4 ms·

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

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.
      - kgwgk 6y ago
        I’m sorry. That is the probability distribution. EE doesn’t change that mathematical fact. The distribution is not “fixed that way”, it is what it is. Now, it’s of course true that the mean of a finite sample taken from that distribution (which has a HUGE variance) will also have a huge variance. Why would that give a meaning to your talking about “deciding what distribution to use” in EUT, I don’t know.
      - kgwgk 6y ago
        Note as well that there is no concentration of measure in your example. You need to express the outcome as a rate of growth per flip for that. And what do I care that the compounded growth over one million flips was close to 0% per flip if I have 49% probability of walking away with more that 50 million dollars?
      - ssivark 6y ago
        HN seems to stop deeper comments, so I'll make one last comment. We've managed to get to the heart of the point, and say everything that needs to be said :-) > I’m sorry. That is the probability distribution. EE doesn’t change that mathematical fact. The distribution is not “fixed that way”, it is what it is. That is where the mistaken assumption lies, and repeating it is simply refusing to engage with the point. > Now, it’s of course true that the mean of a finite sample taken from that distribution (which has a HUGE variance) will also have a huge variance. You're halfway there! The underlying distribution on trajectories has a huge variance like you pointed out, but the mean of a finite sample is likely to have only a (vanishingly) small variance, because of the central limit theorem! (aka concentration of measure for "typical" trajectories) So, averaging using the low variance distribution on the mean of (typical) finite samples is more representative model (corresponds to real world measurements) than averaging using the high-variance distribution on the underlying ensemble (what you assumed the de facto probability distribution). The former is called a "time average" and the latter is called an "ensemble average".

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.
    - kgwgk 6y ago
      I’m sorry. That is the probability distribution. EE doesn’t change that mathematical fact. The distribution is not “fixed that way”, it is what it is. Now, it’s of course true that the mean of a finite sample taken from that distribution (which has a HUGE variance) will also have a huge variance. Why would that give a meaning to your talking about “deciding what distribution to use” in EUT, I don’t know.
    - kgwgk 6y ago
      Note as well that there is no concentration of measure in your example. You need to express the outcome as a rate of growth per flip for that. And what do I care that the compounded growth over one million flips was close to 0% per flip if I have 49% probability of walking away with more that 50 million dollars?
    - ssivark 6y ago
      HN seems to stop deeper comments, so I'll make one last comment. We've managed to get to the heart of the point, and say everything that needs to be said :-) > I’m sorry. That is the probability distribution. EE doesn’t change that mathematical fact. The distribution is not “fixed that way”, it is what it is. That is where the mistaken assumption lies, and repeating it is simply refusing to engage with the point. > Now, it’s of course true that the mean of a finite sample taken from that distribution (which has a HUGE variance) will also have a huge variance. You're halfway there! The underlying distribution on trajectories has a huge variance like you pointed out, but the mean of a finite sample is likely to have only a (vanishingly) small variance, because of the central limit theorem! (aka concentration of measure for "typical" trajectories) So, averaging using the low variance distribution on the mean of (typical) finite samples is more representative model (corresponds to real world measurements) than averaging using the high-variance distribution on the underlying ensemble (what you assumed the de facto probability distribution). The former is called a "time average" and the latter is called an "ensemble average".