Suppose X_1, ..., X_n is a random sample from a normal distribution with parameter μ and σ^2 . If S^2=Σ(X_i - Xbar)^2 /n , where Xbar is the sample mean, determine the distribution of nS^2 /σ^2 .

QUESTION: 

Suppose $X_{1},... ,X_{n}$ is a random sample from a normal distribution with parameter $\mu$ and $\sigma^{2}$. If $S^{2}=\frac{\sum(X_{i}-\bar{X})^{2}}{n}$ , where $\bar{X}$ is the sample  mean, determine the distribution of $\frac{nS^{2}}{\sigma^{2}}$.

ANSWER:

Where $\sum_{i=1}^{n}(x_{i}-\bar{x})=0$ then

$\sum_{i=1}^{n}(x_{i-\mu})^{2}=\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}+n(\bar{x}-\mu)^{2}$

To obtain a random variable with a normal distribution and chi-square, the equation is multiplied by $\frac{1}{\sigma^{2}}$

$\frac{\sum_{i=1}^{n}(x_{i-\mu})^{2}}{\sigma^{2}}=\frac{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}{\sigma^{2}}+\frac{n(\bar{x}-\mu)^{2}}{\sigma^{2}}$

$V=\sum_{i=1}^{n}W_{i}\sim \chi^{2}_{(n)} \rightarrow M_{V}(t)=(1-2t)^{-\frac{n}{2}}$

$Y^{2}=\frac{n(\bar{x}-\mu)^{2}}{\sigma^{2}}\sim \chi^{2}_{(1)} \rightarrow M_{Y^{2}}(t)=(1-2t)^{-\frac{1}{2}}$

Based on the definition of the MGF equation, then:

so   $M_{\sum_{i=1}^{n}(\frac{x_{i-\mu}}{\sigma})^{2}}(t)\sim \chi^{2}_{(n-1)}$

To get the distribution of $\frac {nS^{2}}{\sigma^{2}}$ then it needs to be multiplied by $\frac{n}{n}$ :

$\begin{aligned}\sum_{i=1}^{n}(\frac{x_{i-\mu}}{\sigma})^{2} . \frac{n}{n}&=\frac{n\sum_{i=1}^{n}(\frac{x_{i-\mu}}{\sigma})^{2}}{n}\\&=\frac{n\sum_{i=1}^{n}\frac{(x_{i}-\bar{x})^{2}}{\sigma^{2}}}{n}\\&=n\sum_{i=1}^{n}\frac{\frac{(x_{i}-\bar{x})^{2}}{n}}{\sigma^{2}}\\&=\frac{nS^{2}}{\sigma^{2}}\end{aligned}$

Based on these results, it can be concluded that $\frac{nS^{2}}{\sigma^{2}}$has a chi-square distribution with degrees of freedom $(n-1)$ or  $\frac{nS^{2}}{\sigma^{2}}\sim \chi ^{2}_{(n-1)}$.