5 ms·

Ergodicity Economics and von Neumann-Morgenstern utility

3 points by ssivark 6y ago

kgwgk 6y ago
> I have seen claims that the von Neumann Morgenstern utility theorem (vNM) grounding Expected Utility Theory (EUT) voids Ergodicity Economics (EE) and time averages Really? I've seen the opposite claims. > EE is not in opposition with the vNM framework! I agree. Ole Peters doesn't agree, though. That Nature Physics articule is full of nonsense: > Expected utility theory predicts that people are insensitive to changes in the dynamics. (...) Ergodicity economics predicts something quite different. > A positive result (...) would corroborate ergodicity economics and falsify expected utility theory (insofar as experiments falsify models).
- ssivark 6y ago
  There's a proverbial slip between the cup and the lip. vNM does not (technically) assume a probability distribution, but EUT as commonly applied add one extra piece on top of vNM -- implicitly ignoring dynamics and assuming a distribution which leads to "ensemble averaging".
  - kgwgk 6y ago
    What do you mean? vNM doesn’t assume any particular probability distribution (or any particular utility function). But it does assume the existence of a probability distribution (and a utility function). How could they define an expected value otherwise?
    - ssivark 6y ago
      Sure, no qualms with vNM. But how does an application of EUT decide what distribution to use? (That's where the subtlety/assumption lies) Example situation: 1 million tosses of a fair coin, with multiplicative payoff -- wealth doubles for heads and halves for tails. Say you start with a dollar -- what payoff should you expect? Is "expected payoff" (ensemble averaging) the right calculation to model that?
      - kgwgk 6y ago
        I think you're mixing things up. The distribution is not something that you choose arbitrarily. It's given. It's a part of the problem statement. Starting with $1 the distribution of outcomes after 4 doubles-or-halves flips is P($0.0625) = 1/16, P($0.25) = 4/16, P($1) = 6/16, P($4) = 4/16, P($16) = 1/16. After one million flips it will be more complex (and more skewed to the upside, it will go from close to $0 to $10^301030) but also well defined. When you say "what distribution to use" you may mean "what utility function to use". You're right that one may not one want to use the final payoff to represent the utility of each outcome. That's the whole point of introducing a utility function. The logarithm has been a popular choice of utility function for centuries. Among other reasons, because it maximizes the asymptotic growth rate. The distribution of utility values becomes in the 4 flips case (for simplicity, the million flips case would be similar): P(u = -log(16)) = 1/16, P(u = -log(4)) = 4/16, P(u = 0) = 6/16, P(u = log(4)) = 4/16, P(u = log(16)) = 1/16 The expected utility (the ensemble average of the utilities of the potential outcomes, if you will) is zero. The same that for $1 without any uncertainty. Assuming logarithmic utility, you would be indifferent between keeping your dollar or getting the outcome of the million-flips game. BUT that would be assuming that your whole fortune, present and future, depended on that dollar. Otherwise you need to include the rest of your wealth when calculating the logarithmic utility. Or your preferences may not be well represented by that utility function: you'll be more willing to risk your only dollar today when you know you're getting your paycheck tomorrow. If you're a normal person and you're offered to play the game you will prefer to do so than to keep $1 with certainty. You have 49% probability of getting at least $50m while in the worse case your lose $1.
        ssivark 6y ago
        > The distribution is not something that you choose arbitrarily. It's given. It's a part of the problem statement. Starting with $1 the distribution of outcomes after 4 doubles-or-halves flips is P($0.0625) = 1/16, P($0.25) = 4/16, P($1) = 6/16, P($4) = 4/16, P($16) = 1/16. The fundamental claim of EE is that the distribution cannot be fixed that way to be 1/N for each of the N trajectories in the limit that the number of trajectories far exceeds the number of agents. (Ref: concentration of measure) Whether you agree or disagree, at least we are now at the crux of the matter. I would encourage you re-read my post with this update.

3 points by ssivark 6y ago

> I have seen claims that the von Neumann Morgenstern utility theorem (vNM) grounding Expected Utility Theory (EUT) voids Ergodicity Economics (EE) and time averages Really? I've seen the opposite claims. > EE is not in opposition with the vNM framework! I agree. Ole Peters doesn't agree, though. That Nature Physics articule is full of nonsense: > Expected utility theory predicts that people are insensitive to changes in the dynamics. (...) Ergodicity economics predicts something quite different. > A positive result (...) would corroborate ergodicity economics and falsify expected utility theory (insofar as experiments falsify models).

ssivark 6y ago
There's a proverbial slip between the cup and the lip. vNM does not (technically) assume a probability distribution, but EUT as commonly applied add one extra piece on top of vNM -- implicitly ignoring dynamics and assuming a distribution which leads to "ensemble averaging".
- kgwgk 6y ago
  What do you mean? vNM doesn’t assume any particular probability distribution (or any particular utility function). But it does assume the existence of a probability distribution (and a utility function). How could they define an expected value otherwise?
  - ssivark 6y ago
    Sure, no qualms with vNM. But how does an application of EUT decide what distribution to use? (That's where the subtlety/assumption lies) Example situation: 1 million tosses of a fair coin, with multiplicative payoff -- wealth doubles for heads and halves for tails. Say you start with a dollar -- what payoff should you expect? Is "expected payoff" (ensemble averaging) the right calculation to model that?
    - kgwgk 6y ago
      I think you're mixing things up. The distribution is not something that you choose arbitrarily. It's given. It's a part of the problem statement. Starting with $1 the distribution of outcomes after 4 doubles-or-halves flips is P($0.0625) = 1/16, P($0.25) = 4/16, P($1) = 6/16, P($4) = 4/16, P($16) = 1/16. After one million flips it will be more complex (and more skewed to the upside, it will go from close to $0 to $10^301030) but also well defined. When you say "what distribution to use" you may mean "what utility function to use". You're right that one may not one want to use the final payoff to represent the utility of each outcome. That's the whole point of introducing a utility function. The logarithm has been a popular choice of utility function for centuries. Among other reasons, because it maximizes the asymptotic growth rate. The distribution of utility values becomes in the 4 flips case (for simplicity, the million flips case would be similar): P(u = -log(16)) = 1/16, P(u = -log(4)) = 4/16, P(u = 0) = 6/16, P(u = log(4)) = 4/16, P(u = log(16)) = 1/16 The expected utility (the ensemble average of the utilities of the potential outcomes, if you will) is zero. The same that for $1 without any uncertainty. Assuming logarithmic utility, you would be indifferent between keeping your dollar or getting the outcome of the million-flips game. BUT that would be assuming that your whole fortune, present and future, depended on that dollar. Otherwise you need to include the rest of your wealth when calculating the logarithmic utility. Or your preferences may not be well represented by that utility function: you'll be more willing to risk your only dollar today when you know you're getting your paycheck tomorrow. If you're a normal person and you're offered to play the game you will prefer to do so than to keep $1 with certainty. You have 49% probability of getting at least $50m while in the worse case your lose $1.
      - ssivark 6y ago
        > The distribution is not something that you choose arbitrarily. It's given. It's a part of the problem statement. Starting with $1 the distribution of outcomes after 4 doubles-or-halves flips is P($0.0625) = 1/16, P($0.25) = 4/16, P($1) = 6/16, P($4) = 4/16, P($16) = 1/16. The fundamental claim of EE is that the distribution cannot be fixed that way to be 1/N for each of the N trajectories in the limit that the number of trajectories far exceeds the number of agents. (Ref: concentration of measure) Whether you agree or disagree, at least we are now at the crux of the matter. I would encourage you re-read my post with this update.