Approximating expectation of exponential of Wishart matrix

Question

I am trying to obtain an Approximating expectation of exponential of Wishart matrix $X (N,N)$ with $\operatorname{rank} (X)=K$defined as: \begin{align} J = E[{e^{{v^H}Xv}}] \end{align}
where $v$ is $(1,N)$ given vector.

I have used Wishart distribution defined by: \begin{align} f = {e^{\sum\limits_{i = 1}^K {{\lambda _i}} }}\prod\limits_{i = 1}^K {\lambda _i^{N - K}\prod\limits_{j > i}^K {{{({\lambda _i} - {\lambda _j})}^2}} } \end{align} and using the eigenvector written of $X$ as $ X = U\Sigma {U^H}$ so the problem can be written as: \begin{align} E[{e^{\sum\limits_{i = 1}^K {{b^i}{\lambda _i}} }}] \end{align} For smaller value on $N$ and$K$ I found finite expression bat as function of $b^i$ that are obiened by multiplying the vectors of unitary random matrix $U$ by my vector $t$.

can I say that $E[b_i]=||t||^2 $?

if it's possible can same person help me or give me another approach Thanks

Answer 1

The expectation value diverges, at least in the special case discussed below, but likely in general.

Take $v=(1,0,0,\ldots 0)$, so only a single element of $v$ is nonzero. The distribution of $y=v^H Xv=X_{11}$ follows from the known marginal distribution of Wishart matrix elements,

$$P(y)=\frac{1}{\sqrt{2\pi y}}e^{-y/2},\;\;y>0.$$

Then the desired expectation value $J$ of $e^y$ diverges.