Timeline for What does it mean L1 loss is not differentiable?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2023 at 0:43 | comment | added | IntegrateThis | Alternatively you can sample uniformly from the sub-derivative at 0 (the closed interval [-1, 1]). There are many functions that are not differentiable at certain points for which this approach of sub-gradient sampling helps. | |
| Oct 4, 2019 at 11:39 | comment | added | alexeymosco | @mrgloom, researchers back-door this problem by adding a small value to x when it is zero, I have seen such solutions working in practice. | |
| Oct 3, 2019 at 9:30 | comment | added | mrgloom | Having Fact 1: L1 loss used in practice in regression, Fact 2: L1 loss not differentiable at x=0 what conclusions can we make? Option 1: L1 loss not differentiable at x=0 is not a problem Option 2: In practice people somehow overcome this problem while minimizing L1 loss, i.e. adding epsilon to x, when x is 0? | |
| Oct 2, 2019 at 22:01 | comment | added | whuber♦ | Apart from making the theory based on differentiable loss functions inapplicable, getting the wrong answers, or having your calculations fail, you mean? | |
| Oct 2, 2019 at 21:16 | comment | added | mrgloom | I understand that derivative not exist at x=0, but what practical problems can arise from this fact? | |
| Oct 2, 2019 at 18:38 | comment | added | Dave | Explicitly, the derivative is undefined at $x=0$. | |
| Oct 2, 2019 at 18:36 | history | answered | Tomasz Bartkowiak | CC BY-SA 4.0 |