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The article concludes that a +50%/-40% coin toss on average loses 10% every two tosses because 150% * 60% = 90%, but that ignores the two heads/two tails outcom

by jakell 3y ago

The article concludes that a +50%/-40% coin toss on average loses 10% every two tosses because 150% * 60% = 90%, but that ignores the two heads/two tails outcomes. Including those outcomes, ie AVERAGE(225%, 90%, 90%, 36%) = 110.25%, recovering the more intuitive result that the coin toss gains on average. The author seems to be confusing mode and mean; the modal path does approach zero.

bogtog 3y ago
This was my reaction too, I feel like there is something I'm not getting. I get that the bet is virtually guaranteed to go negative if the number of rounds is high enough (100+ rounds), but the opportunity for huge payouts still pulls the average up.
- jakell 3y ago
  Exactly
- bzax 3y ago
  The "average" of the distribution goes up as you increase the number of rounds, but the probability that you get an average or above value when you sample that distribution once goes to zero as the number of rounds increases.
bzax 3y ago
If you repeat this game n times (as n goes to infinity), you will have Θ(n) pairs of (heads, tails) and O(sqrt(n)) unpaired wins or losses, except for a vanishingly small fraction of the time when the results fall outside of any fixed number of standard deviations. The point is that you as an individual playing a repeated game don't get to meaningfully sample the expected value of the distribution. You only get to sample once, and you will almost surely (i.e. with probability approaching 1 as n goes to infinity) sample a point in the distribution where you lose nearly all of your money.
- jakell 3y ago
  Absolutely. The individual is long-run guaranteed to be wiped out. But I disagree with the original author’s way of concluding that fact (ie, that it arises from “losing 5% per round”, which is just false).
  - deleted 3y ago
    [deleted]
  - bzax 3y ago
    I believe the entire point of the ergodicity question here is "If you apply this process n times, with n approaching infinity, obviously the result may depend on what point in the n-times iterated distribution you sample, but if you choose a volume of vanishingly small measure to exclude, can you make a single concrete statement about what the process is doing without taking an expected value over the different outcomes" And the answer is yes - with probability approaching 1 as n increases (ie excluding a portion of the distribution whose measure decreases to 0), the random process matches a deterministic process which is described by "you lose 5% each round".
    - jakell 3y ago
      Great description of a framing I hadn’t considered before, thanks!
      - bzax 3y ago
        I should admit I'm being very generous to Peters here - I came to the conclusion that this is what he means only because the math of ergodicity (https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorem...) talks a lot about "except on a set of measure zero". He provides no explanation of how he moves from "the time average of values in a particular run of the process" (which is ergodicity) to "what does a typical process round do, with probability 1" (which is perhaps what someone computing a utility function cares about). I asked a friend who is an econ professor "Why does this Peters guy explain this so poorly" and his response was more or less, yes, all of economics has been wondering that too since he first published his Nature Physics paper on this a decade ago.
didgeoridoo 3y ago
You’re taking arithmetic averages of percentages… I don’t think that calculates anything meaningful. Try 225% * 90% * 90% * 36% to get the expected value.
- jakell 3y ago
  You can substitute $ for % in my comment if it helps. If you start with $100, your expected wealth after two throws is the average of $225, $90, $90 and $36.
  - didgeoridoo 3y ago
    That average is still greater than $100, because you haven’t yet hit the Kelly point beyond which the downside catastrophe dominates. Play it out a few more rounds and see where the average heads to. [Edit: delete bad math]
    - bzax 3y ago
      The expected value of this distribution goes up with every iteration, there is no such Kelly point. You could try this with heads: double your money tails: lose all your money in which case the expected value is always $1, as you have a 1/2^n chance of having $2^n dollars after n rounds, and 0 otherwise. The point of discussing ergodicity here, however, is whether you can describe the behavior of the iterated distribution deterministically if you exclude a portion of that distribution which has measure zero.

by jakell 3y ago

bogtog 3y ago

This was my reaction too, I feel like there is something I'm not getting. I get that the bet is virtually guaranteed to go negative if the number of rounds is high enough (100+ rounds), but the opportunity for huge payouts still pulls the average up.

jakell 3y ago
Exactly
bzax 3y ago
The "average" of the distribution goes up as you increase the number of rounds, but the probability that you get an average or above value when you sample that distribution once goes to zero as the number of rounds increases.

bzax 3y ago

If you repeat this game n times (as n goes to infinity), you will have Θ(n) pairs of (heads, tails) and O(sqrt(n)) unpaired wins or losses, except for a vanishingly small fraction of the time when the results fall outside of any fixed number of standard deviations. The point is that you as an individual playing a repeated game don't get to meaningfully sample the expected value of the distribution. You only get to sample once, and you will almost surely (i.e. with probability approaching 1 as n goes to infinity) sample a point in the distribution where you lose nearly all of your money.

jakell 3y ago
Absolutely. The individual is long-run guaranteed to be wiped out. But I disagree with the original author’s way of concluding that fact (ie, that it arises from “losing 5% per round”, which is just false).
- deleted 3y ago
  [deleted]
- bzax 3y ago
  I believe the entire point of the ergodicity question here is "If you apply this process n times, with n approaching infinity, obviously the result may depend on what point in the n-times iterated distribution you sample, but if you choose a volume of vanishingly small measure to exclude, can you make a single concrete statement about what the process is doing without taking an expected value over the different outcomes" And the answer is yes - with probability approaching 1 as n increases (ie excluding a portion of the distribution whose measure decreases to 0), the random process matches a deterministic process which is described by "you lose 5% each round".
  - jakell 3y ago
    Great description of a framing I hadn’t considered before, thanks!
    - bzax 3y ago
      I should admit I'm being very generous to Peters here - I came to the conclusion that this is what he means only because the math of ergodicity (https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorem...) talks a lot about "except on a set of measure zero". He provides no explanation of how he moves from "the time average of values in a particular run of the process" (which is ergodicity) to "what does a typical process round do, with probability 1" (which is perhaps what someone computing a utility function cares about). I asked a friend who is an econ professor "Why does this Peters guy explain this so poorly" and his response was more or less, yes, all of economics has been wondering that too since he first published his Nature Physics paper on this a decade ago.

didgeoridoo 3y ago

You’re taking arithmetic averages of percentages… I don’t think that calculates anything meaningful. Try 225% * 90% * 90% * 36% to get the expected value.

jakell 3y ago
You can substitute $ for % in my comment if it helps. If you start with $100, your expected wealth after two throws is the average of $225, $90, $90 and $36.
- didgeoridoo 3y ago
  That average is still greater than $100, because you haven’t yet hit the Kelly point beyond which the downside catastrophe dominates. Play it out a few more rounds and see where the average heads to. [Edit: delete bad math]
  - bzax 3y ago
    The expected value of this distribution goes up with every iteration, there is no such Kelly point. You could try this with heads: double your money tails: lose all your money in which case the expected value is always $1, as you have a 1/2^n chance of having $2^n dollars after n rounds, and 0 otherwise. The point of discussing ergodicity here, however, is whether you can describe the behavior of the iterated distribution deterministically if you exclude a portion of that distribution which has measure zero.