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

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1 Answer 1

up vote 3 down vote accepted

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.)

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but what if it's random variable ? –  Qbik May 11 '13 at 15:39

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