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Interestingly, since the L^1 discrepancy is convex, it's possible to analytically minimize it by using the subdifferential, and thereby prove that its minimum i
by ogogmad 6y ago
Interestingly, since the L^1 discrepancy is convex, it's possible to analytically minimize it by using the subdifferential, and thereby prove that its minimum is attained at the median. The subdifferential is a set-valued generalisation of the derivative to all convex functions, including non-differentiable ones. See https://en.wikipedia.org/wiki/Subderivative https://en.wikipedia.org/wiki/Subderivative and https://towardsdatascience.com/beyond-the-derivative-subderivatives-1c4e5bf20679 https://towardsdatascience.com/beyond-the-derivative-subderi...
Whenever the subdifferential of a convex function at a point includes 0, that point is a global minimum of the function.
- amitport 6y ago"Whenever the subderivative of a convex function at a point includes 0, that point is a global minimum of the function" * you mean a strictly convex function
- Nesco 6y agoA derivative can also be obtained for the abs function interpreted as a distribution / generalized function. However I don’t think it is helpful for calculus because of the dirac measure that pop-up
- greens 6y agoL^1's derivative is a perfectly good function, it's not defined or continuous at 0, but whatever... it's for the same reason that the median handles even-sized sets in a special way. 0 = d/ds \sum_i | x_i - s | = \sum_i sign(x_i - s) which is true precisely when x_i <= s for half the elements and x_i >= s for the other half (taking sign(0) = 0, sign(+x) = 1, sign(-x) = -1).