> You're halfway there! The underlying distribution on trajectories has a huge variance like you pointed out, but the mean of a finite sample is likely to have

> You're halfway there! The underlying distribution on trajectories has a huge variance like you pointed out, but the mean of a finite sample is likely to have only a (vanishingly) small variance, because of the central limit theorem! (aka concentration of measure for "typical" trajectories) The mean of a finite sample is a random number. This random number has a distribution and has a variance. The variance of the sample mean is the variance of the original distribution divided by N. Huge. Saying that "the mean of a finite sample is likely to have only a small variance" doesn't mean anything. Maybe you meant that the variance of a finite sample, which is also a random number and has a distribution, is much more likely to be low (all the elements in the sample are in the high-density lower part of the original distribution) than high (a few elements in the very skewed upper tail). And, as I said, in your example there is no concentration of measure in the original distribution. Typical trajectories will move around $1 but the probability of having exactly $1 at the end goes down as the horizon grows. As an aside, every trajectory will eventually reach any (positive) level that you may specify. And bounce between any two levels infinite times. Well, those things may not happen for some trajectories but they have probability zero. Infinite stochastic processes are fun. To see that concentration you have to transform the original dollar outcome distribution into a growth rate per flip distribution (which converges to 0 in your example).