An application of Markov property

Question

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 is 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.

Answer 1

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