Calculate risk (under squared error loss) of MSE estimator of normal variance I am given that $X_i\stackrel{\text{iid}}{\sim}N(\mu,\sigma^2)$. As part of a larger problem, I am to calculate the risk function of $\widehat{\sigma^2}_\text{MSE} = \frac1{n-1}\sum_{i=1}^n(X_i-\bar X)^2$, using the loss function $L(\sigma^2,\widehat{\sigma^2}) = (\sigma^2-\widehat{\sigma^2})^2$.
I get stuck just trying to simplify down the horrible expression one gets when applying the definitions of risk and loss functions. Maybe I'm missing some shortcut or something. Any tips would be appreciated. So far I have:
\begin{align*}
R(\sigma^2,\widehat{\sigma^2}_\text{MSE}) = {} & \mathbb E[(\sigma^2 - \frac1{n-1}\sum_{i=1}^n(X_i-\bar X)^2)^2]\\
= {} & \cdots\\
= {} & \sigma^4 - 2\sigma^2\frac1{n-1} \left( \sum_{i=1}^n \mathbb E(X_i^2-\bar X^2) \right) \\ & {} + \frac1{(n-1)^2}\mathbb E\left[ \sum_{i=1}^n(X_i-\bar X)^4 + 2\sum_{i<j} (X_i-\bar X)^2(X_j-\bar X)^2 \right]
\end{align*}
Any attempts to expand out that last big term seem to end in disaster.
 A: Let $\widehat{\sigma}^2 \equiv \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X}_n)^2$. Since $X_i \sim \mbox{ iid } N(\mu, \sigma^2)$ we know that $Y \equiv (n-1)\widehat{\sigma}^2/\sigma^2$ is a $\chi^2$ random variable with $n-1$ degrees of freedom. Now, we can write:
$$\widehat{\sigma}^2 - \sigma^2 = \frac{\sigma^2}{n-1}\left[(n-1)\widehat{\sigma}^2/\sigma^2 - (n-1)\sigma^2/\sigma^2 \right]= \frac{\sigma^2}{n-1}\left[Y - (n-1)\right]$$
Using the preceding equality, we can write the MSE as follows:
$$MSE = \mathbb{E}[(\widehat{\sigma}^2 - \sigma^2)^2] = \mathbb{E}\left[\left\{\left(\frac{\sigma^2}{n-1}\right)\left[Y - (n-1)\right]\right\}^2\right]= \left(\frac{\sigma^2}{n-1}\right)^2\mathbb{E}[\left\{Y - (n-1)\right\}^2]$$
Now, the mean of a $\chi^2(\nu)$ random variable is simply $\nu$ hence $\mathbb{E}[\left\{Y - (n-1)\right\}^2]= Var(\chi^2_{n-1})$. And the variance of a $\chi^2(\nu)$ random variable is simply $2\nu$. Substituting, we have
$$MSE = \mathbb{E}[(\widehat{\sigma}^2 - \sigma^2)^2] = \left(\frac{\sigma^2}{n-1}\right)^2 \times 2(n-1) = \frac{2\sigma^4}{n-1}$$
A: Suppose you know that $\displaystyle \frac 1 {\sigma^2}\sum_{i=1}^n (X_i - \overline X)^2 \sim \chi^2_{n-1}.$
Recall that $\operatorname{E}(\chi^2_{n-1}) = n-1$ and $\operatorname{var}(\chi^2_{n-1})=2(n-1).$
Then you have
\begin{align}
\operatorname{E}\left( \frac 1 {n-1} \sum_{i=1}^n \left(X_i - \overline X \right)^2 \right) & = \sigma^2, \\[10pt]
\operatorname{var}\left( \frac 1 {n-1} \sum_{i=1}^n \left( X_i-\overline X \right)^2 \right) & = \frac{2\sigma^4}{n-1}.
\end{align}
The bias is therefore $0$ and
$$
\text{mean squared error} = \left( \text{bias} \right)^2 + \text{variance} = 0 + \frac{2\sigma^4}{n-1}.
$$
