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Let $M$ be some metric space. We will also say that $M$ is a measurable space.

How do you find a probability measure $P$ on $M$ that maximizes the expected distance between two points? (That is, $P$ maximizes $E[d(X,Y)]$, where $X$ and $Y$ are independent random variables with distribution $P$.)

If $M$ is a discrete metric, then $P$ will the uniform distribution, for example.

Note that we can think of this as a game, and $P$ as a mixed strategy. You would have two players, which are both trying to maximize the distance. This is not a purely game-theoretical situation though, since we need to require the players to pick the same strategy. There is something called superrationality that might apply, but it is not well studied though. Maybe techniques similar to those used in game theory could be used though.

Chain Markov
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Christopher King
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  • You need a measure on $M^2$ (i.e. the joint distribution of 2 random points on $M$). –  Feb 04 '19 at 20:36
  • @d.k.o. wouldn't that just be a product space of $M$ with itself? – Christopher King Feb 04 '19 at 20:38
  • Do you mean the product measure $P\otimes P$? Maybe, if you assume independence. –  Feb 04 '19 at 20:40
  • @d.k.o. yes, I want to assume they are i.d.d. Otherwise, you could just pick to points that are a diameter apart, and always choose them. The point is that you need to choose randomly, because if you don't, the expected distance is guaranteed to be $0$. – Christopher King Feb 04 '19 at 20:42
  • Makes sense. I think you need to include this info in the question. –  Feb 04 '19 at 20:43
  • @d.k.o. Okay, I edited the question. Is it well formed and does it make more sense now? – Christopher King Feb 04 '19 at 20:53
  • $Ed (X,Y)$ can be $\infty$. – Kavi Rama Murthy Feb 04 '19 at 23:39
  • @KaviRamaMurthy Yes – Christopher King Feb 04 '19 at 23:45
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    Very interesting question, but it depends a lot on the details of the metric space. If your space is ${0,\pm1}$, though, with $d(x,y)=|x-y|$, I think the optimum is to pick $\pm1$ uniformly, not to pick all 3 points uniformly, though. – kimchi lover Feb 05 '19 at 16:11
  • @kimchilover - I imagine your observation can be generalized: If the space is a real interval $[a,b]$, and $d(x,y) = |x-y|$, the optimum is to pick $a,b$ uniformly. This seems "obvious" but I can't prove it... Your thoughts? – antkam May 31 '19 at 05:46
  • @kimchilover - But $1/2 > 1/3$ so my proposal works. Sorry if my last comment wasn't clear: I meant to pick among just the two endpoints ${a,b}$ uniformly, i.e. with $1/2$ prob each. (Just like how you picked from $\pm 1$ and ignored the $0$. Indeed if endpoint picking optimizes the $[-1, 1]$ interval case, it also obviously optimizes the ${-1, 0, 1}$ case, which is why I call it a generalization.) – antkam May 31 '19 at 12:06
  • @antkam Sorry: I misread your comment. You might be right. I'll try to think about this. – kimchi lover May 31 '19 at 12:09

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