2

We consider the notations introduced in Durrett's book (5th edition): let $X=(X_k)_{k \in \mathbb{N}}$ be a Markov chain with countable countable state space $S$ and transition matrix $P.$ Let $\mathcal{T}$ be the tail $\sigma$-field of $X:\mathcal{T}=\bigcap_{k \in \mathbb{N}}\sigma(X_k,X_{k+1},...).$

Suppose that for every initial distribution $\mu$ and every $F \in \mathcal{T},P_{\mu}(F) \in \{0,1\}.$ Consider Theorem $5.7.4$ from Durrett's book (picture). How can we deduce using this theorem that in this case every bounded space-time harmonic function is constant?

enter image description here

john
  • 165

2 Answers2

1

If $A\in\mathcal T$ then $C_\mu=\Bbb P_\mu[A]$ is either $0$ or $1$. In fact, either $C_\mu=0$ for all $\mu$ or $C_\mu=1$ for all $\mu$. For suppose that $C_\mu=0$ but $C_\nu=1$ for two initial distributions $\mu$ and $\nu$. Then $C_{(\mu+\nu)/2}=(C_\mu+C_\nu)/2=1/2$, a contradiction.

As a corollary to this observation, the $\Bbb P_\mu$-a.s. constant value of a $\mathcal T$-measurable random variable $Z$ doesn't depend on $\mu$.

If $h$ is a bounded space-time harmonic function then $h(X_n,m+n)$ is a bounded $\Bbb P_\mu$-martingale for each $\mu$ and each non-negative integer $m$ . The limit of this martingale coincides $\Bbb P_\mu$-a.s with the $\mathcal T$-measurable random variable $Z_m:=\liminf_nh(X_n,m+n)$, for each $\mu$. Thus, writing $C_m$ for $\Bbb P_\mu[Z_m]$, we have $h(m,x)=\Bbb E_x[h(X_0,m+0)]=\Bbb E_\mu[\lim_nh(X_n,m+n)]=\Bbb E_\mu[Z_m]=C_m$. Finally, by appling the Markov property at time $1$, you see that $$ C_m=\Bbb E_\mu[Z_m]=\Bbb E_{\nu}[Z_{m+1}]=C_{m+1}, $$ where $\nu(B)=\Bbb P_\mu[X_1\in B]$. Therefore $C_m$ doesn't depend on $m$. It follows that $h$ is constant on $S\times\{0,1,2,\ldots\}$.

John Dawkins
  • 29,845
  • 1
  • 23
  • 39
0

On the $\sigma$-algebra ${\cal T}$ the probability measure $P_\mu$ takes only the values $\{0,1\}\,.$ By this answer every ${\cal T}$-measurable $Z$ is $P_\mu$-almost surely constant.

We conclude that $\mathbb E_\mu[Z|{\cal F}_n]$ is $P_\mu$-a.s. constant. From Durrett's Theorem 5.7.4 it follows now that $h(X_n,n)$ is $P_\mu$-almost surely constant.

Kurt G.
  • 17,136