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Consider two independently distributed sets of complex Gaussian random variables $\{{X}_{i}\}^{N}_{i=1}$ and $\{{Y}_{i}\}^{N}_{i=1}$ with mean zero and variances as $\sigma^{2}_{x}$ and $\sigma^{2}_{y}$ respectively. I have calculated the value as shown below. Can someone please help verifying it.

$E[|\sum^{N}_{i=1}X^{*}_{i}Y_{i}|^{2}]$ =$ E[|\mathbf{X}^{\mathit{H}}\mathbf{Y}|^{2}]$=$N\sigma^{2}_{x}\sigma^{2}_{y}$

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    Please present some analysis in support. :) – 2017-01-08
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    Is $X^*_i$ the same thing as $\mathbf{X}_i$? – 2017-01-08
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    Sorry, corrected the typo – 2017-01-08
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    It looks like you are assuming that both sequences consist of iid random variables. Is this correct? – 2017-01-08
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    Yes, $X$ and $Y$ are IID Gaussian random vectors so trying to find out average of square of absolute value of inner product of IID Gaussian random vectors – 2017-01-08
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    if $X$ and $Y$ are Gaussian iid you should mention this in the question. – 2017-01-08
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    I have mentioned in the title, still edited. – 2017-01-08
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    Can someone tell the reason for downvote ?! – 2017-01-09

1 Answers 1

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I did it this way:
$\begin{align*} E[\sum^{N}_{i=1}|X^{*}_{i}Y_{i}|^{2}] &= E[|\mathbf{X}^{H}\mathbf{Y}|^{2}] =E[{(\mathbf{X}^{H}\mathbf{Y})(\mathbf{Y}^{H}\mathbf{X})}] \\ &=E{[\mathrm{tr}(\mathbf{X}^{H}\mathbf{Y}\mathbf{Y}^{H}\mathbf{X})]} =E{[\mathrm{tr}(\mathbf{Y}\mathbf{Y}^{H}\mathbf{X}\mathbf{X}^{H})]} =\mathrm{tr}[E(\mathbf{Y}\mathbf{Y}^{H}\mathbf{X}\mathbf{X}^{H})] \\&=\mathrm{tr}[E(\mathbf{Y}\mathbf{Y}^{H})E(\mathbf{X}\mathbf{X}^{H})] \\&=\mathrm{tr}(\sigma^{2}_{y}\mathbf{I}_{N}\sigma^{2}_{x}\mathbf{I}_{N}) \\&=\sigma^{2}_{x}\sigma^{2}_{y}\mathrm{tr}(\mathbf{I}_{N}) \\&=N\sigma^{2}_{x}\sigma^{2}_{y} \end{align*} $

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    @zhoraster, @ Therkel would you please check ? – 2017-01-09