Question

You are given a transition matrix P. Find the steady-state distribution vector

P =
0.3 0 0.7
1 0 0
0 0.4 0.6

hello, i don't understand how to the stedy-state vector system works. i have already tried multiplying
P by[ 0.3 0 0.7] that did not work, i also tried multiplying it by [ x y z] but it didn't seem to work.

extra info:
the example that i was looking at in the book did this

[ x y] [.8 .2]
.1 .9
so i assumed that i could do the same with x y z

Answer 1

Conditions

You are given a transition matrix $P$. Find the steady-state distribution vector

P =

0.3 0 0.7

1 0 0

0 0.4 0.6

hello, i don't understand how to the steady-state vector system works. i have already tried multiplying

P by[ 0.3 0 0.7] that did not work, i also tried multiplying it by [ x y z] but it didn't seem to work.

extra info:

the example that i was looking at in the book did this

[ x y ] [ .8 .2 ]

.1 .9

so i assumed that i could do the same with x y z

Solution

The steady state vector $x$ satisfies the equation $Px = x$.

That is, it is an eigenvector for the eigenvalue 1.

We must multiply the matrix $P$-I on $(x,y,z)$

$$\left( \begin{array}{ccc} -0.7 & 0 & 0.7 \\ 1 & -1 & 0 \\ 0 & 0.4 & -0.4 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = 0$$ $$\left\{ \begin{array}{l} -0.7x + 0.7z = 0 \\ x - y = 0 \\ 0.4y - 0.4z = 0 \end{array} \right.$$ $$x = y$$ $$y = z$$ $$z = x$$ $$Let x = \frac{1}{\sqrt{3}}$$

The steady state vector $(x,y,z) = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$