Claim: If $Y=e^X$ has a lognormal distribution whereby $X$ ~ $N(\mu,\sigma^2)$, then the pdf of $Y$ is $g(y)=\frac{1}{y\sqrt{2\pi\sigma^2}}exp[-(lny-\mu)^2/2\sigma^2], 0<y<\infty$.
Suppose $Y=u(X)=e^X$, so that $Y=u(X)=\sum \frac{x^n}{n!}$. To show that $u$ is injective, we consider two cases:
If $X\ge0$, then $e^X$ is clearly positive.
If $X<0$, then $e^X=\frac{1}{e^{-X}}$ is also positive.
Thus, $u: S_x \to S_y=(0,\infty)$, and hence $0<y<\infty$.
Because $e^X$ is its own derivative, we have that $e^X$ has positive derivative everywhere. Thus, $u$ is injective.
As $u$ is injective, its left-inverse function is $v:S_y \to S_x$ defined as $X=v(Y)=lnY$ for which $(v \circ u)(X)=ln(e^X)=X, \forall X \in S_x$.
Now, the pdf of $X$ is $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^-(x-\mu)^2/2\sigma^2$, $-\infty<x<\infty$, so
$G(y)=P(Y\le y)=P(e^X \le y)=P(X \le lny)=\int_0^{lny} f(x)dx$.
By the FTOC, we have $g(y)=G'(y)=\frac{d}{dy}\int_0^{lny} f(x)dx=f[ln(y)] \cdot|\frac{1}{y}|=\frac{1}{y\sqrt{2\pi\sigma^2}}exp[-(lny-\mu)^2/2\sigma^2]$, $0<y<\infty$.
Is this correct? I want to make sure my reasoning here is correct, thanks!
