Let $X$ be a random variable having a normal density and consider the random variable $Y = e^X$. Then $Y$ has a log normal density. Find this density of $Y$.
If $Z$ is the standard normal distribution and $X$ has $\mu$ and $\sigma$:
$P(Y \leq y) = P(X\leq lny)=P(Z \leq \frac{lny-\mu}{\sigma})=F_Z(\frac{lny-\mu}{\sigma})$
Where $F_Z$ is the cumulative distribution function for the standard normal distribution. However, the solution has:
$F_Y(y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{log^2y}{2}}$ for y > 0
I'm not sure if I did something wrong but from my research it seems like the standard normal distribution is difficult to integrate from its PDF. If my answer is correct, I'm not sure how to arrive at the $F_Y$ in the actual solution from $F_Z$
Edit: Per the comment below I should be solving for the probability density, not the CDF:
$$f_Y(y)=\frac{dF_Z(\frac{lny-\mu}{\sigma})}{dy}=f_Z(\frac{lny-\mu}{\sigma})\frac{1}{y\sigma}$$ $$f_Y(y)=\frac{1}{y\sigma\sqrt{2\pi}}e^{-(\frac{lny-\mu}{\sigma})^2/2}$$
This still doesn't match the solution?