Assume that $\mathbf{X}$ is a random positivedefinite 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 $X^{2}$. $\endgroup$– Carlo BeenakkerFeb 28, 2020 at 12:15

$\begingroup$ For example, something like Jensen's inequality. $\mathbb{E}[X^{1}]\geq\mathbb{E}[X]^{1}$. $\endgroup$– Math_YFeb 28, 2020 at 12:34
1 Answer
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 $n\times n$ positivedefinite 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 $P(X=t)=1/2=P(X=2t)$ 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, $$f'(x)=\frac{2x}{x^3},\quad f''(x)=2\frac{x3}{x^4},$$ 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 $x>0$, $$f(x)\le g_a(x):=f(a)+f'(a)(xa).$$ 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 $$Ef(X)\le\min_{a\in(0,2]}B_a(mI) =\begin{cases} \tfrac14\,I &\text{ if }m\ge2,\\ \tfrac{m1}{m^2}\,I &\text{ if }0<m\le2. \end{cases}$$
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 $n=1$, $P(X=t)=1/2=P(X=2t)$, 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