yeah he didn't use a.s. though, he used EV
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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
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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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limsup[n->inf] (1/n log (any-other/ kelly)) <= 0 a.s.
by KKT conditions we have
E[any-other/kelly] <= 1
did not forget a log
and then yea i think E[all-in/kelly] = 0 in particular
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ah ok so I haven't read through the whole paper but I'm guessing that it's assuming that the two strategies both get the same "rolls" of the dice, as opposed to different ones
I was assuming they were different
(i.e. if we simulated on coin flips could I win one that you lose?)
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If you _do_ make this assumption then I think it's proof is correct
also though I think that anything relying on that assumption is very suspect
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i.e.:
you choose kelly, I choose "all in"
on each step each of us flips a coin; if my coin comes up heads I win, otherwise I lose. If yours comes up heads you win, otherwise you lose.
It's an even coin, 3:2 odds.
In this case E[my payout / your payout] = infinity
OK I did the math to check
so E[Sam] = 0.6^N * 2^N = (6/5)^N
Kelly = 1.2^wins * 0.8^losses
so E[1/Kelly]
= E[(5/6)^W (5/4)^L]
= E[(5/6)^N (5/6)^-L (5/4)^L]
= (5/6)^N E[(6/4)^L]
So indeed E[Sam/Kelly] = E[(6/4)^L] -> infinity
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