Question

Determine whether the stochastic matrix P is regular. Then find the steady state matrix X of the Markov chain with matrix of transition probabilities P. P=

0.22 0.20 0.65
0.62 0.60 0.15
0.16 0.20 0.20

Answer 1

Explanation:

Given that:

`P = \left[\begin{array}{ccc}0.22&0.20&0.65\\0.62&0.60&0.15\\0.16&0.20&0.20\end{array}\right]`

For a steady-state of a given matrix
`\bar X`

`\bar X = \left[\begin{array}{c}a\\b\\c\end{array}\right]`

As a result P
`\bar X` =
`\bar X` and a+b+c must be equal to 1

So, if P
`\bar X` =
`\bar X`

Then;

`P = \left[\begin{array}{ccc}0.22&0.20&0.65\\0.62&0.60&0.15\\0.16&0.20&0.20\end{array}\right]\left[\begin{array}{c}a\\b\\c\end{array}\right] =\left[\begin{array}{c}a\\b\\c\end{array}\right]`

`\implies \left\begin{array}{ccc}0.22a+&0.20b+&0.65c\\0.62a+&0.60b+&0.15c\\0.16a+&0.20b+&0.20c\end{array} \right = \left \begin{array}{c}a ---(1)\\b---(2)\\c---(3)\end{array}\right`

Equating both equation (1) and (3)

(0.22a+ 0.2b + 0.65c) - (0.16a + 0.2b + 0.2c) = a - c

0.06a + 0.45c = a - c

collect like terms

0.06a - a = -c - 0.45c

-0.94 a = -1.45 c

0.94 a = 1.45 c

`c =( 0.94)/(1.45)a`

`c =( 94)/(145)a --- (4)`

Using equation (2)

0.62a + 0.60b + 0.15c = b

where;

c = 94/145 a

`0.62a + 0.60b + 0.15((94)/(145)) a= b`

`0.62a + 0.15((94)/(145)) a= -0.60b+b`

`0.62a + ((141)/(1450)) a= 0.40b`

`(0.62+(141)/(1450)) a= 0.40b`

`((62)/(100)+(141)/(1450)) a= 0.40b`

`((1043)/(1450))a= 0.40b`

`((1043)/(1450))a= (4)/(10) b`

`((1043 * 10)/(1450 * 4))a = (4)/(10) * (10)/(4)`

`b = ((1043)/(580)) a --- (5)`

From a + b + c = 1

`a + (1043)/(580)a + (94)/(145) a = 1`

`a + (1043)/(580)a + (94*4)/(145*4) a = 1`

`a + (1043)/(580)a + (376)/(580) a = 1`

`(580+ 1043+376 )/(580) a= 1`

`(1999)/(580) a= 1`

`a = (580)/(1999)`

∴

`b = (1043)/(580) * (580)/(1999)`

`b = (1043)/(1999)`

`c = (94)/(145) * (580)/(1999)`

`c= (376)/(1999)`

∴

The steady matrix of
`\bar X` is:

`\bar X = \left[\begin{array}{c}(580)/(1999) \\ \\ (1043)/(1999)\\ \\ (376)/(1999)\end{array}\right]`