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}

  • $\begingroup$ Are the entries of X iid Gaussian variables? $\endgroup$
    – RaphaelB4
    Oct 13, 2020 at 11:55

1 Answer 1


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 $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$ 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 $M$, $$E\left[\exp(- {{v^T}X{X^T}{v}}) \right]=\int_0^\infty d\lambda_1\cdots\int_0^\infty d\lambda_M \,P(\lambda_1,\ldots\lambda_M)\prod_{m=1}^M(1+2M|v|^2\lambda_m)^{-1/2},$$ 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 $\endgroup$
    – hichem hb
    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 $(1+2M|v|^2\lambda_m)^{-1/2}$; then it remains to integrate over the $\lambda_m$'s, with the joint distribution of the Wishart ensemble. $\endgroup$ Oct 13, 2020 at 17:51
  • $\begingroup$ sir if suppose that $H$ is Complex thus, $ \omega _{_i}^2$ is exponential random variable? $\endgroup$
    – hichem hb
    Nov 13, 2020 at 1:02
  • 1
    $\begingroup$ yes, that would be the case. $\endgroup$ Nov 13, 2020 at 7:42
  • 1
    $\begingroup$ its $1/M$ --- the sum of the $M$ variances equals unity. $\endgroup$ Nov 13, 2020 at 23:33

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