Gauss to Laplace transmutation!

When browsing X validated the other day [translate by procrastinating!], I came upon the strange property that the marginal distribution of a zero mean normal variate with exponential variance is a Laplace distribution. I first thought there was a mistake since we usually take an inverse Gamma on the variance parameter, not a Gamma. But then the marginal is a t distribution. The result is curious and can be expressed in a variety of ways:

– the product of a χ²₁ and of a χ₂ is a χ²₂;
– the determinant of a 2×2 normal matrix is a Laplace variate;
– a difference of exponentials is Laplace…

The OP was asking for a direct proof of the result and I eventually sorted it out by a series of changes of variables, although there exists a much more elegant and general proof by Mike West, then at the University of Warwick, based on characteristic functions (or Fourier transforms). It reminded me that continuous, unimodal [at zero] and symmetric densities were necessary scale mixtures [a wee misnomer] of Gaussians. Mike proves in this paper that exponential power densities [including both the Normal and the Laplace cases] correspond to the variances having an inverse positive stable distribution with half the power. And this is a straightforward consequence of the exponential power density being proportional to the Fourier transform of a stable distribution and of a Fubini inversion. (Incidentally, the processing times of Biometrika were not that impressive at the time, with a 2-page paper submitted in Dec. 1984 published in Sept. 1987!)

This is a very nice and general derivation, but I still miss the intuition as to why it happens that way. But then, I know nothing, and even less about products of random variates!