# Expectation of inverse of random matrices

Assume that $$\mathbf{X}$$ is a random positive-definite matrix. Then, is there any upper or lower bound on the expectation of the following expression $$\mathbb{E}[\mathbf{X}^{-1}]-\alpha\mathbb{E}[\mathbf{X}^{-2}]$$ based on $$\mathbb{E}[\mathbf{X}]$$?

• how might this work? the expectation of $X$ will not give you information on $X$ near zero, which you need to know the expectation of $X^{-1}$ and $X^{-2}$. – Carlo Beenakker Feb 28 '20 at 12:15
• For example, something like Jensen's inequality. $\mathbb{E}[X^{-1}]\geq\mathbb{E}[X]^{-1}$. – Math_Y Feb 28 '20 at 12:34

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 matrix with a given mean $$EX$$ and $$f(x):=\frac1x-\frac1{x^2}$$ for real $$x>0$$.

First of all, there is no finite lower bound here. Indeed, already for $$n=1$$, letting $$P(X=t)=1/2=P(X=2-t)$$ with $$t\downarrow0$$, we get $$Ef(X)\to-\infty$$.

However, we can get an upper bound on $$Ef(X)$$, which will be exact if for some $$b\in(0,2]$$ we have $$P(X=bI)=1$$, where $$I$$ is the $$n\times n$$ identity matrix. Indeed, $$f'(x)=\frac{2-x}{x^3},\quad f''(x)=2\frac{x-3}{x^4},$$ so that $$f$$ is increasing on $$(0,2]$$, decreasing on $$[2,\infty)$$, and concave on $$(0,2]$$. So, for any $$a\in(0,2]$$ and all real $$x>0$$, $$f(x)\le g_a(x):=f(a)+f'(a)(x-a).$$ So, by the spectral decomposition, $$f(X)\le g_a(X)$$, and hence $$Ef(X)\le Eg_a(X)=B_a(EX):=(f(a)-f'(a)a)I+f'(a)EX.$$ The upper bound $$B_a(EX)$$ on $$Ef(X)$$ will be exact if $$P(X=bI)=1$$ for some $$b\in(0,2]$$ and $$a=b$$.

More generally, if e.g. $$EX\le mI$$ for some real $$m>0$$, then $$Ef(X)\le\min_{a\in(0,2]}B_a(mI) =\begin{cases} \tfrac14\,I &\text{ if }m\ge2,\\ \tfrac{m-1}{m^2}\,I &\text{ if }0

If now $$\alpha\le0$$, then the corresponding function $$F(x):=\frac1x-\frac\alpha{x^2}$$ is convex, and hence $$EF(X)\ge F(E(X)$$ by Jensen's inequality. In this case, there is no upper bound, though -- as the above example with $$n=1$$, $$P(X=t)=1/2=P(X=2-t)$$, and $$t\downarrow0$$ shows.

• this would be a lower bound for $\alpha<0$, right? – Carlo Beenakker Feb 28 '20 at 14:48
• @CarloBeenakker : I have added the case $\alpha\le0$ as well. – Iosif Pinelis Feb 28 '20 at 14:56