I've got this density function of a log-normal random variable.
$$f_X(x;\sigma)=\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{\ln(x)^2}{2\sigma^2}}$$
I'm trying to find the density function of $Y = \ln X$ and show that $Y$ is distributed $\operatorname{N}(0,\sigma^2)$.
I know the normal distribution is as follows:
$$f_X(x;\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x – \mu)^2}{2\sigma^2}}$$
But I'm not sure how to proceed from the first equation to the second.
Any help appreciated
Thanks!
Best Answer
Let $Y=\ln X$, hence, $$ F_Y(y) = P(Y\le y)=P(X\le e^y)=F_X(e^y), $$ thus the density can be found as its derivative w.r.t $y$, i.e., $$ f_Y(y)=f_X(e^y)e^y=\frac{e^y}{e^y\sqrt{2\pi \sigma^2}}\exp\{-(\ln e^y)^2/(2\sigma^2)\} = \frac{1}{\sqrt{2\pi \sigma^2}}\exp\{-y^2/(2\sigma^2)\}, $$ hence, $Y \sim N(0,\sigma^2)$.