# Mean Squared Error for point estimation

I am attempting to understand Mean Squared Error when evaluating point estimators for particular parameters of interest. The book we are reading for class states the following:

The mean squared error (MSE) of an estimator $$W$$ of a parameter $$\theta$$ is the function $$\theta$$ defined by $$E_{\theta} (W-\theta)^2 = \text{Var}_{\theta}W + (E_{\theta} W - \theta)^2=\text{Var}_{\theta} W + (\text{Bias}_{\theta} W)^2$$

Where we define the bias of an estimator as follows:

$$\text{Bias}_{\theta} W = E_{\theta}W-\theta$$

An estimator whose bias is identically (in $$\theta$$) equal to $$0$$ is called unbiased and satisfies $$E_{\theta}W=\theta$$ for all $$\theta$$.

For an unbiased estimator, we have $$E_{\theta} (W-\theta)^2 = \text{Var}_{\theta}W$$, and so if an estimator is unbiased its MSE is equal to its variance.

I am confused as to why we are just left with the variance of the estimator when the estimator is unbiased. I think this is due to variability that may come from the sample from which we are computing our estimator, but I am not 100% sure if this is the right thinking.

• I know where it comes from in the formula, but my question is asking more about how the variance part is derived in the MSE formula. Commented Feb 9 at 21:47
• Do you just want a derivation of the identity?
– Dave
Commented Feb 9 at 21:50
• I cannot understand this bit: The mean squared error (MSE) of an estimator $W$ of a parameter $\theta$ is the function $\theta$. How come $\theta$ is both the parameter and the function? There must be a word missing or something. Commented Feb 10 at 8:03

$$\text{MSE}=(\text{bias})^2+\text{var}.$$
An unbiased estimator has $$\text{bias}=0$$.
\begin{align} \text{MSE}&=0^2+\text{var}\\\implies \text{MSE}&=\text{var}.\end{align}