Question

Let α and β be positive constants. Consider a continuous-time Markov chain X(t) with state space S = {0, 1, 2} and jump rates

q(i,i+1) = β   for0≤i≤1

q(j,j−1) = α for1≤j≤2.

Find the stationary probability distribution π = (π0, π1, π2) for this chain.

Answer 1

Answer:

Given Data

Let α and β be positive constants

S = {0, 1, 2}

jump rates

$q(i,i+1) = \beta for 0 \leq i \leq1$

961.j-1) = aforl<i < 2.

X ( t ) be continuous time Markov chain with state space s = { 0 , 1 , 2 }

$q(i,i+1) = \beta for 0 \leq i \leq1$

961.j-1) = aforl<i < 2.

0   1 2

  $P=\begin{bmatrix} 0\\1 \\2 \end{bmatrix}\begin{bmatrix} \alpha &\beta &0 \\ \alpha& 0 &\beta \\ 0 &\alpha &\beta \end{bmatrix}$

V = ΣΥ;P;A

(V. Vi v) = (V V V2) P

$V_{0}+V_{1}+V_{2}=1$

$V_{0}=\alpha V_{0}+\alpha V_{1}$   $\Rightarrow (1-\alpha)V_{0}=\alpha V_{1}$

$V_{1}=\beta V_{0}+\alpha V_{2}$  

$V_{2}=\beta V_{1}+\beta V_{2}$   $\Rightarrow (1-\beta)V_{2}=\beta V_{1}$

$V_{1}=\beta\left ( \frac{\alpha}{1-\alpha} \right )V_{1}+\alpha\left ( \frac{\beta}{1-\beta} \right )V_{1}$

$V_{1}=\left ( \frac{\beta \alpha}{1-\alpha}+\frac{\alpha \beta}{1-\beta} \right )V_{1}$

$V_{0}=\frac{\alpha}{1-\alpha}V_{1}$

$V_{2}=\frac{\beta}{1-\beta}V_{1}$

$\frac{\alpha}{1-\alpha}V_{1}+V_{1}+\frac{\beta}{1-\beta}V_{1}=1$

$\frac{(1-\beta)\alpha V_{1}+(1-\alpha)(1-\beta)V_{1}+(1-\alpha)\beta V_{1}}{(1-\alpha)(1-\beta)}=1$$\alpha V_{1} -\alpha \beta V_{1}+V_{1}-\alpha V_{1}-\beta V_{1}+\alpha \beta V_{1}+\beta V_{1}-\alpha \beta V_{1}=(1-\alpha)(1-\beta)$$V_{1}-\alpha \beta V_{1}= 1-\alpha-\beta -\alpha \beta$

$V_{1}=\frac{(1-\alpha)-\beta(1-\alpha)}{1-\alpha \beta}$

$V_{2}=\frac{\beta}{1-\beta}^{\frac{1}{(1-\alpha \beta)}(1-\alpha)(1-\beta)}$

$V_{0}=\frac{\alpha}{1-\alpha}.\frac{1}{1-\alpha \beta}(1-\alpha)(1-\beta)$

$V_{0}=\frac{\alpha(1-\beta)}{1-\alpha \beta}$

$V_{1}=\frac{(1-\alpha)(1-\beta)}{1-\alpha \beta}$

$V_{2}=\frac{\beta(1-\alpha)}{1-\alpha \beta}$

Stationary distribution is

$\pi _{0}=\frac{\alpha(1-\beta)}{1-\alpha \beta}$

$\pi_{1}=\frac{(1-\alpha)(1-\beta)}{1-\alpha \beta}$

$\pi_{2}=\frac{\beta(1-\alpha)}{1-\alpha \beta}$

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