# PDF of $g = \frac{1}{n} \sum_{k=1}^{n}{|h_k|\exp\left( j \theta_k \right)}$?

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.

• Do you mean $i$ instead of $j$ in the definition of $g$? Dec 8 '19 at 18:28
• @MickyboYakari, that "j" is the representation of $\sqrt{-1}$ Dec 8 '19 at 18:31
• 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? Dec 8 '19 at 18:43
• @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? Dec 8 '19 at 18:51
• @CarloBeenakker, now I see. Thanks. 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.
• Do you think it is possible to find the PDF with $h_k$ following a Nakagami distribution? 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".