The Convoy Problem

Continuing my break from algebra, I’m going to deal with the convoy problem in probability (speaking of which, Echidne’s fourth post on statistics is up). The convoy problem is as follows: you have an infinite number of cars driving one behind the other. The first car is a(1), the second is a(2), the third is a(3)…

Now, each car wants to drive at a certain speed. The important part is that the speed of car a(n), call it v(n), is discrete, so it can only take integer values, and follows the same distribution for all n. In other words, a car can only drive at a whole number of kilometers per hour, and each car chooses its speed from a fixed probability.

Since the cars are one behind the other, no car can actually drive faster than a car ahead of it, so the cars form convoys: the first convoy consists of the first car and all cars behind it that want to drive at an identical or higher speed, the second starts with the first car to drive more slowly, etc.

It makes sense to talk about the length of convoys. So let’s call the length of the first convoy L; obviously, L follows some distribution, based on the distribution of v. The problem is to show that the mean, or expected value, of L is infinite.

The trick is to use discreteness. v can only take integer values, and since cars don’t go in a negative speed, v is nonnegative. So there has to be a least integer m such that v has a nonzero probability of being m. In particular, there is a nonzero probability that v(1) = m.

In that case, all cars have to go at speed m; they can’t go any faster without crashing into a car in front of them, and they can’t go any more slowly because by choice of m, all speeds values less than m are impossible. Then L will be infinite.

We then get that there’s a nonzero probability of L being infinite. Even if in all other cases L = 1, the lowest possible value, averaging finite values and positive infinity still gives positive infinity.

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