I need to know the formula for the pdf of $S^2$.
I know this:
$$ \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \>, $$
but I want to state the correct formula for the pdf of $S^2$, not $(n-1)S^2/\sigma^2$.
Any thoughts?
I need to know the formula for the pdf of $S^2$.
I know this:
$$ \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \>, $$
but I want to state the correct formula for the pdf of $S^2$, not $(n-1)S^2/\sigma^2$.
Any thoughts?
Given $\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \>,$
and the fact that a chi-squared($\nu$) is a Gamma($\frac{\nu}{2},2$), (under the scale parameterization) then
$S^2 = \frac{(n-1)S^2}{\sigma^2}\cdot \frac{\sigma^2}{(n-1)}\sim \text{Gamma}(\frac{(n-1)}{2},\frac{2\sigma^2}{(n-1)})$
If you need a proof, it should suffice to show that the relationship between chi-square and gamma random variables holds and then follow the scaling argument here. This relationship is pretty much verifiable by inspection.
The pdf is as follows:
\begin{equation} f(x) = \frac{\left(\frac{\nu}{2\, \sigma^{2}}\right)^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)}\, x^{\frac{\nu}{2}-1}\, \exp\left\{-x\, \frac{\nu}{2\, \sigma^{2}}\right\} \end{equation}
$\nu \equiv \text{degrees of freedom}= N-1$, where $N$ is the sample size.
$\sigma \equiv \text{standard deviation of the parent distribution}$.