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
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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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and EV(ai) = infinity * EV(kelly) so EV(all in / kelly) = infinity
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(a) What are you indexing over here?
EV(all in / kelly) = sum(ai_n/kelly_n)/N
I think it should be something like EV[A/K] = sum_{n,m} P(A_n, K_m) *a_n / k_m
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(b) "so sum(ai_n / kelly_n)/N > sum(ai_n/average_kelly)/N"
I'm not sure I follow - why is this true?
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er hm ok so one question: are we assuming in E(all in / kelly) that we're we-rolling the dice between them or that they come up the same?
I.e. are they correlated?
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Good question - it looks like in the literature, they are assumed to be the *same* sequence of rolls, i.e correlated
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