Mean absolute error

Related measures

The mean absolute error is one of a number of ways of comparing forecasts with their eventual outcomes. Well-established alternatives are the mean absolute scaled error (MASE), mean absolute log error (MALE), and the mean squared error. These all summarize performance in ways that disregard the direction of over- or under- prediction; a measure that does place emphasis on this is the mean signed difference.

Where a prediction model is to be fitted using a selected performance measure, in the sense that the least squares approach is related to the mean squared error, the equivalent for mean absolute error is least absolute deviations.

MAE is not identical to root-mean square error (RMSE), although some researchers report and interpret it that way. The MAE is conceptually simpler and also easier to interpret than RMSE: it is simply the average absolute vertical or horizontal distance between each point in a scatter plot and the Y=X line. In other words, MAE is the average absolute difference between X and Y. Furthermore, each error contributes to MAE in proportion to the absolute value of the error. This is in contrast to RMSE which involves squaring the differences, so that a few large differences will increase the RMSE to a greater degree than the MAE.

Optimality property

The mean absolute error of a real variable c with respect to the random variable X is$${\displaystyle E(\left|X-c\right|).}$$Provided that the probability distribution of X is such that the above expectation exists, then m is a median of X if and only if m is a minimizer of the mean absolute error with respect to X. In particular, m is a sample median if and only if m minimizes the arithmetic mean of the absolute deviations.

More generally, a median is defined as a minimum of$${\displaystyle E(|X-c|-|X|),}$$as discussed at Multivariate median (and specifically at Spatial median). This optimization-based definition of the median is useful in statistical data-analysis, for example, in k-medians clustering.

Proof of optimality

Statement: The classifier minimising ${\displaystyle \mathbb{E}|y-\hat{y}|}$ is ${\displaystyle \hat{f}(x)=\text{Median}(y|X=x)}$ .

Proof:

The Loss functions for classification is$${\displaystyle \begin{array}{cc}L& =\mathbb{E}[|y-a||X=x]\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|y-a|f_{Y|X}(y)dy\\ & ={\int }_{-\mathrm{\infty }}^a(a-y)f_{Y|X}(y)dy+{\int }_a^{\mathrm{\infty }}(y-a)f_{Y|X}(y)dy.\end{array}}$$Differentiating with respect to ${\displaystyle a}$ gives$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }a}L={\int }_{-\mathrm{\infty }}^a{f}_{Y|X}(y)dy+{\int }_a^{\mathrm{\infty }}-f_{Y|X}(y)dy=0.}$$This means$${\displaystyle {\int }_{-\mathrm{\infty }}^{a}f(y)dy={\int }_a^{\mathrm{\infty }}f(y)dy.}$$Hence,$${\displaystyle F_{Y|X}(a)=\mathrm{0.5.}}$$