# MSE of an estimator

In the book Fundamentals of Statistical Processing, Volume I: Estimation Theory by Steven M. Kay on page 19, it say that the mean square error of an estimator, $$\hat{\theta}$$, given the true value, $$\theta$$, is defined as

\begin{align} \text{mse}(\hat{\theta}) &= E[(\hat{\theta} - \theta)^2] \\ &= E \left[ \left[\left( \hat{\theta} - E[\hat{\theta}] \right) + \left( E[\hat{\theta}] - \theta \right) \right]^2 \right]\\ &= \text{var}(\hat{\theta}) + b^2(\theta) \end{align} where $$b(\theta) = E[\hat{\theta}] - \theta$$ is the bias.

But if I write it out, it doesn't come out to be the same

\begin{align} \text{mse}(\hat{\theta}) &= E[(\hat{\theta} - \theta)^2] \\ &= E \left[ \left[\left( \hat{\theta} - E[\hat{\theta}] \right) + \left( E[\hat{\theta}] - \theta \right) \right]^2 \right]\\ &= E \left[ \left( \hat{\theta} - E[\hat{\theta}] \right)^2 + \left( E[\hat{\theta}] - \theta \right)^2 + 2 \left( \hat{\theta} - E[\hat{\theta}] \right) \left( E[\hat{\theta}] - \theta \right) \right]\\ &= \text{var}(\hat{\theta}) + b^2(\theta) + 2 E \left[\left( \hat{\theta} - E[\hat{\theta}] \right) \left( E[\hat{\theta}] - \theta \right) \right] \end{align}

What am I missing here?

HINT:

$$E\left[\hat \theta - E\left(\hat \theta \right)\right] = E\left(\hat \theta\right) - E\left(\hat \theta\right)$$

• well now i feel stupid :) thanks anyways
– Zero
Dec 30 '20 at 16:32
• It's important that $\theta$ is modeled as a deterministic parameter rather than a random variable. That's why we can take the other factor out of the expectation. Dec 30 '20 at 17:13
• @MattL. yeah...i just didn't even think about expanding further and stopped like in the question...thanks :)
– Zero
Jan 1 at 10:50
• @MattL. It is indeed true that for classical estimation (contrary to Bayesian) $\theta$ is deterministic. Jan 1 at 14:13

It's important to point out here that Kay is talking about classical estimation, which means that the parameters to be estimated are unknown but deterministic. Hence, the term $$E[\hat{\theta}]-\theta$$ is deterministic, and, consequently, the last term in your equation becomes

\begin{align} E \left[\left( \hat{\theta} - E[\hat{\theta}] \right) \left( E[\hat{\theta}] - \theta \right) \right]&= \left(E[\hat{\theta}]-\theta\right)E\left[\hat{\theta}-E[\hat{\theta}]\right]\\&=\left(E[\hat{\theta}]-\theta\right)\left(E[\hat{\theta}]-E[\hat{\theta}]\right) \\&=0\end{align}