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> 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

> 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
  You fixed the rules of the game. It follows that there is a distribution of outcomes of the game (it was guaranteed to end). This distribution is not about time-averaging or ensemble-averaging. There is no time, there is no ensemble. Don’t you agree that there is a probability distribution that describes the outcomes of the game you proposed? If you don’t, that’s where you’re mistaken. Expected utility theory uses that probability distribution of outcomes. In this case it was directly derived from your description of the game. The probability distribution could also represent the probability assigned to the different potential outcomes by the agent, and another agent could have a different probability distribution. In any case, there is no notion of ensemble average over N realizations or time average over N realizations or anything like that in EUT. There is a set of possible outcomes, which have a value called utility attached (representing preferences). For each possible decision, there is a probability distribution over the outcomes. An expected value can be calculated. The optimal decision is the one with highest expectation of the utility. Saying that utility theory has an issue because of typical trajectories and time averages and ensemble averages doesn’t make much sense. All those things are not part of EUT. You could just as well say that it has a problem with the spark plugs. If you say that you may solve some problems differently, that’s ok. That doesn’t make EUT wrong or disallows the use of probability distributions to calculate expected values. And as far as I know, all the decision problems solved by EE have a straightforward correspondence to a EUT solution with some utility function (and with a perfectly standard probability distribution that can be used to calculate expected utilities).
- kgwgk 6y ago
  > 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) The mean of a finite sample is a random number. This random number has a distribution and has a variance. The variance of the sample mean is the variance of the original distribution divided by N. Huge. Saying that "the mean of a finite sample is likely to have only a small variance" doesn't mean anything. Maybe you meant that the variance of a finite sample, which is also a random number and has a distribution, is much more likely to be low (all the elements in the sample are in the high-density lower part of the original distribution) than high (a few elements in the very skewed upper tail). And, as I said, in your example there is no concentration of measure in the original distribution. Typical trajectories will move around $1 but the probability of having exactly $1 at the end goes down as the horizon grows. As an aside, every trajectory will eventually reach any (positive) level that you may specify. And bounce between any two levels infinite times. Well, those things may not happen for some trajectories but they have probability zero. Infinite stochastic processes are fun. To see that concentration you have to transform the original dollar outcome distribution into a growth rate per flip distribution (which converges to 0 in your example).

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
You fixed the rules of the game. It follows that there is a distribution of outcomes of the game (it was guaranteed to end). This distribution is not about time-averaging or ensemble-averaging. There is no time, there is no ensemble. Don’t you agree that there is a probability distribution that describes the outcomes of the game you proposed? If you don’t, that’s where you’re mistaken. Expected utility theory uses that probability distribution of outcomes. In this case it was directly derived from your description of the game. The probability distribution could also represent the probability assigned to the different potential outcomes by the agent, and another agent could have a different probability distribution. In any case, there is no notion of ensemble average over N realizations or time average over N realizations or anything like that in EUT. There is a set of possible outcomes, which have a value called utility attached (representing preferences). For each possible decision, there is a probability distribution over the outcomes. An expected value can be calculated. The optimal decision is the one with highest expectation of the utility. Saying that utility theory has an issue because of typical trajectories and time averages and ensemble averages doesn’t make much sense. All those things are not part of EUT. You could just as well say that it has a problem with the spark plugs. If you say that you may solve some problems differently, that’s ok. That doesn’t make EUT wrong or disallows the use of probability distributions to calculate expected values. And as far as I know, all the decision problems solved by EE have a straightforward correspondence to a EUT solution with some utility function (and with a perfectly standard probability distribution that can be used to calculate expected utilities).
kgwgk 6y ago
> 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) The mean of a finite sample is a random number. This random number has a distribution and has a variance. The variance of the sample mean is the variance of the original distribution divided by N. Huge. Saying that "the mean of a finite sample is likely to have only a small variance" doesn't mean anything. Maybe you meant that the variance of a finite sample, which is also a random number and has a distribution, is much more likely to be low (all the elements in the sample are in the high-density lower part of the original distribution) than high (a few elements in the very skewed upper tail). And, as I said, in your example there is no concentration of measure in the original distribution. Typical trajectories will move around $1 but the probability of having exactly $1 at the end goes down as the horizon grows. As an aside, every trajectory will eventually reach any (positive) level that you may specify. And bounce between any two levels infinite times. Well, those things may not happen for some trajectories but they have probability zero. Infinite stochastic processes are fun. To see that concentration you have to transform the original dollar outcome distribution into a growth rate per flip distribution (which converges to 0 in your example).