# Probability Density Function of a function

• Aug 26th 2013, 01:14 AM
Rafael
Probability Density Function of a function
Dear All,

let us say we have PDF(x) i.e. probability density function of variable x. How to find probability density function for f(x), PDF(f(x))?
Particularly

PDF(x) is

http://mathhelpforum.com/image/png;b...AASUVORK5CYII=

and f(x)=x2

• Aug 26th 2013, 01:16 AM
Rafael
Re: Probability Density Function of a function
Sorry expression for the PDF(x) did not appear it is PDF for Gaussian distribution.
• Aug 26th 2013, 04:41 AM
Rafael
Re: Probability Density Function of a function
I think that

PDF_f(x)(x)=const*PDF(inversf(x))

if PDF(x) is Gaussian and f(x)=x^2;
then PDF_x^2(x)=const*PDF(sqrt(x)) so the x should be replaced by sqrt(x) in Gaussian PDF in case of 0 mean. In case non 0 man x-mean should be replaced by sqrt(x-mean).
• Aug 26th 2013, 12:13 PM
emakarov
Re: Probability Density Function of a function
You are partially right. $F_{f(X)}(y)=P(f(X)\le y$). When can we say that $f(x)\le y\iff x\le f^{-1}(y)$ and therefore $P(f(X)\le y)=P(X\le f^{-1}(y))$? This holds, in particular, when f increases monotonically. Then f is injective, so $f^{-1}$ exists.

If X is non-negative, then $P(X^2\le y)=P(X\le\sqrt{y})$ because $f(x)=x^2$ increases monotonically when $x\ge0$. However, when X is normal, it can assume negative values. In this case, $P(X^2\le y)=P(-\sqrt{y}\le X\le\sqrt{y})= F_X(\sqrt{y})-F_X(-\sqrt{y})$. If $X\sim\mathcal{N}(0,\sigma^2)$, this equals $2(F_X(\sqrt{y})-F(0))$. In general, if $X\sim\mathcal{N}(\mu,\sigma^2)$, you can use the fact that $F_X(y)=\Phi\left(\frac{y-\mu}{\sigma}\right)$ where $\Phi$ is the PDF of $\mathcal{N}(0,1)$.
• Aug 26th 2013, 01:14 PM
Rafael
Re: Probability Density Function of a function
Dear emakarov,

thanks a lot, I would like to understand this in more ditails so could you pleas