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Let $\hat{\theta}_1$, $\hat{\theta}_2$ and $\hat{\theta}_3$ be the estimators of $\theta$. We know that $E(\hat{\theta}_1) = E(\hat{\theta}_2) = \theta$, $E(\hat{\theta}_3) \not= \theta$, $V(\hat{\theta}_1)=12$, $V(\hat{\theta}_2)=10$ and $E[(\hat{\theta}_3 - \theta)]^2 = 6$. Which one of the estimators do you prefer?

Answer: $MSE(\hat{\theta}_1) = 12$, $MSE(\hat{\theta}_2) = 10$ and $MSE(\hat{\theta}_3) = 6$. $\hat{\theta}_3$ is the best

I am very very confused. Why $MSE(\hat{\theta}_3) = 6$? This is true only if $E[(\hat{\theta}_3 - \theta)^2] = 6$, but this is not explicit this is true. For me, $E[(\hat{\theta}_3 - \theta)^2] \not= E[(\hat{\theta}_3 - \theta)]^2$

EDIT:

In my book,

  • $MSE(\hat{\theta}) = E[(\hat{\theta} - \theta)^2] = V(\hat{\theta}) + [Bias(\hat{\theta})]^2$
  • $Bias(\hat{\theta}) = E(\hat{\theta}) - \theta$
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  • $\begingroup$ Are you sure you're not just quoting a typographical error? Surely the original question meant to stipulate that $E[(\hat\theta_3-\theta)^2]=6.$ $\endgroup$ – whuber Nov 11 '20 at 14:53
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    $\begingroup$ Yes, this is what I thought. It is an error in the book $\endgroup$ – David Nov 11 '20 at 14:56

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