The source of disagreement seems to be about how to compute the EV “in the limit of infinite bets”.
I.e given n bets with a 1/2 chance of winning each, where you triple your stake with each bet, the naive EV maximization strategy gives you a total expect value of (3/2)n, which is also the maximum achievable overall EV. Does this entail that the EV at infinite bets is ∞? No, because with probability one, you’ll lose one of the bets and end up with zero money.
I don’t find this argument for Kelly super convincing.
You can’t actually bet an infinite number of times, and any finite bound on the number of bets, even if it’s 1010101010, immediately collapses back to the above situation where naive EV-maximization also maximizes the overall expected value. So this argument doesn’t actually support using Kelly over naive EV maximization in real life.
There are tons of strategies other than Kelly which achieve the goal of infinite EV in the limit. Looking at EV in the limit doesn’t give you a way of choosing between these. You can compare them over finite horizons and notice that Kelly gives you better EV than others here (maximal geometric growth rate).… but then we’re back to the fact that over finite time horizons, naive EV does even better than any of those.
This argument doesn’t work because limits don’t commute with integrals (including expected values). (Since practical situations are finite, this just tells you that the limiting situation is not a good model).
To the extent that the experiment with infinite bets makes sense, it definitely has EV 0.
We can equip the space Ω=∏∞n=1{0,1} with a probability measure corresponding to independent coinflips, then describe the payout using naive EV maximization as a function Ω→[0,∞] - it is ∞ on the point (1,1,…) and 0 everywhere else. The expected value/integral of this function is zero.
EDIT: To make the “limit” thing clear, we can describe the payout after n bets using naive EV maximization as a function fn:Ω→[0,∞], which is 3n if the first n values are 1, and 0 otherwise.
Then E(fn)=(3/2)n, and f=limfn (pointwise), but E(f)=0.
The corresponding functions g,gn:Ω→[0,∞] corresponding to the EV using a Kelly strategy have E(gn)<E(fn) for all n, but E(g)=∞
Oooooh. Neat. Thank you. I guess… how do we know EV(lim(fn))=0? I don’t know enough analysis anymore to remember how to prove this. [reads internet] Well, Wikipedia tells me two functions with the same values everywhere but measure 0 even if those values are +inf have the same integral, so looks good. :D
The source of disagreement seems to be about how to compute the EV “in the limit of infinite bets”. I.e given n bets with a 1/2 chance of winning each, where you triple your stake with each bet, the naive EV maximization strategy gives you a total expect value of (3/2)n, which is also the maximum achievable overall EV. Does this entail that the EV at infinite bets is ∞? No, because with probability one, you’ll lose one of the bets and end up with zero money.
I don’t find this argument for Kelly super convincing.
You can’t actually bet an infinite number of times, and any finite bound on the number of bets, even if it’s 1010101010, immediately collapses back to the above situation where naive EV-maximization also maximizes the overall expected value. So this argument doesn’t actually support using Kelly over naive EV maximization in real life.
There are tons of strategies other than Kelly which achieve the goal of infinite EV in the limit. Looking at EV in the limit doesn’t give you a way of choosing between these. You can compare them over finite horizons and notice that Kelly gives you better EV than others here (maximal geometric growth rate).… but then we’re back to the fact that over finite time horizons, naive EV does even better than any of those.
It’s worse than that. The EV at infinite bets is actually ∞ even for naive EV maximization. WolframAlpha link
This argument doesn’t work because limits don’t commute with integrals (including expected values). (Since practical situations are finite, this just tells you that the limiting situation is not a good model).
To the extent that the experiment with infinite bets makes sense, it definitely has EV 0. We can equip the space Ω=∏∞n=1{0,1} with a probability measure corresponding to independent coinflips, then describe the payout using naive EV maximization as a function Ω→[0,∞] - it is ∞ on the point (1,1,…) and 0 everywhere else. The expected value/integral of this function is zero.
EDIT: To make the “limit” thing clear, we can describe the payout after n bets using naive EV maximization as a function fn:Ω→[0,∞], which is 3n if the first n values are 1, and 0 otherwise. Then E(fn)=(3/2)n, and f=limfn (pointwise), but E(f)=0.
The corresponding functions g,gn:Ω→[0,∞] corresponding to the EV using a Kelly strategy have E(gn)<E(fn) for all n, but E(g)=∞
lim(EV(fn)) != EV(lim(fn))
Oooooh. Neat. Thank you. I guess… how do we know EV(lim(fn))=0? I don’t know enough analysis anymore to remember how to prove this. [reads internet] Well, Wikipedia tells me two functions with the same values everywhere but measure 0 even if those values are +inf have the same integral, so looks good. :D