Is there a way to derive the probability density function of a random variable Z given by $Z = f(X,Y)$ where $X$ and $Y$ are two independent random variables distributed normally and lognormally respectively and $f$ is some arbitrary non-linear function?

I am able to get an idea of the distribution using a Monte-Carlo simulation but I can't seem to find a way to calculate the resulting PDF analytically or at least semi-analytically with some numerical method.

  • $\begingroup$ For an arbitrary function $f$, there is probably no hope of finding of finding the density analytically. $\endgroup$ – Math1000 Dec 21 '19 at 19:41
  • $\begingroup$ @Math1000 Is there not even some expression that could be evaluated numerically? $\endgroup$ – Alex G Dec 21 '19 at 19:45
  • $\begingroup$ @AlexG You may find a good approximation of the resulting PDF of $Z$ as a sum of weighted (and shifted) Gaussians. $\endgroup$ – Nash J. Dec 22 '19 at 0:42

Consider the CDF of $Z$:

$$ \begin{align} G_Z(z) &= \Pr\{Z \leq z\} \\ &= \Pr\{f(X, Y)\leq z\} \\ &= \int_0^{+\infty} \Pr\{f(X, y)\leq z\}g_Y(y)dy \end{align}$$

where $g_Y$ is the pdf of $Y$. Next you would need to find set $h(y, z)$ such that $$ f(X,y) \leq z \iff X \in h(y,z)$$

You may also consider conditional on $X$ in the last step, see if the $h$ is more convenient to solve. To find the pdf $g_Z(z)$, differentiate both sides with respect to $z$:

$$ g_Z(z) = \frac {\partial} {\partial z} G_Z(z) = \int_0^{+\infty} \frac {\partial} {\partial z} \Pr\{X \in h(y,z)\}g_Y(y)dy $$

In general $h$ is a measurable Borel set, so the probability in the integrand can be expressed as a series of CDF of $X$. As long as the set is not singular like Cantor, you should be able to differentiate it. Any the last part is to evaluate the integral numerically (analytically if possible).


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