# $MSE(\hat{\theta}_3) = 6$?

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,

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