Variance with minimal MSE in normal distribution
Solution 1:
Assuming your calculations of the expectation and variance of the estimator $$\hat e = p \sum_{i=1}^n (X_i - \bar X)^2$$ are correct, then you have $$MSE = \operatorname{E}[\hat e - \sigma^2]^2 + \operatorname{Var}[\hat e] = (p(n-1) - 1)^2 (\sigma^2)^2 + 2p^2 \sigma^4 (n-1),$$ or $$MSE = \sigma^4\left((n^2-1)p^2 - 2(n-1)p + 1\right).$$ Then take the derivative with respect to $p$, set it to zero, and solve for the critical point: this gives $$p = (n+1)^{-1}.$$