PDF of $e^X$ where $X$ is normally distributed

Question

I am having a little bit of trouble with the probaility density function of compositions of random variables.

Let $X$ be a normally distributed random variable with with expected value $\mu $ and variance $\sigma^2$. Furthermore let $Y:=e^X$. Determine the probability density function of the random variable Y.

My first idea was to write:

$P(Y\leq t)=P(e^X\leq t)=P(X\leq \log(t))=...$

and then use the fact that X is normally distributed. My questions is, if my idea is right or, if there is another way to determine this. Also I think I have to worry about the fact that this fails if $t\leq0$.

Answer 1

$P(X \leq \log t)=\int_{-\infty}^{\log t} f(x) \, dx$ where $f(x)=\frac 1 {\sqrt {2\pi}}e^{-(x-\mu)^{2}/2\sigma^{2}}$. Differentiate to get the density as $\frac {f(\log t)} t$.

Answer 2

You are on the right track. P(q) is, as you note, shorthand for $P(Q<=q)$ where Q is the RV (you used $t$, which is OK, but it's a lot easier to keep track of what is happening if you use the same letter.)

This is the lognormal return distribution. For lots of explanation and help, see: https://en.wikipedia.org/wiki/Log-normal_distribution