# Is there a 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 $\mu$):

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

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:

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

Thanks!

## 1 Answer

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 $\mathbb{E}[(\widehat{\mu} - \mu)(\widehat{\mu} - \mu)^t]$.