# Derive compound Gaussian distribution consisting of a complex normal distribution with normal distributed variance

We want to derive a distribution from two Gaussian distributions using the definition of a compound distribution $$\qquad \qquad \qquad \quad f(z) = \int g(z|\theta) h(\theta) \,\text{d}\theta \qquad \qquad \qquad (*)$$ More specific, we would like to use $$(*)$$ when $$g(z|\sigma)$$ is the pdf of a 1D complex normal distribution (ie $$Z \sim \mathcal{CN}(0, |\sigma|^2)$$), where $$\Sigma \sim N(0, \tilde{\sigma}^2)$$ with pdf $$h(\sigma)$$. The complex normal distribution will be defined $$g(z|\sigma) = \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2}, \qquad z\in\mathbb{C},$$ where $$\sigma$$ is normally distributed $$h(\sigma) = \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right), \qquad \sigma\in\mathbb{R}.$$

From the definition in $$(*)$$ we then get an integral \begin{align} \int_{-\infty}^\infty \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2} \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right) \, \text{d} \sigma = 2 \int_{0}^\infty \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2} \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right) \, \text{d} \sigma, \end{align} where the last step comes from that the function is even.

Hence, we end up with an integral similar to the one below. $$\int_0^\infty \exp\left(-\frac{1}{\sigma^2} - \sigma^2\right)\frac{1}{\sigma^2}\, \text{d} \sigma.$$ And if we make the substitution $$x = 1/\sigma$$ => $$\text{d}x = -1/\sigma^2 \text{d} \sigma$$, we get $$-\int_{-\infty}^0 \exp\left(-x^2-\frac{1}{x^2}\right)\, \text{d} x$$ According to integral calculator this should have a solution (but not an analytical solution) consisting of erfc factors. I suppose you can rewrite this as a normal distribution but I cant understand how exactly.

• According to Maple the resulting distribution will be a Weibull distribution. Nov 25, 2023 at 20:22
• Maybe I’m misreading, but it seems you’re saying the covariance is normally distributed? Nov 25, 2023 at 20:26
• Probably bad notation from my part to use $\Sigma$. What I mean is that the 1D complex normal distribution a has variance which is normally distributed. Nov 25, 2023 at 20:41
• So the variance can take negative values in your model, right? Hm. Nov 25, 2023 at 22:02
• Aah now I see what you meant by variance can take negative values. Once again, my bad. My previous comment meant to say the 1D complex normal distribution a has standard deviation which is normally distributed Nov 27, 2023 at 20:12

How to evaluate $\int_{0}^{\infty}\exp(-x^2-1/x^2)dx$?
How should I calculate $\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx$
The answer itself is the complex Weibull distribution \begin{align*} f(z) &= \frac{1}{\sqrt{2} \pi \tilde{\sigma}} \frac{1}{|z|} \exp\left(- \frac{\sqrt{2} |z|}{\tilde{\sigma}}\right), \end{align*}