Considering the following complex random vectors (Complex Gaussian random variables): \begin{align} \textbf{h} &= [h_{1}, h_{2}, \ldots, h_{M}]^{T}\ \ \sim \mathcal{CN}(\textbf{0}_{M},d\textbf{I}_{M \times M}), \\ \textbf{w} &= [w_{1}, w_{2}, \ldots, w_{M}]^{T} \sim \mathcal{CN}(\textbf{0}_{M},(\frac{1}{p})\textbf{I}_{M \times M}), \\ \textbf{y} &= [y_{1}, y_{2}, \ldots, y_{M}]^{T}\ \ \ \sim \mathcal{CN}(\textbf{0}_{M},(d + \frac{1}{p})\textbf{I}_{M \times M}), \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 whether it is possible or not to use Mathematica to find the desired expectation.

Would it be possible?

$\textbf{UPDATE 21-04-2018}$: 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

$\textbf{UPDATE 24-04-2018}$: I've found an approximation to the case 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.

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

  • 1
    $\begingroup$ Can you state the problem in Mathematica notation, rather than the notation of your specialty? $\endgroup$
    – John Doty
    Apr 15, 2018 at 16:31
  • $\begingroup$ To somewhat echo @JohnDoty 's comment: Did your updates where you performed the simulations use Mathematica ? Do you have a copy of Mathematica or are you just looking for an analytic solution from someone who has a copy of Mathematica ? $\endgroup$
    – JimB
    Apr 24, 2018 at 21:57
  • $\begingroup$ You've cross-posted this on (at least) 3 Stack Exchange forums simultaneously. Don't do that or at least notify that you've done so. $\endgroup$
    – JimB
    Apr 24, 2018 at 22:02
  • $\begingroup$ @JimB, I had the updates based on matlab results. I do not have Mathematica neither know how to use it to find an analytical solution to the problem. I don't even know if that is possible with Mathematica and that is why I'm here. I need an analytical solution from someone who knows how to solve that with Mathematica. Thanks! $\endgroup$ Apr 25, 2018 at 7:22

1 Answer 1


The following seems to work, at least for numerical estimates of the expectation:

M = 4;
d = 2;
p = 3;
w = Array[ww, 2 M];
h = Array[hh, 2 M];
 (h.(h + w)) (h.(h + w))/ ((h + w).(h + w))^2,
  w \[Distributed] MultinormalDistribution[ConstantArray[0, {2 M}], d IdentityMatrix[2 M]],
  h \[Distributed] MultinormalDistribution[ConstantArray[0, {2 M}], 1/p IdentityMatrix[2 M]]


Of course, I cannot give guarantees for correctness.

  • $\begingroup$ What I need is an analytical equation for the expectation (it can also be an approximated analytical equation). Additionally, it needs to consider complex Gaussian random variables instead of real-valued Gaussians. Thanks! $\endgroup$ Apr 24, 2018 at 21:46

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