Variance of $S^2$ taken from Normal Distribution Suppose that $Y_1,Y_2,...,Y_n$ is a random sample of size $n$ where $Y_i$~$N(0,\sigma^2)$.  Let $\bar Y$ and $S^2$ denote the usual sample mean and variance.
Find the the variance of $S^2$.
In this problem, I'm assuming that $S^2=\frac{\sum_{i=1}^{n}(Y_i-\bar Y)^2}{n-1}$.  I know that the difference of normal random variables is normal (since $\bar Y$ is normal), so I get a chi squared random variable in the numerator and then obtain the following result:
$$V(S^2)=\frac{2n(\sigma^2+\frac{\sigma^2}{n})^2}{(n-1)^2}$$
But I have a lingering suspicion that this isn't correct.  Did I make an assumption somewhere that I shouldn't have?
 A: Note the well known result that:
$$
\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}
$$
(chi-squared distribution with $n-1$ degrees of freedom), and further, the variance of $\chi^2_{n-1}$ is $2(n-1)$.
Now, using this information:
\begin{align*}
2(n-1) = Var \left ( \frac{(n-1)S^2}{\sigma^2} \right) & = \frac{(n-1)^2}{\sigma^4} Var(S^2)
\end{align*}
and so:
$$
Var(S^2) = \frac{2(n-1)\sigma^4}{(n-1)^2} = \frac{2\sigma^4}{n-1}
$$
So your answer is a bit off, perhaps you could share your steps/assumptions for some advice on where you might have gone wrong
Proof of the result:
note the following useful facts:


*

*For standard normal $Z \sim N(0,1)$, $Z^2 \sim \chi^2_{1}$

*$\bar{X} \bot S^2$ (independence)

*The moment generating function of a $\chi_p^2$ is $m_{\chi_p^2}(t)= (1-2t)^{-p/2}$

*If $X_1, \dots, X_n$ are independent and $X_i = \chi^2_{p_i}$ then:


$$
\sum_{i=1}^n X_i \sim \chi^2_{p1 + p2 + \dots + p_n}
$$
The trick for these types of proofs is to use this fact:
\begin{align*}
&\sum (X_i - \mu)^2 = (n-1)S^2 + n(\bar{X} - \mu)^2\\
\implies &\sum \left ( \frac{X_i - \mu}{\sigma} \right )^2  = \frac{(n-1)S^2}{\sigma^2} + n \left ( \frac{\bar{X} - \mu}{\sigma} \right ) ^2\\
\implies &\sum \left ( \frac{X_i - \mu}{\sigma} \right )^2  = \frac{(n-1)S^2}{\sigma^2} +  \left ( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \right ) ^2\\
\implies & A  = B+C
\end{align*}
By fact number 4
$$
A =  \sum \left ( \frac{X_i - \mu}{\sigma} \right )^2 \sim \chi_{n}^2
$$
and by fact number 1
$$
C = \left ( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \right ) ^2 = Z^2 = \chi_1^2
$$
Using facts number 2 and number 3:
\begin{align*}
&m_A(t) = m_B(t) m_C(t)\\
\implies &(1-2t)^{-n/2} = M_B(t) (1-2t)^{-1/2}\\
\implies & M_{B}(t) = \frac{(1-2t)^{-n/2}}{(1-2t)^{-1/2}}\\
\implies & M_{B}(t) = (1-2t)^{-(n-1)/2}\\
\end{align*}
which you should note is the moment generating function of a chi square distributed random variable with degrees of freedom $n-1$. The variance follows easily from there.
