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Events (or random variables) are independent when information on some of them tells you nothing about the probability of occurrence (/ distribution) of the others. Please DO NOT use this tag for independent variable use [predictor] instead.

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Independence: Evaluating $\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$

The generic meaning of $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$$ is the measure of the $n$-dimensional set $A_1\times A_2\times\cdots\times A_n$ under the probability measure $F_n$ associated wit …
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0 votes

When are $f(X)$ and $f'(X)$ independent?

A more modest property, namely a lack of correlation, holds in a series of cases. Consider the case when $f$ is invertible and let the density of $X$ write as $p(f(x))$ (wlog). Further assume (wlog) t …
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8 votes
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Number of independent samples for weighted samples?

This "number of independent samples I really have" is called the effective sample size in simulation books, $N_\text{ess}$. Given a sample $$ x_1,\ldots,x_N \sim g(x) $$ leading to weights $w_i$ $(1 …
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34 votes
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Zero correlation of all functions of random variables implying independence

indicator functions of measurable sets like$$f(x)=\mathbb I_A(x)\quad g(x)=\mathbb I_B(x)$$leads to$$\text{cov}(f(X),g(Y))=\mathbb P(X\in A,Y\in B)-\mathbb P(X\in A)\mathbb P(Y\in B)$$therefore implying independence
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4 votes
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Is the error term a sum of r.v.?

In the second equation, "Supposing that β=1, y∼N(0,1), x∼N(0,1) and x, y are independent" is defining another model and hence another distribution on $u$, which is unrelated with the $u$ introd …
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3 votes
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Normalized subvectors of Dirichlet, mutually independent?

.$$ The result follows from the independence of the $Y_i$'s since $$\bigg(\frac{X_k}{\sum_{j\in\pi_1}X_j}\bigg)_{k\in\pi_1},\bigg(\frac{X_k}{\sum_{j\in\pi_2}X_j}\bigg)_{k\in\pi_2},\bigg(\frac{X_k}{\sum …
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8 votes

Expectation of double quadratic form

Although the solution is essentially and already contained in W. Huber's answer, here is a detailed derivation with a non-zero mean: \begin{align} \mathbb E(\overbrace{\hat{Y_k}'A\hat{Y_l}}^{\text{rea …
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5 votes
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Decomposing dependent Bernoulli random variables into independent Bernoulli random variables

If you decompose the joint distribution of $X_1,\ldots,X_n$ as $$p(x_1)p(x_2|x_1)\cdots p(x_n|x_1,\ldots,x_{n-1})$$ then each term in the product is a Bernoulli distribution, which probability depends …
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1 vote

Is a conditional distribution independent of the variable being conditioned on?

Since the conditioning event is $X=k$, the conditional distribution depends on it (or not when $X$ and $Y$ are independent), but it makes no sense to introduce independence once $X=k$ has happened. …
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2 votes
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Why the distribution $Y/\sigma$ does not depend on $\sigma$?

The confusion stems from not processing the change of variable correctly. If one sets$$Z=\sigma^{-1}Y$$the density $f_Z$ of $Z$ writes as $$f_Z(z)=f_Y(\underbrace{\sigma z}_{y(z)}) \times \underbrace{ …
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2 votes
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Bayesian framework - Prior and Likelihood independence

If the prior is built as the posterior over a sample $(x_1,x_0)$, namely $$\pi_a(\theta)\propto\pi_0(\theta)f_1(x_1|\theta)f_0(x_0|\theta)$$ and if one considers the (final) posterior over a sample $( …
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1 vote

Show That $X_1+X_2$ And $X_1-X_2$ Are Independent If $X_1$ And $X_2$ Be Standard Normal Dist...

Since$$\left(\begin{matrix}X_1\\X_2\end{matrix}\right)\sim\mathcal N_2\left(\left(\begin{matrix}0\\0\end{matrix}\right),\mathbf I_2\right)$$ where $\mathbf I_2$ denote the identity matrix \begin{align …
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2 votes

Gaussian distribution: moments, independence and rotation

Distributions that involve more than two parameters are rarely characterised by their first two moments. An example is the g-and-k distribution, which quantile function $$q_{A,B,c,g,k}(z) = A + B [1 + …
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5 votes
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Does dependency imply an equation?

For simplificity's sake, consider both $X$ and $Y$ to be unidimensional random variables. Assume that $X$ and $Y$ are dependent and denote by $F_{Y|X}(y|x)$ the conditional cdf of $Y$ given $X=x$. The …
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6 votes
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Independence of ratios of independent variates

{\{\varrho \sin(\theta)\}^2}{\{\varrho \sin(\theta)\}^2+\{\varrho \cos(\theta)\}^2}=\sin(\theta)^2$$ and $$Y=\frac{\varrho^2}{\varrho^2+x_3}$$ are indeed functions of different variates (although the independence
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