You’re leaving out geometric growth of successive bets. Kelly maximizes expected geometric growth rate. Therefore over enough bets Kelly maximizes expected, i.e. mean, wealth, not merely median wealth.
As GuySrinavasan says, do the math. It doesn’t work out. Maximizing geometric growth rate is not the same as maximizing mean value. It turns out Kelly favors the first at a severe cost to the second.
This is my big motivator for writing stuff like this: discussions of Kelly usually prove an optimality notion like expected growth rate, and then leave it to the reader to notice that this doesn’t at all imply more usual optimality notions. Most readers don’t notice; it’s very natural to assume that “Kelly maximizes growth rate” entails “Kelly maximizes expected wealth”.
But if Kelly maximized expected wealth, then that would probably have been proved instead of this geometric-growth-rate property. You have to approach mathematics the same way you approach political debates, sometimes. Keep an eye out for when theorems answer something only superficially similar to the question you would have asked.
A bettor who can make an infinite number of expected profitable bets is going to outperform one who can only make a finite number of bets.
(any number between 1 and 0 exclusive)^infinity=0, i.e. for an infinite series of bets, the probability of ruin with naive EV maximization is 1. So, expected value is actually −1x your bet size.
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
Let’s consider a max-expectation bettor on a double-or-nothing bet with an 80% probability of paying out.
My expected value per dollar in this bet is $1.60, whereas the expected value of a dollar in my pocket is $1. So I maximize expected value by putting all my money in. If I start with $100, my expected value after 1 round is $160. The expected value of playing this way for two rounds is $100x1.6x1.6 = $256. In general, the expected value of this strategy is 100 ⋅1.6n.
The Kelly strategy puts 60% of its money down, instead. So in expectation, the Kelly strategy multiplies the money by .6⋅1.6+.4=1.36.
So after one round, the Kelly bettor has $136 in expectation. After two rounds, about $185. In general, the Kelly strategy gets an expected value of 100⋅1.36n.
So, after a large number of rounds, the all-in strategy will very significantly exceed the Kelly strategy in expected value.
I suspect you will object that I’m ignoring the probability of ruin, which is very close to 1 after a large number of rounds. But the expected value doesn’t ignore the probability of ruin. It’s already priced in: the expected value of 1.6 includes the 80% chance of success and the 20% chance of failure: .8⋅2+.2⋅0. Similarly, the $256 expected value for two rounds already accounts for the chance of zero; you can see how by multiplying out 100⋅(.8⋅2+.2⋅0)2 (which shows the three possibilities which have value zero, and the one which doesn’t). Similarly for the nth round: the expected value of 100⋅1.6nalready discounts the winnings by the (tiny) probability of success. (Otherwise, the sum would be $2^n instead.)
I thought something like this the first time I saw abramdemski’s pushback. Then I actually did the math in some simple cases. Try doing the math to find a sequence where Kelly beats naive wealth maximization. You will convince either yourself or abramdemski!
You’re leaving out geometric growth of successive bets. Kelly maximizes expected geometric growth rate. Therefore over enough bets Kelly maximizes expected, i.e. mean, wealth, not merely median wealth.
As GuySrinavasan says, do the math. It doesn’t work out. Maximizing geometric growth rate is not the same as maximizing mean value. It turns out Kelly favors the first at a severe cost to the second.
This is my big motivator for writing stuff like this: discussions of Kelly usually prove an optimality notion like expected growth rate, and then leave it to the reader to notice that this doesn’t at all imply more usual optimality notions. Most readers don’t notice; it’s very natural to assume that “Kelly maximizes growth rate” entails “Kelly maximizes expected wealth”.
But if Kelly maximized expected wealth, then that would probably have been proved instead of this geometric-growth-rate property. You have to approach mathematics the same way you approach political debates, sometimes. Keep an eye out for when theorems answer something only superficially similar to the question you would have asked.
A bettor who can make an infinite number of expected profitable bets is going to outperform one who can only make a finite number of bets.
(any number between 1 and 0 exclusive)^infinity=0, i.e. for an infinite series of bets, the probability of ruin with naive EV maximization is 1. So, expected value is actually −1x your bet size.
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
Let’s consider a max-expectation bettor on a double-or-nothing bet with an 80% probability of paying out.
My expected value per dollar in this bet is $1.60, whereas the expected value of a dollar in my pocket is $1. So I maximize expected value by putting all my money in. If I start with $100, my expected value after 1 round is $160. The expected value of playing this way for two rounds is $100x1.6x1.6 = $256. In general, the expected value of this strategy is 100 ⋅1.6n.
The Kelly strategy puts 60% of its money down, instead. So in expectation, the Kelly strategy multiplies the money by .6⋅1.6+.4=1.36.
So after one round, the Kelly bettor has $136 in expectation. After two rounds, about $185. In general, the Kelly strategy gets an expected value of 100⋅1.36n.
So, after a large number of rounds, the all-in strategy will very significantly exceed the Kelly strategy in expected value.
I suspect you will object that I’m ignoring the probability of ruin, which is very close to 1 after a large number of rounds. But the expected value doesn’t ignore the probability of ruin. It’s already priced in: the expected value of 1.6 includes the 80% chance of success and the 20% chance of failure: .8⋅2+.2⋅0. Similarly, the $256 expected value for two rounds already accounts for the chance of zero; you can see how by multiplying out 100⋅(.8⋅2+.2⋅0)2 (which shows the three possibilities which have value zero, and the one which doesn’t). Similarly for the nth round: the expected value of 100⋅1.6n already discounts the winnings by the (tiny) probability of success. (Otherwise, the sum would be $2^n instead.)
I thought something like this the first time I saw abramdemski’s pushback. Then I actually did the math in some simple cases. Try doing the math to find a sequence where Kelly beats naive wealth maximization. You will convince either yourself or abramdemski!