# Behaviour of a Markov Chain, given a Lyapunov condition

I'm reading this notes from Martin Hairer about convergence of Markov Processes (on a discrete state space $S$ and in continuous time). On page 12, before presenting the so-called "Harris Theorem", the following assumption is made, toghether with an heuristic explanation:

For a fixed $T>0$, there exists a function $V:S\to [0,\infty)$ and constants $K\geqslant 0$ and $\gamma\in (0,1)$ such that $$\mathcal{P}_TV(x)\leqslant \gamma V(x) +K$$ Where $\mathcal{P}_Tf(x)=\mathbb{E}_x(f(X_T))$ , i.e., $\mathcal{P}$ is the Markov semigroup of the MC $(X_t)$ .

The heuristic idea of the above assumption is this one:

[The above assumption] ensures that the dynamic enters the “centre” of the state space regularly with tight control on the length of the excursions from the centre.

So my questions are:

a) Can this idea be mathematically presented?

b) Suppose $S$ is $\mathbb{N}_0^n$, with $n\in \mathbb{N}$ , and let $\mathcal{C}$ be some finite neighbourhood of the origin. Then for $x\notin \mathcal{C}$ and $T_{\mathcal{C}}$ the hitting time of $\mathcal{C}$, can I assure that $$\mathbb{P}_x(T_{\mathcal{C}}<\infty)=1$$

c) What about question b) if I know that $V(x)=||x||_1:=\sum|x_i|$ and that $K=0$?

This question arises in the context of a problem whith this Markov Chain.

As for question c), I've only been able to say that

$$\mathbb{P}_x(||X_T||<||x||_1)\leqslant \gamma$$

with $T$ and $\gamma$ as before.

• You can have a look in the discrete time setting to the book of Mein and Tweedie (Markov Chains and Stochasitc Stability) where a condition looking very close to this one appears to be equivalent to ergodicity theroem 1.3.1 (altogether with 2 others conditions which are interpreted as different forms of stochastic stability). Regards – The Bridge Sep 27 '17 at 14:34
• The basic idea is to rewrite the Lyapunov condition as follows. For $\tilde\gamma\in (0,\gamma)$, there exists a set $\mathcal C$ such that $$\mathcal{P}_T V(x) \leq \tilde\gamma V(x) + K \; \mathbb{1}_{\mathcal{C}}(x).$$ The set is given essentially as a sublevel set of $V$: $\mathcal{C} = \{ x: (\gamma-\tilde \gamma) V(x) \geq K \}$. Then, the hitting probability of the set $\mathcal{C}$ is exponentially. If this is the kind of result, I can look for a reference or write some more explanation as answer. – André Schlichting Oct 2 '17 at 11:12
• @AndréSchlichting yes, I think that may help, thank you in advance. – Max Oct 2 '17 at 19:59