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I've only barely looked into the Kelly Criterion, but can someone explain the intuition behind maximizing the expected value of the log of your wealth? Trying t

I've only barely looked into the Kelly Criterion, but can someone explain the intuition behind maximizing the expected value of the log of your wealth? Trying the same derivation mentioned in the article but without the logarithm: the expected value comes out to 0.5×(1 + 1.1×f) + 0.5×(1 − f) = 1 + 0.05f which would make it seem that betting the entire fraction always maximizes your expected value. But why does this reasoning break down in the long term, and why does maximizing the log seem to make it work?

OscarCunningham 8y ago
The reasoning doesn't break down in the long term. If you repeatedly bet your entire fortune then you will have a very high probability of losing all of it, but you will also have a very small probability of having a huge amount of money. If your utility really is linear then this is worth it.
n4r9 8y ago
Maximising the expected return of a single throw is different to maximising the growth rate over many throws. This is perhaps a bit unintuitive, for the reason that the expectation doesn't tell you a great amount about a distribution. You could imagine a similar game in which one choice of bet results in an enormous return with a tiny probability, or else ruin. Whilst your expected return method would tell you to repeatedly make this choice, doing so is clearly insane.
chrischattin 8y ago
You're right if you only have one coin flip. But, this assumes a series of flips. The way I understand the Kelly Criterion it is it's similar to a fancy Martingale. It's statistically inevitable that X number of coin flips will go against you. So, the smaller % of your capital at risk, the less your risk of ruin. And, if you have an edge in the coin flip, the more flips you get, the better you'll perform in the long run. But, it's not that great of a proxy for real life as explained by other commenters in this thread.
imh 8y ago
Don't look at the expectation of a single throw. Look at the distribution of the outcome after many throws. Relate that to the product of each throw's relative return. Look at how the log expectation of the product of n random variables converges as n gets big, and you see what to optimize.
- imh 8y ago
  (too late for me to edit my post, so I'll reply to it) The logic I posted above will show you that the log of your money X converges in probability to some value x that you can optimize. Optimizing x will optimize a monotonic utility f(x), which seems useful, but isn't. The problem is that just because X converges in probability to x, doesn't mean that f(X) converges in probability to f(x). Probability and convergence are weird.
JoshuaDavid 8y ago
Betting everything maximizes your mean wealth. Using the Kelly Criterion maximizes your median wealth. Let's say there are 1024 (2^10) people with $1 each, and a bet which is 50 / 50 of returning either 3x or 0x (i.e. you bet $1 and either get back $3 or $0). Let's further specify that there will be 10 rounds of betting. Using the "bet everything" strategy, you end up with the following distribution of outcomes: 10 wins, 0 losses -- 1 person: $59,049 1 or more losses -- 1023 people: $0 The average ending balance is $57.67, while the median ending balance is $0. Using the Kelly Criterion, one will bet 25% of the bankroll each time. You will end up with the following distribution of outcomes. 0 wins, 10 losses -- 1 person: $57.66 1 wins, 9 losses -- 10 people: $28.83 2 wins, 8 losses -- 45 people: $14.41 3 wins, 7 losses -- 120 people: $7.20 4 wins, 6 losses -- 210 people: $3.60 5 wins, 5 losses -- 252 people: $1.80 6 wins, 4 losses -- 210 people: $0.90 7 wins, 3 losses -- 120 people: $0.45 8 wins, 2 losses -- 45 people: $0.22 9 wins, 1 losses -- 10 people: $0.11 10 wins, 0 losses -- 1 person: $0.05 The average ending balance is $3.25, while the median ending balance is $1.80. Incidentally, the choice of maximizing the median is somewhat arbitrary. You can also use a similar approach to maximize, say, the 25th percentile outcome, at the expense of average and median outcomes (You would bet 10% of the stake each time, yielding a 25th percentile of $1.08, a median of $1.25, and a mean return of $1.28).
- szemet 8y ago
  "the median is somewhat arbitrary" I don't know if it matters, but: in the bet everything scenario you end up with 1023 people who will be poorer than they would be in the Kelly scenario. In the 25th percentile maximizing case, you end up with 852 people who will be poorer than in the Kelly case. Imagine you are running a bank, then in both of your alternative cases a large majority of people would be upset: "If you used Kelly Criterion we would have been richer"
- OscarCunningham 8y ago
  If you're allowed to bet a different proportion of your money in each turn then you can push the median even higher than the Kelly criterion. For example consider the strategy that bets nothing if it has more than $2.98, whatever is needed to reach $2.98 on the next bet if it has less than $2.98, or bets everything if it can't reach $2.98 on the next bet. This strategy has a greater than 50% chance of ending up with $2.98 since it has a 50% chance of success on the first bet and a nonzero probability of success after that. So its median outcome is $2.98. I don't know which strategy actually maximises the median, but I suspect it involves being conservative after you've been lucky and agressive after you've been unlucky.
Double_Cast 8y ago
Thought experiment: Suppose you bet 100% of your bankroll every round. If at any point you lose a round, your bankroll is now $0. Any money you made from former rounds is for naught. Whoops. bankroll_final = (bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n) $0 = (bankroll_initial)(210% )(210% )(210% )(...)(0% ) Taking the "non-log" Expected Value would be optimal if your bankroll were "renewed" to the same constant each round. Because the outcome of each individual round would be independent from other rounds. Bankroll_final = (bankroll)(round_1) + (bankroll)(round_2) + (...) + (bankroll)(round_n) But since the outcome of each round depends on previous rounds, we want to optimize for Expected Value of Growth. For which we'll use a geometric mean rather than an arithmetic mean. Also, EV[bankroll_final] = EV[(bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n)] is equivalent to EV[bankroll_final] = EV[(bankroll_initial) e^ln(r1 + r2 + r3 + (...) + r_n)] which allows us to discuss wagers in terms of e^(x) and in terms of growthrates.

OscarCunningham 8y ago

The reasoning doesn't break down in the long term. If you repeatedly bet your entire fortune then you will have a very high probability of losing all of it, but you will also have a very small probability of having a huge amount of money. If your utility really is linear then this is worth it.

n4r9 8y ago

Maximising the expected return of a single throw is different to maximising the growth rate over many throws. This is perhaps a bit unintuitive, for the reason that the expectation doesn't tell you a great amount about a distribution. You could imagine a similar game in which one choice of bet results in an enormous return with a tiny probability, or else ruin. Whilst your expected return method would tell you to repeatedly make this choice, doing so is clearly insane.

chrischattin 8y ago

You're right if you only have one coin flip. But, this assumes a series of flips. The way I understand the Kelly Criterion it is it's similar to a fancy Martingale. It's statistically inevitable that X number of coin flips will go against you. So, the smaller % of your capital at risk, the less your risk of ruin. And, if you have an edge in the coin flip, the more flips you get, the better you'll perform in the long run. But, it's not that great of a proxy for real life as explained by other commenters in this thread.

imh 8y ago

Don't look at the expectation of a single throw. Look at the distribution of the outcome after many throws. Relate that to the product of each throw's relative return. Look at how the log expectation of the product of n random variables converges as n gets big, and you see what to optimize.

imh 8y ago
(too late for me to edit my post, so I'll reply to it) The logic I posted above will show you that the log of your money X converges in probability to some value x that you can optimize. Optimizing x will optimize a monotonic utility f(x), which seems useful, but isn't. The problem is that just because X converges in probability to x, doesn't mean that f(X) converges in probability to f(x). Probability and convergence are weird.

JoshuaDavid 8y ago

Betting everything maximizes your mean wealth. Using the Kelly Criterion maximizes your median wealth. Let's say there are 1024 (2^10) people with $1 each, and a bet which is 50 / 50 of returning either 3x or 0x (i.e. you bet $1 and either get back $3 or $0). Let's further specify that there will be 10 rounds of betting. Using the "bet everything" strategy, you end up with the following distribution of outcomes: 10 wins, 0 losses -- 1 person: $59,049 1 or more losses -- 1023 people: $0 The average ending balance is $57.67, while the median ending balance is $0. Using the Kelly Criterion, one will bet 25% of the bankroll each time. You will end up with the following distribution of outcomes. 0 wins, 10 losses -- 1 person: $57.66 1 wins, 9 losses -- 10 people: $28.83 2 wins, 8 losses -- 45 people: $14.41 3 wins, 7 losses -- 120 people: $7.20 4 wins, 6 losses -- 210 people: $3.60 5 wins, 5 losses -- 252 people: $1.80 6 wins, 4 losses -- 210 people: $0.90 7 wins, 3 losses -- 120 people: $0.45 8 wins, 2 losses -- 45 people: $0.22 9 wins, 1 losses -- 10 people: $0.11 10 wins, 0 losses -- 1 person: $0.05 The average ending balance is $3.25, while the median ending balance is $1.80. Incidentally, the choice of maximizing the median is somewhat arbitrary. You can also use a similar approach to maximize, say, the 25th percentile outcome, at the expense of average and median outcomes (You would bet 10% of the stake each time, yielding a 25th percentile of $1.08, a median of $1.25, and a mean return of $1.28).

szemet 8y ago
"the median is somewhat arbitrary" I don't know if it matters, but: in the bet everything scenario you end up with 1023 people who will be poorer than they would be in the Kelly scenario. In the 25th percentile maximizing case, you end up with 852 people who will be poorer than in the Kelly case. Imagine you are running a bank, then in both of your alternative cases a large majority of people would be upset: "If you used Kelly Criterion we would have been richer"
OscarCunningham 8y ago
If you're allowed to bet a different proportion of your money in each turn then you can push the median even higher than the Kelly criterion. For example consider the strategy that bets nothing if it has more than $2.98, whatever is needed to reach $2.98 on the next bet if it has less than $2.98, or bets everything if it can't reach $2.98 on the next bet. This strategy has a greater than 50% chance of ending up with $2.98 since it has a 50% chance of success on the first bet and a nonzero probability of success after that. So its median outcome is $2.98. I don't know which strategy actually maximises the median, but I suspect it involves being conservative after you've been lucky and agressive after you've been unlucky.

Double_Cast 8y ago

Thought experiment: Suppose you bet 100% of your bankroll every round. If at any point you lose a round, your bankroll is now $0. Any money you made from former rounds is for naught. Whoops. bankroll_final = (bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n) $0 = (bankroll_initial)(210% )(210% )(210% )(...)(0% ) Taking the "non-log" Expected Value would be optimal if your bankroll were "renewed" to the same constant each round. Because the outcome of each individual round would be independent from other rounds. Bankroll_final = (bankroll)(round_1) + (bankroll)(round_2) + (...) + (bankroll)(round_n) But since the outcome of each round depends on previous rounds, we want to optimize for Expected Value of Growth. For which we'll use a geometric mean rather than an arithmetic mean. Also, EV[bankroll_final] = EV[(bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n)] is equivalent to EV[bankroll_final] = EV[(bankroll_initial) e^ln(r1 + r2 + r3 + (...) + r_n)] which allows us to discuss wagers in terms of e^(x) and in terms of growthrates.