Easy proof of $\sum_{i=1}^n \left(Z_i - \bar{Z}\right)^2 \sim \chi^2_{n-1}$? Let $Z_1,\cdots,Z_n$ be independent standard normal random variables. There are many (lengthy) proofs out there, showing that
$$ \sum_{i=1}^n \left(Z_i - \frac{1}{n}\sum_{j=1}^n Z_j \right)^2 \sim \chi^2_{n-1} $$
Many proofs are quite long and some of them use induction (e.g. Casella Statistical Inference). I am wondering if there is any easy proof of this result.
 A: Note you say $Z_is$ are iid with standard normal $N(0,1)$, with $\mu=0$ and $\sigma=1$
Then $Z_i^2\sim \chi^2_{(1)}$
Then 
\begin{align}\sum_{i=1}^n Z_i^2&=\sum_{i=1}^n(Z_i-\bar{Z}+\bar{Z})^2=\sum_{i=1}^n(Z_i-\bar{Z})^2+n\bar{Z}^2\\&=\sum_{i=1}^n(Z_i-\bar{Z})^2+\left[\frac{\sqrt{n}(\bar{Z}-0)}{1}\right ]^2 \tag{1}
\end{align}
Note that the left hand side of (1),
$$\sum_{i=1}^n Z_i^2\sim\chi^2_{(n)}$$
and that the second term on the right hand side
$$\left[\frac{\sqrt{n}(\bar{Z}-0)}{1}\right ]^2 \sim\chi^2_{(1)}.$$
Furthermore $\operatorname{Cov}(Z_i-\bar Z,\bar Z)=0$ such that $Z_i-\bar Z$ and $\bar Z$ are independent.  Therefore the two last terms in (1)
 (functions of $Z_i-\bar Z$ and $Z_i$) are also independent.  Their mgfs are therefore related to the mgf of the left hand side of (1) through
$$
M_n(t) = M_{n-1}(t)M_1(t)
$$
where $M_n(t)=(1-2t)^{-n/2}$ and $M_1(t)=(1-2t)^{-1/2}$.  The mgf
of $\sum_{i=1}^n(Z_i-\bar{Z})^2$ is therefore $M_{n-1}(t)=M_n(t)/M_1(t)=(1-2t)^{-(n-1)/2}$.  Thus, $\sum_{i=1}^n(Z_i-\bar{Z})^2$ is chi-square with $n-1$ degrees of freedom.
A: For $k=1, 2, \ldots, n-1$, define
$$X_k = (Z_1 + Z_2 + \cdots + Z_k - kZ_{k+1})/\sqrt{k+k^2}.$$
The $X_k$, being linear transformations of multinormally distributed random variables $Z_i$, also have a multinormal distribution.  Note that

*

*The variance-covariance matrix of $(X_1, X_2, \ldots, X_{n-1})$ is the $n-1\times n-1$ identity matrix.


*$X_1^2 + X_2^2 + \cdots + X_{n-1}^2 = \sum_{i=1}^n (Z_i-\bar Z)^2.$
$(1)$, which is easy to check, directly implies $(2)$ upon observing all the $X_k$ are uncorrelated with $\bar Z.$  The calculations all come down to the fact that $1+1+\cdots+1 - k = 0$, where there are $k$ ones.
Together these show that $\sum_{i=1}^n(Z_i-\bar Z)^2$ has the distribution of the sum of $n-1$ uncorrelated unit-variance Normal variables.  By definition, this is the $\chi^2(n-1)$ distribution, QED.
References

*

*For an explanation of where the construction of $X_k$ comes from, see the beginning of my answer at How to perform isometric log-ratio transformation concerning Helmert matrices.


*This is a simplification of the general demonstration given in ocram's answer at Why is RSS distributed chi square times n-p.  That answer asserts "there exists a matrix" to construct the $X_k$; here, I exhibit such a matrix.
