Finding variance Given a sapmle $(X_1, X_2 , \ldots , X_n)$ from  normal distribution with parameters $(a , \sigma ^ 2)$, find $$ \operatorname{Var}\left( \frac{1}{n} \sum\limits_{n=1}^n(X_i - \overline X)^2\right)$$ where $\overline X$ is the sample mean. I can calculate it when $\overline X$  is replaced with expectation using independence , but in this case I cannot use independence. I want to reduce it to fourth moment of normal distribution but I fail to. 
 A: Hint: $$ \operatorname{Var}\left( \frac{1}{n} \sum\limits_{n=1}^n(X_i - \overline X)^2\right) = E\left[\left( \frac{1}{n} \sum\limits_{n=1}^n(X_i - \overline X)^2\right)^2\right] - \left(E\left[ \frac{1}{n} \sum\limits_{n=1}^n(X_i - \overline X)^2\right]\right)^2 $$
Distribute, Linearity, and Independence (since they are a sample).
A: (Below, I assume $X_1,\ldots,X_n$ are independent.)
If you happen to know that
$$
\frac{\sum_{i=1}^n (X_i-\bar X)^2}{\sigma^2} \sim \chi^2_{n-1}, \tag 1
$$
then this reduces to a problem on the variance of a chi-square distribution.
Here's a quick sketch of a proof of $(1)$:
The vector $(\ldots, X_i,\ldots)$ decomposes as
$$
(\ldots, X_i,\ldots) = \underbrace{(\bar X,\ldots,\bar X)}_{\text{(all entries equal)}} + (\ldots, X_i-\bar X,\ldots).
$$
The second summand is the orthogonal projection of $(X_1,\ldots,X_n)$ onto the space in which the sum of all components is $0$.  That orthogonal projection maps the population mean vector $(a,\ldots,a)$ to $0$.  Since the standard normal distribution is spherically symmetric, the square of the norm of the vector $(\ldots, X_i-\bar X,\ldots)$ has the same distribution as if $a$ had been $0$ and you had simply projected onto the space $(x_1,\ldots,x_{n-1},0)$.  I.e. $\sigma^2$ times a chi-square distribution with $n-1$ degrees of freedom.
