statistical proof chi squared distribution

Question

Lecture book just stated that: $(n-1)^2/\sigma^2$ is a chi squared distribution with $(n-1)$ degrees of freedom where $n$ sample size, $S^2$ sample variance and $\sigma$ population standard deviation,but how do I prove such thing. Thanks, any tips hints would do.

Answer 1

Let $X_1,\dotsc,X_n\sim N(\mu. \sigma^2)$ be i.i.d normally distributed. Then the key result we need is that $(X_1-\bar{X},\dotsc,X_n-\bar{X})$ is independent of $\bar{X}$. Now we observe that $$ \underbrace{\frac{1}{\sigma^2}\sum_{i=1}^n(X_i-\mu)^2}_{\sim \chi^2_{(n)}} =\frac{1}{\sigma^2}\sum_{i=1}^n(X_i-\bar{X})^2+ \underbrace{\frac{n(\bar{X}-\mu)^2}{\sigma^2}}_{\sim \chi^2_{(1)}}\tag{1}. $$ Since $(X_1-\bar{X},\dotsc,X_n-\bar{X})$ is independent of $\bar{X}$, the two summands on the right are independent. Let $M(t)$ be the moment generating function of $(n-1)S^2/\sigma^2$. Equation (1) implies that $$ (1-2t)^{-n/2}=M(t)\times(1-2t)^{-1/2} $$ whence it follows immediately that $$ (n-1)S^2/\sigma^2\sim \chi^2_{(n-1)}. $$