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Given that following two random variables $\textbf{x} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{x}^{2}\textbf{I}_{M})$ and $\textbf{y} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{y}^{2}\textbf{I}_{M})$ are independent, what would be the expectation

$$\mathbb{E} \left[ \left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2 \right], $$ where $\mathcal{CN}(.,.)$ is the complex normal random variable.

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    $\begingroup$ I think you can use the trace trick for that, $|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$ and the trace commutes with the expectation... $\endgroup$
    – user114668
    Commented Feb 16, 2019 at 17:36
  • $\begingroup$ @student, thanks! do you think we can apply that trick by Marsaglia? $\endgroup$
    – Felipe Augusto de Figueiredo
    Commented Feb 16, 2019 at 17:38
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    $\begingroup$ In general, Marsaglia’s trick is only useful to get the $x$-linear expectation $E[x \, f(y)]$ for normal and correlated $x$ and $y$. $\endgroup$
    – user114668
    Commented Feb 16, 2019 at 18:09
  • $\begingroup$ @student, with the trace trick I find the first term equal to 1 and the second equal to $\frac{M\sigma_{y}^2}{(M-1)\sigma_{x}^2}$. The $M$ term should not be in the numerator. Do you think there is something wrong? $\endgroup$
    – Felipe Augusto de Figueiredo
    Commented Feb 16, 2019 at 20:34

2 Answers 2

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Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ and integrating over the independent chi-squared variables $\xi_2$, $\xi_{2M}$, and $\xi_{2M-2}$ I find the expectation value $$\Rightarrow\mathbb{E}(Z)=\frac{1}{M-1}(\sigma_y/\sigma_x)^2.$$ More generally, the $p$-th moment is finite for $M>p$, given by $$\mathbb{E}(Z^p)=\frac{(\sigma_y/\sigma_x)^2}{{M-1}\choose{p}}.$$

This answer looks simple enough, I wonder which distribution $P(Z)$ has reciprocals of binomial coefficients as its moments?

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    $\begingroup$ If you look for a random variable $Z_N$ s.t. $\mathbb{E}(Z_N^p) = {N\choose p}^{-1}$ for $p\leq N$ and undefined for $p \geq N+1$, you can have it with the ratio of an independent $ \boldsymbol{e}\sim $Exp(1) random variable and $ \gamma_{N + 1}\sim$Gamma($N + 1$) random variable. You can check that $ \mathbb{E}((\boldsymbol{e}/\gamma_{N + 1})^p ) = 1/{N\choose p} $. $\endgroup$
    – Synia
    Commented Feb 20, 2019 at 20:34
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To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:

Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace outside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x = \sigma_y^2 \, \Vert x \Vert^2$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$

(This might also be helpful in other situations, since normality is not used until the very end.)

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