# Pdf of a function of a random variable

• May 6th 2011, 06:55 PM
Jame
Pdf of a function of a random variable
I am supposed to fin the pdf f $\displaystyle Y = e^X$ where $\displaystyle X$ is normal r.v. with parameters $\displaystyle \mu, \sigma^2$

The pdf of X is

$\displaystyle \frac{1}{ \sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

(Sorry I can't get this to appear in tex for some reason)

If I integrate this function from negative infinity to ln(y), then differeniate with respect to y, I should get the function I seek.

However I am not sure how to integrate the normal distribution function, could someone please explain to me some way I way go about doing that?

Thank you for your help.
• May 6th 2011, 10:54 PM
Sambit
Let $\displaystyle F_Y(.)$be cdf of $\displaystyle Y$ and $\displaystyle F_X(.)$ be cdf of $\displaystyle X$; $\displaystyle f_Y(.)$ and $\displaystyle f_X(.)$ are pdf of $\displaystyle Y$ and $\displaystyle X$ respectively.

$\displaystyle F_Y(y) = P(Y\leq y) = P(e^X\leq y) = P(X\leq \ln y) = F_X(\ln y)$

Hence,differentiating, $\displaystyle f_Y(y) = \frac{d}{dx}F_X(\ln y) = f_X(\ln y)\frac{d}{dx}\ln y = f_X(\ln y)\frac{1}{y}$
$\displaystyle = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}}\frac{1}{y} = \frac{1}{y\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}$

This is called Log-Normal distribution.
• May 7th 2011, 09:58 PM
matheagle
minor correction, the derivative of F(y) wrt y is f(y)
you have d/dx, you mean d/dy.
Otherwise you would need the chain rule

This is what you meant....

Hence,differentiating, $\displaystyle f_Y(y) = \frac{d}{dy}F_X(\ln y) = f_X(\ln y)\frac{d}{dy}\ln y = f_X(\ln y)\frac{1}{y}$