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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55,701 | Derivation of the distribution of $\hat{\phi}=[\hat{\phi}_1, \cdots, \hat{\phi}_p]$ in AR(p) models | It doesn't have to be thru Yule-Walker. You can use the OLS approach, namely,
$$X_t = \pmb X_{t-1}^T \pmb \phi + n_t , \qquad t = p+1,p+2,\ldots$$
where $\pmb X_{t-1} = [X_{t-1} \ldots X_{t-p} ]$and $\pmb \phi = [\phi_1 \ldots \phi_p] $.
Stacking all such equations, we get
$$\pmb y = \pmb X \pmb \phi + \pmb n$$
where ... | Derivation of the distribution of $\hat{\phi}=[\hat{\phi}_1, \cdots, \hat{\phi}_p]$ in AR(p) models | It doesn't have to be thru Yule-Walker. You can use the OLS approach, namely,
$$X_t = \pmb X_{t-1}^T \pmb \phi + n_t , \qquad t = p+1,p+2,\ldots$$
where $\pmb X_{t-1} = [X_{t-1} \ldots X_{t-p} ]$and $ | Derivation of the distribution of $\hat{\phi}=[\hat{\phi}_1, \cdots, \hat{\phi}_p]$ in AR(p) models
It doesn't have to be thru Yule-Walker. You can use the OLS approach, namely,
$$X_t = \pmb X_{t-1}^T \pmb \phi + n_t , \qquad t = p+1,p+2,\ldots$$
where $\pmb X_{t-1} = [X_{t-1} \ldots X_{t-p} ]$and $\pmb \phi = [\phi_1 ... | Derivation of the distribution of $\hat{\phi}=[\hat{\phi}_1, \cdots, \hat{\phi}_p]$ in AR(p) models
It doesn't have to be thru Yule-Walker. You can use the OLS approach, namely,
$$X_t = \pmb X_{t-1}^T \pmb \phi + n_t , \qquad t = p+1,p+2,\ldots$$
where $\pmb X_{t-1} = [X_{t-1} \ldots X_{t-p} ]$and $ |
55,702 | How to interpret the basics of a logistic regression calibration plot please? | The ticks across the x-axis represent the frequency distribution (may be called a rug plot) of the predicted probabilities. This is a way to see where there is sparsity in your predictions and where there is a relative abundance of predictions in a given area of predicted probabilities.
The "Apparent" line is essentia... | How to interpret the basics of a logistic regression calibration plot please? | The ticks across the x-axis represent the frequency distribution (may be called a rug plot) of the predicted probabilities. This is a way to see where there is sparsity in your predictions and where t | How to interpret the basics of a logistic regression calibration plot please?
The ticks across the x-axis represent the frequency distribution (may be called a rug plot) of the predicted probabilities. This is a way to see where there is sparsity in your predictions and where there is a relative abundance of prediction... | How to interpret the basics of a logistic regression calibration plot please?
The ticks across the x-axis represent the frequency distribution (may be called a rug plot) of the predicted probabilities. This is a way to see where there is sparsity in your predictions and where t |
55,703 | Using Random Forest variable importance for feature selection | You are entirely correct!
A couple of months ago I was in the exact same position when justifying a different feature selection approach in front of my supervisors. I will cite the sentence I used in my thesis, although it has not been published yet.
Since the ordering of the variables depends on all samples, the
sele... | Using Random Forest variable importance for feature selection | You are entirely correct!
A couple of months ago I was in the exact same position when justifying a different feature selection approach in front of my supervisors. I will cite the sentence I used in | Using Random Forest variable importance for feature selection
You are entirely correct!
A couple of months ago I was in the exact same position when justifying a different feature selection approach in front of my supervisors. I will cite the sentence I used in my thesis, although it has not been published yet.
Since ... | Using Random Forest variable importance for feature selection
You are entirely correct!
A couple of months ago I was in the exact same position when justifying a different feature selection approach in front of my supervisors. I will cite the sentence I used in |
55,704 | IPTW for multiple treatments | You'll want to check out McCaffrey et al. (2013) for advice on this, not Austin & Stuart (2015), which is for binary treatments only. It's not clear to me which causal estimand you want, so I'll explain how to get weights for both.
The ATE for any pair of treatments is the effect of moving everyone from one treatment t... | IPTW for multiple treatments | You'll want to check out McCaffrey et al. (2013) for advice on this, not Austin & Stuart (2015), which is for binary treatments only. It's not clear to me which causal estimand you want, so I'll expla | IPTW for multiple treatments
You'll want to check out McCaffrey et al. (2013) for advice on this, not Austin & Stuart (2015), which is for binary treatments only. It's not clear to me which causal estimand you want, so I'll explain how to get weights for both.
The ATE for any pair of treatments is the effect of moving ... | IPTW for multiple treatments
You'll want to check out McCaffrey et al. (2013) for advice on this, not Austin & Stuart (2015), which is for binary treatments only. It's not clear to me which causal estimand you want, so I'll expla |
55,705 | Why do anova(type='marginal') and anova(type='III') yield different results on lmer() models? | This is admittedly confusing, but there are a bunch of differences/limitations between anova() and car::Anova() and between lme and lmer fits. tl;dr probably best to use car::Anova() for consistent results across model types.
anova on lme: allows type="sequential" or type="marginal" (only). type="marginal" should be ... | Why do anova(type='marginal') and anova(type='III') yield different results on lmer() models? | This is admittedly confusing, but there are a bunch of differences/limitations between anova() and car::Anova() and between lme and lmer fits. tl;dr probably best to use car::Anova() for consistent r | Why do anova(type='marginal') and anova(type='III') yield different results on lmer() models?
This is admittedly confusing, but there are a bunch of differences/limitations between anova() and car::Anova() and between lme and lmer fits. tl;dr probably best to use car::Anova() for consistent results across model types.... | Why do anova(type='marginal') and anova(type='III') yield different results on lmer() models?
This is admittedly confusing, but there are a bunch of differences/limitations between anova() and car::Anova() and between lme and lmer fits. tl;dr probably best to use car::Anova() for consistent r |
55,706 | Does online data augmentation make sense? | From an optimization standpoint, repetition is nice (we want to optimize the same function). From a modeling standpoint, repetition can risk memorizing the training data without learning anything generalizable. For image data, online augmentation is motivated by observing that we can translate or add noise or otherwise... | Does online data augmentation make sense? | From an optimization standpoint, repetition is nice (we want to optimize the same function). From a modeling standpoint, repetition can risk memorizing the training data without learning anything gene | Does online data augmentation make sense?
From an optimization standpoint, repetition is nice (we want to optimize the same function). From a modeling standpoint, repetition can risk memorizing the training data without learning anything generalizable. For image data, online augmentation is motivated by observing that ... | Does online data augmentation make sense?
From an optimization standpoint, repetition is nice (we want to optimize the same function). From a modeling standpoint, repetition can risk memorizing the training data without learning anything gene |
55,707 | Is stochastic gradient descent pseudo-stochastic? | Exhausting all $N$ samples before being able to repeat a sample means that the process is not independent. However, the process is still stochastic.
Consider a shuffled deck of cards. You look at the top card and see $\mathsf{A}\spadesuit$ (Ace of Spades), and set it aside. You'll never see another $\mathsf{A}\spadesui... | Is stochastic gradient descent pseudo-stochastic? | Exhausting all $N$ samples before being able to repeat a sample means that the process is not independent. However, the process is still stochastic.
Consider a shuffled deck of cards. You look at the | Is stochastic gradient descent pseudo-stochastic?
Exhausting all $N$ samples before being able to repeat a sample means that the process is not independent. However, the process is still stochastic.
Consider a shuffled deck of cards. You look at the top card and see $\mathsf{A}\spadesuit$ (Ace of Spades), and set it as... | Is stochastic gradient descent pseudo-stochastic?
Exhausting all $N$ samples before being able to repeat a sample means that the process is not independent. However, the process is still stochastic.
Consider a shuffled deck of cards. You look at the |
55,708 | How to predict the next number in a series while having additional series of data that might affect it? | Great Question!
The general approach is called a ARMAX model
The reason for the generality of approach is that it is important to consider the following possible states of nature which not only provide complications BUT opportunities..
The big mac price might be predicted better using previous big mac prices in conju... | How to predict the next number in a series while having additional series of data that might affect | Great Question!
The general approach is called a ARMAX model
The reason for the generality of approach is that it is important to consider the following possible states of nature which not only provi | How to predict the next number in a series while having additional series of data that might affect it?
Great Question!
The general approach is called a ARMAX model
The reason for the generality of approach is that it is important to consider the following possible states of nature which not only provide complications... | How to predict the next number in a series while having additional series of data that might affect
Great Question!
The general approach is called a ARMAX model
The reason for the generality of approach is that it is important to consider the following possible states of nature which not only provi |
55,709 | How to predict the next number in a series while having additional series of data that might affect it? | One of the possible solutions: Support Vector Regression or SVR. Using machine learning programming the solution will look something like this:
var samples = [[2.5, 0], [2.5, 1], [1.6, 2]];
var targets = [2.2, 2.1, 1.5];
var regression->train(samples, targets);
result = var regression->predict([1.8, 3]);
return resul... | How to predict the next number in a series while having additional series of data that might affect | One of the possible solutions: Support Vector Regression or SVR. Using machine learning programming the solution will look something like this:
var samples = [[2.5, 0], [2.5, 1], [1.6, 2]];
var target | How to predict the next number in a series while having additional series of data that might affect it?
One of the possible solutions: Support Vector Regression or SVR. Using machine learning programming the solution will look something like this:
var samples = [[2.5, 0], [2.5, 1], [1.6, 2]];
var targets = [2.2, 2.1, 1... | How to predict the next number in a series while having additional series of data that might affect
One of the possible solutions: Support Vector Regression or SVR. Using machine learning programming the solution will look something like this:
var samples = [[2.5, 0], [2.5, 1], [1.6, 2]];
var target |
55,710 | accuracy and precision in regression vs classification | As you've pointed out, they are not the same, and sometimes refer to wildly different things (i.e., precision is a property of the model in classification, and refers to a measure of variance in regression). Unfortunately, in statistics and i'm sure other disciplines, we tend to abuse notation and use the same word to ... | accuracy and precision in regression vs classification | As you've pointed out, they are not the same, and sometimes refer to wildly different things (i.e., precision is a property of the model in classification, and refers to a measure of variance in regre | accuracy and precision in regression vs classification
As you've pointed out, they are not the same, and sometimes refer to wildly different things (i.e., precision is a property of the model in classification, and refers to a measure of variance in regression). Unfortunately, in statistics and i'm sure other disciplin... | accuracy and precision in regression vs classification
As you've pointed out, they are not the same, and sometimes refer to wildly different things (i.e., precision is a property of the model in classification, and refers to a measure of variance in regre |
55,711 | accuracy and precision in regression vs classification | Accuracy is the overall accuracy of the model.
It is the ratio of
(total number of correct predictions) / (total population)
Precision (Positive Prediction Value) on the other hand is the ratio of
(correctly predicted as positive) / (total number of positive predictions) | accuracy and precision in regression vs classification | Accuracy is the overall accuracy of the model.
It is the ratio of
(total number of correct predictions) / (total population)
Precision (Positive Prediction Value) on the other hand is the ratio of | accuracy and precision in regression vs classification
Accuracy is the overall accuracy of the model.
It is the ratio of
(total number of correct predictions) / (total population)
Precision (Positive Prediction Value) on the other hand is the ratio of
(correctly predicted as positive) / (total number of positive pr... | accuracy and precision in regression vs classification
Accuracy is the overall accuracy of the model.
It is the ratio of
(total number of correct predictions) / (total population)
Precision (Positive Prediction Value) on the other hand is the ratio of |
55,712 | Interpreting GLM with logged variable | Your model implies that the odds of re-offending are given by
$$\frac{p}{1-p} = {\rm precon}^{\beta_1}\times {\rm age}^{\beta_2}$$
where $\beta_1$ and $\beta_2$ are the regression coefficients and $p$ is the probability of re-offending.
BTW, I notice that you removed the intercept from the linear predictor by adding ~ ... | Interpreting GLM with logged variable | Your model implies that the odds of re-offending are given by
$$\frac{p}{1-p} = {\rm precon}^{\beta_1}\times {\rm age}^{\beta_2}$$
where $\beta_1$ and $\beta_2$ are the regression coefficients and $p$ | Interpreting GLM with logged variable
Your model implies that the odds of re-offending are given by
$$\frac{p}{1-p} = {\rm precon}^{\beta_1}\times {\rm age}^{\beta_2}$$
where $\beta_1$ and $\beta_2$ are the regression coefficients and $p$ is the probability of re-offending.
BTW, I notice that you removed the intercept ... | Interpreting GLM with logged variable
Your model implies that the odds of re-offending are given by
$$\frac{p}{1-p} = {\rm precon}^{\beta_1}\times {\rm age}^{\beta_2}$$
where $\beta_1$ and $\beta_2$ are the regression coefficients and $p$ |
55,713 | Creating thinned models in during Dropout process | assume we have n neurons and each neuron has the probability to be disabled.
situation 0: zero neuron remains, n neurons are disabled, C(n,0)
situation 1: only one neuron remains, n-1 neurons are disabled, C(n,1)
situation 2: only two neurons remain, n-2 neurons are disabled, C(n,2)
.
.
.
situation n: n neurons remain,... | Creating thinned models in during Dropout process | assume we have n neurons and each neuron has the probability to be disabled.
situation 0: zero neuron remains, n neurons are disabled, C(n,0)
situation 1: only one neuron remains, n-1 neurons are disa | Creating thinned models in during Dropout process
assume we have n neurons and each neuron has the probability to be disabled.
situation 0: zero neuron remains, n neurons are disabled, C(n,0)
situation 1: only one neuron remains, n-1 neurons are disabled, C(n,1)
situation 2: only two neurons remain, n-2 neurons are dis... | Creating thinned models in during Dropout process
assume we have n neurons and each neuron has the probability to be disabled.
situation 0: zero neuron remains, n neurons are disabled, C(n,0)
situation 1: only one neuron remains, n-1 neurons are disa |
55,714 | Creating thinned models in during Dropout process | The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activati... | Creating thinned models in during Dropout process | The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping ou | Creating thinned models in during Dropout process
The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but t... | Creating thinned models in during Dropout process
The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping ou |
55,715 | Creating thinned models in during Dropout process | I too haven't understood their reasoning, I always assumed it was a typo or something...
The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:
$$
\frac{n!}{r! \cdot (n-r)!}
$$
possible combinations (not $2^n$ as the authors ... | Creating thinned models in during Dropout process | I too haven't understood their reasoning, I always assumed it was a typo or something...
The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dro | Creating thinned models in during Dropout process
I too haven't understood their reasoning, I always assumed it was a typo or something...
The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:
$$
\frac{n!}{r! \cdot (n-r)!}
$... | Creating thinned models in during Dropout process
I too haven't understood their reasoning, I always assumed it was a typo or something...
The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dro |
55,716 | Creating thinned models in during Dropout process | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
This is quite simple. It can be thought as a task of g... | Creating thinned models in during Dropout process | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Creating thinned models in during Dropout process
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
This... | Creating thinned models in during Dropout process
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
55,717 | Is group lasso equivalent to ridge regression when there is 1 group | To see that group LASSO gives a similar solution to ridge regression upto a square in the penalty function, you need to look at subgradient conditions that characterize the solution of the group LASSO estimator. The best reference for this purpose I think is http://statweb.stanford.edu/~tibs/ftp/sparse-grlasso.pdf
Thes... | Is group lasso equivalent to ridge regression when there is 1 group | To see that group LASSO gives a similar solution to ridge regression upto a square in the penalty function, you need to look at subgradient conditions that characterize the solution of the group LASSO | Is group lasso equivalent to ridge regression when there is 1 group
To see that group LASSO gives a similar solution to ridge regression upto a square in the penalty function, you need to look at subgradient conditions that characterize the solution of the group LASSO estimator. The best reference for this purpose I th... | Is group lasso equivalent to ridge regression when there is 1 group
To see that group LASSO gives a similar solution to ridge regression upto a square in the penalty function, you need to look at subgradient conditions that characterize the solution of the group LASSO |
55,718 | What does it mean when I add a new variable to my linear model and the R^2 stays the same? | Seeing little to no change in $R^2$ when you add a variable to a linear model means that the variable has little to no additional explanatory power to the response over what is already in your model. As you note, this can be either because it tells you almost nothing about the response or it explains the same variation... | What does it mean when I add a new variable to my linear model and the R^2 stays the same? | Seeing little to no change in $R^2$ when you add a variable to a linear model means that the variable has little to no additional explanatory power to the response over what is already in your model. | What does it mean when I add a new variable to my linear model and the R^2 stays the same?
Seeing little to no change in $R^2$ when you add a variable to a linear model means that the variable has little to no additional explanatory power to the response over what is already in your model. As you note, this can be eith... | What does it mean when I add a new variable to my linear model and the R^2 stays the same?
Seeing little to no change in $R^2$ when you add a variable to a linear model means that the variable has little to no additional explanatory power to the response over what is already in your model. |
55,719 | What does it mean when I add a new variable to my linear model and the R^2 stays the same? | As others have alluded, seeing no change in $R^2$ when you add a variable to your regression is unusual. In finite samples, this should only happen when your new variable is a linear combination of variables already present. In this case, most standard regression routines simply exclude that variable from the regressio... | What does it mean when I add a new variable to my linear model and the R^2 stays the same? | As others have alluded, seeing no change in $R^2$ when you add a variable to your regression is unusual. In finite samples, this should only happen when your new variable is a linear combination of va | What does it mean when I add a new variable to my linear model and the R^2 stays the same?
As others have alluded, seeing no change in $R^2$ when you add a variable to your regression is unusual. In finite samples, this should only happen when your new variable is a linear combination of variables already present. In t... | What does it mean when I add a new variable to my linear model and the R^2 stays the same?
As others have alluded, seeing no change in $R^2$ when you add a variable to your regression is unusual. In finite samples, this should only happen when your new variable is a linear combination of va |
55,720 | how to understand random factor and fixed factor interaction? | This interaction between a fixed and a random factor allows for differences in behavior of the fixed factor among the random factors. Let's run that code on the data set, available in the MASS package in R. (I kept the short variable names provided in that copy of the data.)
BVmodel <- lmer(Y ~ V + (1|B/V), data=oats)
... | how to understand random factor and fixed factor interaction? | This interaction between a fixed and a random factor allows for differences in behavior of the fixed factor among the random factors. Let's run that code on the data set, available in the MASS package | how to understand random factor and fixed factor interaction?
This interaction between a fixed and a random factor allows for differences in behavior of the fixed factor among the random factors. Let's run that code on the data set, available in the MASS package in R. (I kept the short variable names provided in that c... | how to understand random factor and fixed factor interaction?
This interaction between a fixed and a random factor allows for differences in behavior of the fixed factor among the random factors. Let's run that code on the data set, available in the MASS package |
55,721 | Likelihood modification in Metropolis Hastings ratio for transformed parameter | You should notice that what you denote $p(y|f(\theta))$ is actually the same as $p(y|\theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.
... | Likelihood modification in Metropolis Hastings ratio for transformed parameter | You should notice that what you denote $p(y|f(\theta))$ is actually the same as $p(y|\theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not | Likelihood modification in Metropolis Hastings ratio for transformed parameter
You should notice that what you denote $p(y|f(\theta))$ is actually the same as $p(y|\theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable... | Likelihood modification in Metropolis Hastings ratio for transformed parameter
You should notice that what you denote $p(y|f(\theta))$ is actually the same as $p(y|\theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not |
55,722 | Generating vector image from a hand drawn picture. Machine Learning | It depends a little on the exact problem.
If you are interested only in text, then two fields come to mind:
Optical character recognition (OCR), which is specifically about text recognition from images. However, these methods tend to focus on "nice" documents, and may not be applicable to harder case (i.e., generaliz... | Generating vector image from a hand drawn picture. Machine Learning | It depends a little on the exact problem.
If you are interested only in text, then two fields come to mind:
Optical character recognition (OCR), which is specifically about text recognition from ima | Generating vector image from a hand drawn picture. Machine Learning
It depends a little on the exact problem.
If you are interested only in text, then two fields come to mind:
Optical character recognition (OCR), which is specifically about text recognition from images. However, these methods tend to focus on "nice" ... | Generating vector image from a hand drawn picture. Machine Learning
It depends a little on the exact problem.
If you are interested only in text, then two fields come to mind:
Optical character recognition (OCR), which is specifically about text recognition from ima |
55,723 | Generating vector image from a hand drawn picture. Machine Learning | In addition to user3658307's comprehensive answer I'd like to mention Generative Adversarial Networks (image to image translation GANs specifically) - they should come in handy if you can give them enough training data. The advantage over OCR-based methods would be simplicity - you don't need to build any pipeline, the... | Generating vector image from a hand drawn picture. Machine Learning | In addition to user3658307's comprehensive answer I'd like to mention Generative Adversarial Networks (image to image translation GANs specifically) - they should come in handy if you can give them en | Generating vector image from a hand drawn picture. Machine Learning
In addition to user3658307's comprehensive answer I'd like to mention Generative Adversarial Networks (image to image translation GANs specifically) - they should come in handy if you can give them enough training data. The advantage over OCR-based met... | Generating vector image from a hand drawn picture. Machine Learning
In addition to user3658307's comprehensive answer I'd like to mention Generative Adversarial Networks (image to image translation GANs specifically) - they should come in handy if you can give them en |
55,724 | What is the correct terminology for repeating groups of coin flips multiple times in a simulation? | In your simulation you want to tally up the number of successes (heads) for a series of 5 independent Bernoulli trials from 3 realization (or draws, or observations).
In R this can be done using the rbinom(n, size, prob) function, where n is the number of observations, size is the number of trials and prob the probabil... | What is the correct terminology for repeating groups of coin flips multiple times in a simulation? | In your simulation you want to tally up the number of successes (heads) for a series of 5 independent Bernoulli trials from 3 realization (or draws, or observations).
In R this can be done using the r | What is the correct terminology for repeating groups of coin flips multiple times in a simulation?
In your simulation you want to tally up the number of successes (heads) for a series of 5 independent Bernoulli trials from 3 realization (or draws, or observations).
In R this can be done using the rbinom(n, size, prob) ... | What is the correct terminology for repeating groups of coin flips multiple times in a simulation?
In your simulation you want to tally up the number of successes (heads) for a series of 5 independent Bernoulli trials from 3 realization (or draws, or observations).
In R this can be done using the r |
55,725 | What is the correct terminology for repeating groups of coin flips multiple times in a simulation? | I understand your confusion. Things become more clear once you think in terms of random variables, outcome sets, and realizations/observations of your random process/variable.
Flipping a coin
A random variable is a function mapping outcomes of a random experiment to numbers. In the case of flipping a coin, we can defin... | What is the correct terminology for repeating groups of coin flips multiple times in a simulation? | I understand your confusion. Things become more clear once you think in terms of random variables, outcome sets, and realizations/observations of your random process/variable.
Flipping a coin
A random | What is the correct terminology for repeating groups of coin flips multiple times in a simulation?
I understand your confusion. Things become more clear once you think in terms of random variables, outcome sets, and realizations/observations of your random process/variable.
Flipping a coin
A random variable is a functi... | What is the correct terminology for repeating groups of coin flips multiple times in a simulation?
I understand your confusion. Things become more clear once you think in terms of random variables, outcome sets, and realizations/observations of your random process/variable.
Flipping a coin
A random |
55,726 | Distribution of min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)? | Your notations do not help. Rephrase the problem as
$$X_1,\ldots,X_4\stackrel{\text{iid}}{\sim} f(x)$$
and
$$Y=\min\{X_1+X_2,X_1+X_3,X_1+X_4,X_2+X_3,X_2+X_4,X_3+X_4\}$$
You can then express $Y$ in terms of the order statistics
$$X_{(1)}\le X_{(2)}\le X_{(3)}\le X_{(4)}$$
and deduce that $Y$ is a specific sum of two of ... | Distribution of min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)? | Your notations do not help. Rephrase the problem as
$$X_1,\ldots,X_4\stackrel{\text{iid}}{\sim} f(x)$$
and
$$Y=\min\{X_1+X_2,X_1+X_3,X_1+X_4,X_2+X_3,X_2+X_4,X_3+X_4\}$$
You can then express $Y$ in ter | Distribution of min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)?
Your notations do not help. Rephrase the problem as
$$X_1,\ldots,X_4\stackrel{\text{iid}}{\sim} f(x)$$
and
$$Y=\min\{X_1+X_2,X_1+X_3,X_1+X_4,X_2+X_3,X_2+X_4,X_3+X_4\}$$
You can then express $Y$ in terms of the order statistics
$$X_{(1)}\le X_{(2)}\le X_{(3)}\le X_{(4)}$$
an... | Distribution of min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)?
Your notations do not help. Rephrase the problem as
$$X_1,\ldots,X_4\stackrel{\text{iid}}{\sim} f(x)$$
and
$$Y=\min\{X_1+X_2,X_1+X_3,X_1+X_4,X_2+X_3,X_2+X_4,X_3+X_4\}$$
You can then express $Y$ in ter |
55,727 | How much of a problem are autocorrelated residuals of a binary GAM (Generalized Additive model)? | The autocorrelation is going to affect any statistical inference you try to do with the model, such as testing is smooths are significant.
It is trivial to include random effects and spatio-temporal smooths in the GAM. You'll need to expand on what features you want to include in this model but, for example:
An isotop... | How much of a problem are autocorrelated residuals of a binary GAM (Generalized Additive model)? | The autocorrelation is going to affect any statistical inference you try to do with the model, such as testing is smooths are significant.
It is trivial to include random effects and spatio-temporal s | How much of a problem are autocorrelated residuals of a binary GAM (Generalized Additive model)?
The autocorrelation is going to affect any statistical inference you try to do with the model, such as testing is smooths are significant.
It is trivial to include random effects and spatio-temporal smooths in the GAM. You'... | How much of a problem are autocorrelated residuals of a binary GAM (Generalized Additive model)?
The autocorrelation is going to affect any statistical inference you try to do with the model, such as testing is smooths are significant.
It is trivial to include random effects and spatio-temporal s |
55,728 | Identically distributed vs P(X > Y) = P(Y > X) | This answer is written under the assumption that $\mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$\mathbb{P}(X<Y)=\mathbb{P}(Y<X)\tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And ob... | Identically distributed vs P(X > Y) = P(Y > X) | This answer is written under the assumption that $\mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$\mathbb{P}(X<Y)=\mathbb{P}(Y<X)\ | Identically distributed vs P(X > Y) = P(Y > X)
This answer is written under the assumption that $\mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$\mathbb{P}(X<Y)=\mathbb{P}(Y<X)\tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $... | Identically distributed vs P(X > Y) = P(Y > X)
This answer is written under the assumption that $\mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$\mathbb{P}(X<Y)=\mathbb{P}(Y<X)\ |
55,729 | Identically distributed vs P(X > Y) = P(Y > X) | I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[X\in A], [Y\in B]$ are independent. In particular for any two numbers $x,y$ the events $[X\le x], [Y\... | Identically distributed vs P(X > Y) = P(Y > X) | I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets | Identically distributed vs P(X > Y) = P(Y > X)
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[X\in A], [Y\in B]$ are independent. In particular for ... | Identically distributed vs P(X > Y) = P(Y > X)
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets |
55,730 | Multivariate conditional entropy | Definition of conditional entropy
Suppose $A,B,C,D,E$ are random variables with joint distribution $p(a,b,c,d,e)$. The conditional entropy of $E$ given $A,B,C,D$ is defined as the expected value (over the joint distribution) of $-\log p(e \mid a,b,c,d)$:
$$H(E \mid A, B, C, D) =
-\mathbb{E}_{p(a,b,c,d,e)} \big[ \log p(... | Multivariate conditional entropy | Definition of conditional entropy
Suppose $A,B,C,D,E$ are random variables with joint distribution $p(a,b,c,d,e)$. The conditional entropy of $E$ given $A,B,C,D$ is defined as the expected value (over | Multivariate conditional entropy
Definition of conditional entropy
Suppose $A,B,C,D,E$ are random variables with joint distribution $p(a,b,c,d,e)$. The conditional entropy of $E$ given $A,B,C,D$ is defined as the expected value (over the joint distribution) of $-\log p(e \mid a,b,c,d)$:
$$H(E \mid A, B, C, D) =
-\mathb... | Multivariate conditional entropy
Definition of conditional entropy
Suppose $A,B,C,D,E$ are random variables with joint distribution $p(a,b,c,d,e)$. The conditional entropy of $E$ given $A,B,C,D$ is defined as the expected value (over |
55,731 | Is there any tool to test the tendency toward/away from stationarity? | There are many such tests.
Generally, they fall into two categories:
Tests for unit root (I.e. tests of stationarity)
$H_{0}:$ Time series is/are stationary.
$H_{A}:$ Time series has/have unit root (i.e. is non-stationary).
Example: The Hadri-Lagrange multiplier test.
Tests for stationarity (I.e. tests of unit root)
$... | Is there any tool to test the tendency toward/away from stationarity? | There are many such tests.
Generally, they fall into two categories:
Tests for unit root (I.e. tests of stationarity)
$H_{0}:$ Time series is/are stationary.
$H_{A}:$ Time series has/have unit root ( | Is there any tool to test the tendency toward/away from stationarity?
There are many such tests.
Generally, they fall into two categories:
Tests for unit root (I.e. tests of stationarity)
$H_{0}:$ Time series is/are stationary.
$H_{A}:$ Time series has/have unit root (i.e. is non-stationary).
Example: The Hadri-Lagran... | Is there any tool to test the tendency toward/away from stationarity?
There are many such tests.
Generally, they fall into two categories:
Tests for unit root (I.e. tests of stationarity)
$H_{0}:$ Time series is/are stationary.
$H_{A}:$ Time series has/have unit root ( |
55,732 | Is there any tool to test the tendency toward/away from stationarity? | You can plot the time-series and inspect it visually. A stationary time-series will have constant variance and have a constant expected value (edit to incorporate @Richard Hardy's comment). plot(Value~Time)
You could look at the ACF and PACF plots. They should die down if the series is stationary.
There is also the Aug... | Is there any tool to test the tendency toward/away from stationarity? | You can plot the time-series and inspect it visually. A stationary time-series will have constant variance and have a constant expected value (edit to incorporate @Richard Hardy's comment). plot(Value | Is there any tool to test the tendency toward/away from stationarity?
You can plot the time-series and inspect it visually. A stationary time-series will have constant variance and have a constant expected value (edit to incorporate @Richard Hardy's comment). plot(Value~Time)
You could look at the ACF and PACF plots. T... | Is there any tool to test the tendency toward/away from stationarity?
You can plot the time-series and inspect it visually. A stationary time-series will have constant variance and have a constant expected value (edit to incorporate @Richard Hardy's comment). plot(Value |
55,733 | Can a Bernoulli distribution be approximated by a Normal distribution? | Let's analyze the error.
The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show where the functions differ appreciably.
(Why plot distribution functions instead of density functions? Because a Bernoul... | Can a Bernoulli distribution be approximated by a Normal distribution? | Let's analyze the error.
The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show wh | Can a Bernoulli distribution be approximated by a Normal distribution?
Let's analyze the error.
The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show where the functions differ appreciably.
(Why plot ... | Can a Bernoulli distribution be approximated by a Normal distribution?
Let's analyze the error.
The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show wh |
55,734 | Can a Bernoulli distribution be approximated by a Normal distribution? | I don't think you can conclude that N(p,p(1−p)) could represent an approximation of bernoulli(p). First of all, for a bernoulli variable, a random sample could only be 0 or 1, on the other hand, the range of normal variable could be from -inf to inf. Secondly, If we have a random distribution with mean p, and variance... | Can a Bernoulli distribution be approximated by a Normal distribution? | I don't think you can conclude that N(p,p(1−p)) could represent an approximation of bernoulli(p). First of all, for a bernoulli variable, a random sample could only be 0 or 1, on the other hand, the r | Can a Bernoulli distribution be approximated by a Normal distribution?
I don't think you can conclude that N(p,p(1−p)) could represent an approximation of bernoulli(p). First of all, for a bernoulli variable, a random sample could only be 0 or 1, on the other hand, the range of normal variable could be from -inf to inf... | Can a Bernoulli distribution be approximated by a Normal distribution?
I don't think you can conclude that N(p,p(1−p)) could represent an approximation of bernoulli(p). First of all, for a bernoulli variable, a random sample could only be 0 or 1, on the other hand, the r |
55,735 | u-shape for logistic regression? | Yes. Include a quadratic term for wine units consumed. The statistical significance of this term may indicate the presence of an inflection point, at which point the linear trend pivots. It may also indicate an "acceleration" effect, where sequentially higher or lower doses may have escalating trends with the outcome r... | u-shape for logistic regression? | Yes. Include a quadratic term for wine units consumed. The statistical significance of this term may indicate the presence of an inflection point, at which point the linear trend pivots. It may also i | u-shape for logistic regression?
Yes. Include a quadratic term for wine units consumed. The statistical significance of this term may indicate the presence of an inflection point, at which point the linear trend pivots. It may also indicate an "acceleration" effect, where sequentially higher or lower doses may have esc... | u-shape for logistic regression?
Yes. Include a quadratic term for wine units consumed. The statistical significance of this term may indicate the presence of an inflection point, at which point the linear trend pivots. It may also i |
55,736 | How is an RNN (or any neural network) a parametric model? | A neural network is defined by the weights on its connections, which are its parameters. It doesn't matter what data the network was trained upon, once you have a set of weights, you can throw away your training dataset without repercussion. If you want to classify a new sample, you can do it with only the parameterize... | How is an RNN (or any neural network) a parametric model? | A neural network is defined by the weights on its connections, which are its parameters. It doesn't matter what data the network was trained upon, once you have a set of weights, you can throw away yo | How is an RNN (or any neural network) a parametric model?
A neural network is defined by the weights on its connections, which are its parameters. It doesn't matter what data the network was trained upon, once you have a set of weights, you can throw away your training dataset without repercussion. If you want to class... | How is an RNN (or any neural network) a parametric model?
A neural network is defined by the weights on its connections, which are its parameters. It doesn't matter what data the network was trained upon, once you have a set of weights, you can throw away yo |
55,737 | Classification when Learning is not Feasible | A decision tree will always try to find splits and fit the data, whether it is "noise" or not. The nodes will end up being more pure than the initial full distribution. However, you'll likely get very close to a score that represents the distribution of the class labels.
The only way to get a single node is if all th... | Classification when Learning is not Feasible | A decision tree will always try to find splits and fit the data, whether it is "noise" or not. The nodes will end up being more pure than the initial full distribution. However, you'll likely get ve | Classification when Learning is not Feasible
A decision tree will always try to find splits and fit the data, whether it is "noise" or not. The nodes will end up being more pure than the initial full distribution. However, you'll likely get very close to a score that represents the distribution of the class labels.
T... | Classification when Learning is not Feasible
A decision tree will always try to find splits and fit the data, whether it is "noise" or not. The nodes will end up being more pure than the initial full distribution. However, you'll likely get ve |
55,738 | Classification when Learning is not Feasible | Usually, the "empty" decision tree you are referring to is a single node (e.g: in WEKA). It contains all the data (no splits). Other ml algorithms have some kind of "empty" model as well.
On real-life data, this will not happen, and model complexity in this setting will be "more than 0".
Class distributions will vary ... | Classification when Learning is not Feasible | Usually, the "empty" decision tree you are referring to is a single node (e.g: in WEKA). It contains all the data (no splits). Other ml algorithms have some kind of "empty" model as well.
On real-life | Classification when Learning is not Feasible
Usually, the "empty" decision tree you are referring to is a single node (e.g: in WEKA). It contains all the data (no splits). Other ml algorithms have some kind of "empty" model as well.
On real-life data, this will not happen, and model complexity in this setting will be "... | Classification when Learning is not Feasible
Usually, the "empty" decision tree you are referring to is a single node (e.g: in WEKA). It contains all the data (no splits). Other ml algorithms have some kind of "empty" model as well.
On real-life |
55,739 | Classification when Learning is not Feasible | This really depends how you define "noise".
Assume you have training data and there exists a relationship between your target and predictors over some partition of the data. Then your tree will find a way in which to split and make predictions it believes to be better than the mean. You then apply your model to a test... | Classification when Learning is not Feasible | This really depends how you define "noise".
Assume you have training data and there exists a relationship between your target and predictors over some partition of the data. Then your tree will find | Classification when Learning is not Feasible
This really depends how you define "noise".
Assume you have training data and there exists a relationship between your target and predictors over some partition of the data. Then your tree will find a way in which to split and make predictions it believes to be better than ... | Classification when Learning is not Feasible
This really depends how you define "noise".
Assume you have training data and there exists a relationship between your target and predictors over some partition of the data. Then your tree will find |
55,740 | Classification when Learning is not Feasible | This is my first time answering a question in this site.
I think this setting violates the requisite universal in almost all machine learning algorithms, that is the examples $(\mathbf{x}_i,y_i)$ have to be independently drawn from same underlying distribution $\mathcal{D}: \mathcal{X}\times \mathcal{Y}$. In this sense... | Classification when Learning is not Feasible | This is my first time answering a question in this site.
I think this setting violates the requisite universal in almost all machine learning algorithms, that is the examples $(\mathbf{x}_i,y_i)$ have | Classification when Learning is not Feasible
This is my first time answering a question in this site.
I think this setting violates the requisite universal in almost all machine learning algorithms, that is the examples $(\mathbf{x}_i,y_i)$ have to be independently drawn from same underlying distribution $\mathcal{D}: ... | Classification when Learning is not Feasible
This is my first time answering a question in this site.
I think this setting violates the requisite universal in almost all machine learning algorithms, that is the examples $(\mathbf{x}_i,y_i)$ have |
55,741 | How can I mix image and data into a CNN | You need to define sub-modules of the network and then somehow merge them and do further processing on the whole data. This is usually done by creating smaller neural networks within the bigger one. For example, you have one sub-network that processes images (say convolutional network) and another one that processes ta... | How can I mix image and data into a CNN | You need to define sub-modules of the network and then somehow merge them and do further processing on the whole data. This is usually done by creating smaller neural networks within the bigger one. F | How can I mix image and data into a CNN
You need to define sub-modules of the network and then somehow merge them and do further processing on the whole data. This is usually done by creating smaller neural networks within the bigger one. For example, you have one sub-network that processes images (say convolutional ne... | How can I mix image and data into a CNN
You need to define sub-modules of the network and then somehow merge them and do further processing on the whole data. This is usually done by creating smaller neural networks within the bigger one. F |
55,742 | Infinitesimal independence | If the $X$ and $Y$ rv's have a joint density $f(x,y)$, then you can compute the marginals of $X$ and $Y$, say $g(x), h(y)$ and compare the joint density to the independence density $g(x)h(y)$, for example via the likelihood ratio
$$
\frac{f(x,y)}{g(x)h(y)}
$$
and plot that. In the following we have done this for th... | Infinitesimal independence | If the $X$ and $Y$ rv's have a joint density $f(x,y)$, then you can compute the marginals of $X$ and $Y$, say $g(x), h(y)$ and compare the joint density to the independence density $g(x)h(y)$, for exa | Infinitesimal independence
If the $X$ and $Y$ rv's have a joint density $f(x,y)$, then you can compute the marginals of $X$ and $Y$, say $g(x), h(y)$ and compare the joint density to the independence density $g(x)h(y)$, for example via the likelihood ratio
$$
\frac{f(x,y)}{g(x)h(y)}
$$
and plot that. In the followi... | Infinitesimal independence
If the $X$ and $Y$ rv's have a joint density $f(x,y)$, then you can compute the marginals of $X$ and $Y$, say $g(x), h(y)$ and compare the joint density to the independence density $g(x)h(y)$, for exa |
55,743 | Infinitesimal independence | In fact I realized my definition with the small interval around points was weird: I realized there exist the Pointwise Mutual Information (https://en.wikipedia.org/wiki/Pointwise_mutual_information), that will simply measure the independence between two random variables at a particular point x, y. I guess it was what I... | Infinitesimal independence | In fact I realized my definition with the small interval around points was weird: I realized there exist the Pointwise Mutual Information (https://en.wikipedia.org/wiki/Pointwise_mutual_information), | Infinitesimal independence
In fact I realized my definition with the small interval around points was weird: I realized there exist the Pointwise Mutual Information (https://en.wikipedia.org/wiki/Pointwise_mutual_information), that will simply measure the independence between two random variables at a particular point ... | Infinitesimal independence
In fact I realized my definition with the small interval around points was weird: I realized there exist the Pointwise Mutual Information (https://en.wikipedia.org/wiki/Pointwise_mutual_information), |
55,744 | Collinearity of features and random forest | Actually, the blog post does not say that there is an issue with correlated features. It says only that the feature importances that they calculated did not yield correct answer. Now, this does not have to be a problem with random forest itself, but with the feature importance algorithm they used (for example, the defa... | Collinearity of features and random forest | Actually, the blog post does not say that there is an issue with correlated features. It says only that the feature importances that they calculated did not yield correct answer. Now, this does not ha | Collinearity of features and random forest
Actually, the blog post does not say that there is an issue with correlated features. It says only that the feature importances that they calculated did not yield correct answer. Now, this does not have to be a problem with random forest itself, but with the feature importance... | Collinearity of features and random forest
Actually, the blog post does not say that there is an issue with correlated features. It says only that the feature importances that they calculated did not yield correct answer. Now, this does not ha |
55,745 | Collinearity of features and random forest | As pointed out in the comments you should clarify what is your purpose and the data at hands. Several models such as partial least squares take advantage of correlated predictors.
The findCorrelation function in the caret package could serve to detect and remove correlated predictors according to prespecified treshold... | Collinearity of features and random forest | As pointed out in the comments you should clarify what is your purpose and the data at hands. Several models such as partial least squares take advantage of correlated predictors.
The findCorrelation | Collinearity of features and random forest
As pointed out in the comments you should clarify what is your purpose and the data at hands. Several models such as partial least squares take advantage of correlated predictors.
The findCorrelation function in the caret package could serve to detect and remove correlated pr... | Collinearity of features and random forest
As pointed out in the comments you should clarify what is your purpose and the data at hands. Several models such as partial least squares take advantage of correlated predictors.
The findCorrelation |
55,746 | Collinearity of features and random forest | @Tim made some excellent points. One method of determining what (if anything) colinearity is doing to your results is to use the perturb package in R. This adds random noise to the variables and then sees what happens to the results. It requires "some kind of model object" but that need not be a regression. I don't kno... | Collinearity of features and random forest | @Tim made some excellent points. One method of determining what (if anything) colinearity is doing to your results is to use the perturb package in R. This adds random noise to the variables and then | Collinearity of features and random forest
@Tim made some excellent points. One method of determining what (if anything) colinearity is doing to your results is to use the perturb package in R. This adds random noise to the variables and then sees what happens to the results. It requires "some kind of model object" but... | Collinearity of features and random forest
@Tim made some excellent points. One method of determining what (if anything) colinearity is doing to your results is to use the perturb package in R. This adds random noise to the variables and then |
55,747 | Proof of Convergence in Distribution with unbounded moment | Firstly define $Y_{k,n} = X_k 1\{ |X_k| \leq n \}$. Then it is easy to see that $Var(Y_{k,n}) = 2 \log n$ and that
$$Var (T_n ) = Var \left( \sum_{k=1}^n Y_{k,n} \right) = 2n \log n$$
Letting $S_n = \sum_{k=1}^n X_k$ we also see that
$$P(S_n \neq T_n) \leq P(\cup_k X_k \neq Y_{k,n}) \leq n P(X_k > n) = \frac{n}{2n^2} \... | Proof of Convergence in Distribution with unbounded moment | Firstly define $Y_{k,n} = X_k 1\{ |X_k| \leq n \}$. Then it is easy to see that $Var(Y_{k,n}) = 2 \log n$ and that
$$Var (T_n ) = Var \left( \sum_{k=1}^n Y_{k,n} \right) = 2n \log n$$
Letting $S_n = \ | Proof of Convergence in Distribution with unbounded moment
Firstly define $Y_{k,n} = X_k 1\{ |X_k| \leq n \}$. Then it is easy to see that $Var(Y_{k,n}) = 2 \log n$ and that
$$Var (T_n ) = Var \left( \sum_{k=1}^n Y_{k,n} \right) = 2n \log n$$
Letting $S_n = \sum_{k=1}^n X_k$ we also see that
$$P(S_n \neq T_n) \leq P(\c... | Proof of Convergence in Distribution with unbounded moment
Firstly define $Y_{k,n} = X_k 1\{ |X_k| \leq n \}$. Then it is easy to see that $Var(Y_{k,n}) = 2 \log n$ and that
$$Var (T_n ) = Var \left( \sum_{k=1}^n Y_{k,n} \right) = 2n \log n$$
Letting $S_n = \ |
55,748 | Proof of Convergence in Distribution with unbounded moment | This is a proof by c.f. approach:
The c.f. of $X_i$ is
$$
\phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx.
$$
Hence, for $Y_n = (X_1+X_2+\dots+X_n)(\sqrt{n\log n})^{-1}$, we have
\begin{align*}
\phi_{Y_n}(t) =& \phi_i\left(\frac{t}{\sqr... | Proof of Convergence in Distribution with unbounded moment | This is a proof by c.f. approach:
The c.f. of $X_i$ is
$$
\phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx.
$$
Hence, for $Y_n | Proof of Convergence in Distribution with unbounded moment
This is a proof by c.f. approach:
The c.f. of $X_i$ is
$$
\phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx.
$$
Hence, for $Y_n = (X_1+X_2+\dots+X_n)(\sqrt{n\log n})^{-1}$, we have
... | Proof of Convergence in Distribution with unbounded moment
This is a proof by c.f. approach:
The c.f. of $X_i$ is
$$
\phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx.
$$
Hence, for $Y_n |
55,749 | Proof of Convergence in Distribution with unbounded moment | To see if this gets us anywhere useful, I'm going to go some of the way along the lines suggested by Glen_b in the comments. The characteristic function of the underlying random variables is:
$$\begin{equation} \begin{aligned}
\varphi_X(t) = \mathbb{E}(\exp(itX))
&= \int \limits_{\mathbb{R}} \exp(itx) f_X(x) dx \\[6pt... | Proof of Convergence in Distribution with unbounded moment | To see if this gets us anywhere useful, I'm going to go some of the way along the lines suggested by Glen_b in the comments. The characteristic function of the underlying random variables is:
$$\begi | Proof of Convergence in Distribution with unbounded moment
To see if this gets us anywhere useful, I'm going to go some of the way along the lines suggested by Glen_b in the comments. The characteristic function of the underlying random variables is:
$$\begin{equation} \begin{aligned}
\varphi_X(t) = \mathbb{E}(\exp(it... | Proof of Convergence in Distribution with unbounded moment
To see if this gets us anywhere useful, I'm going to go some of the way along the lines suggested by Glen_b in the comments. The characteristic function of the underlying random variables is:
$$\begi |
55,750 | Proof of Convergence in Distribution with unbounded moment | When I started reading this question I was a bit confused. This factor $\sqrt{n\log n}$ is not intuitive to me. It is not the typical expression in the CLT.
Below I am trying to view your question in a more intuitive way without resorting to characteristic functions and looking at the limits of higher moments (which wo... | Proof of Convergence in Distribution with unbounded moment | When I started reading this question I was a bit confused. This factor $\sqrt{n\log n}$ is not intuitive to me. It is not the typical expression in the CLT.
Below I am trying to view your question in | Proof of Convergence in Distribution with unbounded moment
When I started reading this question I was a bit confused. This factor $\sqrt{n\log n}$ is not intuitive to me. It is not the typical expression in the CLT.
Below I am trying to view your question in a more intuitive way without resorting to characteristic func... | Proof of Convergence in Distribution with unbounded moment
When I started reading this question I was a bit confused. This factor $\sqrt{n\log n}$ is not intuitive to me. It is not the typical expression in the CLT.
Below I am trying to view your question in |
55,751 | Why do my XGboosted trees all look the same? | When a gradient boosting machine fits a tree $f(x)$ to the target variable $y$, it calculates the error (used in the next iteration to fit the next tree) as:
$$e = y - \epsilon f(x)$$
where in this case $\epsilon = 0.1$, the learning rate. Now imagine that there is a fairly simple tree $g(x)$ that actually fits the da... | Why do my XGboosted trees all look the same? | When a gradient boosting machine fits a tree $f(x)$ to the target variable $y$, it calculates the error (used in the next iteration to fit the next tree) as:
$$e = y - \epsilon f(x)$$
where in this ca | Why do my XGboosted trees all look the same?
When a gradient boosting machine fits a tree $f(x)$ to the target variable $y$, it calculates the error (used in the next iteration to fit the next tree) as:
$$e = y - \epsilon f(x)$$
where in this case $\epsilon = 0.1$, the learning rate. Now imagine that there is a fairly... | Why do my XGboosted trees all look the same?
When a gradient boosting machine fits a tree $f(x)$ to the target variable $y$, it calculates the error (used in the next iteration to fit the next tree) as:
$$e = y - \epsilon f(x)$$
where in this ca |
55,752 | Uniqueness of Reproducing Kernel Hilbert Spaces | Prof. Dino Sejdinovic, author of the slides my question referred to, was kind enough to send me the following answer:
Feature spaces are not unique and indeed they are all isomorphic. RKHS
of a given kernel is unique as a space of functions, i.e. the one
that contains functions of the form $k(\cdot , x)$. If these... | Uniqueness of Reproducing Kernel Hilbert Spaces | Prof. Dino Sejdinovic, author of the slides my question referred to, was kind enough to send me the following answer:
Feature spaces are not unique and indeed they are all isomorphic. RKHS
of a giv | Uniqueness of Reproducing Kernel Hilbert Spaces
Prof. Dino Sejdinovic, author of the slides my question referred to, was kind enough to send me the following answer:
Feature spaces are not unique and indeed they are all isomorphic. RKHS
of a given kernel is unique as a space of functions, i.e. the one
that contain... | Uniqueness of Reproducing Kernel Hilbert Spaces
Prof. Dino Sejdinovic, author of the slides my question referred to, was kind enough to send me the following answer:
Feature spaces are not unique and indeed they are all isomorphic. RKHS
of a giv |
55,753 | Estimating the MLE where the parameter is also the constraint | Your density function is:
$$p_X(x|\alpha,\beta) = \frac{\alpha}{\beta} \Big( \frac{x}{\beta} \Big)^{\alpha-1} \quad \quad \quad \text{for } 0 \leqslant x \leqslant \beta.$$
Hence, your log-likelihood function is:
$$\ell_\mathbf{x}(\alpha, \beta) = n \ln \alpha - n \alpha \ln \beta + (\alpha-1) \sum_{i=1}^n \ln x_i \qua... | Estimating the MLE where the parameter is also the constraint | Your density function is:
$$p_X(x|\alpha,\beta) = \frac{\alpha}{\beta} \Big( \frac{x}{\beta} \Big)^{\alpha-1} \quad \quad \quad \text{for } 0 \leqslant x \leqslant \beta.$$
Hence, your log-likelihood | Estimating the MLE where the parameter is also the constraint
Your density function is:
$$p_X(x|\alpha,\beta) = \frac{\alpha}{\beta} \Big( \frac{x}{\beta} \Big)^{\alpha-1} \quad \quad \quad \text{for } 0 \leqslant x \leqslant \beta.$$
Hence, your log-likelihood function is:
$$\ell_\mathbf{x}(\alpha, \beta) = n \ln \alp... | Estimating the MLE where the parameter is also the constraint
Your density function is:
$$p_X(x|\alpha,\beta) = \frac{\alpha}{\beta} \Big( \frac{x}{\beta} \Big)^{\alpha-1} \quad \quad \quad \text{for } 0 \leqslant x \leqslant \beta.$$
Hence, your log-likelihood |
55,754 | Estimating the MLE where the parameter is also the constraint | Looks like both $\alpha$ and $\beta$ are unknown here. So our parameter is $\theta=(\alpha,\beta)$.
The population pdf is $$f_{\theta}(x)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}\mathbf1_{0<x<\beta}\quad,\,\alpha>0$$
So, given the sample $(x_1,x_2,\ldots,x_n)$, likelihood function of $\theta$ is
\begin{align}
L(\theta... | Estimating the MLE where the parameter is also the constraint | Looks like both $\alpha$ and $\beta$ are unknown here. So our parameter is $\theta=(\alpha,\beta)$.
The population pdf is $$f_{\theta}(x)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}\mathbf1_{0<x<\beta}\ | Estimating the MLE where the parameter is also the constraint
Looks like both $\alpha$ and $\beta$ are unknown here. So our parameter is $\theta=(\alpha,\beta)$.
The population pdf is $$f_{\theta}(x)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}\mathbf1_{0<x<\beta}\quad,\,\alpha>0$$
So, given the sample $(x_1,x_2,\ldots,x_... | Estimating the MLE where the parameter is also the constraint
Looks like both $\alpha$ and $\beta$ are unknown here. So our parameter is $\theta=(\alpha,\beta)$.
The population pdf is $$f_{\theta}(x)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}\mathbf1_{0<x<\beta}\ |
55,755 | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Let $\gamma_{ij} \sim \texttt{gamma}(A a_{ij})$ independently. Recall that construction of the Dirichlet distribution as a normalization of gamma random variables. Then the Dirichlet distribution we start with is equal in distribution to
$$
\left(\frac{\gamma_{00}}{\sum \gamma_{ij}}, \ldots, \frac{\gamma_{11}}{\sum \g... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Let $\gamma_{ij} \sim \texttt{gamma}(A a_{ij})$ independently. Recall that construction of the Dirichlet distribution as a normalization of gamma random variables. Then the Dirichlet distribution we s | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Let $\gamma_{ij} \sim \texttt{gamma}(A a_{ij})$ independently. Recall that construction of the Dirichlet distribution as a normalization of gamma random variables. Then the Dirichlet distribution we start with is equal in distrib... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Let $\gamma_{ij} \sim \texttt{gamma}(A a_{ij})$ independently. Recall that construction of the Dirichlet distribution as a normalization of gamma random variables. Then the Dirichlet distribution we s |
55,756 | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Edit:
(I decided not to delete this answer, as it contains a proof of the distributive property of the Dirichlet Distribution. I have however now managed to answer the original post, which I put in a separate answer)
I generally think about these problems using the fundamental theorem of calculus. You make a new variab... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Edit:
(I decided not to delete this answer, as it contains a proof of the distributive property of the Dirichlet Distribution. I have however now managed to answer the original post, which I put in a | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Edit:
(I decided not to delete this answer, as it contains a proof of the distributive property of the Dirichlet Distribution. I have however now managed to answer the original post, which I put in a separate answer)
I generally ... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Edit:
(I decided not to delete this answer, as it contains a proof of the distributive property of the Dirichlet Distribution. I have however now managed to answer the original post, which I put in a |
55,757 | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Right, I've finally solved this, and decided to add another answer rather than edit my original one as I've edited it a lot of times and it's getting messy.
From the joint distribution $P(\underline{q})=\frac{q_{1}^{\alpha_{1}-1}q_{2}^{\alpha_{2}-1}q_{3}^{\alpha_{3}-1}q_{4}^{\alpha_{4}-1}}{B(\underline{\alpha})}$
def... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution | Right, I've finally solved this, and decided to add another answer rather than edit my original one as I've edited it a lot of times and it's getting messy.
From the joint distribution $P(\underline{ | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Right, I've finally solved this, and decided to add another answer rather than edit my original one as I've edited it a lot of times and it's getting messy.
From the joint distribution $P(\underline{q})=\frac{q_{1}^{\alpha_{1}-1... | Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution
Right, I've finally solved this, and decided to add another answer rather than edit my original one as I've edited it a lot of times and it's getting messy.
From the joint distribution $P(\underline{ |
55,758 | What are global sensitivity and local sensitivity in differential privacy? | I have read the reference to the definition in Zhu's book. The paper is 《Smooth Sensitivity and Sampling in Private Data Analysis》, which gives more distinct definitons of global sensitivity and local sensitivity[1].
Obviously, $GS_f=max_{x}LS_f(x)$
[1] K. Nissim, S. Raskhodnikova, and A. Smith, “Smooth sensitivity a... | What are global sensitivity and local sensitivity in differential privacy? | I have read the reference to the definition in Zhu's book. The paper is 《Smooth Sensitivity and Sampling in Private Data Analysis》, which gives more distinct definitons of global sensitivity and local | What are global sensitivity and local sensitivity in differential privacy?
I have read the reference to the definition in Zhu's book. The paper is 《Smooth Sensitivity and Sampling in Private Data Analysis》, which gives more distinct definitons of global sensitivity and local sensitivity[1].
Obviously, $GS_f=max_{x}LS... | What are global sensitivity and local sensitivity in differential privacy?
I have read the reference to the definition in Zhu's book. The paper is 《Smooth Sensitivity and Sampling in Private Data Analysis》, which gives more distinct definitons of global sensitivity and local |
55,759 | What are global sensitivity and local sensitivity in differential privacy? | Global Sensitivity (GS) depends only on the function f. When trying to figure out the GS of function f, we examine all possible pairs of neighboring datasets in the domain of the function f.
Local Sensitivity (LS) depends on both the function f and the data set at hand D (known as an instance). When trying to figure ou... | What are global sensitivity and local sensitivity in differential privacy? | Global Sensitivity (GS) depends only on the function f. When trying to figure out the GS of function f, we examine all possible pairs of neighboring datasets in the domain of the function f.
Local Sen | What are global sensitivity and local sensitivity in differential privacy?
Global Sensitivity (GS) depends only on the function f. When trying to figure out the GS of function f, we examine all possible pairs of neighboring datasets in the domain of the function f.
Local Sensitivity (LS) depends on both the function f ... | What are global sensitivity and local sensitivity in differential privacy?
Global Sensitivity (GS) depends only on the function f. When trying to figure out the GS of function f, we examine all possible pairs of neighboring datasets in the domain of the function f.
Local Sen |
55,760 | Binary predictor with highly skewed distribution | In general, it's not an issue; you should keep it if it makes sense to be in the model, which presumably it does or it wouldn't be there to begin with.
Consider, for example, a model for weekly sales of chayote squash in the New Orleans area (see https://en.wikipedia.org/wiki/Chayote, down in the "Americas" section.)... | Binary predictor with highly skewed distribution | In general, it's not an issue; you should keep it if it makes sense to be in the model, which presumably it does or it wouldn't be there to begin with.
Consider, for example, a model for weekly sale | Binary predictor with highly skewed distribution
In general, it's not an issue; you should keep it if it makes sense to be in the model, which presumably it does or it wouldn't be there to begin with.
Consider, for example, a model for weekly sales of chayote squash in the New Orleans area (see https://en.wikipedia.o... | Binary predictor with highly skewed distribution
In general, it's not an issue; you should keep it if it makes sense to be in the model, which presumably it does or it wouldn't be there to begin with.
Consider, for example, a model for weekly sale |
55,761 | Practical limits to collinearity problems? | Why am I relatively unlikely to get, say, an x_1 coefficient of -105 and an x_2 coefficient of 110? Those add up to 5, but there is something pushing the results toward 2.5, 2.5.
Linear combinations of your two variables $x_1 = x + \epsilon_1$ and $x_2 = x+\epsilon_2$ can be described like:
$$ \frac{a+b}{2} (x + \epsil... | Practical limits to collinearity problems? | Why am I relatively unlikely to get, say, an x_1 coefficient of -105 and an x_2 coefficient of 110? Those add up to 5, but there is something pushing the results toward 2.5, 2.5.
Linear combinations o | Practical limits to collinearity problems?
Why am I relatively unlikely to get, say, an x_1 coefficient of -105 and an x_2 coefficient of 110? Those add up to 5, but there is something pushing the results toward 2.5, 2.5.
Linear combinations of your two variables $x_1 = x + \epsilon_1$ and $x_2 = x+\epsilon_2$ can be d... | Practical limits to collinearity problems?
Why am I relatively unlikely to get, say, an x_1 coefficient of -105 and an x_2 coefficient of 110? Those add up to 5, but there is something pushing the results toward 2.5, 2.5.
Linear combinations o |
55,762 | Practical limits to collinearity problems? | In the way you've added noise, you could write $x' = x +\epsilon$ (where $\epsilon$ is a normally distributed variable representing noise). Furthermore, $y=5x+4+\eta$, where $\eta$ is another Gaussian noise term.
You are trying to fit a regression of the form $Ax + Bx' +C$, and you know that you're fitting it to a ta... | Practical limits to collinearity problems? | In the way you've added noise, you could write $x' = x +\epsilon$ (where $\epsilon$ is a normally distributed variable representing noise). Furthermore, $y=5x+4+\eta$, where $\eta$ is another Gaussian | Practical limits to collinearity problems?
In the way you've added noise, you could write $x' = x +\epsilon$ (where $\epsilon$ is a normally distributed variable representing noise). Furthermore, $y=5x+4+\eta$, where $\eta$ is another Gaussian noise term.
You are trying to fit a regression of the form $Ax + Bx' +C$, ... | Practical limits to collinearity problems?
In the way you've added noise, you could write $x' = x +\epsilon$ (where $\epsilon$ is a normally distributed variable representing noise). Furthermore, $y=5x+4+\eta$, where $\eta$ is another Gaussian |
55,763 | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$ | I provided a proof of the Eckart-Young theorem in my answer to What norm of the reconstruction error is minimized by the low-rank approximation matrix obtained with PCA?. Let's take a look at the first steps of this proof:
We want to find matrix $A$ of rank $k$ that minimizes $\|X-A\|^2_F$. We can factorize $A=BW^\top... | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$ | I provided a proof of the Eckart-Young theorem in my answer to What norm of the reconstruction error is minimized by the low-rank approximation matrix obtained with PCA?. Let's take a look at the firs | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$
I provided a proof of the Eckart-Young theorem in my answer to What norm of the reconstruction error is minimized by the low-rank approximation matrix obtained with PCA?. Let's take a look at the first steps of this proof:
We want to find matrix $A$ of rank $k$ that ... | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$
I provided a proof of the Eckart-Young theorem in my answer to What norm of the reconstruction error is minimized by the low-rank approximation matrix obtained with PCA?. Let's take a look at the firs |
55,764 | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$ | The purpose of this answer is to show (a) the result is far more general and (b) when you look at it the right way, it is obvious: because the matrices of rank less than $k$ occupy a negligible portion of the space of matrices of ranks less than or equal to $k$ (their complement is dense), you can never do any worse wh... | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$ | The purpose of this answer is to show (a) the result is far more general and (b) when you look at it the right way, it is obvious: because the matrices of rank less than $k$ occupy a negligible portio | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$
The purpose of this answer is to show (a) the result is far more general and (b) when you look at it the right way, it is obvious: because the matrices of rank less than $k$ occupy a negligible portion of the space of matrices of ranks less than or equal to $k$ (their... | Eckart-Young-Mirsky theorem: rank $≤k$ or rank $=k$
The purpose of this answer is to show (a) the result is far more general and (b) when you look at it the right way, it is obvious: because the matrices of rank less than $k$ occupy a negligible portio |
55,765 | Bootstrap intervals for predictions, how to interpret it? | Yes, it is perfectly sensible.
For a quick interpretation, I like the one provided by Davison: Assuming $T$ is an estimator of a parameter $\psi$ based on a random sample $Y_1, . . . , Y_n$, $V_T^{0.5}$ is the standard error of $T$, $n \rightarrow \infty$ and $\zeta_\alpha$ is the $\alpha$-th quantile of a standard no... | Bootstrap intervals for predictions, how to interpret it? | Yes, it is perfectly sensible.
For a quick interpretation, I like the one provided by Davison: Assuming $T$ is an estimator of a parameter $\psi$ based on a random sample $Y_1, . . . , Y_n$, $V_T^{0.5 | Bootstrap intervals for predictions, how to interpret it?
Yes, it is perfectly sensible.
For a quick interpretation, I like the one provided by Davison: Assuming $T$ is an estimator of a parameter $\psi$ based on a random sample $Y_1, . . . , Y_n$, $V_T^{0.5}$ is the standard error of $T$, $n \rightarrow \infty$ and $... | Bootstrap intervals for predictions, how to interpret it?
Yes, it is perfectly sensible.
For a quick interpretation, I like the one provided by Davison: Assuming $T$ is an estimator of a parameter $\psi$ based on a random sample $Y_1, . . . , Y_n$, $V_T^{0.5 |
55,766 | Why use parametric test at all if non parametric tests are 'less strict' | It is true that precisely normal populations are rare in the real world.
However, some very useful procedures are 'robust' against mild non-normality.
Perhaps the most important of them is the t test, which performs remarkably well with samples of moderate or large size that are not exactly normal.
Also, some
tests th... | Why use parametric test at all if non parametric tests are 'less strict' | It is true that precisely normal populations are rare in the real world.
However, some very useful procedures are 'robust' against mild non-normality.
Perhaps the most important of them is the t test, | Why use parametric test at all if non parametric tests are 'less strict'
It is true that precisely normal populations are rare in the real world.
However, some very useful procedures are 'robust' against mild non-normality.
Perhaps the most important of them is the t test, which performs remarkably well with samples of... | Why use parametric test at all if non parametric tests are 'less strict'
It is true that precisely normal populations are rare in the real world.
However, some very useful procedures are 'robust' against mild non-normality.
Perhaps the most important of them is the t test, |
55,767 | Why use parametric test at all if non parametric tests are 'less strict' | All computation is based on model building. When you count two apples you imply, that there are either two identical things or a precise definition of what an apple exactly is and on what an apple exactly not is. The truth is: There is no sharp and precise definition of what an apple is, but our usual feeling, whether ... | Why use parametric test at all if non parametric tests are 'less strict' | All computation is based on model building. When you count two apples you imply, that there are either two identical things or a precise definition of what an apple exactly is and on what an apple exa | Why use parametric test at all if non parametric tests are 'less strict'
All computation is based on model building. When you count two apples you imply, that there are either two identical things or a precise definition of what an apple exactly is and on what an apple exactly not is. The truth is: There is no sharp an... | Why use parametric test at all if non parametric tests are 'less strict'
All computation is based on model building. When you count two apples you imply, that there are either two identical things or a precise definition of what an apple exactly is and on what an apple exa |
55,768 | Why use parametric test at all if non parametric tests are 'less strict' | Tests such as t-tests don't actually require the data to be normal. What they require is that the distribution of the sample mean of the data (under the null hypothesis) follows a normal distribution (or very close to it). This will cause the t-statistic to follow a t-distribution, as it should. This happens when the d... | Why use parametric test at all if non parametric tests are 'less strict' | Tests such as t-tests don't actually require the data to be normal. What they require is that the distribution of the sample mean of the data (under the null hypothesis) follows a normal distribution | Why use parametric test at all if non parametric tests are 'less strict'
Tests such as t-tests don't actually require the data to be normal. What they require is that the distribution of the sample mean of the data (under the null hypothesis) follows a normal distribution (or very close to it). This will cause the t-st... | Why use parametric test at all if non parametric tests are 'less strict'
Tests such as t-tests don't actually require the data to be normal. What they require is that the distribution of the sample mean of the data (under the null hypothesis) follows a normal distribution |
55,769 | Why use parametric test at all if non parametric tests are 'less strict' | I think I can summarise the other answers as follows: in many non-normal cases (if the breach of normality is not too big), parametric tests still have good power, while (important!) the actual type I-error rate is close to the nominal $\alpha$-level, so why not use them?
I would like to add that parametric tests offe... | Why use parametric test at all if non parametric tests are 'less strict' | I think I can summarise the other answers as follows: in many non-normal cases (if the breach of normality is not too big), parametric tests still have good power, while (important!) the actual type I | Why use parametric test at all if non parametric tests are 'less strict'
I think I can summarise the other answers as follows: in many non-normal cases (if the breach of normality is not too big), parametric tests still have good power, while (important!) the actual type I-error rate is close to the nominal $\alpha$-le... | Why use parametric test at all if non parametric tests are 'less strict'
I think I can summarise the other answers as follows: in many non-normal cases (if the breach of normality is not too big), parametric tests still have good power, while (important!) the actual type I |
55,770 | When to switch off the continuity correction in chisq.test function? | The test statistic is approximately $\chi^2$ distributed. According to Agresti in Categorical Data Analysis, Yates said Fisher recommended the hypergeometric distribution for an exact test. And Yates proposed this correction so the continuity corrected $\chi^2$ test approximates the result of the exact test. But nowada... | When to switch off the continuity correction in chisq.test function? | The test statistic is approximately $\chi^2$ distributed. According to Agresti in Categorical Data Analysis, Yates said Fisher recommended the hypergeometric distribution for an exact test. And Yates | When to switch off the continuity correction in chisq.test function?
The test statistic is approximately $\chi^2$ distributed. According to Agresti in Categorical Data Analysis, Yates said Fisher recommended the hypergeometric distribution for an exact test. And Yates proposed this correction so the continuity correcte... | When to switch off the continuity correction in chisq.test function?
The test statistic is approximately $\chi^2$ distributed. According to Agresti in Categorical Data Analysis, Yates said Fisher recommended the hypergeometric distribution for an exact test. And Yates |
55,771 | Can I run PCA on a 4-tensor? [closed] | PCA won't work on a 4D tensor, but you could use an auto-encoder.
Note that PCA will take a 2D dataset and reduce the number of columns in it (say 100 columns to 10).
With a 4D dataset, you could use an autoencoder to either reduce it to a 4D dataset with fewer "columns" or reduce it to a 3D dataset. | Can I run PCA on a 4-tensor? [closed] | PCA won't work on a 4D tensor, but you could use an auto-encoder.
Note that PCA will take a 2D dataset and reduce the number of columns in it (say 100 columns to 10).
With a 4D dataset, you could use | Can I run PCA on a 4-tensor? [closed]
PCA won't work on a 4D tensor, but you could use an auto-encoder.
Note that PCA will take a 2D dataset and reduce the number of columns in it (say 100 columns to 10).
With a 4D dataset, you could use an autoencoder to either reduce it to a 4D dataset with fewer "columns" or reduce ... | Can I run PCA on a 4-tensor? [closed]
PCA won't work on a 4D tensor, but you could use an auto-encoder.
Note that PCA will take a 2D dataset and reduce the number of columns in it (say 100 columns to 10).
With a 4D dataset, you could use |
55,772 | Can I run PCA on a 4-tensor? [closed] | There are actually a few generalizations of PCA to higher-order tensors:
The Tucker decomposition used in "higher-order singular value decomposition".
PARAFAC aka CANDECOMP, is in some ways a special case of Tucker decompositions.
Another, more recent variant uses the tensor train decomposition. | Can I run PCA on a 4-tensor? [closed] | There are actually a few generalizations of PCA to higher-order tensors:
The Tucker decomposition used in "higher-order singular value decomposition".
PARAFAC aka CANDECOMP, is in some ways a special | Can I run PCA on a 4-tensor? [closed]
There are actually a few generalizations of PCA to higher-order tensors:
The Tucker decomposition used in "higher-order singular value decomposition".
PARAFAC aka CANDECOMP, is in some ways a special case of Tucker decompositions.
Another, more recent variant uses the tensor train... | Can I run PCA on a 4-tensor? [closed]
There are actually a few generalizations of PCA to higher-order tensors:
The Tucker decomposition used in "higher-order singular value decomposition".
PARAFAC aka CANDECOMP, is in some ways a special |
55,773 | Can I run PCA on a 4-tensor? [closed] | The answer to the question in the title is yes, you can perform a PCA on any set of data described by a coordinate system with any number of axes. The result is a new set of axes, called principal coordinates. The PCA produces as many PC as there are original axes, but the new coordinate system as different properties,... | Can I run PCA on a 4-tensor? [closed] | The answer to the question in the title is yes, you can perform a PCA on any set of data described by a coordinate system with any number of axes. The result is a new set of axes, called principal coo | Can I run PCA on a 4-tensor? [closed]
The answer to the question in the title is yes, you can perform a PCA on any set of data described by a coordinate system with any number of axes. The result is a new set of axes, called principal coordinates. The PCA produces as many PC as there are original axes, but the new coor... | Can I run PCA on a 4-tensor? [closed]
The answer to the question in the title is yes, you can perform a PCA on any set of data described by a coordinate system with any number of axes. The result is a new set of axes, called principal coo |
55,774 | What are the different states in Open AI Taxi Environment? | From the paper Hierarchical Reinforcement Learning: Learning sub-goals and state-abstraction
In terms of state space there are 500 possible states:
25 squares, 5 locations for the passenger (counting the four starting locations and the taxi), and 4 destinations.
When working with enumerated states, then the count o... | What are the different states in Open AI Taxi Environment? | From the paper Hierarchical Reinforcement Learning: Learning sub-goals and state-abstraction
In terms of state space there are 500 possible states:
25 squares, 5 locations for the passenger (counti | What are the different states in Open AI Taxi Environment?
From the paper Hierarchical Reinforcement Learning: Learning sub-goals and state-abstraction
In terms of state space there are 500 possible states:
25 squares, 5 locations for the passenger (counting the four starting locations and the taxi), and 4 destinati... | What are the different states in Open AI Taxi Environment?
From the paper Hierarchical Reinforcement Learning: Learning sub-goals and state-abstraction
In terms of state space there are 500 possible states:
25 squares, 5 locations for the passenger (counti |
55,775 | Choice of hyper-parameters for Recursive Feature Elimination (SVM) | I will answer my own question for posterity.
In the excellent book Applied Predictive Modeling by Kjell Johnson and Max Kuhn, the RFE algorithm is stated very clearly. It is not stated so clearly (in my opinion) in Guyon et al's original paper. Here it is:
Apparently, the correct procedure is to fully tune and train... | Choice of hyper-parameters for Recursive Feature Elimination (SVM) | I will answer my own question for posterity.
In the excellent book Applied Predictive Modeling by Kjell Johnson and Max Kuhn, the RFE algorithm is stated very clearly. It is not stated so clearly (in | Choice of hyper-parameters for Recursive Feature Elimination (SVM)
I will answer my own question for posterity.
In the excellent book Applied Predictive Modeling by Kjell Johnson and Max Kuhn, the RFE algorithm is stated very clearly. It is not stated so clearly (in my opinion) in Guyon et al's original paper. Here i... | Choice of hyper-parameters for Recursive Feature Elimination (SVM)
I will answer my own question for posterity.
In the excellent book Applied Predictive Modeling by Kjell Johnson and Max Kuhn, the RFE algorithm is stated very clearly. It is not stated so clearly (in |
55,776 | Deriving $\chi^2$ density from the standard normal $Z$ density | Let us first find the density of the absolute value of $X\sim N(0,1)$, $Y=|X|$, $g(y)$. For $y>0$,
\begin{eqnarray*}
G(y)\equiv\Pr(Y\leqslant y)&=&\Pr(|X|\leqslant y)\\
&=&\Pr(-y\leqslant X\leqslant y)\\
&=&F(y)-F(-y)
\end{eqnarray*}
and hence
$$g(y)=G'(y)=f(y)+f(-y)$$ (note the second inner derivative equals $-1$). Si... | Deriving $\chi^2$ density from the standard normal $Z$ density | Let us first find the density of the absolute value of $X\sim N(0,1)$, $Y=|X|$, $g(y)$. For $y>0$,
\begin{eqnarray*}
G(y)\equiv\Pr(Y\leqslant y)&=&\Pr(|X|\leqslant y)\\
&=&\Pr(-y\leqslant X\leqslant y | Deriving $\chi^2$ density from the standard normal $Z$ density
Let us first find the density of the absolute value of $X\sim N(0,1)$, $Y=|X|$, $g(y)$. For $y>0$,
\begin{eqnarray*}
G(y)\equiv\Pr(Y\leqslant y)&=&\Pr(|X|\leqslant y)\\
&=&\Pr(-y\leqslant X\leqslant y)\\
&=&F(y)-F(-y)
\end{eqnarray*}
and hence
$$g(y)=G'(y)=... | Deriving $\chi^2$ density from the standard normal $Z$ density
Let us first find the density of the absolute value of $X\sim N(0,1)$, $Y=|X|$, $g(y)$. For $y>0$,
\begin{eqnarray*}
G(y)\equiv\Pr(Y\leqslant y)&=&\Pr(|X|\leqslant y)\\
&=&\Pr(-y\leqslant X\leqslant y |
55,777 | Deriving $\chi^2$ density from the standard normal $Z$ density | As pointed out in the comments, your error here is that your density transformation does not take account of the nonlinearity of the transformation. I will show you a better empirical demonstration of the distributional equivalence, where we don't attempt the transform at all, but simply compare the kernel density of ... | Deriving $\chi^2$ density from the standard normal $Z$ density | As pointed out in the comments, your error here is that your density transformation does not take account of the nonlinearity of the transformation. I will show you a better empirical demonstration o | Deriving $\chi^2$ density from the standard normal $Z$ density
As pointed out in the comments, your error here is that your density transformation does not take account of the nonlinearity of the transformation. I will show you a better empirical demonstration of the distributional equivalence, where we don't attempt ... | Deriving $\chi^2$ density from the standard normal $Z$ density
As pointed out in the comments, your error here is that your density transformation does not take account of the nonlinearity of the transformation. I will show you a better empirical demonstration o |
55,778 | Coordinate descent with constraints | The bland answer is that to solve constrained optimization problems, you need an algorithm that knows and uses the constraints - simply applying unconstrained optimization algorithms will not work.
Box constraints
Taking the algorithm you described in your answer, it will ''not work'' for box constraints, depending on ... | Coordinate descent with constraints | The bland answer is that to solve constrained optimization problems, you need an algorithm that knows and uses the constraints - simply applying unconstrained optimization algorithms will not work.
Bo | Coordinate descent with constraints
The bland answer is that to solve constrained optimization problems, you need an algorithm that knows and uses the constraints - simply applying unconstrained optimization algorithms will not work.
Box constraints
Taking the algorithm you described in your answer, it will ''not work'... | Coordinate descent with constraints
The bland answer is that to solve constrained optimization problems, you need an algorithm that knows and uses the constraints - simply applying unconstrained optimization algorithms will not work.
Bo |
55,779 | Coordinate descent with constraints | Perhaps this question is more difficult than I thought, or would be better asked on math.stackexchange.com. But since it arose in the context of SVM I will leave it here.
There are some papers discussing coordinate descent and constrained optimization with the following key words.
Linear and box constraints and bloc... | Coordinate descent with constraints | Perhaps this question is more difficult than I thought, or would be better asked on math.stackexchange.com. But since it arose in the context of SVM I will leave it here.
There are some papers discus | Coordinate descent with constraints
Perhaps this question is more difficult than I thought, or would be better asked on math.stackexchange.com. But since it arose in the context of SVM I will leave it here.
There are some papers discussing coordinate descent and constrained optimization with the following key words.
... | Coordinate descent with constraints
Perhaps this question is more difficult than I thought, or would be better asked on math.stackexchange.com. But since it arose in the context of SVM I will leave it here.
There are some papers discus |
55,780 | Regularized parameter overfitting the data (example) | Regularization does not guarantee to reduce overfit.
Regularization reduces overfit in many cases because in these cases the real data model (e.g., physics models) have small weights. Reguarlization is a way to inject this knowledge in our model. It weeds out those models that have large weights, which tend to be mod... | Regularized parameter overfitting the data (example) | Regularization does not guarantee to reduce overfit.
Regularization reduces overfit in many cases because in these cases the real data model (e.g., physics models) have small weights. Reguarlization | Regularized parameter overfitting the data (example)
Regularization does not guarantee to reduce overfit.
Regularization reduces overfit in many cases because in these cases the real data model (e.g., physics models) have small weights. Reguarlization is a way to inject this knowledge in our model. It weeds out those... | Regularized parameter overfitting the data (example)
Regularization does not guarantee to reduce overfit.
Regularization reduces overfit in many cases because in these cases the real data model (e.g., physics models) have small weights. Reguarlization |
55,781 | Regularized parameter overfitting the data (example) | Regularization cost for magenta classifier is low but still it seems to overfit the data and vice-versa for the black classifier. What's going on? L2 regularization tend to make the coefficients close to zero. But how does that helps in reducing overfitting?
Overfitting isn't about what happens to the training data al... | Regularized parameter overfitting the data (example) | Regularization cost for magenta classifier is low but still it seems to overfit the data and vice-versa for the black classifier. What's going on? L2 regularization tend to make the coefficients close | Regularized parameter overfitting the data (example)
Regularization cost for magenta classifier is low but still it seems to overfit the data and vice-versa for the black classifier. What's going on? L2 regularization tend to make the coefficients close to zero. But how does that helps in reducing overfitting?
Overfit... | Regularized parameter overfitting the data (example)
Regularization cost for magenta classifier is low but still it seems to overfit the data and vice-versa for the black classifier. What's going on? L2 regularization tend to make the coefficients close |
55,782 | In Bayesian statistics, what does this notation formally mean? | You are correct to say that the statement is not mathematically precise; it is a shorthand that is sometimes used to set out a hierarchical model. I'm not a fan of this notation personally, since it is an unnecessary abuse of notation. (I prefer to specify independence separately.)
Presumably the intention of such st... | In Bayesian statistics, what does this notation formally mean? | You are correct to say that the statement is not mathematically precise; it is a shorthand that is sometimes used to set out a hierarchical model. I'm not a fan of this notation personally, since it | In Bayesian statistics, what does this notation formally mean?
You are correct to say that the statement is not mathematically precise; it is a shorthand that is sometimes used to set out a hierarchical model. I'm not a fan of this notation personally, since it is an unnecessary abuse of notation. (I prefer to specif... | In Bayesian statistics, what does this notation formally mean?
You are correct to say that the statement is not mathematically precise; it is a shorthand that is sometimes used to set out a hierarchical model. I'm not a fan of this notation personally, since it |
55,783 | In Bayesian statistics, what does this notation formally mean? | Taking the simpler case
$$
v_i \overset{ind}{\sim} g_i(v_i)
$$
presumably means that "$v_i$'s are independently distributed according to $g_i$ distributions each". The notation is a shortcut for describing each $i$-th variable as separate case. There's often trade-off between simplicity and formality.
But the notation ... | In Bayesian statistics, what does this notation formally mean? | Taking the simpler case
$$
v_i \overset{ind}{\sim} g_i(v_i)
$$
presumably means that "$v_i$'s are independently distributed according to $g_i$ distributions each". The notation is a shortcut for descr | In Bayesian statistics, what does this notation formally mean?
Taking the simpler case
$$
v_i \overset{ind}{\sim} g_i(v_i)
$$
presumably means that "$v_i$'s are independently distributed according to $g_i$ distributions each". The notation is a shortcut for describing each $i$-th variable as separate case. There's ofte... | In Bayesian statistics, what does this notation formally mean?
Taking the simpler case
$$
v_i \overset{ind}{\sim} g_i(v_i)
$$
presumably means that "$v_i$'s are independently distributed according to $g_i$ distributions each". The notation is a shortcut for descr |
55,784 | Modelling nonnegative integer time series | Use a model for integer time series, not ARIMA or Holt-Winters' method. You could try an INAR for example, or an autoregressive conditional Poisson model, or a GLARMA model. See Fokianos (2012) for a comprehensive review. Some possibilities in R are described in Liboschik et al (2017). | Modelling nonnegative integer time series | Use a model for integer time series, not ARIMA or Holt-Winters' method. You could try an INAR for example, or an autoregressive conditional Poisson model, or a GLARMA model. See Fokianos (2012) for a | Modelling nonnegative integer time series
Use a model for integer time series, not ARIMA or Holt-Winters' method. You could try an INAR for example, or an autoregressive conditional Poisson model, or a GLARMA model. See Fokianos (2012) for a comprehensive review. Some possibilities in R are described in Liboschik et al... | Modelling nonnegative integer time series
Use a model for integer time series, not ARIMA or Holt-Winters' method. You could try an INAR for example, or an autoregressive conditional Poisson model, or a GLARMA model. See Fokianos (2012) for a |
55,785 | Teaching (very elementary) statistical modelling | Not shure, whether this is worth an answer or just a comment, but I want the room this forum gives for answers only.
Have look at this golf putting example, that Andrew Gelman gave in a webinar about stan. Forget about the Bayes-aspect of it. It just shows a standard model compared with an informed model and how the r... | Teaching (very elementary) statistical modelling | Not shure, whether this is worth an answer or just a comment, but I want the room this forum gives for answers only.
Have look at this golf putting example, that Andrew Gelman gave in a webinar about | Teaching (very elementary) statistical modelling
Not shure, whether this is worth an answer or just a comment, but I want the room this forum gives for answers only.
Have look at this golf putting example, that Andrew Gelman gave in a webinar about stan. Forget about the Bayes-aspect of it. It just shows a standard mo... | Teaching (very elementary) statistical modelling
Not shure, whether this is worth an answer or just a comment, but I want the room this forum gives for answers only.
Have look at this golf putting example, that Andrew Gelman gave in a webinar about |
55,786 | Teaching (very elementary) statistical modelling | But missing is seemingly why we might choose the Poisson distribution over other choices
This is indeed a frequent trait of probability textbooks – and even of research works in statistics. Somewhat of a taboo it seems. It was pointed out by the statistician A. P. Dawid (1982, § 4, p. 220):
Where do probability model... | Teaching (very elementary) statistical modelling | But missing is seemingly why we might choose the Poisson distribution over other choices
This is indeed a frequent trait of probability textbooks – and even of research works in statistics. Somewhat | Teaching (very elementary) statistical modelling
But missing is seemingly why we might choose the Poisson distribution over other choices
This is indeed a frequent trait of probability textbooks – and even of research works in statistics. Somewhat of a taboo it seems. It was pointed out by the statistician A. P. Dawid... | Teaching (very elementary) statistical modelling
But missing is seemingly why we might choose the Poisson distribution over other choices
This is indeed a frequent trait of probability textbooks – and even of research works in statistics. Somewhat |
55,787 | Find the PDF from quantiles | Your approach is valid, if you use the integral of a cubic B-spline to interpolate the quantiles with the condition that all B-spline coefficients are nonnegative.
https://en.m.wikipedia.org/wiki/B-spline
This preserves the monotonicty of the quantiles in the interpolating function, while normal cubic splines do not.
T... | Find the PDF from quantiles | Your approach is valid, if you use the integral of a cubic B-spline to interpolate the quantiles with the condition that all B-spline coefficients are nonnegative.
https://en.m.wikipedia.org/wiki/B-sp | Find the PDF from quantiles
Your approach is valid, if you use the integral of a cubic B-spline to interpolate the quantiles with the condition that all B-spline coefficients are nonnegative.
https://en.m.wikipedia.org/wiki/B-spline
This preserves the monotonicty of the quantiles in the interpolating function, while no... | Find the PDF from quantiles
Your approach is valid, if you use the integral of a cubic B-spline to interpolate the quantiles with the condition that all B-spline coefficients are nonnegative.
https://en.m.wikipedia.org/wiki/B-sp |
55,788 | Prove that $ \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y] = \int_{- \infty}^\infty \int_{- \infty}^\infty (F(x,y)-F_X(x) F_Y(y)) \,dx\,dy,$ | For the second equality, think about the integral $$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} [I(u,X_1)-I(u,X_2)][I(v,Y_1)-I(v,Y_2)]du dv $$ which is a random variable. The structure of the integrand allows the integral to be rewritten as $$ \int_{- \infty}^{\infty} [I(u,X_1)-I(u,X_2)]du \int_{- \infty}^{\infty... | Prove that $ \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y] = \int_{- \infty}^\infty \int_{- \infty}^\ | For the second equality, think about the integral $$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} [I(u,X_1)-I(u,X_2)][I(v,Y_1)-I(v,Y_2)]du dv $$ which is a random variable. The structure of the in | Prove that $ \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y] = \int_{- \infty}^\infty \int_{- \infty}^\infty (F(x,y)-F_X(x) F_Y(y)) \,dx\,dy,$
For the second equality, think about the integral $$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} [I(u,X_1)-I(u,X_2)][I(v,Y_1)-I(v,Y_2)]du dv $$ which is a random variable. The... | Prove that $ \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y] = \int_{- \infty}^\infty \int_{- \infty}^\
For the second equality, think about the integral $$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} [I(u,X_1)-I(u,X_2)][I(v,Y_1)-I(v,Y_2)]du dv $$ which is a random variable. The structure of the in |
55,789 | Do "splits" in scatterplots indicate anything? | When I see a plot like that, my thinking immediately goes toward sample subgroups. It seems like you have two different sample groups, each with a visually apparent trend line. Are all your samples similar, or are there some categorical differences between them? Suppose one of your divergent graphs represents price vs.... | Do "splits" in scatterplots indicate anything? | When I see a plot like that, my thinking immediately goes toward sample subgroups. It seems like you have two different sample groups, each with a visually apparent trend line. Are all your samples si | Do "splits" in scatterplots indicate anything?
When I see a plot like that, my thinking immediately goes toward sample subgroups. It seems like you have two different sample groups, each with a visually apparent trend line. Are all your samples similar, or are there some categorical differences between them? Suppose on... | Do "splits" in scatterplots indicate anything?
When I see a plot like that, my thinking immediately goes toward sample subgroups. It seems like you have two different sample groups, each with a visually apparent trend line. Are all your samples si |
55,790 | Do "splits" in scatterplots indicate anything? | There is at least one discrete variable, parm3 and it's possible that there are other un-labeled groupings. I'd start by redo that graphic while labeling the parm3 values with color coding. Ten you can quickly see whether one or two colors fall into the "splits" that are apparent.
That appears to be a "pairs" plot. In... | Do "splits" in scatterplots indicate anything? | There is at least one discrete variable, parm3 and it's possible that there are other un-labeled groupings. I'd start by redo that graphic while labeling the parm3 values with color coding. Ten you c | Do "splits" in scatterplots indicate anything?
There is at least one discrete variable, parm3 and it's possible that there are other un-labeled groupings. I'd start by redo that graphic while labeling the parm3 values with color coding. Ten you can quickly see whether one or two colors fall into the "splits" that are ... | Do "splits" in scatterplots indicate anything?
There is at least one discrete variable, parm3 and it's possible that there are other un-labeled groupings. I'd start by redo that graphic while labeling the parm3 values with color coding. Ten you c |
55,791 | One-sided McNemar's test | I have described the gist of McNemar's test rather extensively here and here, it may help you to read those. Briefly, McNemar's test assesses the balance of the off-diagonal counts. If people were as likely to transition from approval to disapproval as from disapproval to approval, then the off-diagonal values should... | One-sided McNemar's test | I have described the gist of McNemar's test rather extensively here and here, it may help you to read those. Briefly, McNemar's test assesses the balance of the off-diagonal counts. If people were a | One-sided McNemar's test
I have described the gist of McNemar's test rather extensively here and here, it may help you to read those. Briefly, McNemar's test assesses the balance of the off-diagonal counts. If people were as likely to transition from approval to disapproval as from disapproval to approval, then the o... | One-sided McNemar's test
I have described the gist of McNemar's test rather extensively here and here, it may help you to read those. Briefly, McNemar's test assesses the balance of the off-diagonal counts. If people were a |
55,792 | Residual diagnostics for negative binomial regression | Check out the DHARMa package in R. It uses a simulation based approach with quantile residuals to generate the type of residuals you may be interested in. And it works with glm.nb from MASS.
The essential idea is explained here and goes in three steps:
Simulate plausible responses for each case. You can use the distri... | Residual diagnostics for negative binomial regression | Check out the DHARMa package in R. It uses a simulation based approach with quantile residuals to generate the type of residuals you may be interested in. And it works with glm.nb from MASS.
The essen | Residual diagnostics for negative binomial regression
Check out the DHARMa package in R. It uses a simulation based approach with quantile residuals to generate the type of residuals you may be interested in. And it works with glm.nb from MASS.
The essential idea is explained here and goes in three steps:
Simulate pla... | Residual diagnostics for negative binomial regression
Check out the DHARMa package in R. It uses a simulation based approach with quantile residuals to generate the type of residuals you may be interested in. And it works with glm.nb from MASS.
The essen |
55,793 | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value? | You are correct, in the sense that the uniformly most powerful (UMP) test for equivalence for the one-sample, paired, and two-sample $T$-test does involve a non-central $t$-distribution.
The two one-sided tests (TOST) procedure with both rejection regions with size $\alpha$, or equivalently the $1 - 2 \alpha$ confidenc... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the | You are correct, in the sense that the uniformly most powerful (UMP) test for equivalence for the one-sample, paired, and two-sample $T$-test does involve a non-central $t$-distribution.
The two one-s | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value?
You are correct, in the sense that the uniformly most powerful (UMP) test for equivalence for the one-sample, paired, and two-sample $T$-test does involve a non-central $t$-distribution.
The two one-sided tes... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the
You are correct, in the sense that the uniformly most powerful (UMP) test for equivalence for the one-sample, paired, and two-sample $T$-test does involve a non-central $t$-distribution.
The two one-s |
55,794 | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value? | What you are proposing does not appear to be any different from what actually occurs in this test. Remember that in a classical hypothesis test, the p-value is calculated using the null distribution of the test statistic. This calculation assumes that the null hypothesis is true. You suggest that the calculation of ... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the | What you are proposing does not appear to be any different from what actually occurs in this test. Remember that in a classical hypothesis test, the p-value is calculated using the null distribution | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value?
What you are proposing does not appear to be any different from what actually occurs in this test. Remember that in a classical hypothesis test, the p-value is calculated using the null distribution of the t... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the
What you are proposing does not appear to be any different from what actually occurs in this test. Remember that in a classical hypothesis test, the p-value is calculated using the null distribution |
55,795 | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value? | I want to expand upon / comment on the other answers.
First, note that if you are using the noncentral t-distribution, you need to specify your equivalence bounds as a standardized mean difference $\delta$, since the noncentrality parameter will be $\delta\sqrt{n}$. If you are using TOST (i.e. central t-distribution), ... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the | I want to expand upon / comment on the other answers.
First, note that if you are using the noncentral t-distribution, you need to specify your equivalence bounds as a standardized mean difference $\d | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the $p$ value?
I want to expand upon / comment on the other answers.
First, note that if you are using the noncentral t-distribution, you need to specify your equivalence bounds as a standardized mean difference $\delta$, s... | Why doesn't the TOST equivalence testing procedure use non-central $t$ distribution to determine the
I want to expand upon / comment on the other answers.
First, note that if you are using the noncentral t-distribution, you need to specify your equivalence bounds as a standardized mean difference $\d |
55,796 | Is assuming a binomial distribution appropriate when the number of possible successes is fixed? | In general, it's not appropriate.
The remaining question is, wether the practical impact in your application is relevant to your usecase. To assess this, you could for example perturbate the applications in the training set (let people apply in different groups than they did in originally) and check how that imacts you... | Is assuming a binomial distribution appropriate when the number of possible successes is fixed? | In general, it's not appropriate.
The remaining question is, wether the practical impact in your application is relevant to your usecase. To assess this, you could for example perturbate the applicati | Is assuming a binomial distribution appropriate when the number of possible successes is fixed?
In general, it's not appropriate.
The remaining question is, wether the practical impact in your application is relevant to your usecase. To assess this, you could for example perturbate the applications in the training set ... | Is assuming a binomial distribution appropriate when the number of possible successes is fixed?
In general, it's not appropriate.
The remaining question is, wether the practical impact in your application is relevant to your usecase. To assess this, you could for example perturbate the applicati |
55,797 | Is assuming a binomial distribution appropriate when the number of possible successes is fixed? | As mentioned by the existing answer, it is in princple not appropriate to use logistic regression in this kind of situation. It could be that you would still get decent results, but I would go for a different approach.
I assume you will find an extensive literature on this topic, if you look into statistical models for... | Is assuming a binomial distribution appropriate when the number of possible successes is fixed? | As mentioned by the existing answer, it is in princple not appropriate to use logistic regression in this kind of situation. It could be that you would still get decent results, but I would go for a d | Is assuming a binomial distribution appropriate when the number of possible successes is fixed?
As mentioned by the existing answer, it is in princple not appropriate to use logistic regression in this kind of situation. It could be that you would still get decent results, but I would go for a different approach.
I ass... | Is assuming a binomial distribution appropriate when the number of possible successes is fixed?
As mentioned by the existing answer, it is in princple not appropriate to use logistic regression in this kind of situation. It could be that you would still get decent results, but I would go for a d |
55,798 | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitting? | Using a kernelized SVM is equivalent to mapping the data into feature space, then using a linear SVM in feature space. The feature space mapping is defined implicitly by the kernel function, which computes the inner product between data points in feature space. That is:
$$\kappa(x_i, x_j) = \langle \Phi(x_i), \Phi(x_j)... | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitti | Using a kernelized SVM is equivalent to mapping the data into feature space, then using a linear SVM in feature space. The feature space mapping is defined implicitly by the kernel function, which com | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitting?
Using a kernelized SVM is equivalent to mapping the data into feature space, then using a linear SVM in feature space. The feature space mapping is defined implicitly by the kernel function, which computes the inner ... | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitti
Using a kernelized SVM is equivalent to mapping the data into feature space, then using a linear SVM in feature space. The feature space mapping is defined implicitly by the kernel function, which com |
55,799 | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitting? | Varying the gamma values indicates how the squared distance between the any 2 observations are changing. In radial kernel functions nearer observations have more effects on test observations. So upon increasing the gamma value we are classifying the nearest observation as farthest. So we are neglecting the effects of m... | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitti | Varying the gamma values indicates how the squared distance between the any 2 observations are changing. In radial kernel functions nearer observations have more effects on test observations. So upon | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitting?
Varying the gamma values indicates how the squared distance between the any 2 observations are changing. In radial kernel functions nearer observations have more effects on test observations. So upon increasing the g... | Why a large gamma in the RBF kernel of SVM leads to a wiggly decision boundary and causes over-fitti
Varying the gamma values indicates how the squared distance between the any 2 observations are changing. In radial kernel functions nearer observations have more effects on test observations. So upon |
55,800 | Is there a specific name for this plot? | This is ultimately just a scatterplot. I don't think there is a special name for it. I don't see this as meaningfully related to correspondence analysis except in that you can make a scatterplot of the results from a correspondence analysis, and this is also a scatterplot. Notably, this does not have much to do with... | Is there a specific name for this plot? | This is ultimately just a scatterplot. I don't think there is a special name for it. I don't see this as meaningfully related to correspondence analysis except in that you can make a scatterplot of | Is there a specific name for this plot?
This is ultimately just a scatterplot. I don't think there is a special name for it. I don't see this as meaningfully related to correspondence analysis except in that you can make a scatterplot of the results from a correspondence analysis, and this is also a scatterplot. Not... | Is there a specific name for this plot?
This is ultimately just a scatterplot. I don't think there is a special name for it. I don't see this as meaningfully related to correspondence analysis except in that you can make a scatterplot of |
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