Let $X_1,...,X_n\sim \mathcal{N}(\mu, \theta)$ be iid. Define $Y=\frac{1}{n}\sum_{i=1}^n (X_i-\overline X)^2$ where $\overline X = \frac{1}{n}\sum_{i=1}^n X_i$
I want to compute $\mathbb{E}[Y]$ and $\mathbb{E}[Y^2]$ and cannot seem to get to a reasonable result. Can something about how $Y$ must be distributed be said? I don't think I can compute the expectation purely algebraically.
EDIT: I could now proceed and I am almost done.
Since we have that $\mathbb{E}[S_n^2]=\theta$ for $S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline X)$
it follows that $\mathbb{E}[Y]=\frac{n-1}{n}\theta$ which is correct.
For $\mathbb{E}[Y^2]$ I will try a similar approach using the translation theorem $\mathbb{E}[Y^2]=Var[Y]+(\mathbb{E}[Y])^2$.
I know that $Var[S_n^2]=1/n(3\theta^2 -\frac{n-3}{n-1}\theta^2)$ This follows from here: StackExchange Post
We then have that
$\mathbb{E}[Y^2]= 1/n(3\theta^2 -\frac{n-3}{n-1}\theta^2) + \theta^2=\frac{\theta^2}{n^2}(\frac{n^2(n+1)}{n-1})$
Because I can transform between $S_n^2$ and $Y$ by applying a factor $(n-1)/n$.
However, this is incorrect, it should be $\frac{\theta^2}{n^2}(n^2-1)$