# Steady state distribution for Markov pure jump process

Assuming an irreducible, positive recurrent Markov Pure jump markov process with state space, $S={0,1}$

The embedded Markov Chain which is doubly stochastic (i.e) columns and rows of the transition matrix sum up to 1.

Find the steady state distribution (SSD).

I know that for a Markov chain that has a doubly stochastic transition matrix will have the uniform distribution as the SSD. In other words, 1/(number of states in the state space).

I could solve for the SSD by solving $\pi q=0$, where q is the generator.Since the answer is not just 1/(number of states in the state space), would like to ask why does it differ? and is there a way to relate them both?

Thanks heaps!

I'm not totally clear on what you are asking, but the uniform distribution is the steady state here. Since there are only $2$ states, your generator is of the form $$q = \left( \begin{array}{cc} -t & t \\ s & -s \\ \end{array} \right)$$ for some $s,t > 0$ and consequently the embedded chain has kernel $$K = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right).$$ The steady state for this chain, and therefore $q$, is $\pi = [1/2, 1/2]$.