Timeline for Minimising the f-divergence to a conditional probability constraint
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2018 at 13:06 | vote | accept | King Kong | ||
| Feb 19, 2018 at 12:17 | comment | added | Iosif Pinelis | @BenEva : You can always replace $f$ by $\tilde f:=f-c$, where $c$ is any constant; in particular, you can take $c=f(1)$, thus making $\tilde f(1)=0$, if desired. Then $D_f(P||Q)$ will get replaced by $D_{f-c}(P||Q)=D_f(P||Q)-c$, which clearly will not affect the minimizing $Q$ at all. | |
| Feb 19, 2018 at 11:26 | comment | added | King Kong | Thanks a lot for the very helpful answer (and sorry if the question wasn't clearly phrased). I just have one more question. Isn't it normally assumed that f(1) = 0 for any f-divergence? This doesn't seem to be satisfied by the counterexample? | |
| Feb 18, 2018 at 5:42 | comment | added | Iosif Pinelis | I have added a remark about how the above counterexample was obtained. | |
| Feb 18, 2018 at 4:51 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 18, 2018 at 4:39 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 16, 2018 at 21:35 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 16, 2018 at 21:24 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |