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Aug 22 at 13:55 history edited Daniele Tampieri CC BY-SA 4.0
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Aug 22 at 11:42 history edited Aryeh Kontorovich CC BY-SA 4.0
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Aug 16 at 8:38 comment added Aryeh Kontorovich Induction -- as well as all other "local" methods seem to fail badly here. I can prove the minimizer claim for $n=2$, but not that it's unique. The minimizer for odd $n$ has all $p_i$s equal and all $q_i$s too but otherwise doesn't have the nice, simple form that even $n$ does.
Aug 15 at 18:25 comment added Thomas Steinke I'm tempted to try proving this by induction on $n$, but the fact that it only applies to even $n$ seems like a big barrier. Is there a form of the conjecture that holds for all $n$? Can you at least prove it for $n=2$?
Aug 13 at 18:02 history edited Aryeh Kontorovich CC BY-SA 4.0
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Aug 13 at 17:41 comment added Yury Let us continue this discussion in chat.
Aug 13 at 17:25 history edited Aryeh Kontorovich CC BY-SA 4.0
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Aug 13 at 17:25 comment added Aryeh Kontorovich you're right -- it's not convex! but the claim about the minimum still appears to be true
Aug 13 at 16:56 comment added Yury My plot shows that it is not a convex function of $x$ even when all $p_i = x + 0.1$ and $q_i = x$, $n = 3$.
Aug 13 at 16:48 comment added Aryeh Kontorovich Actually now less sure.
Aug 13 at 16:45 comment added Yury Are you sure that the objective is convex?
Aug 13 at 15:58 comment added Aryeh Kontorovich Note that this is false for odd n, so where does your argument fail there?
Aug 13 at 15:49 comment added Yury Once we prove that the objective is convex, the rest follows easily from symmetry. The optimal solution must be invariant under the permutation group (acting on the indices of $p_i$ and $q_i$) and "reflections": $(p, q) \mapsto (1 - q, 1- p)$. The only fixed point is your solution.
Aug 13 at 14:57 history edited Aryeh Kontorovich CC BY-SA 4.0
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Aug 13 at 11:29 history asked Aryeh Kontorovich CC BY-SA 4.0