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 Q3 both parts here as;

Q3) a) Let the equilibrium transition probabilities here be X, Y and Z respectively.

From first row, we get:
X = 0.5X + Y/9 + 0.3Z
X = 2Y/9 + 0.6Z

From second row, we get here:
Y = X/3 + Y/2 + Z/5
0.5Y = X/3 + 0.2Z
1.5Y = X + 0.6Z

Therefore, 1.5Y = 2Y/9 + 1.2Z
(3Y/2) - (2Y/9) = 1.2Z
(18 - 4)Y / 18 = 1.2Z
7Y/9 = 1.2Z
Y = 9*1.2Z / 7 = 10.8Z / 7
X = 1.5Y - 0.6Z = 16.2Z / 7 - 0.6Z = 12Z / 7

Also, we have here:
X + Y + Z = 1
(12Z/7) + (10.8Z/7) + Z = 1
29.8Z = 7
Z = 0.2349
Y = 10.8Z / 7 = 0.3624
X = 12Z / 7 = 0.4027

This is the required equilibrium transition probabilities here.

b) Let the probabilities here be X, Y and Z respectively.

From first row, we have:
X = 0.4X + 0.75Z
0.6X = 0.75Z

From second row, we get:
Y = 2Y/3 + 0.25Z
Y = 0.75Z
X = 1.25Z

Therefore, X + Y + Z = 1
1.25Z + 0.75Z + Z = 1
Z = 1/3 = 0.3333
X = 0.4167
Y = 0.75Z = 0.25

Therefore the equilibrium transition probabilities here are given as:
(0.4167, 0.25, 0.3333)