# Compounding a Gaussian distribution with variance distributed according to the absolute value of another Gaussian distribution

Have there been earlier descriptions of the following compound distribution?

Compounding a Gaussian distribution with variance distributed according to the absolute value or square of another Gaussian distribution:

$$f(y) = \int \phi(x)\frac{\phi(y/\vert x\vert)}{\vert x \vert} dx$$

This distribution would look like this:

and it possibly relates to a solution for this question Limiting distribution of $\frac1n \sum_{k=1}^{n}|S_{k-1}|(X_k^2 - 1)$ where $X_k$ are i.i.d standard normal

compGaus <- function(x) {
dy <- 0.1
y <- seq(-16,16,dy)
out <- 0
for (ys in y) {
var = ys^2
dens = dnorm(ys,0,1)
if (var>0) {
out = out + dens*dnorm(x/var^0.5,0,1)/var^0.5*dy
}
}
out
}
compGaus <- Vectorize(compGaus)

x <- seq(-2,2,0.1)
plot(x,compGaus(x),log="")

• This reminds me of Bruno Lecoutre's "lambda prime" distribution (more generally "k prime"). – steveo'america Aug 21 '19 at 17:14
• @Glen_b I'll correct that it is the absolute value of the other Gaussian (or a half-Gaussian). – Sextus Empiricus Aug 22 '19 at 6:22
• rather, the square root of a chi-square with one degree of freedom. The "lambda prime" subsumes this case for general degrees of freedom in the denominator (and admits a non-zero gaussian in the numerator via a non-centrality parameter). – steveo'america Aug 22 '19 at 16:28
• hmmm, now that I think about it, I thought this was a ratio, but it isn't. Effectively your $y$ is the product of $|x|$ for normal $x$ and some normal $z$. If both $x, z$ are standard Gaussians, you can ignore the absolute value on the $x$ and this is a standard product normal distribution. – steveo'america Aug 29 '19 at 19:54
• Indeed, you are right. It is a product distribution, besides a compound distribution. – Sextus Empiricus Aug 29 '19 at 20:55

You can treat $$y$$ as $$x z$$, where $$x$$ and $$z$$ are standard Gaussian variables. Thus $$y$$ follows the Product Normal distribution.