Non-Ergodic Economics — Or, Multiplicative Games are Bad

Here's a simple game, motivated by a talk by Ole Peters. Suppose we start with \(R_{0}\) dollars. We flip a fair coin, and on each flip of the coin, we win 50% if the coin comes up heads, and lose 40% if the coin comes up tails. This sounds like a good game: on the first flip, our expected winnings are \[ \frac{1}{2} 1.5 R_{0} + \frac{1}{2} 0.6 R_{0} = 1.05 R_{0},\] so it looks like we'll do well. But now, as Peters puts it, lets add in a dynamic: let's see if we want to keep playing this game. This seems like a reasonable thing to do, since when people invest, they don't tend to put their money into the stock market for a day and then pull it out.

In this case, we have ourselves a stochastic process \(\{R_{t}\}_{t = 0}^{\infty}\) where \(R_{t}\) is the amount of money we have after \(t\) time steps. It's clear from the description of the game that the process will evolve according to \[R_{t} = \left\{\begin{array}{cc} 1.5 R_{t-1} &: \text{with probability } \frac{1}{2} \\ 0.6 R_{t-1} &: \text{with probability } \frac{1}{2} \end{array}\right.\] This is a nice way to look at the process procedurally (for instance, it shows us the process is Markov, and gives us an easy way to simulate from it). But it will get unwieldy if we want to do anything analytical. Instead, notice that to determine \(R_{t}\), we are flipping a coin \(t\) times, and counting up the number of heads and tails1, which tells us how many times we need to multiply \(R_{0}\) by either 1.5 or 0.6, respectively. Coin flips are governed by the binomial distribution, and we are multiplying, so this stochastic process is called the binomial multiplicative process2. After \(t\) flips, the only values that \(R_{t}\) can take are \(R_{0} 1.5^{i} 0.6^{t-i}\), \(i = 0, \ldots, t\), since we can have had zero heads, one heads, two heads, ..., up to \(t\) heads, and it takes those values with probability \(\binom{t}{i} \left(\frac{1}{2}\right)^{i} \left(\frac{1}{2}\right)^{t - i} = \binom{t}{i} \left(\frac{1}{2}\right)^{t}\). Thus, we've fully specified the distribution of the process for any length of time, and, in principle, we're done. For example, if we want to compute the mean of the process, we group terms and apply the binomial theorem, giving us \[ m(t) = E[R_{t}] = R_{0} 1.05^{t}.\] Thus, over time, the expected value grows exponentially. This again makes it seem like we want to play the game. Finally, we can compute the variance as a function of time, giving us \[ V(t) = \text{Var}(R_{t}) = R_{0} (1.305^{t} - 1.1025^{t}),\] which again increases with time. (This is our first sign that perhaps we don't want to play the game, if the winnings of people get more and more spread out as the game goes on.) We can plot both \(m(t)\) and \(V(t)\) as a function of time,

for the mean and

for the variance. The black dots were computed using a sample of ten thousand players. The red dots are the analytical values. As anticipated, we get good agreement between the sample values and the population values. Finally, note that the plots are semi-log. This is commonly done without mention. But I think it's worth mentioning.

So far, we've looked at measures of center and variation. But again, we have the entire distribution available to us, so why not use it? Let's look at the distribution of wealths for a society of people playing this game. First, here's a schematic of how the game plays out: This looks very much like a (biased) random walk, and for good reason: if we take the logarithm of a multiplicative process, it becomes an additive process, and we in fact find ourselves with a random walk. This explains the use of logarithms throughout. Logarithms also makes the computer much happier. If we shaded the lines in by the number of sample paths that traverse them, we would see that most people follow along the central trajectory, with less and less people following along the further out branches. (We see this explicitly in the most extreme trajectories, which don't get traced out with only 10000 sample paths.) We can of course compute the proportion of walks that end up at node, first for \(t = 1\)

and then for \(t = 10\)

Again, not that the x-axis is scaled to have logarithmic spacing. As expected, this gives us a binomial distributions. Recalling that a binomial process (a series of fair coin flips!) underlies the dynamics we observe, this shouldn't be too surprising. What we do see is that most people have very little money ($50 or less), while a few people have much more money ($5000). If we keep playing this game out to, say, \(t = 1000\), we see

In this case, most people have $\(1.322 \times 10^{-23}\) or less, while a few have $\(1.233841 \times 10^{176}\), an astronomical amount of money. This explains the strange discrepancy we've seen between the expected value as a function of time (which continues to increase), and what Peters points out at the beginning of his lectures (and I have so far failed to do), namely that any particular person is very likely to lose a great deal of money if they keep playing this game. In particular, here's a single realization played out over the course of 10000 goes at the game:

This person has clearly lost out. And they are the norm. Another reminder that average values don't necessarily capture what you would like to know about a distribution.

  1. Since scalar multiplication is commutative, it doesn't matter what order the heads and tails occurred.

  2. Ole only gave this name in passing during his lecture. Again, I've noticed that mathematicians don't feel comfortable with something until it has a name, while physicists don't seem to mind.