# An application of Markov property

I'm a new learner on Markov chain, and I was confused about a small point when I use the Markov property to solve exercise 6.3.1 in Durrett's PTE(2010):

Let $$A\in\sigma(X_0, \cdots, X_n)$$ and $$B\in\sigma(X_n, X_{n+1}, \cdots)$$. Use the Markov property to show that for any initial distribution $$\mu$$ $$P_\mu(A\cap B|X_n)=P_\mu(A|X_n)P_\mu(B|X_n).$$

I have tried an idea:

LHS=$$E_\mu(1_A1_B|X_n)$$=$$E_\mu(E_\mu(1_A1_B|\mathcal{F}_n)|X_n)$$=$$E_\mu(1_AE_\mu(1_B|\mathcal{F}_n)|X_n)$$=$$E_\mu(1_AE_\mu(1_B|X_n)|X_n)$$=$$E_\mu(1_B|X_n)E_\mu(1_A|X_n)$$=RHS.

But I don't know how to prove the equality $$E_\mu(1_B|\mathcal{F}_n)=E_\mu(1_B|X_n)$$ using Markov property though it coincide well with the intuition.

Here the Markov property refers to Theorem 6.3.1 in Durrett's book:

Let $$Y:\Omega_0\rightarrow\mathbb{R}$$ be bounded and measurable. $$E_\mu(Y\circ\theta_m|\mathcal{F}_m)=E_{X_m}Y.$$

I just can't figure out how to choose the $$Y$$ in Theorem 6.3.1. Any help will be appreciated.

## 1 Answer

Taking the conditional expectation (with respect to $$X_m$$) at both sides of the Markov property

$$E_{\mu}(Y \circ \theta_m \mid \mathcal{F}_m) = E_{X_m}(Y) \tag{1}$$

we get

$$E_{\mu}(Y \circ \theta_m \mid X_m) = E_{X_m}(Y). \tag{2}$$

Combining $$(1)$$ and $$(2)$$ shows

$$E_{\mu}(Y \circ \theta_m \mid \mathcal{F}_m) = E_{\mu}(Y \circ \theta_m \mid X_m) \tag{3}$$

for any bounded measurable random variable $$Y$$. In particular, this identity holds for

$$Y = \prod_{j=1}^k 1_{C_j}(X_j)$$

where $$C_j$$ are measurable sets. Then

$$Y \circ \theta_m = \prod_{j=1}^k 1_{C_j}(X_{j+m}),$$

and so $$(3)$$ shows

$$E_{\mu} \left( \prod_{j=1}^k 1_{C_j}(X_{j+m}) \mid \mathcal{F}_m \right) = E_{\mu} \left( \prod_{j=1}^k 1_{C_j}(X_{j+m}) \mid X_m \right). \tag{4}$$

Since sets of the form $$\bigcap_{j=1}^k \{X_{j+m} \in C_j\}$$ are a $$\cap$$-stable generator of $$\sigma(X_m,X_{m+1},\ldots)$$, this implies

$$E_{\mu}(1_B \mid \mathcal{F}_m) = E_{\mu}(1_B \mid X_m)$$

for any $$B \in \sigma(X_m,X_{m+1},\ldots)$$.

• Very clear answer. Thanks a lot! By the way, I want to get much more knowledge about Markov chains defined in this abstract way, could you please recommend some books or notes which I can get access to through the Internet? – Feng Shao Mar 31 at 6:51
• @FengShao I don't know wich books you have access to. There are plenty of threads which give recommendations for literature on Markov chains, see e.g. this question, this question, this question or this question – saz Mar 31 at 7:23