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
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