# Transforming random variable and finding PDF and Expected Value

• Jun 6th 2010, 05:53 PM
acevipa
Transforming random variable and finding PDF and Expected Value
Suppose X is a continuous random variable with probability density $\displaystyle f$, and $\displaystyle \phi$ is a continuous increasing differentiable function.

1) Show that the probability density of $\displaystyle g$ of $\displaystyle Y=\phi(X)$ is given by:

$\displaystyle g(y)=f(\phi^{-1}(y))\psi(y)$

where $\displaystyle \phi^{-1}$ is the inverse of $\displaystyle \phi$, and

$\displaystyle \psi(y)=\frac{d}{dy}\phi^{-1}(y)$

2) Show that the expected value of Y is:

$\displaystyle E(Y)=\int_{-\infty}^{\infty} \phi(x)f(x) dx$
• Jun 6th 2010, 09:31 PM
matheagle
(1) Just differentiate the CDFs...

$\displaystyle g(y)= {d\over dy} F_Y(y)={d\over dy}P(Y\le y)={d\over dy}P(\phi(X)\le y)$

$\displaystyle ={d\over dy}P(X\le \phi^{-1}(y))={d\over dy}F_X(\phi^{-1}(y))$

Now use the chain rule..

$\displaystyle =f(\phi^{-1}(y)) \frac{d}{dy}\phi^{-1}(y)=f(\phi^{-1}(y))\psi(y)$

This works as well if $\displaystyle \phi$ is decreasing.

(2) Is obvious if you just replace $\displaystyle Y$ with $\displaystyle \phi(X)$