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?
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.
