E.g. two portfolios each facing a different bet with probability of winning p. The amount to bet is different than a single portfolio.
You may be arguing that they should have access to the same bets, but then you're conceding you can't use Kelly in an isolated case.
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Yes! I'm also curious if there's a closed form for general n? (or even specific n>2?)
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Replying to @danrobinson @elliot_olds and 2 others
This seems not true?
Take two pots with wealth x,y
Give them independent coins with p=0.75 of heads
Let m = 2p-1
Pot-Kelly says bet mx, my in each pot
Portfolio-Kelly says bet min(x, (x+y)*m/(1+m^2)) in the first pot
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This is not true in the limit as t approaches infinity. You can’t beat the expected asymptotic geometric growth rate of per-pot Kelly in that case (assuming you can’t rebalance between pots). Agree?
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You're claiming this, even when I can provide each pot with *independent* opportunities that *are not* available to any other pot?
Do you have a proof/argument for this?
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I think so, at least if each pot’s opportunities don’t change over time. (If they do, it might still hold in some cases but it depends on my expectations about how those opportunities change over time, so it becomes pretty complicated)
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If each pot’s opportunities don’t change over time, then one of them has a higher log return than the other. (Or they are equal.) That pot is the only pot that matters asymptotically
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If each pot has the same quality of opportunity (coins with same bias, but different coins), my intuition is you want to bet more than Kelly because the risk of a string of bad luck putting you into a deep hole is lower.
The more coins you have the more certain outcomes are.
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The issue with that logic is that you can’t rebalance between pots. So when your pot gets in a hole, it is stuck in it
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My intuition agrees with - note that you can bet *more* than Kelly and still have positive growth rate (but less than Kelly)
So maybe we can just bet Kelly fraction + epsilon in each pot, and while the *individual* pots have lower growth rate, their *sum* might not
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i.e. E[log(X)], E[log(Y)] < E[log(Z)] does not imply that E[log((X+Y)/2)] < E[log(Z)]
Let me see if I can turn this into a proof either way
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OK so I think:
1) is wrong given his assumptions
2) if you *also* assume that there aren't two pots which have coins with *exactly* the same growth rate, is right given his assumptions
3) his assumptions are stupid
4) given good assumptions, he's wrong
If there are two pots with the same growth rate, and you can’t rebalance between them, you think you can beat the growth rate?
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(2) here feels unnecessary
either two pots with the same growth rate r can have their sum have growth rate higher than r, in which case is wrong even with (2)
or any finite sum has growth rate r anyway, in which case is right even without (2)