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
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Doesn't lim X_n = infinity a.s. imply that lim 1/X_n = 0 a.s.?
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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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I don't think EV(all in) > EV(kelly) implies EV(kelly / all in) -> 0
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ugh sorry try this on for size:
compare Kelly to "all in every time".
a.s. Kelly > all in
but in EV all in > Kelly
so as t --> inf, EV(all in/ kelly) --> infinity even though probability(kelly > all in) --> 1
I'm not convinced EV(all in / kelly) -> infinity
I think it can be consistent that
(a) EV[all in] > EV[kelly]
(b) kelly / all in -> infinity almost surely
(c) EV[all in / kelly] -> 0
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pretty sure it's infinity
try this on:
a) EV(all in / kelly) = sum(ai_n/kelly_n)/N
b) ai and kelly are uncorrelated
c) so sum(ai_n / kelly_n)/N > sum(ai_n/average_kelly)/N
d) so you can pull out the average_kelly from the sum, and get that EV(all in/kelly) > EV(AI)/EV(kelly)
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