If $X \sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:
(i) Find the CDF of of $Y$ and then differentiate. Use the notation $\Phi(x)$ and $\phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $\phi(x) = \Phi'(x)$.
So I'm not sure how to differentiate $\Phi\big(\dfrac{\ln x-\mu}{\sigma} \bigg)$ to get $\dfrac{1}{x\sigma\sqrt{2\pi}}e^-\frac{(\ln x-\mu)^2}{2\sigma^2}$
(ii) Use the transformation formula.
I'm not sure where to even begin with this one.
1 Answer
If $Y=e^X$, then $\varphi^{-1}(Y)= \log Y$. Hence, $f_X (y) =\frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}}$ and differentiate $|\frac{d \varphi^{-1}(Y)}{dy} | = |\frac{1}{Y}|$. Hence, the pdf of $Y$ is $$ h_{Y}(y) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}}\frac{1}{|y|} $$
