Judge a special positive definite matrix in probability Assume $\mathbf{x}$ is a random vector. The question is to judge whether
$$E \{ (\mathbf{xx'})^{-1}  \}- E\{(\mathbf{xx'})\}^{-1}$$
is positive definite or not.
I have no idea how to do it. Could anyone give me some hint or reference to solve it?
Any work will be appreciated! Thanks a lot!
 A: Let $X:=\mathbf x$ and $b^T:=b'$. As was pointed out in Rodrigo de Azevedo's comment, the matrix $XX^T$ is of rank $1$, and so, $(XX^T)^{-1}$ does not exist unless the height of the column $X$ is $1$. So, instead of $XX^T$, let us use $XX^T+C$, for an arbitrary positive-definite matrix $C$, which e.g. can be of the form $\epsilon I$, with a however small $\epsilon>0$.
For each nonrandom vector $a$, the real-valued function $g$ defined on the (convex) set of all symmetric positive-definite matrices by the formula $g(A):=a^TA^{-1}a$ is convex. So, by Jensen's inequality,
$$a^TE(XX^T+C)^{-1}a=Ea^T(XX^T+C)^{-1}a=Eg(XX^T+C)\ge g(EXX^T+C)=a^T(EXX^T+C)^{-1}a.$$
So,
$$E(XX^T+C)^{-1}\ge(EXX^T+C)^{-1},$$
as desired.

The convexity of the function $g$ is proved as follows: For any symmetric positive-definite matrix $A$, any symmetric matrix $B$ of the same dimensions, and any real number $t$ close enough to $0$, we have $(A+tB)^{-1}-A^{-1}=(A+tB)^{-1}(I-(A+tB)A^{-1})=(A+tB)^{-1}(-tB)A^{-1}$, whence for $f(t):=(A+tB)^{-1}$ we have $f'(0)=-A^{-1}BA^{-1}$ and hence $f''(0)=2A^{-1}BA^{-1}BA^{-1}=2(BA^{-1})^TA^{-1}(BA^{-1})\ge0$, since $A^{-1}>0$. So, for $h(t):=g(A+tB)$, we have $h''(0)=a^Tf''(0)a\ge0$.
