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Considering the following random vectors:

\begin{align} \textbf{h} &= [h_{1}, h_{2}, \ldots, h_{M}]^{T} \sim \mathcal{CN}\left(\textbf{0}_{M},d\textbf{I}_{M \times M}\right), \\[8pt] \textbf{w} &= [w_{1}, w_{2}, \ldots, w_{M}]^{T} \sim \mathcal{CN}\left(\textbf{0}_{M},\frac{1}{p}\textbf{I}_{M \times M}\right), \\[8pt] \textbf{y} &= [y_{1}, y_{2}, \ldots, y_{M}]^{T} \sim \mathcal{CN}\left(\textbf{0}_{M},\left(d + \frac{1}{p}\right)\textbf{I}_{M \times M}\right), \end{align}

where $\textbf{y} = \textbf{h} + \textbf{w}$ and therefore, $\textbf{y}$ and $\textbf{h}$ are not independent.

I'm trying to find the following expectation:

$$\mathbb{E} \left[ \frac{\textbf{h}^{H} \textbf{y}\textbf{y}^{H} \textbf{h}}{ \| \textbf{y} \|^{4} } \right],$$

where $\| \textbf{y} \|^{4} = (\textbf{y}^{H} \textbf{y}) (\textbf{y}^{H} \textbf{y}$).

In order to find the desired expectation, I'm applying the following approximation:

$$\mathbb{E} \left[ \frac{\textbf{x}}{\textbf{z}} \right] \approx \frac{\mathbb{E}[\textbf{x}]}{\mathbb{E}[\textbf{z}]} - \frac{\text{cov}(\textbf{x},\textbf{z})}{\mathbb{E}[\textbf{z}]^{2}} + \frac{\mathbb{E}[\textbf{x}]}{\mathbb{E}[\textbf{z}]^{3}}\text{var}(\mathbb{E}[\textbf{z}]).$$

However, applying this approximation to the desired expectation is time consuming and prone to errors as it involves expansions with lots of terms .

I have been wondering if there is a more direct/smarter way of finding the desired expectation.

$\textbf{UPDATE 21-04-2018}$: I've created a simulation in order to identify the pdf shape of the ratio inside of the expectation operator and as can be seen below it seems much like the pdf of a Gaussian random variable. Additionally, I've also noticed that the ratio results in real valued terms, the imaginary part is always equal to zero.

Is there another kind of approximation that can be used to find the expectation (one analytical/closed form result and not only the simulated value of the expection) given that the pdf looks like a Gaussian and probably can be approximated as such?

pdf of the ratio inside the expectation operator

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  • $\begingroup$ Could it be solved with Mathematica? $\endgroup$ – Felipe Augusto de Figueiredo Apr 15 '18 at 18:40
  • $\begingroup$ I tried the following on Mathematica: [Mu] = {0, 0}; [CapitalSigma] = IdentityMatrix[2]; Expectation[(((h[ConjugateTranspose].(h + w)).((h + w)[ConjugateTranspose] h))/((h + w)[ConjugateTranspose].(h + w).(h + w)[ConjugateTranspose].(h + w))), {h [Distributed] MultinormalDistribution[[Mu], d IdentityMatrix[2]].{1., 1. I}, w [Distributed] MultinormalDistribution[[Mu], (1/p) IdentityMatrix[2]].{1., 1. I}}], however, it returns odd expressions that doesn't make sense to me. I've never used that, so I dont really know if that is right. $\endgroup$ – Felipe Augusto de Figueiredo Apr 15 '18 at 18:44
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    $\begingroup$ Do not crosspost. math.stackexchange.com/questions/2738096/…. $\endgroup$ – StubbornAtom Apr 15 '18 at 20:30
  • $\begingroup$ Does $\mathcal{C}\mathcal{N}$ just mean the usual multivariate normal distribution? $\endgroup$ – wij Apr 20 '18 at 11:18
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    $\begingroup$ You've cross-posted on (at least) 3 Stack Exchange forums. That's not good form especially when you don't notify that you've done so. $\endgroup$ – JimB Apr 24 '18 at 22:00
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I've found approximations for both cases, i.e., independent and dependent cases.

Case (1) where $\textbf{h}$ and $\textbf{y}$ are independent.

$$\mathbb{E} \left[ \frac{\textbf{h}^{H}_{l} \textbf{y}_{k} \textbf{y}^{H} _{k} \textbf{h}_{l} }{ \| \textbf{y}_{k} \|^{4} } \right] = \frac{d_{l}[(M+1)(M-2)+4M+6]}{\zeta_{k}M(M+1)^{2}}$$

where $\zeta_{k} = d_{k} + \frac{1}{p}$, $\textbf{h}_{l} \sim \mathcal{CN}\left(\textbf{0}_{M},d_{l}\textbf{I}_{M \times M}\right)$ and $\textbf{h}_{k} \sim \mathcal{CN}\left(\textbf{0}_{M},d_{k}\textbf{I}_{M \times M}\right)$. Note that $\textbf{y}_{k} = \textbf{h}_{k} + w$ and that $\textbf{h}_{k}$ and $\textbf{h}_{l}$ are independent.

Case (2) where $\textbf{h}$ and $\textbf{y}$ are dependent.

$$\mathbb{E} \left[ \frac{\textbf{h}^{H}_{k} \textbf{y}_{k} \textbf{y}^{H} _{k} \textbf{h}_{k} }{ \| \textbf{y}_{k} \|^{4} } \right] = \frac{pd_{k}[pd_{k}M(M+1)^2 + M^2+3M+4]}{(pd_{k}+1)^2 M(M+1)^2}$$

where $\textbf{h}_{k} \sim \mathcal{CN}\left(\textbf{0}_{M},d_{k}\textbf{I}_{M \times M}\right)$ and $\textbf{y}_{k} \sim \mathcal{CN}\left(\textbf{0}_{M}, \left(d_{k} + \frac{1}{p}\right)\textbf{I}_{M \times M}\right)$. Note that $\textbf{y}_{k} = \textbf{h}_{k} + w$ and therefore, $\textbf{h}_{k}$ and $\textbf{y}_{k}$ are not independent.

I have used the following approximation in both updates: $$\mathbb{E} \left[ \frac{\textbf{x}}{\textbf{z}} \right] \approx \frac{\mathbb{E}[\textbf{x}]}{\mathbb{E}[\textbf{z}]} - \frac{\text{cov}(\textbf{x},\textbf{z})}{\mathbb{E}[\textbf{z}]^{2}} + \frac{\mathbb{E}[\textbf{x}]}{\mathbb{E}[\textbf{z}]^{3}}\text{var}(\mathbb{E}[\textbf{z}]).$$

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