Generating a Laplace prior from a normal random variable with Rayleigh standard deviation. I read on Wikipedia Laplace distribution that the following is true:
If $X|Y \sim N(\mu,\sigma=Y)$ with $Y \sim \text{Rayleigh}(b)$, then $X \sim \text{Laplace}(\mu, b)$.
However, there doesn't seem to be a reference for it, and I can't manage to prove it myself. Can someone show me why this is true?  
 A: You can use the identity 
$$
  I(u,\,v) := \int_0^\infty e^{-u y^2 - v /y^2 } \text{d}y = 
  \frac{\sqrt{\pi}}{2 \sqrt{u}}\, e^{-2 \sqrt{uv}} \qquad (u >0, \, v>0). 
$$
which relates (after the change of variable $t:=y^2$) to the Laplace
transform of the function $t \mapsto t^{-1/2} e^{-v / t}$ and its
value at $s = u$, see N.N. Lebedev Special functions and their
applications,
footnote of p. 118. The Laplace transform
can be found in the famous book by 
Abramowitz and Stegun as formula (29.3.84) p. 1026.
Now
$$
  p(x) = \int_0^\infty p(x \vert y) \, p(y)\,\text{d}y =
  \int_0^\infty \frac{1}{y \sqrt{2 \pi}}\, e^{-(x-\mu)^2/(2y^2)} 
      \times \frac{y}{b^2} \, e^{-y^2/(2b^2)} \,\text{d}y 
$$
and thus $p(x) = I(u,\,v)/(b^2 \sqrt{2 \pi})$ where $I(u,\,v)$ is the
integral above for $u:= 1/(2b^2)$ and $v:= (x - \mu)^2/2$. Using the
value of $I(u,\,v)$ and rearranging, we find $p(x) = \exp\{-|x -
\mu|/b\} / (2b)$ as you claimed.  I do not know how the Laplace transform 
was computed, nor if a simpler derivation exists.
