Proof that the sampling distribution of the sample variance from $N(0,1)\sim \chi_{n-1}^2$

Is this true? How to verify it? From the definition of chi square I can not judge whether it is chi square.

• You can find the proof here – Antoni Parellada Oct 18 '16 at 1:27
• Independence of $X_{i}$ is required – L.V.Rao Oct 18 '16 at 2:16
• It is $\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}$ that follows $\chi_{n-1}^{2}$. – L.V.Rao Oct 18 '16 at 3:54

Let $X_{i}$ for $i=1,2,\cdots n$ are independent $N(0,1)$ random variables. Then, the distribution of $\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\sim \chi_{n-1}^{2}$.\

Let $\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}=(n-1)S^{2}$.

\begin{eqnarray} \sum_{i=1}^{n}(X_{i}-\bar{X})^{2}&=&\underbrace{\sum_{i=1}^{n} X_{i}^{2}}_{\chi_{n}^{2}}-n\bar{X}^{2}\nonumber\\ (n-1)S^{2}&=&\sum_{i=1}^{n} X_{i}^{2}-n\bar{X}^{2}\nonumber\\ \sum_{i=1}^{n} X_{i}^{2}&=& (n-1)S^{2} + n\bar{X}^{2} \end{eqnarray} Now consider, \begin{eqnarray*} X_{i}\sim N(0,1)\\ \bar{X}\sim N(0,\frac{1}{n})\\ \frac{\bar{X}}{1/\sqrt{n}}\sim N(0,1)\\ n\bar{X}^{2}\sim \chi_{1}^{2}\\ \end{eqnarray*} Identifying the distribution of the terms in the earlier equation, it can be expressed as $$\chi_{n}^{2} = (n-1)S^{2}+\chi_{1}^{2}$$

By a property of chi-squared distribution, $S^{2}$ and $\bar{X}$ are independent. That is, right hand side of the above equation is sum of two independent random variables. Now consider the moment generating function on both sides. The left hand side random variable being a $\chi_{n}^{2}$ random variable, its MGF is $(1-2t)^{-n/2}$.

Substituting the MGF values and simplifying,

\begin{eqnarray*} M_{\chi_{n}^{2}}(t) = M_{(n-1)S^{2}}(t)\cdot M_{\chi_{1}^{2}}(t)\\ (1-2t)^{-n/2} = M_{(n-1)S^{2}}(t) \cdot (1-2t)^{-1/2}\\ M_{(n-1)S^{2}}(t) = (1-2t)^{-n/2} \cdot (1-2t)^{-1/2}\\ M_{(n-1)S^{2}}(t) = (1-2t)^{-(n-1)/2} \end{eqnarray*} which is the MGF of a $\chi_{n-1}^{2}$ random variable. Hence, $(n-1)S^{2}\sim\chi_{n-1}^{2}$.

• The independence of $S^2$ and $\bar X$ seems like a critical step. Can you provide any details about that step? – littleO Jan 27 '19 at 1:24
• @littleO you are correct. For details, you may refer to the book Statistical inference by Casella and Berger. – L.V.Rao Jan 27 '19 at 6:00
• @littleO please have a look at the reference provided by Antoni Parellada in one of the comments above. – L.V.Rao Jan 27 '19 at 7:06
• Hmm, the page Antoni Parellada linked to skips the proof that $S^2$ and $\bar X$ are independent. I'll check out Casella and Berger. Thanks! – littleO Jan 29 '19 at 5:51
• @littleO It can be shown using Basu's Theorem. – Lei Hao May 20 at 4:43