Cfrgtkky

I would like to ask whether anyone would mind providing me with some direction on how to proceed with this proof.

The question asked me to use the theorem below to prove that, for random
sample of size n, from a normal population with the variance $sigma^2$
, the sampling distribution of $S^2$ has the mean $sigma$ and the variance
$2sigma^4/(n-1)$

The theorem to use is:
If $bar{X}$ and $S^2$ are the mean and the variance of a random sample of size n from a normal population with the mean $mu$ and the standard deviation $sigma$, then

1. 
   $bar{X}$ and $S^2$ are independent.


2. the random variable $(n-1)times S^2/sigma^2$ has a chi-square distribution with n-1 degree of freedom.

I totally understand the theorem above that I should have supposed to be using but I am having trouble understand where I should have started on to apply the theorem to prove the variance of $S^2$.

Thank you for any help and your reading. Appreciated!

Equivalently, you want to prove $S^2$ has mean $nu$ and variance $2nu$, where $nu:=n-1$ is the number of degrees of freedom viz. $S^2simchi_nu^2$. Since means are additive, and so are variances for uncorrelated variables, we only need to check the case $nu=1$. In that case we want to show $S^2,,S^4$ have respective means $1,,1^2+2=3$, i.e. that these are the respective means of $Z^2,,Z^4$ for $Zsim N(0,,1)$. The first result follows from $Z$ having mean $0$ and variance $1$, so the hard part is proving $mathbb{E}Z^4=3$. There are several ways to do this, but note that$$int_{Bbb R}exp -alpha x^2dx=sqrt{pi}alpha^{-1/2}impliesint_{Bbb R}x^4exp -alpha x^2 dx=frac{3}{4}sqrt{2pi}alpha^{-5/2}$$(by applying $partial_alpha^2$), so$$mathbb{E}Z^4=int_{Bbb R}frac{1}{sqrt{2pi}}x^4exp -frac{x^2}{2}dx=frac{3}{4 (1/2)^2}=3.$$You could also use characteristic or moment- or cumulant-generating functions.