What is the mean of this exponential random variable?

Question

I have read a paper that says that the following is exponentially distributed

$$ Y= \bigl| \sum_{i=1}^n \gamma_i^{-\frac{1}{2}} h_i \bigl|^2$$

where $\gamma_i$ are non-negative constants and

$$h_i \sim \mathcal{CN} (0,1)\,,$$

where $\mathcal{CN}(\mu,\Gamma)$ denotes a circularly symmetric complex normal distribution.

The authors claim that the mean of this exponential distributed $Y$ is $$\sum_{i=1}^n{\gamma_i}^{-1}$$.

Does anyone know why?

My thoughts are the following

1- The sum of Gaussian is also Gaussian

2- Magnitude of Gaussian is Rayleigh distributed

3- Taking the square of Rayleigh is exponential

But how do we get that the mean ?

My second part of this question is

What happens if $$h_{i} \sim \text{Nakagami } m $$

Answer 1

Just addressing the first part for now:

Your line of thinking about why it's exponential is essentially along the right lines, after small modifications to get some details correct (you'll need independence in your step 2 for example, to invoke the Rayleigh).

For constant $c_i$, I think we have that $c_i h_i \sim \mathcal{CN}(0,c_i^2I)$ (where here $\mathcal{CN}$ with two arguments refers to the circularly symmetric complex normal).

This along with linearity of expectation explains where the $\sum \gamma_i^{-1}$ comes from.

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Answer to Q from comments:

how does the scale parameter $γ^{−1}_i$ change the distribution of $|h_i|^2$?

If $h_i^2 \sim \text{Gamma}(\alpha, \beta)$ (where here I mean the shape-scale parameterization, not the shape-rate parameterization), then

$γ^{−1}_i\,h_i^2\sim\text{Gamma}(\alpha, γ^{−1}_i\beta)$.

(If you want the shape-rate parameterization, you multiply the rate parameter by $\gamma_i$)

Answer 2

A standard circularly symmetric complex normal random variable is of the form $Z = X + iY$ where $X$ and $Y$ are independent zero-mean normal random variables with variance $\frac 12$. Here $i = \sqrt{-1}$. We write $Z \sim \mathcal{CN}(0,1)$ and $X, Y \sim \mathcal{N}(0,\frac 12)$. Note that $$E[|Z|^2] = E[X^2+Y^2]= \frac 12 + \frac 12 =1.$$ Note also that $|Z|^2$ is an exponential random variable with parameter (and hence mean) equal to $1$. From this, it is easy to deduce that $\sigma Z \sim \mathcal{CN}(0,\sigma^2)$ and that $|\sigma Z|^2$ is an exponential random variable with mean $\sigma^2$ and hence parameter $\frac{1}{\sigma^2}$.

Thus, since the $H_k$ are independent $\mathcal{CN}(0,1)$ random variables, meaning that $H_k = X_k + i Y_k$ where $X_k$ and $Y_k$ are independent $\mathcal N(0,\frac 12)$ random variables, $\gamma_k^{-1/2}H_k$ is a $\mathcal{CN}(0,\gamma_k^{-1})$ random variable, and so $$Z = \sum_{k=1}^N \gamma_k^{-1/2}H_k = \sum_{k=1}^N\gamma_k^{-1/2}(X_k+iY_k) = \bar{X} + i\bar{Y} \sim \mathcal{CN}\left(0,\sum_k \gamma_k^{-1}\right).$$ It follows that $|Z|^2 = \displaystyle \left|\sum_{k=1}^N \gamma_k^{-1/2}H_k\right|$ is an exponential random variable with mean $\sum_k \gamma_k^{-1}$.

About the only claim that does not follow from simple applications of general rules is that if $X$ and $Y$ are independent zero-mean normal random variables with common variance $\frac 12\sigma^2$, then $X^2+Y^2$ has an exponential distribution. Note that the usual method changing to polar coordinates gives $$P\{X^2+Y^2 > z\} =\int_{\sqrt{z}}^\infty \int_0^{2\pi} \frac{1}{\pi\sigma^2} r\exp(-r^2/\sigma^2) \,\mathrm d\theta \, \mathrm dr = \exp(-z/\sigma^2)$$ showing that $X^2+Y^2$ has an exponential distribution with parameter $\sigma^{-2}$ and hence mean $\sigma^2$.