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?
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$\begingroup$ Perhaps "0.5" should be "50"? $\endgroup$– whuber ♦Commented Dec 10, 2019 at 16:26
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$\begingroup$ I couldn't find out is it my answer,@S.Kolassa-ReinstateMonica $\endgroup$– Farshid ShekariCommented Dec 10, 2019 at 17:13
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1 Answer
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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.
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$\begingroup$ 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. $\endgroup$ Commented Dec 12, 2019 at 5:18
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$\begingroup$ Sorry I corrected, It was $1-\gamma$ and not $\gamma -1$ $\endgroup$ Commented Dec 12, 2019 at 6:49
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$\begingroup$ 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 $\endgroup$ Commented Dec 12, 2019 at 9:54
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$\begingroup$ Can I use this loss to compare it with MAE? $\endgroup$ Commented Dec 15, 2019 at 5:14