Does $\operatorname{MSE}(\hat{\theta}) = \operatorname{Var}(\theta)+ \left(\operatorname{Bias}(\hat{\theta},\theta)\right)^2$?

Question

We know that
$\operatorname{MSE}(\hat{\theta})=\operatorname{E}\left[(\hat{\theta}-\theta)^2\right]$ and
$\operatorname{MSE}(\hat{\theta})=\operatorname{Var}(\hat{\theta})+ \left(\operatorname{Bias}(\hat{\theta},\theta)\right)^2$
Is it true that :
$\operatorname{MSE}(\hat{\theta})=\operatorname{Var}(\theta)+ \left(\operatorname{Bias}(\hat{\theta},\theta)\right)^2$
?

Answer 1

If that were true, then one would have $\operatorname{var}(\hat\theta))=\operatorname{var}(\theta)$. The first two equalities above are identities in frequentist estimation, and in that context, $\theta$ is constant rather than random, so $\operatorname{var}(\theta)=0$.

(In other words: No.)