How determine the distribution of $S_{n-1}^{2}$?

Question

If $ X_1,\dots,X_n $ is a random sample with $X_i \sim N(\mu,\sigma^2) $, how determinate the distribution of $S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 $?

I tried to apply the method of moments as follow:

$S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 = \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 $

$$ M_{S_{n-1}^2 }(t) = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 ) } ] = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{t(- \frac{n}{n-1} \bar{X}^2 ) } ] = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] $$ $$= E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] = \prod_{i=1}^n E[ e^{t ( \frac{1}{n-1} X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ]$$

But in this point I'm lose. Should I apply the following?

$$ X_i \sim N( \mu,\sigma^2 ) \Rightarrow X_i = \sigma Z + \mu $$ $$ Y = (\sigma Z+ \mu )^2 \Rightarrow M_Y(t) = \frac{1}{\sqrt{1-2t\sigma^2}} \exp\left( \dfrac{\mu^2t}{1-2t\sigma^2}\right) , \ \text{if } \ 2\operatorname{Re}(t\sigma^2) < 1$$

Or isn't this the simplest way to determinate $S_{n-1}^2$?

Answer 1

The term $S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 $ is called sample variance and its distribution is $$S_{n-1}^2 = \frac{\sigma^2}{n-1} \chi^2(n-1)$$ where $\chi^2(n-1)$ the chi squared distribution of $n-1$ degrees of freedom. One of the proofs can be found here: https://online.stat.psu.edu/stat414/lesson/26/26.3

Answer 2

Here's a different way from "NN2"'s answer. $$ \left[ \begin{array}{c} X_1 \\ \vdots \\ X_n \end{array} \right] \sim \operatorname N_n\left( \left[ \begin{array}c \mu \\ \vdots \\ \mu \end{array} \right], \, \sigma^2 I_n \right). $$ The mapping $$ \left[ \begin{array}{c} X_1 \\ \vdots \\ X_n \end{array} \right] \mapsto \left[ \begin{array}{c} X_1 - \overline X \\ \vdots \\ X_n -\overline X \end{array} \right] $$ is an orthogonal projection onto an $(n-1)$-dimensional subspace and it takes $\left[ \begin{array}c \mu \\ \vdots \\ \mu \end{array} \right]$ to $\left[ \begin{array}c 0 \\ \vdots \\ 0 \end{array} \right].$

The distribution of the vector above is spherically symmetric about the mean, i.e. the density is constant on spheres centered at the vector that is the expected value.

Consequently, the distribution of the orthogonal projection of our random vector onto that $(n-1)$-dimensional subspace is spherically symmetric. We can express in terms of an orthonormal basis of that $(n-1)$-dimensional subspace, as $(U_1,\ldots,U_{n-1})^\top.$

So now we have $$ (X_1-\overline X)^2 + \cdots + (X_n - \overline X)^2 = U_1^2 + \cdots + U_{n-1}^2 $$ and $U_1,\ldots,U_n \sim\text{i.i.d.}\operatorname N(0,\sigma^2).$ Hence $$ \frac{U_1^2 + \cdots + U_{n-1}^2}{\sigma^2} \sim \chi^2_{n-1}. $$

Answer 3

Using the fact that $\overline X$ and $S_{n-1}^2$ are independent in normal distribution.

$\sum_{i}(X_i-\mu)^2=\sum_{i}(X_i-\overline X)^2+n(\overline X-\mu)^2$

$\underbrace{\dfrac{\sum_{i}(X_i-\mu)^2}{\sigma^2}}_W=\underbrace{\dfrac{\sum_{i}(X_i-\overline X)^2}{\sigma^2}}_{W_1}+\underbrace{\dfrac{n(\overline X-\mu)^2}{\sigma^2}}_{W_2}$

$W=W_1+W_2$

$W\sim \chi_{n}^2 ,\ \ \ W_2\sim\chi_{1}^2 ; \text{Let me know if you didnt get this, I will elaborate.}$

$W_2$ is a function $\overline X$ and $W_1$ is function $S_{n-1}^2$ so $W_1$ amd $W_2$ are independent.

Therefore,

$M_W{(t)}=M_{W_{1}}(t)\cdot M_{W_{2}}(t)$

$M_{W_{1}}(t)=\dfrac{M_{W}(t)}{M_{W_{2}}(t)}=\dfrac{(1-2t)^{\frac{-n}{2}}}{1-2t)^{\frac{-1}{2}}}=(1-2t)^{\frac{-(n-1)}{2}}\ \ \ ,t< \frac{1}{2}$

$W_1=\dfrac{\sum_{i}(X_i-\overline X)^2}{\sigma^2}=\dfrac{(n-1)S_{n-1}^2}{\sigma^2} \sim \chi_{n-1}^2$