4 ms·

I think your n_subjects is too low. You need that to be high enough or you'll miss those low-probability winners that bring up the average.

by pcmonk 5y ago

nonameiguess 5y ago
I did it with more subjects and it doesn't make a difference. The only reason I reduced to 100 is because the plot is unreadable otherwise. Looking at the Julia code, I think what he is doing wrong is making all wins worth $.50 and all losses worth $.40, but the bet computes a win or loss based on your current wealth, not your starting wealth. His formula would work if you were always betting $1 no matter what your bankroll was, but that isn't what the actual post stipulates.
- maximilianroos 5y ago
  You're so fixed to your conclusion that you're now reading code wrong. If you don't trust the Julia code, try running with the same parameters in Python.
  - nonameiguess 5y ago
    I don't understand why you think changing the number of participants changes the ensemble average. I just ran it with 1,000,000 participants and 1,000 trials and ended up with an ensemble average of $0.07 on the 1,000th trial, trending toward 0. The only difference is the simulation took longer. The curve looks exactly the same. The ensemble average trends upward until about 500 trials, then trends downward and keeps doing so forever. My code is right up there and you can run it. You can even just run the OP's notebook that he provided but increase the number of trials. Change the "num_flips_per_sim" parameter he provides in cell 6 to anything over 500 and you will always get sum(count_lose_capital) == everyone.
    - maximilianroos 5y ago
      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 5y ago
        [deleted]
      - concreteblock 5y 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 5y 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 5y 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 5y 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?
    - mrow84 5y ago
      The number of possible outcomes grows exponentially with time, and so does the ensemble size required to capture the extremal behaviour. Repeated losses bring you closer to zero, which is relatively well sampled by many realisations, but repeated wins produce exponentially larger returns, and so missing out on these realisations catastrophically affects the ensemble average. A shorter run (say 100 steps) would be more likely to capture enough realisations to produce a reasonable estimate. You could assess this behaviour yourself, for very low step numbers, by calculating the variability in a sampled ensemble average, relative to the exhaustive (i.e. true) ensemble average. This particular problem is another consequence of the properties dynamical system being examined, but not quite the same as the issues caused by its non-ergodicity.
      - mrow84 5y ago
        I was interested in seeing the results myself, so here is some python: import numpy as np import itertools from matplotlib import pyplot as plt def ensemble_mean(outcomes): # Assume we are given a (K, T) array of outcomes, and compute the ensemble average # for T+1 time steps, starting with 1 wealth. K, T = outcomes.shape X = np.ones((K, T+1), dtype=np.float64) X[:, 1:] = np.where(outcomes, 1.5, 0.6) Z = np.cumprod(X, axis=1) return Z.mean(axis=0) time_steps = 20 all_outcomes = np.array(list(itertools.product([0, 1], repeat=time_steps-1))) exhaustive_mean = ensemble_mean(all_outcomes) ensemble_size = 100 ensemble_samples = 10000 ensemble_means = np.zeros((time_steps, ensemble_samples)) for i in range(ensemble_samples): print(i) # generate ensembles as though we were sampling (i.e. with replacement) J = np.random.choice(all_outcomes.shape[0], size=ensemble_size, replace=True) ensemble_means[:, i] = ensemble_mean(all_outcomes[J, :]) plt.hist(ensemble_means[-1], bins=1000, histtype='step') plt.axvline(exhaustive_mean[-1]) plt.title("Modal sampled ensemble mean is below true ensemble mean") plt.show()
- jackcosgrove 5y ago
  So if you get lucky you de-risk and make small bets comparable to your initial winning bet, rather than betting it all. Sounds like a good strategy whether in Vegas or Wall Street.

I think your n_subjects is too low. You need that to be high enough or you'll miss those low-probability winners that bring up the average.

by pcmonk 5y ago

I think your n_subjects is too low. You need that to be high enough or you'll miss those low-probability winners that bring up the average.

nonameiguess 5y ago

I did it with more subjects and it doesn't make a difference. The only reason I reduced to 100 is because the plot is unreadable otherwise. Looking at the Julia code, I think what he is doing wrong is making all wins worth $.50 and all losses worth $.40, but the bet computes a win or loss based on your current wealth, not your starting wealth. His formula would work if you were always betting $1 no matter what your bankroll was, but that isn't what the actual post stipulates.

maximilianroos 5y ago
You're so fixed to your conclusion that you're now reading code wrong. If you don't trust the Julia code, try running with the same parameters in Python.
- nonameiguess 5y ago
  I don't understand why you think changing the number of participants changes the ensemble average. I just ran it with 1,000,000 participants and 1,000 trials and ended up with an ensemble average of $0.07 on the 1,000th trial, trending toward 0. The only difference is the simulation took longer. The curve looks exactly the same. The ensemble average trends upward until about 500 trials, then trends downward and keeps doing so forever. My code is right up there and you can run it. You can even just run the OP's notebook that he provided but increase the number of trials. Change the "num_flips_per_sim" parameter he provides in cell 6 to anything over 500 and you will always get sum(count_lose_capital) == everyone.
  - maximilianroos 5y ago
    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 5y ago
      [deleted]
    - concreteblock 5y 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 5y 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 5y 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 5y 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?
  - mrow84 5y ago
    The number of possible outcomes grows exponentially with time, and so does the ensemble size required to capture the extremal behaviour. Repeated losses bring you closer to zero, which is relatively well sampled by many realisations, but repeated wins produce exponentially larger returns, and so missing out on these realisations catastrophically affects the ensemble average. A shorter run (say 100 steps) would be more likely to capture enough realisations to produce a reasonable estimate. You could assess this behaviour yourself, for very low step numbers, by calculating the variability in a sampled ensemble average, relative to the exhaustive (i.e. true) ensemble average. This particular problem is another consequence of the properties dynamical system being examined, but not quite the same as the issues caused by its non-ergodicity.
    - mrow84 5y ago
      I was interested in seeing the results myself, so here is some python: import numpy as np import itertools from matplotlib import pyplot as plt def ensemble_mean(outcomes): # Assume we are given a (K, T) array of outcomes, and compute the ensemble average # for T+1 time steps, starting with 1 wealth. K, T = outcomes.shape X = np.ones((K, T+1), dtype=np.float64) X[:, 1:] = np.where(outcomes, 1.5, 0.6) Z = np.cumprod(X, axis=1) return Z.mean(axis=0) time_steps = 20 all_outcomes = np.array(list(itertools.product([0, 1], repeat=time_steps-1))) exhaustive_mean = ensemble_mean(all_outcomes) ensemble_size = 100 ensemble_samples = 10000 ensemble_means = np.zeros((time_steps, ensemble_samples)) for i in range(ensemble_samples): print(i) # generate ensembles as though we were sampling (i.e. with replacement) J = np.random.choice(all_outcomes.shape[0], size=ensemble_size, replace=True) ensemble_means[:, i] = ensemble_mean(all_outcomes[J, :]) plt.hist(ensemble_means[-1], bins=1000, histtype='step') plt.axvline(exhaustive_mean[-1]) plt.title("Modal sampled ensemble mean is below true ensemble mean") plt.show()
jackcosgrove 5y ago
So if you get lucky you de-risk and make small bets comparable to your initial winning bet, rather than betting it all. Sounds like a good strategy whether in Vegas or Wall Street.