Show that $\mathbb{P}((X_{n+1},...,X_N)\in F|X_n\in A, (X_{n-1},...,X_0)\in G)=\mathbb{P}((X_{n+1,}...,X_N)\in F|X_n\in A)$

Question

In our reading we had the following Theorem concerning Markov chains:

Take $0<n<N$ and $(X_n)_{n\in\mathbb{N}}$ a Markov chain. Then for all $a_n\in E$ (where $E$ is the state space) and all subsets $G\subseteq E^n, F\subseteq E^{N-n}$ we have that $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a). $$

Now the following question:

Does the statement of the theorem remain true if $\left\{X_n=a\right\}$ is replaced by $\left\{X_n\in A\right\}$ for some arbitrary $A\subseteq E$?

I tried to prove that in the following way using the theorem and the $\sigma$-additicity of $\mathbb{P}$ and $\left\{X_n\in A\right\}=\biguplus_{a_n\in A}(X_n=a_n)$:

$$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n\in A,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|\biguplus_{a_n\in A}(X_n=a_n)\cap(X_{n-1},...,X_0)\in G)\\ =\mathbb{P}((X_{n+1},...,X_N)\in F|\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])\\=\frac{\mathbb{P}[(X_{n+1},...,X_N)\in F]\cap\biguplus_{a_n\in A}[(X_n=a)\cap (X_{n-1},...,X_0)\in G])}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}\\=\frac{\sum_{a_n\in A}\mathbb{P}([(X_{n+1},...,X_N)\in F]\cap [ (X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])}\\=\frac{\sum_{a_n\in A}\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\cdot\mathbb{P}(X_n=a_n,(X_{n-1},...,X_0)\in G)}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])} $$ Now using the theorem I can write the summands as $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\cdot\mathbb{P}(X_n=a,(X_{n-1},...,X_0)\in G) =\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n)\mathbb{P}(X_n=a,(X_{n-1},...,X_0)\in G), $$

so I now have $$ =\frac{\sum_{a_n\in A}\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n)\cdot \mathbb{P}(X_n=a_n, (X_{n-1},...,X_0)\in G)}{\mathbb{P}(\biguplus_{a_n\in A}[(X_n=a_n)\cap (X_{n-1},...,X_0)\in G])} $$

Do not know how to continue...

Maybe I am not right up to here or maybe the statement does not remain true..

Answer 1

Take $0<n<N$ and $(X_n)_{n\in\mathbb{N}}$ a Markov chain. Then for all $a_n\in E$ (where $E$ is the state space) and all subsets $G\subseteq E^n, F\subseteq E^{N-n}$ we have that $$ \mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a_n,(X_{n-1},...,X_0)\in G)\\=\mathbb{P}((X_{n+1},...,X_N)\in F|X_n=a). $$ Does the statement of the theorem remain true if $\{X_n = a\}$ is replaced by $\{X_n \in A\}$ for $A \subset E$?

This is, perhaps surprisingly, not true, even in trivial cases.

Consider the Markov chain on $\{0,1\}$ with transition kernel given by $$P(X_{n+1} = 1 \;|\ X_n = 0) = 1, \quad P(X_{n+1} = 0 \;|\ X_n = 1) = 1.$$ Let $P(X_0 = 0) = P(X_0 = 1) = 1/2$. Let $A = \{0, 1\}$. Then \begin{gather*} P(X_2 \in \{0\} \;|\ X_1 \in A, X_0 \in \{0\}) = 1, \end{gather*} since $X_0 = 0$ implies $X_1 = 1$ and $X_2 = 0$. On the other hand, \begin{gather*} P(X_2 \in \{0\} \;|\ X_1 \in A) < 1. \end{gather*}