Suppose I have a discrete time Markov chain $\boldsymbol{X}$ with state space $\mathbb{R}^+$. The chain is $\psi$-irreducible, aperiodic, atomless and has an invariant measure $\pi$.

If $\pi$ is finite, does that imply this measure is unique and that it is the stationary distribution of the Markov chain? I.e. does it imply $\boldsymbol{X}$ is Harris?

  • $\begingroup$ A finite invariant measure yields a stationary distribution when normalized to have total measure 1. $\endgroup$ – charles.y.zheng Aug 29 '14 at 15:41
  • $\begingroup$ I'm confused. Then what is the difference between being positive recurrent and being postive Harris recurrent? $\endgroup$ – dff Sep 1 '14 at 8:44
  • $\begingroup$ Recurrent is the condition $\sum P^n(x,A) = \infty$, Harris recurrent is the condition $P_x(X_n \in A i.o) = 1$ for all x. $\endgroup$ – charles.y.zheng Sep 2 '14 at 17:22

Yes. Let $T$ denote the transition kernel. Suppose $\pi$ and $\nu$ are distinct stationary distributions. Let $\tau'$ be defined by $\tau'(A) = \min(\pi(A),\nu(A))$: then $\tau'$ is a finite measure. Let $p = \tau'(\mathbb{R}^+)$; since $\pi \neq \nu$, we have $p < 1$, and since the chain is irreducible, $p > 0$. Let $\tau = \tau'/p$, $\tilde{\pi} = (\pi-\tau')/(1-p)$ and $\tilde{\nu} = (\nu - \tau')/(1-p)$, so that

$\pi = p\tau + (1-p)\tilde{\pi}$


$\nu = p\tau + (1-p)\tilde{\nu}$.

Note that $\tilde{\pi} = \max(0, \pi-\nu)/(1-p)$ and $\tilde{\nu} = \max(0,\nu-\pi)/(1-p)$, so that $\max(\tilde{\pi},\tilde{\nu}) = 0$

Now we show that $\tau$ is also a stationary distribution of the chain. Observe that $\pi= T\pi = pT\tau + (1-p)T\tilde{\pi}$ so $\pi \geq pT\tau$. Similarly, $\nu \geq pT\tau$. Therefore, $pT\tau \leq \min(\pi,\nu) = p\tau$. Combined with the fact that $\tau(\mathbb{R}^+) = T\tau(\mathbb{R}^+)$, we have $T\tau = \tau$.

This, in turn, implies that $\tilde{\pi}$ and $\tilde{\nu}$ are also stationary distributions. However, the fact that $\max(\tilde{\pi},\tilde{\nu}) = 0$ then implies that the chain is reducible--a contradiction.

  • $\begingroup$ Thanks. Does this also immediatelly imply the chain is ergodic? $\endgroup$ – dff Sep 1 '14 at 9:08
  • 1
    $\begingroup$ Yes, see sciencedirect.com/science/article/pii/0304414975900332 $\endgroup$ – charles.y.zheng Sep 2 '14 at 17:06
  • $\begingroup$ Now I am confused. I've been reading "Markov Chains and Stochastic Stability" by Tweedie (probability.ca/MT/BOOK.pdf). If I look at Theorem 13.0.1, then you're saying I could drop the word Harris everywhere and the theorem would still hold? $\endgroup$ – dff Sep 2 '14 at 21:21
  • $\begingroup$ Maybe not--that is not obvious. All I proved is that if a stationary distribution for an irreducible chain exists, then it is unique. $\endgroup$ – charles.y.zheng Sep 3 '14 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.