# how to calculate the posterior distribution with known mean when variance has exponential distribution?

Here's the problem:

$Y\sim\operatorname{Exp} (1)$, $X|Y\sim N(0,Y)$. The joint distribution of $X$ and $Y$: $f(x,y) = \frac{1}{\sqrt{2\pi y}} e^{-\frac{x^2}{2y} + y}$.

I'm asked to derive the marginal density of $X$. I began with the MGF of $X$ and got $\mathbf{E}\left[e^{tx}\right] = \frac{1}{t^2/2 - 1}$, but I couldn't identify what distribution this is. Any idea? Thanks!

• if this is HW please add the self-study tag – Antoine Aug 1 '16 at 17:39
• it's not HW. it's the review problem for an exam – Amber Xue Aug 1 '16 at 17:55
• The rules for self-study questions apply to old exam problems, too. Please see stats.stackexchange.com/tags/self-study/info. – Juho Kokkala Aug 1 '16 at 18:06

According to Wikipedia, the Laplace Distribution with parameters $\mu$ and $b$ has the following Moment Generating Function: $$M(t) = \frac{e^{\mu t}}{1 - b^2t^2}$$ Thus if you set $\mu = 0$ and $b = \frac{1}{\sqrt{2}}$, then: $$M(t) = \frac{1}{1 - t^2/2}$$ However, this differs from your answer by a factor of -1