How to prove Laplace distribution is scale mixture of Gaussians?? How does one prove the Laplace distribution is a scale mixture of Gaussians?
I.e, how does one show that $X \sim \text{Laplace}(\lambda)$ is a scale mixture of Normal
$Y \sim N(0,\tau)$ and exponential $\tau \sim \text{Exp}(λ^2/2)$?
 A: If $X$ is a Laplace random variable, it can be thought of as a normal rv whose variance is distributed exponentially. I think of it in relation to a random walk. If a particle walks for a fixed length of time, its final location is given by a Gaussian (in a limiting sense). If it walks for an exponential length of time, its final location is Laplace.
Let $\tau\sim\text{Exp}(\alpha)$, and let $X\sim N(0,2D\tau)$; $\tau$ is the length of time the particle randomly wanders around before settling on its final location $X$, and $D$ is its diffusion coefficient (which can be thought of as how fast it walks). I've studied this in the context of biological dispersal, say a parent plant at location $0$ reproduces and its seeds randomly disperse by the wind (assuming no bias in wind direction).
$$
\begin{aligned}
f_X(x)
&=\int_{0}^\infty f_X(x|\tau=s)dP(\tau=s) \\
&=\int_0^\infty \frac{1}{\sqrt{4\pi D s}}e^{-\frac{x^2}{4Ds}}\alpha e^{-\alpha s} \ ds.
\end{aligned}
$$
This integral involves error functions, so we'll turn to looking at moment generating functions:
$$
\begin{aligned}
M_X(t)=\int_{-\infty}^\infty e^{tx}f_X(x)\ dx
&=\int_{-\infty}^\infty e^{tx}\int_0^\infty \frac{1}{\sqrt{4\pi D s}}e^{-\frac{x^2}{4Ds}}\alpha e^{-\alpha s} \ ds \ dx \\
&=\int_0^\infty \alpha e^{-\alpha s} \int_{-\infty}^\infty e^{tx}\frac{1}{\sqrt{4\pi D s}}e^{-\frac{x^2}{4Ds}} \ dx \ ds \\
&=\int_0^\infty \alpha e^{-\alpha s} e^{Dst^2} \ ds \\
&=\left[\frac{\alpha}{\alpha-Dt^2} e^{(-\alpha +Dt^2)s} \right]_0^\infty \\
&=\frac{1}{1-\frac{D}{\alpha} t^2}, \ \text{ as long as } \ t^2<\sqrt{\frac{D}{\alpha}}.
\end{aligned}
$$
This is exactly the moment generating function for the mean zero Laplace distribution with rate parameter $\sqrt{\frac{\alpha}{D}}$. Thus 
$$
f_X(x)=\frac{1}{2}\sqrt{\frac{\alpha}{D}} e^{−\sqrt{\frac{\alpha}{D}} \ |x|}.
$$
For your parameters, $D\lambda^2=\alpha$ and $D=1/2$, thus you get $2D\tau=\tau$ for the variance of your normal distribution. The technique here should also work if the mean isn't zero.
If you want to do the calculation to get the Laplace distribution explicitly, a contour integral in the complex plane works. The calculation is nearly identical to this one for the Cauchy distribution.
