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Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that \begin{equation} \gamma n^2\to a \end{equation} (as $n\to\infty$) …

$\newcommand{\R}{\mathbb{R}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$ In view of the spherical symmetry of the distribution of the $l$-dimensiona …

$\newcommand\si{\sigma}$ $\newcommand\Si{\Sigma}$ $\newcommand\R{\mathbb R}$ Let $\Si:=\{\pm 1\}^n$. The map $$\R^{n\times n}\ni w\mapsto f(w):=(w_\si)_{\si\in\Si}\in\R^\Si, $$ where $w_\si:=\si^T w\ …

A greater and more general lower bound holds: $$(*)\qquad E f\Big(\sum_{i=1}^d \lambda_i g_i^2\Big) \ge E f(X_\lambda),$$ where $\lambda:=\lambda_1+\dots+\lambda_d$, $X_\lambda$ has the $\chi^2$ dis …

$\newcommand{\C}{\mathbb C}$$\newcommand{\R}{\mathbb R}$Any linear isometry of $\C^d$ is a unitary transformation. Therefore, the distribution of the random vector $V:=(X_1,Y_1,\dots,X_d,Y_d)$ is unif …

Let us assume that $\alpha>0$. Then, by rescaling, without loss of generality $\alpha=1$. So, we have to provide an upper or lower bound on $Ef(X)$, where $X$ is a random $n\times n$ positive-definite …

The $A_i$'s are independent zero-mean random vectors in $\mathbb{R}^{d \times d}$, which is a Hilbert space with respect to the Frobenius norm $\|\cdot\|:=\|\cdot\|_F$. So, by a vector version of Rose …

$\newcommand{\tr}{\operatorname{tr}}$ Let $x=[x^1,\dots,x^p]^T:=x_1$, $y:=xx^T$, $\mu:=Ey$, $w:=y-\mu$, and $s:=\sum_1^n x_ix_i^T$. Then, by the appropriate laws of large numbers, $s/n\to Ey$ almost s …

Assume that $\det\Sigma\ne0$. Then the random matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the Moore--Penrose inverse of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence $$X^+X=X^\top(XX^\top)^ …

$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$Your first displayed inequality for all positive definite random matrices $Q$ means exactly that the function $$P\mapsto\frac\l …

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sig …

$\newcommand\Si{\Sigma}$ $\newcommand\X{\mathbf X}$ The answer is no. Indeed, let $p_i:=P_i$, $p:=P$, $\X:=(X_1,\dots,X_n)$, and $\Si_\X:=\Si(p)$. At least one of the vectors $\Si^{-1}X_j$ is nonzero, …

The answer is no. Indeed, first of all, to make sense of the question, we need to deal with an infinite sequence of iid $N(0,1)$ random variables (r.v.'s) $X_1,X_2,\dots$. Next, for $n=1,2,\dots,\inft …

It is apparently assumed that the $Z_{ij}$'s are independent, as we will do here -- since otherwise hardly anything can be said. Suppose also that $m\ge2$ and $0<p<1$. The $ab$-entry of the matrix $Y: …

$\newcommand\ep\epsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$We have \begin{equation*} P((1-\ep)\|x\|\le\|Ax\|\le(1+\ep)\|x\|)\ge\de \tag{1}\label{1} \end{equation*} for some $\ep,\de …