# Prove the sampling distribution of $S^2$ has the mean $\sigma^2$ and the variance $2\sigma^4/(n-1)$

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 $$2\sigma^4/(n-1)$$

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 $$S^2$$.

• You have the distribution of $\frac{n-1}{\sigma^2}S^2\sim\chi^2_{n-1}$ by (2), so basically compute the expected value and variance of $\chi^2_{n-1}$. Nov 13, 2018 at 0:51
• I think there is a typo. $\mu$ should be replaced by $\sigma^ 2$ as in the title in the second paragraph. Nov 13, 2018 at 0:55
• Thank you guys!! Yes, I need to make the correction. Thanks!!
– Chen
Nov 13, 2018 at 21:03

Equivalently, you want to prove $$S^2$$ has mean $$\nu$$ and variance $$2\nu$$, where $$\nu:=n-1$$ is the number of degrees of freedom viz. $$S^2\sim\chi_\nu^2$$. 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 $$Z\sim N(0,\,1)$$. 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$$\int_{\Bbb R}\exp -\alpha x^2dx=\sqrt{\pi}\alpha^{-1/2}\implies\int_{\Bbb R}x^4\exp -\alpha x^2 dx=\frac{3}{4}\sqrt{2\pi}\alpha^{-5/2}$$(by applying $$\partial_\alpha^2$$), so$$\mathbb{E}Z^4=\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}x^4\exp -\frac{x^2}{2}dx=\frac{3}{4 (1/2)^2}=3.$$You could also use characteristic or moment- or cumulant-generating functions.

• Thank you very much! May I confirm with you that "$exp-\alpha x^2dx$" is $exp^{-\alpha x^2}$ right? Thank you again for your help!!!
– Chen
Nov 14, 2018 at 17:55
• @Chen $e^y$ is often written $\exp y$, but as far as I know if never denoted $\exp^y$.
– J.G.
Nov 14, 2018 at 19:12
• Oh thank you and I am sorry for my typo. My point was to make sure the power of the exponential is ${-ax^2}$ as a whole, not \$exp{-a} \times x^2}. Thank you very much!
– Chen
Nov 14, 2018 at 19:21
• @Chen Oh yes, that part's right.
– J.G.
Nov 14, 2018 at 19:28
• May I ask what did that "apply" mean? Any typo? Thank you!
– Chen
Nov 14, 2018 at 21:11