Given the following function of random variables

$$g = \frac{1}{n} \sum_{k=1}^{n}{|h_k|\exp\left( j \theta_k \right)},$$ where $h_1, \cdots, h_n$ are i.i.d. random variables following the complex Gaussian distribution $\mathcal{CN}(0,\beta)$ and $\theta_1, \cdots, \theta_n$ are i.i.d. random variables with probability density function (PDF) given by $\frac{1}{2\pi}$ ($i.e.$, the uniform distribution). Additionally, we assume that $h_k$ and $\theta_k$ are independent for all values of $k$.

What would be the PDF of $g$ for small $n$ ($e.g.$, n = 2) and for the case where $n \gg 1$?

This problem arises from the study on wireless communications channels and is of great importance to the research community.

  • $\begingroup$ Do you mean $i$ instead of $j$ in the definition of $g$? $\endgroup$ Dec 8 '19 at 18:28
  • $\begingroup$ @MickyboYakari, that "j" is the representation of $\sqrt{-1}$ $\endgroup$ Dec 8 '19 at 18:31
  • $\begingroup$ doesn't $|h_k|e^{i\theta_k}$ have the same distribution as $h_k$ --- so you just have a sum of i.i.d. complex Gaussians, which is again Gaussian? $\endgroup$ Dec 8 '19 at 18:43
  • $\begingroup$ @CarloBeenakker, in fact, $h_k$ has Nakagami distribution, however, I thought it would be better to start with a complex Gaussian. Sorry, but I'm failling to see how $|h_k|e^{j\theta_k}$ is equal to $h_k$. $\theta_k$ is uniform. Could you shed some light on it, please? $\endgroup$ Dec 8 '19 at 18:51
  • $\begingroup$ @CarloBeenakker, now I see. Thanks. $\endgroup$ Dec 8 '19 at 20:00

For the complex Gaussian distribution the real and imaginary parts of $h_k$ are i.i.d. with a normal distribution; the absolute value $|h_k|$ has distribution $P(|h_k|)=|h_k|\exp(-|h_k|^2/2)$ and the argument $\phi_k={\rm arg}\,h_k$ is uniformly distributed in $(0,2\pi)$, independently of $|h_k|$. So to generate the random variable $G_k=|h_k|e^{i\phi_k}$ you may either draw a real random number $|h_k|$ with distribution $P(|h_k|)$ and a second independent random number $\phi_k$ uniformly in $(0,2\pi)$, or equivalently draw the complex variable $G_k$ directly from a complex Gaussian.

Then $g$ is a sum of independent complex Gaussians $G_k$, which is again a complex Gaussian.

  • $\begingroup$ Do you think it is possible to find the PDF with $h_k$ following a Nakagami distribution? $\endgroup$ Dec 8 '19 at 20:01

For the general case of Nakagami distribuitons, it's best to go back to the original source. Suppose that the $h_k$ have Nakagami distribution with parameters $(m_k,\Omega_k)$. In the article "The $m$-Distribution-The General Formula of Intensity Distribution of Rapid Fading", Nakagami works out the pdf for the the distribution of the amplitude of $H =\sum_{k=1}^{n}{|h_k|\exp\left( j \theta_k \right)}$ as $$p_{|H|}(r)=r\int_0^{\infty}\prod_{k=1}^n {}_1F_{1}\left(m_k;1;-\frac{\Omega_k}{4m_k}x^2\right)J_0(rx)xdx$$ where ${}_1F_1$ is the confluent hypergeometric function, and $J_0$ is the zeroth order Bessel function. It is also remarked that this can actually be approximated reasonably well by a Nakagami distribution itself, taking new parameters $\tilde\Omega=\sum_{k=1}^n \Omega_k$ and $$\tilde m=\frac{(\sum_{k=1}^n \Omega_k)^2}{\sum_{k=1}^n\frac{\Omega_k^2}{m_k}+\sum_{k=1}^n\sum_{l\neq k}\Omega_k\Omega_l}.$$ So, as a corollary $|H|^2$ can be approximated by a Gamma distribution with these parameters $\left(\tilde m, \frac{\tilde \Omega}{\tilde m}\right)$. For a rigorous derivation of the exact pdf above, as well as a discussion of how good the approximation is in various cases see the article "Accurate Error-Rate Performance Analysis of OFDM on Frequency-Selective Nakagami-m Fading Channels".


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