Does Markov chain imply that probability in moment n depends only on moment k even in cases like $P(X_n = x| X_k \in A, X_{k_1} = a_1, \dots)$?

Question

I'm trying to figure out is event in moment n dependent only on moment k even in cases where we consider multiple possible values for moment k? In formula I want to know if $P(X_n = x| X_k \in A, X_{k_1} = a_1, \dots) = P(X_n = x| X_k \in A)$ where ($X_n,n \in \mathbb{Z}_+$) is Markov chain, A is subset of X, $n > k > k_1 > \dots > 0$ natural numbers, $\{x, a_1, a_2, \dots\} \subseteq X$.

I was trying to prove it in this post but still didn't manage to do this. Is this statement even true?

Answer 1

Try and solve it in a more simply case first.

Suppose that $X$ takes values in a countable space and consider something like $$ \mathbb P(X_n = x | X_{n-2} = x_{n-2}, X_{n-3} = x_{n-3}, ...). $$ In your case the $X_{n-2}$ is replaced by the more general $X_k$. So right now we want to prove that only $X_{n-2}$ is relevant, we can do that by using the fact that $\mathbb P(A_2 | A_1) = \mathbb P(A_2 | X, A_1)\mathbb P(X | A_1) + \mathbb P(A_2 | X^c, A_1)\mathbb P(X^c | A_1)$ together with the Law of Total Probability as follows:

$$ \begin{align} \mathbb P(X_n = x | X_{n-2} = x_{n-2}, ...)=&\sum_{\alpha\in\Omega} \mathbb P(X_n = x | X_{n-1} = \alpha, X_{n-2} = x_{n-2}, ...) \mathbb P(X_{n-1}=\alpha|X_{n-2}=\alpha, ...) + \mathbb P(X_n = x | X_{n-1} \neq \alpha, X_{n-2} = x_{n-2},...)\mathbb P(X_{n-1}\neq\alpha|X_{n-2}=x_{n-2},...)\\ =& \sum_{\alpha\in\Omega} \mathbb P(X_n = x | X_{n-1} = \alpha, X_{n-2} = x_{n-2}) \mathbb P(X_{n-1}=\alpha|X_{n-2}=\alpha) + \mathbb P(X_n = x | X_{n-1} \neq \alpha, X_{n-2} = x_{n-2})\mathbb P(X_{n-1}\neq\alpha|X_{n-2}=x_{n-2})\\ =&\mathbb P(X_n=x|X_{n-2}=x_{n-2}) \end{align} $$

Do you see how you can generalise this?