to be explicit: are you saying that you are always trying to maximize ev(log(wealth)), or only in the case of infinitely repeating log-identical bets where you can only choose a single % of your wealth to bet each time?
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If the outcomes are i.i.d., maximizing ev(log(wealth))) yields an optimal fixed fraction strategy.
If they are not i.i.d., with no restrictions on the distribution of the process, maximizing the conditionally expected log given current information is asymptotically optimal.
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what do you mean by "optimal"?
I think (?) the following strategy will beat your proposed strategy more than 50% of the time:
a) bet 1.1x Kelly until you're in the 75th percentile outcome that Kelly would be at or the 25th percentile
b) afterwards, do Kelly
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Optimal in that it induces terminal wealth W* such that E(W/W*) <=1 for all other strategies W (Cover's "Elements of Info Theory" 16.33).
Then by Markov's inequality, P(W>= tW*) <= 1/t, t>=1: an opponent can't outperform W* by a factor t with p > 1/t.
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I think you and are using different notions of optimality
These statements look true to me
But "an opponent can't outperform W* by a factor t with p > 1/t" doesn't mean much, maybe when the opponent outperforms by a factor of t, it outperforms by *much* more than t
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er to be clear does E(W/W*) mean "the linear expected value of ($ made by SBF's strategy / $ made by Kelly)"?
Because if so you can get that to be > 1
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Yes, I think's the correct interpretation
The Thorpe paper claims
"Given a strategy A which maximizes E[log X_n(A)] and any other “essentially different” strategy B (not necessarily a fixed fractional betting strategy), then
lim n→∞ X_n(A)/X_n(B) = ∞ almost surely"
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he had b/a < 1 which is false even if a/b = infinity is true
Doesn't lim X_n = infinity a.s. imply that lim 1/X_n = 0 a.s.?
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yeah he didn't use a.s. though, he used EV
But doesn't almost sure convergence imply convergence in expectation?
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nope!
compare Kelly to "all in every time".
a.s. Kelly > all in
but in EV all in > Kelly
so as t --> inf, EV(kelly / all in) --> 0 even though probability(kelly > all in) --> 1
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