How to find the PDF for Y=e$^x$

Question

How do I find the PDF for $Y$ = e$^X$ when $X$ is $N$(μ,σ$^2$)

I have seen the problem where $X$ is $N$(0,1), but I am curious on how to find it given just these parameters?

Answer 1

You're looking for the pdf of a lognormal distribution. I would look at the cdf and thus start with $P(Y<z) = P(\exp(X)<z) = P(X < \ln(z))$ for $z>0$, then use the well-known cdf of $N(\mu,\sigma^2)$ and an appropriate integral transformation. Hope this helps.

Answer 2

Let $f(x)$ be the pdf of $N(0,1)$, while $\Phi(x)$ its cdf. Then we have

$$P(Y\leq M) = P(e^X\leq M) = P(X\leq \ln M)=P(\frac{X-\mu}{\sigma}\leq \frac{\ln M-\mu}{\sigma})$$

As $\frac{X-\mu}{\sigma}\sim N(0,1)$, we have

$$P(Y\leq M) = \Phi(\frac{\ln M-\mu}{\sigma})=\int_{-\infty}^{\frac{\ln M-\mu}{\sigma}}f(x)dx$$

By taking the derivative with respect $M$, we have

$$f_Y(M) = \dfrac{1}{\sigma M}f(\frac{\ln M-\mu}{\sigma})$$

Answer 3

Since $Y = \exp\{X\}$ is one-to-one on $(-\infty,\infty)$, you can apply a one-to-one change of variable: $$f_Y(y) = \frac{f_X(\log y)}{\left|\frac{dy}{dx}\right|_{\log y}} = f_X(\log y)\left|\frac{dx}{dy}\right|_{\log y} = \frac{1}{\sqrt{2\pi} \sigma y}\exp\left\{\frac{1}{2}\left(\frac{\log y-\mu}{\sigma}\right)^2\right\}.$$

This is a log-normal distribution.