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?

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    $\begingroup$ Short answer: no. $\endgroup$
    – Glen_b
    May 11, 2014 at 23:28
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    $\begingroup$ @Glen_b thank you for your reply. Would you say it is a function of bias and variance? From an intuitive point of view, MSE is similar to MADE in terms of what it is measuring. $\endgroup$
    – Kian
    May 15, 2014 at 20:00
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    $\begingroup$ No, it's not a function of bias and variance, except perhaps in restricted circumstances (restricting consideration to particular families, for example). $\endgroup$
    – Glen_b
    May 15, 2014 at 20:02
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    $\begingroup$ @Glen_b Would it be fair to say that MAE is a function of bias, variance, and maybe some other terms, depending on the situation? $\endgroup$
    – Dave
    Jul 25, 2019 at 21:27
  • $\begingroup$ That's rather vague. In the broadest possible sense it's trivially true but perhaps not in any clearly useful way. To simplify the discussion, imagine bias is 0. They both relate to the distribution of $\hat\theta$ but you could move MAE and variance together, or either one could go up while the other went down, or one could stay constant while the other changed. $\endgroup$
    – Glen_b
    Jul 25, 2019 at 23:08


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