Timeline for Optimal Kelly criterion for process with N discrete outcomes
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8, 2023 at 22:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 10, 2023 at 19:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 8, 2023 at 15:34 | answer | added | David E Speyer | timeline score: 1 | |
| Feb 8, 2023 at 7:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 9, 2023 at 5:28 | answer | added | apcris | timeline score: 0 | |
| Aug 3, 2022 at 8:52 | comment | added | lotuspaperboy | An approximation I came up with is: $$\sum_{n=1}^N{p_nb_n\over1+b_n f}=0$$ $$\sum_{n=1}^N{p_nb_n}(1+b_nf)^{-1}=0$$ $$\sum_{n=1}^N{p_nb_n}(1-b_nf+(b_nf)^2-(b_nf)^3+(b_nf)^4-....)=0$$ Cutting things off afters the third term of each expansion and solving the polynomial gives: $$f \approx{1\over2}\left({\sum_{n=1}^Np_nb_n^2\over\sum_{n=1}^Np_nb_n^3}-\sqrt{\left({\sum_{n=1}^Np_nb_n^2\over\sum_{n=1}^Np_nb_n^3}\right)^2-4\left({\sum_{n=1}^Np_nb_n\over\sum_{n=1}^Np_nb_n^3}\right)}\right) $$ It's unwieldy but does seem to work. An exact solution would be great though! | |
| S Aug 1, 2022 at 20:15 | review | First questions | |||
| Aug 2, 2022 at 6:20 | |||||
| S Aug 1, 2022 at 20:15 | history | asked | lotuspaperboy | CC BY-SA 4.0 |