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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 …

$\newcommand\la\lambda$The answer is: yes, of course. Indeed, let $X_{N,i}:=N^{2/3}(\la_i-2)$. By the limit theorem you cited and (say) Example 2.3, p. 18, the $k$-tuple $(X_{N,N},\dots,X_{N,N-k+1})$ …

Your formula $$\int_{\{X\in A\}}f\ dx = P[X\in A]\tag{1}$$ (I guess you wanted to say it should hold for all Borel $A\subseteq\mathbb R$) can be rewritten as $\int_B f\ dx = P(B)$ for all $B$ in $\si …

The subgaussian norm of a real-valued random variable $Y$ is $$\|Y\|:=\|Y\|_{\psi_2}:=\inf\{t>0\colon Ee^{Y^2/t^2}\le2\}.$$ If $Y$ is such that $P(Y=0)<1$ and $Ee^{Y^2/t^2}<\infty$ for all real $t>0$ …

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 …

Your desired conclusion holds, even without the additional assumption that $\frac1{\sqrt n}(|y|^2-E|y|^2)$ is a sub-exponential random variable with a sub-exponential norm not dependent on $n$. We onl …

$\newcommand\si{\sigma}$ $\newcommand\Si{\Sigma}$ $\newcommand\R{\mathbb R}$ 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\ …

No, of course not. Indeed, consider a simplest case when $m=n=1$, and $X:=\mathbf X$ and $y:=\mathbf y$ are iid standard normal random variables. Then $$E(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mat …

$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z_n:=\sqrt n\,X_n$, so that $Z_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$ $$ P …

(It will be assumed here that $A$ and $b$ are independent. Of course, without an assumption on the joint distribution of $A$ and $b$, hardly anything can be said about the joint distribution of $A$ an …

(It will be assumed here that $A$ and $b$ are independent. Of course, without an assumption on the joint distribution of $A$ and $b$, hardly anything can be said about the joint distribution of $A$ an …

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)^ …

$\newcommand\R{\Bbb R}\newcommand\Si{\Sigma}$The answer to your both questions is yes, regardless of what the covariance matrix $\Sigma$ is. Indeed, let $P_X:=X(X^\top X)^{-1}X^\top$, the orthoproject …

$\newcommand{\Om}{\Omega}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\De}{\Delta}$The question makes sense only if $A^{-1}$ exists, which will be assumed in what follows. For $x\in\O …

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 …