(5.31) Find the probability density function of $Y=\e{X}$ when $X$ is normally distributed with parameters $\mu$ and $\sigma^2$. The random variable $Y$ is said to have a lognormal distribution (since log $Y$ has a normal distribution) with parameters $\mu$ and $\sigma^2$.
Solution
$$ f_X(x)=\ndn $$
Theormem 7.1 gives the following for a strictly increasing/decreasing function $Y=g(X)$:
$$ f_Y(y)=\cases{f_X(g^{-1}(y))\normB{\wderv{y}{g^{-1}(y)}}&y=g(x)\text{ for some }x\\0&y\neq g(x)\text{ for any }x} $$
Let $Y=g(X)=\e{X}$. Then $g$ is strictly increasing on $\wR$. Since $g:\wR\mapsto(0,\infty)$, for every $y>0$ there exists some $x\in\wR$ such that $y=g(x)$ and $g^{-1}(y)=\ln(y)$. But for $y\leq0$, there’s no $x\in\wR$ such that $y=g(x)$. Hence for $y>0$ we have
$$ f_Y(y)=f_X\bop\ln(y)\bcp\normB{\wderv{y}{\ln(y)}} $$
$$ =\ndna{\mu}{\sigma}{\ln(y)}\normB{\frac1y} $$
$$ =\ndna{\mu}{\sigma}{\ln(y)}\normB{\frac1y} $$
$$ =\frac1y\ndna{\mu}{\sigma}{\ln(y)} $$
And
$$ f_Y(y)=\cases{\frac1y\ndna{\mu}{\sigma}{\ln(y)}&y>0\\0&y\leq0} \tag{5.31.1} $$
Alternative solution: for $y>0$, we have
$$ \cdfa{y}{Y}=\pr{Y\leq y}=\pr{\e{X}\leq y}=\pr{X\leq\ln{y}}=\cdfa{\ln{y}}{X} $$
Take the derivative of the distribution to get the density:
$$ \pdfa{y}{Y}=\wderiv{\cdfa{y}{Y}}{y}=\wderiv{\prn{\cdfa{\ln{y}}{X}}}{y}=\wderiv{\cdfa{x}{X}}{x}\bigbar_{x=\ln{y}}\wts\wderiv{\ln{y}}{y} $$
$$ =\pdfa{\ln{y}}{X}\wts\frac1y=\frac1{\sqrt{2\pi}\sigma}\e{-\frac{(\ln{y}-\mu)^2}{2\sigma^2}}\wts\frac1y $$
For $y\leq0$, we have
$$ \cdfa{y}{Y}=\pr{Y\leq y}=\pr{\e{X}\leq y}=0 $$
Take the derivative for $y\leq0$:
$$ \pdfa{y}{Y}=0\dq\text{for }y\leq0 $$
So we get the same result as 5.31.1.