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 $\mathbb{E}[\mathbf{X}]$?

  • 1
    $\begingroup$ 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 $X^{-2}$. $\endgroup$ Feb 28, 2020 at 12:15
  • $\begingroup$ For example, something like Jensen's inequality. $\mathbb{E}[X^{-1}]\geq\mathbb{E}[X]^{-1}$. $\endgroup$
    – Math_Y
    Feb 28, 2020 at 12:34

1 Answer 1


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 $$Ef(X)\le\min_{a\in(0,2]}B_a(mI) =\begin{cases} \tfrac14\,I &\text{ if }m\ge2,\\ \tfrac{m-1}{m^2}\,I &\text{ if }0<m\le2. \end{cases}$$

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.

  • $\begingroup$ this would be a lower bound for $\alpha<0$, right? $\endgroup$ Feb 28, 2020 at 14:48
  • $\begingroup$ @CarloBeenakker : I have added the case $\alpha\le0$ as well. $\endgroup$ Feb 28, 2020 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.