5 ms·

Take the outside view here — 3-4 people have commented, all disagreeing with you. One of them has offered an explanation of why you're experiment is poorly desi

Take the outside view here — 3-4 people have commented, all disagreeing with you. One of them has offered an explanation of why you're experiment is poorly designed, and I've given you code which produces a different result. The appropriate response to that is introspection, not repetition.

deleted 6y ago
[deleted]
concreteblock 6y ago
No matter how many test subjects you use, if you run the experiment for a very long time, everyone goes bankrupt and will never recover. More precisely there is a finite time after which no-one ever passes above $0.0000000000000001. That is a mathematical theorem. This doesn’t depend on the number of test subjects, and you can add as many zeroes as you want. Therefore in the long run the mean outcome is 0. Forgive me if I have misinterpreted what you are are trying to say. Edit: I’ve just realized that I have indeed missed your point.
- kgwgk 6y ago
  No matter for how long you run the experiment if you use enough subjects some of them will win an absurdly large amount of money and the sample mean will converge to the mean of the distribution (which grows exponentially with time). It’s a mathematical theorem. (I would be curious to see a proof of your theorem, by the way.)
  - concreteblock 6y ago
    The mean converges exponentially to zero with time. It doesn’t grow exponentially. So the theorem you cited also goes in the same direction of my statement.
    - kgwgk 6y ago
      > The mean converges to zero with time. It doesn’t grow exponentially. t=0 mean(w) = 1 t=1 mean(w) = 1/2*1.5 + 1/2*0.6 = 1.05 t=2 mean(w) = 1/4*1.5*1.5 + 1/2*1.5*0.6 + 1/4*0.6*0.6 = 1.1025 .... t mean(w) = 1.05^t Don’t you agree?
      - concreteblock 6y ago
        You’re right. I made a mistake - I thought you were trying to contradict the theorem I stated. I’ve just realized you were saying something orthogonal. As for the proof of my theorem, By taking logarithms, the process becomes an additive random walk with negative drift (log 1.6 + log 0.5 < 0). This is well known to converge to negative infinity almost surely. After exponentiating to undo the logarithm, this is exactly the statement I made. It does not matter how many test subjects there are ( as long as there’s finitely many) because, informally speaking , you can just wait for each of them to become irrevocably bankrupt in turn.
        kgwgk 6y ago
        I think we agree then: - for a fixed sample size we can find a time large enough that the probability of the sample mean being above $1 is as low as we want - for a fixed time we can find a sample size large enough that the probability of the sample mean being below $1 is as low as we want - when both the sample size and the horizon grow without limit which effect dominates will depend on how we make it happen Adding "almost surely" to "everyone goes bankrupt and will never recover" or "there is a finite time after which no-one ever passes above $0.0000000000000001" is a subtle change but it's enough to allow for someone to go to infinity with infinitesimal probability. This is why the distribution mean can grow exponentially, it wouldn't be possible if the everyone and no-one in those quotes were strictly true.
        concreteblock 6y ago
        I agree with the first 3 statements. Just to confirm, I am using 'almost surely' in the technical sense, which means 'with probability 1.' Consider the following statement: If you keep flipping a fair coin every day, it is almost sure that after some day you will have gotten a tails. This is the same 'almost surely' that I am referring to.
nonameiguess 6y ago
Look at my other reply, which was above but is now below. The number of participants required to be likely to find any who stay above water gets very high eventually, much higher than 1,000,000. This can just be calculated. After 500 trials, you need 279 heads to stay above $1 net wealth. 1.5^278 + 0.6^222 = 0.50 and 1.5^279 + 0.6^221 = 1.26, so that's your breakeven point. The probability of getting at least 279 heads in 500 coin flips is 0.005364, so with 1,000 participants, you expect to see about 5 still above water. At 1000 trials, the breakeven point becomes 558 and the probability of getting at least that many heads in 1000 flips is 0.00013614. So the expected number of people who stay above water in a pool of 1000 participants is 0. Out of 1,000,000, it is 13, so you're right, there are some, but at that point it's not nearly enough and we're not sampling the ones whose wealth is enough to actually bring the mean back up, so it keeps trending to 0 in any sample of a practical trial size. This is a pretty interesting property of this problem, really. It's not related to ergodicity, but just the relative proportion of probability mass represented by above 1 and below one itself trending asymptotically toward 0 even though the analytical expectation trends toward infinity. I don't know that there is even a word for that, but seemingly which of those moves faster toward its limit would determine what sample ensemble average you really see when the number of realized states is far less than the number of possible states. This probably has some implications for Pascal's Mugger type problems in decision theory. If some course of action has potentially infinite future payoff and destroys expected utility calculations because of that, but the expected number of possible universes in which a positive outcome happens at all trends toward 0 faster than the expectation trends toward infinity, that gives a decision rule. In this specific case, don't take this bet, at least not in an indefinitely repeating form.
- maximilianroos 6y ago
  Thanks for the thoughtful reply. The breakeven analysis is good!

deleted 6y ago

[deleted]

concreteblock 6y ago

No matter how many test subjects you use, if you run the experiment for a very long time, everyone goes bankrupt and will never recover. More precisely there is a finite time after which no-one ever passes above $0.0000000000000001. That is a mathematical theorem. This doesn’t depend on the number of test subjects, and you can add as many zeroes as you want. Therefore in the long run the mean outcome is 0. Forgive me if I have misinterpreted what you are are trying to say. Edit: I’ve just realized that I have indeed missed your point.

kgwgk 6y ago
No matter for how long you run the experiment if you use enough subjects some of them will win an absurdly large amount of money and the sample mean will converge to the mean of the distribution (which grows exponentially with time). It’s a mathematical theorem. (I would be curious to see a proof of your theorem, by the way.)
- concreteblock 6y ago
  The mean converges exponentially to zero with time. It doesn’t grow exponentially. So the theorem you cited also goes in the same direction of my statement.
  - kgwgk 6y ago
    > The mean converges to zero with time. It doesn’t grow exponentially. t=0 mean(w) = 1 t=1 mean(w) = 1/2*1.5 + 1/2*0.6 = 1.05 t=2 mean(w) = 1/4*1.5*1.5 + 1/2*1.5*0.6 + 1/4*0.6*0.6 = 1.1025 .... t mean(w) = 1.05^t Don’t you agree?
    - concreteblock 6y ago
      You’re right. I made a mistake - I thought you were trying to contradict the theorem I stated. I’ve just realized you were saying something orthogonal. As for the proof of my theorem, By taking logarithms, the process becomes an additive random walk with negative drift (log 1.6 + log 0.5 < 0). This is well known to converge to negative infinity almost surely. After exponentiating to undo the logarithm, this is exactly the statement I made. It does not matter how many test subjects there are ( as long as there’s finitely many) because, informally speaking , you can just wait for each of them to become irrevocably bankrupt in turn.
      - kgwgk 6y ago
        I think we agree then: - for a fixed sample size we can find a time large enough that the probability of the sample mean being above $1 is as low as we want - for a fixed time we can find a sample size large enough that the probability of the sample mean being below $1 is as low as we want - when both the sample size and the horizon grow without limit which effect dominates will depend on how we make it happen Adding "almost surely" to "everyone goes bankrupt and will never recover" or "there is a finite time after which no-one ever passes above $0.0000000000000001" is a subtle change but it's enough to allow for someone to go to infinity with infinitesimal probability. This is why the distribution mean can grow exponentially, it wouldn't be possible if the everyone and no-one in those quotes were strictly true.
        concreteblock 6y ago
        I agree with the first 3 statements. Just to confirm, I am using 'almost surely' in the technical sense, which means 'with probability 1.' Consider the following statement: If you keep flipping a fair coin every day, it is almost sure that after some day you will have gotten a tails. This is the same 'almost surely' that I am referring to.

nonameiguess 6y ago

Look at my other reply, which was above but is now below. The number of participants required to be likely to find any who stay above water gets very high eventually, much higher than 1,000,000. This can just be calculated. After 500 trials, you need 279 heads to stay above $1 net wealth. 1.5^278 + 0.6^222 = 0.50 and 1.5^279 + 0.6^221 = 1.26, so that's your breakeven point. The probability of getting at least 279 heads in 500 coin flips is 0.005364, so with 1,000 participants, you expect to see about 5 still above water. At 1000 trials, the breakeven point becomes 558 and the probability of getting at least that many heads in 1000 flips is 0.00013614. So the expected number of people who stay above water in a pool of 1000 participants is 0. Out of 1,000,000, it is 13, so you're right, there are some, but at that point it's not nearly enough and we're not sampling the ones whose wealth is enough to actually bring the mean back up, so it keeps trending to 0 in any sample of a practical trial size. This is a pretty interesting property of this problem, really. It's not related to ergodicity, but just the relative proportion of probability mass represented by above 1 and below one itself trending asymptotically toward 0 even though the analytical expectation trends toward infinity. I don't know that there is even a word for that, but seemingly which of those moves faster toward its limit would determine what sample ensemble average you really see when the number of realized states is far less than the number of possible states. This probably has some implications for Pascal's Mugger type problems in decision theory. If some course of action has potentially infinite future payoff and destroys expected utility calculations because of that, but the expected number of possible universes in which a positive outcome happens at all trends toward 0 faster than the expectation trends toward infinity, that gives a decision rule. In this specific case, don't take this bet, at least not in an indefinitely repeating form.

maximilianroos 6y ago
Thanks for the thoughtful reply. The breakeven analysis is good!