There is a recent paper[1] and an epic twitter thread[2] well worth the reading
[1] https://arxiv.org/pdf/1906.04652.pdf
[2] https://twitter.com/hulme_oliver/status/1139148255969906689
[1] https://arxiv.org/pdf/1906.04652.pdf
[2] https://twitter.com/hulme_oliver/status/1139148255969906689
"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.
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...
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.
Maybe a reference to this paper [1] ?
[1] The time resolution of the St. Petersburg paradox, by Ole Peters
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?
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
If you earn 10% and then lose 10% you are worse off than before. Start with 10: (11-1.1 < 10).
I agree that expectation may not be the most meaningful number here, in the long run is almost sure that we lose everything but there is a zero probability of infinite gains that makes up for it!
Why do you say almost sure to lose everything?
In fact, an even stronger statement is possible: If you start with $1MM, eventually you will fall below $1 and never go above $1 again no matter how long you keep playing.
And all of that not just on a fraction of sequences, but with probability 1.
It's just a regular sub-martingale with 'positive' expectation as you say.
As he says in the Twitter thread anyway, they are coining this name.
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.
($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.
The optimal bet is not the same as the critical bet separating the positive growth and negative growth regions.
50% of ($1.25 + $.90) for a $1 bet seems like a great game to play.
You would need around 26% more heads than tails just break even (0.6*(1.5^1.26)=1). But even if you were lucky at first, you will always "regress toward the means" in terms of head to tails ratio if you kept playing.
2.25 * .36 = .36 * 2.25 = .81
You started with $1 and now have 81 cents. I recently learned that a lot of early computer-era academic finance actually made the same error of averaging returns, as described in Finding Alpha:
http://falkenblog.blogspot.com/2016/08/finding-alpha-pdf.htm...
set.seed(123)
# simulating multiplicative growth
n <- 1e4
init_wealth <- 1
x <- sample(c(0.6, 1.5), n, replace=TRUE)
init_wealth * prod(x)
# analytic model
n1 <- length(x[x==0.6])/n
n2 <- length(x[x==1.5])/n
rate <- (0.6^n1) * (1.5^n2)
init_wealth * (rate^n)
(Edit: formatting, and wording for clarity)
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).
H = [1.5 or 0.6] * [1.5 or 0.6] * ... } for N factors
log H = log([1.5 or 0.6]) + log([1.5 or 0.6]) + ...
E log H = Sum{1..N} E log([1.5 or 0.6])
= Sum{1..N} [-0.023]
= -0.023 N
< 0
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.