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 |
|---|---|---|---|---|---|---|
39,601 | Why do we say EM is a partially non-Bayesian method? | EM is based on a demarginalisation of the (standard or observed) likelihood
$$L^\text{o}(\theta|\mathbf x)=\int_{\mathfrak Z} L^\text{c}(\theta|\mathbf x,\mathbf z)\,\text d\mathbf z \tag{1}$$
introducing a latent variable $\mathbf Z$ to simplify the representation of the (observed) likelihood$$L^\text{o}(\theta|\mathb... | Why do we say EM is a partially non-Bayesian method? | EM is based on a demarginalisation of the (standard or observed) likelihood
$$L^\text{o}(\theta|\mathbf x)=\int_{\mathfrak Z} L^\text{c}(\theta|\mathbf x,\mathbf z)\,\text d\mathbf z \tag{1}$$
introdu | Why do we say EM is a partially non-Bayesian method?
EM is based on a demarginalisation of the (standard or observed) likelihood
$$L^\text{o}(\theta|\mathbf x)=\int_{\mathfrak Z} L^\text{c}(\theta|\mathbf x,\mathbf z)\,\text d\mathbf z \tag{1}$$
introducing a latent variable $\mathbf Z$ to simplify the representation o... | Why do we say EM is a partially non-Bayesian method?
EM is based on a demarginalisation of the (standard or observed) likelihood
$$L^\text{o}(\theta|\mathbf x)=\int_{\mathfrak Z} L^\text{c}(\theta|\mathbf x,\mathbf z)\,\text d\mathbf z \tag{1}$$
introdu |
39,602 | Why do we say EM is a partially non-Bayesian method? | In full Bayesian inference, you go from a prior distribution over parameter values, $P(\theta)$ to a posterior distribution given the data $P(\theta | x)$. With Expectation Maximisation, you go from a prior distribution to an estimate $\hat \theta$ of the most likely posterior value of $\theta$ given the data and the p... | Why do we say EM is a partially non-Bayesian method? | In full Bayesian inference, you go from a prior distribution over parameter values, $P(\theta)$ to a posterior distribution given the data $P(\theta | x)$. With Expectation Maximisation, you go from a | Why do we say EM is a partially non-Bayesian method?
In full Bayesian inference, you go from a prior distribution over parameter values, $P(\theta)$ to a posterior distribution given the data $P(\theta | x)$. With Expectation Maximisation, you go from a prior distribution to an estimate $\hat \theta$ of the most likely... | Why do we say EM is a partially non-Bayesian method?
In full Bayesian inference, you go from a prior distribution over parameter values, $P(\theta)$ to a posterior distribution given the data $P(\theta | x)$. With Expectation Maximisation, you go from a |
39,603 | Statistics Help: Difference between Differences? | It's a little risky to answer without better understanding your use case, but assuming iid nested data within columns and testing the hypotheses:
$$
H_0: \mu_A - \mu_B = \mu_C - \mu_D, \\
H_1: \mu_A - \mu_B \ne \mu_C - \mu_D,
$$
you could appeal to the central limit theorem and construct a Wald test, comparing
$$
Z = \... | Statistics Help: Difference between Differences? | It's a little risky to answer without better understanding your use case, but assuming iid nested data within columns and testing the hypotheses:
$$
H_0: \mu_A - \mu_B = \mu_C - \mu_D, \\
H_1: \mu_A - | Statistics Help: Difference between Differences?
It's a little risky to answer without better understanding your use case, but assuming iid nested data within columns and testing the hypotheses:
$$
H_0: \mu_A - \mu_B = \mu_C - \mu_D, \\
H_1: \mu_A - \mu_B \ne \mu_C - \mu_D,
$$
you could appeal to the central limit theo... | Statistics Help: Difference between Differences?
It's a little risky to answer without better understanding your use case, but assuming iid nested data within columns and testing the hypotheses:
$$
H_0: \mu_A - \mu_B = \mu_C - \mu_D, \\
H_1: \mu_A - |
39,604 | Understanding different Monte Carlo approximation notations | Yes, the formula you provide should converge to a true answer for arbitrary probability distribution $g(x)$ given infinite sample points. The problem is that you don't want to wait infinitely long. So instead a more interesting question is whether it is likely to converge to a value close to the true value given a fini... | Understanding different Monte Carlo approximation notations | Yes, the formula you provide should converge to a true answer for arbitrary probability distribution $g(x)$ given infinite sample points. The problem is that you don't want to wait infinitely long. So | Understanding different Monte Carlo approximation notations
Yes, the formula you provide should converge to a true answer for arbitrary probability distribution $g(x)$ given infinite sample points. The problem is that you don't want to wait infinitely long. So instead a more interesting question is whether it is likely... | Understanding different Monte Carlo approximation notations
Yes, the formula you provide should converge to a true answer for arbitrary probability distribution $g(x)$ given infinite sample points. The problem is that you don't want to wait infinitely long. So |
39,605 | Understanding different Monte Carlo approximation notations | In probabilistic terms, the Monte Carlo method (or its justification) is called the Law of Large Numbers. The convergence
$$\frac{1}{N} \sum_{i=1}^N f(X_i) \stackrel{\text{a.s.}}{\to} \mathbb E_g[f(X)]\tag{1}$$ does not assume anything but iid-ness of the $X_i$'s and the existence of the expectation.
A more precise cha... | Understanding different Monte Carlo approximation notations | In probabilistic terms, the Monte Carlo method (or its justification) is called the Law of Large Numbers. The convergence
$$\frac{1}{N} \sum_{i=1}^N f(X_i) \stackrel{\text{a.s.}}{\to} \mathbb E_g[f(X) | Understanding different Monte Carlo approximation notations
In probabilistic terms, the Monte Carlo method (or its justification) is called the Law of Large Numbers. The convergence
$$\frac{1}{N} \sum_{i=1}^N f(X_i) \stackrel{\text{a.s.}}{\to} \mathbb E_g[f(X)]\tag{1}$$ does not assume anything but iid-ness of the $X_i... | Understanding different Monte Carlo approximation notations
In probabilistic terms, the Monte Carlo method (or its justification) is called the Law of Large Numbers. The convergence
$$\frac{1}{N} \sum_{i=1}^N f(X_i) \stackrel{\text{a.s.}}{\to} \mathbb E_g[f(X) |
39,606 | Estimating Pi using Monte Carlo simulations but without using fractional numbers | Trivial -- but magical:
BBP <- function(n = 13) {
sum(sapply(seq_len(n), function(k) {
((sample.int(8*k+1, 1) <= 4) -
(sample.int(8*k+4, 1) <= 2) -
(sample.int(8*k+5, 1) <= 1) -
(sample.int(8*k+6, 1) <= 1)) / 16^k
})) + (4 - 2/4 - 1/5 - 1/6)
}
As you can see in this R code, only rational a... | Estimating Pi using Monte Carlo simulations but without using fractional numbers | Trivial -- but magical:
BBP <- function(n = 13) {
sum(sapply(seq_len(n), function(k) {
((sample.int(8*k+1, 1) <= 4) -
(sample.int(8*k+4, 1) <= 2) -
(sample.int(8*k+5, 1) <= 1) -
| Estimating Pi using Monte Carlo simulations but without using fractional numbers
Trivial -- but magical:
BBP <- function(n = 13) {
sum(sapply(seq_len(n), function(k) {
((sample.int(8*k+1, 1) <= 4) -
(sample.int(8*k+4, 1) <= 2) -
(sample.int(8*k+5, 1) <= 1) -
(sample.int(8*k+6, 1) <= 1)) / 16^... | Estimating Pi using Monte Carlo simulations but without using fractional numbers
Trivial -- but magical:
BBP <- function(n = 13) {
sum(sapply(seq_len(n), function(k) {
((sample.int(8*k+1, 1) <= 4) -
(sample.int(8*k+4, 1) <= 2) -
(sample.int(8*k+5, 1) <= 1) -
|
39,607 | Estimating Pi using Monte Carlo simulations but without using fractional numbers | A simple method would be generating a pair of integers from $[0,M)$ and see if it's inside the quadrant, i.e. let the numbers be denoted as $m_1, m_2$. If $m_1^2+m_2^2<M^2$, the point is inside the quadrant. If the numbers were continuous, the probability would have been $\pi/4$. So, increasing $M$, increases the preci... | Estimating Pi using Monte Carlo simulations but without using fractional numbers | A simple method would be generating a pair of integers from $[0,M)$ and see if it's inside the quadrant, i.e. let the numbers be denoted as $m_1, m_2$. If $m_1^2+m_2^2<M^2$, the point is inside the qu | Estimating Pi using Monte Carlo simulations but without using fractional numbers
A simple method would be generating a pair of integers from $[0,M)$ and see if it's inside the quadrant, i.e. let the numbers be denoted as $m_1, m_2$. If $m_1^2+m_2^2<M^2$, the point is inside the quadrant. If the numbers were continuous,... | Estimating Pi using Monte Carlo simulations but without using fractional numbers
A simple method would be generating a pair of integers from $[0,M)$ and see if it's inside the quadrant, i.e. let the numbers be denoted as $m_1, m_2$. If $m_1^2+m_2^2<M^2$, the point is inside the qu |
39,608 | Estimating Pi using Monte Carlo simulations but without using fractional numbers | To produce the probability $1/\pi$, the following algorithm can be used (Flajolet et al. 2010), which is based on a series expansion by Ramanujan:
Set $t$ to 0.
Flip two fair coins. If both show heads, add 1 to $t$ and repeat this step. Otherwise, go to step 3.
Flip two fair coins. If both show heads, add 1 to $t$ and... | Estimating Pi using Monte Carlo simulations but without using fractional numbers | To produce the probability $1/\pi$, the following algorithm can be used (Flajolet et al. 2010), which is based on a series expansion by Ramanujan:
Set $t$ to 0.
Flip two fair coins. If both show head | Estimating Pi using Monte Carlo simulations but without using fractional numbers
To produce the probability $1/\pi$, the following algorithm can be used (Flajolet et al. 2010), which is based on a series expansion by Ramanujan:
Set $t$ to 0.
Flip two fair coins. If both show heads, add 1 to $t$ and repeat this step. O... | Estimating Pi using Monte Carlo simulations but without using fractional numbers
To produce the probability $1/\pi$, the following algorithm can be used (Flajolet et al. 2010), which is based on a series expansion by Ramanujan:
Set $t$ to 0.
Flip two fair coins. If both show head |
39,609 | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \mathcal{N}(0,1)$ | TLDR; $s(z)$ is asymptotically normal, and its variance is $\frac {12} {n-1}$ according to CLT for Markov chains. It can be shown that the distribution is a special case of generalized $\chi^2$ distribution.
Markov Chain approach, asymptotics and variance
The sequence $z_i$ is Markov chain because once you know $z_i$ t... | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \ | TLDR; $s(z)$ is asymptotically normal, and its variance is $\frac {12} {n-1}$ according to CLT for Markov chains. It can be shown that the distribution is a special case of generalized $\chi^2$ distri | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \mathcal{N}(0,1)$
TLDR; $s(z)$ is asymptotically normal, and its variance is $\frac {12} {n-1}$ according to CLT for Markov chains. It can be shown that the distribution is a special case of generalized $\chi^2$ distribut... | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \
TLDR; $s(z)$ is asymptotically normal, and its variance is $\frac {12} {n-1}$ according to CLT for Markov chains. It can be shown that the distribution is a special case of generalized $\chi^2$ distri |
39,610 | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \mathcal{N}(0,1)$ | Small pedantic note: Below I changed the coefficient into $1/\sqrt{n-1}$ otherwise the limiting distribution will be a degenerate distribution (zero variance). In that case one would also need to subtract the mean of the $z_i^2$. That means, only a scaled and shifted sum like $\sum_{i=1}^{n-1} \frac{(x_{i+1}-x_i)^2-1}{... | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \ | Small pedantic note: Below I changed the coefficient into $1/\sqrt{n-1}$ otherwise the limiting distribution will be a degenerate distribution (zero variance). In that case one would also need to subt | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \mathcal{N}(0,1)$
Small pedantic note: Below I changed the coefficient into $1/\sqrt{n-1}$ otherwise the limiting distribution will be a degenerate distribution (zero variance). In that case one would also need to subtrac... | Variance and asymptotic normality of $\frac{1}{n-1}\sum_{i=1}^{n-1}(x_{i+1}-x_i)^2$, where $X \sim \
Small pedantic note: Below I changed the coefficient into $1/\sqrt{n-1}$ otherwise the limiting distribution will be a degenerate distribution (zero variance). In that case one would also need to subt |
39,611 | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function? | The likelihood function $L(\theta)$ provides us with both, a point estimator and a confidence set estimator for $\theta$. In particular, to make things simple, let $X_1,\ldots, X_n$ be an i.i.d. random sample from a model having pdf $f_\theta$, with $\theta\in \mathbb{R}$.
The Maximum Likelihood Estimator is defined as... | How to create an interval estimate based on being within some percentage of the maximum value of the | The likelihood function $L(\theta)$ provides us with both, a point estimator and a confidence set estimator for $\theta$. In particular, to make things simple, let $X_1,\ldots, X_n$ be an i.i.d. rando | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function?
The likelihood function $L(\theta)$ provides us with both, a point estimator and a confidence set estimator for $\theta$. In particular, to make things simple, let $X_1,\ldots, X_n$ be an i.i.d. ran... | How to create an interval estimate based on being within some percentage of the maximum value of the
The likelihood function $L(\theta)$ provides us with both, a point estimator and a confidence set estimator for $\theta$. In particular, to make things simple, let $X_1,\ldots, X_n$ be an i.i.d. rando |
39,612 | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function? | Under some conditions you can show that the MLE of a parameter has the following distribution: $\sqrt n (\hat\theta-\theta)\sim\mathcal N(0,\hat\sigma^2)$, where $\hat\sigma^2$ is the corresponding estimate of the variance. The variance estimate $\hat\sigma^2/n$ can be used to estimate the confidence interval of the pa... | How to create an interval estimate based on being within some percentage of the maximum value of the | Under some conditions you can show that the MLE of a parameter has the following distribution: $\sqrt n (\hat\theta-\theta)\sim\mathcal N(0,\hat\sigma^2)$, where $\hat\sigma^2$ is the corresponding es | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function?
Under some conditions you can show that the MLE of a parameter has the following distribution: $\sqrt n (\hat\theta-\theta)\sim\mathcal N(0,\hat\sigma^2)$, where $\hat\sigma^2$ is the corresponding ... | How to create an interval estimate based on being within some percentage of the maximum value of the
Under some conditions you can show that the MLE of a parameter has the following distribution: $\sqrt n (\hat\theta-\theta)\sim\mathcal N(0,\hat\sigma^2)$, where $\hat\sigma^2$ is the corresponding es |
39,613 | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function? | You have likelihood intervals/regions and confidence intervals/regions. They are different.
Confidence intervals stem from the fiducial distribution, which is a different thing from the likelihood function.
If we form a confidence interval based on inverting an $\alpha$-level likelihood ratio test, is that basically ... | How to create an interval estimate based on being within some percentage of the maximum value of the | You have likelihood intervals/regions and confidence intervals/regions. They are different.
Confidence intervals stem from the fiducial distribution, which is a different thing from the likelihood fun | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function?
You have likelihood intervals/regions and confidence intervals/regions. They are different.
Confidence intervals stem from the fiducial distribution, which is a different thing from the likelihood f... | How to create an interval estimate based on being within some percentage of the maximum value of the
You have likelihood intervals/regions and confidence intervals/regions. They are different.
Confidence intervals stem from the fiducial distribution, which is a different thing from the likelihood fun |
39,614 | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function? | TL;DR I use an example — inference for the rate $\theta$ in a Poisson model — to show how to construct a likelihood interval. I've borrowed this example from the book In All Likelihood: Statistical Modelling And Inference Using Likelihood by Yudi Pawitan.
Say we observe $x = 3$ from $\operatorname{Poisson}(\theta)$. Th... | How to create an interval estimate based on being within some percentage of the maximum value of the | TL;DR I use an example — inference for the rate $\theta$ in a Poisson model — to show how to construct a likelihood interval. I've borrowed this example from the book In All Likelihood: Statistical Mo | How to create an interval estimate based on being within some percentage of the maximum value of the likelihood function?
TL;DR I use an example — inference for the rate $\theta$ in a Poisson model — to show how to construct a likelihood interval. I've borrowed this example from the book In All Likelihood: Statistical ... | How to create an interval estimate based on being within some percentage of the maximum value of the
TL;DR I use an example — inference for the rate $\theta$ in a Poisson model — to show how to construct a likelihood interval. I've borrowed this example from the book In All Likelihood: Statistical Mo |
39,615 | Is there ever a statistical reason NOT to use Satterthwaite's method to account for unequal variances? | For 2-sample t tests. For two-sample t tests, I think it is now standard practice
to use the Welch two-sample t test, unless there is strong
prior evidence (say, from data of the same type) that population
variances are equal. In some statistical software packages the
Welch test is the default 2-sample t test, so that ... | Is there ever a statistical reason NOT to use Satterthwaite's method to account for unequal variance | For 2-sample t tests. For two-sample t tests, I think it is now standard practice
to use the Welch two-sample t test, unless there is strong
prior evidence (say, from data of the same type) that popul | Is there ever a statistical reason NOT to use Satterthwaite's method to account for unequal variances?
For 2-sample t tests. For two-sample t tests, I think it is now standard practice
to use the Welch two-sample t test, unless there is strong
prior evidence (say, from data of the same type) that population
variances a... | Is there ever a statistical reason NOT to use Satterthwaite's method to account for unequal variance
For 2-sample t tests. For two-sample t tests, I think it is now standard practice
to use the Welch two-sample t test, unless there is strong
prior evidence (say, from data of the same type) that popul |
39,616 | How does LightGBM deals with incremental learning (and concept drift)? | LightGBM will add more trees if we update it through continued training (e.g. through BoosterUpdateOneIter). Assuming we use refit we will be using existing tree structures to update the output of the leaves based on the new data. It is faster than re-training from scratch, since we do not have to re-discover the optim... | How does LightGBM deals with incremental learning (and concept drift)? | LightGBM will add more trees if we update it through continued training (e.g. through BoosterUpdateOneIter). Assuming we use refit we will be using existing tree structures to update the output of the | How does LightGBM deals with incremental learning (and concept drift)?
LightGBM will add more trees if we update it through continued training (e.g. through BoosterUpdateOneIter). Assuming we use refit we will be using existing tree structures to update the output of the leaves based on the new data. It is faster than ... | How does LightGBM deals with incremental learning (and concept drift)?
LightGBM will add more trees if we update it through continued training (e.g. through BoosterUpdateOneIter). Assuming we use refit we will be using existing tree structures to update the output of the |
39,617 | What is the intuition behind what makes dice coefficient handle imbalanced data? | Dice score measures the relative overlap between the prediction and the ground truth (intersection over union). It has the same value for small and large objects both: Did you guess a half of the object correctly? Great, your loss is 1/2. I don't care if the object was 10 or 1000 pixels large.
On the other hand, cross-... | What is the intuition behind what makes dice coefficient handle imbalanced data? | Dice score measures the relative overlap between the prediction and the ground truth (intersection over union). It has the same value for small and large objects both: Did you guess a half of the obje | What is the intuition behind what makes dice coefficient handle imbalanced data?
Dice score measures the relative overlap between the prediction and the ground truth (intersection over union). It has the same value for small and large objects both: Did you guess a half of the object correctly? Great, your loss is 1/2. ... | What is the intuition behind what makes dice coefficient handle imbalanced data?
Dice score measures the relative overlap between the prediction and the ground truth (intersection over union). It has the same value for small and large objects both: Did you guess a half of the obje |
39,618 | What is the intuition behind what makes dice coefficient handle imbalanced data? | Dealing with a two-class problem is intuitive, it calculates the overlap between foreground regions (while background is not of interest). In a multi-class scenario, when provided with probability maps on one end and one-hot encoded labels on the other, it effectively performs multiple two-class problems. In practice, ... | What is the intuition behind what makes dice coefficient handle imbalanced data? | Dealing with a two-class problem is intuitive, it calculates the overlap between foreground regions (while background is not of interest). In a multi-class scenario, when provided with probability map | What is the intuition behind what makes dice coefficient handle imbalanced data?
Dealing with a two-class problem is intuitive, it calculates the overlap between foreground regions (while background is not of interest). In a multi-class scenario, when provided with probability maps on one end and one-hot encoded labels... | What is the intuition behind what makes dice coefficient handle imbalanced data?
Dealing with a two-class problem is intuitive, it calculates the overlap between foreground regions (while background is not of interest). In a multi-class scenario, when provided with probability map |
39,619 | Why does the likelihood function of a binomial distribution not include the combinatorics term? [duplicate] | We often don’t care about the likelihood, just the value for which the likelihood is maximized.
When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maximum whether you include constants out front or not. Sure, that maximum value will be different, but that is ... | Why does the likelihood function of a binomial distribution not include the combinatorics term? [dup | We often don’t care about the likelihood, just the value for which the likelihood is maximized.
When you use the likelihood function to find a maximum likelihood estimator, you get the same point giv | Why does the likelihood function of a binomial distribution not include the combinatorics term? [duplicate]
We often don’t care about the likelihood, just the value for which the likelihood is maximized.
When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maxi... | Why does the likelihood function of a binomial distribution not include the combinatorics term? [dup
We often don’t care about the likelihood, just the value for which the likelihood is maximized.
When you use the likelihood function to find a maximum likelihood estimator, you get the same point giv |
39,620 | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent lme4 model? | I'm going to answer that myself in case that others may be interested at some point. In the end I did follow the approach mentioned above. The code would look like this:
library(lme4)
library(robustlmm)
data(Dyestuff, package = "lme4")
# fit a mixed model
model <- lmer(Yield ~ 1|Batch, data=Dyestuff)
# fit the robust... | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent l | I'm going to answer that myself in case that others may be interested at some point. In the end I did follow the approach mentioned above. The code would look like this:
library(lme4)
library(robustlm | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent lme4 model?
I'm going to answer that myself in case that others may be interested at some point. In the end I did follow the approach mentioned above. The code would look like this:
library(lme4)
library(robustlmm)
data(... | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent l
I'm going to answer that myself in case that others may be interested at some point. In the end I did follow the approach mentioned above. The code would look like this:
library(lme4)
library(robustlm |
39,621 | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent lme4 model? | I would like to add for future reference, that the tab_model function of sjPlot package can also calculate p values based on Satterthwaite-approximated dfs. | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent l | I would like to add for future reference, that the tab_model function of sjPlot package can also calculate p values based on Satterthwaite-approximated dfs. | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent lme4 model?
I would like to add for future reference, that the tab_model function of sjPlot package can also calculate p values based on Satterthwaite-approximated dfs. | Obtaining p-values in a robustlmm mixed model via Satterthwaite-approximated DFs of the equivalent l
I would like to add for future reference, that the tab_model function of sjPlot package can also calculate p values based on Satterthwaite-approximated dfs. |
39,622 | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$ | You're almost there. As pointed out by @whuber, the trick is to recognise that $\phi(x)^2$ is another Gaussian that integrates to one after normalisation:
\begin{align}
\int \phi(x)^2 dx
&= \int_{-\infty}^\infty \left(\frac1{\sqrt{2\pi}} e^{-\frac12x^2}\right)^2dx
\\&= \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \frac1... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi( | You're almost there. As pointed out by @whuber, the trick is to recognise that $\phi(x)^2$ is another Gaussian that integrates to one after normalisation:
\begin{align}
\int \phi(x)^2 dx
&= \int_{- | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$
You're almost there. As pointed out by @whuber, the trick is to recognise that $\phi(x)^2$ is another Gaussian that integrates to one after normalisation:
\begin{align}
\int \phi(x)^2 dx
&= \int_{-\infty}^\infty \... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(
You're almost there. As pointed out by @whuber, the trick is to recognise that $\phi(x)^2$ is another Gaussian that integrates to one after normalisation:
\begin{align}
\int \phi(x)^2 dx
&= \int_{- |
39,623 | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$ | If you want to check your work, it only takes a few seconds with a computer algebra system. In your case, $X \sim N(0,1)$ with pdf $f(x)$:
The cdf is:
where Erf denotes the error function.
Then the desired correlation can be found immediately with:
... where I am using the Corr function from the mathStatica package... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi( | If you want to check your work, it only takes a few seconds with a computer algebra system. In your case, $X \sim N(0,1)$ with pdf $f(x)$:
The cdf is:
where Erf denotes the error function.
Then the | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$
If you want to check your work, it only takes a few seconds with a computer algebra system. In your case, $X \sim N(0,1)$ with pdf $f(x)$:
The cdf is:
where Erf denotes the error function.
Then the desired correla... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(
If you want to check your work, it only takes a few seconds with a computer algebra system. In your case, $X \sim N(0,1)$ with pdf $f(x)$:
The cdf is:
where Erf denotes the error function.
Then the |
39,624 | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$ | For a general case with $a,b$ we can derive the result of following integral,
\begin{eqnarray}
\int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left(z_{1}-bx\leq -a,z_{2}-bx\leq -a\right)\\&=&
\mathcal{MVN}\left(x=\{-a,-a\},\mu=\{0,0\},\Sigma=\begin{bmatrix}b^{2}+1 & ... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi( | For a general case with $a,b$ we can derive the result of following integral,
\begin{eqnarray}
\int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$
For a general case with $a,b$ we can derive the result of following integral,
\begin{eqnarray}
\int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left(z_{1}-bx\leq -a... | If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(
For a general case with $a,b$ we can derive the result of following integral,
\begin{eqnarray}
\int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left |
39,625 | Two sample t-test to show equality of the two means | You cannot use the first test in the way you describe, because failure to reject in the first test only says that you were unable to reject $H_0$ nothing more than that. It is like only being given the information that "the prosecutor was unable to provide the jury with enough evidence to secure a conviction" - that do... | Two sample t-test to show equality of the two means | You cannot use the first test in the way you describe, because failure to reject in the first test only says that you were unable to reject $H_0$ nothing more than that. It is like only being given th | Two sample t-test to show equality of the two means
You cannot use the first test in the way you describe, because failure to reject in the first test only says that you were unable to reject $H_0$ nothing more than that. It is like only being given the information that "the prosecutor was unable to provide the jury wi... | Two sample t-test to show equality of the two means
You cannot use the first test in the way you describe, because failure to reject in the first test only says that you were unable to reject $H_0$ nothing more than that. It is like only being given th |
39,626 | Two sample t-test to show equality of the two means | I learned in school that the null hypothesis should always represent the "common" belief and the alternative hypothesis should represent the change that I would like to show.
That is not accurate explanation of the null hypothesis. The null hypothesis is simply a hypothesis that consists of a specific distribution fro... | Two sample t-test to show equality of the two means | I learned in school that the null hypothesis should always represent the "common" belief and the alternative hypothesis should represent the change that I would like to show.
That is not accurate exp | Two sample t-test to show equality of the two means
I learned in school that the null hypothesis should always represent the "common" belief and the alternative hypothesis should represent the change that I would like to show.
That is not accurate explanation of the null hypothesis. The null hypothesis is simply a hyp... | Two sample t-test to show equality of the two means
I learned in school that the null hypothesis should always represent the "common" belief and the alternative hypothesis should represent the change that I would like to show.
That is not accurate exp |
39,627 | Statistical significance when comparing two models for classification | Simply speaking, the performance metrics used are statistics derived from our test set. We can go ahead and compute confidence intervals around these statistics as we would do in a classical setting.
For example, let's say we use Accuracy (which is not good metric for classification), i.e. the proportion of correctly ... | Statistical significance when comparing two models for classification | Simply speaking, the performance metrics used are statistics derived from our test set. We can go ahead and compute confidence intervals around these statistics as we would do in a classical setting. | Statistical significance when comparing two models for classification
Simply speaking, the performance metrics used are statistics derived from our test set. We can go ahead and compute confidence intervals around these statistics as we would do in a classical setting.
For example, let's say we use Accuracy (which is ... | Statistical significance when comparing two models for classification
Simply speaking, the performance metrics used are statistics derived from our test set. We can go ahead and compute confidence intervals around these statistics as we would do in a classical setting. |
39,628 | Does confidence interval for odds ratio assume log-normal distribution? | Given the comments, I have included the proof of the floating equation at the bottom of the response.
Given a two-by-two contingency table where the OR is $\frac{a/b}{c/d}$, if you take the log of both sides, the log odds ratio is $\log(a) - \log(b) - (\log(c) - \log(d))$. Hence the formula for the log odds ratio is a... | Does confidence interval for odds ratio assume log-normal distribution? | Given the comments, I have included the proof of the floating equation at the bottom of the response.
Given a two-by-two contingency table where the OR is $\frac{a/b}{c/d}$, if you take the log of bo | Does confidence interval for odds ratio assume log-normal distribution?
Given the comments, I have included the proof of the floating equation at the bottom of the response.
Given a two-by-two contingency table where the OR is $\frac{a/b}{c/d}$, if you take the log of both sides, the log odds ratio is $\log(a) - \log(... | Does confidence interval for odds ratio assume log-normal distribution?
Given the comments, I have included the proof of the floating equation at the bottom of the response.
Given a two-by-two contingency table where the OR is $\frac{a/b}{c/d}$, if you take the log of bo |
39,629 | Identifying outliers in the data | The method is correct. The problem with outlier detection is: there is not a general solution and real thresholds. It should be used additionally but not primary to assess your data. You find in different literature different threshold. The inventor of this method suggests your threshold with D > 1 1. Your method (4/s... | Identifying outliers in the data | The method is correct. The problem with outlier detection is: there is not a general solution and real thresholds. It should be used additionally but not primary to assess your data. You find in diff | Identifying outliers in the data
The method is correct. The problem with outlier detection is: there is not a general solution and real thresholds. It should be used additionally but not primary to assess your data. You find in different literature different threshold. The inventor of this method suggests your thresho... | Identifying outliers in the data
The method is correct. The problem with outlier detection is: there is not a general solution and real thresholds. It should be used additionally but not primary to assess your data. You find in diff |
39,630 | Identifying outliers in the data | The method is correct IN THIS CASE but not in general . You need a sufficient model to detect an anomaly ( read underlying signal and latent deterministic structure ).
You approach assumes the kind of model i.e. a model with time and versions of time as exogenous predictors premising no pulses/level shifts/seasonal pul... | Identifying outliers in the data | The method is correct IN THIS CASE but not in general . You need a sufficient model to detect an anomaly ( read underlying signal and latent deterministic structure ).
You approach assumes the kind of | Identifying outliers in the data
The method is correct IN THIS CASE but not in general . You need a sufficient model to detect an anomaly ( read underlying signal and latent deterministic structure ).
You approach assumes the kind of model i.e. a model with time and versions of time as exogenous predictors premising no... | Identifying outliers in the data
The method is correct IN THIS CASE but not in general . You need a sufficient model to detect an anomaly ( read underlying signal and latent deterministic structure ).
You approach assumes the kind of |
39,631 | Is the sum of p-value and specificity 1 | There's some confusion between $\alpha$, the type I error rate, and the $p$-value. I'll try to explain these two concepts before turning to specificity.
You consider a test statistic $X$, with left deviations of $X$ going against the null hypothesis $H_0$ (this is a bit unusual, in general right deviations go againt $H... | Is the sum of p-value and specificity 1 | There's some confusion between $\alpha$, the type I error rate, and the $p$-value. I'll try to explain these two concepts before turning to specificity.
You consider a test statistic $X$, with left de | Is the sum of p-value and specificity 1
There's some confusion between $\alpha$, the type I error rate, and the $p$-value. I'll try to explain these two concepts before turning to specificity.
You consider a test statistic $X$, with left deviations of $X$ going against the null hypothesis $H_0$ (this is a bit unusual, ... | Is the sum of p-value and specificity 1
There's some confusion between $\alpha$, the type I error rate, and the $p$-value. I'll try to explain these two concepts before turning to specificity.
You consider a test statistic $X$, with left de |
39,632 | Is the sum of p-value and specificity 1 | You seem to be referring to type I and type II errors, that are among the fundamental concepts of null hypothesis testing. If $H_0$ is true, then we expect to see $\alpha$ proportion of false positives (incorrectly reject), we call this type I error. On another hand, if $H_0$ is false, we expect to see $\beta$ false ne... | Is the sum of p-value and specificity 1 | You seem to be referring to type I and type II errors, that are among the fundamental concepts of null hypothesis testing. If $H_0$ is true, then we expect to see $\alpha$ proportion of false positive | Is the sum of p-value and specificity 1
You seem to be referring to type I and type II errors, that are among the fundamental concepts of null hypothesis testing. If $H_0$ is true, then we expect to see $\alpha$ proportion of false positives (incorrectly reject), we call this type I error. On another hand, if $H_0$ is ... | Is the sum of p-value and specificity 1
You seem to be referring to type I and type II errors, that are among the fundamental concepts of null hypothesis testing. If $H_0$ is true, then we expect to see $\alpha$ proportion of false positive |
39,633 | Is this a typo/error in Bishop's book | Indeed, indeed, there is a typo in (9.16) and it should be
$$0=\sum_{n=1}^N \gamma(z_{nk}) {\mathbf \Sigma}_k^{-1} ({\mathbf x}_n-{\mathbf \mu}_k)$$Fortunately, this does not impact the next equation (9.17).
As for deriving the conditional MLE of the covariance matrix ${\mathbf \Sigma}_k$, the result and the method are... | Is this a typo/error in Bishop's book | Indeed, indeed, there is a typo in (9.16) and it should be
$$0=\sum_{n=1}^N \gamma(z_{nk}) {\mathbf \Sigma}_k^{-1} ({\mathbf x}_n-{\mathbf \mu}_k)$$Fortunately, this does not impact the next equation | Is this a typo/error in Bishop's book
Indeed, indeed, there is a typo in (9.16) and it should be
$$0=\sum_{n=1}^N \gamma(z_{nk}) {\mathbf \Sigma}_k^{-1} ({\mathbf x}_n-{\mathbf \mu}_k)$$Fortunately, this does not impact the next equation (9.17).
As for deriving the conditional MLE of the covariance matrix ${\mathbf \Si... | Is this a typo/error in Bishop's book
Indeed, indeed, there is a typo in (9.16) and it should be
$$0=\sum_{n=1}^N \gamma(z_{nk}) {\mathbf \Sigma}_k^{-1} ({\mathbf x}_n-{\mathbf \mu}_k)$$Fortunately, this does not impact the next equation |
39,634 | What is wrong with one tailed z-tests for a proportion? | My comment was specifically about your articulation of the appropriate one-sample one-sided (aka one-tailed) null hypothesis (not about one-sided tests per se) which, for proportions, should either be $H_{0}: p \ge p_{0}$ with $H_{1}: p < p_{0}$, or $H_{0}: p \le p_{0}$ with $H_{1}: p > p_{0}$. Bear in mind that null h... | What is wrong with one tailed z-tests for a proportion? | My comment was specifically about your articulation of the appropriate one-sample one-sided (aka one-tailed) null hypothesis (not about one-sided tests per se) which, for proportions, should either be | What is wrong with one tailed z-tests for a proportion?
My comment was specifically about your articulation of the appropriate one-sample one-sided (aka one-tailed) null hypothesis (not about one-sided tests per se) which, for proportions, should either be $H_{0}: p \ge p_{0}$ with $H_{1}: p < p_{0}$, or $H_{0}: p \le ... | What is wrong with one tailed z-tests for a proportion?
My comment was specifically about your articulation of the appropriate one-sample one-sided (aka one-tailed) null hypothesis (not about one-sided tests per se) which, for proportions, should either be |
39,635 | What is wrong with one tailed z-tests for a proportion? | One problem with your suggestion is that the normal approximation for proportions is only reasonable in certain circumstances. A large sample and the observed proportion well away from the boundaries of 0 and 1. Neither of those circumstances is specified in the original question. See this question: Testing equality of... | What is wrong with one tailed z-tests for a proportion? | One problem with your suggestion is that the normal approximation for proportions is only reasonable in certain circumstances. A large sample and the observed proportion well away from the boundaries | What is wrong with one tailed z-tests for a proportion?
One problem with your suggestion is that the normal approximation for proportions is only reasonable in certain circumstances. A large sample and the observed proportion well away from the boundaries of 0 and 1. Neither of those circumstances is specified in the o... | What is wrong with one tailed z-tests for a proportion?
One problem with your suggestion is that the normal approximation for proportions is only reasonable in certain circumstances. A large sample and the observed proportion well away from the boundaries |
39,636 | Confused about Dropout implementations in Tensorflow | This is because you're using the Adam optimizer. The Adam optimizer is a kind of momentum optimizer (specifically, it tracks the first and second moments of the updates), so an update will still occur for all model parameters even though dropout is present. | Confused about Dropout implementations in Tensorflow | This is because you're using the Adam optimizer. The Adam optimizer is a kind of momentum optimizer (specifically, it tracks the first and second moments of the updates), so an update will still occur | Confused about Dropout implementations in Tensorflow
This is because you're using the Adam optimizer. The Adam optimizer is a kind of momentum optimizer (specifically, it tracks the first and second moments of the updates), so an update will still occur for all model parameters even though dropout is present. | Confused about Dropout implementations in Tensorflow
This is because you're using the Adam optimizer. The Adam optimizer is a kind of momentum optimizer (specifically, it tracks the first and second moments of the updates), so an update will still occur |
39,637 | Confused about Dropout implementations in Tensorflow | Dropout:
Dropout in Tensorflow is implemented slightly different than in the original paper: instead of scaling the weights by 1/(1-p) after updating the weights (where p is the dropout rate), the neuron outputs (e.g., the outputs from ReLUs) are scaled by 1/(1-p) during the forward and backward passes. In this manner,... | Confused about Dropout implementations in Tensorflow | Dropout:
Dropout in Tensorflow is implemented slightly different than in the original paper: instead of scaling the weights by 1/(1-p) after updating the weights (where p is the dropout rate), the neu | Confused about Dropout implementations in Tensorflow
Dropout:
Dropout in Tensorflow is implemented slightly different than in the original paper: instead of scaling the weights by 1/(1-p) after updating the weights (where p is the dropout rate), the neuron outputs (e.g., the outputs from ReLUs) are scaled by 1/(1-p) du... | Confused about Dropout implementations in Tensorflow
Dropout:
Dropout in Tensorflow is implemented slightly different than in the original paper: instead of scaling the weights by 1/(1-p) after updating the weights (where p is the dropout rate), the neu |
39,638 | Berry Esseen Theorem for Infinite Absolute Third Moment | The Berry -Esseen theorem requires the finiteness of the third moment. A generalization of Berry-Esseen inequality that doesn't, can be found as Theorem 5 on page 112 of
Petrov, V. V. (1975). Sums of independent random variables (Vol. 82). Springer Science & Business Media., doi:10.1007/978-3-642-65809-9.
For the cas... | Berry Esseen Theorem for Infinite Absolute Third Moment | The Berry -Esseen theorem requires the finiteness of the third moment. A generalization of Berry-Esseen inequality that doesn't, can be found as Theorem 5 on page 112 of
Petrov, V. V. (1975). Sums of | Berry Esseen Theorem for Infinite Absolute Third Moment
The Berry -Esseen theorem requires the finiteness of the third moment. A generalization of Berry-Esseen inequality that doesn't, can be found as Theorem 5 on page 112 of
Petrov, V. V. (1975). Sums of independent random variables (Vol. 82). Springer Science & Busi... | Berry Esseen Theorem for Infinite Absolute Third Moment
The Berry -Esseen theorem requires the finiteness of the third moment. A generalization of Berry-Esseen inequality that doesn't, can be found as Theorem 5 on page 112 of
Petrov, V. V. (1975). Sums of |
39,639 | What is "unit" standard deviation? | It means you are converting your data features from its original units (miles, dollars, elapsed time,...) to units of standard deviation. As you requested follows a very simple example:
Supose you want to predict house prices from two features: number of bedrooms (integer unit) and size (in squared meters unit), lik... | What is "unit" standard deviation? | It means you are converting your data features from its original units (miles, dollars, elapsed time,...) to units of standard deviation. As you requested follows a very simple example:
Supose you w | What is "unit" standard deviation?
It means you are converting your data features from its original units (miles, dollars, elapsed time,...) to units of standard deviation. As you requested follows a very simple example:
Supose you want to predict house prices from two features: number of bedrooms (integer unit) and... | What is "unit" standard deviation?
It means you are converting your data features from its original units (miles, dollars, elapsed time,...) to units of standard deviation. As you requested follows a very simple example:
Supose you w |
39,640 | Warnings during WAIC computation: how to proceed? | I'm happy that you care about these diagnostic messages. I'm responsible for that specific warning message. WAIC doesn't have a good way to diagnose the reliability and that 0.4 threshold is empirically chosen. Pareto k diagnostic in PSIS-LOO is much better. You did not mention the specific k values, but if they are la... | Warnings during WAIC computation: how to proceed? | I'm happy that you care about these diagnostic messages. I'm responsible for that specific warning message. WAIC doesn't have a good way to diagnose the reliability and that 0.4 threshold is empirical | Warnings during WAIC computation: how to proceed?
I'm happy that you care about these diagnostic messages. I'm responsible for that specific warning message. WAIC doesn't have a good way to diagnose the reliability and that 0.4 threshold is empirically chosen. Pareto k diagnostic in PSIS-LOO is much better. You did not... | Warnings during WAIC computation: how to proceed?
I'm happy that you care about these diagnostic messages. I'm responsible for that specific warning message. WAIC doesn't have a good way to diagnose the reliability and that 0.4 threshold is empirical |
39,641 | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$ | We have
\begin{align}
\operatorname{Var}(x^T(A+B)x)&=\operatorname{Var}(x^TAx+x^TBx)
\\&=\operatorname{Var}(x^TAx)+\operatorname{Var}(x^TBx)+2\operatorname{Cov}(x^TAx,x^TBx)
\end{align}
So, $$\operatorname{Cov}(x^TAx,x^TBx)=\frac12\left[\operatorname{Var}(x^T(A+B)x)-\operatorname{Var}(x^TAx)-\operatorname{Var}(x^TBx)\r... | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$ | We have
\begin{align}
\operatorname{Var}(x^T(A+B)x)&=\operatorname{Var}(x^TAx+x^TBx)
\\&=\operatorname{Var}(x^TAx)+\operatorname{Var}(x^TBx)+2\operatorname{Cov}(x^TAx,x^TBx)
\end{align}
So, $$\operato | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$
We have
\begin{align}
\operatorname{Var}(x^T(A+B)x)&=\operatorname{Var}(x^TAx+x^TBx)
\\&=\operatorname{Var}(x^TAx)+\operatorname{Var}(x^TBx)+2\operatorname{Cov}(x^TAx,x^TBx)
\end{align}
So, $$\operatorname{Cov}(x^TAx,x^TBx... | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$
We have
\begin{align}
\operatorname{Var}(x^T(A+B)x)&=\operatorname{Var}(x^TAx+x^TBx)
\\&=\operatorname{Var}(x^TAx)+\operatorname{Var}(x^TBx)+2\operatorname{Cov}(x^TAx,x^TBx)
\end{align}
So, $$\operato |
39,642 | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$ | Just trying to provide a sketch for the proof, the key here is to show the following equations: $$\begin{align}\operatorname{E}[x'Ax]&=\operatorname{tr}(A\Sigma)+\mu'A\mu, \label{1}\tag{1}\\\operatorname{E}[x'Axx'Bx]&=2\operatorname{tr}(A\Sigma B\Sigma)+4\mu'A\Sigma B\mu+(\operatorname{tr}(A\Sigma)+\mu'A\mu)(\operatorn... | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$ | Just trying to provide a sketch for the proof, the key here is to show the following equations: $$\begin{align}\operatorname{E}[x'Ax]&=\operatorname{tr}(A\Sigma)+\mu'A\mu, \label{1}\tag{1}\\\operatorn | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$
Just trying to provide a sketch for the proof, the key here is to show the following equations: $$\begin{align}\operatorname{E}[x'Ax]&=\operatorname{tr}(A\Sigma)+\mu'A\mu, \label{1}\tag{1}\\\operatorname{E}[x'Axx'Bx]&=2\op... | Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$
Just trying to provide a sketch for the proof, the key here is to show the following equations: $$\begin{align}\operatorname{E}[x'Ax]&=\operatorname{tr}(A\Sigma)+\mu'A\mu, \label{1}\tag{1}\\\operatorn |
39,643 | Kernels in Gaussian Processes | Notation / setting
We are considering a GP regression model:
\begin{equation}
y_i = f(x_i) + \epsilon_i
\end{equation}
where $y_i\in \mathbb{R}$,$x_i \in \mathbb{R}^d$, $f$ a Gaussian process (whose realizations are functions $f:\mathbb{R}^d\rightarrow \mathbb{R}$),
\begin{equation}
f \sim \mathrm{GP}(m(x_i), \kappa(x... | Kernels in Gaussian Processes | Notation / setting
We are considering a GP regression model:
\begin{equation}
y_i = f(x_i) + \epsilon_i
\end{equation}
where $y_i\in \mathbb{R}$,$x_i \in \mathbb{R}^d$, $f$ a Gaussian process (whose r | Kernels in Gaussian Processes
Notation / setting
We are considering a GP regression model:
\begin{equation}
y_i = f(x_i) + \epsilon_i
\end{equation}
where $y_i\in \mathbb{R}$,$x_i \in \mathbb{R}^d$, $f$ a Gaussian process (whose realizations are functions $f:\mathbb{R}^d\rightarrow \mathbb{R}$),
\begin{equation}
f \si... | Kernels in Gaussian Processes
Notation / setting
We are considering a GP regression model:
\begin{equation}
y_i = f(x_i) + \epsilon_i
\end{equation}
where $y_i\in \mathbb{R}$,$x_i \in \mathbb{R}^d$, $f$ a Gaussian process (whose r |
39,644 | Joint distribution in layman's terms | As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, i.e. every possible outcome is equally likely). I can use these to find e.g. P(coin lands heads) or P(I roll a six).
The... | Joint distribution in layman's terms | As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, | Joint distribution in layman's terms
As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, i.e. every possible outcome is equally likely). I can use these to find e.g. P(coin... | Joint distribution in layman's terms
As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, |
39,645 | What parameters to optimize in KNN? | As you correctly recognise almost always people focus on optimising against $k$ when in comes to $k$-NN applications. A standard paper on the matter is Complete Cross-Validation for Nearest Neighbor Classifiers by Mullin and Sukthankar.
Having said that, this is half of the story (or one third actually if you take $k$... | What parameters to optimize in KNN? | As you correctly recognise almost always people focus on optimising against $k$ when in comes to $k$-NN applications. A standard paper on the matter is Complete Cross-Validation for Nearest Neighbor C | What parameters to optimize in KNN?
As you correctly recognise almost always people focus on optimising against $k$ when in comes to $k$-NN applications. A standard paper on the matter is Complete Cross-Validation for Nearest Neighbor Classifiers by Mullin and Sukthankar.
Having said that, this is half of the story (o... | What parameters to optimize in KNN?
As you correctly recognise almost always people focus on optimising against $k$ when in comes to $k$-NN applications. A standard paper on the matter is Complete Cross-Validation for Nearest Neighbor C |
39,646 | What parameters to optimize in KNN? | For better results, normalizing data on the same scale is highly recommended. Generally, the normalization range considered between 0 and 1. KNN is not suitable for the large dimensional data. In such cases, dimension needs to reduce to improve the performance. Also, handling missing values will help us in improving re... | What parameters to optimize in KNN? | For better results, normalizing data on the same scale is highly recommended. Generally, the normalization range considered between 0 and 1. KNN is not suitable for the large dimensional data. In such | What parameters to optimize in KNN?
For better results, normalizing data on the same scale is highly recommended. Generally, the normalization range considered between 0 and 1. KNN is not suitable for the large dimensional data. In such cases, dimension needs to reduce to improve the performance. Also, handling missing... | What parameters to optimize in KNN?
For better results, normalizing data on the same scale is highly recommended. Generally, the normalization range considered between 0 and 1. KNN is not suitable for the large dimensional data. In such |
39,647 | MAR vs. MNAR: how can I decide? | MAR means that the probability of becoming missing on variable $x$ is independent of the (possibly unobserved) variable $x$. So to empirically differentiate between MAR and NMAR your need to know the values of $x$ when $x$ is missing, which we obviously do not have. Sometimes, there is something about the design of the... | MAR vs. MNAR: how can I decide? | MAR means that the probability of becoming missing on variable $x$ is independent of the (possibly unobserved) variable $x$. So to empirically differentiate between MAR and NMAR your need to know the | MAR vs. MNAR: how can I decide?
MAR means that the probability of becoming missing on variable $x$ is independent of the (possibly unobserved) variable $x$. So to empirically differentiate between MAR and NMAR your need to know the values of $x$ when $x$ is missing, which we obviously do not have. Sometimes, there is s... | MAR vs. MNAR: how can I decide?
MAR means that the probability of becoming missing on variable $x$ is independent of the (possibly unobserved) variable $x$. So to empirically differentiate between MAR and NMAR your need to know the |
39,648 | MAR vs. MNAR: how can I decide? | To Maarten's answer, I would add that MAR and MNAR are not really hard distinctions. Rather, it may be better to think of a continuum from MAR to MNAR. There will almost always be some degree of dependence between the probability of missingness on $x$ and the values of $x$ itself: the question is really how problematic... | MAR vs. MNAR: how can I decide? | To Maarten's answer, I would add that MAR and MNAR are not really hard distinctions. Rather, it may be better to think of a continuum from MAR to MNAR. There will almost always be some degree of depen | MAR vs. MNAR: how can I decide?
To Maarten's answer, I would add that MAR and MNAR are not really hard distinctions. Rather, it may be better to think of a continuum from MAR to MNAR. There will almost always be some degree of dependence between the probability of missingness on $x$ and the values of $x$ itself: the qu... | MAR vs. MNAR: how can I decide?
To Maarten's answer, I would add that MAR and MNAR are not really hard distinctions. Rather, it may be better to think of a continuum from MAR to MNAR. There will almost always be some degree of depen |
39,649 | How to determine significant associations in a mosaic plot? | The formula for the standardized residuals is:
$$\begin{align}\text{Pearson's residuals}\,&=\,\frac{\text{Observed - Expected}}{ \sqrt{\text{Expected}}}\\
d_{ij}&=\frac{n_{ij}-m_{ij}}{\sqrt{m_{ij}}}
\end{align}$$
where $m_{ij} = E( f_{ij})$ is the expected frequency of the $i$-th row and the $j$-th column.
The sum of ... | How to determine significant associations in a mosaic plot? | The formula for the standardized residuals is:
$$\begin{align}\text{Pearson's residuals}\,&=\,\frac{\text{Observed - Expected}}{ \sqrt{\text{Expected}}}\\
d_{ij}&=\frac{n_{ij}-m_{ij}}{\sqrt{m_{ij}}}
\ | How to determine significant associations in a mosaic plot?
The formula for the standardized residuals is:
$$\begin{align}\text{Pearson's residuals}\,&=\,\frac{\text{Observed - Expected}}{ \sqrt{\text{Expected}}}\\
d_{ij}&=\frac{n_{ij}-m_{ij}}{\sqrt{m_{ij}}}
\end{align}$$
where $m_{ij} = E( f_{ij})$ is the expected fre... | How to determine significant associations in a mosaic plot?
The formula for the standardized residuals is:
$$\begin{align}\text{Pearson's residuals}\,&=\,\frac{\text{Observed - Expected}}{ \sqrt{\text{Expected}}}\\
d_{ij}&=\frac{n_{ij}-m_{ij}}{\sqrt{m_{ij}}}
\ |
39,650 | How to determine significant associations in a mosaic plot? | This is best interpreted using some specific language. Within the 40+ age group (in the plot, labeled "40-") there is a significant association between the variables memory and attitude. We cite associations between variables, not between values or categories within them (such as "moderate" or "no").
A more specifi... | How to determine significant associations in a mosaic plot? | This is best interpreted using some specific language. Within the 40+ age group (in the plot, labeled "40-") there is a significant association between the variables memory and attitude. We cite ass | How to determine significant associations in a mosaic plot?
This is best interpreted using some specific language. Within the 40+ age group (in the plot, labeled "40-") there is a significant association between the variables memory and attitude. We cite associations between variables, not between values or categorie... | How to determine significant associations in a mosaic plot?
This is best interpreted using some specific language. Within the 40+ age group (in the plot, labeled "40-") there is a significant association between the variables memory and attitude. We cite ass |
39,651 | Theoretical: Minimum Number of Support Vectors | Yes. The minimum number of support vectors is two for your scenario. You don't need more than two here.
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or size of data set, the number of support vectors could be as little as 2.
Reference: https://stackoverflow.com/questions... | Theoretical: Minimum Number of Support Vectors | Yes. The minimum number of support vectors is two for your scenario. You don't need more than two here.
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or | Theoretical: Minimum Number of Support Vectors
Yes. The minimum number of support vectors is two for your scenario. You don't need more than two here.
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or size of data set, the number of support vectors could be as little as 2.... | Theoretical: Minimum Number of Support Vectors
Yes. The minimum number of support vectors is two for your scenario. You don't need more than two here.
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or |
39,652 | Theoretical: Minimum Number of Support Vectors | Unfortunately the figure provided in the answer by @SmallChess still has 3 support vectors. The answer is correct - the minimum number of support vectors is 2. The best way to understand this issue is the convex hull model of SVM (for instance http://www.robots.ox.ac.uk/~cvrg/bennett00duality.pdf). The minimum case wil... | Theoretical: Minimum Number of Support Vectors | Unfortunately the figure provided in the answer by @SmallChess still has 3 support vectors. The answer is correct - the minimum number of support vectors is 2. The best way to understand this issue is | Theoretical: Minimum Number of Support Vectors
Unfortunately the figure provided in the answer by @SmallChess still has 3 support vectors. The answer is correct - the minimum number of support vectors is 2. The best way to understand this issue is the convex hull model of SVM (for instance http://www.robots.ox.ac.uk/~c... | Theoretical: Minimum Number of Support Vectors
Unfortunately the figure provided in the answer by @SmallChess still has 3 support vectors. The answer is correct - the minimum number of support vectors is 2. The best way to understand this issue is |
39,653 | Theoretical: Minimum Number of Support Vectors | If the training data involves only one class (which is a trivial problem) then there is no support vector at all.
Reference: Learning From Data - A Short Course, by Yaser S. Abu-Mostafa, Malik Magdon-Ismail and Hsuan-Tien Lin. e-Chapter 8, Exercise 8.12 and Page 31. | Theoretical: Minimum Number of Support Vectors | If the training data involves only one class (which is a trivial problem) then there is no support vector at all.
Reference: Learning From Data - A Short Course, by Yaser S. Abu-Mostafa, Malik Magdon- | Theoretical: Minimum Number of Support Vectors
If the training data involves only one class (which is a trivial problem) then there is no support vector at all.
Reference: Learning From Data - A Short Course, by Yaser S. Abu-Mostafa, Malik Magdon-Ismail and Hsuan-Tien Lin. e-Chapter 8, Exercise 8.12 and Page 31. | Theoretical: Minimum Number of Support Vectors
If the training data involves only one class (which is a trivial problem) then there is no support vector at all.
Reference: Learning From Data - A Short Course, by Yaser S. Abu-Mostafa, Malik Magdon- |
39,654 | Does variance only work on normally distributed data (as a measure of dispersion)? | Your question is a little vague, but no, variance isn't used because of its association with the normal distribution. Most distributions have at least a mean and a variance. Some do not have a variance. Some can either have or not have a variance. Some have no mean and so do not have a variance.
Just for mental cla... | Does variance only work on normally distributed data (as a measure of dispersion)? | Your question is a little vague, but no, variance isn't used because of its association with the normal distribution. Most distributions have at least a mean and a variance. Some do not have a varia | Does variance only work on normally distributed data (as a measure of dispersion)?
Your question is a little vague, but no, variance isn't used because of its association with the normal distribution. Most distributions have at least a mean and a variance. Some do not have a variance. Some can either have or not hav... | Does variance only work on normally distributed data (as a measure of dispersion)?
Your question is a little vague, but no, variance isn't used because of its association with the normal distribution. Most distributions have at least a mean and a variance. Some do not have a varia |
39,655 | Does variance only work on normally distributed data (as a measure of dispersion)? | First we need to be clear about the distinction between a measure of the variability of a distribution (such as its standard deviation or its mean deviation or its range) and the best way to estimate that measure from a sample. For example, if your distribution is uniform, the best sample estimate of the population mea... | Does variance only work on normally distributed data (as a measure of dispersion)? | First we need to be clear about the distinction between a measure of the variability of a distribution (such as its standard deviation or its mean deviation or its range) and the best way to estimate | Does variance only work on normally distributed data (as a measure of dispersion)?
First we need to be clear about the distinction between a measure of the variability of a distribution (such as its standard deviation or its mean deviation or its range) and the best way to estimate that measure from a sample. For examp... | Does variance only work on normally distributed data (as a measure of dispersion)?
First we need to be clear about the distinction between a measure of the variability of a distribution (such as its standard deviation or its mean deviation or its range) and the best way to estimate |
39,656 | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$? | This is a nice problem for testing the development code in the next version of mathStatica.
Note that $\text{Cov}(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment), which is why I am requesting the {1,1} CentralMoment of {X, X Y} ... when the variables are {X,Y}:
where $\mu_{2,1} ... | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$? | This is a nice problem for testing the development code in the next version of mathStatica.
Note that $\text{Cov}(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment) | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$?
This is a nice problem for testing the development code in the next version of mathStatica.
Note that $\text{Cov}(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment), which is why I am requesting the {1,1} CentralMoment of {X, X... | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$?
This is a nice problem for testing the development code in the next version of mathStatica.
Note that $\text{Cov}(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment) |
39,657 | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$? | My first reaction is that you won't be able to find this value without knowing the dependence structure between $X$ and $Y$. This is further verified by using law of total covariance as shown below,
\begin{align*}
Cov(X,XY)& = E[Cov(X,XY\mid X)] + Cov(E[X\mid X],E[XY \mid X])\\
& = E[X^2Cov(1,Y \mid X)] + Cov(X, X\, E... | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$? | My first reaction is that you won't be able to find this value without knowing the dependence structure between $X$ and $Y$. This is further verified by using law of total covariance as shown below,
\ | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$?
My first reaction is that you won't be able to find this value without knowing the dependence structure between $X$ and $Y$. This is further verified by using law of total covariance as shown below,
\begin{align*}
Cov(X,XY)& = E[Cov(X,XY\mid X)] + Cov(E[X\mid X]... | Is there a way to calculate easily $\mathrm{Cov}(X,XY)$?
My first reaction is that you won't be able to find this value without knowing the dependence structure between $X$ and $Y$. This is further verified by using law of total covariance as shown below,
\ |
39,658 | Association rules - support, confidence and lift | It depends on your task. But usually you want all three to be high.
high support: should apply to a large amount of cases
high confidence: should be correct often
high lift: indicates it is not just a coincidence
Consider e.g. "rain" and "day". Assuming we live in a very unfortunate place at the Equator, where it is ... | Association rules - support, confidence and lift | It depends on your task. But usually you want all three to be high.
high support: should apply to a large amount of cases
high confidence: should be correct often
high lift: indicates it is not just | Association rules - support, confidence and lift
It depends on your task. But usually you want all three to be high.
high support: should apply to a large amount of cases
high confidence: should be correct often
high lift: indicates it is not just a coincidence
Consider e.g. "rain" and "day". Assuming we live in a ve... | Association rules - support, confidence and lift
It depends on your task. But usually you want all three to be high.
high support: should apply to a large amount of cases
high confidence: should be correct often
high lift: indicates it is not just |
39,659 | Interpretation of the entropy with a coding length? | Code:
AAA as 0
AAB,AAC,ABA,ACA,BAA,CAA as 1000...1101
12 triplets with 1 letter "A" as 1110000...1111011
8 triplets without "A" as 11111000...11111111
Each triplet takes on average 0.512*1+0.384*4+0.096*7+0.008*8=2.784 bits, or 0.928 bits per character. With coding 4-character, 5-character etc. groups, you can furthe... | Interpretation of the entropy with a coding length? | Code:
AAA as 0
AAB,AAC,ABA,ACA,BAA,CAA as 1000...1101
12 triplets with 1 letter "A" as 1110000...1111011
8 triplets without "A" as 11111000...11111111
Each triplet takes on average 0.512*1+0.384*4+0. | Interpretation of the entropy with a coding length?
Code:
AAA as 0
AAB,AAC,ABA,ACA,BAA,CAA as 1000...1101
12 triplets with 1 letter "A" as 1110000...1111011
8 triplets without "A" as 11111000...11111111
Each triplet takes on average 0.512*1+0.384*4+0.096*7+0.008*8=2.784 bits, or 0.928 bits per character. With coding ... | Interpretation of the entropy with a coding length?
Code:
AAA as 0
AAB,AAC,ABA,ACA,BAA,CAA as 1000...1101
12 triplets with 1 letter "A" as 1110000...1111011
8 triplets without "A" as 11111000...11111111
Each triplet takes on average 0.512*1+0.384*4+0. |
39,660 | GLMs must be 'linear in the parameters' | Your example model can be reëxpressed to be linear in the parameters $\alpha=\beta_1\beta_2$ & $\zeta=\exp\beta_3$:
$$g(\operatorname{E} Y) = \beta_0 + \alpha x_1 + \zeta x_2^2$$
(Clearly $\beta_1$ & $\beta_2$ aren't separately estimable; a non-linear model wouldn't help there. And note that $\hat\zeta$ must be constra... | GLMs must be 'linear in the parameters' | Your example model can be reëxpressed to be linear in the parameters $\alpha=\beta_1\beta_2$ & $\zeta=\exp\beta_3$:
$$g(\operatorname{E} Y) = \beta_0 + \alpha x_1 + \zeta x_2^2$$
(Clearly $\beta_1$ & | GLMs must be 'linear in the parameters'
Your example model can be reëxpressed to be linear in the parameters $\alpha=\beta_1\beta_2$ & $\zeta=\exp\beta_3$:
$$g(\operatorname{E} Y) = \beta_0 + \alpha x_1 + \zeta x_2^2$$
(Clearly $\beta_1$ & $\beta_2$ aren't separately estimable; a non-linear model wouldn't help there. A... | GLMs must be 'linear in the parameters'
Your example model can be reëxpressed to be linear in the parameters $\alpha=\beta_1\beta_2$ & $\zeta=\exp\beta_3$:
$$g(\operatorname{E} Y) = \beta_0 + \alpha x_1 + \zeta x_2^2$$
(Clearly $\beta_1$ & |
39,661 | GLMs must be 'linear in the parameters' | Linear in the parameters means that you can write your prediction as
$$\beta_0+\sum_{j=1}^px_{ij}\beta_j $$
For some definition of $x_{ij} $. But these x's need not be linear functions of your data. For example, ploynomial fitting of a time series has $x_{ij}=t_i^j $ where $t_i $ is the time associated with data poin... | GLMs must be 'linear in the parameters' | Linear in the parameters means that you can write your prediction as
$$\beta_0+\sum_{j=1}^px_{ij}\beta_j $$
For some definition of $x_{ij} $. But these x's need not be linear functions of your data. | GLMs must be 'linear in the parameters'
Linear in the parameters means that you can write your prediction as
$$\beta_0+\sum_{j=1}^px_{ij}\beta_j $$
For some definition of $x_{ij} $. But these x's need not be linear functions of your data. For example, ploynomial fitting of a time series has $x_{ij}=t_i^j $ where $t_i... | GLMs must be 'linear in the parameters'
Linear in the parameters means that you can write your prediction as
$$\beta_0+\sum_{j=1}^px_{ij}\beta_j $$
For some definition of $x_{ij} $. But these x's need not be linear functions of your data. |
39,662 | GLMs must be 'linear in the parameters' | I think it's better for you to understand the three components of the GLM. Esp, you need understand how link function is defined.
You can refer to the page 7 in the slides below. 'linear in the parameters' is true after being transformed by the link function.
enter link description here | GLMs must be 'linear in the parameters' | I think it's better for you to understand the three components of the GLM. Esp, you need understand how link function is defined.
You can refer to the page 7 in the slides below. 'linear in the param | GLMs must be 'linear in the parameters'
I think it's better for you to understand the three components of the GLM. Esp, you need understand how link function is defined.
You can refer to the page 7 in the slides below. 'linear in the parameters' is true after being transformed by the link function.
enter link descript... | GLMs must be 'linear in the parameters'
I think it's better for you to understand the three components of the GLM. Esp, you need understand how link function is defined.
You can refer to the page 7 in the slides below. 'linear in the param |
39,663 | Sampling from Gaussian Process Posterior | You want to sample posterior using the data and model given.
In this case you can:
sample from posterior normal distribution with given mean and covariance matrix - use model.predict with full_covariance=True in case;
use built-in function model.posterior_samples_f that does the job for you.
A sample code is below:
i... | Sampling from Gaussian Process Posterior | You want to sample posterior using the data and model given.
In this case you can:
sample from posterior normal distribution with given mean and covariance matrix - use model.predict with full_covari | Sampling from Gaussian Process Posterior
You want to sample posterior using the data and model given.
In this case you can:
sample from posterior normal distribution with given mean and covariance matrix - use model.predict with full_covariance=True in case;
use built-in function model.posterior_samples_f that does th... | Sampling from Gaussian Process Posterior
You want to sample posterior using the data and model given.
In this case you can:
sample from posterior normal distribution with given mean and covariance matrix - use model.predict with full_covari |
39,664 | Do bookmakers have an "underhead" for non-favourites? [closed] | Usually the opposite in fact. It tends to be the case that when gambling people are attracted to the small probability of a large win, rather than a higher probability of a moderate win. This effect can be neatly modelled by Prospect Theory.
This results in a Favourite-longshot bias that has been studied many times (... | Do bookmakers have an "underhead" for non-favourites? [closed] | Usually the opposite in fact. It tends to be the case that when gambling people are attracted to the small probability of a large win, rather than a higher probability of a moderate win. This effect | Do bookmakers have an "underhead" for non-favourites? [closed]
Usually the opposite in fact. It tends to be the case that when gambling people are attracted to the small probability of a large win, rather than a higher probability of a moderate win. This effect can be neatly modelled by Prospect Theory.
This results ... | Do bookmakers have an "underhead" for non-favourites? [closed]
Usually the opposite in fact. It tends to be the case that when gambling people are attracted to the small probability of a large win, rather than a higher probability of a moderate win. This effect |
39,665 | "Inverse" Q-Q plot? | You've re-discovered the P-P plot. For an introduction, see here.
I'll add a slightly droll comment from one text, to the effect that if you want to be, or to appear, optimistic about fit, you use a P-P plot, whereas if you want to be (appear) pessimistic, you use a Q-Q plot.
Your example is a case in point. The P-P ... | "Inverse" Q-Q plot? | You've re-discovered the P-P plot. For an introduction, see here.
I'll add a slightly droll comment from one text, to the effect that if you want to be, or to appear, optimistic about fit, you use a | "Inverse" Q-Q plot?
You've re-discovered the P-P plot. For an introduction, see here.
I'll add a slightly droll comment from one text, to the effect that if you want to be, or to appear, optimistic about fit, you use a P-P plot, whereas if you want to be (appear) pessimistic, you use a Q-Q plot.
Your example is a cas... | "Inverse" Q-Q plot?
You've re-discovered the P-P plot. For an introduction, see here.
I'll add a slightly droll comment from one text, to the effect that if you want to be, or to appear, optimistic about fit, you use a |
39,666 | Length of Time-Series for Forecasting Modeling | No matter what is model, generally the more data you have the better. If you want to make a forecast you want your sample to be representative enough for the population as it changes. In most cases you have only partial knowledge about the changes in the past and no knowledge about the future. Gathering more data helps... | Length of Time-Series for Forecasting Modeling | No matter what is model, generally the more data you have the better. If you want to make a forecast you want your sample to be representative enough for the population as it changes. In most cases yo | Length of Time-Series for Forecasting Modeling
No matter what is model, generally the more data you have the better. If you want to make a forecast you want your sample to be representative enough for the population as it changes. In most cases you have only partial knowledge about the changes in the past and no knowle... | Length of Time-Series for Forecasting Modeling
No matter what is model, generally the more data you have the better. If you want to make a forecast you want your sample to be representative enough for the population as it changes. In most cases yo |
39,667 | Length of Time-Series for Forecasting Modeling | If you are trying to model weekly data in order to make a 26 week out forecast , I would recommend that you consider using more than 2 years of data (104 points) . A four year history base would give you the opportunity to extract seasonal structure be it auto-regressive or fixed effects and any level shifts/time trend... | Length of Time-Series for Forecasting Modeling | If you are trying to model weekly data in order to make a 26 week out forecast , I would recommend that you consider using more than 2 years of data (104 points) . A four year history base would give | Length of Time-Series for Forecasting Modeling
If you are trying to model weekly data in order to make a 26 week out forecast , I would recommend that you consider using more than 2 years of data (104 points) . A four year history base would give you the opportunity to extract seasonal structure be it auto-regressive o... | Length of Time-Series for Forecasting Modeling
If you are trying to model weekly data in order to make a 26 week out forecast , I would recommend that you consider using more than 2 years of data (104 points) . A four year history base would give |
39,668 | How to use PCA in regression? | Say your predictor matrix is $X$ and your response vector is $y$. PCA is concerned only with the (co)variance within the predictor matrix $X$ itself, while a regression model is (also) concerned with the covariance between $X$ and the response $y$. If there is no relationship between these concepts, dimension reducti... | How to use PCA in regression? | Say your predictor matrix is $X$ and your response vector is $y$. PCA is concerned only with the (co)variance within the predictor matrix $X$ itself, while a regression model is (also) concerned with | How to use PCA in regression?
Say your predictor matrix is $X$ and your response vector is $y$. PCA is concerned only with the (co)variance within the predictor matrix $X$ itself, while a regression model is (also) concerned with the covariance between $X$ and the response $y$. If there is no relationship between the... | How to use PCA in regression?
Say your predictor matrix is $X$ and your response vector is $y$. PCA is concerned only with the (co)variance within the predictor matrix $X$ itself, while a regression model is (also) concerned with |
39,669 | Probability of failure | Let $X$ denote the resistance and $Y$ the load. Then,
\begin{align}
P\{Y > X\} &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X,Y}(x,y)
\,\mathrm dx \,\mathrm dy\\
&= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X}(x)f_{Y}(y)
\,\mathrm dx \,\mathrm dy & \scriptstyle{\text{because}~X~\text{and}
~Y~\text{are independe... | Probability of failure | Let $X$ denote the resistance and $Y$ the load. Then,
\begin{align}
P\{Y > X\} &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X,Y}(x,y)
\,\mathrm dx \,\mathrm dy\\
&= \int_{y=-\infty}^\infty \int_{x | Probability of failure
Let $X$ denote the resistance and $Y$ the load. Then,
\begin{align}
P\{Y > X\} &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X,Y}(x,y)
\,\mathrm dx \,\mathrm dy\\
&= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X}(x)f_{Y}(y)
\,\mathrm dx \,\mathrm dy & \scriptstyle{\text{because}~X~\text{and}... | Probability of failure
Let $X$ denote the resistance and $Y$ the load. Then,
\begin{align}
P\{Y > X\} &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X,Y}(x,y)
\,\mathrm dx \,\mathrm dy\\
&= \int_{y=-\infty}^\infty \int_{x |
39,670 | Probability of failure | Rephrased, the probability of failure is equivalent to the probability that resistance - load is less than zero. What you're looking for is the distribution of the difference of random variables.
Since these are independent, you can use convolution to solve for their difference. But it's applied to the densities, not a... | Probability of failure | Rephrased, the probability of failure is equivalent to the probability that resistance - load is less than zero. What you're looking for is the distribution of the difference of random variables.
Sinc | Probability of failure
Rephrased, the probability of failure is equivalent to the probability that resistance - load is less than zero. What you're looking for is the distribution of the difference of random variables.
Since these are independent, you can use convolution to solve for their difference. But it's applied ... | Probability of failure
Rephrased, the probability of failure is equivalent to the probability that resistance - load is less than zero. What you're looking for is the distribution of the difference of random variables.
Sinc |
39,671 | Why is asymptotic normality important for an estimator? | Why is asymptotic normality important for an estimator?
I wouldn't say it's important, really, but when it happens, it can be convenient, and the plain fact is, it happens a lot -- for many popular estimators in commonly used models, it is the case that the distribution of an appropriately standardized estimator will ... | Why is asymptotic normality important for an estimator? | Why is asymptotic normality important for an estimator?
I wouldn't say it's important, really, but when it happens, it can be convenient, and the plain fact is, it happens a lot -- for many popular e | Why is asymptotic normality important for an estimator?
Why is asymptotic normality important for an estimator?
I wouldn't say it's important, really, but when it happens, it can be convenient, and the plain fact is, it happens a lot -- for many popular estimators in commonly used models, it is the case that the distr... | Why is asymptotic normality important for an estimator?
Why is asymptotic normality important for an estimator?
I wouldn't say it's important, really, but when it happens, it can be convenient, and the plain fact is, it happens a lot -- for many popular e |
39,672 | How to train a Gaussian mixture hidden Markov model? | In the reference at the bottom $^*$, I see the training involves the following:
Initialize the HMM & GMM parameters (randomly or using prior assumptions).
Then repeat the following until convergence criteria are satisfied:
Do a forward pass and backwards pass to find probabilities associated with the training sequence... | How to train a Gaussian mixture hidden Markov model? | In the reference at the bottom $^*$, I see the training involves the following:
Initialize the HMM & GMM parameters (randomly or using prior assumptions).
Then repeat the following until convergence | How to train a Gaussian mixture hidden Markov model?
In the reference at the bottom $^*$, I see the training involves the following:
Initialize the HMM & GMM parameters (randomly or using prior assumptions).
Then repeat the following until convergence criteria are satisfied:
Do a forward pass and backwards pass to fin... | How to train a Gaussian mixture hidden Markov model?
In the reference at the bottom $^*$, I see the training involves the following:
Initialize the HMM & GMM parameters (randomly or using prior assumptions).
Then repeat the following until convergence |
39,673 | How to train a Gaussian mixture hidden Markov model? | This paper[1] is absolute classic and has the whole HMM machinery for gaussian mixture laid out for you. I think it's fair to say Rabiner made the first important step in speech recognition with GMM in 1980s.
[1] Rabiner, L. R. (1989). A tutorial on hidden Markov models and selected applications in speech recognition. ... | How to train a Gaussian mixture hidden Markov model? | This paper[1] is absolute classic and has the whole HMM machinery for gaussian mixture laid out for you. I think it's fair to say Rabiner made the first important step in speech recognition with GMM i | How to train a Gaussian mixture hidden Markov model?
This paper[1] is absolute classic and has the whole HMM machinery for gaussian mixture laid out for you. I think it's fair to say Rabiner made the first important step in speech recognition with GMM in 1980s.
[1] Rabiner, L. R. (1989). A tutorial on hidden Markov mod... | How to train a Gaussian mixture hidden Markov model?
This paper[1] is absolute classic and has the whole HMM machinery for gaussian mixture laid out for you. I think it's fair to say Rabiner made the first important step in speech recognition with GMM i |
39,674 | How to train a Gaussian mixture hidden Markov model? | pomegranate is another python library that provides GMM and HMM with even better documents than hmmlearn. Currently I prepare transfer from hmmlearn to it.
http://pomegranate.readthedocs.io/en/latest/GeneralMixtureModel.html | How to train a Gaussian mixture hidden Markov model? | pomegranate is another python library that provides GMM and HMM with even better documents than hmmlearn. Currently I prepare transfer from hmmlearn to it.
http://pomegranate.readthedocs.io/en/latest/ | How to train a Gaussian mixture hidden Markov model?
pomegranate is another python library that provides GMM and HMM with even better documents than hmmlearn. Currently I prepare transfer from hmmlearn to it.
http://pomegranate.readthedocs.io/en/latest/GeneralMixtureModel.html | How to train a Gaussian mixture hidden Markov model?
pomegranate is another python library that provides GMM and HMM with even better documents than hmmlearn. Currently I prepare transfer from hmmlearn to it.
http://pomegranate.readthedocs.io/en/latest/ |
39,675 | How to train a Gaussian mixture hidden Markov model? | Assuming your HMM uses Gaussian Mixture, for parameters estimation, you perform forward and backward pass and update the parameters. The difference is that you need to include normal pdf mixture as the probability of observation given a state. So, for transition probability estimation, you do it just like a discrete ob... | How to train a Gaussian mixture hidden Markov model? | Assuming your HMM uses Gaussian Mixture, for parameters estimation, you perform forward and backward pass and update the parameters. The difference is that you need to include normal pdf mixture as th | How to train a Gaussian mixture hidden Markov model?
Assuming your HMM uses Gaussian Mixture, for parameters estimation, you perform forward and backward pass and update the parameters. The difference is that you need to include normal pdf mixture as the probability of observation given a state. So, for transition prob... | How to train a Gaussian mixture hidden Markov model?
Assuming your HMM uses Gaussian Mixture, for parameters estimation, you perform forward and backward pass and update the parameters. The difference is that you need to include normal pdf mixture as th |
39,676 | Test for randomness - randtests - fails | At least for the "Mann-Kendall Rank Test", the problem seems to be in the testing package you're using, and not in the data.
Specifically, the Mann-Kendall test is supposed to detect monotone trends in the data by calculating the Kendall rank correlation coefficient between the data points and their position in the inp... | Test for randomness - randtests - fails | At least for the "Mann-Kendall Rank Test", the problem seems to be in the testing package you're using, and not in the data.
Specifically, the Mann-Kendall test is supposed to detect monotone trends i | Test for randomness - randtests - fails
At least for the "Mann-Kendall Rank Test", the problem seems to be in the testing package you're using, and not in the data.
Specifically, the Mann-Kendall test is supposed to detect monotone trends in the data by calculating the Kendall rank correlation coefficient between the d... | Test for randomness - randtests - fails
At least for the "Mann-Kendall Rank Test", the problem seems to be in the testing package you're using, and not in the data.
Specifically, the Mann-Kendall test is supposed to detect monotone trends i |
39,677 | Test for randomness - randtests - fails | One can not substantiate (much less assert) the correctness of a Cryptographically Secure Pseudo Random Number Generator using tests of its output (because any useful test fits the definition of a break of the CSPRNG); that seems to be what's attempted in the question (since it is about "check the correctness of the PR... | Test for randomness - randtests - fails | One can not substantiate (much less assert) the correctness of a Cryptographically Secure Pseudo Random Number Generator using tests of its output (because any useful test fits the definition of a bre | Test for randomness - randtests - fails
One can not substantiate (much less assert) the correctness of a Cryptographically Secure Pseudo Random Number Generator using tests of its output (because any useful test fits the definition of a break of the CSPRNG); that seems to be what's attempted in the question (since it i... | Test for randomness - randtests - fails
One can not substantiate (much less assert) the correctness of a Cryptographically Secure Pseudo Random Number Generator using tests of its output (because any useful test fits the definition of a bre |
39,678 | Number of samples needed in Monte Carlo simulation: how good is this approximation? | The approximation could be poor when $p$ is close to zero or one, but when $p = 1/2$ it holds exactly.
The idea here is that we want to estimate the probability of an event by using a sample proportion across many Monte Carlo trials, and we want to know how accurate of an estimate that proportion is of the true probabi... | Number of samples needed in Monte Carlo simulation: how good is this approximation? | The approximation could be poor when $p$ is close to zero or one, but when $p = 1/2$ it holds exactly.
The idea here is that we want to estimate the probability of an event by using a sample proportio | Number of samples needed in Monte Carlo simulation: how good is this approximation?
The approximation could be poor when $p$ is close to zero or one, but when $p = 1/2$ it holds exactly.
The idea here is that we want to estimate the probability of an event by using a sample proportion across many Monte Carlo trials, an... | Number of samples needed in Monte Carlo simulation: how good is this approximation?
The approximation could be poor when $p$ is close to zero or one, but when $p = 1/2$ it holds exactly.
The idea here is that we want to estimate the probability of an event by using a sample proportio |
39,679 | Number of samples needed in Monte Carlo simulation: how good is this approximation? | This approximation is called Wald confidence interval. It's based on normal approximation of Binomial. How good is this approximation? There are two answers: when sample size is at least 30 and "it depends".
The "30" answer is very popular and has been propagated from book to book until it became an axiom, pretty much.... | Number of samples needed in Monte Carlo simulation: how good is this approximation? | This approximation is called Wald confidence interval. It's based on normal approximation of Binomial. How good is this approximation? There are two answers: when sample size is at least 30 and "it de | Number of samples needed in Monte Carlo simulation: how good is this approximation?
This approximation is called Wald confidence interval. It's based on normal approximation of Binomial. How good is this approximation? There are two answers: when sample size is at least 30 and "it depends".
The "30" answer is very popu... | Number of samples needed in Monte Carlo simulation: how good is this approximation?
This approximation is called Wald confidence interval. It's based on normal approximation of Binomial. How good is this approximation? There are two answers: when sample size is at least 30 and "it de |
39,680 | Credible set for beta distribution | A level $\alpha$ "Highest Posterior Density" (HPD) interval for a (posterior) distribution $F$ with (continuous) density $f$ and mode $\mu$ is an interval $I=[x,y]$ containing $\mu$ for which
$1-\alpha$ of the probability is in the interval: $F(I) = F(y) - F(x) = 1-\alpha$.
The densities are the same at either end: $f... | Credible set for beta distribution | A level $\alpha$ "Highest Posterior Density" (HPD) interval for a (posterior) distribution $F$ with (continuous) density $f$ and mode $\mu$ is an interval $I=[x,y]$ containing $\mu$ for which
$1-\alp | Credible set for beta distribution
A level $\alpha$ "Highest Posterior Density" (HPD) interval for a (posterior) distribution $F$ with (continuous) density $f$ and mode $\mu$ is an interval $I=[x,y]$ containing $\mu$ for which
$1-\alpha$ of the probability is in the interval: $F(I) = F(y) - F(x) = 1-\alpha$.
The densi... | Credible set for beta distribution
A level $\alpha$ "Highest Posterior Density" (HPD) interval for a (posterior) distribution $F$ with (continuous) density $f$ and mode $\mu$ is an interval $I=[x,y]$ containing $\mu$ for which
$1-\alp |
39,681 | GLMnet - "Unstandardizing" Linear Regression Coefficients | This is mostly a case of carefully working out the math. I'll handle the two predictor + intercept case, it should be clear how to generalize it.
The standardized elastic net model results in the following relationship:
$$\frac{y - \mu(y)}{\sigma(y)} = \beta_1 \frac{x_1 - \mu(x_1)}{\sigma(x_1)} + \beta_2 \frac{x_1 - \... | GLMnet - "Unstandardizing" Linear Regression Coefficients | This is mostly a case of carefully working out the math. I'll handle the two predictor + intercept case, it should be clear how to generalize it.
The standardized elastic net model results in the fol | GLMnet - "Unstandardizing" Linear Regression Coefficients
This is mostly a case of carefully working out the math. I'll handle the two predictor + intercept case, it should be clear how to generalize it.
The standardized elastic net model results in the following relationship:
$$\frac{y - \mu(y)}{\sigma(y)} = \beta_1 ... | GLMnet - "Unstandardizing" Linear Regression Coefficients
This is mostly a case of carefully working out the math. I'll handle the two predictor + intercept case, it should be clear how to generalize it.
The standardized elastic net model results in the fol |
39,682 | GLMnet - "Unstandardizing" Linear Regression Coefficients | The process GLMnet is following when it calculates coefficients for linear regression seem to be as follows:
Standardise each $x$ by subtracting the mean and dividing by the standard deviation (in calculating the standard deviation, divide by $N$, not $N-1$).
Standardise $y$ by dividing by its standard deviation (agai... | GLMnet - "Unstandardizing" Linear Regression Coefficients | The process GLMnet is following when it calculates coefficients for linear regression seem to be as follows:
Standardise each $x$ by subtracting the mean and dividing by the standard deviation (in cal | GLMnet - "Unstandardizing" Linear Regression Coefficients
The process GLMnet is following when it calculates coefficients for linear regression seem to be as follows:
Standardise each $x$ by subtracting the mean and dividing by the standard deviation (in calculating the standard deviation, divide by $N$, not $N-1$).
S... | GLMnet - "Unstandardizing" Linear Regression Coefficients
The process GLMnet is following when it calculates coefficients for linear regression seem to be as follows:
Standardise each $x$ by subtracting the mean and dividing by the standard deviation (in cal |
39,683 | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | The approach in the question seems to be correct as long as the variables of concern are continuous or binary. Categorical variables with three or more levels cannot be multiplied as stated.
The standardized interaction term should be the standardized version of the product of the two original variables, not the produc... | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | The approach in the question seems to be correct as long as the variables of concern are continuous or binary. Categorical variables with three or more levels cannot be multiplied as stated.
The stand | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
The approach in the question seems to be correct as long as the variables of concern are continuous or binary. Categorical variables with three or more levels cannot be multiplied as stated.
The standardized interaction term should be the stand... | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
The approach in the question seems to be correct as long as the variables of concern are continuous or binary. Categorical variables with three or more levels cannot be multiplied as stated.
The stand |
39,684 | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | Standardization is not a requirement, but is an option. Mean-centering (a part of standardization) makes the lower order terms more interpretable. Penguin_Knight showed that standarizing after forming the interaction term rather than before gives you the same results as the unstandardized model. Note that this is a con... | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | Standardization is not a requirement, but is an option. Mean-centering (a part of standardization) makes the lower order terms more interpretable. Penguin_Knight showed that standarizing after forming | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
Standardization is not a requirement, but is an option. Mean-centering (a part of standardization) makes the lower order terms more interpretable. Penguin_Knight showed that standarizing after forming the interaction term rather than before giv... | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
Standardization is not a requirement, but is an option. Mean-centering (a part of standardization) makes the lower order terms more interpretable. Penguin_Knight showed that standarizing after forming |
39,685 | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | You don't need to standardize anything unless the scales are vastly different. Even in this case you don't necessarily need to standardize, a simple unit of measure (scale) will work fine. Once you changes the scale, then the interactions would be on scaled variables too. | Adding Interaction Terms to Multiple Linear Regression, how to standardize? | You don't need to standardize anything unless the scales are vastly different. Even in this case you don't necessarily need to standardize, a simple unit of measure (scale) will work fine. Once you ch | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
You don't need to standardize anything unless the scales are vastly different. Even in this case you don't necessarily need to standardize, a simple unit of measure (scale) will work fine. Once you changes the scale, then the interactions would... | Adding Interaction Terms to Multiple Linear Regression, how to standardize?
You don't need to standardize anything unless the scales are vastly different. Even in this case you don't necessarily need to standardize, a simple unit of measure (scale) will work fine. Once you ch |
39,686 | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights) | (I'm the creator of lavaan.survey)
As Stas already indicated, the combination (multiple imputation * complex sampling) can be tricky business. The main papers are Kott (1995) and Kim, Brick & Fuller (2006).
Here are some considerations:
As mentioned by Stas, all the usual best practices of MI apply. Considering the be... | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights) | (I'm the creator of lavaan.survey)
As Stas already indicated, the combination (multiple imputation * complex sampling) can be tricky business. The main papers are Kott (1995) and Kim, Brick & Fuller ( | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights)
(I'm the creator of lavaan.survey)
As Stas already indicated, the combination (multiple imputation * complex sampling) can be tricky business. The main papers are Kott (1995) and Kim, Brick & Fuller (2006).
Here are som... | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights)
(I'm the creator of lavaan.survey)
As Stas already indicated, the combination (multiple imputation * complex sampling) can be tricky business. The main papers are Kott (1995) and Kim, Brick & Fuller ( |
39,687 | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights) | If I were dealing with this in my project, and I am grateful that I don't have to, this is what I would have done.
Take a survey bootstrap sample that respects my survey design -- see Rao and Wu 1988.
For each bootstrap replicate, impute the missing data once, see Shao and Sitter 1996.
Within each imputation, follow t... | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights) | If I were dealing with this in my project, and I am grateful that I don't have to, this is what I would have done.
Take a survey bootstrap sample that respects my survey design -- see Rao and Wu 1988 | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights)
If I were dealing with this in my project, and I am grateful that I don't have to, this is what I would have done.
Take a survey bootstrap sample that respects my survey design -- see Rao and Wu 1988.
For each bootstra... | Questions on multiple imputation with MICE for a multigroup-SEM-analysis? (including survey weights)
If I were dealing with this in my project, and I am grateful that I don't have to, this is what I would have done.
Take a survey bootstrap sample that respects my survey design -- see Rao and Wu 1988 |
39,688 | Maximum of Independent Gamma random variables? [closed] | The gamma distribution is in the Gumbel domain of attraction. You can
refer to the book by L. de Haan and A Ferreira, Extreme Values Theory,
an Introduction.
See therein theorem 1.1.8 and exercise 1.7 for its application to the
gamma distribution. Another very useful book even provides explicit
values for the two sequ... | Maximum of Independent Gamma random variables? [closed] | The gamma distribution is in the Gumbel domain of attraction. You can
refer to the book by L. de Haan and A Ferreira, Extreme Values Theory,
an Introduction.
See therein theorem 1.1.8 and exercise 1. | Maximum of Independent Gamma random variables? [closed]
The gamma distribution is in the Gumbel domain of attraction. You can
refer to the book by L. de Haan and A Ferreira, Extreme Values Theory,
an Introduction.
See therein theorem 1.1.8 and exercise 1.7 for its application to the
gamma distribution. Another very us... | Maximum of Independent Gamma random variables? [closed]
The gamma distribution is in the Gumbel domain of attraction. You can
refer to the book by L. de Haan and A Ferreira, Extreme Values Theory,
an Introduction.
See therein theorem 1.1.8 and exercise 1. |
39,689 | Maximum of Independent Gamma random variables? [closed] | Taken from
David, H. A., & Nagaraja, H. N. (2003). Order statistics 3d ed., ch 10. p 296
The "two references that give the most complete and rigorous discussion of the problem" are | Maximum of Independent Gamma random variables? [closed] | Taken from
David, H. A., & Nagaraja, H. N. (2003). Order statistics 3d ed., ch 10. p 296
The "two references that give the most complete and rigorous discussion of the problem" are | Maximum of Independent Gamma random variables? [closed]
Taken from
David, H. A., & Nagaraja, H. N. (2003). Order statistics 3d ed., ch 10. p 296
The "two references that give the most complete and rigorous discussion of the problem" are | Maximum of Independent Gamma random variables? [closed]
Taken from
David, H. A., & Nagaraja, H. N. (2003). Order statistics 3d ed., ch 10. p 296
The "two references that give the most complete and rigorous discussion of the problem" are |
39,690 | Why do one-versus-all multi class SVMs need to be calibrated? | Setup
Recall that an SVM can be viewed as a weight vector $w$ and an intercept $b$, and that the output function for a test input $x$ is is $\langle w, x \rangle + b$. To get a binary prediction, we take $f(x) = \mathrm{sign}(\langle w, x \rangle + b)$.
(I'm going to use some primal notations here, but use $\langle \c... | Why do one-versus-all multi class SVMs need to be calibrated? | Setup
Recall that an SVM can be viewed as a weight vector $w$ and an intercept $b$, and that the output function for a test input $x$ is is $\langle w, x \rangle + b$. To get a binary prediction, we t | Why do one-versus-all multi class SVMs need to be calibrated?
Setup
Recall that an SVM can be viewed as a weight vector $w$ and an intercept $b$, and that the output function for a test input $x$ is is $\langle w, x \rangle + b$. To get a binary prediction, we take $f(x) = \mathrm{sign}(\langle w, x \rangle + b)$.
(I'... | Why do one-versus-all multi class SVMs need to be calibrated?
Setup
Recall that an SVM can be viewed as a weight vector $w$ and an intercept $b$, and that the output function for a test input $x$ is is $\langle w, x \rangle + b$. To get a binary prediction, we t |
39,691 | Why do one-versus-all multi class SVMs need to be calibrated? | This is because each individual one-vs-all classifier corresponds to different support vectors and their respective alphas in the decomposition. The score output for a test data point is in no way bounded and should be normalized (e.g. using Platt's scheme from scores to probabilities) in order to be comparable. | Why do one-versus-all multi class SVMs need to be calibrated? | This is because each individual one-vs-all classifier corresponds to different support vectors and their respective alphas in the decomposition. The score output for a test data point is in no way bou | Why do one-versus-all multi class SVMs need to be calibrated?
This is because each individual one-vs-all classifier corresponds to different support vectors and their respective alphas in the decomposition. The score output for a test data point is in no way bounded and should be normalized (e.g. using Platt's scheme f... | Why do one-versus-all multi class SVMs need to be calibrated?
This is because each individual one-vs-all classifier corresponds to different support vectors and their respective alphas in the decomposition. The score output for a test data point is in no way bou |
39,692 | How to Assess the Fit of Thousands of Distributions? | I'll suggest using representative plots. Pull 16 or 20 subjects, and show their QQ-plots in 4x4 or 4x5 chart. Sometimes, you can plot several subjects in the same plot. This doesn't substitute other ways of representing the fits, but on the other hand I don't think you can avoid this step either. It's used a lot in pan... | How to Assess the Fit of Thousands of Distributions? | I'll suggest using representative plots. Pull 16 or 20 subjects, and show their QQ-plots in 4x4 or 4x5 chart. Sometimes, you can plot several subjects in the same plot. This doesn't substitute other w | How to Assess the Fit of Thousands of Distributions?
I'll suggest using representative plots. Pull 16 or 20 subjects, and show their QQ-plots in 4x4 or 4x5 chart. Sometimes, you can plot several subjects in the same plot. This doesn't substitute other ways of representing the fits, but on the other hand I don't think y... | How to Assess the Fit of Thousands of Distributions?
I'll suggest using representative plots. Pull 16 or 20 subjects, and show their QQ-plots in 4x4 or 4x5 chart. Sometimes, you can plot several subjects in the same plot. This doesn't substitute other w |
39,693 | How to Assess the Fit of Thousands of Distributions? | I hope that understood your situation and the question correctly. Considering your data set's number of distributions, visual exploratory approaches (such as QQ plots, which you mentioned) are not feasible in this case. Therefore, you have to resort to analytical approaches, such as goodness-of-fit (GoF) tests, as some... | How to Assess the Fit of Thousands of Distributions? | I hope that understood your situation and the question correctly. Considering your data set's number of distributions, visual exploratory approaches (such as QQ plots, which you mentioned) are not fea | How to Assess the Fit of Thousands of Distributions?
I hope that understood your situation and the question correctly. Considering your data set's number of distributions, visual exploratory approaches (such as QQ plots, which you mentioned) are not feasible in this case. Therefore, you have to resort to analytical app... | How to Assess the Fit of Thousands of Distributions?
I hope that understood your situation and the question correctly. Considering your data set's number of distributions, visual exploratory approaches (such as QQ plots, which you mentioned) are not fea |
39,694 | (Feed-Forward) Neural Networks keep converging to mean | Problem is identified in chat discussion. The logistic function in the hidden layer is saturated with large input values.
It is recommended to normalize the input values to [-1,1] range as standard NN practice.
Question owner acknowledged that it resolved his problem. | (Feed-Forward) Neural Networks keep converging to mean | Problem is identified in chat discussion. The logistic function in the hidden layer is saturated with large input values.
It is recommended to normalize the input values to [-1,1] range as standard NN | (Feed-Forward) Neural Networks keep converging to mean
Problem is identified in chat discussion. The logistic function in the hidden layer is saturated with large input values.
It is recommended to normalize the input values to [-1,1] range as standard NN practice.
Question owner acknowledged that it resolved his probl... | (Feed-Forward) Neural Networks keep converging to mean
Problem is identified in chat discussion. The logistic function in the hidden layer is saturated with large input values.
It is recommended to normalize the input values to [-1,1] range as standard NN |
39,695 | (Feed-Forward) Neural Networks keep converging to mean | (1) Some of your predictors are on wildly different scales. Most texts (e.g., Ripley, 96; Hastie, Tibshirani, Friedman 08) recommend preprocessing the predictors by scaling the range to [0,1].
(2) In my experience, fit is very sensitive to tuning parameter values. Again, most authors recommend searching over an extensi... | (Feed-Forward) Neural Networks keep converging to mean | (1) Some of your predictors are on wildly different scales. Most texts (e.g., Ripley, 96; Hastie, Tibshirani, Friedman 08) recommend preprocessing the predictors by scaling the range to [0,1].
(2) In | (Feed-Forward) Neural Networks keep converging to mean
(1) Some of your predictors are on wildly different scales. Most texts (e.g., Ripley, 96; Hastie, Tibshirani, Friedman 08) recommend preprocessing the predictors by scaling the range to [0,1].
(2) In my experience, fit is very sensitive to tuning parameter values. ... | (Feed-Forward) Neural Networks keep converging to mean
(1) Some of your predictors are on wildly different scales. Most texts (e.g., Ripley, 96; Hastie, Tibshirani, Friedman 08) recommend preprocessing the predictors by scaling the range to [0,1].
(2) In |
39,696 | (Feed-Forward) Neural Networks keep converging to mean | I had this problem before.
I think in the general case the average problem can happen when.
Your final hidden layer blew up and all values are activated all the time. When all the values are 1 or 0 all the time. You basically just have the bias unit doing the lifting. And the bias unit can only predict the average at ... | (Feed-Forward) Neural Networks keep converging to mean | I had this problem before.
I think in the general case the average problem can happen when.
Your final hidden layer blew up and all values are activated all the time. When all the values are 1 or 0 a | (Feed-Forward) Neural Networks keep converging to mean
I had this problem before.
I think in the general case the average problem can happen when.
Your final hidden layer blew up and all values are activated all the time. When all the values are 1 or 0 all the time. You basically just have the bias unit doing the lift... | (Feed-Forward) Neural Networks keep converging to mean
I had this problem before.
I think in the general case the average problem can happen when.
Your final hidden layer blew up and all values are activated all the time. When all the values are 1 or 0 a |
39,697 | How to prevent collinearity? | As far as I understand, collinearity or multicollinearity (hereafter referred to simply as collinearity) cannot be prevented/avoided during data analysis, because collinearity is a built-in "feature" of data. Therefore, a particular data set has certain levels of collinearity (or the lack of). However, collinearity can... | How to prevent collinearity? | As far as I understand, collinearity or multicollinearity (hereafter referred to simply as collinearity) cannot be prevented/avoided during data analysis, because collinearity is a built-in "feature" | How to prevent collinearity?
As far as I understand, collinearity or multicollinearity (hereafter referred to simply as collinearity) cannot be prevented/avoided during data analysis, because collinearity is a built-in "feature" of data. Therefore, a particular data set has certain levels of collinearity (or the lack o... | How to prevent collinearity?
As far as I understand, collinearity or multicollinearity (hereafter referred to simply as collinearity) cannot be prevented/avoided during data analysis, because collinearity is a built-in "feature" |
39,698 | How to prevent collinearity? | There are several sampling techniques that can be used to reduce collinearity, so I'll mention just one of them: a stratified random sampling plan can eliminate some of the problem. Here's an example using the scenario you described above: If you find that certain trees are collinear with altitude or slope of the tr... | How to prevent collinearity? | There are several sampling techniques that can be used to reduce collinearity, so I'll mention just one of them: a stratified random sampling plan can eliminate some of the problem. Here's an exampl | How to prevent collinearity?
There are several sampling techniques that can be used to reduce collinearity, so I'll mention just one of them: a stratified random sampling plan can eliminate some of the problem. Here's an example using the scenario you described above: If you find that certain trees are collinear wit... | How to prevent collinearity?
There are several sampling techniques that can be used to reduce collinearity, so I'll mention just one of them: a stratified random sampling plan can eliminate some of the problem. Here's an exampl |
39,699 | How to prevent collinearity? | Obviously, I can't speak for Leno or Zuur. I am also not an ecologist. However, from the description, I think they just mean walk more than one path. In essence, multicollinearity just means that your variables are correlated with each other.
If you walk only one path up the mountain, every variable you measure at... | How to prevent collinearity? | Obviously, I can't speak for Leno or Zuur. I am also not an ecologist. However, from the description, I think they just mean walk more than one path. In essence, multicollinearity just means that y | How to prevent collinearity?
Obviously, I can't speak for Leno or Zuur. I am also not an ecologist. However, from the description, I think they just mean walk more than one path. In essence, multicollinearity just means that your variables are correlated with each other.
If you walk only one path up the mountain, ... | How to prevent collinearity?
Obviously, I can't speak for Leno or Zuur. I am also not an ecologist. However, from the description, I think they just mean walk more than one path. In essence, multicollinearity just means that y |
39,700 | How to prevent collinearity? | Similar to @StatsStudent,
You need replication of the other variables within values of the same slope. E.g. be also doing transcets around the mountain (sideways) starting at different elevations/slopes.
If there is truly no variaiton in tree trype with slope, then you will never be able to seperate the two using natur... | How to prevent collinearity? | Similar to @StatsStudent,
You need replication of the other variables within values of the same slope. E.g. be also doing transcets around the mountain (sideways) starting at different elevations/slop | How to prevent collinearity?
Similar to @StatsStudent,
You need replication of the other variables within values of the same slope. E.g. be also doing transcets around the mountain (sideways) starting at different elevations/slopes.
If there is truly no variaiton in tree trype with slope, then you will never be able to... | How to prevent collinearity?
Similar to @StatsStudent,
You need replication of the other variables within values of the same slope. E.g. be also doing transcets around the mountain (sideways) starting at different elevations/slop |
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