The mean squared error of an estimator $\hat{\theta}$ with respect to an unknown parameter $\theta$ is defined as $$ MSE(\hat{\theta})=E[(\hat{\theta}-\theta)^{2}]. $$ It is well known that there is the following bias-variance decomposition:
$$ MSE(\hat{\theta})=Var(\hat{\theta})+\left(Bias(\hat{\theta},\theta)\right)^{2}. $$ My question: does the mean absolute deviation error $$ MADE(\hat{\theta})=E(|\hat{\theta}-\theta|) $$ have a similar decomposition?
I realise this is an old question but I was thinking about this recently. If you're ok having an inequality, you can use the triangle inequality to get:
$$ MAE(\hat{\theta}) = E(|\hat{\theta}-\theta|) $$ $$ \le E(|\hat{\theta} - E[\hat{\theta}]| + |E[\hat{\theta}] - \theta|) $$ $$ = E(|\hat{\theta} - E[\hat{\theta}]|) + E(|E[\hat{\theta}] - \theta|) $$ $$ = MAD(\hat{\theta}) + |Bias(\hat{\theta}, \theta)|. $$
The MSE bias-variance decomposition tells us we can achieve low error by reducing bias and variance. The upshot of this inequality is the same since it's $ \le $.
You could also go the other way and get
$$ MAE(\hat{\theta}) \ge \left| MAD(\hat{\theta}) - |Bias(\hat{\theta}, \theta)| \right|. $$
