Show that a scale mixtures of normals is a power exponential

Question

I'm trying to show that a scale mixture of normals yields a Laplace distribution. I've gotten to the point where I have $\int N(0,\tau)\times Ga(\tau\:;\:1,\frac{\lambda^{2}}{2}) \:d\tau$ should equal a Laplace. It's unclear to me how to solve the integral.

Alternatively, I've seen a derivation of the scale of mixture normals as a power exponential here, with certain parameter values yielding the Laplace. However, to be honest it's hard to follow. Can someone please help decipher this derivation or provide a clue as to how to solve the integral?

Answer 1

The marginal distribution of $\beta$ associated with $$\beta|\tau\sim\mathcal{N}(0,\tau)\quad\tau\sim\mathcal{E}(\lambda^2/2)$$ [with the convention that $\tau$ is the variance] has density \begin{align*}\mathfrak{t}(\beta) &=\int_0^\infty \tau^{-1/2}\varphi(\beta/\sqrt{\tau})\frac{\lambda^2}{2}\exp\{-\lambda^2\tau/2\}\text{d}\tau\\ &=\frac{\lambda^2 e^{-\lambda|\beta|}}{2\sqrt{2\pi}}\int_0^\infty \tau^{-1/2}\exp\left\{-\frac{1}{2}\left(\frac{|\beta|}{\sqrt{\tau}}-\lambda\sqrt{\tau}\right)^2\right\}\text{d}\tau\\ \end{align*} [where the $\lambda|\beta|$ appears by creating a perfect square in the exponential]. This suggests the change of variable $\nu=\sqrt{\tau}$ and leads to $$\mathfrak{t}(\beta) = \frac{\lambda^2 e^{-\lambda|\beta|}}{2\sqrt{2\pi}}\int_0^\infty \exp\left\{-\frac{1}{2}\left(\frac{|\beta|}{\nu}-\lambda\nu\right)^2\right\}\text{d}\nu$$ [since $\tau^{-1/2}\text{d}\tau=2\text{d}\nu$]. This further suggests the change of variable $$\zeta=\frac{|\beta|}{\nu}-\lambda\nu$$ with its inverse $$\nu=\left\{-\zeta+\sqrt{\zeta^2+4\lambda|\beta|} \right\}\big/2\lambda$$ [obtained by solving a second degree polynomial equation] and the Jacobian $$\frac{\text{d}\nu}{\text{d}\zeta}=\left\{-1+\frac{\zeta}{\sqrt{\zeta^2+4\lambda|\beta|}} \right\}\big/2\lambda$$ which is always negative. Hence \begin{align*}\mathfrak{t}(\beta)&=\frac{\lambda^2 e^{-\lambda|\beta|}}{4\lambda\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left\{-\frac{\zeta^2}{2}\right\}\left\{1-\frac{\zeta}{\sqrt{\zeta^2+4\lambda|\beta|}} \right\}\text{d}\zeta\\ &=\frac{\lambda e^{-\lambda|\beta|}}{2}\int_{-\infty}^\infty \left\{1-\frac{\zeta}{\sqrt{\zeta^2+4\lambda|\beta|}} \right\}\varphi(\zeta)\text{d}\zeta\\ &=\frac{\lambda e^{-\lambda|\beta|}}{2}\left\{1-\int_{-\infty}^\infty\frac{\zeta\varphi(\zeta)}{\sqrt{\zeta^2+4\lambda|\beta|}}\text{d}\zeta\right\}=\frac{\lambda e^{-\lambda|\beta|}}{2}\end{align*} [since the integrand is an odd function of $\zeta$ in the last integral]. This establishes [without complex calculus] that the marginal distribution of $\beta$ is indeed a Laplace or double-exponential distribution.

Answer 2

I've always found the direct integration in this case to be a complicated integral. The Moment Generating Function (MGF) approach works too.

The MGF: $\beta | \tau \sim N(0, \tau)$ and then

$M_{\beta|\tau}(t)=e^{\frac{\tau t^2}{2}}.$ Now to get the MGF of $\beta$ marginally, take the expectation with respect to $\tau$.

$$\mathbb{E}(M_{\beta|\tau}(t)) = \int_0^\infty e^{\frac{\tau t^2}{2}} \frac{\lambda^2}{2}e^{-\tau \frac{\lambda^2}{2}}d\tau =\int_0^\infty \frac{\lambda^2}{2} e^{-\tau \left(-\frac{t^2}{2} +\frac{\lambda^2}{2}\right)} d\tau =\frac{\lambda^2/2}{\lambda^2/2 - t^2/2} =\frac{1}{1 - \frac{t^2}{\lambda^2}}, $$ Now you can recognize this last function as the MGF of a Laplace (double exponential) distribution.