Let $X$ be an $(N, M)$ random Gaussian matrix where $M<N$. For a given vector $v$, I want to estimate the expectation of: \begin{align} E\left[ {{v^T}X{X^T}{v}} \right] \end{align} and \begin{align} E\left[ {{e^{  {v^T}X{X^T}{v}}}} \right] \end{align}
1 Answer
Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=v^{1} Ov$, then $$v^T XX^T v =v^2 \sum_{m=1}^M \lambda_m w_m^2.$$ The matrix $XX^T$ has a Wishart distribution, with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{\lambda_m/2}\lambda_m^{(NM1)/2}\prod_{i<j}\lambda_i\lambda_j,$$ with $E[\sum_{m}\lambda_m]=NM$.
This gives $$E\left[ {{v^T}X{X^T}{v}} \right]=v^2 N.$$ The expectation $E\left[\exp( {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $M$, $$E\left[\exp( {{v^T}X{X^T}{v}}) \right]=\int_0^\infty d\lambda_1\cdots\int_0^\infty d\lambda_M \,P(\lambda_1,\ldots\lambda_M)\prod_{m=1}^M(1+2Mv^2\lambda_m)^{1/2},$$ for large $M$ it tends to $e^{v^2 N}$.

$\begingroup$ sir I didn't understand how did you find the last equation if you give me more detail if is possible $\endgroup$ Oct 13, 2020 at 17:31

1$\begingroup$ for the last equation I averaged $e^{v^2w_m^2\lambda_m}$ over the Gaussian variable $w_m$, which gives the factor $(1+2Mv^2\lambda_m)^{1/2}$; then it remains to integrate over the $\lambda_m$'s, with the joint distribution of the Wishart ensemble. $\endgroup$ Oct 13, 2020 at 17:51

$\begingroup$ sir if suppose that $H$ is Complex thus, $ \omega _{_i}^2$ is exponential random variable? $\endgroup$ Nov 13, 2020 at 1:02

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1$\begingroup$ its $1/M$  the sum of the $M$ variances equals unity. $\endgroup$ Nov 13, 2020 at 23:33