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Reading this answer as to why minimizing MAE results in median forecasts, I did not fully understand why MAE is not unique!

  • What is exactly meant by this?
  • Can someone give a specific example of situation and model where MAE is not unique?
  • Can someone explain how to deal with this situation?

Thanks in advance.

jj_coder
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    You already know that a value minimizing the MAE is called a median. By consulting this site (or other places) about medians you will find it is almost never uniquely defined when the amount of data is even. Consider, therefore, the dataset $0,1.$ For any median $m\in[0,1],$ compute the MAE as $(|m-0|+|m-1|)/2=1/2$ and notice it does not vary with $m.$ For details, please visit my discussion of "Absolute $(L_1)$ Loss" at https://stats.stackexchange.com/a/114363/919. – whuber Apr 27 '22 at 15:16
  • Thank you for your answer. However I do not understand your conclusion, are you saying MAE is indeed not unique or MAE is unique?

    From my point of view MAE is defined independently of the median. Whether minimizing MAE provides an estimation for the median is another thing.

    What I do not understand is why the post in my original question claims that MAE is not uniquely defined.

     – jj_coder Apr 27 '22 at 17:12
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    "Does not vary with $m$" implies $m$ is not uniquely determined. The median can be defined (not merely estimated!) as a value minimizing the MAE: see https://stats.stackexchange.com/questions/251600 for a general account. Thus, to be clear, the smallest value attained by the MAE is definite, but the MAE itself (namely, any $m$ minimizing the MAE) is usually not unique. – whuber Apr 27 '22 at 17:16
  • Thank you for your answer, let me reflect upon it for a while...! – jj_coder Apr 27 '22 at 18:33
  • Consider a statement like "the lowest scoring player on the team has hit 2 home runs"; that minimum value of "2" is unique (you can't have it be both 2 and 3, for example),, but there may well be several players who scored only 2 home runs. – Glen_b Apr 28 '22 at 10:33
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    Suppose you had the observations $(1,100), (2,500), (3,900), (4,300)$. It looks as if higher values of $x$ seem to give higher values of $y$ though with big fluctuations. Perhaps something like $y_i=100x_i+200+\epsilon_i$ would be a reasonable fit, and it is in fact MAE with $\sum |\epsilon_i|=1000$. But so too are $y_i=-50x_i+550+\epsilon_i$ with the slope reversed or $y_i=300+\epsilon_i$ with no slope or many other possibilities, all MAE with $\sum |\epsilon_i|=1000$ – Henry Jun 09 '22 at 16:26

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