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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?

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  • $\begingroup$ First note that the function required by the question, anb the solution given here is the pdf, not the CDF, Once you obtain the CDF of $Y$, you differentiate with respect to the argument of it, i.e. the dummy variable $y$, by applying the chain rule. Of course in the process you will need to know the pdf of standard normal $Z$ as well, which is the derivative of $F_Z$. $\endgroup$ – BGM May 6 at 8:17
  • $\begingroup$ @BGM I attempted that (in my edit) and it still doesn't match the solution. I'm not sure where I went wrong. $\endgroup$ – Yandle May 7 at 3:40
  • $\begingroup$ What you've written seems to be correct, as per the wiki article en.wikipedia.org/wiki/Log-normal_distribution $\endgroup$ – rubikscube09 May 7 at 3:45

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