Very simple question about trace

Question

In MSE in econometrics,
$$\mathrm{MSE}=E\| \hat{\theta}-\theta \|^2$ $=E(\hat{\theta}-\theta)'(\hat{\theta}-\theta)$$

$$=E(\hat{\theta}-E\hat{\theta}+E\hat{\theta}-\theta)'(\hat{\theta}-E\hat{\theta}+E\hat{\theta}-\theta)$$

$$=E(\hat{\theta}-E\hat{\theta})'(\hat{\theta}-E\hat{\theta})+(E\hat{\theta}-\theta)'(E\hat{\theta}-\theta)$$

$$=\operatorname{tr}(\operatorname{Var}(\hat{\theta}))+\| E\hat{\theta}-\theta \|^2$$

Here, why the last trace term appeared?

@Vishal MSE=Mean Squared Error, $\theta$=parameter, $\hat{\theta}$=estimator. Let me know if you need more information! — signup, Sep 16 '13 at 17:12

score 0 · Accepted Answer · answered Sep 16 '13 at 18:19

Note that: $$ E(\hat{\theta}-E(\hat{\theta}))^\prime E(\hat{\theta}-E(\hat{\theta}))=E\begin{bmatrix}\hat{\theta}_1-E(\hat{\theta}_1),&\dots,&\hat{\theta}_n-E(\hat{\theta}_n)\end{bmatrix}\begin{bmatrix}\hat{\theta_1}-E(\hat{\theta}_1)\\\vdots\\ \hat{\theta}_n-E(\hat{\theta}_n)\end{bmatrix}=\sum_{i=1}^nV(\hat{\theta}_i) $$

This is the diagonal of the variance covariance matrix of the coefficients. The trace operator gives you just that - the sum of the diagonal elements.

Very simple question about trace

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