Markov DTMC and CTMC: How to build a $Q$ generating matrix using given $2$ state $q$ probabilities? Find Steady State Probabilities

Question

I know that $$q_{(i,j)(i+1,j)}=\lambda_1$$ $$q_{(i,j)(i-1,j+1)}=\lambda_2$$ $$q_{(i,j)(i+1,j-1)}=0.5\lambda_3$$ $$q_{(i,j)(i,j-1)}=0.5\lambda_3$$ where $\lambda_1 , \lambda_2 , \lambda_3 $ are rates of exponential distribution.

I'm trying to write down the $Q$ matrix to solve for stationary probabilities using: $(\pi_0,\pi_1 ... \pi_n)Q=0$

I don't know how to write this $Q$ matrix down. I'm not sure how to define these $\pi_i$ , do I need to have it in $2$ variables like $\pi_{i,j}$?

Answer 1

A CTMC on a finite state space can be presented in two equivalent ways:

1. As a generator matrix $Q$
2. As a transition probability matrix $P$ and average holding times $\tau_i$ for each $i \in S$.

This $P$ is the transition probability matrix of a certain DTMC, which is called the jump chain of the original CTMC.

To go from 1 to 2, you do one of two things for each $i$.

1. If $Q_{ii}=0$ then $P_{ii}=1,P_{ij}=0$ for $j \neq i$ and $\tau_i=+\infty$.
2. If $Q_{ii} \neq 0$ then $P_{ii}=0,P_{ij}=-\frac{Q_{ij}}{Q_{ii}}$ and $\tau_i=-\frac{1}{Q_{ii}}$.

To go from 2 to 1, take $Q_{ii}=-\frac{1}{\tau_i}$ and $Q_{ij}=-P_{ij} Q_{ii}$.

The stationary distribution $\pi$ is not yours to define, it is the solution of a system of equations.