Random variable as source of variance for a second random variable?

Question

I am trying to obtain the pdf of a random variable with a Gaussian distribution $X \sim \mathcal{N}(0, A)$, whose variance A is determined by another random variable A with an exponential distribution with $\lambda=1$.

So the process would be the following:

1. Obtain a particular value $a$ from random variable $A$
2. Obtain a particular value $x$ from random variable $X$ with variance $a$

I have found that the correct pdf for random variable X is:

$$ \frac{1}{\sqrt{2}}\exp\left(-\sqrt{2}|x|\right) $$

And I have also checked that this formula is correct doing a simulation. But, how can I compute this pdf analitically?

Simulated histogram + plotted pdf

Thanks!

Answer 1

Given $A$, $X$ has same distribution as $\sqrt A Z$ where $Z$ has standard normal distribution. So the coniditional density of $X$ is $\frac 1 {\sqrt A} \phi (\frac x {\sqrt A})$ where $\phi$ is the standard normal density. Finally the density of $X$ is $\int_0^{\infty} \frac 1 {\sqrt a} \phi (\frac x {\sqrt a})e^{-a}da$. I will let you carry out this integration.