Problem 15.4 Easy Free

Proportions in an Infinite Line

On an infinite sidewalk, every animal is a cat or a dog, and each animal is followed by the next animal in line. Six out of every seven cats are immediately followed by a dog, while one out of every four dogs is immediately followed by a cat. Modelling the sequence of species as a stationary two-state Markov chain, find the long-run proportion of dogs among all animals on the sidewalk.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let cc and dd be the long-run proportions of cats and dogs, so c+d=1c+d=1. The transition probabilities are

P(C→D)=67,P(D→C)=14.\mathbb{P}(\text{C}\to\text{D})=\tfrac67,\qquad \mathbb{P}(\text{D}\to\text{C})=\tfrac14 .

Balance of flows. In a stationary sequence, the proportion of consecutive pairs of the form (cat, dog) is 67c\tfrac67c. The proportion of pairs of the form (dog, cat) is 14d\tfrac14d.

Along the line, cat-blocks and dog-blocks alternate. Hence every switch from cats to dogs is followed by a switch back, and the two kinds of switch occur equally often in the long run:

67 c=14 d⟹d=247 c.\tfrac67\,c=\tfrac14\,d \quad\Longrightarrow\quad d=\tfrac{24}{7}\,c .

This is exactly the stationarity equation πP=π\pi P=\pi for a two-state chain.

Substituting into c+d=1c+d=1 gives 317c=1\tfrac{31}{7}c=1. Hence c=731c=\tfrac{7}{31} and d=2431d=\tfrac{24}{31}.

2431≈0.774\boxed{\dfrac{24}{31}\approx 0.774}

Generalisation. Suppose a cat is followed by a dog with probability pp, and a dog by a cat with probability qq. Then pc=qdpc=qd, so the proportion of dogs is pp+q\dfrac{p}{p+q}.