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As long as $a<b$, the distribution of $Z:=Z_i$ does not depend on $a,b$. This follows because for $Y_k:=(X_k-a)/(b-a)$ we have $Y_k\overset{iid}\sim U[0,1]$ and $Z_i=\dfrac{Y_{(i)}-Y_{(1)}}{Y_{(n)}-Y_ …

The inequality $1- \exp(-D_{\infty}(P,Q)) \leq D_{\infty}(P,Q)$ is just an instance of the elementary inequality $e^u\ge1+u$ for all real $u$. Let us show an improved version of the inequality $D_{t …

$\newcommand{\eD}{\overset{\text{D}}\to}$ Let $X_n$ and $X$ be any random vectors with distributions $\mu_n$ and $\mu$, respectively. Let $Y_n$ be an independent copy of $X_n$, for each $n$, and let $ …

In order to find the distribution of $\phi$, you need to know the joint distribution of $S,A,C$. Then you can use the standard transformation technique, as described e.g. in Section 4.2; see also bibl …

$\newcommand{\R}{\mathbb{R}}$ Welcome to MathOverflow! Your conditions on $X$, $Y$, $Z$, $A$, as I understood them, imply that $X$, $Y$, $Z$, $A$ are independent nonnegative random variables (r.v.'s …

The exact distribution is very unlikely to exist in closed form. However, one can see that this distribution is $\approx \mathcal N(0,1/N)$ for large $N$. Moreover, an explicit Berry--Esseen-type bou …

Yes, this is a special case of Slepian's inequality.

$\newcommand\Si\Sigma$First here, there is no such thing as an orthonormal matrix. So, let us assume that you meant an orthogonal matrix instead. Then we have this question: Suppose that $X\sim N(\mu …

The function $\mathbb R\ni t\mapsto g(t):=f(it)$ is not the characteristic function of any probability distribution. Indeed, if it were, then we would have $|g(t)|\le1$ for all real $t$. However, in f …

$\newcommand\R{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand\la\lambda\newcommand{\be}{\beta}\newcommand{\al}{\alpha}\newcommand\La\Lambda$It is given that the random variable (r.v.) $\La$ has the g …

Yes, this is true. Indeed, the first circular moment for a probability density $f$ on the interval $[0,2\pi]$ is $$m_1(f):=\int_0^{2\pi}e^{ix}f(x)\,dx.$$ So, $$\begin{aligned}&|m_1(f)| \\ &=\max_{u\i …

$\newcommand\si{\sigma}\newcommand\la{\lambda}\newcommand\th{\theta}\newcommand\Th{\Theta}$Your family $(f_\th)$, where $\th:=(\lambda,\sigma)\in\Th:=(0,\infty)^2$, is not exponential. Indeed, suppose …

There is no such thing as a "cumulative probability density function". You seem to get confused between the notions of (i) the (cumulative) probability distribution function, (ii) the probability dens …

$\newcommand\p\partial\newcommand\ga\gamma$By the substitution formula for multiple integrals, the joint pdf of $|g_1|^2,\dots,|g_N|^2$ is given by $$P_{|g_1|^2,\dots,|g_N|^2}(s_1,\dots,s_N)=P_{|g_1|, …

If you want your "reshaping" transformation function to be monotonic, then the answer is no. Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$. Suppose t …