Markov DTMC and CTMC: How to build a $Q$ generating matrix using given $2$ state $q$ probabilities? Find Steady State Probabilities 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 $\pi_{i,j}$?

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


*

*As a generator matrix $Q$

*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 $i$.


*

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

*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.
