# Quantile loss 50th is MAE, is it? [duplicate]

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?

• Perhaps "0.5" should be "50"? – whuber Dec 10 '19 at 16:26
• I couldn't find out is it my answer,@S.Kolassa-ReinstateMonica – Farshid Shekari Dec 10 '19 at 17:13

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.

• 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 5:18
• Sorry I corrected, It was $1-\gamma$ and not $\gamma -1$ – Davide ND Dec 12 '19 at 6:49
• 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 9:54
• Can I use this loss to compare it with MAE? – Farshid Shekari Dec 15 '19 at 5:14