Question

Given the transition matrix P for a Markov chain, find the stable vector W. Write entries as fractions in lowest terms.

P= 0.5 0 0.5
    0.2 0.2 0.6
      0    1     0

Answer 1

Given the transition matrix P for a Markov chain ,

Let the stable vector be W = [X , Y , Z]

stable vector must satisfy

Wp = W

and X +Y+Z =1 eq (1)

i.e [X,Y,Z] = [X,Y,Z]

0.5X +0.2Y = X (2)

0.2Y+Z = Y   (3)

0.5X+0.6Y =Z (4)

now from equation (2)

0.5 X +0.2Y = X

0.2Y = X-0.5X

0.2Y = 0.5X    (5)

0.2Y +Z =Y

Z = Y -0.2Y

= 0.8Y

Put the equation (4)in equation (3)

0.5X +0.6 Y = 0.8Y

0.5X = 0.8Y - 0.6Y

0.5X = 0.2Y

X = 0.2Y / 0.5

X = 0.4 Y (6)

put the X value in equation (5)

0.2Y = 0.5X

0.2Y = 0.5(0.4Y)

0.2Y = 0.2Y

Y = 1

Now put the Y value in equation (6)

X = 0.4 Y

= 0.4(1)

= 0.4

now keep the X value and Y value in equation (4)

0.5 X + 0.6 Y = Z

0.5(0.4) +0.6 (1) = Z

0.2 +0.6 = Z

Z = 0.8

and W =[X, Y , Z ]

the value of W is

W = [0.4 , 1 , 0.8 ]