Lets say $Y=\frac{1}{n}XX^\intercal$ and $X$ is a $n\times m$ random matrix whose entries are i.i.d gaussian. We know when $n$ and $m$ go to infinity with a fixed ratio, the singular values of $Y$ follow Marchenko–Pastur distribution. I was wondering how to compute $$ \operatorname{Tr}\left((a Y_1+bY_2)^{-1}(cY_1+dY_2)\right). $$ where $a,b,c,d$ are constants, $Y_1$ and $Y_2$ are two independent realizations of Y.
1 Answer
I doubt there is an exact closed-form expression, but you could make progress numerically. Note that
$$\mathbb{E}\left[\mathrm{Tr}\left((a Y_1+bY_2)^{-1}(cY_1+dY_2)\right)\right]=(c/a)\bigl( n+(d/c-b/a)f(b/a)\bigr),$$ $$f(x)=\mathbb{E}\left[\mathrm{Tr}\left(Y_2(Y_1+xY_2)^{-1}\right)\right].$$ This function $f(x)$ satisfies $$f(x)=n/x-x^{-2}f(1/x),$$ so you only need to determine it in the interval $[0,1]$, which should be doable numerically.
This relation implies that $f(1)=n/2$, hence $$\mathbb{E}\left[\mathrm{Tr}\left(( Y_1+Y_2)^{-1}(cY_1+dY_2)\right)\right]=\tfrac{1}{2}n( c+d).$$
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$\begingroup$ Great thanks! I guess, if $2n=m$ and then $Y_2$ is invertible, it might be possible to get a closed form of $f(x)$ by free convolution? $\endgroup$ Commented Jul 24, 2022 at 13:16