= Solution
The estimator is unbiased because $\mathbb E_\theta X=n\theta$, and
$$
\operatorname{Var}_\theta(\widehat\theta)
=\frac1{n^2}\operatorname{Var}_\theta(X)
=\frac{\theta(1-\theta)}n.
$$
The <bias-variance decomposition of mean squared error> therefore gives
$$
\boxed{f(\theta)
=\mathbb E_\theta[(\widehat\theta-\theta)^2]
=\frac{\theta(1-\theta)}n}.
$$
The quadratic $\theta(1-\theta)$ is maximized at $\theta=1/2$, so
$$
\boxed{\sup_{0<\theta<1}f(\theta)=\frac1{4n}}.
$$
Back to article page
