show that MSE$(\hat{\theta}) = E[(\hat{\theta} − θ)^2] = V(\hat{\theta}) + (B(\hat{\theta}))^2$.

Question

Using the identity $(\hat{\theta} − θ) = [\hat{\theta} − E(\hat{\theta})] + [E(\hat{\theta}) − θ] = [\hat{\theta} − E(\hat{\theta})] + B(\hat{\theta})$,

I need to show that MSE$(\hat{\theta}) = E[(\hat{\theta} − θ)^2] = V(\hat{\theta}) + (B(\hat{\theta}))^2$.

I got MSE$(\hat{\theta}) = E[(\hat{\theta} − θ)^2] = E[([(\hat{\theta}-E(\hat{\theta})]+[E(\hat{\theta})-{\theta}])^2] = E[([\hat{\theta} - E(\hat{\theta})] + B(\hat{\theta}))^2] = E[[\hat{\theta} - E(\hat{\theta})]^2+2B(\hat{\theta})[\hat{\theta}-E(\hat{\theta})]+B(\hat{\theta})^2] = E[[\hat{\theta} - E(\hat{\theta})]^2+2(E(\hat{\theta}) - \theta)E[\hat{\theta}-E(\hat{\theta})]+E[(E(\hat{\theta}) - \theta)^2] $

I am struct here. Can anyone help

Answer 1

It's really a lot simpler than that. I would begin with the RHS instead: $$\begin{align*} {\rm Var}[\hat \theta] + {\rm Bias}[\hat\theta]^2 &= {\rm E}[\hat \theta^2] - {\rm E}[\hat \theta]^2 + ({\rm E}[\hat \theta] - \theta)^2 \\ &= {\rm E}[\hat \theta^2] - 2 \theta {\rm E}[\hat \theta] + \theta^2 \\ &= {\rm E}[\hat \theta^2 - 2 \theta \hat \theta + \theta^2] \\ &= {\rm E}[(\hat\theta - \theta)^2] \\ &= {\rm MSE}[\hat \theta]. \end{align*}$$ If you want to prove it using the hint, the above computation should help you see how to do it.

Answer 2

Expand the square, $$ \begin{align} (\hat\theta-\theta)^2&=(\hat\theta-{\rm E}[\hat\theta]+{\rm E}[\hat\theta]-\theta)^2\\ &=(\hat\theta-{\rm E}[\hat\theta])^2+({\rm E}[\hat\theta]-\theta)^2+2(\hat\theta-{\rm E}[\hat\theta])({\rm E}[\hat\theta]-\theta) \end{align} $$ and take expectation to get $$ {\rm E}[(\hat\theta-\theta)^2]=\mathrm{V}(\hat\theta)+\mathrm{B}(\hat\theta)^2 $$ since $$ {\rm E}[2(\hat\theta-{\rm E}[\hat\theta])({\rm E}[\hat\theta]-\theta)]=0. $$