MSE of an estimator

Question

$R\sim Bi(n,\theta)$

i. The MLE of $\theta$ is $t_1(R) = R/n$. Find the MSE of the estimator

ii. prove that $t_2(R) = R/(n+1)$. Show that the $t_2$ has a lower MSE for $0<\theta<$ $2n+1\over 3n+1$

iii. Explain which estimator is better

I calculated that the MSE of $t_1$ is $$ \mbox{MSE}(t_1) = Var(R/n) = \frac{1}{n^2}Var(R) = \frac{\theta (1-\theta)}{n} . $$ Then $$ \mbox{MSE}(t_2)=\frac{n\theta (1-\theta)}{(n+1)^2}. $$

Edit: I think it would be smaller for all values of $\theta$ not just that range

Answer 1

The bias of $t_2$ is $$ \mbox{Bias}(t_2) = E\left (\frac{R}{n+1}\right)-\theta = \frac{n\theta}{n+1}-\theta = \theta\left(\frac{n}{n+1}-1\right) . $$ So $$ \mbox{MSE}(t_2) =\frac{n\theta (1-\theta)}{(n+1)^2} +\left\{\theta\left(\frac n{n+1}-1\right)\right\}^2 . $$ Then $$ \frac{\theta (1-\theta)}n-\frac{n\theta (1-\theta)}{(n+1)^2} -\left\{\theta\left(\frac n{n+1}-1\right)\right\}^2>0 $$

$$ \frac{\theta(-\theta-3\theta n + 2n+1)}{n(n+1)^2}>0 $$

$$ -\theta-3\theta n + 2n+1>0 $$

$$ \theta<\frac{2n+1}{3n+1} $$

iii. $t_2$ would be better for all $\theta<\frac{2n+1}{3n+1}$ as it has the lower MSE. Otherwise $t_1$ is better.