Suppose I have a family of countable state-space, discrete-time Markov chains, indexed by a parameter $r \in \mathbb{R}$. The state space is the same for all values of $r$; the transition probabilities depend on $r$ continuously. Is there some result like:


If the chain obtained for $r=r\_0$ is positive recurrent, then the same is true for all values of $r$ in some open neighborhood of $r\_0$; and as $r \to r\_0$, the associated invariant distributions converge. (I assume that if the first statement is true, then so is the second one.)


In terms of simple linear algebra: I have a countable set of linear equations (defining the invariant distribution in terms of the transition probabilities of the Markov chain), which depend continuously on $r$. At some value $r=r\_0$ the system has a unique positive solution. Is there a neighborhood of $r\_0$ for which this will still be true? (What conditions do I need to impose on my set of equations to get this?)



Continuous family of Markov chains
No it's not true. Consider Markov chains on $\mathbb N$ where the only transitions are from $n$ to $n\pm 1$.


For $n>1$, set $P\_{n,n+1}=r+\frac12(1-\frac1n)$ and $P\_{n,n-1}=-r+\frac12(1+\frac1n)$.


For $r=0$, solving the detailed balance equation, we get $\pi\_{n+1}n(n+2)=\pi\_n(n-1)(n+1)$ which has solutions $\pi\_n=C/[(n-1)(n+1)]$. Hence this is positive recurrent.


For $r>0$, there is a rightwards drift and hence it is transient.


