Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 05 (c)/Solution 1

We can find the equilibrium probability distribution by solving for ${\mathbf {x}}=[a,b,c,d]^{T}$ below:

$P{\mathbf {x}}={\begin{bmatrix}1/2&1/6&1/6&0\\1/4&1/2&1/6&1/4\\1/4&1/6&1/2&1/4\\0&1/6&1/6&1/2\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\mathbf {x}}$

This is equivalent to the following set of equations

$1/2a+1/6b+1/6c+0d=a$

$1/4a+1/2b+1/6c+1/4d=b$

$1/4a+1/6b+1/2c+1/4d=c$

$0a+1/6b+1/6c+1/2d=d$

$a+b+c+d=1$

We can simplify these equations to

$(1)\qquad b+c=3a$

$(2)\qquad 3a+2c+3d=6b$

$(3)\qquad 3a+2b+3d=6c$

$(4)\qquad b+c=3d$

$(5)\qquad a+b+c+d=1$

Equating (1) and (4), we get

$a=d.$

Substracting (3) from (2) we get

$2b-2c=0.$

Hence,

$b=c.$

Plugging this into equation (1), we get

$2b=2c=3d=3a$

Using the last condition, that the sum of all variables is equal to 1 we get:

answer: $\color {blue}{\begin{bmatrix}1/5&3/10&3/10&1/5\end{bmatrix}}$