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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:

example

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="")
Alexis
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Sextus Empiricus
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    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 06:22
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    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
  • @SextusEmpiricus I don't understand the following part of the question: "variance distributed according to the absolute value or square of another Gaussian distribution". The absolute value will be a half-Normal, whereas the square will be a Chi-squared. Also, are you assuming that the mean is zero? If the mean is not zero, then the result cannot be a Product Normal (as stated in the answer below), as the Product Normal assumes zero means. – wolfies Nov 09 '19 at 16:21
  • @wolfies indeed the half normal would be the same. The mean was assumed to be zero (for my particular [case](https://stats.stackexchange.com/questions/422897) which was a failure anyway). But with nonzero mean you could have still used the product normal (and apply a shift afterwards). – Sextus Empiricus Nov 09 '19 at 17:10
  • I used absolute value and square, both, to generalize the question. They were nor supposed to refer to the same distribution. – Sextus Empiricus Nov 09 '19 at 17:14

1 Answers1

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

steveo'america
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