# Solved – corresponding bias-variance decomposition of MSE for vectors

I know from standard theory that the bias-variance decomposition for Mean Squared Error is (for an estimator \$hat{mu}\$ of \$mu\$):

\$\$
Eleft[left(hat{mu}-muright)^2right] = Var(hat{mu}) + left(Eleft[muright]-muright)^2 = Var(hat{mu}) +Bias(hat{mu}, mu)^2
\$\$

However, if today we are talking about vectors, where \$boldsymbol{hat{mu}}\$ is an estimator of \$boldsymbol{mu}\$, both of which are \$ntimes 1\$ vectors, I was wondering if there is a corresponding nice decomposition as in the scalar case above for:

\$\$
Eleft[||boldsymbol{hat{mu}}-boldsymbol{mu}||^2right]
\$\$
?

Thanks!

Contents

Simply note that

\$\$|| widehat{mu} – mu ||^2 = sumlimits_{i = 1}^{n} (widehat{mu}_{i} – mu_{i})^2\$\$

Then, the answer is given by the decomposition you gave earlier:

\$\$ mathbb{E}[(widehat{mu}_{i} – mu_{i})^2] = Var[widehat{mu}_{i}] + [Bias(widehat{mu}_{i}, mu_{i} )]^2 \$\$

Summing all up, we get

\$\$ mathbb{E}[||widehat{mu} – mu||^2] = sumlimits_{i = 1}^{n} Var[widehat{mu}_{i}] + [Bias(widehat{mu}_{i}, mu_{i} )]^2 \$\$

Another issue, totally different, is the covariance matrix \$mathbb{E}[(widehat{mu} – mu)(widehat{mu} – mu)^t]\$.

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