12 ms·

There is a recent paper[1] and an epic twitter thread[2] well worth the reading [1] https://arxiv.org/pdf/1906.04652.pdf https://arxiv.org/pdf/1906.04652.pdf

by fasteo 7y ago

There is a recent paper[1] and an epic twitter thread[2] well worth the reading [1] https://arxiv.org/pdf/1906.04652.pdf https://arxiv.org/pdf/1906.04652.pdf [2] https://twitter.com/hulme_oliver/status/1139148255969906689 https://twitter.com/hulme_oliver/status/1139148255969906689

mturmon 7y ago
The twitter thread gives a nice example of a failure of ergodicity that I was not aware of: "Now consider a gamble with multiplicative dynamics. Win 50% of your current wealth for heads, lose 40% of your current wealth for tails. Changes in wealth now are non-ergodic, so calculating the expectation value is not informative of the time average growth rate of wealth. This exact gamble has a positive expectation value, but it has a negative time average growth rate. We call it the Peters coin game." Clearly the expectation of return at a fixed time is > 0, but they give an example time series with obvious negative drift.
- dr_dshiv 7y ago
  I'm confused. My takeaway was that dice rolls and coin tosses are ergodic because flipping one coin 20 times or flipping 20 coins is meaningfully equivalent. Whereas following an individual person for 1000 days is different from following 1000 people for 1 day. Is that not the right heuristic? Also -- why is it called ergodicity? I keep thinking it has something to do with work or thermodynamics, but I'm missing it...
  - ska 7y ago
    You are remembering the ergodic hypothesis in thermodynamics, whereI think the name originated and then was generalized.
    - dr_dshiv 7y ago
      Is there a relationship?
      - ska 7y ago
        In a sense. Ergodic hypothesis is basically that the phase space is explored "equally" (in a precise sense), over time. Ergodic random processes are pretty much independent of their initial state. So at a hand-wavy level its a very similar idea.
- heavenlyblue 7y ago
  They're simply calculating expectation incorrectly. Expectation is a probability-weighted sum over every attempt. While the game of 50/40 is based on a probability-weighted product over every attempt. Basically "average expectation" for that game is 0.95 (and they give a false name to it). Tl;dr: it's got nothing to do with ergodicity. More importantly, they quote "Peters coin game" which has no results in Google whatsoever.
  - fasteo 7y ago
    >>> More importantly, they quote "Peters coin game" which has no results in Google whatsoever. Maybe a reference to this paper [1] ? [1] The time resolution of the St. Petersburg paradox, by Ole Peters https://arxiv.org/pdf/1011.4404.pdf https://arxiv.org/pdf/1011.4404.pdf
  - Sahhaese 7y ago
    The expectation of a single flip is 'positive' ( > 1 ), the expectation of n flips is 'negative' ( < 1). My layman reading of this is that this 'appears' to be a violation of martingale theory because this 'should' be a sub-martingale but martingale theory requires finite variance so I suspect the variance of this must go infinite as n tends to infinity?
    - kgwgk 7y ago
      Why do you say that the expectation of n flips is negative? For 2 flips there are 4 equiprobable cases with gains +125%, -10%, -10%, -64% and the expectation is 10.25%. The expected gain for n flips is 1.05^n-1
      - pfortuny 7y ago
        AFAIK (am reading quickly) mean values of COMPOUND interest rates (which is what you are doing) are not useful. If you earn 10% and then lose 10% you are worse off than before. Start with 10: (11-1.1 < 10).
        7y ago
- gowld 7y ago
  How does this work? By binomial theorem, the expectation at time t is ((1.5 + 0.6)^n)/2^n) = 1.05^n so the "annualized" return per unit time is 1.05, exactly the same as the expectation. What is "time average growth rate"? Web search turns up lots of references do it by Peters, but no definitions.
  - yellowstuff 7y ago
    You have the wrong model. You're betting a different amount each time. Say you start with $1 and get a heads then a tails: ($1 * 1.5) * .6 = $.90 What if you get a tails then a heads? ($1 * .6) * 1.5 = $.90 Doesn't seem like such a great game now, does it? The Kelly Criterion sets a limit for how much of your total wealth you should bet when the odds favor you. If you bet more than that limit you increase your odds of losing everything without improving your expected return. So imagine a better game where you could bet any amount, and you still got paid $5 for every $4 you risked and had a 50% chance of winning. The correct amount to bet is 10% of your bankroll. If you bet more than that in the long run you will go broke.
    - kgwgk 7y ago
      > If you bet more than that in the long run you will go broke. The optimal bet is not the same as the critical bet separating the positive growth and negative growth regions.
      - yellowstuff 7y ago
        You're correct. Betting more than double the Kelly amount results in negative expected growth: https://wizardofodds.com/gambling/kelly-criterion/ https://wizardofodds.com/gambling/kelly-criterion/
    - gowld 7y ago
      You are ignoring half of the possible outcomes. If you get HH or TT, that's $2.25 and 0.36, which averages to $1.25. 50% of ($1.25 + $.90) for a $1 bet seems like a great game to play.
- mr_toad 7y ago
  >calculating the expectation value is not informative Well, only if you use a linear estimator, i.e. the arithmetic mean. Any accountant would know to use a geometric mean (the annualised rate of return).
  - mturmon 7y ago
    It is still an interesting example. Following your suggestion, which is a good one, the player's holdings at time N, starting with $1, are: H = [1.5 or 0.6] * [1.5 or 0.6] * ... } for N factors To make it into a more familiar process, take logs: log H = log([1.5 or 0.6]) + log([1.5 or 0.6]) + ... Now we have a familiar random walk with drift, and our intuition is much better-developed for this process. In particular, the mean E log H = Sum{1..N} E log([1.5 or 0.6]) = Sum{1..N} [-0.023] = -0.023 N < 0 So the expectation of the log earnings is negative. In fact, because of the drift, sample paths of log H will be unbounded below with probability 1. This nontrivial fact (from the basic theory of random walk) will determine how the process sample paths (even outside the log domain) "look" - they will always eventually squash towards 0. Every single one. But the expectation of the earnings themselves (as several comments have shown) is > 1. Apparently, without the compression given by the log, the upward excursions given by lucky bets outweigh the unlucky ones. We can look at this discrepancy ("good game in original units, bad game in log units") the other way. Even though the sample paths eventually go to zero, in that case, you only lose $1. Whereas many other sample paths will make large excursions upward before coming to zero. So the expectation of the non-logged process adds up over many losses of $1, and a few low-probability gains of much more.

by fasteo 7y ago

mturmon 7y ago

The twitter thread gives a nice example of a failure of ergodicity that I was not aware of: "Now consider a gamble with multiplicative dynamics. Win 50% of your current wealth for heads, lose 40% of your current wealth for tails. Changes in wealth now are non-ergodic, so calculating the expectation value is not informative of the time average growth rate of wealth. This exact gamble has a positive expectation value, but it has a negative time average growth rate. We call it the Peters coin game." Clearly the expectation of return at a fixed time is > 0, but they give an example time series with obvious negative drift.

dr_dshiv 7y ago
I'm confused. My takeaway was that dice rolls and coin tosses are ergodic because flipping one coin 20 times or flipping 20 coins is meaningfully equivalent. Whereas following an individual person for 1000 days is different from following 1000 people for 1 day. Is that not the right heuristic? Also -- why is it called ergodicity? I keep thinking it has something to do with work or thermodynamics, but I'm missing it...
- ska 7y ago
  You are remembering the ergodic hypothesis in thermodynamics, whereI think the name originated and then was generalized.
  - dr_dshiv 7y ago
    Is there a relationship?
    - ska 7y ago
      In a sense. Ergodic hypothesis is basically that the phase space is explored "equally" (in a precise sense), over time. Ergodic random processes are pretty much independent of their initial state. So at a hand-wavy level its a very similar idea.
heavenlyblue 7y ago
They're simply calculating expectation incorrectly. Expectation is a probability-weighted sum over every attempt. While the game of 50/40 is based on a probability-weighted product over every attempt. Basically "average expectation" for that game is 0.95 (and they give a false name to it). Tl;dr: it's got nothing to do with ergodicity. More importantly, they quote "Peters coin game" which has no results in Google whatsoever.
- fasteo 7y ago
  >>> More importantly, they quote "Peters coin game" which has no results in Google whatsoever. Maybe a reference to this paper [1] ? [1] The time resolution of the St. Petersburg paradox, by Ole Peters https://arxiv.org/pdf/1011.4404.pdf https://arxiv.org/pdf/1011.4404.pdf
- Sahhaese 7y ago
  The expectation of a single flip is 'positive' ( > 1 ), the expectation of n flips is 'negative' ( < 1). My layman reading of this is that this 'appears' to be a violation of martingale theory because this 'should' be a sub-martingale but martingale theory requires finite variance so I suspect the variance of this must go infinite as n tends to infinity?
  - kgwgk 7y ago
    Why do you say that the expectation of n flips is negative? For 2 flips there are 4 equiprobable cases with gains +125%, -10%, -10%, -64% and the expectation is 10.25%. The expected gain for n flips is 1.05^n-1
    - pfortuny 7y ago
      AFAIK (am reading quickly) mean values of COMPOUND interest rates (which is what you are doing) are not useful. If you earn 10% and then lose 10% you are worse off than before. Start with 10: (11-1.1 < 10).
      - 7y ago
gowld 7y ago
How does this work? By binomial theorem, the expectation at time t is ((1.5 + 0.6)^n)/2^n) = 1.05^n so the "annualized" return per unit time is 1.05, exactly the same as the expectation. What is "time average growth rate"? Web search turns up lots of references do it by Peters, but no definitions.
- yellowstuff 7y ago
  You have the wrong model. You're betting a different amount each time. Say you start with $1 and get a heads then a tails: ($1 * 1.5) * .6 = $.90 What if you get a tails then a heads? ($1 * .6) * 1.5 = $.90 Doesn't seem like such a great game now, does it? The Kelly Criterion sets a limit for how much of your total wealth you should bet when the odds favor you. If you bet more than that limit you increase your odds of losing everything without improving your expected return. So imagine a better game where you could bet any amount, and you still got paid $5 for every $4 you risked and had a 50% chance of winning. The correct amount to bet is 10% of your bankroll. If you bet more than that in the long run you will go broke.
  - kgwgk 7y ago
    > If you bet more than that in the long run you will go broke. The optimal bet is not the same as the critical bet separating the positive growth and negative growth regions.
    - yellowstuff 7y ago
      You're correct. Betting more than double the Kelly amount results in negative expected growth: https://wizardofodds.com/gambling/kelly-criterion/ https://wizardofodds.com/gambling/kelly-criterion/
  - gowld 7y ago
    You are ignoring half of the possible outcomes. If you get HH or TT, that's $2.25 and 0.36, which averages to $1.25. 50% of ($1.25 + $.90) for a $1 bet seems like a great game to play.
mr_toad 7y ago
>calculating the expectation value is not informative Well, only if you use a linear estimator, i.e. the arithmetic mean. Any accountant would know to use a geometric mean (the annualised rate of return).
- mturmon 7y ago
  It is still an interesting example. Following your suggestion, which is a good one, the player's holdings at time N, starting with $1, are: H = [1.5 or 0.6] * [1.5 or 0.6] * ... } for N factors To make it into a more familiar process, take logs: log H = log([1.5 or 0.6]) + log([1.5 or 0.6]) + ... Now we have a familiar random walk with drift, and our intuition is much better-developed for this process. In particular, the mean E log H = Sum{1..N} E log([1.5 or 0.6]) = Sum{1..N} [-0.023] = -0.023 N < 0 So the expectation of the log earnings is negative. In fact, because of the drift, sample paths of log H will be unbounded below with probability 1. This nontrivial fact (from the basic theory of random walk) will determine how the process sample paths (even outside the log domain) "look" - they will always eventually squash towards 0. Every single one. But the expectation of the earnings themselves (as several comments have shown) is > 1. Apparently, without the compression given by the log, the upward excursions given by lucky bets outweigh the unlucky ones. We can look at this discrepancy ("good game in original units, bad game in log units") the other way. Even though the sample paths eventually go to zero, in that case, you only lose $1. Whereas many other sample paths will make large excursions upward before coming to zero. So the expectation of the non-logged process adds up over many losses of $1, and a few low-probability gains of much more.