the smooth-then-derive method is OK. However there are ways to avoid the 2 steps, I think. For example, one could model the derivative of the trend directly. The idea is the following:
W |
48,948 | Does the sum of squared "dependent" Gaussian variables still exhibit a Chi-squared distribution? | Suppose we have $X_1 = u_1$ and for $i>1$ $X_i = \rho X_{i-1} + u_i$. I'll set $\sigma^2_u = 1$ for simplicity. Then
$$
\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n\end{bmatrix} = \begin{bmatrix}
1 & 0 & 0 & 0 & \dots & 0 \\
\rho & 1 & 0 & 0 & \dots & 0 \\
\rho^2 & \rho & 1 & 0 & \dots & 0 \\
&&&\vdots&&\\
\rho^{n-1} & ... | Does the sum of squared "dependent" Gaussian variables still exhibit a Chi-squared distribution? | Suppose we have $X_1 = u_1$ and for $i>1$ $X_i = \rho X_{i-1} + u_i$. I'll set $\sigma^2_u = 1$ for simplicity. Then
$$
\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n\end{bmatrix} = \begin{bmatrix}
1 & 0 | Does the sum of squared "dependent" Gaussian variables still exhibit a Chi-squared distribution?
Suppose we have $X_1 = u_1$ and for $i>1$ $X_i = \rho X_{i-1} + u_i$. I'll set $\sigma^2_u = 1$ for simplicity. Then
$$
\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n\end{bmatrix} = \begin{bmatrix}
1 & 0 & 0 & 0 & \dots & 0 \\... | Does the sum of squared "dependent" Gaussian variables still exhibit a Chi-squared distribution?
Suppose we have $X_1 = u_1$ and for $i>1$ $X_i = \rho X_{i-1} + u_i$. I'll set $\sigma^2_u = 1$ for simplicity. Then
$$
\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n\end{bmatrix} = \begin{bmatrix}
1 & 0 |
48,949 | Precision and recall estimated with stratified sampling | I'm not sure I'd call this stratified sampling, but your calculations for precision and recall make sense. It's as if you cluster negative examples into 1K groups and choose a representative among them and calculate approximated statistics based on these ones.
Of course, the effect of this procedure on the training is ... | Precision and recall estimated with stratified sampling | I'm not sure I'd call this stratified sampling, but your calculations for precision and recall make sense. It's as if you cluster negative examples into 1K groups and choose a representative among the | Precision and recall estimated with stratified sampling
I'm not sure I'd call this stratified sampling, but your calculations for precision and recall make sense. It's as if you cluster negative examples into 1K groups and choose a representative among them and calculate approximated statistics based on these ones.
Of ... | Precision and recall estimated with stratified sampling
I'm not sure I'd call this stratified sampling, but your calculations for precision and recall make sense. It's as if you cluster negative examples into 1K groups and choose a representative among the |
48,950 | Are These Conjectures Regarding Sufficient Statistics True? | Nice conjectures you got there; shame if something were to happen to them
(A) is false for all $k>1$ for the simple reason that any vector of real numbers $T_1,...,T_s$ can be reduced to a single real number without loss of information (i.e., $\mathbb{R}^s$ is cardinally equivalent to $\mathbb{R}$; see e.g., here). Fo... | Are These Conjectures Regarding Sufficient Statistics True? | Nice conjectures you got there; shame if something were to happen to them
(A) is false for all $k>1$ for the simple reason that any vector of real numbers $T_1,...,T_s$ can be reduced to a single real | Are These Conjectures Regarding Sufficient Statistics True?
Nice conjectures you got there; shame if something were to happen to them
(A) is false for all $k>1$ for the simple reason that any vector of real numbers $T_1,...,T_s$ can be reduced to a single real number without loss of information (i.e., $\mathbb{R}^s$ is... | Are These Conjectures Regarding Sufficient Statistics True?
Nice conjectures you got there; shame if something were to happen to them
(A) is false for all $k>1$ for the simple reason that any vector of real numbers $T_1,...,T_s$ can be reduced to a single real |
48,951 | Are These Conjectures Regarding Sufficient Statistics True? | $U[0,\theta]$ is a counterexample to C. $T_1=\max_i X_i$ is sufficient for $\theta$ but it is not complete because $$E\left[\frac{n+1}{n}T_1\right]=\theta=E[2\bar X]$$
(I believe A and B are true.) | Are These Conjectures Regarding Sufficient Statistics True? | $U[0,\theta]$ is a counterexample to C. $T_1=\max_i X_i$ is sufficient for $\theta$ but it is not complete because $$E\left[\frac{n+1}{n}T_1\right]=\theta=E[2\bar X]$$
(I believe A and B are true.) | Are These Conjectures Regarding Sufficient Statistics True?
$U[0,\theta]$ is a counterexample to C. $T_1=\max_i X_i$ is sufficient for $\theta$ but it is not complete because $$E\left[\frac{n+1}{n}T_1\right]=\theta=E[2\bar X]$$
(I believe A and B are true.) | Are These Conjectures Regarding Sufficient Statistics True?
$U[0,\theta]$ is a counterexample to C. $T_1=\max_i X_i$ is sufficient for $\theta$ but it is not complete because $$E\left[\frac{n+1}{n}T_1\right]=\theta=E[2\bar X]$$
(I believe A and B are true.) |
48,952 | Role of regression model fit in causal analysis | A DAG is a non-parametric model of the causal relations among a set a variables. The DAG tells you that covariate and state should be adjusted for, but it tells you nothing about how to adjust. Covariate adjustment is one approach, and stratification is another. Here you are proposing three covariate adjustment models:... | Role of regression model fit in causal analysis | A DAG is a non-parametric model of the causal relations among a set a variables. The DAG tells you that covariate and state should be adjusted for, but it tells you nothing about how to adjust. Covari | Role of regression model fit in causal analysis
A DAG is a non-parametric model of the causal relations among a set a variables. The DAG tells you that covariate and state should be adjusted for, but it tells you nothing about how to adjust. Covariate adjustment is one approach, and stratification is another. Here you ... | Role of regression model fit in causal analysis
A DAG is a non-parametric model of the causal relations among a set a variables. The DAG tells you that covariate and state should be adjusted for, but it tells you nothing about how to adjust. Covari |
48,953 | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | We can show this without having to deal with integrals, or the normal density function. It arises from a symmetry property, and the normal density of $x$ decreasing as $|x|$ increases.
First, note that this is trivial for $k \leq 0$, since then $\mathbb{P}(|X| > k) = 1$ for all $\mu$. We therefore only need to examine ... | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | We can show this without having to deal with integrals, or the normal density function. It arises from a symmetry property, and the normal density of $x$ decreasing as $|x|$ increases.
First, note tha | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
We can show this without having to deal with integrals, or the normal density function. It arises from a symmetry property, and the normal density of $x$ decreasing as $|x|$ increases.
First, note that this is trivial for $k \leq... | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
We can show this without having to deal with integrals, or the normal density function. It arises from a symmetry property, and the normal density of $x$ decreasing as $|x|$ increases.
First, note tha |
48,954 | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | A possible approach: wlog we may assume $\mu_1 \ge 0$, $k \ge 0$.
$$
P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt
$$
and it would suffice to show that the integrand is nonnegative for all $t\ge0$. | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | A possible approach: wlog we may assume $\mu_1 \ge 0$, $k \ge 0$.
$$
P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt
$$
and it would suffice to show that | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
A possible approach: wlog we may assume $\mu_1 \ge 0$, $k \ge 0$.
$$
P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt
$$
and it would suffice to show that the integrand is nonnegative ... | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
A possible approach: wlog we may assume $\mu_1 \ge 0$, $k \ge 0$.
$$
P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt
$$
and it would suffice to show that |
48,955 | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | You can reduce your question to the case you understand. Let $s_i$ be the sign of $\mu_i$, that is, $s_i \in \{-1,1\}$ and $s_i \mu_i = |\mu_i|$. Since the question is about the marginals of $(X_1,X_2)$ we can use any coupling between them.
Take $X_1 = \mu_1 + Z$ and $X_2 = \mu_2 + s_1 Z$ where $Z \sim N(0,\sigma^2)$. ... | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? | You can reduce your question to the case you understand. Let $s_i$ be the sign of $\mu_i$, that is, $s_i \in \{-1,1\}$ and $s_i \mu_i = |\mu_i|$. Since the question is about the marginals of $(X_1,X_2 | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
You can reduce your question to the case you understand. Let $s_i$ be the sign of $\mu_i$, that is, $s_i \in \{-1,1\}$ and $s_i \mu_i = |\mu_i|$. Since the question is about the marginals of $(X_1,X_2)$ we can use any coupling be... | Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?
You can reduce your question to the case you understand. Let $s_i$ be the sign of $\mu_i$, that is, $s_i \in \{-1,1\}$ and $s_i \mu_i = |\mu_i|$. Since the question is about the marginals of $(X_1,X_2 |
48,956 | Can you use Lasso for variable selection of fixed effects, then run a mixed model using the fixed effects from Lasso? | There doesn't appear to be a consensus on how to perform variable selection on both fixed and random effects. There are technical papers proposing solutions to this problem, like this paper from Fan and Li.
Bondell et al. argue against separating the fixed and random when performing variable selection, as the structure... | Can you use Lasso for variable selection of fixed effects, then run a mixed model using the fixed ef | There doesn't appear to be a consensus on how to perform variable selection on both fixed and random effects. There are technical papers proposing solutions to this problem, like this paper from Fan a | Can you use Lasso for variable selection of fixed effects, then run a mixed model using the fixed effects from Lasso?
There doesn't appear to be a consensus on how to perform variable selection on both fixed and random effects. There are technical papers proposing solutions to this problem, like this paper from Fan and... | Can you use Lasso for variable selection of fixed effects, then run a mixed model using the fixed ef
There doesn't appear to be a consensus on how to perform variable selection on both fixed and random effects. There are technical papers proposing solutions to this problem, like this paper from Fan a |
48,957 | Bounding sum of quartic deviations from sample mean | So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!
W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \mu$ and show $\sum_i(Z_i-\bar Z)^4 \leq 16 \sum_i Z_i^4 $).
Using the binomial theorem, we obtain
\begin{align} \tag{1}... | Bounding sum of quartic deviations from sample mean | So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!
W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \ | Bounding sum of quartic deviations from sample mean
So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!
W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \mu$ and show $\sum_i(Z_i-\bar Z)^4 \leq 16 \sum_i Z_i^4 $).
Using t... | Bounding sum of quartic deviations from sample mean
So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!
W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \ |
48,958 | Bounding sum of quartic deviations from sample mean | Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$
$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X)\right)$
$=\sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i+b_i\right)\left(2a_i-b_i\right)$, by letting $X_i-\... | Bounding sum of quartic deviations from sample mean | Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$
$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X | Bounding sum of quartic deviations from sample mean
Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$
$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X)\right)$
$=\sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i... | Bounding sum of quartic deviations from sample mean
Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$
$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X |
48,959 | Is it possible Fisher information matrix be indefinite? | From https://en.wikipedia.org/wiki/Observed_information the observed Fisher information matrix is just the negative Hessian of the log likelihood function. If your log likelihood function is not convex then the hessian will not be positive-definite (and thus indefinite).
This is the case if you are using the "Observed ... | Is it possible Fisher information matrix be indefinite? | From https://en.wikipedia.org/wiki/Observed_information the observed Fisher information matrix is just the negative Hessian of the log likelihood function. If your log likelihood function is not conve | Is it possible Fisher information matrix be indefinite?
From https://en.wikipedia.org/wiki/Observed_information the observed Fisher information matrix is just the negative Hessian of the log likelihood function. If your log likelihood function is not convex then the hessian will not be positive-definite (and thus indef... | Is it possible Fisher information matrix be indefinite?
From https://en.wikipedia.org/wiki/Observed_information the observed Fisher information matrix is just the negative Hessian of the log likelihood function. If your log likelihood function is not conve |
48,960 | Best way to construct a QQ-plot | Your normal Q-Q plots look fine. For a sample of size $n=50,$ it seems you may be striving for more precision than Q-Q plots usually provide.
Several styles of normal Q-Q plots are
used to judge normality of a possibly normal sample. Some Q-Q plots put data on the horizontal axis and some on the vertical axis. (Preferr... | Best way to construct a QQ-plot | Your normal Q-Q plots look fine. For a sample of size $n=50,$ it seems you may be striving for more precision than Q-Q plots usually provide.
Several styles of normal Q-Q plots are
used to judge norma | Best way to construct a QQ-plot
Your normal Q-Q plots look fine. For a sample of size $n=50,$ it seems you may be striving for more precision than Q-Q plots usually provide.
Several styles of normal Q-Q plots are
used to judge normality of a possibly normal sample. Some Q-Q plots put data on the horizontal axis and som... | Best way to construct a QQ-plot
Your normal Q-Q plots look fine. For a sample of size $n=50,$ it seems you may be striving for more precision than Q-Q plots usually provide.
Several styles of normal Q-Q plots are
used to judge norma |
48,961 | Deriving the limiting distribution of the Hodges-Le Cam estimator in Bickel and Doksum (2015) | Yes, that's for $\theta>0$, but the argument for $\theta<0$ is exactly symmetric.
The key random variable is the indicator $A_n=\{|\bar X_n|\leq n^{-1/4}\}$. Since $\bar X_n\sim N(0,1/n)$, we know $P(A_n)$ exactly. Asymptotically, if $\theta=0$, $P(A_n)\to 1$. If $\theta\neq 0$, $P(A_n)\to 0$.
Now,
$$\tilde\theta_n=... | Deriving the limiting distribution of the Hodges-Le Cam estimator in Bickel and Doksum (2015) | Yes, that's for $\theta>0$, but the argument for $\theta<0$ is exactly symmetric.
The key random variable is the indicator $A_n=\{|\bar X_n|\leq n^{-1/4}\}$. Since $\bar X_n\sim N(0,1/n)$, we know $P | Deriving the limiting distribution of the Hodges-Le Cam estimator in Bickel and Doksum (2015)
Yes, that's for $\theta>0$, but the argument for $\theta<0$ is exactly symmetric.
The key random variable is the indicator $A_n=\{|\bar X_n|\leq n^{-1/4}\}$. Since $\bar X_n\sim N(0,1/n)$, we know $P(A_n)$ exactly. Asymptotic... | Deriving the limiting distribution of the Hodges-Le Cam estimator in Bickel and Doksum (2015)
Yes, that's for $\theta>0$, but the argument for $\theta<0$ is exactly symmetric.
The key random variable is the indicator $A_n=\{|\bar X_n|\leq n^{-1/4}\}$. Since $\bar X_n\sim N(0,1/n)$, we know $P |
48,962 | Simulate an example showing difference between fixed effects and mixed effects models | You are not doing anything wrong.
Fitting fixed effects for a grouping variable, instead of random intercepts, is a perfectly valid approach, so the results you get will be comparable.
So you might wonder: why bother fitting a mixed effects model at all ? The answer to that is when the number of groups becomes large yo... | Simulate an example showing difference between fixed effects and mixed effects models | You are not doing anything wrong.
Fitting fixed effects for a grouping variable, instead of random intercepts, is a perfectly valid approach, so the results you get will be comparable.
So you might wo | Simulate an example showing difference between fixed effects and mixed effects models
You are not doing anything wrong.
Fitting fixed effects for a grouping variable, instead of random intercepts, is a perfectly valid approach, so the results you get will be comparable.
So you might wonder: why bother fitting a mixed e... | Simulate an example showing difference between fixed effects and mixed effects models
You are not doing anything wrong.
Fitting fixed effects for a grouping variable, instead of random intercepts, is a perfectly valid approach, so the results you get will be comparable.
So you might wo |
48,963 | A proof that the median is a nonlinear statistical functional | Here is a very simple and almost-rigorous argument.
The medians of the two components are $m_1=\mu_1$ and $m_2=\mu_2$. In particular, they are independent of the components' standard deviations $\sigma_1$ and $\sigma_2$. So any weighted linear combination of the components' medians will also be independent of $\sigma_1... | A proof that the median is a nonlinear statistical functional | Here is a very simple and almost-rigorous argument.
The medians of the two components are $m_1=\mu_1$ and $m_2=\mu_2$. In particular, they are independent of the components' standard deviations $\sigm | A proof that the median is a nonlinear statistical functional
Here is a very simple and almost-rigorous argument.
The medians of the two components are $m_1=\mu_1$ and $m_2=\mu_2$. In particular, they are independent of the components' standard deviations $\sigma_1$ and $\sigma_2$. So any weighted linear combination of... | A proof that the median is a nonlinear statistical functional
Here is a very simple and almost-rigorous argument.
The medians of the two components are $m_1=\mu_1$ and $m_2=\mu_2$. In particular, they are independent of the components' standard deviations $\sigm |
48,964 | Trying to understand the logic behind the p-value in hypothesis testing [duplicate] | For a continuous real-valued distribution $P(X = c)=0$, where $c$ is some real-valued constant. So with continuous distributions we tend to talk about the probability of events which are ranges of values, such as $P(X \ge c)$ which covers the range from $c$ to $\infty$, or $P(X\leq c)$ which covers the range from $-\in... | Trying to understand the logic behind the p-value in hypothesis testing [duplicate] | For a continuous real-valued distribution $P(X = c)=0$, where $c$ is some real-valued constant. So with continuous distributions we tend to talk about the probability of events which are ranges of val | Trying to understand the logic behind the p-value in hypothesis testing [duplicate]
For a continuous real-valued distribution $P(X = c)=0$, where $c$ is some real-valued constant. So with continuous distributions we tend to talk about the probability of events which are ranges of values, such as $P(X \ge c)$ which cove... | Trying to understand the logic behind the p-value in hypothesis testing [duplicate]
For a continuous real-valued distribution $P(X = c)=0$, where $c$ is some real-valued constant. So with continuous distributions we tend to talk about the probability of events which are ranges of val |
48,965 | Trying to understand the logic behind the p-value in hypothesis testing [duplicate] | Suppose you have normal data and wonder whether they are consistent
with $H_0: \mu = 50$ or whether to reject $H_0$ in favor of
$H_a: \mu > 50.$ A sample x of size $n = 20$ has mean $\bar X = 51.25,$
and standard deviation $S = 2.954.$
So the sample mean is greater than $50.$ The question is whether it is
sufficiently ... | Trying to understand the logic behind the p-value in hypothesis testing [duplicate] | Suppose you have normal data and wonder whether they are consistent
with $H_0: \mu = 50$ or whether to reject $H_0$ in favor of
$H_a: \mu > 50.$ A sample x of size $n = 20$ has mean $\bar X = 51.25,$
| Trying to understand the logic behind the p-value in hypothesis testing [duplicate]
Suppose you have normal data and wonder whether they are consistent
with $H_0: \mu = 50$ or whether to reject $H_0$ in favor of
$H_a: \mu > 50.$ A sample x of size $n = 20$ has mean $\bar X = 51.25,$
and standard deviation $S = 2.954.$
... | Trying to understand the logic behind the p-value in hypothesis testing [duplicate]
Suppose you have normal data and wonder whether they are consistent
with $H_0: \mu = 50$ or whether to reject $H_0$ in favor of
$H_a: \mu > 50.$ A sample x of size $n = 20$ has mean $\bar X = 51.25,$
|
48,966 | Example where the posterior from Jags and Stan are really different and have real impacts on decisions using the model | Whenever I want to get started with understanding a new statistical topic, I start by reading articles about it. In this case, I'd start with Carpenter et al. "Stan: A Probabilistic Programming Language," in the Journal of Statistical Software which introduces Stan. The first paragraph is enough to get us started.
Th... | Example where the posterior from Jags and Stan are really different and have real impacts on decisio | Whenever I want to get started with understanding a new statistical topic, I start by reading articles about it. In this case, I'd start with Carpenter et al. "Stan: A Probabilistic Programming Langua | Example where the posterior from Jags and Stan are really different and have real impacts on decisions using the model
Whenever I want to get started with understanding a new statistical topic, I start by reading articles about it. In this case, I'd start with Carpenter et al. "Stan: A Probabilistic Programming Languag... | Example where the posterior from Jags and Stan are really different and have real impacts on decisio
Whenever I want to get started with understanding a new statistical topic, I start by reading articles about it. In this case, I'd start with Carpenter et al. "Stan: A Probabilistic Programming Langua |
48,967 | Example where the posterior from Jags and Stan are really different and have real impacts on decisions using the model | Currently there are no obvious real world examples in which using Stan vs JAGS matters (that is, the posterior would be so different that it would induce a substantively different decisions or conclusions). This is backed by personal experience and it was further evidenced by the lack of answers to this very simple que... | Example where the posterior from Jags and Stan are really different and have real impacts on decisio | Currently there are no obvious real world examples in which using Stan vs JAGS matters (that is, the posterior would be so different that it would induce a substantively different decisions or conclus | Example where the posterior from Jags and Stan are really different and have real impacts on decisions using the model
Currently there are no obvious real world examples in which using Stan vs JAGS matters (that is, the posterior would be so different that it would induce a substantively different decisions or conclusi... | Example where the posterior from Jags and Stan are really different and have real impacts on decisio
Currently there are no obvious real world examples in which using Stan vs JAGS matters (that is, the posterior would be so different that it would induce a substantively different decisions or conclus |
48,968 | Origin of terms "sensitivity" and "specificity" | Binney, N., C. Hyde and P. M. Bossuyt. ‘On the origin of sensitivity and specificity’. Annals of Internal Medicine, 174 (2021): 401–7. 10.7326/M20-5028. | Origin of terms "sensitivity" and "specificity" | Binney, N., C. Hyde and P. M. Bossuyt. ‘On the origin of sensitivity and specificity’. Annals of Internal Medicine, 174 (2021): 401–7. 10.7326/M20-5028. | Origin of terms "sensitivity" and "specificity"
Binney, N., C. Hyde and P. M. Bossuyt. ‘On the origin of sensitivity and specificity’. Annals of Internal Medicine, 174 (2021): 401–7. 10.7326/M20-5028. | Origin of terms "sensitivity" and "specificity"
Binney, N., C. Hyde and P. M. Bossuyt. ‘On the origin of sensitivity and specificity’. Annals of Internal Medicine, 174 (2021): 401–7. 10.7326/M20-5028. |
48,969 | Origin of terms "sensitivity" and "specificity" | tl;dr (this answer is a bit of a book report): The terms sensitivity and specificity have their origin in the roots of immunology in the first years of the 20th century, and work towards antibody testing for syphilis.
As @NickCox and others in the published literature have suggested (e.g., Lilienfield, 2007), Yerushal... | Origin of terms "sensitivity" and "specificity" | tl;dr (this answer is a bit of a book report): The terms sensitivity and specificity have their origin in the roots of immunology in the first years of the 20th century, and work towards antibody test | Origin of terms "sensitivity" and "specificity"
tl;dr (this answer is a bit of a book report): The terms sensitivity and specificity have their origin in the roots of immunology in the first years of the 20th century, and work towards antibody testing for syphilis.
As @NickCox and others in the published literature ha... | Origin of terms "sensitivity" and "specificity"
tl;dr (this answer is a bit of a book report): The terms sensitivity and specificity have their origin in the roots of immunology in the first years of the 20th century, and work towards antibody test |
48,970 | Trouble with finding equation for predicting y based on data provided | It is important not to over-interpret these plots. The first plot of residuals vs fitted values is a little misleading in my opinion if you only focus on the red line, partly due to the fairly small sample size.
Yes, the red line has a curved shape, but looking at the data points, it is not clear at all that there is ... | Trouble with finding equation for predicting y based on data provided | It is important not to over-interpret these plots. The first plot of residuals vs fitted values is a little misleading in my opinion if you only focus on the red line, partly due to the fairly small s | Trouble with finding equation for predicting y based on data provided
It is important not to over-interpret these plots. The first plot of residuals vs fitted values is a little misleading in my opinion if you only focus on the red line, partly due to the fairly small sample size.
Yes, the red line has a curved shape,... | Trouble with finding equation for predicting y based on data provided
It is important not to over-interpret these plots. The first plot of residuals vs fitted values is a little misleading in my opinion if you only focus on the red line, partly due to the fairly small s |
48,971 | Viewing a neural net loss function as a Gaussian Process | Disclaimer: I just glimpsed at the linked articles. My answer focuses solely on Gaussian processes per se.
Your example
If you understand your example $f(x)=x^2+Y$ as a random function of $x$, then it is a Gaussian process, if and only if $Y$ is a Gaussian random variable. If it has any other distribution, it is still ... | Viewing a neural net loss function as a Gaussian Process | Disclaimer: I just glimpsed at the linked articles. My answer focuses solely on Gaussian processes per se.
Your example
If you understand your example $f(x)=x^2+Y$ as a random function of $x$, then it | Viewing a neural net loss function as a Gaussian Process
Disclaimer: I just glimpsed at the linked articles. My answer focuses solely on Gaussian processes per se.
Your example
If you understand your example $f(x)=x^2+Y$ as a random function of $x$, then it is a Gaussian process, if and only if $Y$ is a Gaussian random... | Viewing a neural net loss function as a Gaussian Process
Disclaimer: I just glimpsed at the linked articles. My answer focuses solely on Gaussian processes per se.
Your example
If you understand your example $f(x)=x^2+Y$ as a random function of $x$, then it |
48,972 | Viewing a neural net loss function as a Gaussian Process | Neural Networks as Gaussian Processes
Consider a neural network with only one layer (i.e. no hidden layers, i.e. logistic regression): $$\operatorname{reg}: \mathbb{R}^N \to \mathbb{R}^M : \boldsymbol{x} \mapsto \boldsymbol{s} = \boldsymbol{W} \boldsymbol{x}.$$
If we replace the entries in $\boldsymbol{W} \in \mathbb{R... | Viewing a neural net loss function as a Gaussian Process | Neural Networks as Gaussian Processes
Consider a neural network with only one layer (i.e. no hidden layers, i.e. logistic regression): $$\operatorname{reg}: \mathbb{R}^N \to \mathbb{R}^M : \boldsymbol | Viewing a neural net loss function as a Gaussian Process
Neural Networks as Gaussian Processes
Consider a neural network with only one layer (i.e. no hidden layers, i.e. logistic regression): $$\operatorname{reg}: \mathbb{R}^N \to \mathbb{R}^M : \boldsymbol{x} \mapsto \boldsymbol{s} = \boldsymbol{W} \boldsymbol{x}.$$
I... | Viewing a neural net loss function as a Gaussian Process
Neural Networks as Gaussian Processes
Consider a neural network with only one layer (i.e. no hidden layers, i.e. logistic regression): $$\operatorname{reg}: \mathbb{R}^N \to \mathbb{R}^M : \boldsymbol |
48,973 | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$ | I think you have made the right attempt, since the method of zero estimator is the necessary and sufficient condition for a UMVUE to exist. And here is my proof:
Let's first assume there exist a UMVUE $\hat\theta $ for $\theta$. Then, by the theorem, any zero estimator $\eta$ will have:
$E_{\theta}\eta = 0$, for any $... | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$ | I think you have made the right attempt, since the method of zero estimator is the necessary and sufficient condition for a UMVUE to exist. And here is my proof:
Let's first assume there exist a UMVUE | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$
I think you have made the right attempt, since the method of zero estimator is the necessary and sufficient condition for a UMVUE to exist. And here is my proof:
Let's first assume there exist a UMVUE $\hat\theta $ for $\theta$. Then, by ... | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$
I think you have made the right attempt, since the method of zero estimator is the necessary and sufficient condition for a UMVUE to exist. And here is my proof:
Let's first assume there exist a UMVUE |
48,974 | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$ | Let's approach this from first principles.
An estimator $t$ assigns some guess of $\theta$ to any possible outcome $X.$ Since the possible outcomes are integers, write this guess as $t_i$ when $X=i$ for any integer $i.$
We are hoping to find estimators that tend to be close to $\theta$ when $\theta$ is the parameter. ... | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$ | Let's approach this from first principles.
An estimator $t$ assigns some guess of $\theta$ to any possible outcome $X.$ Since the possible outcomes are integers, write this guess as $t_i$ when $X=i$ | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$
Let's approach this from first principles.
An estimator $t$ assigns some guess of $\theta$ to any possible outcome $X.$ Since the possible outcomes are integers, write this guess as $t_i$ when $X=i$ for any integer $i.$
We are hoping to ... | Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$
Let's approach this from first principles.
An estimator $t$ assigns some guess of $\theta$ to any possible outcome $X.$ Since the possible outcomes are integers, write this guess as $t_i$ when $X=i$ |
48,975 | Why are language modeling pre-training objectives considered unsupervised? | Your argument is right. From the perspective of the language model, you have well-defined target labels and use supervise learning methods to teach the model to predict the labels.
Calling it unsupervised pre-training is certainly sort of paper-publishing marketing, but it is not entirely wrong. It is unsupervised from... | Why are language modeling pre-training objectives considered unsupervised? | Your argument is right. From the perspective of the language model, you have well-defined target labels and use supervise learning methods to teach the model to predict the labels.
Calling it unsuperv | Why are language modeling pre-training objectives considered unsupervised?
Your argument is right. From the perspective of the language model, you have well-defined target labels and use supervise learning methods to teach the model to predict the labels.
Calling it unsupervised pre-training is certainly sort of paper-... | Why are language modeling pre-training objectives considered unsupervised?
Your argument is right. From the perspective of the language model, you have well-defined target labels and use supervise learning methods to teach the model to predict the labels.
Calling it unsuperv |
48,976 | 2-dimensional minimal sufficient statistic for $U(-k\theta+k,k\theta+k)$ | Your own attempted answer incorrectly treats $k$ as a fixed value, rather than an index. This gives you an incorrect likelihood function, which means that your subsequent work is also incorrect. We first observe the event equivalence:
$$-\theta k + k \leqslant x_k \leqslant \theta k + k
\quad \quad \quad \iff \quad \... | 2-dimensional minimal sufficient statistic for $U(-k\theta+k,k\theta+k)$ | Your own attempted answer incorrectly treats $k$ as a fixed value, rather than an index. This gives you an incorrect likelihood function, which means that your subsequent work is also incorrect. We | 2-dimensional minimal sufficient statistic for $U(-k\theta+k,k\theta+k)$
Your own attempted answer incorrectly treats $k$ as a fixed value, rather than an index. This gives you an incorrect likelihood function, which means that your subsequent work is also incorrect. We first observe the event equivalence:
$$-\theta ... | 2-dimensional minimal sufficient statistic for $U(-k\theta+k,k\theta+k)$
Your own attempted answer incorrectly treats $k$ as a fixed value, rather than an index. This gives you an incorrect likelihood function, which means that your subsequent work is also incorrect. We |
48,977 | Can adding features to linear regression increase model bias? | I wrote a paper about this. An early version of it is on arxiv here: https://arxiv.org/abs/2003.08449 (edit: The full published version is now available open access here: https://journals.sagepub.com/doi/full/10.1177/0962280221995963).
The short of it is yes under some conditions. Namely, there has to be some bias alre... | Can adding features to linear regression increase model bias? | I wrote a paper about this. An early version of it is on arxiv here: https://arxiv.org/abs/2003.08449 (edit: The full published version is now available open access here: https://journals.sagepub.com/ | Can adding features to linear regression increase model bias?
I wrote a paper about this. An early version of it is on arxiv here: https://arxiv.org/abs/2003.08449 (edit: The full published version is now available open access here: https://journals.sagepub.com/doi/full/10.1177/0962280221995963).
The short of it is yes... | Can adding features to linear regression increase model bias?
I wrote a paper about this. An early version of it is on arxiv here: https://arxiv.org/abs/2003.08449 (edit: The full published version is now available open access here: https://journals.sagepub.com/ |
48,978 | How to deduce the function of each layer in a neural network without being explicitly told beforehand? | I got an answer from my supervisor, so I thought I would post it here since a few people upvoted.
Firstly, always good to be remember that "layer" is a bit or an overload term. For example, there might be many low-level layers in each high-level layer of SRCNN.
At a high level, it appears that SRCNN maps the raw pix... | How to deduce the function of each layer in a neural network without being explicitly told beforehan | I got an answer from my supervisor, so I thought I would post it here since a few people upvoted.
Firstly, always good to be remember that "layer" is a bit or an overload term. For example, there mig | How to deduce the function of each layer in a neural network without being explicitly told beforehand?
I got an answer from my supervisor, so I thought I would post it here since a few people upvoted.
Firstly, always good to be remember that "layer" is a bit or an overload term. For example, there might be many low-le... | How to deduce the function of each layer in a neural network without being explicitly told beforehan
I got an answer from my supervisor, so I thought I would post it here since a few people upvoted.
Firstly, always good to be remember that "layer" is a bit or an overload term. For example, there mig |
48,979 | Making sense of the $\phi_k$ correlation coefficient | I'm one of the $\phi_k$ authors, so happy to help out.
Indeed there is no closed-form formula for phi_k, but it boils down to interpreting the Pearson $\chi^2$ value between two (binned) variables as coming from a tilted bivariate normal distribution. The $\chi^2$ needs to pass a certain noise pedestal, else $\phi_k$ i... | Making sense of the $\phi_k$ correlation coefficient | I'm one of the $\phi_k$ authors, so happy to help out.
Indeed there is no closed-form formula for phi_k, but it boils down to interpreting the Pearson $\chi^2$ value between two (binned) variables as | Making sense of the $\phi_k$ correlation coefficient
I'm one of the $\phi_k$ authors, so happy to help out.
Indeed there is no closed-form formula for phi_k, but it boils down to interpreting the Pearson $\chi^2$ value between two (binned) variables as coming from a tilted bivariate normal distribution. The $\chi^2$ ne... | Making sense of the $\phi_k$ correlation coefficient
I'm one of the $\phi_k$ authors, so happy to help out.
Indeed there is no closed-form formula for phi_k, but it boils down to interpreting the Pearson $\chi^2$ value between two (binned) variables as |
48,980 | Effect size calculation for comparison between medians | Cohen's d is probably not the best effect size statistic for a comparison of medians, since it is based on the mean and standard deviation(s), and is related to the calculations used in a t test.
I haven't seen this anywhere that I remember, but I suppose you could use an analogous statistic that uses the differences i... | Effect size calculation for comparison between medians | Cohen's d is probably not the best effect size statistic for a comparison of medians, since it is based on the mean and standard deviation(s), and is related to the calculations used in a t test.
I ha | Effect size calculation for comparison between medians
Cohen's d is probably not the best effect size statistic for a comparison of medians, since it is based on the mean and standard deviation(s), and is related to the calculations used in a t test.
I haven't seen this anywhere that I remember, but I suppose you could... | Effect size calculation for comparison between medians
Cohen's d is probably not the best effect size statistic for a comparison of medians, since it is based on the mean and standard deviation(s), and is related to the calculations used in a t test.
I ha |
48,981 | Effect size calculation for comparison between medians | Mood's test is really just a chi-square test in disguise (https://en.wikipedia.org/wiki/Median_test).
Hence, I recommend using the measure of effect size for chi-square tests. There are three standards used: phi, the odds ratio, and Cramer's V. The first two are only defined when comparing two samples, while Cramer's V... | Effect size calculation for comparison between medians | Mood's test is really just a chi-square test in disguise (https://en.wikipedia.org/wiki/Median_test).
Hence, I recommend using the measure of effect size for chi-square tests. There are three standard | Effect size calculation for comparison between medians
Mood's test is really just a chi-square test in disguise (https://en.wikipedia.org/wiki/Median_test).
Hence, I recommend using the measure of effect size for chi-square tests. There are three standards used: phi, the odds ratio, and Cramer's V. The first two are on... | Effect size calculation for comparison between medians
Mood's test is really just a chi-square test in disguise (https://en.wikipedia.org/wiki/Median_test).
Hence, I recommend using the measure of effect size for chi-square tests. There are three standard |
48,982 | Markov chain ( Absorption) | It would be overkill to solve this problem using Markov Chain theory: but the underlying concepts will help you frame it an a way that admits a simple solution.
Formulating the problem
The most fundamental concept is that of a state: we may model this situation in terms of 41 distinct positions, or "states," situated a... | Markov chain ( Absorption) | It would be overkill to solve this problem using Markov Chain theory: but the underlying concepts will help you frame it an a way that admits a simple solution.
Formulating the problem
The most fundam | Markov chain ( Absorption)
It would be overkill to solve this problem using Markov Chain theory: but the underlying concepts will help you frame it an a way that admits a simple solution.
Formulating the problem
The most fundamental concept is that of a state: we may model this situation in terms of 41 distinct positio... | Markov chain ( Absorption)
It would be overkill to solve this problem using Markov Chain theory: but the underlying concepts will help you frame it an a way that admits a simple solution.
Formulating the problem
The most fundam |
48,983 | Markov chain ( Absorption) | Imagine that the hill-climbing journey consists of 41 states, one for each meter possible, so states 0, 1, 3, ...., 40. The transition probability matrix then becomes a 41x41 matrix, representing the different probabilities of going from one state to another. It looks like the following:
0 1 2 -- 40
0 0... | Markov chain ( Absorption) | Imagine that the hill-climbing journey consists of 41 states, one for each meter possible, so states 0, 1, 3, ...., 40. The transition probability matrix then becomes a 41x41 matrix, representing the | Markov chain ( Absorption)
Imagine that the hill-climbing journey consists of 41 states, one for each meter possible, so states 0, 1, 3, ...., 40. The transition probability matrix then becomes a 41x41 matrix, representing the different probabilities of going from one state to another. It looks like the following:
0... | Markov chain ( Absorption)
Imagine that the hill-climbing journey consists of 41 states, one for each meter possible, so states 0, 1, 3, ...., 40. The transition probability matrix then becomes a 41x41 matrix, representing the |
48,984 | Confidence interval from summary function | The p value is for a test of the null hypothesis that the estimate is equal to zero. It literally means the probability of observing these data (or data even further from zero), if the parameter for this estimate IS actually zero. Since this probability is extremely small (0.00000003) this is very strong evidence that ... | Confidence interval from summary function | The p value is for a test of the null hypothesis that the estimate is equal to zero. It literally means the probability of observing these data (or data even further from zero), if the parameter for t | Confidence interval from summary function
The p value is for a test of the null hypothesis that the estimate is equal to zero. It literally means the probability of observing these data (or data even further from zero), if the parameter for this estimate IS actually zero. Since this probability is extremely small (0.00... | Confidence interval from summary function
The p value is for a test of the null hypothesis that the estimate is equal to zero. It literally means the probability of observing these data (or data even further from zero), if the parameter for t |
48,985 | Count data with largely varying sample sizes. Can I get anything from the data? | In principle, it seems you'd want to use a chi-squared test
to see if the two groups tend to have the same
distribution of category counts.
In practice, sparse
data as in the last few categories of your first
dataset make it impossible to do a 'standard'
chi-squared test. In particular, several expected cell
counts are... | Count data with largely varying sample sizes. Can I get anything from the data? | In principle, it seems you'd want to use a chi-squared test
to see if the two groups tend to have the same
distribution of category counts.
In practice, sparse
data as in the last few categories of yo | Count data with largely varying sample sizes. Can I get anything from the data?
In principle, it seems you'd want to use a chi-squared test
to see if the two groups tend to have the same
distribution of category counts.
In practice, sparse
data as in the last few categories of your first
dataset make it impossible to d... | Count data with largely varying sample sizes. Can I get anything from the data?
In principle, it seems you'd want to use a chi-squared test
to see if the two groups tend to have the same
distribution of category counts.
In practice, sparse
data as in the last few categories of yo |
48,986 | Can we solve overfitting by adding more parameters? | According to recent works on the Double Descent phenomena, specially Belkin's, yes, you may be able to fix overfitting with more parameters.
That happens because, according to their hypothesis, if you have just enough parameters to interpolate training data, the solution space becomes constrained, precluding you from a... | Can we solve overfitting by adding more parameters? | According to recent works on the Double Descent phenomena, specially Belkin's, yes, you may be able to fix overfitting with more parameters.
That happens because, according to their hypothesis, if you | Can we solve overfitting by adding more parameters?
According to recent works on the Double Descent phenomena, specially Belkin's, yes, you may be able to fix overfitting with more parameters.
That happens because, according to their hypothesis, if you have just enough parameters to interpolate training data, the solut... | Can we solve overfitting by adding more parameters?
According to recent works on the Double Descent phenomena, specially Belkin's, yes, you may be able to fix overfitting with more parameters.
That happens because, according to their hypothesis, if you |
48,987 | Can we solve overfitting by adding more parameters? | Adding parameters will lead to more overfitting. The more parameters, the more models you can represent. The more models, the more likely you'll find one that fits your training data exactly.
To avoid overfitting, choose the simplest model that does not underfit, and use cross-validation to make sure. | Can we solve overfitting by adding more parameters? | Adding parameters will lead to more overfitting. The more parameters, the more models you can represent. The more models, the more likely you'll find one that fits your training data exactly.
To avoid | Can we solve overfitting by adding more parameters?
Adding parameters will lead to more overfitting. The more parameters, the more models you can represent. The more models, the more likely you'll find one that fits your training data exactly.
To avoid overfitting, choose the simplest model that does not underfit, and ... | Can we solve overfitting by adding more parameters?
Adding parameters will lead to more overfitting. The more parameters, the more models you can represent. The more models, the more likely you'll find one that fits your training data exactly.
To avoid |
48,988 | Clustering with large number of clusters | The solution I used, in the end, was my implementation of batched K-Means.
Usual implementations of batched K-Means do both the expectation and the maximization step on a single batch. This is not possible in this case becase the data bach must be smaller than the number of clusters.
The solution is to do the expectati... | Clustering with large number of clusters | The solution I used, in the end, was my implementation of batched K-Means.
Usual implementations of batched K-Means do both the expectation and the maximization step on a single batch. This is not pos | Clustering with large number of clusters
The solution I used, in the end, was my implementation of batched K-Means.
Usual implementations of batched K-Means do both the expectation and the maximization step on a single batch. This is not possible in this case becase the data bach must be smaller than the number of clus... | Clustering with large number of clusters
The solution I used, in the end, was my implementation of batched K-Means.
Usual implementations of batched K-Means do both the expectation and the maximization step on a single batch. This is not pos |
48,989 | two times square in distance calculation on one example? | It's a typo, the kernel is
$$K(x_1,x_2)=\exp\left(-||x_1-x_2||^2 \over 2\sigma^2\right)$$
Note that if you do the calculation w/o squaring again, the result rounded upto two digits will be the same. | two times square in distance calculation on one example? | It's a typo, the kernel is
$$K(x_1,x_2)=\exp\left(-||x_1-x_2||^2 \over 2\sigma^2\right)$$
Note that if you do the calculation w/o squaring again, the result rounded upto two digits will be the same. | two times square in distance calculation on one example?
It's a typo, the kernel is
$$K(x_1,x_2)=\exp\left(-||x_1-x_2||^2 \over 2\sigma^2\right)$$
Note that if you do the calculation w/o squaring again, the result rounded upto two digits will be the same. | two times square in distance calculation on one example?
It's a typo, the kernel is
$$K(x_1,x_2)=\exp\left(-||x_1-x_2||^2 \over 2\sigma^2\right)$$
Note that if you do the calculation w/o squaring again, the result rounded upto two digits will be the same. |
48,990 | Finding the Q function for the EM algorithm | To quote verbatim from Wikipedia
The EM algorithm is used to find (local) maximum likelihood parameters
of a statistical model in cases where the equations cannot be solved
directly. Typically these models involve latent variables in addition
to unknown parameters and known data observations. That is, either
missing v... | Finding the Q function for the EM algorithm | To quote verbatim from Wikipedia
The EM algorithm is used to find (local) maximum likelihood parameters
of a statistical model in cases where the equations cannot be solved
directly. Typically these | Finding the Q function for the EM algorithm
To quote verbatim from Wikipedia
The EM algorithm is used to find (local) maximum likelihood parameters
of a statistical model in cases where the equations cannot be solved
directly. Typically these models involve latent variables in addition
to unknown parameters and known ... | Finding the Q function for the EM algorithm
To quote verbatim from Wikipedia
The EM algorithm is used to find (local) maximum likelihood parameters
of a statistical model in cases where the equations cannot be solved
directly. Typically these |
48,991 | Better confidence intervals for weighted average | An estimator that is obviously better in some ways is
$$\hat\mu= \frac{\sum_{\textrm{observed }k} n_kx_k}{\sum_{\textrm{observed }k} n_k}$$
In particular, if $|J|$ is large enough that all $K$ distinct items
will be observed at least once (with probability going to 1) and the
error of $\hat\mu$ will be exactly zero, wh... | Better confidence intervals for weighted average | An estimator that is obviously better in some ways is
$$\hat\mu= \frac{\sum_{\textrm{observed }k} n_kx_k}{\sum_{\textrm{observed }k} n_k}$$
In particular, if $|J|$ is large enough that all $K$ distinc | Better confidence intervals for weighted average
An estimator that is obviously better in some ways is
$$\hat\mu= \frac{\sum_{\textrm{observed }k} n_kx_k}{\sum_{\textrm{observed }k} n_k}$$
In particular, if $|J|$ is large enough that all $K$ distinct items
will be observed at least once (with probability going to 1) an... | Better confidence intervals for weighted average
An estimator that is obviously better in some ways is
$$\hat\mu= \frac{\sum_{\textrm{observed }k} n_kx_k}{\sum_{\textrm{observed }k} n_k}$$
In particular, if $|J|$ is large enough that all $K$ distinc |
48,992 | Better confidence intervals for weighted average | estimate the variance of the estimator using the usual CLT-based approach.
...
Can I use this information to produce estimates with smaller confidence intervals?
Yes, you can. (This is true in general. In many cases, you can do better than a normal approximation, especially when the distribution is not really a normal... | Better confidence intervals for weighted average | estimate the variance of the estimator using the usual CLT-based approach.
...
Can I use this information to produce estimates with smaller confidence intervals?
Yes, you can. (This is true in genera | Better confidence intervals for weighted average
estimate the variance of the estimator using the usual CLT-based approach.
...
Can I use this information to produce estimates with smaller confidence intervals?
Yes, you can. (This is true in general. In many cases, you can do better than a normal approximation, especi... | Better confidence intervals for weighted average
estimate the variance of the estimator using the usual CLT-based approach.
...
Can I use this information to produce estimates with smaller confidence intervals?
Yes, you can. (This is true in genera |
48,993 | Using n linear regression models for different subsets vs using one model for the entire dataset | Are there any mathematical drawbacks with this idea?
Yes, you will lose statistical power as well as running into multiple testing problems. Don't split the dataset.
How do I compare the resulting n models to my other model? I am unsure on how to interpret the n resulting metrics (i.e., a cross-validated r squared s... | Using n linear regression models for different subsets vs using one model for the entire dataset | Are there any mathematical drawbacks with this idea?
Yes, you will lose statistical power as well as running into multiple testing problems. Don't split the dataset.
How do I compare the resulting | Using n linear regression models for different subsets vs using one model for the entire dataset
Are there any mathematical drawbacks with this idea?
Yes, you will lose statistical power as well as running into multiple testing problems. Don't split the dataset.
How do I compare the resulting n models to my other mo... | Using n linear regression models for different subsets vs using one model for the entire dataset
Are there any mathematical drawbacks with this idea?
Yes, you will lose statistical power as well as running into multiple testing problems. Don't split the dataset.
How do I compare the resulting |
48,994 | Interpreting a zero-inflation negative binomial model | What does the zero-inflation model actually represent?
this is a model for the occurance of non zeros vs zeros. It can be interpreted in the same was as a logistic regression model where success means a non-zero count and you are modelling the probability of obtaining a non-zero count.
Are these p-values sufficient... | Interpreting a zero-inflation negative binomial model | What does the zero-inflation model actually represent?
this is a model for the occurance of non zeros vs zeros. It can be interpreted in the same was as a logistic regression model where success mea | Interpreting a zero-inflation negative binomial model
What does the zero-inflation model actually represent?
this is a model for the occurance of non zeros vs zeros. It can be interpreted in the same was as a logistic regression model where success means a non-zero count and you are modelling the probability of obtai... | Interpreting a zero-inflation negative binomial model
What does the zero-inflation model actually represent?
this is a model for the occurance of non zeros vs zeros. It can be interpreted in the same was as a logistic regression model where success mea |
48,995 | Does Linear regression needs target variable to be normally distributed. (GLM context)? | It depends on what you’re doing. If you just want to predict, then it doesn’t matter, and the Gauss-Markov theorem does not say anything about a normal error term.
However, when the error term is normal, then the OLS estimator $\hat{\beta}$ is the maximum likelihood estimator. If you don’t know about MLEs, you’ll see t... | Does Linear regression needs target variable to be normally distributed. (GLM context)? | It depends on what you’re doing. If you just want to predict, then it doesn’t matter, and the Gauss-Markov theorem does not say anything about a normal error term.
However, when the error term is norm | Does Linear regression needs target variable to be normally distributed. (GLM context)?
It depends on what you’re doing. If you just want to predict, then it doesn’t matter, and the Gauss-Markov theorem does not say anything about a normal error term.
However, when the error term is normal, then the OLS estimator $\hat... | Does Linear regression needs target variable to be normally distributed. (GLM context)?
It depends on what you’re doing. If you just want to predict, then it doesn’t matter, and the Gauss-Markov theorem does not say anything about a normal error term.
However, when the error term is norm |
48,996 | Does Linear regression needs target variable to be normally distributed. (GLM context)? | The only normality assumption in linear regression if you intend to do any testing is that the residuals be normally distributed. In simple linear regression with only one variable in the model, this implies that the independent variable must also be normally distributed. In multiple regression, however, the residual... | Does Linear regression needs target variable to be normally distributed. (GLM context)? | The only normality assumption in linear regression if you intend to do any testing is that the residuals be normally distributed. In simple linear regression with only one variable in the model, this | Does Linear regression needs target variable to be normally distributed. (GLM context)?
The only normality assumption in linear regression if you intend to do any testing is that the residuals be normally distributed. In simple linear regression with only one variable in the model, this implies that the independent va... | Does Linear regression needs target variable to be normally distributed. (GLM context)?
The only normality assumption in linear regression if you intend to do any testing is that the residuals be normally distributed. In simple linear regression with only one variable in the model, this |
48,997 | Hypothesis testing for logistic regression with categorical predictor | In R you can use glht (generalized linear hypothesis test) command in the multcomp package to define the contrasts of interest and test them, though I would advise against much reliance on p value "significance" | Hypothesis testing for logistic regression with categorical predictor | In R you can use glht (generalized linear hypothesis test) command in the multcomp package to define the contrasts of interest and test them, though I would advise against much reliance on p value "si | Hypothesis testing for logistic regression with categorical predictor
In R you can use glht (generalized linear hypothesis test) command in the multcomp package to define the contrasts of interest and test them, though I would advise against much reliance on p value "significance" | Hypothesis testing for logistic regression with categorical predictor
In R you can use glht (generalized linear hypothesis test) command in the multcomp package to define the contrasts of interest and test them, though I would advise against much reliance on p value "si |
48,998 | Hypothesis testing for logistic regression with categorical predictor | Use the likelihood ratio test (LRT) to compare the different β's as described here: https://courses.washington.edu/b515/l13.pdf
pp. 20-24. And if you don't follow it, start from page 1. | Hypothesis testing for logistic regression with categorical predictor | Use the likelihood ratio test (LRT) to compare the different β's as described here: https://courses.washington.edu/b515/l13.pdf
pp. 20-24. And if you don't follow it, start from page 1. | Hypothesis testing for logistic regression with categorical predictor
Use the likelihood ratio test (LRT) to compare the different β's as described here: https://courses.washington.edu/b515/l13.pdf
pp. 20-24. And if you don't follow it, start from page 1. | Hypothesis testing for logistic regression with categorical predictor
Use the likelihood ratio test (LRT) to compare the different β's as described here: https://courses.washington.edu/b515/l13.pdf
pp. 20-24. And if you don't follow it, start from page 1. |
48,999 | Is the regressor (sometimes called "independent" variable) actually independent of the response from a probabilistic perspective? | The "dependent" and "independent" terminology for the variables is unfortunate terminology, which is best avoided. Statistical dependence is always bidirectional ---i.e., if a variable is statistically dependent on another variable, then that second variable is also statistically dependent with the first variable. In... | Is the regressor (sometimes called "independent" variable) actually independent of the response from | The "dependent" and "independent" terminology for the variables is unfortunate terminology, which is best avoided. Statistical dependence is always bidirectional ---i.e., if a variable is statistical | Is the regressor (sometimes called "independent" variable) actually independent of the response from a probabilistic perspective?
The "dependent" and "independent" terminology for the variables is unfortunate terminology, which is best avoided. Statistical dependence is always bidirectional ---i.e., if a variable is s... | Is the regressor (sometimes called "independent" variable) actually independent of the response from
The "dependent" and "independent" terminology for the variables is unfortunate terminology, which is best avoided. Statistical dependence is always bidirectional ---i.e., if a variable is statistical |
49,000 | Confidence interval example in Computer Age Statistical Inference | A confidence interval estimate relates to an interval that has $\alpha \%$ probability to contain the parameter, conditional on the parameter.
This contrasts with a credible interval, which has $\alpha \%$ probability to contain the parameter, conditional on the observation.
See the example from this question/answer wh... | Confidence interval example in Computer Age Statistical Inference | A confidence interval estimate relates to an interval that has $\alpha \%$ probability to contain the parameter, conditional on the parameter.
This contrasts with a credible interval, which has $\alph | Confidence interval example in Computer Age Statistical Inference
A confidence interval estimate relates to an interval that has $\alpha \%$ probability to contain the parameter, conditional on the parameter.
This contrasts with a credible interval, which has $\alpha \%$ probability to contain the parameter, conditiona... | Confidence interval example in Computer Age Statistical Inference
A confidence interval estimate relates to an interval that has $\alpha \%$ probability to contain the parameter, conditional on the parameter.
This contrasts with a credible interval, which has $\alph |
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