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The probability that, of the $m$ random points on the $n$-sphere (that is, on the unit sphere in $\mathbb R^n$), $m/2$ points lie in the all-positive orthant and the other $m/2$ lie in the all-negativ …

$\newcommand{\R}{\mathbb{R}} \newcommand{\x}{\mathbf{x}} \newcommand{\X}{\mathbf{X}}$ This is to present a formalization of the answer by Carlo Beenakker, without explicit use of the delta function. …

The answer is no. E.g., let $n=3$ and $z_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$. This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $ …

Let $G_1,\dots,G_n$ be iid standard normal random variables. Then the random vector \begin{equation*} (Y_1,\dots,Y_n):=\Big(\frac{G_1}{\sqrt{\sum_1^n G_j^2}},\dots, \frac{G_n}{\sqrt{\sum_1^n G …

$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\la …

$\newcommand{\R}{\mathbb{R}} \newcommand{\x}{\mathbf{x}} \newcommand{\X}{\mathbf{X}}$ Here is yet another solution, which is partly informal but I think not hard to completely formalize. Its advantage …

A quick way to generate a random point uniformly distributed in a bounded region $D$ is to generate a random point $P$ uniformly distributed in a rectangle $R$ containing $D$ and, if $P\notin R$, then …

Unconditional? Certainly not. E.g., suppose that $\mu$ is the standard Gaussian measure on $\mathbb R^d$ and $B$ is the ball of radius $r>0$ centered at $0$. Then for any $\delta\in(0,1)$, by the law …

This approach is of course well known. Clearly, it just says that $$P(X\in A)=P((N_1,\dots,N_d)\in C_A),$$ where $A$ is a Borel subset of the unit sphere $S^{d-1}$ and $C_A:=\mathbb R_+A$ is the corr …

This is to show rigorously that the uniform rectangular grid does not work -- cf. the answer by mike. As in the answer by Dieter Kadelka, suppose that the $cn$ vertical lines and the $cn$ horizontal l …

$\newcommand\Z{\mathbf Z}\newcommand\ep\varepsilon\newcommand\tP{\tilde P}\newcommand\de\delta\newcommand\R{\mathbb R}$Note that the random point $\Z/|\Z|$ is uniformly distributed on the sphere $S^{n …

Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0, …

We have $f=e^g$, $g$ is concave, $\int f=1$, $\int x f(x)\,dx=0$, and $\int x^2 f(x)\,dx=1$. As you noted, then $f(0)\ge 1/8$ and hence $$g(0)\ge-a,\tag{0}\label{0}$$ where $a:=\ln8$. We have to show …

$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$The answer is yes. Indeed, it is easy to see (cf. e.g. Proposition 7 or the beginning of its proof) that the Wasserstein distance between …

The intuition in the comment by Ori Gurel-Gurevich appears to be correct. Indeed, let us show that $\pi-\phi_n$ is on the order of $1/n^3$ in probability. Let $T$ denote the set of all triangles w …