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. 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 ofE(|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 \mathbb{E}|y-\hat{y}| is \hat{f}(x)=\text{Median}(y|X=x) . Proof: The
Loss functions for classification is\begin{align} L &= \mathbb{E}[|y-a||X=x]\\ &= \int_{-\infty}^{\infty} |y-a| f_{Y|X}(y) \, dy \\ &= \int_{-\infty}^a (a-y)f_{Y|X}(y)\, dy+\int_a^{\infty}(y-a)f_{Y|X}(y)\, dy.\\ \end{align}Differentiating with respect to a gives\frac{\partial }{\partial a}L = \int_{-\infty}^af_{Y|X}(y)\, dy+\int_a^{\infty}-f_{Y|X}(y)\, dy=0 .This means\int_{-\infty}^a f(y)\, dy = \int_a^{\infty} f(y)\, dy .Hence,F_{Y|X}(a)=0.5 . ==See also==