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Given the 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, therefore, as I have calculated,

\begin{equation}\label{eq:lemma7} \mathbb{E} \left\lbrace \Re \left[ \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right] \right\rbrace = \int_{0}^{\infty} \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dwdu, \end{equation} where $\mathcal{CN}(.,.)$ is the complex normal random variable, $\Re(.)$ is the real part, $k \triangleq \sqrt{\frac{\sigma_{x}^{2}}{\sigma_{y}^{2}}}$, $f_{W}(w)$ and $f_{U}(u)$ are defined below.

\begin{equation} f_{W}(w) = \frac{M}{\pi} {B}\left( \frac{1}{2}, M \right) \left( 1 -w^{2} \right)^{M - \frac{1}{2}}, \;\; |w| < 1. \end{equation}

\begin{equation}\label{eq:lemma6} f_{U}(u) = \frac{2\Gamma(2M)}{(\Gamma(M))^{2}} \frac{u^{2M-1}}{(u^{2} + 1)^{2M}}, \;\; u > 0. \end{equation} where ${B}(.,.)$ is the Beta function and $M > 1$.

I have been using the integral in my calculations, however, I'd like to know if a closed form expression is possible for this double integral. I've tried Mathematica, however, it never finishes running. Follows below the Mathematica code I used.

fw = (M/Pi)*Beta[1/2, M]*((1 - (w^2))^(M - (1/2)));
fu = (2*Gamma[2*M])/((Gamma[M])^2)*(u^(2*M - 1))/(((u^2) + 1)^(2*M));
f = ((a*u + w)*fu*fw)/(a*u + (1/(a*u)) + 2*w);
ii = Integrate[f, {w, -1, 1}, Assumptions -> (a > 0 && M > 2 && 0 < u < \[Infinity])]
Integrate[ii, {u, 0, \[Infinity]}, Assumptions -> (a > 0 && M > 2)]

Follows a link to a matlab Monte Carlo simulation comparison with the proposed closed-form solution.

Matlab script

UPDATE 11/02/2019

By applying the following approximation to the expectation above: $$\mathbb{E} \left[ \frac{\textbf{w}}{\textbf{z}} \right] \approx \frac{\mathbb{E}[\textbf{w}]}{\mathbb{E}[\textbf{z}]} - \frac{\text{cov}(\textbf{w},\textbf{z})}{\mathbb{E}[\textbf{z}]^{2}} + \frac{\mathbb{E}[\textbf{w}]}{\mathbb{E}[\textbf{z}]^{3}}\text{var}(\mathbb{E}[\textbf{z}]).$$

I found

$$\mathbb{E} \left\lbrace \Re \left[ \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right] \right\rbrace \approx \frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}$$.

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    $\begingroup$ What is $\mathcal{CN}$? My first guess was "complex normal" but that doesn't seem to make sense in context. $\endgroup$ – Nate Eldredge Feb 10 at 19:55
  • $\begingroup$ @NateEldredge, yes, it is the complex normal r.v. $\endgroup$ – Felipe Augusto de Figueiredo Feb 10 at 20:00
  • $\begingroup$ There was an error on the title and description, which is now fixed. $\endgroup$ – Felipe Augusto de Figueiredo Feb 10 at 20:14
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    $\begingroup$ the integral over $w$ evaluates to an algebraic function of $u$ times a hypergeometric function; the remaining integral over $u$ does not seem to have a closed-form expression for finite $M$; perhaps you are interested in the large-$M$ limit? $\endgroup$ – Carlo Beenakker Feb 10 at 21:43
  • $\begingroup$ @CarloBeenakker, thanks for your comment. The large-M limit is also one of the things I would like to know as my work is somehow related to that. $\endgroup$ – Felipe Augusto de Figueiredo Feb 10 at 22:23
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My guess in the comments was based on the fact that the expectation depends on $\sigma_x^2/\sigma_y^2$ only, and must be $1/2$ for $\sigma_x^2 = \sigma_y^2$, with limits of $0$ for $\sigma_x^2 \ll \sigma_y^2$ and $1$ for $\sigma_x^2 \gg \sigma_y^2$.

However, I believe this can be shown with an old trick by Marsaglia, although you should make sure there are no problems with the vectors being complex normal.

Write $\Sigma_x = E[x x^H]$ for the covariance matrix. Let $z = x + y$, and $S = \Sigma_x \, (\Sigma_x + \Sigma_y)^{-1}$. We have $$E[(x - S z) \, z^H] = 0 = E[x - S z] \, E[z]^H \;,$$ so that $x - S z$ and $z$ are uncorrelated and jointly normal, hence independent. Then $$E\left[\frac{z^H x}{\Vert z \Vert^2}\right] = E\left[\frac{z^H (x - S z)}{\Vert z \Vert^2}\right] + E\left[\frac{z^H S z}{\Vert z \Vert^2}\right] \\ = E\left[\frac{z^H}{\Vert z \Vert^2}\right] E\left[x - S z\right] + E\left[\frac{z^H S z}{\Vert z \Vert^2}\right] \\ = E\left[\frac{z^H S z}{\Vert z \Vert^2}\right] \;.$$ Since $S = \frac{\sigma_x^2}{\sigma_x^2 + \sigma_y^2} I$, the result follows.

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    $\begingroup$ While it does not matter for a real result, if you do take the real part beforehand, you can make real vectors $$\tilde x = \begin{bmatrix}\mathrm{Re} \, x \\ \mathrm{Im} \, x\end{bmatrix}, \quad \tilde y = \begin{bmatrix}\mathrm{Re} \, y \\ \mathrm{Im} \, y\end{bmatrix}$$ with $\Sigma_{\tilde{x}} = \frac{\sigma_x^2}{2} I_{2M}$ and $\Sigma_{\tilde{y}} = \frac{\sigma_y^2}{2} I_{2M}$, and the expectation is simply $$E \, \left[\frac{(\tilde{x} + \tilde{y})^T \tilde{x}}{\Vert \tilde{x} + \tilde{y} \Vert^2}\right] \;.$$ In that case, you will not have to worry about complex random variables. $\endgroup$ – student Feb 11 at 11:55
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    $\begingroup$ The trick to take a sum of normal r.v.s that is independent of the normal r.v. in the denominator was originally used in (Marsaglia 1965), and explicitly explained in (Marsaglia 2006, Section 2). $\endgroup$ – student Feb 11 at 12:22
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    $\begingroup$ thanks a lot for reference. $\endgroup$ – Felipe Augusto de Figueiredo Feb 11 at 12:23
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    $\begingroup$ if I am not mistaken, the expectation value diverges for $M=1$ (so when $x$ and $y$ are complex scalars); I don't quite see how the argument breaks down for $M=1$. $\endgroup$ – Carlo Beenakker Feb 11 at 12:48
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    $\begingroup$ @CarloBeenakker Not sure why it would. For $M = 1$, you can write out the integral and do a change of variables $z = x + y, w = x - s z$, where now simply $s = \sigma_x^2/(\sigma_x^2 + \sigma_y^2)$. You can then evaluate the integrals over $w$ first, which gets rid of the problematic terms in $z$. $\endgroup$ – student Feb 11 at 13:13
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Mathematica evaluates the integral over $w$ in terms of a hypergeometric function, $$ \int_{-1}^{1} \frac{(ku + w)f_{U}(u)f_{W}(w)}{ku+\frac{1}{ku}+2w} dw=$$ $$\frac{k^2 4^M u^{2 M+1}\Gamma(M+\tfrac{1}{2})}{\sqrt{\pi } \left(k^2 u^2-1\right)^3\left(u^2+1\right)^{2 M}} \left[\frac{1}{(M-1)!} \left(k^4 u^4+4 M k^2 u^2+4M+3\right)-\frac{4 M}{(M+1)!} \left(k^2 u^2+1\right) (\tfrac{3}{4}+m^2+2m)\, \, _2{F}_1\left(-\tfrac{1}{2},1;M+2;\frac{4 k^2 u^2}{\left(k^2 u^2+1\right)^2}\right)\right].$$ The integral $\int_0^\infty du$ of the first term between square brackets has a closed form expression (again involving a hypergeometric function), but the integral of the second term does not.

In the large-$M$ limit we may average numerator and denominator separately, so $$\mathbb{E} \left\lbrace \frac{(\textbf{x} + \textbf{y})^{H}\textbf{x}}{\| \textbf{x} + \textbf{y}\|^{2}} \right\rbrace \rightarrow\frac{\sigma^2_x}{\sigma_x^2+\sigma_y^2}, \;\;M\rightarrow\infty.$$

In the accepted answer @student has now elegantly shown this holds actually for all $M$ (quite an impressive "student" !)

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  • $\begingroup$ Do you think there is a closed form (i.e., a limit) when M tends to the infinity? If so, I can change the title of the question. $\endgroup$ – Felipe Augusto de Figueiredo Feb 10 at 23:04

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