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
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