# Tag Info

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Let us consider first an example with only $n=3$ random variables. So, we will have $X_{1},X_{2},X_{3}$ are a multiplicative system. And we will prove that $$\mathbb{E}[\prod_{i=1}^{3}(a_{i}X_{i}+b_{i})] =\prod_{i=1}^{3}b_{i}$$ We have the product $(a_{1}X_{1}+b_{1})*(a_{2}X_{2}+b_{2})*(a_{3}X_{3}+b_{3})$, where if we expand it we will have that it is equal ...

2

Sampling just means selecting (and then observing) items from a larger population; in statistics the sampling is usually randomised, and we are interested in the stochastic properties of random samples. In any case, the notion of "sampling" is a general concept that goes far beyond consideration of the binomial distribution. In some cases it will ...

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Comment continued with examples: $\mathsf{Binom}(n = 64, p = 1/2)$ [blue bars] compared with $\mathsf{Norm}(\mu=32,\sigma=4)$ [red curve]. x = 0:64; pdf = dbinom(x, 64, .5) plot(x, pdf, type="h", lwd=3, col="blue") curve(dnorm(x, 32, 4), add=T, lwd=2, col="red") abline(h=0, col="green2") $\mathsf{Binom}(n = 128, p =... 2 Apart from the usual mathematical techniques of checking your derivation, checking for consistency (e.g., differentiating the CDF to see whether it equals the PDF), and considering whether your answer makes sense, whenever there is a question of determining the distribution of a random variable (or collection of variables) you almost always have available ... 8 In the case of$Z=XY\sim \text{Beta}(\alpha,1)$, the moment generating function (mgf) of$-\ln(XY)=-\ln X-\ln Yis \begin{align} M_{-\ln(XY)}(t) &= E(e^{-t\ln Z}) \\ &=E(Z^{-t}) \\ &=\int_0^1 z^{-t}\alpha z^{\alpha-1}dz \\ &= \frac{1}{1-t/\alpha} \end{align} which is the mgf of an exponentially distributed random variable with rate ... 1 First you have to specify what distribution you are drawing the the variable from, e.g.\mathcal{N}(0, 1)$. Then, denote the variable by a letter, typically a capital one, like$X$. Finally, you can say$X \sim \mathcal{N}(0, 1)$. PS: your question deals with the concept of "realization" of random variable. Referring to this[1], you can say "$...

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You ask for a general method. Here is one. When $X^p$ has a Beta$(\alpha,\beta)$ distribution for $p\gt 0,$ this means for all $0\lt y \lt 1$ that $$F_X(y^{1/p}) = \Pr(X \le y^{1/p}) = \Pr(X^p \le y) = \frac{1}{B(\alpha,\beta)}\int_0^y t^{\alpha-1}(1-t)^{\beta-1}\,\mathrm{d}t.$$ Differentiating with respect to $y$ via the Chain Rule (at the left) and ...

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If $X^{2}\sim\operatorname{Beta}(1,1)$ (which is a uniform distribution), then $X^p\sim\operatorname{Kumaraswamy}(1/p, 1)$ (see the Wikipedia page). The PDF of the resulting Kumaraswamy distribution is given by $$f(x)=\frac{x^{\frac{1}{p}-1}}{p}\qquad 0<x<1$$ So for the example with $p=1/2$ we have $(X^2)^{1/2}=X\sim\operatorname{Kumaraswamy}(2, 1)$ ...

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Suppose $(X_1,\ldots,X_d)$ is a positive continuous random variable. Write its joint probability element as $$f(x_1,\ldots,x_d)\,\mathrm{d}x_1 \cdots \mathrm{d}x_{d-1} \mathrm{d}x_d$$ so that the density function is $f(x_1,\ldots, x_d).$ Leaving the case of the $L^\infty$ norm for later (it is special), consider $0\le p\lt \infty.$ You ask about the ...

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Provided $X$ is non-degenerate (that is, it is not almost surely constant) and has finite variance (without which it's impossible to have any correlation), you can always find such a $Y.$ One method begins by taking any random variable $Y_0$ for which $E[Y_0]=0$ but $E[XY_0]\ne 0$ and $E[|X|\,Y_0^2] \lt \infty.$ There always exists such a variable when $X$ ...

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Let's start with the first covariance: $$\text{Cov}(A\cdot B,C)=\mathbb{E}[A\cdot B\cdot C]-\mathbb{E}[A\cdot B]\mathbb{E}[C]$$ You know the means of $A,B,C$, namely $\mu_A,\mu_B,\mu_C$, and the pairwise correlation between $A,B$, which I will denote as $\rho_{A,B}$. In case you also have access to the standard deviations of $A,B$, written $\sigma_A,\sigma_B$...

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If $X=Y$ and Rademacher distributed, the product will be a constant and have zero covariance (uncorrelated) with $X$ or $Y$. Is it always possible? My previous example was incorrect.

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It depends on the nature of the assumptions on your $n$ random variables $Y_1, ..., Y_n$, which in turn will influence the kind of parametric statistical model that is specified. Here is a coin flipping example in context of maximum likelihood estimation that might be useful. Suppose you have a single biased coin, and you flip it $n$ times, with a view to ...

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If you have a model with a single parameter, then you are looking to find the single parameter, if it has multiple parameter, you are looking to find multiple parameters. With maximum likelihood, you find the parameters by maximizing the likelihood that is a function of the parameters. So it can be a function of one or more parameters, depending on the ...

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Since the wrapped Cauchy distribution is made of the superposition of truncated scaled Cauchy distributions$^1$ translated by $2n\pi$ for all $n$'s $(n\in\mathbb Z)$: $$f_{WC}(\theta;\gamma)=\sum_{n=-\infty}^\infty \frac{\gamma}{\pi(\gamma^2+(\theta+2\pi n)^2)}\,\mathbb I_{(-\pi,\pi)}(\theta)\qquad \gamma>0$$ it can be simulated by Simulating a regular ...

2

You are correct: for a transformation $T$ we have $$P(T(X) = a) = P(X \in T^{-1}(\{a\})) = \sum_{x \in T^{-1}(a)} f_X(x).$$ If $T$ is invertible then $$P(T(X)=a) = f_X(T^{-1}(a)) = (f_X\circ T^{-1})(a)$$ so we are indeed just plugging $T^{-1}(a)$ into the density of $X$. As to why the continuous and discrete cases are different, in one sentence I'd say ...

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The conditional density of $X$ given $Y$ is defined almost everywhere as $$f_{X|Y}(x|y)=\dfrac{f_{XY}(x,y)}{f_Y(y)}$$This means that for a measurable set A and almost every $y$ $$\mathbb P(X\in A|Y=y)=\int_A \dfrac{f_{XY}(x,y)}{f_Y(y)}\,\text dx$$ which satisfies the marginal equation $$\mathbb E^Y[\mathbb P(X\in A|Y)] = \mathbb P(X\in A)$$ (also seen as the ...

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