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Let $S$ be a countable set and $(X_n)$ an $S$-valued discrete Markov chain. Then $$ \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_0=x_0] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n] $$ for all $x_0, \ldots, x_{n+1} \in S$. This is called the Markov property. Previously, I proved that

Theorem 1 $$ \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_m=x_m] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n] $$ for all $m \le n$ and $x_0, \ldots, x_{n+1} \in S$.

Could you provide some hints on how to prove below result?

Theorem 2 $$ \mathbb P [X_{n} = x_{n} |X_m = x_m, \ldots, X_0=x_0] = \mathbb P [X_{n} = x_{n} |X_m = x_m] $$ for all $m \le n$ and $x_0, \ldots, x_{n} \in S$.

Akira
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1 Answers1

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Taking for example $m = n-2$, we have \begin{align} &P(X_n = x_n, X_{n-2}=x_{n-2}, \ldots, X_0=x_0) \\ &= \sum_{x_{n-1}} P(X_n = x_n, X_{n-1}=x_{n-1}, X_{n-2}=x_{n-2}, \ldots, X_0=x_0) \\ &= \sum_{x_{n-1}} P(X_n = x_n \mid X_{n-1} = x_{n-1}, \ldots, X_0 = x_0) P(X_{n-1} = x_{n-1} \mid X_{n-2} = x_{n-2},\ldots, X_0=x_0) P(X_{n-2}=x_{n-2},\ldots, X_0=x_0) \\ &= P(X_{n-2}=x_{n-2},\ldots, X_0=x_0) \sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2}), \end{align} where the last step is by your definition of Markov chain. Dividing both sides by $P(X_{n-2}=x_{n-2}, \ldots, X_0=x_0)$ yields \begin{align} &P(X_n = x_n \mid X_{n-2}=x_{n-2}, \ldots, X_0=x_0) \\ &= \sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2}) \\ &= \sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1}, X_{n-2} = x_{n-2})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2}) \\ &= \sum_{x_{n-1}} P(X_n=x_n, X_{n-1}=x_{n-1} \mid X_{n-2}=x_{n-2}) \\ &= P(X_n=x_n \mid X_{n-2} = x_{n-2}) \end{align} where the second equality is due to Theorem 1.

The proof for general $m \le n$ should be similar.

angryavian
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  • It seems you want to divide both sides by $P(X_n = x_n, X_{n-2}=x_{n-2}, \ldots, X_0=x_0)$ rather than by $P(X_n = x_n \mid X_{n-2}=x_{n-2}, \ldots, X_0=x_0)$. I'm surprised that this theorem generalizes both Theorem 1 and 2, but its proof is much simpler. – Akira Jan 04 '23 at 09:50
  • @Akira Thanks for pointing out the typo. I think my proof here is longer because I am only using properties that have been explicitly mentioned in your question. – angryavian Jan 04 '23 at 17:24