Concentration Occurs on the Log-scale

Wouter M. Koolen

2023-06-14

Abstract

On being rich in expectation yet bankrupt in practise

Introduction

Let’s play a game. You start with \(\$1\). We play for \(n=1000\) rounds. Each round, we flip an independent coin with bias \(\theta=0.9\) of heads. If it’s heads, you multiply your capital by \(a=1.1\), while if it’s tails you multiply it by \(b=0.25\). The expected multiplication factor is in your favour: in each round it is \[\theta a + (1-\theta) b ~=~ 1.015\] Would you like to play this game?

By independence, your expected payoff is \[(\theta a + (1-\theta) b)^n ~\approx~ \text{\$$3$ million}\] This game is a money fountain! Or is it? Let’s see what will happen by inspecting the distribution of your payoff after the \(n\) rounds. The following figure displays its survival function. The survival function at \(x\) is the probability that the payoff is at least \(x\). Note the logarithmic scale(s):

Survival function of your payoff, on log-linear (left) and log-log (right) scale.

In conclusion, that \(3\) million expected value is dominated by large but terribly unlikely outcomes (these would be to the far right in the graph). In particular, we see that the probability that you end up with at least your initial capital of \(\$1\) is about one in 27 thousand. From this perspective this game looks hardly interesting.

In conclusion, this game is better than a lottery (since here you win big in expectation) but you still lose in practise. More generally, a product \(S\) of i.i.d. random variables \[S ~=~ \prod_{i=1}^n X_i\] does not concentrate around its expectation \(\operatorname*{\mathbb E}[X]^n\). Instead, concentration occurs after taking logarithms: \[\frac{1}{n} \ln S ~=~ \frac{1}{n} \sum_{i=1}^n \ln X_i ~\to~ \operatorname*{\mathbb E}[\ln X]\] which indicates that \(S\) concentrates around \(e^{n \operatorname*{\mathbb E}[\ln X]}\). In our example \(\operatorname*{\mathbb E}[X] > 1\) but \(\operatorname*{\mathbb E}[\ln X] < 0\). So while the expected value of \(S\) tends to infinity with \(n\), its practical outcome will be certain bankruptcy.