If a random variable $W$ is Normally distributed, then $\exp(W)$ is Log-Normally distributed.
However, the pdfs of these two random variables differ by a factor of $\exp(W)^{-1}$.
The Normal pdf for $W$ is
$$P(w \in W) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(w – \mu)^2}{2 \sigma^2}}$$
but the Log-Normal pdf for $\exp(W)$ is
$$P(\exp(w) \in \exp(W)) = \frac{1}{\exp(w)\sigma\sqrt{2\pi}}\ e^{-\frac{\left(w-\mu\right)^2}{2\sigma^2}}$$
Why is this the case? Shouldn't they be equivalent?
Best Answer
I'd like to provide a simple and straightforward demonstration to where this factor comes from.
Let $p(x)$ be the PDF of a normal distribution $\mathcal{N}\left(\mu,\sigma^2\right)$. Then, of course:
$$\int\limits_{-\infty}^{+\infty} p(x)dx = \frac{1}{\sqrt{2\pi}\sigma} {\rm e}^ {-\frac{\left( x-\mu \right)^2}{2\sigma^2} } {\rm d}x = 1$$
If you now make a substitution $x = \log y$, i.e. $x$ is distributed normally, and $y$ has a log-normal distribution, you'll get
$$\int\limits_{-\infty}^{+\infty} p(\log y){\rm d} (\log y) = 1 = \frac{1}{\sqrt{2\pi}\sigma} {\rm e}^ {-\frac{\left( \log y-\mu \right)^2}{2\sigma^2} } \boxed{{\rm d}(\log y)} $$
and the boxed term is $\frac{{\rm d}y}{y}$, showing where does the $\frac{1}{y}$ come from. So that the PDF of $y$ is
$$\frac{1}{\sqrt{2\pi}\sigma y} e^ {-\frac{\left( \log y-\mu \right)^2}{2\sigma^2} }$$