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.

  • $\begingroup$ According to Maple the resulting distribution will be a Weibull distribution. $\endgroup$ Nov 25, 2023 at 20:22
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    $\begingroup$ Maybe I’m misreading, but it seems you’re saying the covariance is normally distributed? $\endgroup$ Nov 25, 2023 at 20:26
  • $\begingroup$ 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. $\endgroup$ Nov 25, 2023 at 20:41
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    $\begingroup$ So the variance can take negative values in your model, right? Hm. $\endgroup$ Nov 25, 2023 at 22:02
  • $\begingroup$ 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 $\endgroup$ Nov 27, 2023 at 20:12

1 Answer 1


I was unable to find how to link stack exchange posts to each other, but thanks to Did and Tony in

How to evaluate $\int_{0}^{\infty}\exp(-x^2-1/x^2)dx$?

and Matthew Zhukov and xpaul in

How should I calculate $\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx$

for helping us find the answer. If you have time please upvote them.

The answer itself is the complex Weibull distribution $$ \begin{align*} f(z) &= \frac{1}{\sqrt{2} \pi \tilde{\sigma}} \frac{1}{|z|} \exp\left(- \frac{\sqrt{2} |z|}{\tilde{\sigma}}\right), \end{align*} $$


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