Timeline for A functional equation involving the inverse function
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2021 at 3:23 | vote | accept | Iosif Pinelis | ||
| Jan 27, 2021 at 13:57 | answer | added | mike | timeline score: 0 | |
| Jan 24, 2021 at 16:59 | answer | added | fedja | timeline score: 4 | |
| Jan 24, 2021 at 4:07 | comment | added | Iosif Pinelis | @MateuszKwaśnicki : Thank you for the clarification. | |
| Jan 22, 2021 at 21:41 | comment | added | Mateusz Kwaśnicki | @IosifPinelis: I meant that if $p$ is the p.d.f. of the uniform distribution over $[a,b]$, then $\epsilon p = g - g^{-1}$ can only work if $\epsilon = (b-a)^2 / n$ for $n = 1, 2, \ldots$ In this case, the area between the graphs of $g$ and $g^{-1}$ is the union of $n$ squares. | |
| Jan 22, 2021 at 15:00 | comment | added | Iosif Pinelis | @MateuszKwaśnicki : Thank you for your comment. I am not quite sure what you mean by "for a sequence of $\epsilon$s going to zero", but I guess this was done in the answer linked to the question posted on this page. I don't know an example for the posted version of goodness, with the phrase "for each small enough $\epsilon>0$" in the definition. Now I have added a relaxed version of goodness, with "for some real $\epsilon>0$" in place of "for each small enough $\epsilon>0$". For the relaxed version, there are of course trivial examples. | |
| Jan 22, 2021 at 14:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 22, 2021 at 8:50 | comment | added | Mateusz Kwaśnicki | Characterisation might be problematic. If you extend the definition slightly to allow for $g$ merely non-decreasing, then uniform distributions are not "good", but they satisfy the definition of being "good" for a sequence of $\epsilon$s going to zero. This suggests the problem has some algebraic ingredient in it, and for this reason I would not expect any nice answer. (Do we know any example of a "good" distribution?) | |
| Jan 22, 2021 at 4:07 | comment | added | Iosif Pinelis | @fedja : Thank you for your comment. This would answer the question at least in the case when $p$ is unimodal. At this point, I don't see how to prove your observations -- will be looking forward to your answer. | |
| Jan 22, 2021 at 3:48 | comment | added | Iosif Pinelis | @FedorPetrov : Thank you for your comment. I have fixed this. | |
| Jan 22, 2021 at 3:47 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 21, 2021 at 23:33 | comment | added | fedja | The situation is more or less governed by two observations: 1) If $p(x)\to 0$ at $-\infty$, then the increasing function $g(x)\ge x$ solving $g(x)-g^{-1}(x)=p(x)$ on $(-\infty,0]$ is unique. 2) If, in addition, $p$ is non-decreasing on $(-\infty,0]$, then the solution on $(-\infty,0]$ exists. The same can be done looking right instead of left, so for the triangular function on the line the answer is "no" because 2 easy solutions from the left and from the right fail to glue together properly. I hope somebody will figure it out before I find time to make a proper post... :) | |
| Jan 21, 2021 at 22:15 | comment | added | Fedor Petrov | maybe $g(x)\geqslant x$, not $g(x)>x$? | |
| Jan 21, 2021 at 15:47 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 21, 2021 at 13:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 21, 2021 at 13:37 | comment | added | Iosif Pinelis | @mike : No, I don't know that. | |
| Jan 21, 2021 at 10:30 | comment | added | mike | do you know that a piece-wise linear g won't work for the triangular pdf ? | |
| Jan 21, 2021 at 6:14 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 21, 2021 at 3:35 | comment | added | Iosif Pinelis | @AnthonyQuas : Yes. I have just added this clarification. | |
| Jan 21, 2021 at 3:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 21, 2021 at 3:20 | comment | added | Anthony Quas | I guess $g^{-1}$ is the compositional inverse of $g$; not $1/g(x)$? | |
| Jan 21, 2021 at 1:36 | history | asked | Iosif Pinelis | CC BY-SA 4.0 |