5
$\begingroup$

Given the independent and complex Gaussian random variables $h$ and $w$, how does one can find the following expectation?

$$\mathbb{E} \left[ \frac{|h|^4}{|h+w|^2} \right] = \int_{\mathbb{C}}\int_{\mathbb{C}}{\frac{|h|^4}{|h+w|^2}} f(h)g(w)d_{h}d_{w},$$

where $h \sim \mathcal{CN}\left(0,d\right)$ and $w \sim \mathcal{CN}\left(0,p\right)$. The pdf of $h$ and $w$ are defined as

$$f(h) = \frac{1}{\pi d} \text{e}^{-\frac{|h|^2}{d}},$$

$$g(w) = \frac{1}{\pi p} \text{e}^{-\frac{|w|^2}{p}}.$$

I've tried to change the variables like: $|h|^2 = r^2$ and $|w|^2 = n^2$, however, I could not apply this change of variables to the denominator: $|h + w|^2$. Note that $h$ and $w$ are complex random variables that can be written in rectangular form like: $h = a + i*b$ and $w = c + i*d$, where $i = \sqrt{-1}$.

$\textbf{UPDATE}$: After running some simulations, it seems as $p \to \infty$ the expectation above tends to $d$. It can be checked with the following matlab simulation script: https://pastebin.com/U48fcMZ9

$\endgroup$
  • $\begingroup$ It looks divergent, due to the relatively high chance that $|h+w|\approx 0$ whenever $|h|\ne 0.$ $\endgroup$ – whuber Jun 12 '18 at 22:21
  • $\begingroup$ @whuber, thanks for your comment. I checked with a matlab simulation and it does not diverge. Please, have a look at the simulation script: pastebin.com/U48fcMZ9. Thanks. $\endgroup$ – Felipe Augusto de Figueiredo Jun 13 '18 at 6:03
  • 2
    $\begingroup$ Simulations rarely, if ever, diverge. The best they can do (as far as indicating divergence goes) is to jump erratically as the sample size gets larger and larger. The problem is that the jumps depend on simulating values of $w+h$ that are sufficiently close to zero that the denominator $1/|w+h|^2$ overwhelms the collective results of the previous simulation. That's not easy to do unless you construct the simulation specifically to focus on that region. Why not choose simple values for $p$ and $d$, condition on $H$, and compute the resulting integral? Matlab should tell you it diverges. $\endgroup$ – whuber Jun 13 '18 at 12:55
8
$\begingroup$

The expectation is infinite.

One way to see this is to condition on $H$. Preliminary changes of variable (merely involving rescaling $H$ and $W$ and then shifting to a new origin) reduce the conditional expectation to a positive constant times a two-dimensional integral of the form

$$\mathcal{I}(\lambda)=\iint_{\mathbb{C}}\ \frac{1}{|z|^2} e^{-\lambda |z-1|^2}\ dz d\bar z$$

with $\lambda \gt 0.$

In polar coordinates $(r,\theta),$ $|z|^2 = r^2$ and $|z-1|^2 = r^2 - 2r\cos(\theta)+1,$ and the area element is $dzd\bar{z} = r dr d\theta,$ giving

$$\mathcal{I}(\lambda) = e^{-\lambda}\int_0^{2\pi}d\theta \int_0^\infty \frac{1}{r^2}e^{-\lambda(r^2 - 2r\cos\theta)}\ r\, drd.$$

For $0 \le r \le \sqrt{1 + 1/\lambda} -1 = u(\lambda)\gt 0,$ the expression in the exponent exceeds $-1,$ so we may underestimate this integral by replacing the exponential by $e^{-1}$ and limiting $r$ to this range:

$$\mathcal{I}(\lambda) \ge e^{-\lambda-1}\int_0^{2\pi}d\theta \int_0^{u(\lambda)}\frac{1}{r}dr = 2\pi e^{-\lambda-1} \lim_{\epsilon\to 0} \int_\epsilon^{u(\lambda)} \frac{dr}{r}\ \propto\ \lim_{\epsilon\to 0}\log(u(\lambda)) - \log(\epsilon),$$

which diverges to $+\infty.$

Since all conditional expectations are infinite, the expectation must be infinite.

A simulation bears this out. For simplicity I chose $H$ and $W$ to have independent standard (Complex) Normal normal distributions, generated twenty million realizations $(h,w),$ and computed the running mean of $|h|^4/|h+w|^2.$ The periodic large jumps are characteristic of a divergent expectation: no matter how far out you run this simulation, these jumps will recur (whenever a tiny value of $|w+h|$ is generated compared to $|h|^2$) and its mean will never converge.

enter image description here

This plot shows the running mean "Mean" as a function of the number of simulated values "N" for $n=10^4$ through $n=2\times 10^7.$ Colors highlight the largest jumps. Evidently one could be fooled by relying on a simulation to estimate the mean: notice how the purple segment from $N\approx 508,000$ to $N\approx 9,300,000$ seems to settle down--only to be followed by a large jump. This indicates that the simulation-based estimate depends entirely on when you choose to end the simulation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ With all due respect, this kind of felt like a homework problem, yet you gave a full answer. I thought this site was supposed to be for problems you actually encounter while doing statistical work, but the OP provided no context and was not asked to. $\endgroup$ – within_person Jun 13 '18 at 14:31
  • 1
    $\begingroup$ @within I can see your point of view. However, the absence of any comments to that effect or votes to close after 43 views suggests it might be a small minority opinion. What I found most interesting about the question was the claim by the OP in a comment about simulation results: for me, that sufficed to establish a statistical component. That, I hope, also explains why I provided only a sketch of the theoretical answer. If this is homework then the OP still has considerable work to do but, if as I suspect, it is not homework then he will have no problems filling in the gaps. $\endgroup$ – whuber Jun 13 '18 at 14:52
  • 1
    $\begingroup$ @within_person Sorry, I might have failed to give details. The problem is related to a research problem on channel capacity for massive mimo systems. This is just a small part of the whole problem. $\endgroup$ – Felipe Augusto de Figueiredo Jun 13 '18 at 17:33
  • $\begingroup$ @whuber, It takes almost no effort to identify questions on this site that are not aligned with stated rules, but still lack close votes or commenting indicating a problem. So I don't know if that appeal to authority/populist argument there really means much. Speaking of authority, you might underestimate your own power to give credibility to a question that undermines site rules simply by answering it. I don't spend a lot of time here but clearly you're the site's standard bearer. Anyway, I don't mean to start an argument. This is a good answer. $\endgroup$ – within_person Jun 13 '18 at 18:58
  • $\begingroup$ @Within I would hope to be the last person to appeal to authority and I respect individual opinions especially when they are well supported. I have to accept that the site's users make the rules: that's the thrust of the argument about lack of comments or votes. I have little other means of gauging user sentiment overall. And I would be the first to recognize that my efforts at interpreting such sentiment are imperfect. We do have a Cross Validated Meta site for raising and debating general issues of this nature. $\endgroup$ – whuber Jun 13 '18 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.