Question

Problem 3:

Find the equilibrium distribution for each transition matrix.

a)

1/2 1/9 3/10

1/3 1/2 1/5

1/6 7/18 1/2

b)

2/5 0   3/4

0 2/3 1/4

3/5 1/3 0

Problem 4:

For either transition matrix in problem 3, find the other two eigenvalues with corresponding eigenvectors.

Answer 1

We would be looking at Question 3 both parts here as:

Question 3:

a) Let the equilibrium distribution of being in 3 states here be: X, Y and Z respectively.

From first row, we get here:
X = 0.5X + (1/9)*Y + 0.3Z
X = (2/9)Y + 0.6Z

From second row, we get here:
Y = (1/3)X + 0.5Y + 0.2Z
1.5Y = X + 0.6Z
X = 1.5Y - 0.6Z

Putting this value of X in the first equation, we get here:

1.5Y = (2/9)Y + 1.2Z
[(3/2) - (2/9)] Y = 1.2Z
(27 - 4)Y / 18 = 1.2Z
23Y = 21.6Z
Y = 21.6Z / 23

X = 1.5Y - 0.6Z = 1.5*21.6Z / 23 - 0.6Z = 0.8087 Z

Also sum of all probabilities should be 1. Therefore we get here:

X + Y + Z = 1
0.8087 Z + 0.9391 Z + Z = 1
Z = 0.3639
Y = 0.9391 Z = 0.3418
X = 0.8087 Z = 0.2943

Therefore the equilibrium distribution here is given as:

(0.2943, 0.3418, 0.3639)

b) Again let the equilibrium distribution be X, Y and Z respectively.

From first row, we get here:
X = 0.4X + 0.75Z
X = 1.25 Z

From second row, we get here:
Y = (2/3)Y + 0.25Z
Y = 0.75 Z

X + Y + Z = 1
1.25Z + 0.75Z + Z = 1
Z = 1/3

Therefore Y = 0.75/3 = 0.25
X = 1.25Z = 0.4167

Therefore the equilibrium distribution here is given as:

(0.4167, 0.25, 1/3)