Let $ X $ be a single observation from Bernoulli density $ f(x;\theta) = \theta^{x}(1-\theta)^{1-x}$, $ x\in \lbrace 0,1\rbrace $, $ 0<\theta<1 $. Let $ t_{1}(X)=X $ and $ t_{2}(X)=\dfrac{1}{2} $. Compare de mean-squared error of $ t_{1}(x) $ with that of $ t_{2}(x) $.
My attempt:
$ MSE(t_{1}(X)) = E(X-\theta)^{2} = Var(X) = \theta(1-\theta) $
$ MSE(t_{2}(X)) = E(\dfrac{1}{2}-\theta)^{2} = (\dfrac{1}{2}-\theta)^{2} $
Here I get stuck. I do not know how to make the comparison. I know they are two parables in the interval $ (0,1) $, and when I form the quotient $ \dfrac{MSE(t_{1}(X))}{MSE(t_{2}(X))} $ I do not get anywhere. I also know that $ t_{1} $ is unbiased while $ t_{2} $ is not unbiased. I will appreciate any help.
It is possible that I do not understand what really the problem is. The $MSE(t_1(X))=\theta(1-\theta)$ is zero at $\theta=0\vee 1$ and takes its maximal value $\frac14$ at $\theta=\frac12$. Vice versa, $MSE(t_2(X))=\left(\frac12-\theta\right)^2$ is zero at $\theta=\frac12$ and takes its maximal value $\frac14$ at $\theta=0\vee 1$. So, this estimators are not comparable: for some values of $\theta$ the first mean square error is less than the second one, for some values the opposite inequality holds. It means that neither $t_1(X)$ dominates $t_2(X)$ for this loss function nor vice versa.
We can also note that
$$ \left(\frac12-\theta\right)^2 < \theta(1-\theta) \iff \left|\,\theta-\frac12\right|<\frac1{2\sqrt{2}} $$ but this information is useless until some prior knowledge about the parameter value does not restrict the domain of the parameter to the interval $\left(\frac12-\frac1{2\sqrt{2}},\,\frac12-\frac1{2\sqrt{2}}\right)$ or to its complement.