How many birds sit on the north wire on average?

Question

Five birds perch on two wires, north and south. Every minute, one of the five birds is chosen at random and flies to the other wire. The state is the number of birds on the north wire, from 0 to 5. In the long run, what is the average number of birds on the north wire?

Accepted Answer

This is the Ehrenfest model with N = 5. From a state with k birds on north, the count rises with probability \frac{N-k}{N} (a south bird flies over) and falls with probability \frac{k}{N} (a north bird flies away). Detailed balance for this birth-death chain gives \frac{\pi{k+1}}{\pik} = \frac{N-k}{k+1}, so the long-run shares are proportional to \binom{5}{k}, and the stationary distribution is Binomial(5, 	frac12): \pik = \binom{5}{k}/32. We want the average of the count under this distribution. The mean of a Binomial(N, 	frac12) is N 	imes 	frac12, so 	ext{average number on north} = 5 …

How many birds sit on the north wire on average?

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