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It's a standard result that given $X_1,\cdots ,X_n $ random sample from $N(\mu,\sigma^2)$, the random variable $$\frac{(n-1)S^2}{\sigma^2}$$ has a chi-square distribution with $(n-1)$ degrees of freedom, where $$S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.$$ I would like help in proving the above result.

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You ought to include your definition of S. – Emre Jun 22 '11 at 22:04
where $S^2 = \dfrac{\sum_i (X_i - \bar{X})^2}{n-1}$ I assume – Henry Jun 22 '11 at 22:09
@Emre: I forgot to define it. – Nana Jun 22 '11 at 22:16
@Henry: Yes, you are right. – Nana Jun 22 '11 at 22:16
up vote 16 down vote accepted

A standard proof goes something like this. It assumes you already know the following.

  1. $\bar{X}$ (the sample mean) and $S^2$ are independent.
  2. If $Z \sim N(0,1)$ then $Z^2 \sim \chi^2(1)$.
  3. If $X_i \sim \chi^2(1)$ and the $X_i$ are independent then $\sum_{i=1}^n X_i \sim \chi^2(n)$.
  4. A $\chi^2(n)$ random variable has the moment generating function $(1-2t)^{-n/2}$.

With some algebra, you can show, by adding $-\bar{X} + \bar{X}$ inside the parentheses and grouping appropriately, that $\sum_{i=1}^n (X_i - \mu)^2 = \sum_{i=1}^n (X_i - \bar{X})^2 + n(\bar{X} - \mu)^2$. Then, dividing through by $\sigma^2$ yields $$ \sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n \left(\frac{X_i - \bar{X}}{\sigma}\right)^2 + \left(\frac{\bar{X} - \mu}{\sigma/\sqrt{n}}\right)^2.$$
Denote these expressions by $U, V$, and $W$, respectively, so that the formula reads $U = V+W$. By facts (2) and (3) above, $U \sim \chi^2(n)$ and $W \sim \chi^2(1)$. Also, $V = \frac{(n-1)S^2}{\sigma^2}$.

Since $\bar{X}$ and $S^2$ are independent, so are $V$ and $W$. Thus $M_U(t) = M_V(t) M_W(t)$, where $M_X(t)$ denotes the moment generating function of the random variable $X$. By fact (4) above, this says that $$\frac{1}{(1-2t)^{n/2}} = M_V(t) \frac{1}{(1-2t)^{1/2}}.$$ Thus $$M_V(t) = \frac{1}{(1-2t)^{(n-1)/2}},$$ and therefore $V \sim \chi^2(n-1)$.

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Very nice! Thanks Mike:) – Nana Jun 22 '11 at 22:58
Why does the independence of $\bar{X}$ and $S^2$ imply the independence of $V$ and $W.$ Thanks. – Nana Jul 27 '11 at 5:18
@Nana: Because $V$ is a function of $S^2$ and $W$ is a function of $\bar{X}$. – Mike Spivey Jul 27 '11 at 5:21
Oh I see...Thanks once again. – Nana Jul 27 '11 at 5:33

I disagree with the characterization of the proof in Mike Spivey's answer as the standard proof. It's the proof for people who don't know about projections in linear algebra.

Notice that the mapping $(X_1,\dots,X_n) \mapsto (X_1-\overline{X},\dots,X_n - \overline{X})$ is a projection onto a space of dimension $n-1$. Notice also that its expected value is $0$. Then remember that the probability distribution of the vector $(X_1,\dots,X_n)$ is spherically symmetric. Therefore so is the distribution of its projection onto a space of dimension one less. Hence the square of the norm of that projection is just the square of the norm of a normal random vector with a spherically symmetric distribution centered at the origin. The square of the norm therefore has a chi-square distribution with degrees of freedom equal to the dimension of that space.

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Criticism accepted, and answer upvoted. I will change the article from definite to indefinite. – Mike Spivey Aug 19 '11 at 18:40

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