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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?

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  • $\begingroup$ 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
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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

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

and

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

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.

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  • $\begingroup$ Thanks. Does this also immediatelly imply the chain is ergodic? $\endgroup$ – dff Sep 1 '14 at 9:08
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    $\begingroup$ 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

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