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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.

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  • $\begingroup$ 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. $\endgroup$
    – Harry Lofi
    Commented Feb 9 at 21:47
  • $\begingroup$ Do you just want a derivation of the identity? $\endgroup$
    – Dave
    Commented Feb 9 at 21:50
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    $\begingroup$ 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. $\endgroup$ Commented Feb 10 at 8:03

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$$ \text{MSE}=(\text{bias})^2+\text{var}. $$

An unbiased estimator has $\text{bias}=0$.

$$\begin{align} \text{MSE}&=0^2+\text{var}\\\implies \text{MSE}&=\text{var}.\end{align} $$

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  • $\begingroup$ I know I could count on the CV Crew to clean up the TeX I didn’t type out on mobile! $\endgroup$
    – Dave
    Commented Feb 10 at 16:37

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