Timeline for Is KL divergence $D(P||Q)$ strongly convex over $P$ in infinite dimension
Current License: CC BY-SA 4.0
31 events
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| Aug 2, 2018 at 1:01 | vote | accept | Maziar Sanjabi | ||
| Aug 2, 2018 at 1:01 | history | bounty ended | Maziar Sanjabi | ||
| Aug 2, 2018 at 1:01 | comment | added | Maziar Sanjabi | oh, sorry. You are absolutely right. In my computations, I was using a wrong definition for $f_t$. Thank you so much for your help. | |
| Aug 2, 2018 at 0:55 | comment | added | Iosif Pinelis | Yes, the equality holds, with the details just described in my previous comment. | |
| Aug 2, 2018 at 0:47 | comment | added | Maziar Sanjabi | Sorry, I meant equality. My question was: Does, in this case, the left-hand side value for $\delta$ equal the right-hand side value, i.e. the integral? | |
| Aug 2, 2018 at 0:33 | comment | added | Iosif Pinelis | I don't see any lower bound on $\delta$ in the beginning of the proof, or anywhere in the proof. The only inequalities in the proof occur only at the very end of it, providing a lower bound (equal $\|P_1-P_0\|^2$) on $I(u)$ and, in particular, on $I(u_j(t,s))$. Anyhow, on the set where $f_1=0$ but $f_0>0$, we have $\delta=(1-t)t\,f_0^2\,\int_0^1\Big(\frac t{(1-u_0(t,s))f_0}+\frac{1-t}{(1-u_1(t,s))f_0}\Big)(1-s)\,ds=-(1-t)\ln(1-t)\,f_0$; I don't know if this helps in understanding the proof; please let me know. | |
| Aug 1, 2018 at 23:30 | comment | added | Maziar Sanjabi | Thanks. Now I understand the proof much better. One naive question though: at the beginning of the proof what happens to the lower bound on $\delta$ when $f_1=0$ but $f_0>0$. Does this cause any problem in the bound and the following results? | |
| Aug 1, 2018 at 12:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 12:20 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 12:09 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 12:06 | comment | added | Iosif Pinelis | And, yes, the strong convexity does hold for probability measures on any measurable spaces, finite or not, and it holds whether or not $P_0$ and $P_1$ are absolutely continuous w.r. to each other. Indeed, none of these conditions were needed or used in the proof. | |
| Aug 1, 2018 at 12:01 | comment | added | Iosif Pinelis | Concerning your questions: In the original proof, I had to deal with $\int\frac{(f_1-f_0)^2}{f_0}\,dQ$ and $\int\frac{(f_1-f_0)^2}{f_1}\,dQ$, and therefore I then made the technical assumption that $P_0$ and $P_1$ are absolutely continuous w.r. to each other. Now I only deal with $\int\frac{(f_1-f_u)^2}{f_u}\,dQ$ with $u\in(0,1)$, and for such $u$ the measure $P_1$ is always absolutely continuous w.r. to $P_u$. This is now made explicit in the answer as well. I don't see a way (or need) to further simplify the proof, by using Pinsker's inequality or by other means. | |
| Aug 1, 2018 at 12:00 | comment | added | Iosif Pinelis | The proof is now further simplified, distilled just to the original key ideas. | |
| Aug 1, 2018 at 11:45 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 11:40 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 4:43 | comment | added | Maziar Sanjabi | Are you sure about not needing the absolute continuity condition of $P_1$ and $P_0$ with respect to each other? Where was it exactly used previously and how is it not needed anymore? It is a bit curious as it is not clear to me if the strong convexity holds in the finite dimensions when $P_1$ and $P_2$ are not absolutely continuous wrt each other. Also could the end of the proof be simplified using Pinsker inequality? | |
| Aug 1, 2018 at 3:41 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 3:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 3:34 | comment | added | Iosif Pinelis | The previous, better (and optimal) bound is now restored. The proof is greatly simplified; computer algebra is no longer needed. Also, the mutual absolute continuity condition for $P_0$ and $P_1$ is no longer needed or used. | |
| Aug 1, 2018 at 3:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2018 at 1:25 | comment | added | Iosif Pinelis | The inequality for $u(t)$ went the wrong way. Now this is fixed, albeit the new lower bound is slightly worse, and the proof is now more complicated. As for the mutual absolute continuity condition for $P_0$ and $P_1$, I believe it can be rather easily removed by approximation, say by adding small positive constants to the densities $f_0$ and $f_1$ and then renormalizing to keep the condition $\int f_j\,dQ=1$. | |
| Aug 1, 2018 at 1:16 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 20:27 | comment | added | Maziar Sanjabi | And also why is this inequality true: $u(t):=\frac1{f_{2t/3}}+\frac1{f_{(2t+1)/3}}\ge(1-t)u(0)+tu(1)$ | |
| Jul 31, 2018 at 20:25 | comment | added | Maziar Sanjabi | So the absolute continuity of $dP_1$ and $dP_2$ w.r. to each other seems to be absolutely necessary for the proof (without it the Taylor expansion would fail, right?). In the problem that motivated my question on this, $dP_1$ and $dP_2$ are optimal solutions to the following optimization (assuming everything is well behaved): $P_i = \arg\min_{P} \int (C_i~dP) + \lambda KL(P||Q)$ (subject to some other constraints on the marginals of $P$). And I am not sure if it would be possible to make sure the $P_1$ and $P_2$ would be absolutely continuous w.r.t each other. | |
| Jul 31, 2018 at 20:22 | comment | added | Iosif Pinelis | I have added a remark on the optimality of the lower bound. | |
| Jul 31, 2018 at 20:21 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 15:12 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 14:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 14:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 14:13 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2018 at 14:05 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |