# Mean Squared Error

The Mean Square Error (MSE) of an estimator $T$ of $\psi (\theta )$ is $MSE_{\theta }(T)=E_{\theta }$ {$[T-\psi (\theta )]^{2}$ }

• Two estimates of $\psi (\theta )$ : $T_{1}$ and $T_{1}$ If $MSE_{\theta }(T_{1})\leq {MSE}_{\theta }(T_{2})$ for every $\theta \in \Omega$ ,

then

$T_{1}$ is better than $T_{2}$ .

• $MSE_{\hat {\theta }}(T)$ ### Decomposition of MSE

$MSE_{\theta }(T)=var_{\theta }(T)+[E_{\theta }(T)-\psi (\theta )]^{2}$ PROOF

{\begin{aligned}E_{\theta }{\Bigl (}[T-\psi (\theta )]^{2}{\Bigr )}&=E_{\theta }{\Bigl (}[T-E_{\theta }(T)+E_{\theta }(T)-\psi (\theta )]^{2}{\Bigr )}\\&=E_{\theta }{\Bigl (}[T-E_{\theta }(T)]^{2}{\Bigr )}+E_{\theta }{\Bigl (}[E_{\theta }(T)-\psi (\theta )]^{2}{\Bigr )}+2E_{\theta }{\Bigl (}[T-E_{\theta }(T)][E_{\theta }(T)-\psi (\theta )]{\Bigr )}\\&={\text{var}}_{\theta }(T)+[E_{\theta }(T)-\psi (\theta )]^{2}+2[E_{\theta }(T)-\psi (\theta )]\overbrace {E_{\theta }{\Bigl (}[T-E_{\theta }(T)]{\Bigr )}} ^{0}\\&={\text{var}}_{\theta }(T)+[E_{\theta }(T)-\psi (\theta )]^{2}\end{aligned}} 