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
and f(x)=x^{2 Thanks in advance. }
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).
You are partially right. $\displaystyle F_{f(X)}(y)=P(f(X)\le y$). When can we say that $\displaystyle f(x)\le y\iff x\le f^{-1}(y)$ and therefore $\displaystyle P(f(X)\le y)=P(X\le f^{-1}(y))$? This holds, in particular, when f increases monotonically. Then f is injective, so $\displaystyle f^{-1}$ exists.
If X is non-negative, then $\displaystyle P(X^2\le y)=P(X\le\sqrt{y})$ because $\displaystyle f(x)=x^2$ increases monotonically when $\displaystyle x\ge0$. However, when X is normal, it can assume negative values. In this case, $\displaystyle P(X^2\le y)=P(-\sqrt{y}\le X\le\sqrt{y})= F_X(\sqrt{y})-F_X(-\sqrt{y})$. If $\displaystyle X\sim\mathcal{N}(0,\sigma^2)$, this equals $\displaystyle 2(F_X(\sqrt{y})-F(0))$. In general, if $\displaystyle X\sim\mathcal{N}(\mu,\sigma^2)$, you can use the fact that $\displaystyle F_X(y)=\Phi\left(\frac{y-\mu}{\sigma}\right)$ where $\displaystyle \Phi$ is the PDF of $\displaystyle \mathcal{N}(0,1)$.
Sorry, I have not been studying or teaching probability lately, so I don't know good textbooks. But this is very basic stuff: just the definition of PDF and properties of normal distribution. I think that every textbook on probability should cover this.