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

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

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

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



1 Answer 1


Simply note that

$$|| \widehat{\mu} - \mu ||^2 = \sum\limits_{i = 1}^{n} (\widehat{\mu}_{i} - \mu_{i})^2$$

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

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

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


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