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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46,001 | How to mathematically denote common elements | If it's reasonable to treat them as sets (e.g. each distinct element can only ever appear once, order doesn't matter, etc), then you can use set-intersection to denote the elements in common:
$A_1 = \{8,1,5,6\}$
$A_2 = \{6,8\}$
... etc
Then $A_1\cap A_2\cap A_3\cap A_4\cap A_5 = \{6,8\}$
If you have a large number of s... | How to mathematically denote common elements | If it's reasonable to treat them as sets (e.g. each distinct element can only ever appear once, order doesn't matter, etc), then you can use set-intersection to denote the elements in common:
$A_1 = \ | How to mathematically denote common elements
If it's reasonable to treat them as sets (e.g. each distinct element can only ever appear once, order doesn't matter, etc), then you can use set-intersection to denote the elements in common:
$A_1 = \{8,1,5,6\}$
$A_2 = \{6,8\}$
... etc
Then $A_1\cap A_2\cap A_3\cap A_4\cap A... | How to mathematically denote common elements
If it's reasonable to treat them as sets (e.g. each distinct element can only ever appear once, order doesn't matter, etc), then you can use set-intersection to denote the elements in common:
$A_1 = \ |
46,002 | What is the loss function used for CNN? | In most cases CNNs use a cross-entropy loss on the one-hot encoded output. For a single image the cross entropy loss looks like this:
$$
- \sum_{c=1}^M{(y_c \cdot \log{\hat y_c})}
$$
where $M$ is the number of classes (i.e. $1000$ in ImageNet) and $\hat y_c$ is the model's prediction for that class (i.e. the output of ... | What is the loss function used for CNN? | In most cases CNNs use a cross-entropy loss on the one-hot encoded output. For a single image the cross entropy loss looks like this:
$$
- \sum_{c=1}^M{(y_c \cdot \log{\hat y_c})}
$$
where $M$ is the | What is the loss function used for CNN?
In most cases CNNs use a cross-entropy loss on the one-hot encoded output. For a single image the cross entropy loss looks like this:
$$
- \sum_{c=1}^M{(y_c \cdot \log{\hat y_c})}
$$
where $M$ is the number of classes (i.e. $1000$ in ImageNet) and $\hat y_c$ is the model's predic... | What is the loss function used for CNN?
In most cases CNNs use a cross-entropy loss on the one-hot encoded output. For a single image the cross entropy loss looks like this:
$$
- \sum_{c=1}^M{(y_c \cdot \log{\hat y_c})}
$$
where $M$ is the |
46,003 | What is the loss function used for CNN? | As Jan says in a comment, AlexNet uses cross entropy as the loss function.
It's important to note, though, that a Convolutional Neural Network describes the architecture of the network, not the goal of the network. It is the goal of a network that determines the loss function.
CNN architectures can be used for many tas... | What is the loss function used for CNN? | As Jan says in a comment, AlexNet uses cross entropy as the loss function.
It's important to note, though, that a Convolutional Neural Network describes the architecture of the network, not the goal o | What is the loss function used for CNN?
As Jan says in a comment, AlexNet uses cross entropy as the loss function.
It's important to note, though, that a Convolutional Neural Network describes the architecture of the network, not the goal of the network. It is the goal of a network that determines the loss function.
CN... | What is the loss function used for CNN?
As Jan says in a comment, AlexNet uses cross entropy as the loss function.
It's important to note, though, that a Convolutional Neural Network describes the architecture of the network, not the goal o |
46,004 | Behaviour of Welch's t-test with unequal group sizes | This question is sufficiently broad that a comprehensive answer would require
a simulation study (some key parts of which have undoubtedly been done), going far beyond
our usual style of answers.
A Welch 2-sample t test can't be exactly as good as a pooled t test if
we know the two populations have the same variance.
(... | Behaviour of Welch's t-test with unequal group sizes | This question is sufficiently broad that a comprehensive answer would require
a simulation study (some key parts of which have undoubtedly been done), going far beyond
our usual style of answers.
A We | Behaviour of Welch's t-test with unequal group sizes
This question is sufficiently broad that a comprehensive answer would require
a simulation study (some key parts of which have undoubtedly been done), going far beyond
our usual style of answers.
A Welch 2-sample t test can't be exactly as good as a pooled t test if
... | Behaviour of Welch's t-test with unequal group sizes
This question is sufficiently broad that a comprehensive answer would require
a simulation study (some key parts of which have undoubtedly been done), going far beyond
our usual style of answers.
A We |
46,005 | Behaviour of Welch's t-test with unequal group sizes | The issue here does not appear to be due to imbalanced sample sizes per se, but rather, non-uniformity of the p-value seems to be occurring due to using very small sample sizes for one of the samples. Below I show high-resolution histograms showing simulated p-values for a comparison of one sample with $n_X = 100$ sta... | Behaviour of Welch's t-test with unequal group sizes | The issue here does not appear to be due to imbalanced sample sizes per se, but rather, non-uniformity of the p-value seems to be occurring due to using very small sample sizes for one of the samples. | Behaviour of Welch's t-test with unequal group sizes
The issue here does not appear to be due to imbalanced sample sizes per se, but rather, non-uniformity of the p-value seems to be occurring due to using very small sample sizes for one of the samples. Below I show high-resolution histograms showing simulated p-value... | Behaviour of Welch's t-test with unequal group sizes
The issue here does not appear to be due to imbalanced sample sizes per se, but rather, non-uniformity of the p-value seems to be occurring due to using very small sample sizes for one of the samples. |
46,006 | Regression model with aggregated targets | Here's an approach for solving this type of problem using latent variable models. It's not a specific model, but a general way to formulate a model by breaking the description of the system into two parts: the relationship between individual inputs and (unobserved) individual outputs, and the relationship between indiv... | Regression model with aggregated targets | Here's an approach for solving this type of problem using latent variable models. It's not a specific model, but a general way to formulate a model by breaking the description of the system into two p | Regression model with aggregated targets
Here's an approach for solving this type of problem using latent variable models. It's not a specific model, but a general way to formulate a model by breaking the description of the system into two parts: the relationship between individual inputs and (unobserved) individual ou... | Regression model with aggregated targets
Here's an approach for solving this type of problem using latent variable models. It's not a specific model, but a general way to formulate a model by breaking the description of the system into two p |
46,007 | Regression model with aggregated targets | To verify the solution suggested in the great answer by @user20160 I prepared a toy example that demonstrates it. As suggested by @user20160, I am posting the code as a supplement to the answer. For explanations of this approach, check the other answer.
First, let's generate the independent variable and append the colu... | Regression model with aggregated targets | To verify the solution suggested in the great answer by @user20160 I prepared a toy example that demonstrates it. As suggested by @user20160, I am posting the code as a supplement to the answer. For e | Regression model with aggregated targets
To verify the solution suggested in the great answer by @user20160 I prepared a toy example that demonstrates it. As suggested by @user20160, I am posting the code as a supplement to the answer. For explanations of this approach, check the other answer.
First, let's generate the... | Regression model with aggregated targets
To verify the solution suggested in the great answer by @user20160 I prepared a toy example that demonstrates it. As suggested by @user20160, I am posting the code as a supplement to the answer. For e |
46,008 | Regression model with aggregated targets | Different approaches could be appropriate depending on your goal. I'll describe one approach in case your goal is group-level prediction.
You could use the individual-level features to build a bunch of aggregated features for each group (mean, std, median, max, min, ...). You now have richer features for each group whi... | Regression model with aggregated targets | Different approaches could be appropriate depending on your goal. I'll describe one approach in case your goal is group-level prediction.
You could use the individual-level features to build a bunch o | Regression model with aggregated targets
Different approaches could be appropriate depending on your goal. I'll describe one approach in case your goal is group-level prediction.
You could use the individual-level features to build a bunch of aggregated features for each group (mean, std, median, max, min, ...). You no... | Regression model with aggregated targets
Different approaches could be appropriate depending on your goal. I'll describe one approach in case your goal is group-level prediction.
You could use the individual-level features to build a bunch o |
46,009 | Why does a positively correlated variable have a negative coefficient in a multiple regression? | You are comparing two very different things. In the first case, you are making pairwise comparisons when calculating the correlation coefficient between BodyFat and Weight. In the second, you are doing a multiple regression that also accounts for the variation in BodyFat that is explained by all your other variables.
... | Why does a positively correlated variable have a negative coefficient in a multiple regression? | You are comparing two very different things. In the first case, you are making pairwise comparisons when calculating the correlation coefficient between BodyFat and Weight. In the second, you are doin | Why does a positively correlated variable have a negative coefficient in a multiple regression?
You are comparing two very different things. In the first case, you are making pairwise comparisons when calculating the correlation coefficient between BodyFat and Weight. In the second, you are doing a multiple regression ... | Why does a positively correlated variable have a negative coefficient in a multiple regression?
You are comparing two very different things. In the first case, you are making pairwise comparisons when calculating the correlation coefficient between BodyFat and Weight. In the second, you are doin |
46,010 | Why does a positively correlated variable have a negative coefficient in a multiple regression? | To add to @mkt 's answer, which does capture all the most critical mathematical aspects, a few observations:
The intercept CI spans from -70 to -33. Assuming that body fat is a percentage, then this means that the baseline amount of fat in the cohort is very variable.
If the distribution of BodyFat is left or right t... | Why does a positively correlated variable have a negative coefficient in a multiple regression? | To add to @mkt 's answer, which does capture all the most critical mathematical aspects, a few observations:
The intercept CI spans from -70 to -33. Assuming that body fat is a percentage, then this | Why does a positively correlated variable have a negative coefficient in a multiple regression?
To add to @mkt 's answer, which does capture all the most critical mathematical aspects, a few observations:
The intercept CI spans from -70 to -33. Assuming that body fat is a percentage, then this means that the baseline ... | Why does a positively correlated variable have a negative coefficient in a multiple regression?
To add to @mkt 's answer, which does capture all the most critical mathematical aspects, a few observations:
The intercept CI spans from -70 to -33. Assuming that body fat is a percentage, then this |
46,011 | Questions regarding proof of probability integral transform | The notation is getting in the way, so let's simplify it.
Let $X:\Omega\to\mathbb{R}$ be a random variable with a distribution function $F_X$ defined by
$$F_X(x) = \Pr(X \le x) = \Pr(\{\omega\in\Omega\mid X(\omega)\le x\})$$
for all real numbers $x.$ The axioms of probability imply $F_X$ is non-decreasing and at any ... | Questions regarding proof of probability integral transform | The notation is getting in the way, so let's simplify it.
Let $X:\Omega\to\mathbb{R}$ be a random variable with a distribution function $F_X$ defined by
$$F_X(x) = \Pr(X \le x) = \Pr(\{\omega\in\Omeg | Questions regarding proof of probability integral transform
The notation is getting in the way, so let's simplify it.
Let $X:\Omega\to\mathbb{R}$ be a random variable with a distribution function $F_X$ defined by
$$F_X(x) = \Pr(X \le x) = \Pr(\{\omega\in\Omega\mid X(\omega)\le x\})$$
for all real numbers $x.$ The axi... | Questions regarding proof of probability integral transform
The notation is getting in the way, so let's simplify it.
Let $X:\Omega\to\mathbb{R}$ be a random variable with a distribution function $F_X$ defined by
$$F_X(x) = \Pr(X \le x) = \Pr(\{\omega\in\Omeg |
46,012 | Questions regarding proof of probability integral transform | After having done some homework on the subject, I think I've got a better handle on the proof that I find in [1]. I wanted to take an opportunity to set down my understanding for pedagogic purposes.
Scope:
I'm going to limit this answer to the case of a strictly monotonic cumulative distribution function. Its my under... | Questions regarding proof of probability integral transform | After having done some homework on the subject, I think I've got a better handle on the proof that I find in [1]. I wanted to take an opportunity to set down my understanding for pedagogic purposes.
| Questions regarding proof of probability integral transform
After having done some homework on the subject, I think I've got a better handle on the proof that I find in [1]. I wanted to take an opportunity to set down my understanding for pedagogic purposes.
Scope:
I'm going to limit this answer to the case of a stric... | Questions regarding proof of probability integral transform
After having done some homework on the subject, I think I've got a better handle on the proof that I find in [1]. I wanted to take an opportunity to set down my understanding for pedagogic purposes.
|
46,013 | Probability distribution of the number of infected people in a room | Let $\mathbf{K} = \{ K_h | h \in \mathbb{N}_{0+} \}$ denote the stochastic time-series showing the number of infected people after each handshake, and let $K_0 = 1$ at the start of the series. This is a Markov chain that falls within the category of discrete "pure birth" processes". A single random handshake gives th... | Probability distribution of the number of infected people in a room | Let $\mathbf{K} = \{ K_h | h \in \mathbb{N}_{0+} \}$ denote the stochastic time-series showing the number of infected people after each handshake, and let $K_0 = 1$ at the start of the series. This i | Probability distribution of the number of infected people in a room
Let $\mathbf{K} = \{ K_h | h \in \mathbb{N}_{0+} \}$ denote the stochastic time-series showing the number of infected people after each handshake, and let $K_0 = 1$ at the start of the series. This is a Markov chain that falls within the category of d... | Probability distribution of the number of infected people in a room
Let $\mathbf{K} = \{ K_h | h \in \mathbb{N}_{0+} \}$ denote the stochastic time-series showing the number of infected people after each handshake, and let $K_0 = 1$ at the start of the series. This i |
46,014 | Probability distribution of the number of infected people in a room | Here is an alternative to Ben's answer using simulations in R, using his parameters.
Edit: fixed the bug.
N=40 #number of people
h=80 #handshakes
k=1 #number of infected people at the start
n=1e5 #number of simulations
result=rep(NA,n)
for (r in 1:n) {
initial=rep(0,N) #N healthy people
initial[1:k]=1 #k infected
... | Probability distribution of the number of infected people in a room | Here is an alternative to Ben's answer using simulations in R, using his parameters.
Edit: fixed the bug.
N=40 #number of people
h=80 #handshakes
k=1 #number of infected people at the start
n=1e5 #num | Probability distribution of the number of infected people in a room
Here is an alternative to Ben's answer using simulations in R, using his parameters.
Edit: fixed the bug.
N=40 #number of people
h=80 #handshakes
k=1 #number of infected people at the start
n=1e5 #number of simulations
result=rep(NA,n)
for (r in 1:n) ... | Probability distribution of the number of infected people in a room
Here is an alternative to Ben's answer using simulations in R, using his parameters.
Edit: fixed the bug.
N=40 #number of people
h=80 #handshakes
k=1 #number of infected people at the start
n=1e5 #num |
46,015 | use MCMC posterior as prior for future inference | Strictly speaking, you have to rerun your MCMC algorithm from scratch to approximate the new posterior. MCMC algorithms are not sequential, which means that you cannot update their output with new data to update your estimate of the posterior. You just have to redo it.
However, you can use importance sampling to recur... | use MCMC posterior as prior for future inference | Strictly speaking, you have to rerun your MCMC algorithm from scratch to approximate the new posterior. MCMC algorithms are not sequential, which means that you cannot update their output with new dat | use MCMC posterior as prior for future inference
Strictly speaking, you have to rerun your MCMC algorithm from scratch to approximate the new posterior. MCMC algorithms are not sequential, which means that you cannot update their output with new data to update your estimate of the posterior. You just have to redo it.
... | use MCMC posterior as prior for future inference
Strictly speaking, you have to rerun your MCMC algorithm from scratch to approximate the new posterior. MCMC algorithms are not sequential, which means that you cannot update their output with new dat |
46,016 | Are the errors in this formulation of the simple linear regression model random variables? | I looked up your citation (4th edition, page 21) because I found it very alarming and was relieved to find is actually given as:
$$ \hat{e}_i = y_i − \widehat{E}(Y|X=x_i) = y_i - (\hat{\beta}_0 + \hat{\beta}_1) \tag{2.3} $$
Which is still confusing, I grant you, and the difference isn't actually germane to your questio... | Are the errors in this formulation of the simple linear regression model random variables? | I looked up your citation (4th edition, page 21) because I found it very alarming and was relieved to find is actually given as:
$$ \hat{e}_i = y_i − \widehat{E}(Y|X=x_i) = y_i - (\hat{\beta}_0 + \hat | Are the errors in this formulation of the simple linear regression model random variables?
I looked up your citation (4th edition, page 21) because I found it very alarming and was relieved to find is actually given as:
$$ \hat{e}_i = y_i − \widehat{E}(Y|X=x_i) = y_i - (\hat{\beta}_0 + \hat{\beta}_1) \tag{2.3} $$
Which... | Are the errors in this formulation of the simple linear regression model random variables?
I looked up your citation (4th edition, page 21) because I found it very alarming and was relieved to find is actually given as:
$$ \hat{e}_i = y_i − \widehat{E}(Y|X=x_i) = y_i - (\hat{\beta}_0 + \hat |
46,017 | Are the errors in this formulation of the simple linear regression model random variables? | In simple linear regression, we assume that the observations are randomly perturbed from the conditional expected value, i.e. $E[Y|X=x_i]$; so, each of your observations are assumed to be generated from a model of the form: $$Y=\beta_0+\beta_1X+\epsilon \ \ , \epsilon\sim N(0,\sigma^2)$$
This makes each $\epsilon_i$ a ... | Are the errors in this formulation of the simple linear regression model random variables? | In simple linear regression, we assume that the observations are randomly perturbed from the conditional expected value, i.e. $E[Y|X=x_i]$; so, each of your observations are assumed to be generated fr | Are the errors in this formulation of the simple linear regression model random variables?
In simple linear regression, we assume that the observations are randomly perturbed from the conditional expected value, i.e. $E[Y|X=x_i]$; so, each of your observations are assumed to be generated from a model of the form: $$Y=\... | Are the errors in this formulation of the simple linear regression model random variables?
In simple linear regression, we assume that the observations are randomly perturbed from the conditional expected value, i.e. $E[Y|X=x_i]$; so, each of your observations are assumed to be generated fr |
46,018 | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont get it | I tried with the glmmADMB package, an alternative of lme4 for linear mixed modelling. You can install this package with this code:
install.packages("R2admb")
install.packages("glmmADMB",
repos=c("http://glmmadmb.r-forge.r-project.org/repos",
getOption("repos")),
type="source")
Then you go:
library(glmmADMB)
h... | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont | I tried with the glmmADMB package, an alternative of lme4 for linear mixed modelling. You can install this package with this code:
install.packages("R2admb")
install.packages("glmmADMB",
repos=c("htt | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont get it
I tried with the glmmADMB package, an alternative of lme4 for linear mixed modelling. You can install this package with this code:
install.packages("R2admb")
install.packages("glmmADMB",
repos=c("http://glmmadmb... | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont
I tried with the glmmADMB package, an alternative of lme4 for linear mixed modelling. You can install this package with this code:
install.packages("R2admb")
install.packages("glmmADMB",
repos=c("htt |
46,019 | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont get it | Function glmer() uses by default the Laplace approximation, which is not optimal for dichotomous data. A better alternative is the adaptive Gaussian quadrature. You can use this method by setting argument nAGQ of glmer() to a higher number (e.g., 11 or 15) or alternatively using the GLMMadaptive package. In your exampl... | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont | Function glmer() uses by default the Laplace approximation, which is not optimal for dichotomous data. A better alternative is the adaptive Gaussian quadrature. You can use this method by setting argu | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont get it
Function glmer() uses by default the Laplace approximation, which is not optimal for dichotomous data. A better alternative is the adaptive Gaussian quadrature. You can use this method by setting argument nAGQ of... | Don't understand why glmm random effect variance is zero. Have reviewed similar questions still dont
Function glmer() uses by default the Laplace approximation, which is not optimal for dichotomous data. A better alternative is the adaptive Gaussian quadrature. You can use this method by setting argu |
46,020 | Using indistinguishable subjects as predictors/random effects | Perhaps something it is not clear to me, but the following simulation for your setting seems to work with the random effects structure you suggested, i.e.,
set.seed(2019)
N <- 50 # number of subjects
# id indicators for pairs
ids <- t(combn(N, 2))
id_i <- ids[, 1]
id_j <- ids[, 2]
# random effects
b_i <- rnorm(N, sd ... | Using indistinguishable subjects as predictors/random effects | Perhaps something it is not clear to me, but the following simulation for your setting seems to work with the random effects structure you suggested, i.e.,
set.seed(2019)
N <- 50 # number of subjects
| Using indistinguishable subjects as predictors/random effects
Perhaps something it is not clear to me, but the following simulation for your setting seems to work with the random effects structure you suggested, i.e.,
set.seed(2019)
N <- 50 # number of subjects
# id indicators for pairs
ids <- t(combn(N, 2))
id_i <- i... | Using indistinguishable subjects as predictors/random effects
Perhaps something it is not clear to me, but the following simulation for your setting seems to work with the random effects structure you suggested, i.e.,
set.seed(2019)
N <- 50 # number of subjects
|
46,021 | Using indistinguishable subjects as predictors/random effects | For people looking for a more little theory, Peter Hoff at the Duke/University of Washington has worked on this.
His 2005 JASA paper "Bilinear Mixed-Effects Models for Dyadic Data" (pdf and code) describes an MCMC approach that is very similar to @Dimitris Rizopoulos’ answer above).
That rough code was made into a mor... | Using indistinguishable subjects as predictors/random effects | For people looking for a more little theory, Peter Hoff at the Duke/University of Washington has worked on this.
His 2005 JASA paper "Bilinear Mixed-Effects Models for Dyadic Data" (pdf and code) desc | Using indistinguishable subjects as predictors/random effects
For people looking for a more little theory, Peter Hoff at the Duke/University of Washington has worked on this.
His 2005 JASA paper "Bilinear Mixed-Effects Models for Dyadic Data" (pdf and code) describes an MCMC approach that is very similar to @Dimitris R... | Using indistinguishable subjects as predictors/random effects
For people looking for a more little theory, Peter Hoff at the Duke/University of Washington has worked on this.
His 2005 JASA paper "Bilinear Mixed-Effects Models for Dyadic Data" (pdf and code) desc |
46,022 | Interpretation of standardized (z-score rescaled) linear model coefficients | Noting for readers who might have missed it that you standardized (i.e. rescaled by z-score) only the predictors and not your response variable.
The linear model coefficients can be interpreted as the change in the response (i.e. dependent variable) for a 1 standard deviation increase in the predictor (i.e. independent... | Interpretation of standardized (z-score rescaled) linear model coefficients | Noting for readers who might have missed it that you standardized (i.e. rescaled by z-score) only the predictors and not your response variable.
The linear model coefficients can be interpreted as the | Interpretation of standardized (z-score rescaled) linear model coefficients
Noting for readers who might have missed it that you standardized (i.e. rescaled by z-score) only the predictors and not your response variable.
The linear model coefficients can be interpreted as the change in the response (i.e. dependent vari... | Interpretation of standardized (z-score rescaled) linear model coefficients
Noting for readers who might have missed it that you standardized (i.e. rescaled by z-score) only the predictors and not your response variable.
The linear model coefficients can be interpreted as the |
46,023 | What is the difference between probabilistic forecasting and quantile forecasting? | In a sense, you are right: if we generate forecasts for the 0.001, 0.002, ..., 0.998 and 0.999 quantile, then we pretty much already have a full probabilistic forecast. Essentially, the predicted density would be a histogram with 998 bins.
However, I have rarely seen this. (One of the rare examples is the GEFCom2014 co... | What is the difference between probabilistic forecasting and quantile forecasting? | In a sense, you are right: if we generate forecasts for the 0.001, 0.002, ..., 0.998 and 0.999 quantile, then we pretty much already have a full probabilistic forecast. Essentially, the predicted dens | What is the difference between probabilistic forecasting and quantile forecasting?
In a sense, you are right: if we generate forecasts for the 0.001, 0.002, ..., 0.998 and 0.999 quantile, then we pretty much already have a full probabilistic forecast. Essentially, the predicted density would be a histogram with 998 bin... | What is the difference between probabilistic forecasting and quantile forecasting?
In a sense, you are right: if we generate forecasts for the 0.001, 0.002, ..., 0.998 and 0.999 quantile, then we pretty much already have a full probabilistic forecast. Essentially, the predicted dens |
46,024 | How to fit a longitudinal GAM mixed model (GAMM) | What @Roland is getting at is to use a random spline basis for the time by id random part. So your model would become:
m <- gam(y ~ s(V1) + V2 + s(time, id, bs = 'fs'),
family=gaussian, data=dat, method = "REML")
This model says that the effect of time is smooth and varies by id, with a separate smooth being ... | How to fit a longitudinal GAM mixed model (GAMM) | What @Roland is getting at is to use a random spline basis for the time by id random part. So your model would become:
m <- gam(y ~ s(V1) + V2 + s(time, id, bs = 'fs'),
family=gaussian, data= | How to fit a longitudinal GAM mixed model (GAMM)
What @Roland is getting at is to use a random spline basis for the time by id random part. So your model would become:
m <- gam(y ~ s(V1) + V2 + s(time, id, bs = 'fs'),
family=gaussian, data=dat, method = "REML")
This model says that the effect of time is smoot... | How to fit a longitudinal GAM mixed model (GAMM)
What @Roland is getting at is to use a random spline basis for the time by id random part. So your model would become:
m <- gam(y ~ s(V1) + V2 + s(time, id, bs = 'fs'),
family=gaussian, data= |
46,025 | PCA principal components in sklearn not matching eigen-vectors of covariance calculated by numpy | While this is a pure python related question which is not fitted here for CrossValidated, let me help you anyway. Both procedures find the correct eigenvectors. The difference is in its representation. While PCA() lists the entries of an eigenvectors rowwise, np.linalg.eig() lists the entries of the eigenvectors column... | PCA principal components in sklearn not matching eigen-vectors of covariance calculated by numpy | While this is a pure python related question which is not fitted here for CrossValidated, let me help you anyway. Both procedures find the correct eigenvectors. The difference is in its representation | PCA principal components in sklearn not matching eigen-vectors of covariance calculated by numpy
While this is a pure python related question which is not fitted here for CrossValidated, let me help you anyway. Both procedures find the correct eigenvectors. The difference is in its representation. While PCA() lists the... | PCA principal components in sklearn not matching eigen-vectors of covariance calculated by numpy
While this is a pure python related question which is not fitted here for CrossValidated, let me help you anyway. Both procedures find the correct eigenvectors. The difference is in its representation |
46,026 | What is the definition of the geometric mean of a random variable? | You can define the geometric mean of a strictly positive random variable readily enough; an easy method would be to take:
$$GM(X)=\exp(E[\log(X)])$$
For a discrete variable you can write the geometric mean as $\prod_i x_i^{p_i}$ where $p_i={p(X=x_i)}$. | What is the definition of the geometric mean of a random variable? | You can define the geometric mean of a strictly positive random variable readily enough; an easy method would be to take:
$$GM(X)=\exp(E[\log(X)])$$
For a discrete variable you can write the geometric | What is the definition of the geometric mean of a random variable?
You can define the geometric mean of a strictly positive random variable readily enough; an easy method would be to take:
$$GM(X)=\exp(E[\log(X)])$$
For a discrete variable you can write the geometric mean as $\prod_i x_i^{p_i}$ where $p_i={p(X=x_i)}$. | What is the definition of the geometric mean of a random variable?
You can define the geometric mean of a strictly positive random variable readily enough; an easy method would be to take:
$$GM(X)=\exp(E[\log(X)])$$
For a discrete variable you can write the geometric |
46,027 | What is the definition of the geometric mean of a random variable? | I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the continuous version of the arithmetic mean, right? Well the arithmetic mean of a finite number of elements can be defined as:... | What is the definition of the geometric mean of a random variable? | I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the cont | What is the definition of the geometric mean of a random variable?
I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the continuous version of the arithmetic mean, right? Well th... | What is the definition of the geometric mean of a random variable?
I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the cont |
46,028 | Unbiased estimator of binomial PMF | Since, for a Binomial $\text{B}(n,p)$ variable $X$, and $k\le n$, the factorial moment is given by
$$\mathbb{E}_p[X(X-1)\cdots(X-k+1)] = n(n-1)\cdots(n-k+1)p^k,$$
the $s$ Bernoulli rvs $\lbrace X_i\rbrace_{i=1}^{s}$ can easily return independent unbiased estimates of both $p^k$ and $(1-p)^{n-k}$ if $k+(n-k)\le s$, that... | Unbiased estimator of binomial PMF | Since, for a Binomial $\text{B}(n,p)$ variable $X$, and $k\le n$, the factorial moment is given by
$$\mathbb{E}_p[X(X-1)\cdots(X-k+1)] = n(n-1)\cdots(n-k+1)p^k,$$
the $s$ Bernoulli rvs $\lbrace X_i\rb | Unbiased estimator of binomial PMF
Since, for a Binomial $\text{B}(n,p)$ variable $X$, and $k\le n$, the factorial moment is given by
$$\mathbb{E}_p[X(X-1)\cdots(X-k+1)] = n(n-1)\cdots(n-k+1)p^k,$$
the $s$ Bernoulli rvs $\lbrace X_i\rbrace_{i=1}^{s}$ can easily return independent unbiased estimates of both $p^k$ and $(... | Unbiased estimator of binomial PMF
Since, for a Binomial $\text{B}(n,p)$ variable $X$, and $k\le n$, the factorial moment is given by
$$\mathbb{E}_p[X(X-1)\cdots(X-k+1)] = n(n-1)\cdots(n-k+1)p^k,$$
the $s$ Bernoulli rvs $\lbrace X_i\rb |
46,029 | Why does Naive Bayes work better when the number of features >> sample size compared to more sophisticated ML algorithms? | What the author is getting at is that Naive Bayes implicitly treats all features as being independent of one another, and therefore the sorts of curse-of-dimensionality problems which typically rear their head when dealing with high-dimensional data do not apply.
If your data has $k$ dimensions, then a fully general ML... | Why does Naive Bayes work better when the number of features >> sample size compared to more sophist | What the author is getting at is that Naive Bayes implicitly treats all features as being independent of one another, and therefore the sorts of curse-of-dimensionality problems which typically rear t | Why does Naive Bayes work better when the number of features >> sample size compared to more sophisticated ML algorithms?
What the author is getting at is that Naive Bayes implicitly treats all features as being independent of one another, and therefore the sorts of curse-of-dimensionality problems which typically rear... | Why does Naive Bayes work better when the number of features >> sample size compared to more sophist
What the author is getting at is that Naive Bayes implicitly treats all features as being independent of one another, and therefore the sorts of curse-of-dimensionality problems which typically rear t |
46,030 | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics | The the joint probability density function (pdf) of bivariate normal distribution is:
$$f(x_1,x_2)=\frac 1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp[-\frac z{2(1-\rho^2)}], $$
where
$$z=\frac{(x_1-\mu_1)^2}{\sigma_1^2}-\frac{2\rho(x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2}+\frac{(x_2-\mu_2)^2}{\sigma_2^2}.$$
When $\rh... | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics | The the joint probability density function (pdf) of bivariate normal distribution is:
$$f(x_1,x_2)=\frac 1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp[-\frac z{2(1-\rho^2)}], $$
where
$$z=\frac{(x_1-\ | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics
The the joint probability density function (pdf) of bivariate normal distribution is:
$$f(x_1,x_2)=\frac 1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp[-\frac z{2(1-\rho^2)}], $$
where
$$z=\frac{(x_1-\mu_1)^2}{\sigma_1^2}-\frac{2\rho(x_1-\mu... | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics
The the joint probability density function (pdf) of bivariate normal distribution is:
$$f(x_1,x_2)=\frac 1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp[-\frac z{2(1-\rho^2)}], $$
where
$$z=\frac{(x_1-\ |
46,031 | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics | Joint normality of two random variables $X,Y$ can be characterized in either of two simple ways:
For every pair $a,b$ of (non-random) real numbers, $aX+bY$ has a univariate normal distribution.
There are random variables $Z_1,Z_2\sim\operatorname{\text{i.i.d.}} \operatorname N(0,1)$ and real numbers $a,b,c,d$ such tha... | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics | Joint normality of two random variables $X,Y$ can be characterized in either of two simple ways:
For every pair $a,b$ of (non-random) real numbers, $aX+bY$ has a univariate normal distribution.
There | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics
Joint normality of two random variables $X,Y$ can be characterized in either of two simple ways:
For every pair $a,b$ of (non-random) real numbers, $aX+bY$ has a univariate normal distribution.
There are random variables $Z_1,Z_2\sim\opera... | Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics
Joint normality of two random variables $X,Y$ can be characterized in either of two simple ways:
For every pair $a,b$ of (non-random) real numbers, $aX+bY$ has a univariate normal distribution.
There |
46,032 | Scaling separately in train and test set? [duplicate] | Any kind of transformation of the data representation that "takes" information from the data should only be "fitted" on the training data. This is because:
If you were using all data you would have a information leakage from the validation or test (also called: holdout) data into your model. This is forbidden! As a re... | Scaling separately in train and test set? [duplicate] | Any kind of transformation of the data representation that "takes" information from the data should only be "fitted" on the training data. This is because:
If you were using all data you would have a | Scaling separately in train and test set? [duplicate]
Any kind of transformation of the data representation that "takes" information from the data should only be "fitted" on the training data. This is because:
If you were using all data you would have a information leakage from the validation or test (also called: hol... | Scaling separately in train and test set? [duplicate]
Any kind of transformation of the data representation that "takes" information from the data should only be "fitted" on the training data. This is because:
If you were using all data you would have a |
46,033 | Scaling separately in train and test set? [duplicate] | You should use train data's mean and std. deviation. For example, in scikit-learn library, StdScaler class, you'll first use fit() function with training data and transform() function with the test data, which transforms the given test data using fitted m and s values calculated from training data you have. If you have... | Scaling separately in train and test set? [duplicate] | You should use train data's mean and std. deviation. For example, in scikit-learn library, StdScaler class, you'll first use fit() function with training data and transform() function with the test da | Scaling separately in train and test set? [duplicate]
You should use train data's mean and std. deviation. For example, in scikit-learn library, StdScaler class, you'll first use fit() function with training data and transform() function with the test data, which transforms the given test data using fitted m and s valu... | Scaling separately in train and test set? [duplicate]
You should use train data's mean and std. deviation. For example, in scikit-learn library, StdScaler class, you'll first use fit() function with training data and transform() function with the test da |
46,034 | Convergence to a Uniform Distribution | I want to make two points. One is that although the concepts in this exercise are relatively sophisticated, a completely elementary simple solution is available. The other is that an appropriate visualization of the problem can lead us in natural steps through a rigorous proof. I will demonstrate that, by pointing o... | Convergence to a Uniform Distribution | I want to make two points. One is that although the concepts in this exercise are relatively sophisticated, a completely elementary simple solution is available. The other is that an appropriate vis | Convergence to a Uniform Distribution
I want to make two points. One is that although the concepts in this exercise are relatively sophisticated, a completely elementary simple solution is available. The other is that an appropriate visualization of the problem can lead us in natural steps through a rigorous proof. ... | Convergence to a Uniform Distribution
I want to make two points. One is that although the concepts in this exercise are relatively sophisticated, a completely elementary simple solution is available. The other is that an appropriate vis |
46,035 | Convergence to a Uniform Distribution | By the way the information is given, namely that it tells us that the probability of $X_n$ taking a specific value is strictly positive, we learn that $X_n$ is certainly a discrete random variable, otherwise we would have $P(X_n = i/n)=0$ (and I guess we can assume that mixed distributions, part discrete part continuou... | Convergence to a Uniform Distribution | By the way the information is given, namely that it tells us that the probability of $X_n$ taking a specific value is strictly positive, we learn that $X_n$ is certainly a discrete random variable, ot | Convergence to a Uniform Distribution
By the way the information is given, namely that it tells us that the probability of $X_n$ taking a specific value is strictly positive, we learn that $X_n$ is certainly a discrete random variable, otherwise we would have $P(X_n = i/n)=0$ (and I guess we can assume that mixed distr... | Convergence to a Uniform Distribution
By the way the information is given, namely that it tells us that the probability of $X_n$ taking a specific value is strictly positive, we learn that $X_n$ is certainly a discrete random variable, ot |
46,036 | Can an optimal weighted average ever have negative weights? | In addition to the other excellent answer. Yes, it can give negative weights, in some cases, even if that can look counterintuitive. Let us see. First, I will go through solving the minimization problem (and I will simplify your notation, replacing $\vec{w}$ with $w$ and so on).
Introducing the Lagrange multiplier $\l... | Can an optimal weighted average ever have negative weights? | In addition to the other excellent answer. Yes, it can give negative weights, in some cases, even if that can look counterintuitive. Let us see. First, I will go through solving the minimization prob | Can an optimal weighted average ever have negative weights?
In addition to the other excellent answer. Yes, it can give negative weights, in some cases, even if that can look counterintuitive. Let us see. First, I will go through solving the minimization problem (and I will simplify your notation, replacing $\vec{w}$ ... | Can an optimal weighted average ever have negative weights?
In addition to the other excellent answer. Yes, it can give negative weights, in some cases, even if that can look counterintuitive. Let us see. First, I will go through solving the minimization prob |
46,037 | Can an optimal weighted average ever have negative weights? | In abstract theory, yes, because of the correlation structure.
If you had uncorrelated measurements (with positive variance), then the weights could only be positive.
Example:
Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \m... | Can an optimal weighted average ever have negative weights? | In abstract theory, yes, because of the correlation structure.
If you had uncorrelated measurements (with positive variance), then the weights could only be positive.
Example:
Let $\mu$ denote the tru | Can an optimal weighted average ever have negative weights?
In abstract theory, yes, because of the correlation structure.
If you had uncorrelated measurements (with positive variance), then the weights could only be positive.
Example:
Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)... | Can an optimal weighted average ever have negative weights?
In abstract theory, yes, because of the correlation structure.
If you had uncorrelated measurements (with positive variance), then the weights could only be positive.
Example:
Let $\mu$ denote the tru |
46,038 | An X% trimmed mean means? | Neither is "right" or "wrong"; it's just that usage is not universal. However, I've seen Wilcox's definition used more than the other. Wikipedia agrees with him, as do several other sites I browsed to, and so do SAS, and R. | An X% trimmed mean means? | Neither is "right" or "wrong"; it's just that usage is not universal. However, I've seen Wilcox's definition used more than the other. Wikipedia agrees with him, as do several other sites I browsed t | An X% trimmed mean means?
Neither is "right" or "wrong"; it's just that usage is not universal. However, I've seen Wilcox's definition used more than the other. Wikipedia agrees with him, as do several other sites I browsed to, and so do SAS, and R. | An X% trimmed mean means?
Neither is "right" or "wrong"; it's just that usage is not universal. However, I've seen Wilcox's definition used more than the other. Wikipedia agrees with him, as do several other sites I browsed t |
46,039 | An X% trimmed mean means? | As Peter correctly points out, the conventions on usage of this term differ, and the definition used by Wilcox seems (unfortunately) to be the more common. I disagree with the view that neither is right or wrong. The definition that removes X% from each side of the ordered data vector, but refers to this as an "X% tr... | An X% trimmed mean means? | As Peter correctly points out, the conventions on usage of this term differ, and the definition used by Wilcox seems (unfortunately) to be the more common. I disagree with the view that neither is ri | An X% trimmed mean means?
As Peter correctly points out, the conventions on usage of this term differ, and the definition used by Wilcox seems (unfortunately) to be the more common. I disagree with the view that neither is right or wrong. The definition that removes X% from each side of the ordered data vector, but r... | An X% trimmed mean means?
As Peter correctly points out, the conventions on usage of this term differ, and the definition used by Wilcox seems (unfortunately) to be the more common. I disagree with the view that neither is ri |
46,040 | Fitting a GARCH(1, 1) model | I explain how to get the log-likelihood function for the GARCH(1,1) model in the answer to this question.
The GARCH model is specified in a particular way, but notation may differ between papers and applications. The log-likelihood may differ due to constants being omitted (they are irrelevant when maximizing).
The M... | Fitting a GARCH(1, 1) model | I explain how to get the log-likelihood function for the GARCH(1,1) model in the answer to this question.
The GARCH model is specified in a particular way, but notation may differ between papers and | Fitting a GARCH(1, 1) model
I explain how to get the log-likelihood function for the GARCH(1,1) model in the answer to this question.
The GARCH model is specified in a particular way, but notation may differ between papers and applications. The log-likelihood may differ due to constants being omitted (they are irrelev... | Fitting a GARCH(1, 1) model
I explain how to get the log-likelihood function for the GARCH(1,1) model in the answer to this question.
The GARCH model is specified in a particular way, but notation may differ between papers and |
46,041 | Bayesian Statistics. Please help me to find an example where posterior variance is greater than prior variance | Gamma-Poisson: Suppose your prior for Poisson data is $\lambda \sim \mathsf{Gamma}(\text{shape}=4, \text{rate}=1/4).$
This distribution has mean 16 and variance 64. It's 95th percentile is about 31.
qgamma(.95, 4, .25)
[1] 31.01463
But your first Poisson observation is $x = 500.$ Then your posterior
distribution is $\... | Bayesian Statistics. Please help me to find an example where posterior variance is greater than prio | Gamma-Poisson: Suppose your prior for Poisson data is $\lambda \sim \mathsf{Gamma}(\text{shape}=4, \text{rate}=1/4).$
This distribution has mean 16 and variance 64. It's 95th percentile is about 31.
q | Bayesian Statistics. Please help me to find an example where posterior variance is greater than prior variance
Gamma-Poisson: Suppose your prior for Poisson data is $\lambda \sim \mathsf{Gamma}(\text{shape}=4, \text{rate}=1/4).$
This distribution has mean 16 and variance 64. It's 95th percentile is about 31.
qgamma(.95... | Bayesian Statistics. Please help me to find an example where posterior variance is greater than prio
Gamma-Poisson: Suppose your prior for Poisson data is $\lambda \sim \mathsf{Gamma}(\text{shape}=4, \text{rate}=1/4).$
This distribution has mean 16 and variance 64. It's 95th percentile is about 31.
q |
46,042 | DQN with XGBoost | The downside of using XGBoost compared to a neural network, is that a neural network can be trained partially whereas an XGBoost regression model will have to be trained from scratch for every update. This is because an XGBoost model uses sequential trees fitted on the residuals of the previous trees so iterative updat... | DQN with XGBoost | The downside of using XGBoost compared to a neural network, is that a neural network can be trained partially whereas an XGBoost regression model will have to be trained from scratch for every update. | DQN with XGBoost
The downside of using XGBoost compared to a neural network, is that a neural network can be trained partially whereas an XGBoost regression model will have to be trained from scratch for every update. This is because an XGBoost model uses sequential trees fitted on the residuals of the previous trees s... | DQN with XGBoost
The downside of using XGBoost compared to a neural network, is that a neural network can be trained partially whereas an XGBoost regression model will have to be trained from scratch for every update. |
46,043 | DQN with XGBoost | In Q learning, it is possible to use practically any regression model that can be updated incrementally.
In fitted Q learning, any regression model can be used, including tree-based approaches, see e.g. page 70 of this book: https://orbi.uliege.be/bitstream/2268/27963/1/book-FA-RL-DP.pdf
However, we can hardly speak a... | DQN with XGBoost | In Q learning, it is possible to use practically any regression model that can be updated incrementally.
In fitted Q learning, any regression model can be used, including tree-based approaches, see e. | DQN with XGBoost
In Q learning, it is possible to use practically any regression model that can be updated incrementally.
In fitted Q learning, any regression model can be used, including tree-based approaches, see e.g. page 70 of this book: https://orbi.uliege.be/bitstream/2268/27963/1/book-FA-RL-DP.pdf
However, we c... | DQN with XGBoost
In Q learning, it is possible to use practically any regression model that can be updated incrementally.
In fitted Q learning, any regression model can be used, including tree-based approaches, see e. |
46,044 | Regression as a way to determine variable importance | Does anyone have any thoughts on the validity of this method?
The main issue I can see is that the linear regression proposes to split the sample space in two (the two "sides" of a hyperplane), whereas the nearest neighbors will split it in regions (maybe a lot), depending on how the different classes are sampled on th... | Regression as a way to determine variable importance | Does anyone have any thoughts on the validity of this method?
The main issue I can see is that the linear regression proposes to split the sample space in two (the two "sides" of a hyperplane), wherea | Regression as a way to determine variable importance
Does anyone have any thoughts on the validity of this method?
The main issue I can see is that the linear regression proposes to split the sample space in two (the two "sides" of a hyperplane), whereas the nearest neighbors will split it in regions (maybe a lot), dep... | Regression as a way to determine variable importance
Does anyone have any thoughts on the validity of this method?
The main issue I can see is that the linear regression proposes to split the sample space in two (the two "sides" of a hyperplane), wherea |
46,045 | Regression as a way to determine variable importance | In the classical regression situation the problem has given rise to a number of solutions. In an article outlining her software for calculating a number of these methods entitled "Relative importance for linear regression in R: the package relaimpo" available here Ulricke Grömping gives a number of different metrics fo... | Regression as a way to determine variable importance | In the classical regression situation the problem has given rise to a number of solutions. In an article outlining her software for calculating a number of these methods entitled "Relative importance | Regression as a way to determine variable importance
In the classical regression situation the problem has given rise to a number of solutions. In an article outlining her software for calculating a number of these methods entitled "Relative importance for linear regression in R: the package relaimpo" available here Ul... | Regression as a way to determine variable importance
In the classical regression situation the problem has given rise to a number of solutions. In an article outlining her software for calculating a number of these methods entitled "Relative importance |
46,046 | Cox PH linearity assumption: reading martingal residual plots | Quoting from Harrell's Regression Modeling Strategies, second edition, page 494:
When correlations among predictors are mild, plots of estimated predictor transformations without adjustment for other predictors (i.e., marginal transformations) may be useful. Martingale residuals may be obtained quickly by fixing $\hat... | Cox PH linearity assumption: reading martingal residual plots | Quoting from Harrell's Regression Modeling Strategies, second edition, page 494:
When correlations among predictors are mild, plots of estimated predictor transformations without adjustment for other | Cox PH linearity assumption: reading martingal residual plots
Quoting from Harrell's Regression Modeling Strategies, second edition, page 494:
When correlations among predictors are mild, plots of estimated predictor transformations without adjustment for other predictors (i.e., marginal transformations) may be useful... | Cox PH linearity assumption: reading martingal residual plots
Quoting from Harrell's Regression Modeling Strategies, second edition, page 494:
When correlations among predictors are mild, plots of estimated predictor transformations without adjustment for other |
46,047 | Joint credible regions from MCMC draws | Indeed, the two approaches you mention are the most straightforward; I believe for example that the function HPDregionplot in the R package emdbook uses a kernel density estimator.
Another option (and this is just a suggestion) would be to find the mode or centre of your distribution, compute the distance of each point... | Joint credible regions from MCMC draws | Indeed, the two approaches you mention are the most straightforward; I believe for example that the function HPDregionplot in the R package emdbook uses a kernel density estimator.
Another option (and | Joint credible regions from MCMC draws
Indeed, the two approaches you mention are the most straightforward; I believe for example that the function HPDregionplot in the R package emdbook uses a kernel density estimator.
Another option (and this is just a suggestion) would be to find the mode or centre of your distribut... | Joint credible regions from MCMC draws
Indeed, the two approaches you mention are the most straightforward; I believe for example that the function HPDregionplot in the R package emdbook uses a kernel density estimator.
Another option (and |
46,048 | Joint credible regions from MCMC draws | If you don't mind having a rectangular confidence region, and you know things are unimodal and symmetric, you could use the approach taken by credible.region() in the bayesSurv R package:
https://rdrr.io/cran/bayesSurv/src/R/credible.region.R
I believe this is the idea behind how it works:
We have $n$ MCMC samples for... | Joint credible regions from MCMC draws | If you don't mind having a rectangular confidence region, and you know things are unimodal and symmetric, you could use the approach taken by credible.region() in the bayesSurv R package:
https://rdrr | Joint credible regions from MCMC draws
If you don't mind having a rectangular confidence region, and you know things are unimodal and symmetric, you could use the approach taken by credible.region() in the bayesSurv R package:
https://rdrr.io/cran/bayesSurv/src/R/credible.region.R
I believe this is the idea behind how... | Joint credible regions from MCMC draws
If you don't mind having a rectangular confidence region, and you know things are unimodal and symmetric, you could use the approach taken by credible.region() in the bayesSurv R package:
https://rdrr |
46,049 | Joint credible regions from MCMC draws | I played around with a few different options, but I figured I'd share the one that I found worked best. Note: In my application, the posterior is well approximated as a multivariate normal. In other applications, the HPDregionplot approach may be more flexible.
There are essentially 3 steps...
Rotate the data along it... | Joint credible regions from MCMC draws | I played around with a few different options, but I figured I'd share the one that I found worked best. Note: In my application, the posterior is well approximated as a multivariate normal. In other a | Joint credible regions from MCMC draws
I played around with a few different options, but I figured I'd share the one that I found worked best. Note: In my application, the posterior is well approximated as a multivariate normal. In other applications, the HPDregionplot approach may be more flexible.
There are essential... | Joint credible regions from MCMC draws
I played around with a few different options, but I figured I'd share the one that I found worked best. Note: In my application, the posterior is well approximated as a multivariate normal. In other a |
46,050 | Why are exponential smoothing models not considered auto-regressive? | For an autoregressive model, non-linear or linear, the number of lags must be finite. An ETS(A,N,N) model can be written as an AR($\infty$) model, but not as an autoregressive model with finite lags. A few other exponential smoothing models can be written in AR($\infty$) form, but none can be written as an autoregressi... | Why are exponential smoothing models not considered auto-regressive? | For an autoregressive model, non-linear or linear, the number of lags must be finite. An ETS(A,N,N) model can be written as an AR($\infty$) model, but not as an autoregressive model with finite lags. | Why are exponential smoothing models not considered auto-regressive?
For an autoregressive model, non-linear or linear, the number of lags must be finite. An ETS(A,N,N) model can be written as an AR($\infty$) model, but not as an autoregressive model with finite lags. A few other exponential smoothing models can be wri... | Why are exponential smoothing models not considered auto-regressive?
For an autoregressive model, non-linear or linear, the number of lags must be finite. An ETS(A,N,N) model can be written as an AR($\infty$) model, but not as an autoregressive model with finite lags. |
46,051 | Checking the proportional hazard assumption | The global test of proportional hazards is not well-calibrated. You haven't controlled for multiple comparisons. It's difficult to gauge power of the test. $\alpha=0.05$ is probably too lax in most sample sizes. The test is arbitrarily powerful in large sample sizes. It's possible that the covariate you identify is a s... | Checking the proportional hazard assumption | The global test of proportional hazards is not well-calibrated. You haven't controlled for multiple comparisons. It's difficult to gauge power of the test. $\alpha=0.05$ is probably too lax in most sa | Checking the proportional hazard assumption
The global test of proportional hazards is not well-calibrated. You haven't controlled for multiple comparisons. It's difficult to gauge power of the test. $\alpha=0.05$ is probably too lax in most sample sizes. The test is arbitrarily powerful in large sample sizes. It's pos... | Checking the proportional hazard assumption
The global test of proportional hazards is not well-calibrated. You haven't controlled for multiple comparisons. It's difficult to gauge power of the test. $\alpha=0.05$ is probably too lax in most sa |
46,052 | Checking the proportional hazard assumption | If hypothesis testing is your main goal, you should not do anything at all and stick with the model you had anticipated using before seeing the data.
Changing the pre-defined model based on the results of the cox.zph() test can lead to biased estimates and invalid p-values.
See SiM paper: https://harlanhappydog.github.... | Checking the proportional hazard assumption | If hypothesis testing is your main goal, you should not do anything at all and stick with the model you had anticipated using before seeing the data.
Changing the pre-defined model based on the result | Checking the proportional hazard assumption
If hypothesis testing is your main goal, you should not do anything at all and stick with the model you had anticipated using before seeing the data.
Changing the pre-defined model based on the results of the cox.zph() test can lead to biased estimates and invalid p-values.
S... | Checking the proportional hazard assumption
If hypothesis testing is your main goal, you should not do anything at all and stick with the model you had anticipated using before seeing the data.
Changing the pre-defined model based on the result |
46,053 | Sampling from a categorical distribution | Let me unravel your question to remove all the not-so-relevant fluff around it.
Given a tuple of $n$ values of the form $(p_i)_{1\leq i\leq n}$, where each $p_i\in(0,1)$, the question is to characterize the distribution of the following process:
Draw independent random variables $U_1,\dots,U_n$ uniformly distributed ... | Sampling from a categorical distribution | Let me unravel your question to remove all the not-so-relevant fluff around it.
Given a tuple of $n$ values of the form $(p_i)_{1\leq i\leq n}$, where each $p_i\in(0,1)$, the question is to character | Sampling from a categorical distribution
Let me unravel your question to remove all the not-so-relevant fluff around it.
Given a tuple of $n$ values of the form $(p_i)_{1\leq i\leq n}$, where each $p_i\in(0,1)$, the question is to characterize the distribution of the following process:
Draw independent random variabl... | Sampling from a categorical distribution
Let me unravel your question to remove all the not-so-relevant fluff around it.
Given a tuple of $n$ values of the form $(p_i)_{1\leq i\leq n}$, where each $p_i\in(0,1)$, the question is to character |
46,054 | Is it possible to do Fizz Buzz in deep learning? | As the author of the linked blog post, I am happy to say that (with the correct choice of hyperparameters and a little luck) Fizz Buzz can be completely learned by a neural network with one hidden layer.
I spent some time investigating why it works, and the reason is somewhat interesting. It hinges upon the binary repr... | Is it possible to do Fizz Buzz in deep learning? | As the author of the linked blog post, I am happy to say that (with the correct choice of hyperparameters and a little luck) Fizz Buzz can be completely learned by a neural network with one hidden lay | Is it possible to do Fizz Buzz in deep learning?
As the author of the linked blog post, I am happy to say that (with the correct choice of hyperparameters and a little luck) Fizz Buzz can be completely learned by a neural network with one hidden layer.
I spent some time investigating why it works, and the reason is som... | Is it possible to do Fizz Buzz in deep learning?
As the author of the linked blog post, I am happy to say that (with the correct choice of hyperparameters and a little luck) Fizz Buzz can be completely learned by a neural network with one hidden lay |
46,055 | Is it possible to do Fizz Buzz in deep learning? | Let me answer the question in a meme way.
Why (always) deep learning? All neural nets do is linear regression (x*w+b) with some non-linearity around the (intermediate) response.
Lets talk machine learning, better yet, optimization. The obvious general class of problems you are referring to is function approximation (n... | Is it possible to do Fizz Buzz in deep learning? | Let me answer the question in a meme way.
Why (always) deep learning? All neural nets do is linear regression (x*w+b) with some non-linearity around the (intermediate) response.
Lets talk machine lea | Is it possible to do Fizz Buzz in deep learning?
Let me answer the question in a meme way.
Why (always) deep learning? All neural nets do is linear regression (x*w+b) with some non-linearity around the (intermediate) response.
Lets talk machine learning, better yet, optimization. The obvious general class of problems ... | Is it possible to do Fizz Buzz in deep learning?
Let me answer the question in a meme way.
Why (always) deep learning? All neural nets do is linear regression (x*w+b) with some non-linearity around the (intermediate) response.
Lets talk machine lea |
46,056 | Interpreting glm.diag.plots | Diagnostic plots for GLMs are very similar to those for LMs, on the grounds that the residuals of GLMs should be homoscedastic, independent of the mean, and asymptotically approach normality, i.e. in the case of large numbers of counts for a Poisson or binomial (this means that such plots are much less useful for Berno... | Interpreting glm.diag.plots | Diagnostic plots for GLMs are very similar to those for LMs, on the grounds that the residuals of GLMs should be homoscedastic, independent of the mean, and asymptotically approach normality, i.e. in | Interpreting glm.diag.plots
Diagnostic plots for GLMs are very similar to those for LMs, on the grounds that the residuals of GLMs should be homoscedastic, independent of the mean, and asymptotically approach normality, i.e. in the case of large numbers of counts for a Poisson or binomial (this means that such plots ar... | Interpreting glm.diag.plots
Diagnostic plots for GLMs are very similar to those for LMs, on the grounds that the residuals of GLMs should be homoscedastic, independent of the mean, and asymptotically approach normality, i.e. in |
46,057 | General approaches and techniques for developing good explanatory models for nonlinear data | I think it is worth considering the use of generalised additive models (GAMs). GAMs are able to encapsulate non-linear relations between the response variable and the outcome variables and are straight-forward to explain. They are well-understood and widely used within the Statistics community.
In totally informal mann... | General approaches and techniques for developing good explanatory models for nonlinear data | I think it is worth considering the use of generalised additive models (GAMs). GAMs are able to encapsulate non-linear relations between the response variable and the outcome variables and are straigh | General approaches and techniques for developing good explanatory models for nonlinear data
I think it is worth considering the use of generalised additive models (GAMs). GAMs are able to encapsulate non-linear relations between the response variable and the outcome variables and are straight-forward to explain. They a... | General approaches and techniques for developing good explanatory models for nonlinear data
I think it is worth considering the use of generalised additive models (GAMs). GAMs are able to encapsulate non-linear relations between the response variable and the outcome variables and are straigh |
46,058 | General approaches and techniques for developing good explanatory models for nonlinear data | I will spin the question to terminology that I think clarifies the issue at hand - how can we increase model flexibility while maintaining interpretability and not substantially increasing model variance (in the sense of bias/variance trade off)?
Applying many transformations to features & interactions is a reasonable ... | General approaches and techniques for developing good explanatory models for nonlinear data | I will spin the question to terminology that I think clarifies the issue at hand - how can we increase model flexibility while maintaining interpretability and not substantially increasing model varia | General approaches and techniques for developing good explanatory models for nonlinear data
I will spin the question to terminology that I think clarifies the issue at hand - how can we increase model flexibility while maintaining interpretability and not substantially increasing model variance (in the sense of bias/va... | General approaches and techniques for developing good explanatory models for nonlinear data
I will spin the question to terminology that I think clarifies the issue at hand - how can we increase model flexibility while maintaining interpretability and not substantially increasing model varia |
46,059 | General approaches and techniques for developing good explanatory models for nonlinear data | As a engineer from the physics field, I understand how this matter can be crucial.
If you need multi-dimensional "smooth" fits, gaussian process regression usually works quite well. The analysis of your data can then be done through projection plots (for 2 or maybe 3 input parameters at the same time, at most). You don... | General approaches and techniques for developing good explanatory models for nonlinear data | As a engineer from the physics field, I understand how this matter can be crucial.
If you need multi-dimensional "smooth" fits, gaussian process regression usually works quite well. The analysis of yo | General approaches and techniques for developing good explanatory models for nonlinear data
As a engineer from the physics field, I understand how this matter can be crucial.
If you need multi-dimensional "smooth" fits, gaussian process regression usually works quite well. The analysis of your data can then be done thr... | General approaches and techniques for developing good explanatory models for nonlinear data
As a engineer from the physics field, I understand how this matter can be crucial.
If you need multi-dimensional "smooth" fits, gaussian process regression usually works quite well. The analysis of yo |
46,060 | How can a deep neural network learn anything? | You're absolutely right, it makes no sense, and it's a huge problem.
The loss function is locally linear. You are effectively saying that $L(w+h) = L(w)+h\cdot dL(w)$, where $L$ is your loss function, $w$ your vector of weights, and $dL$ is the gradient vector. In this sense, the contribution of changes of each weight ... | How can a deep neural network learn anything? | You're absolutely right, it makes no sense, and it's a huge problem.
The loss function is locally linear. You are effectively saying that $L(w+h) = L(w)+h\cdot dL(w)$, where $L$ is your loss function, | How can a deep neural network learn anything?
You're absolutely right, it makes no sense, and it's a huge problem.
The loss function is locally linear. You are effectively saying that $L(w+h) = L(w)+h\cdot dL(w)$, where $L$ is your loss function, $w$ your vector of weights, and $dL$ is the gradient vector. In this sens... | How can a deep neural network learn anything?
You're absolutely right, it makes no sense, and it's a huge problem.
The loss function is locally linear. You are effectively saying that $L(w+h) = L(w)+h\cdot dL(w)$, where $L$ is your loss function, |
46,061 | How can a deep neural network learn anything? | A recent paper addresses this topic: "Doing the impossible: Why neural networks can be trained at all" by Nathan Hodas, Panos Stinis
As deep neural networks grow in size, from thousands to millions to billions of weights, the performance of those networks becomes limited by our ability to accurately train them. A comm... | How can a deep neural network learn anything? | A recent paper addresses this topic: "Doing the impossible: Why neural networks can be trained at all" by Nathan Hodas, Panos Stinis
As deep neural networks grow in size, from thousands to millions t | How can a deep neural network learn anything?
A recent paper addresses this topic: "Doing the impossible: Why neural networks can be trained at all" by Nathan Hodas, Panos Stinis
As deep neural networks grow in size, from thousands to millions to billions of weights, the performance of those networks becomes limited b... | How can a deep neural network learn anything?
A recent paper addresses this topic: "Doing the impossible: Why neural networks can be trained at all" by Nathan Hodas, Panos Stinis
As deep neural networks grow in size, from thousands to millions t |
46,062 | How can a deep neural network learn anything? | This is why learning rates, despite being "as large as can be" are still quite small. On the order of $10^{-3}$ and $10^{-4}$ is common. But you are right that this can be an issue.
Batch normalization is one partial solution to this problem which often allows for faster training: Normalizing the activations after eac... | How can a deep neural network learn anything? | This is why learning rates, despite being "as large as can be" are still quite small. On the order of $10^{-3}$ and $10^{-4}$ is common. But you are right that this can be an issue.
Batch normalizati | How can a deep neural network learn anything?
This is why learning rates, despite being "as large as can be" are still quite small. On the order of $10^{-3}$ and $10^{-4}$ is common. But you are right that this can be an issue.
Batch normalization is one partial solution to this problem which often allows for faster t... | How can a deep neural network learn anything?
This is why learning rates, despite being "as large as can be" are still quite small. On the order of $10^{-3}$ and $10^{-4}$ is common. But you are right that this can be an issue.
Batch normalizati |
46,063 | Serial Mediation in R - how to setup the model? | This should probably be migrated to StackOverflow since it is about software, but:
You could do this in the R package lavaan. In your model, you would first specify models for M1, M2, and Y. We will want to label all the paths, as well. I will label c' as cp, for "c-prime":
M1 ~ a1 * X
M2 ~ a2 * X + d21 * M1
Y ~ cp * ... | Serial Mediation in R - how to setup the model? | This should probably be migrated to StackOverflow since it is about software, but:
You could do this in the R package lavaan. In your model, you would first specify models for M1, M2, and Y. We will w | Serial Mediation in R - how to setup the model?
This should probably be migrated to StackOverflow since it is about software, but:
You could do this in the R package lavaan. In your model, you would first specify models for M1, M2, and Y. We will want to label all the paths, as well. I will label c' as cp, for "c-prime... | Serial Mediation in R - how to setup the model?
This should probably be migrated to StackOverflow since it is about software, but:
You could do this in the R package lavaan. In your model, you would first specify models for M1, M2, and Y. We will w |
46,064 | Validating cluster tendency using Hopkins statistic | I have also been confused about this contradictory information regarding Hopkins statistics.
In http://www.sthda.com/english/articles/29-cluster-validation-essentials/95-assessing-clustering-tendency-essentials/ it is said that
We can conduct the Hopkins Statistic test iteratively, using 0.5 as
the threshold to re... | Validating cluster tendency using Hopkins statistic | I have also been confused about this contradictory information regarding Hopkins statistics.
In http://www.sthda.com/english/articles/29-cluster-validation-essentials/95-assessing-clustering-tendency | Validating cluster tendency using Hopkins statistic
I have also been confused about this contradictory information regarding Hopkins statistics.
In http://www.sthda.com/english/articles/29-cluster-validation-essentials/95-assessing-clustering-tendency-essentials/ it is said that
We can conduct the Hopkins Statistic ... | Validating cluster tendency using Hopkins statistic
I have also been confused about this contradictory information regarding Hopkins statistics.
In http://www.sthda.com/english/articles/29-cluster-validation-essentials/95-assessing-clustering-tendency |
46,065 | Validating cluster tendency using Hopkins statistic | Hopkins is a pretty extreme test for uniform distributions.
It's naive to assume that data will cluster, just because it has a tendency - the test is mostly useful to detect uniform data.
The problem is that it doesn't imply a multimodal distribution. A single Gaussian will have a "clustering tendency" according to Hop... | Validating cluster tendency using Hopkins statistic | Hopkins is a pretty extreme test for uniform distributions.
It's naive to assume that data will cluster, just because it has a tendency - the test is mostly useful to detect uniform data.
The problem | Validating cluster tendency using Hopkins statistic
Hopkins is a pretty extreme test for uniform distributions.
It's naive to assume that data will cluster, just because it has a tendency - the test is mostly useful to detect uniform data.
The problem is that it doesn't imply a multimodal distribution. A single Gaussia... | Validating cluster tendency using Hopkins statistic
Hopkins is a pretty extreme test for uniform distributions.
It's naive to assume that data will cluster, just because it has a tendency - the test is mostly useful to detect uniform data.
The problem |
46,066 | Validating cluster tendency using Hopkins statistic | Based on the source code
for get_clust_tendency() it appear's that Iden's answer is correct and for some versions of the factoextra package, the hopkin's stat may be calculated as (1-H).
Check the code for the get_clust_tendency() function in your version of factoextra, if you see that the hopkin's stat is calculated ... | Validating cluster tendency using Hopkins statistic | Based on the source code
for get_clust_tendency() it appear's that Iden's answer is correct and for some versions of the factoextra package, the hopkin's stat may be calculated as (1-H).
Check the co | Validating cluster tendency using Hopkins statistic
Based on the source code
for get_clust_tendency() it appear's that Iden's answer is correct and for some versions of the factoextra package, the hopkin's stat may be calculated as (1-H).
Check the code for the get_clust_tendency() function in your version of factoext... | Validating cluster tendency using Hopkins statistic
Based on the source code
for get_clust_tendency() it appear's that Iden's answer is correct and for some versions of the factoextra package, the hopkin's stat may be calculated as (1-H).
Check the co |
46,067 | Validating cluster tendency using Hopkins statistic | Actually, as already mentioned in the previous answers, it is a matter of how Hopkins statistic has been calculated.
The literature is has clear information on what value is expected. For example, Lawson and Jurs (1990) and Banerjee & Dave (2004) explains that you may expect 3 different results:
1) H = 0.5 (the datase... | Validating cluster tendency using Hopkins statistic | Actually, as already mentioned in the previous answers, it is a matter of how Hopkins statistic has been calculated.
The literature is has clear information on what value is expected. For example, La | Validating cluster tendency using Hopkins statistic
Actually, as already mentioned in the previous answers, it is a matter of how Hopkins statistic has been calculated.
The literature is has clear information on what value is expected. For example, Lawson and Jurs (1990) and Banerjee & Dave (2004) explains that you ma... | Validating cluster tendency using Hopkins statistic
Actually, as already mentioned in the previous answers, it is a matter of how Hopkins statistic has been calculated.
The literature is has clear information on what value is expected. For example, La |
46,068 | Validating cluster tendency using Hopkins statistic | I have been pretty confused too with contradicting information available online. However if you are using pyclusterend library, the documentation says if hopkins value is less than 0.3, there are possibilities of cluster. To quote:
Hopkins test A statistical test which allow to guess if the data
follow an uniform dist... | Validating cluster tendency using Hopkins statistic | I have been pretty confused too with contradicting information available online. However if you are using pyclusterend library, the documentation says if hopkins value is less than 0.3, there are poss | Validating cluster tendency using Hopkins statistic
I have been pretty confused too with contradicting information available online. However if you are using pyclusterend library, the documentation says if hopkins value is less than 0.3, there are possibilities of cluster. To quote:
Hopkins test A statistical test whi... | Validating cluster tendency using Hopkins statistic
I have been pretty confused too with contradicting information available online. However if you are using pyclusterend library, the documentation says if hopkins value is less than 0.3, there are poss |
46,069 | What exactly happens when I do a feature cross? | Your data is not linearly separable in the original space.
But it seems like it actually is separable with a circle/ellipse (let's say it's inside a circle to simplify the problem): it seems reasonable to have hypothesis that, for some $c$ if $x^2 + y^2< c$ then a point is blue.
That means that if you use $x^2, y^2$ as... | What exactly happens when I do a feature cross? | Your data is not linearly separable in the original space.
But it seems like it actually is separable with a circle/ellipse (let's say it's inside a circle to simplify the problem): it seems reasonabl | What exactly happens when I do a feature cross?
Your data is not linearly separable in the original space.
But it seems like it actually is separable with a circle/ellipse (let's say it's inside a circle to simplify the problem): it seems reasonable to have hypothesis that, for some $c$ if $x^2 + y^2< c$ then a point i... | What exactly happens when I do a feature cross?
Your data is not linearly separable in the original space.
But it seems like it actually is separable with a circle/ellipse (let's say it's inside a circle to simplify the problem): it seems reasonabl |
46,070 | Defining Conditional Likelihood | Usually one assumes that there is a distribution
$$p_{\text{data}}(y,x)$$
that somewhat defines not only the distributions of $x$ and $y$ but also their dependency (i.e. if $y_i = f(x_i)$ then we can estimate $f$ by computing the conditional expectation $E[y|X=x]$ with respect to this common probability distribution ... | Defining Conditional Likelihood | Usually one assumes that there is a distribution
$$p_{\text{data}}(y,x)$$
that somewhat defines not only the distributions of $x$ and $y$ but also their dependency (i.e. if $y_i = f(x_i)$ then we ca | Defining Conditional Likelihood
Usually one assumes that there is a distribution
$$p_{\text{data}}(y,x)$$
that somewhat defines not only the distributions of $x$ and $y$ but also their dependency (i.e. if $y_i = f(x_i)$ then we can estimate $f$ by computing the conditional expectation $E[y|X=x]$ with respect to this ... | Defining Conditional Likelihood
Usually one assumes that there is a distribution
$$p_{\text{data}}(y,x)$$
that somewhat defines not only the distributions of $x$ and $y$ but also their dependency (i.e. if $y_i = f(x_i)$ then we ca |
46,071 | Why Poisson and Binomial distribution are giving different results for the same problem? | Both yield nearly the same result:
> dpois(2,2)
[1] 0.2706706
> dbinom(2,100,.02)
[1] 0.2734139
Both results would get more similar as n tended to infinite and p tended to zero, but n=100 is large but a lot smaller than infinite, so you can get an accurate result up to a couple of significant digits.
Edit in response ... | Why Poisson and Binomial distribution are giving different results for the same problem? | Both yield nearly the same result:
> dpois(2,2)
[1] 0.2706706
> dbinom(2,100,.02)
[1] 0.2734139
Both results would get more similar as n tended to infinite and p tended to zero, but n=100 is large bu | Why Poisson and Binomial distribution are giving different results for the same problem?
Both yield nearly the same result:
> dpois(2,2)
[1] 0.2706706
> dbinom(2,100,.02)
[1] 0.2734139
Both results would get more similar as n tended to infinite and p tended to zero, but n=100 is large but a lot smaller than infinite, ... | Why Poisson and Binomial distribution are giving different results for the same problem?
Both yield nearly the same result:
> dpois(2,2)
[1] 0.2706706
> dbinom(2,100,.02)
[1] 0.2734139
Both results would get more similar as n tended to infinite and p tended to zero, but n=100 is large bu |
46,072 | Inferring the number of topics for gensim's LDA - perplexity, CM, AIC, and BIC | Counter intuitively, it appears that the log_perplexity function doesn't output a $perplexity$ after all (the documentation of the function wasn't clear enough for me personally), but a likelihood $bound$ which must be utilised in the perplexity's lower bound equation thus (Taken from this paper - Online Learning for L... | Inferring the number of topics for gensim's LDA - perplexity, CM, AIC, and BIC | Counter intuitively, it appears that the log_perplexity function doesn't output a $perplexity$ after all (the documentation of the function wasn't clear enough for me personally), but a likelihood $bo | Inferring the number of topics for gensim's LDA - perplexity, CM, AIC, and BIC
Counter intuitively, it appears that the log_perplexity function doesn't output a $perplexity$ after all (the documentation of the function wasn't clear enough for me personally), but a likelihood $bound$ which must be utilised in the perple... | Inferring the number of topics for gensim's LDA - perplexity, CM, AIC, and BIC
Counter intuitively, it appears that the log_perplexity function doesn't output a $perplexity$ after all (the documentation of the function wasn't clear enough for me personally), but a likelihood $bo |
46,073 | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Here is the general intuition behind shrinking co-efficients in linear regression. Borrowing figures and equations from Pattern Recognition and Machine Learning by Bishop.
Imagine that you have to approximate the function, $y = sin(2\pi x)$ from $N$ observations. You can do this using linear regression, which approxim... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Here is the general intuition behind shrinking co-efficients in linear regression. Borrowing figures and equations from Pattern Recognition and Machine Learning by Bishop.
Imagine that you have to ap | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Here is the general intuition behind shrinking co-efficients in linear regression. Borrowing figures and equations from Pattern Recognition and Machine Learning by Bishop.
Imagine that you have to approximate the function, $y = sin(2\pi x)... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Here is the general intuition behind shrinking co-efficients in linear regression. Borrowing figures and equations from Pattern Recognition and Machine Learning by Bishop.
Imagine that you have to ap |
46,074 | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | The question is what we really mean by "fit the best line." Yes, a standard unpenalized regression will be the "best line" to fit the data sample that you have. It might not, however, be the "best line" for a new sample from the population, as that standard unpenalized regression might pick up quirks that are peculiar ... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | The question is what we really mean by "fit the best line." Yes, a standard unpenalized regression will be the "best line" to fit the data sample that you have. It might not, however, be the "best lin | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
The question is what we really mean by "fit the best line." Yes, a standard unpenalized regression will be the "best line" to fit the data sample that you have. It might not, however, be the "best line" for a new sample from the population,... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
The question is what we really mean by "fit the best line." Yes, a standard unpenalized regression will be the "best line" to fit the data sample that you have. It might not, however, be the "best lin |
46,075 | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Shrinkage method will limit the value of regression coefficients, which will avoid model overfitting.
From Elements of Statistical Learning
When there are many correlated variables in a linear regression model,
their coefficients can become poorly determined and exhibit high variance.
A wildly large positive coe... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Shrinkage method will limit the value of regression coefficients, which will avoid model overfitting.
From Elements of Statistical Learning
When there are many correlated variables in a linear regr | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Shrinkage method will limit the value of regression coefficients, which will avoid model overfitting.
From Elements of Statistical Learning
When there are many correlated variables in a linear regression model,
their coefficients can b... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Shrinkage method will limit the value of regression coefficients, which will avoid model overfitting.
From Elements of Statistical Learning
When there are many correlated variables in a linear regr |
46,076 | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Ridge regression is mathematically equivalent to a Bayesian regression with a gaussian prior with $\beta$ centered around 0 (see page 11 of the Gaussian Process book here). I think that's helpful as a way to understand when to deploy ridge regression.
If you are worried that the data may generate unnaturally large $\be... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression | Ridge regression is mathematically equivalent to a Bayesian regression with a gaussian prior with $\beta$ centered around 0 (see page 11 of the Gaussian Process book here). I think that's helpful as a | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Ridge regression is mathematically equivalent to a Bayesian regression with a gaussian prior with $\beta$ centered around 0 (see page 11 of the Gaussian Process book here). I think that's helpful as a way to understand when to deploy ridge ... | Rationale behind shrinking regression coefficients in Ridge or LASSO regression
Ridge regression is mathematically equivalent to a Bayesian regression with a gaussian prior with $\beta$ centered around 0 (see page 11 of the Gaussian Process book here). I think that's helpful as a |
46,077 | What is the difference between a Spine, Register, and Index? | I couldn't find any single resource that would answer all parts of your question. Here's what I could piece together from multiple sources.
A register-based census is a type of census that relies on already existing population registers (according to Google's dictionary function, a register is simply "an official list ... | What is the difference between a Spine, Register, and Index? | I couldn't find any single resource that would answer all parts of your question. Here's what I could piece together from multiple sources.
A register-based census is a type of census that relies on a | What is the difference between a Spine, Register, and Index?
I couldn't find any single resource that would answer all parts of your question. Here's what I could piece together from multiple sources.
A register-based census is a type of census that relies on already existing population registers (according to Google's... | What is the difference between a Spine, Register, and Index?
I couldn't find any single resource that would answer all parts of your question. Here's what I could piece together from multiple sources.
A register-based census is a type of census that relies on a |
46,078 | disadvantages variational inference | Further disadvantages:
The outcome tends to depend heavily on the starting point for the optimization. Example: this paper which is heavily cited but known to have severe problems (software packages based on it were later withdrawn, etc.)
The calculations required to figure out what you are optimizing are often very c... | disadvantages variational inference | Further disadvantages:
The outcome tends to depend heavily on the starting point for the optimization. Example: this paper which is heavily cited but known to have severe problems (software packages | disadvantages variational inference
Further disadvantages:
The outcome tends to depend heavily on the starting point for the optimization. Example: this paper which is heavily cited but known to have severe problems (software packages based on it were later withdrawn, etc.)
The calculations required to figure out what... | disadvantages variational inference
Further disadvantages:
The outcome tends to depend heavily on the starting point for the optimization. Example: this paper which is heavily cited but known to have severe problems (software packages |
46,079 | disadvantages variational inference | Briefly:
Disadvantages: approximate, very little theory around it
Advantages: speed, scalability, novelty
There isn't much theory around variational inference. However you define "optimal" (qv below), you probably can't expect to obtain it.
VI is a method for approximating a difficult-to-compute probability density, ... | disadvantages variational inference | Briefly:
Disadvantages: approximate, very little theory around it
Advantages: speed, scalability, novelty
There isn't much theory around variational inference. However you define "optimal" (qv below | disadvantages variational inference
Briefly:
Disadvantages: approximate, very little theory around it
Advantages: speed, scalability, novelty
There isn't much theory around variational inference. However you define "optimal" (qv below), you probably can't expect to obtain it.
VI is a method for approximating a diffic... | disadvantages variational inference
Briefly:
Disadvantages: approximate, very little theory around it
Advantages: speed, scalability, novelty
There isn't much theory around variational inference. However you define "optimal" (qv below |
46,080 | Why is ROC curve used in assessing how 'good' a logistic regression model is? | A logistic regression doesn't "agree" with anything because the nature of the outcome is 0/1 and the nature of the prediction is a continuous probability. Agreement requires comparable scales: 0.999 does not equal 1. One way of developing a classifier from a probability is by dichotomizing at a threshold. The obvious l... | Why is ROC curve used in assessing how 'good' a logistic regression model is? | A logistic regression doesn't "agree" with anything because the nature of the outcome is 0/1 and the nature of the prediction is a continuous probability. Agreement requires comparable scales: 0.999 d | Why is ROC curve used in assessing how 'good' a logistic regression model is?
A logistic regression doesn't "agree" with anything because the nature of the outcome is 0/1 and the nature of the prediction is a continuous probability. Agreement requires comparable scales: 0.999 does not equal 1. One way of developing a c... | Why is ROC curve used in assessing how 'good' a logistic regression model is?
A logistic regression doesn't "agree" with anything because the nature of the outcome is 0/1 and the nature of the prediction is a continuous probability. Agreement requires comparable scales: 0.999 d |
46,081 | Can Convolution Neural Network be useful with encoded categorical features? | I would hazard a guess that in this case your convolution kernels start memorizing groups of features and useful correlations between consecutive pairs of features. As a concrete example, imagine we have three features: (age, gender, education,race). You can then use, say, 1x2 convolution kernels with weights $w_1,w_2$... | Can Convolution Neural Network be useful with encoded categorical features? | I would hazard a guess that in this case your convolution kernels start memorizing groups of features and useful correlations between consecutive pairs of features. As a concrete example, imagine we h | Can Convolution Neural Network be useful with encoded categorical features?
I would hazard a guess that in this case your convolution kernels start memorizing groups of features and useful correlations between consecutive pairs of features. As a concrete example, imagine we have three features: (age, gender, education,... | Can Convolution Neural Network be useful with encoded categorical features?
I would hazard a guess that in this case your convolution kernels start memorizing groups of features and useful correlations between consecutive pairs of features. As a concrete example, imagine we h |
46,082 | Can Convolution Neural Network be useful with encoded categorical features? | I think you can simply implement Convolution neural nets on categorical data. As we know there are two representation of Texts ( think of it as a huge categorical data): one-hot and embedding representation. you can merge all of your categorical data into one feature. for example by merging 3 genders and and 8 ethnicit... | Can Convolution Neural Network be useful with encoded categorical features? | I think you can simply implement Convolution neural nets on categorical data. As we know there are two representation of Texts ( think of it as a huge categorical data): one-hot and embedding represen | Can Convolution Neural Network be useful with encoded categorical features?
I think you can simply implement Convolution neural nets on categorical data. As we know there are two representation of Texts ( think of it as a huge categorical data): one-hot and embedding representation. you can merge all of your categorica... | Can Convolution Neural Network be useful with encoded categorical features?
I think you can simply implement Convolution neural nets on categorical data. As we know there are two representation of Texts ( think of it as a huge categorical data): one-hot and embedding represen |
46,083 | When a one-tailed test passes but a two-tailed test does not | Case 1: Entertaining the hypothesis that average height may have increased or decreased, we cannot reject the null hypothesis that neither has happened.
Case 2: Entertaining the hypothesis that average height may have increased, we reject the null hypothesis that it hasn't.
Both examined at the same accepted Type I e... | When a one-tailed test passes but a two-tailed test does not | Case 1: Entertaining the hypothesis that average height may have increased or decreased, we cannot reject the null hypothesis that neither has happened.
Case 2: Entertaining the hypothesis that avera | When a one-tailed test passes but a two-tailed test does not
Case 1: Entertaining the hypothesis that average height may have increased or decreased, we cannot reject the null hypothesis that neither has happened.
Case 2: Entertaining the hypothesis that average height may have increased, we reject the null hypothesis... | When a one-tailed test passes but a two-tailed test does not
Case 1: Entertaining the hypothesis that average height may have increased or decreased, we cannot reject the null hypothesis that neither has happened.
Case 2: Entertaining the hypothesis that avera |
46,084 | When a one-tailed test passes but a two-tailed test does not | As @Aksakal says, there is nothing weird about this: it is easy to see that the significance level (for a continuous random variable) is equal to the probability of a type I error.
So your one-sided and two sided test have the same type I error probability. What differs is the power of the two tests. If you know t... | When a one-tailed test passes but a two-tailed test does not | As @Aksakal says, there is nothing weird about this: it is easy to see that the significance level (for a continuous random variable) is equal to the probability of a type I error.
So your one-side | When a one-tailed test passes but a two-tailed test does not
As @Aksakal says, there is nothing weird about this: it is easy to see that the significance level (for a continuous random variable) is equal to the probability of a type I error.
So your one-sided and two sided test have the same type I error probability... | When a one-tailed test passes but a two-tailed test does not
As @Aksakal says, there is nothing weird about this: it is easy to see that the significance level (for a continuous random variable) is equal to the probability of a type I error.
So your one-side |
46,085 | When a one-tailed test passes but a two-tailed test does not | There is nothing weird about this results. The reason why this result looks weird to you is because you're using the same significance level.
The hypo 1 includes the possibility that the height either went down or up, while hypo 2 only includes the increase. So, intuitively (but not precisely) you need to compare the ... | When a one-tailed test passes but a two-tailed test does not | There is nothing weird about this results. The reason why this result looks weird to you is because you're using the same significance level.
The hypo 1 includes the possibility that the height eithe | When a one-tailed test passes but a two-tailed test does not
There is nothing weird about this results. The reason why this result looks weird to you is because you're using the same significance level.
The hypo 1 includes the possibility that the height either went down or up, while hypo 2 only includes the increase.... | When a one-tailed test passes but a two-tailed test does not
There is nothing weird about this results. The reason why this result looks weird to you is because you're using the same significance level.
The hypo 1 includes the possibility that the height eithe |
46,086 | When a one-tailed test passes but a two-tailed test does not | In many fields (e.g. medical statistics where you may be comparing a new drug vs. an old one) the convention is that one-sided tests are done at 2.5% by default (vs. two-sided ones at 5%). And that, even if you only hypothesize an effect in direction, two-sided tests are done. This convention has in part developed to p... | When a one-tailed test passes but a two-tailed test does not | In many fields (e.g. medical statistics where you may be comparing a new drug vs. an old one) the convention is that one-sided tests are done at 2.5% by default (vs. two-sided ones at 5%). And that, e | When a one-tailed test passes but a two-tailed test does not
In many fields (e.g. medical statistics where you may be comparing a new drug vs. an old one) the convention is that one-sided tests are done at 2.5% by default (vs. two-sided ones at 5%). And that, even if you only hypothesize an effect in direction, two-sid... | When a one-tailed test passes but a two-tailed test does not
In many fields (e.g. medical statistics where you may be comparing a new drug vs. an old one) the convention is that one-sided tests are done at 2.5% by default (vs. two-sided ones at 5%). And that, e |
46,087 | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models) | Seems as thought I figured this out. I'm posting it here just in case it helps someone.
Well, it turns out that I wasn't making an error with my "by-hand" calculation, but it was indeed a problem with the pgmpy package. For some reason, it was inferring independence between $A$ and $B$ given $C$ (they should be indepen... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi | Seems as thought I figured this out. I'm posting it here just in case it helps someone.
Well, it turns out that I wasn't making an error with my "by-hand" calculation, but it was indeed a problem with | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models)
Seems as thought I figured this out. I'm posting it here just in case it helps someone.
Well, it turns out that I wasn't making an error with my "by-hand" calculation, but it was indeed a problem with the pgm... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi
Seems as thought I figured this out. I'm posting it here just in case it helps someone.
Well, it turns out that I wasn't making an error with my "by-hand" calculation, but it was indeed a problem with |
46,088 | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models) | I tried installing the new package and the bug remains. So i looked at the Bayesian network code itself and found the bug.
Its in line 530 of BayesianModel.py when finding the descendents. Because
neighbors is an iterator, it is empty after visit.extend and so descendants.extend does nothing and the result is empty. ... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi | I tried installing the new package and the bug remains. So i looked at the Bayesian network code itself and found the bug.
Its in line 530 of BayesianModel.py when finding the descendents. Because
ne | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models)
I tried installing the new package and the bug remains. So i looked at the Bayesian network code itself and found the bug.
Its in line 530 of BayesianModel.py when finding the descendents. Because
neighbors ... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi
I tried installing the new package and the bug remains. So i looked at the Bayesian network code itself and found the bug.
Its in line 530 of BayesianModel.py when finding the descendents. Because
ne |
46,089 | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models) | Confirmed this works well (matches the hand calculations) in version 0.1.14. Note I had to make changes to the code, to fix both errors in how pgmpy was being used (incorrect shapes) and for Python 3. Here is the updated code.
from pgmpy.models import BayesianModel
from pgmpy.factors.discrete import TabularCPD
# Defin... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi | Confirmed this works well (matches the hand calculations) in version 0.1.14. Note I had to make changes to the code, to fix both errors in how pgmpy was being used (incorrect shapes) and for Python 3. | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphical Models)
Confirmed this works well (matches the hand calculations) in version 0.1.14. Note I had to make changes to the code, to fix both errors in how pgmpy was being used (incorrect shapes) and for Python 3. Here is... | Inconsistencies between conditional probability calculations by hand and with pgmpy (Bayesian Graphi
Confirmed this works well (matches the hand calculations) in version 0.1.14. Note I had to make changes to the code, to fix both errors in how pgmpy was being used (incorrect shapes) and for Python 3. |
46,090 | Variable transformation on Kaggle titanic problem | What people are doing is trying different things until they find something that appears to work. There is no reason in particular why transforming a variable that can take many values into a variable that can only take two values should improve matters.
Try to think of it this way. Somewhere out there is data generatin... | Variable transformation on Kaggle titanic problem | What people are doing is trying different things until they find something that appears to work. There is no reason in particular why transforming a variable that can take many values into a variable | Variable transformation on Kaggle titanic problem
What people are doing is trying different things until they find something that appears to work. There is no reason in particular why transforming a variable that can take many values into a variable that can only take two values should improve matters.
Try to think of ... | Variable transformation on Kaggle titanic problem
What people are doing is trying different things until they find something that appears to work. There is no reason in particular why transforming a variable that can take many values into a variable |
46,091 | Variable transformation on Kaggle titanic problem | Note that the IBM Watson solution to this problem was utterly wrong, and that the dataset is too small by a factor of 20 for split-sample validation to work properly. The original source dataset is at https://hbiostat.org/data . Please don't refer to a copy of the original dataset.
sibsp and parch cannot be interpret... | Variable transformation on Kaggle titanic problem | Note that the IBM Watson solution to this problem was utterly wrong, and that the dataset is too small by a factor of 20 for split-sample validation to work properly. The original source dataset is a | Variable transformation on Kaggle titanic problem
Note that the IBM Watson solution to this problem was utterly wrong, and that the dataset is too small by a factor of 20 for split-sample validation to work properly. The original source dataset is at https://hbiostat.org/data . Please don't refer to a copy of the ori... | Variable transformation on Kaggle titanic problem
Note that the IBM Watson solution to this problem was utterly wrong, and that the dataset is too small by a factor of 20 for split-sample validation to work properly. The original source dataset is a |
46,092 | Why is $n < p$ a problem for OLS regression? | I can use gradient descent on the quadratic loss function and get a
solution…
Sure, but you have fewer constraints than unknowns, so your loss function is something like a parabola in three-space: $L =(w_1+w_2-1)^2$. There is a whole space of solutions. In this example, the solutions lie on the line: $w_1+w_2=1$.
... | Why is $n < p$ a problem for OLS regression? | I can use gradient descent on the quadratic loss function and get a
solution…
Sure, but you have fewer constraints than unknowns, so your loss function is something like a parabola in three-space: | Why is $n < p$ a problem for OLS regression?
I can use gradient descent on the quadratic loss function and get a
solution…
Sure, but you have fewer constraints than unknowns, so your loss function is something like a parabola in three-space: $L =(w_1+w_2-1)^2$. There is a whole space of solutions. In this example,... | Why is $n < p$ a problem for OLS regression?
I can use gradient descent on the quadratic loss function and get a
solution…
Sure, but you have fewer constraints than unknowns, so your loss function is something like a parabola in three-space: |
46,093 | Why is $n < p$ a problem for OLS regression? | Here is a little specific example to illustrate the issue:
Suppose you want to fit a regression of $y_i$ on $x_i$, $x_i^2$ and a constant, i.e.
$$ y_i = a x_i + b x_i^2 + c + u_i $$
or, in matrix notation,
\begin{align*}
\mathbf{y} =
\begin{pmatrix}
y_1 \\
\vdots \\
y_n
\end{pmatrix}, \quad
\mathbf{X} =
\b... | Why is $n < p$ a problem for OLS regression? | Here is a little specific example to illustrate the issue:
Suppose you want to fit a regression of $y_i$ on $x_i$, $x_i^2$ and a constant, i.e.
$$ y_i = a x_i + b x_i^2 + c + u_i $$
or, in matrix not | Why is $n < p$ a problem for OLS regression?
Here is a little specific example to illustrate the issue:
Suppose you want to fit a regression of $y_i$ on $x_i$, $x_i^2$ and a constant, i.e.
$$ y_i = a x_i + b x_i^2 + c + u_i $$
or, in matrix notation,
\begin{align*}
\mathbf{y} =
\begin{pmatrix}
y_1 \\
\vdots \\
... | Why is $n < p$ a problem for OLS regression?
Here is a little specific example to illustrate the issue:
Suppose you want to fit a regression of $y_i$ on $x_i$, $x_i^2$ and a constant, i.e.
$$ y_i = a x_i + b x_i^2 + c + u_i $$
or, in matrix not |
46,094 | Why is $n < p$ a problem for OLS regression? | If $X'X$ is not invertable, the estimation $\hat \beta$ does not exists.
Check this post, @mpiktas's answer for details ("Existence" section)
What is a complete list of the usual assumptions for linear regression? | Why is $n < p$ a problem for OLS regression? | If $X'X$ is not invertable, the estimation $\hat \beta$ does not exists.
Check this post, @mpiktas's answer for details ("Existence" section)
What is a complete list of the usual assumptions for linea | Why is $n < p$ a problem for OLS regression?
If $X'X$ is not invertable, the estimation $\hat \beta$ does not exists.
Check this post, @mpiktas's answer for details ("Existence" section)
What is a complete list of the usual assumptions for linear regression? | Why is $n < p$ a problem for OLS regression?
If $X'X$ is not invertable, the estimation $\hat \beta$ does not exists.
Check this post, @mpiktas's answer for details ("Existence" section)
What is a complete list of the usual assumptions for linea |
46,095 | Convolution of random variables: unimodality of the likelihood function | The short answer is no, it is not a unique global maximum, but there is a longer answer as well.
Consider the sum of two Geometric variates with parameters $p_1 = 0.1,\space p_2 = 0.9$. If you have access to the individual variates, all is well, but if you only have access to the sum, it should be intuitively clear th... | Convolution of random variables: unimodality of the likelihood function | The short answer is no, it is not a unique global maximum, but there is a longer answer as well.
Consider the sum of two Geometric variates with parameters $p_1 = 0.1,\space p_2 = 0.9$. If you have a | Convolution of random variables: unimodality of the likelihood function
The short answer is no, it is not a unique global maximum, but there is a longer answer as well.
Consider the sum of two Geometric variates with parameters $p_1 = 0.1,\space p_2 = 0.9$. If you have access to the individual variates, all is well, b... | Convolution of random variables: unimodality of the likelihood function
The short answer is no, it is not a unique global maximum, but there is a longer answer as well.
Consider the sum of two Geometric variates with parameters $p_1 = 0.1,\space p_2 = 0.9$. If you have a |
46,096 | Convolution of random variables: unimodality of the likelihood function | jbowman provided an answer regarding the label-switching problem. But putting that aside it made me wonder about the proof for the unique global maximum given the constraint of $p_1 < p_2 < etc$. How can we be sure that the set of p-values for the maximum $\mathcal{L}$ is unique?
I could get an answer in the case of a ... | Convolution of random variables: unimodality of the likelihood function | jbowman provided an answer regarding the label-switching problem. But putting that aside it made me wonder about the proof for the unique global maximum given the constraint of $p_1 < p_2 < etc$. How | Convolution of random variables: unimodality of the likelihood function
jbowman provided an answer regarding the label-switching problem. But putting that aside it made me wonder about the proof for the unique global maximum given the constraint of $p_1 < p_2 < etc$. How can we be sure that the set of p-values for the ... | Convolution of random variables: unimodality of the likelihood function
jbowman provided an answer regarding the label-switching problem. But putting that aside it made me wonder about the proof for the unique global maximum given the constraint of $p_1 < p_2 < etc$. How |
46,097 | Is there any statistical difference between 10 Bernoulli trials and 1 binomial trial with parameter n = 10? | Yes they're different in a simple, fairly obvious sense.
Ten Bernoulli trials (presumed to be independent with with common parameter $p$) is ten 0/1 values, so an observation on it looks like "$(0,0,0,1,0,1,1,1,0,0)$".
The sum of such a vector is distributed as $\text{binomial}(10,p)$; an observation on that looks like... | Is there any statistical difference between 10 Bernoulli trials and 1 binomial trial with parameter | Yes they're different in a simple, fairly obvious sense.
Ten Bernoulli trials (presumed to be independent with with common parameter $p$) is ten 0/1 values, so an observation on it looks like "$(0,0,0 | Is there any statistical difference between 10 Bernoulli trials and 1 binomial trial with parameter n = 10?
Yes they're different in a simple, fairly obvious sense.
Ten Bernoulli trials (presumed to be independent with with common parameter $p$) is ten 0/1 values, so an observation on it looks like "$(0,0,0,1,0,1,1,1,0... | Is there any statistical difference between 10 Bernoulli trials and 1 binomial trial with parameter
Yes they're different in a simple, fairly obvious sense.
Ten Bernoulli trials (presumed to be independent with with common parameter $p$) is ten 0/1 values, so an observation on it looks like "$(0,0,0 |
46,098 | regression with constraints | Logistic regression with box constraints served me well in the past for problems similar to yours. You are interested in prediction, not in inference, so, as long a suitable estimate of generalization error (for example, 5-fold cross-validation error) is low enough, you should be ok.
Let's consider a logit link functi... | regression with constraints | Logistic regression with box constraints served me well in the past for problems similar to yours. You are interested in prediction, not in inference, so, as long a suitable estimate of generalization | regression with constraints
Logistic regression with box constraints served me well in the past for problems similar to yours. You are interested in prediction, not in inference, so, as long a suitable estimate of generalization error (for example, 5-fold cross-validation error) is low enough, you should be ok.
Let's ... | regression with constraints
Logistic regression with box constraints served me well in the past for problems similar to yours. You are interested in prediction, not in inference, so, as long a suitable estimate of generalization |
46,099 | regression with constraints | For constraint 3: dummy code $x_2$ so that you can model all categories within one regression function.
EDIT: I can think of two ways to enforce similarity between coefficients of those dummy variables but they are rather cumbersome and I do not know available implementations. If you want to go down that road, you cou... | regression with constraints | For constraint 3: dummy code $x_2$ so that you can model all categories within one regression function.
EDIT: I can think of two ways to enforce similarity between coefficients of those dummy variabl | regression with constraints
For constraint 3: dummy code $x_2$ so that you can model all categories within one regression function.
EDIT: I can think of two ways to enforce similarity between coefficients of those dummy variables but they are rather cumbersome and I do not know available implementations. If you want t... | regression with constraints
For constraint 3: dummy code $x_2$ so that you can model all categories within one regression function.
EDIT: I can think of two ways to enforce similarity between coefficients of those dummy variabl |
46,100 | regression with constraints | The only obvious tool based on the constraints is some form of Bayesian logistic regression. The reason is that your constraints would define the prior and the likelihood. For example, by assuming $\partial{f}/\partial{x_1}$ is positive, you are assuming that there is a zero probability that the $\hat{\beta}\le{0}$, ... | regression with constraints | The only obvious tool based on the constraints is some form of Bayesian logistic regression. The reason is that your constraints would define the prior and the likelihood. For example, by assuming $ | regression with constraints
The only obvious tool based on the constraints is some form of Bayesian logistic regression. The reason is that your constraints would define the prior and the likelihood. For example, by assuming $\partial{f}/\partial{x_1}$ is positive, you are assuming that there is a zero probability th... | regression with constraints
The only obvious tool based on the constraints is some form of Bayesian logistic regression. The reason is that your constraints would define the prior and the likelihood. For example, by assuming $ |
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