Distribution of sum of squared deviations

Question

So, given $X_1, \ldots, X_n \stackrel{iid}{\sim} N(\mu, \sigma^2)$, I want to find the joint distribution of $\bar{X} = \frac{1}{n}\sum_{i=1}^n x_i$ and $S^2 = \sum_{i=1} (X_i - \bar{X})^2$ (it is intentional that there is no 1/n or 1/(n-1) in this expression). Now, I know that $\bar{X}$ and $S^2$ are independent, that $\bar{X} \sim N(\mu, \sigma^2/n)$ and $S^2/\sigma^2 \sim \chi^2(n-1)$. So it would be easy if I wanted to find the joint distribution of $\bar{X}$ and $S^2/\sigma^2$, but I don't want that.

I know I can write down $Y = S^2/\sigma^2$ and then say $$f_Y(y) = \frac{1}{2^{(n-1)/2}\Gamma(\frac{n-1}{2})} y^{(n-1)/2} e^{-y/2}.$$

So then how do I get the distribution of $S^2$ from this? Is it as easy as writing $$f_{S^2}(s) = \frac{1}{2^{(n-1)/2}\Gamma(\frac{n-1}{2})} (s^2/\sigma^2)^{(n-1)/2} e^{-(s^2/\sigma^2)/2}$$

or is there more that I have to do?

Then, once I have this pdf I know I can just multiply the two PDFs to get the joint distribution. But is this distribution for $S^2$ correct? Do I have to multiply by a factor of $|\frac{d}{ds} \frac{s^2}{\sigma^2}|$? This is one thing that I thought I may have to do, but I am not sure.

Answer 1

Let's ease the notation a bit and say we want the PDF of $Z=\alpha Y$, ($\alpha\geq 0)$, and we know $f_Y(y)$. Then, $$F_Z(z)=P(Z\leq z)=P(Y\leq z/\alpha)=F_Y(z/\alpha)$$

If we differentiate both sides by $z$, LHS will be $f_Z(z)$, and RHS will be $\frac{f_Y(z/\alpha)}{\alpha}$.

In this question, $\alpha=\sigma^2$, $Z=S^2$, and you already calculated $f_Y(z/\alpha)$, but forgot about the factor $1/\alpha=1/\sigma^2$.