Find the density function of $Y=e^Z$ where $Z \sim N(\mu, \sigma^2)$ Find the density function of $Y=e^Z$  ,where $Z \sim N(\mu, \sigma^2)$ This is called the lognormal density , Since $\log Y$ is normally distributed.
 A: We describe a method that is not the most efficient, but lets us keep good control over what's happening. 
Let $g(x)$ be the density function of our normal. You probably know that $g(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp((x-\mu)^2/2\sigma^2))$.
We want the density function $f_Y(y)$ of $Y$. First we find an expression for the cumulative distribution function $F_Y(y)$ of $Y$.  
Note that $Y$ is always positive. So if $y\le 0$, we have $F_Y(y)=0$. Suppose now that $y\gt 0$. We have
$$\Pr(Y\le y)=\Pr(e^Z\le y)=\Pr(Z\le \ln y)=\int_{-\infty}^{\ln y} g(x)\,dx.$$
Thus, if $y\gt 0$, 
$$F_Y(y)=\int_{-\infty}^{\ln y} g(x)\,dx.\tag{1}$$
Now to find the density function $f_Y(y)$, we differentiate $F_Y(y)$. If $y\le 0$, then $f_Y(y)=0$. For $y\gt 0$, we use the expression (1).
To do the differentiation, we use the Fundamental Theorem of Calculus. Because in (1) we are integrating up to $\ln y$, we will also have to use the Chain Rule. We get, for $y\gt 0$,
$$f_Y(y)=\frac{1}{y}g(\ln y).$$
A: Use the chain rule:
$$
f_Y(y)=\frac{d}{dy}\Pr(Y\le y)=\frac{d}{dy}\Pr(e^Z\le y)=\frac{d}{dy}\Pr(Z\le \log y)
$$
$$
=\frac{d}{dy}\Phi(\log y) = \varphi(\log y)\cdot\frac{d}{dy}\log y = \frac{1}{\sqrt{2\pi}}e^{-(\log y)^2/2}\cdot \frac1y.
$$
