Sampling distribution and estimators?

Question

Consider the following question (original image):

Now consider the estimator $$S^2 = \frac{1}{n-1} \sum_{i=1}^n ( X_i - \overline{X} )^2$$ where $\overline{X}$ is the mean of $X_1 , X_2 , \ldots , X_n$. You are given the following results.

1. The sampling distribution of $S^2$ is $\frac{\sigma^2}{n-1} \chi^2_{n-1}$.
2. The mean of a $\chi^2_{n-1}$ random variable is $n-1$.
3. The variance of a $\chi^2_{n-1}$ random variable is $2(n-1)$.

You are not required to prove these results.)

(iv) Use these results to show that $S^2$ is an unbiased estimator of $\sigma^2$ and that its variance is $\frac{2\sigma^4}{n-1}$.

On question (iv) the only way I can think of doing it is by using the sampling distribution. If I am correct the sampling distribution of an estimator is just its pdf. But the way i did this was as follows: $$ {\rm E}[S^2]=\frac{\sigma^2}{n-1}{\rm E}[\chi_{n-1}^2] $$ But i am sure this is not valid as I don't think you can use the sampling distribution like this. How is the proper way i would do this question and use the sample distribution to find the mean. (I know how to find the mean from a probability density function.)

Answer 1

The formula you wrote is correct, and that is exactly how you are supposed to solve the problem.

Answer 2

We have $$\mathbb{E}(S^2)=\frac{\sigma^2}{n-1}\mathbb{E}(X)=\frac{\sigma^2}{n-1}(n-1)=\sigma^2 \,\,$$ for $X \sim \chi_{n-1}^2 \,$ and where the second equality came from the given result ($2$), so that $S^2$ is unbiased for $\sigma^2$. In regards to computing variance, we have $$\begin{align}\text{Var}(S^2) &= \left(\frac{\sigma^2}{n-1}\right)^2\text{Var}(X) \\&= \left(\frac{\sigma^2}{n-1}\right)^2 2(n-1) \\&= \frac{2\sigma^4}{n-1} \,\,,\end{align}$$ again for $X \sim \chi_{n-1}^2$ and where the second equality here came from the given result ($3$).