# Unbiased estimator of variance of normal distribution

I am struggling with the following question about unbiased estimators. I don't know if its the wording, but I would appreciate any guidance.

Question: Let $X_1,...,X_n$ denote a random sample from a normal distribution with mean zero and variance $\nu$, $0<\nu<\infty$. Show that $$S^2=\frac{1}{n} \sum_{i=1}^{n}{X_i^2}$$ is an (unbiased) estimator for a certain quantity $\sigma^2$. Find $\sigma^2$ and the variance of this estimator for $\sigma^2.$

Edit: I know that to show it is an unbiased estimator, I must show that its expectation is the variance, but I'm having trouble manipulating the variables. The $\frac{1}{n}$ doesn't seem to work itself out. And I am a bit lost as to what its variance is.

Work: If the $X_i$ are IID then: $$E[S^2]=E[\frac{1}{n} \sum_{i=1}^{n}{X_i^2}]= \frac{1}{n}E[\sum_{i=1}^{n}{X_i^2}] = \frac{1}{n}\sum_{i=1}^{n}{E[X_i^2]}=\frac{1}{n}nE[X_i^2]=\sigma^2$$ which proves that this is an unbiased estimator.

The variance of this estimator is $$Var[S^2] = E[S^2]-E[S]^2 = \sigma^2 - E[S]^2 ??$$

From here, I am not sure how to proceed.

• Because this is a homework-type question, you should tell us what you've tried so far and where you get stuck. You should also add the self-study tag to your question. Please take a look at stats.stackexchange.com/tags/self-study/info – Patrick Coulombe Oct 13 '15 at 2:18
• Hint: $Var(X)=E(X^2)-E(X)^2$. You have $X=S^2$. What happens if you substitute? – Christoph Hanck Oct 13 '15 at 4:25
• So, $Var[S^2] = E[S^4] - E[S^2]^2$. Then, $V[S^2] = \frac{1}{n^2}(E[X_i^4] - E[X_i^2]^2) = \frac{1}{n^2}n(3\sigma^4-(\sigma^2)^2) = \frac{2\sigma^4}{n}$? I did the calculations in more detail on paper. This seems right. Thank you so much guys! – Lucky Oct 14 '15 at 0:49