Transforming random variable and finding PDF and Expected Value

**acevipa** · June 6th 2010, 05:53 PM

Suppose X is a continuous random variable with probability density $f$ , and $\phi$ is a continuous increasing differentiable function.

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

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

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

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

2) Show that the expected value of Y is:

$E(Y)=\int_{-\infty}^{\infty} \phi(x)f(x) dx$

**matheagle** · June 6th 2010, 09:31 PM

(1) Just differentiate the CDFs...

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

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

Now use the chain rule..

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

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

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

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