Question

Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

P = [0.31 0.69
0.18 0.82]
P^4 = ______
(Type an integer or decimal for each matrix element. Round to four decimal places as needed.)
Continue taking powers of P until S can be determined
S = ______
(Type an integer or decimal for each matrix element. Round to four decimal places as needed.)

Answer 1

S = [0.2069,0.7931]

Explanation:

Transition Matrix:

`P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

`P^(2) =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(2) =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]`

Transition matrix P raised to the power 3 (at k = 3)

`P^(3) =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(3) =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(3) =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]`

Transition matrix P raised to the power 4 (at k = 4)

`P^(4) =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(4) =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(4) =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]`

Transition matrix P raised to the power 5 (at k = 5)

`P^(5) =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]`

`P^(5) =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]`

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]