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Mean of values (x1, ... xi,...xn) will minimize Mean Square Error, Median of values will minimize Mean Absolute Error.

I would intuite Median would also minimize Mean Absolute Relative Error (MARE) where:

$MARE(x)=\sum|\frac{x-x_i}{x_i}|$

... but at the end of the day I'm not able to prove it, and I didn't find any demonstration on this topic.

Many thanks for your help.


Following @Dap's answer I've drawn a short experiment to visually figure out the problem. This illustrates the weighted median minimizes the Mean Absolute Relative Error and the difference between the 3 metrics (MAE, MSE and MARE).

Minization of MAE, MSE, MARE for 12 points

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  • $\begingroup$ Did you mean Mean Relative Absolute Error (MRAE)? $\endgroup$ Jun 18, 2020 at 21:18

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This is a special case of the (discrete) weighted medians with weights $1/x_i,$ which are a special case of the medians of a measure.

They're not the same as the usual medians. For example $1,1000,1001$ has unique median $1000$ but the MARE is minimized by $1.$

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