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


  • $\begingroup$ Your question is not clear to me. It looks like you've already found the analytical solution, which is $\exp\left(-\sqrt{2}|x|\right)/\sqrt{2}$. $\endgroup$ – Wood May 13 at 22:31
  • $\begingroup$ I have found the final solution and checked that indeed it is correct, but I would like to know how to derive this pdf by myself. $\endgroup$ – Iyán May 13 at 22:34

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.

  • $\begingroup$ Thanks for your answer Kavi! Just one small additional question: why is $\frac{1}{\sqrt{A}}\Phi\left(\frac{x}{\sqrt{A}}\right)$ instead of $\frac{1}{A}\Phi\left(\frac{x}{A}\right)$, since A is the variance? $\endgroup$ – Iyán May 14 at 10:14
  • $\begingroup$ The density of $tX$ is $\frac 1 t f_X(\frac x t)$. Note that $\frac 1 A \Phi(\frac x {\sqrt A})$ does not even inetgrate to $1$ so it is not a density funtion. $\endgroup$ – Kabo Murphy May 14 at 10:19

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