4 ms·

I think the Kelly criterion doesn't apply as widely as people think it does. Its derivation is based on on maximising the growth rate of your fortune. But this

I think the Kelly criterion doesn't apply as widely as people think it does. Its derivation is based on on maximising the growth rate of your fortune. But this is equivalent to assuming that money has logarithmic utility for you. If you don't value money in a logarithmic way then you shouldn't use the Kelly criterion. Personally I feel that my utility function is sublogarithmic. If I'm just spending on myself then beyond a certain point additional money makes me absolutely no happier. Note that the usual justification of progressive taxation also assumes sublogarithmic utility. So based on this we should be more conservative than Kelly. On the other hand, if I plan to give money to charity then my utility function is almost linear. Big charites can absorb a lot of money without becoming less effective. So in this case you should be maximally aggressive, betting everything at every opportunity. Sometimes people say that because the Kelly criterion maximises growth rate it will be the best "in the long run" even if your utility function isn't logarithmic. But I've never seen any evidence of this. Does anybody know of a toy model where you can prove the Kelly criterion is optimal even if your utility is linear?

whitepoplar 8y ago
Not sure I follow. Unless I'm misunderstanding, wouldn't Kelly apply especially for your case? As in, Kelly is concerned with avoiding absorbing barriers and increasing bets when they're on "house money" (given a known edge, of course).
- OscarCunningham 8y ago
  Kelly isn't about either of those things. It's a consequence of Kelly that it never hits the absorbing barrier, but lots of other strategies also have that property. What I'm saying is that among all of those strategies, Kelly isn't optimal.
  - whitepoplar 8y ago
    I misunderstood then, gotcha.
tim333 8y ago
The other thing is that in real life your investment options are basically never a series of biased coin flips.
- phkahler 8y ago
  Coin flipping was an example. One must estimate the probabilities for their actual bets and run the numbers for a specific opportunity. You'll very quickly realize that getting probabilities in real-world situations is difficult. But that doesn't negate the lessons from the concept.
evrydayhustling 8y ago
You are misunderstanding. Optimizing log wealth isn't because you have logarithmic utility... It's because doing so maximizes growth, which is more important that immediate return in a game of repeated investment. Kelly maximizes your long term absolute wealth in such games. Edit: I see you last paragraph acknowledging but doubting this claim. Demonstrating an optimal strategy in a stochastic game is hard to make toy-like, because you have to somehow describe both the probability distributions and the entire range of strategies someone could take. In lieu, the end of this article does a good job of listing simple explanations of what that (provably) optimal strategy property entails: https://www.stat.berkeley.edu/~aldous/157/Papers/Good_Bad_Kelly.pdf https://www.stat.berkeley.edu/~aldous/157/Papers/Good_Bad_Ke...
imh 8y ago
>Sometimes people say that because the Kelly criterion maximises growth rate it will be the best "in the long run" even if your utility function isn't logarithmic. But I've never seen any evidence of this. Does anybody know of a toy model where you can prove the Kelly criterion is optimal even if your utility is linear? Here's one https://greek0.net/blog/2018/04/18/kelly_criterion3/ https://greek0.net/blog/2018/04/18/kelly_criterion3/ Basically it says that if you are making bets where money_(i+1) = f_i(money_i, x_i), such that your money always remains above zero, then you can apply the product form of the law of large numbers http://www.jams.or.jp/scm/contents/e-2006-6/2006-60.pdf http://www.jams.or.jp/scm/contents/e-2006-6/2006-60.pdf That means that over a long enough period any betting strategy that maximizes the geometric mean of the rates will beat any other bet with probability approaching 1. p(money(optimal strategy) > money(other strategy)) --> 1. If your utility is monotonic (x > y implies that u(x) > u(y)) then I think this also implies that p(utility(money(optimal strategy)) > utility(money(other strategy))) --> 1. Basically, you are eventually almost sure to have more money with this strategy than any other. If more money implies more utility, then you are eventually almost sure to have more utility with this strategy than any other.
- OscarCunningham 8y ago
  > this also implies that p(utility(money(optimal strategy)) > utility(money(other strategy))) --> 1 That's true (at least as long as the other strategy also bets a constant proportion of wealth each turn). But it doesn't mean that Kelly is optimal. It could be that in the (increasingly unlikely) cases when the other strategy beats Kelly, the utility produced by the other strategy is much greater than that produced by Kelly. Then the other strategy could still be better overall. In other words, even though we have p(utility(money(Kelly strategy)) > utility(money(other strategy))) --> 1 we also have Expectation[utility(money(Kelly strategy)) - utility(money(other strategy))] < 0 Here's a simplified example of what's going on: consider the bet where you get a penny with probability 1-1/n, and otherwise you lose $(2^n). I think this bet gets worse and worse larger n gets, but the probability of having higher utility if you take the bet tends to 1.
  - imh 8y ago

whitepoplar 8y ago

Not sure I follow. Unless I'm misunderstanding, wouldn't Kelly apply especially for your case? As in, Kelly is concerned with avoiding absorbing barriers and increasing bets when they're on "house money" (given a known edge, of course).

OscarCunningham 8y ago
Kelly isn't about either of those things. It's a consequence of Kelly that it never hits the absorbing barrier, but lots of other strategies also have that property. What I'm saying is that among all of those strategies, Kelly isn't optimal.
- whitepoplar 8y ago
  I misunderstood then, gotcha.

tim333 8y ago

The other thing is that in real life your investment options are basically never a series of biased coin flips.

phkahler 8y ago
Coin flipping was an example. One must estimate the probabilities for their actual bets and run the numbers for a specific opportunity. You'll very quickly realize that getting probabilities in real-world situations is difficult. But that doesn't negate the lessons from the concept.

evrydayhustling 8y ago

You are misunderstanding. Optimizing log wealth isn't because you have logarithmic utility... It's because doing so maximizes growth, which is more important that immediate return in a game of repeated investment. Kelly maximizes your long term absolute wealth in such games. Edit: I see you last paragraph acknowledging but doubting this claim. Demonstrating an optimal strategy in a stochastic game is hard to make toy-like, because you have to somehow describe both the probability distributions and the entire range of strategies someone could take. In lieu, the end of this article does a good job of listing simple explanations of what that (provably) optimal strategy property entails: https://www.stat.berkeley.edu/~aldous/157/Papers/Good_Bad_Kelly.pdf https://www.stat.berkeley.edu/~aldous/157/Papers/Good_Bad_Ke...

imh 8y ago

>Sometimes people say that because the Kelly criterion maximises growth rate it will be the best "in the long run" even if your utility function isn't logarithmic. But I've never seen any evidence of this. Does anybody know of a toy model where you can prove the Kelly criterion is optimal even if your utility is linear? Here's one https://greek0.net/blog/2018/04/18/kelly_criterion3/ https://greek0.net/blog/2018/04/18/kelly_criterion3/ Basically it says that if you are making bets where money_(i+1) = f_i(money_i, x_i), such that your money always remains above zero, then you can apply the product form of the law of large numbers http://www.jams.or.jp/scm/contents/e-2006-6/2006-60.pdf http://www.jams.or.jp/scm/contents/e-2006-6/2006-60.pdf That means that over a long enough period any betting strategy that maximizes the geometric mean of the rates will beat any other bet with probability approaching 1. p(money(optimal strategy) > money(other strategy)) --> 1. If your utility is monotonic (x > y implies that u(x) > u(y)) then I think this also implies that p(utility(money(optimal strategy)) > utility(money(other strategy))) --> 1. Basically, you are eventually almost sure to have more money with this strategy than any other. If more money implies more utility, then you are eventually almost sure to have more utility with this strategy than any other.

OscarCunningham 8y ago
> this also implies that p(utility(money(optimal strategy)) > utility(money(other strategy))) --> 1 That's true (at least as long as the other strategy also bets a constant proportion of wealth each turn). But it doesn't mean that Kelly is optimal. It could be that in the (increasingly unlikely) cases when the other strategy beats Kelly, the utility produced by the other strategy is much greater than that produced by Kelly. Then the other strategy could still be better overall. In other words, even though we have p(utility(money(Kelly strategy)) > utility(money(other strategy))) --> 1 we also have Expectation[utility(money(Kelly strategy)) - utility(money(other strategy))] < 0 Here's a simplified example of what's going on: consider the bet where you get a penny with probability 1-1/n, and otherwise you lose $(2^n). I think this bet gets worse and worse larger n gets, but the probability of having higher utility if you take the bet tends to 1.
- imh 8y ago