Suppose my estimator estimate the number of heads obtained by fliiping a coin with probability $p$ as $\frac{H}{N}$. Then I need to find the MSE of this estimator. My calculation is as follows:
\begin{equation} \begin{aligned} % \begin{split} bias(\hat{\theta}_{MLE}) &= \mathbb{E}[\hat{\theta}_{MLE}] - \theta \\ &= \mathbb{E}[\frac{H}{N}] - p \\ &= \frac{\mathbb{E}[H]}{N} - p \\ &=\frac{Np}{N} - p \\ &= p - p = 0 \\ \end{aligned} \end{equation}
\begin{equation} \begin{aligned} var(\hat{\theta}_{MLE}) &= \mathbb{E}[(\hat{\theta}_{MLE})^2] - (\mathbb{E}[\theta])^2 \\ &= \mathbb{E}[(\frac{H}{N})^2] - p^2 \\ &= \frac{\mathbb{E}[H^2]}{N^2} - p^2 \\ &= \frac{N^2p^2 - Np^2 + Np}{N^2} - p^2 \\ &= p^2 + \frac{Np - Np^2}{N^2} - p^2 \\ &= \frac{p-p^2}{N} \\ \end{aligned} \end{equation}
\begin{equation} \begin{aligned} MSE(\hat{\theta}_{MLE}) &= var(\hat{\theta}_{MLE}) + (bias(\hat{\theta}_{MLE}))^2 \\ &= \frac{p-p^2}{N} + 0 \\ &= \frac{p-p^2}{N} \end{aligned} \end{equation}
I am not sure if it is correct, inparticular, I am not sure if i can take $E[\frac{H}{N}] = \frac{E[H]}{N}$.
Hope that someone can verify my solution.
It is perfectly fine, as you take $N$ to be constant. Note that basically your estimator is the sample mean $\bar{X}_N$ where $X \sim Ber (1, p)$, hence $$ Var(H/N) = Var(\bar{X}_N)=\frac{1}{N^2}\sum_{i=1}^NVar(X_i)=\frac{Np(1-p)}{N^2}=\frac{p(1-p)}{N}. $$