Question

7.3 A three-state Markov chain has distinct holding time parameters a, b, and c From each state, the process is equally likely to transition to the other two states. Exhibit the generator matrix and find the stationary distribution.

Answer 1

Here given three state Markov chain with time parameters a, b, c is as follows Matrix

$p= \begin{bmatrix} 0 & 1/2 & 1/2\\ 1/2& 0 & 1/2\\ 1/2& 1/2 & 0 \end{bmatrix}$

Let we find stationary distribution is as

$\begin{bmatrix} v_{0} &v_{1} & v_{2} \end{bmatrix}=\begin{bmatrix} v_{0} &v_{1} & v_{2} \end{bmatrix} \begin{bmatrix} 0 & 1/2 & 1/2\\ 1/2& 0 & 1/2\\ 1/2& 1/2 & 0 \end{bmatrix}$

$\begin{bmatrix} v_{0} &v_{1} & v_{2} \end{bmatrix}= \begin{bmatrix} \frac{v_{1}}{2}+\frac{v_{2}}{2} &\frac{v_{0}}{2}+\frac{v_{2}}{2} & \frac{v_{0}}{2}+\frac{v_{1}}{2} \end{bmatrix}$

that is

$v_{0}= \frac{v_{1}}{2}+\frac{v_{2}}{2}$ ........................................ (1)

$v_{1}= \frac{v_{0}}{2}+\frac{v_{2}}{2}$ .........................................(2)

$v_{2}=\frac{v_{0}}{2}+\frac{v_{1}}{2}$ .......................................... (3)

and V₀+ V₁ + V₂ = 1 ..................................... (a)

From equation (2)

V₁ = 0.5(V₀ + V₂ ) ......................................... (4)

Equation (1) becomes

V₀ = 0.5 * (0.5(V₀ + V₂ )) + 0.5 * V₂

V₀ = 0.25*V₀ +0.25*V₂ + 0.5 * V₂

0.75*V₀ = 0.75 * V₂

V₀ = V₂

Therefore equation (4) becomes

V₁ = V₀

That is equation (a) becomes

V₀ + V₀ + V₀ = 1

V₀ = 1/3

That is Stationary distribution is as

$\begin{bmatrix} v_{0} &v_{1} & v_{2} \end{bmatrix}= \begin{bmatrix} \frac{1}{3} &\frac{1}{3} & \frac{1}{3} \end{bmatrix}$