Question

A cat and a mouse move independently back and forth between two rooms. At each time step, the cat moves from the current room to the other room with probability 0.8. Starting from room 1 , the mouse moves to room 2 with probability 0.3 (and remains otherwise). Starting from room 2 , the mouse moves to room 1 with probability 0.6 (and remains otherwise). a) Find the stationary distributions of the cat chain and of the mouse chain.

Answer 1

The stationary distribution of the mouse chain is [0.66666667, 1.33333333], which means that the mouse is more likely to be in room 2 than in room 1 after a long time.

Stationary distribution of the cat chain

The cat chain has two states, which we can label as "room 1" and "room 2." The probability transition matrix for the cat chain is:

P = [[0.2, 0.8], [0.8, 0.2]]

To find the stationary distribution of the cat chain, we need to solve the equation:

πP = π

where π is the stationary distribution vector. Multiplying both sides of the equation by the left eigenvector of P with eigenvalue 1, we get:

π = [1, 1]

Therefore, the stationary distribution of the cat chain is [1, 1], which means that the cat is equally likely to be in either room after a long time.

Stationary distribution of the mouse chain

The mouse chain also has two states, which we can label as "room 1" and "room 2." The probability transition matrix for the mouse chain is:

P = [[0.7, 0.3], [0.4, 0.6]]

To find the stationary distribution of the mouse chain, we need to solve the equation:

πP = π

where π is the stationary distribution vector. Multiplying both sides of the equation by the left eigenvector of P with eigenvalue 1, we get:

π = [0.66666667, 1.33333333]