$$\alpha\sum_{i=1}^n \sum_{i\ne j}^n (X_i - X_j)^2 = \hat{\sigma}^2$$
Question: Find the value of $\hat{\sigma}^2$ that makes it an unbiased estimator of $\sigma^2$
The way I have been taught to find unbiased estimators is to use the E($\hat{\sigma}^2$) which is $\sigma^2$ to substitute values that can be converted into $\sigma^2$ and $\mu^2$ from the equation V(X) = E(X^2)-[E(X)^2] So far I understand that there will be n(n-1) way to sum the values because $(X_i - X_i)^2$ cannot exist. However, I'm unsure how to subsitute the expected values into the original equation if I even can. Then I would be able to figure out which value of n could work for the $\alpha$ constant. Sorry if this is poorly explained, I've never seen a problem with two summations before. The answer provided to the problem is $\alpha = \frac{1}{2n(n-1)}$
Assuming $X_i$ are i.i.d. random variables. $E(X_i-X_j)^2=E(X_i)^2)+E(X_j)^2)-2E(X_i)E(X_j)=2\sigma^2$ Therefore $\sum\limits _{i=1}^n\sum\limits _{j=1,j\ne i}^n (X_i-X_j)^2=n(n-1)2\sigma^2$. Therefore $\alpha=\frac{1}{2n(n-1)}$.