# Expectation of exponential of Gaussian random matrix

Let $$X$$ be an $$(N, M)$$ random Gaussian matrix where $$M. For a given vector $$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}

• Are the entries of X iid Gaussian variables? Oct 13, 2020 at 11:55

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 $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i with $$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+2M|v|^2\lambda_m)^{-1/2},$$ for large $$M$$ it tends to $$e^{-|v|^2 N}$$.
• for the last equation I averaged $e^{-|v|^2w_m^2\lambda_m}$ over the Gaussian variable $w_m$, which gives the factor $(1+2M|v|^2\lambda_m)^{-1/2}$; then it remains to integrate over the $\lambda_m$'s, with the joint distribution of the Wishart ensemble. Oct 13, 2020 at 17:51
• sir if suppose that $H$ is Complex thus, $\omega _{_i}^2$ is exponential random variable? Nov 13, 2020 at 1:02
• its $1/M$ --- the sum of the $M$ variances equals unity. Nov 13, 2020 at 23:33