Proving Chi-squared Distribution I have some problems solving the following problem:
Let $X = (X_1, X_2,\ldots, X_n)$, be random sample , where $X_{i} \sim N(\mu, \sigma^{2})$. Show that:
$$U:= \frac{n-1}{\sigma^2} S^2 \sim \chi^2 (n-1); \text{ where } S^2:= \frac{1}{n-1}\sum_{i=1}^n (X_i - \overline{X})^2.$$
Do I need to prove first that the statistic, the sample mean, is sufficient? Or show the independence through Basu's Theorem?
Thanks for any help!
 A: If $\sigma^2$ is “known”, then the sample mean is sufficient.  In other words, for the family of distributions you get by letting $\mu$ vary with $\sigma^2$ fixed, the sample mean is sufficient.  But Basu's theorem relies on completeness, so you'd have to prove completeness before you know that Basu's theorem is applicable.  I think it could be done that way, but I haven't worked out the details.
But to show independence of $\overline X$ and $S^2$, you can first work on proving these two random variables are independent of each other:
$$
\overline X \quad \text{and} \quad 
\left[ \begin{array}{c} X_1 - \overline X \\  \vdots \\  X_n - \overline X \end{array} \right]. \tag 1
$$
That doesn't require Basu's theorem.  Try finding covariances.
But your question as initially phrased asks about something other than independence, namely the distribution of $S^2$.  I don't know why you would think Basu's theorem could help with that.
The mapping that takes the vector
$$
\left[ \begin{array}{c} X_1 \\  \vdots \\  X_n \end{array} \right] \tag 2
$$
to the vector in $(1)$ is the orthogonal projection onto an $(n-1)$-dimensional subspace, and it maps the expected value
$$
\left[ \begin{array}{c} \mu \\  \vdots \\ \mu \end{array} \right]
$$
to a vector of $n$ zeros.  The distribution of $(2)$ is spherically symmetric in $n$ space; therefore it is the same as the distribution of $[ U_1,\ldots, U_n]^T$, where that vector expresses the vector $(1)$ with respect to a certain alternative basis of $\mathbb R^n$.  That alternative basis consists of $n-1$ mutually orthogonal unit vectors in the space onto which one is projecting, and one unit vector at a right angle to that.  With respect to that basis, the projection is
$$
\left[ \begin{array}{c} U_1 \\ U_2 \\ U_3 \\ \vdots \\ U_n \end{array} \right] \mapsto \left[ \begin{array}{c} 0 \\ U_2 \\ U_3 \\ \vdots \\ U_n \end{array} \right].
$$
The first component simply becomes $0$ and the rest are unchanged.  Therefore we have
$$
(X_1-\overline X)^2 + \cdots + (X_n - \overline X)^2 = \underbrace{U_2^2 + \cdots + U_n^2}_\text{starting with 2, not with 1}.
$$
So
$$
\frac 1 {\sigma^2} (U_2^2 + \cdots + U_n^2) \sim \chi^2_{n-1}.
$$
