Variance of empirical mean squared error of binomial distribution

Question

Given $m$ binomially distributed samples $X_1,X_2,\ldots,X_m\sim\text{Bin}(n,p)$. Let $\hat{p}_i=X_i/n$ be the corresponding estimates for $p$. The empirical mean squared error is given by $Z= \frac{1}{m}\sum_{i=1}^m (\hat{p}_i-p)^2$. I am looking for the expected value and the variance of $Z$. $E(Z) = p(1-p)/n$ follows directly from the variance of the binomial distribution. Is there a simple way to also get $\text{Var(Z)}$?

Answer 1

I'm not sure I'd call it "simple", but it's nevertheless fairly straightforward: \begin{align} E\left(Z^2\right)&=\frac{1}{m^2}\left(\sum_{i=1}^mE\left(\left(\hat{p}_i-p\right)^4\right)+2\sum_{i=2}^m\sum_{j=1}^{i-1} E\left(\left(\hat{p}_i-p\right)^2\left(\hat{p}_j-p\right)^2\right)\right)\\ &= \frac{\mu_4}{mn^4}+\frac{(m-1)p^2(1-p)^2}{mn^2}\ , \end{align} where $\ \mu_4=E\left(\left(X_i-np\right)^4\right)\ $ is the $\ 4$-th central moment of $\ X_i\ $—which Wolfram Mathworld gives as $\ \mu_4=3n(n-2)p^2(1-p)^2 + np(1-p)\ $. Therefore: \begin{align} \text{Var}(Z)&=E\left(Z^2\right)-E(Z)^2\\ &= \frac{\mu_4}{mn^4}+\frac{(m-1)p^2(1-p)^2}{mn^2}-\frac{p^2(1-p)^2}{n^2}\\ &= \frac{\mu_4}{mn^4}-\frac{p^2(1-p)^2}{mn^2}\\ &=\frac{2(n-3)p^2(1-p)^2+p(1-p)}{mn^3} \end{align}