Quantile loss 50th is MAE, is it?

Question

I'm not sure the above sentence is true, but I read it here, here and here that quantile loss function percentile 0.5 is MAE (mean absolute error). Is it true (yes or no)? And How?

I couldn't find out is it my answer,@S.Kolassa-ReinstateMonica — Farshid Shekari, Dec 10 '19 at 17:13

Davide ND · Answer 1 · 2019-12-12T06:48:33.130

9

Well mathematically speaking, quantile loss for quantile $\gamma$ is defined as:

$L_{\gamma}(y,y^p) = \sum_{i:y_i<\hat{y}_i}(1-\gamma)|y_i-\hat{y}_i| + \sum_{i:y_i\geq \hat{y}_i}(\gamma)|y_i-\hat{y}_i|$

For $\gamma=0.5$ (median), this becomes:

$L_{0.5}(y,y^p) = \sum_{i:y_i<\hat{y}_i}\frac{1}{2}|y_i-\hat{y}_i| + \sum_{i:y_i\geq \hat{y}_i}\frac{1}{2}|y_i-\hat{y}_i| = \sum\frac{1}{2}|y_i-\hat{y}_i|$

Considering that the MAE is:

$MAE = \frac{1}{n}\sum|y_i-\hat{y}_i|$

the two only differ by a constant, and they are therefore equivalent in terms of optimization.

edited Dec 12 '19 at 06:48

answered Dec 10 '19 at 16:38

Davide ND

2,930

according to the first equation, after replacing 0.5 in that, one of the coefficients is -0.5 and the second one is 0.5, and after summation, these coefficients disappear. – Farshid Shekari Dec 12 '19 at 05:18
Sorry I corrected, It was $1-\gamma$ and not $\gamma -1$ – Davide ND Dec 12 '19 at 06:49
I realized it might not have been clear - I just passed the two coefficients in fraction form because it was easier to compare to MAE – Davide ND Dec 12 '19 at 09:54
Can I use this loss to compare it with MAE? – Farshid Shekari Dec 15 '19 at 05:14

Quantile loss 50th is MAE, is it?

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