General Markov property

Question

Let $X_0,X_1,\ldots$ be a Markov chain with state space $\mathbb{Z}$. Now I found in a textbook that we have $$ \mathbb{P}(X_{n+1} \in A \mid X_n = i, (X_{0},X_1,\ldots,X_{n-1}) \in B) = \mathbb{P}(X_{n+1} \in A \mid X_n = i) $$ for all $n \in \mathbb{N}$, $i \in \mathbb{Z}$, $A \subset \mathbb{Z}$, $B \subset \mathbb{Z}^n$. This was given as a Proposition without a proof? How can we prove it and what is the difference to the original Markov property for Markov chain?

Answer 1

A similar proposition is proved in chapter 1 of “Introduction to Stochastic Processes” by Hoel (page 12), using the solution of exercise 4 of the very same chapter. I will solve the exercise and show how it can be used to prove your proposition.

Let's consider a probability space $(\Omega,F,\mathbb{P})$, and assume the sets mentioned belong to $F$. We want to prove that if $D_k$ are disjoint and $\mathbb{P}(C \mid D_k) = p$ independently of k, then $\mathbb{P}(C \mid \cup_{k}D_k) = p$.

Note that: $$\mathbb{P}(C \mid \cup_{k}D_k) = \frac{\mathbb{P}(C \cap (\cup_{k}D_k))}{\mathbb{P}(\cup_{k}D_k)}= \frac{\sum_{k}{\mathbb{P}(C \cap D_k)}}{\sum_{k}{\mathbb{P}(D_k)}}= \frac{\sum_{k}{\mathbb{P}(C \mid D_k)\mathbb{P}(D_k)}}{\sum_{k}{\mathbb{P}(D_k)}} = \frac{p\sum_{k}{\mathbb{P}(D_k)}}{\sum_{k}{\mathbb{P}(D_k)}} =p $$

Now that we proved that, let's go back to your initial proposition.

Let $B_k = (X_o=i^{(k)}_0,...,X_{n-1} =i^{(k)}_{n-1})$, such that $B = \cup_{k}{B_k}$, and $B_k$ are disjoint. Since we have a Markov chain:

$$ \mathbb{P}(X_{n+1} \in A \mid \{{ X_n =i\}} \cap B_k)=\mathbb{P}(X_{n+1} \in A \mid X_o=i^{(k)}_0,...,X_n=i) =\mathbb{P}(X_{n+1} \in A \mid X_n = i) = p~,$$ independently of $k$.

So what we have to show is that:
\begin{align} \mathbb{P}(X_{n+1} \in A \mid \{{ X_n =i\}} \cap (\cup_{k}{B_k})) &= \mathbb{P}(X_{n+1} \in A \mid X_n=i,(X_o,...,X_{n-1}) \in B) \\ &= \mathbb{P}(X_{n+1} \in A \mid X_n=i) \end{align}

Therefore, the proof to your proposition is achieved by making $C = \{{X_{n+1}\in A\}}$ and $D_k = \{{\{{ X_n =i\}} \cap {B_k}\}}$, and applying the solution previously demonstrated.