Calculating $\mathbb{E}\left(\tfrac{XX^T}{\|AX\|^2}\right)$ for isotropic random vectors $X$

Question

$\newcommand{\EE}{\mathbb{E}}$ Let $A\in\mathbb{R}^{m\times n}$ and $X$ be an isotropic random vector in $\mathbb{R}^n$, i.e. it holds that $\EE(XX^T) = I_n$.

How to calculate $$M = \EE\left(\tfrac{XX^T}{\|AX\|^2}\right)$$ and/or $$AMA^T = \EE\left(\tfrac{AXX^TA^T}{\|AX\|^2}\right) = \EE\left(\tfrac{AXX^TA^T}{X^TA^TAX}\right)?$$

For $X$ being a (scaled) random unit coordinate vector $\sqrt{n}e_k$ where $k$ is uniformly distributed on $\{1,\dots,n\}$ that's simple to do, but would be interested in other isotropic distributions, e.g.

• normally distributed $X$,
• Rademacher vectors $X$ ($X_i = \pm 1$ with equal probability and independently)
• $X$ uniformly distributed on the sphere of radius $\sqrt{n}$.

I would also be happy with (bounds on) the smallest and largest eigenvalues of $M$ and $AMA^T$.

Answer 1

Assuming $X\sim N(0,I)$, writing $A^TA=\sum_i L_i u_i u_i^T$, since $sign(u_i^TX)$ and $sign(u_k^TX)$ are independent and mean-zero, $u_i^TMu_k=0$ if $i\le k$ because the denominator $\|AX\|^2=\sum_i L_i Z_i^2$ is independent of the signs, where $Z_i=u_i^TX$ are iid $N(0,1)$. This shows that $M$ is diagonalizable in the same basis as $A^TA$. It remains to compute the diagonal terms in the basis $(u_i)$, and $$ u_i^TMu_i = E\Big[ \frac{Z_i^2}{\sum_k L_k Z_k^2}\Big] $$ where $Z_i=u_i^TX$ are iid $N(0,1)$. Unfortunately, I do not think explicit expressions are available for general $L_k$ for the expectation in the right-hand side.