$\newcommand{\EE}{\mathbb{E}}$
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 $AMA^T$.



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


