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Let $\{X_n \}$ be a discrete time discrete space Markov chain. We know that $$ P(X_{n+1}=j|X_0=x_0,\ldots,X_{n-1}=x_{n-1},X_n=i)=P(X_{n+1}=j|X_n=i), $$ for all $x_0,\ldots,x_{n-1},i,j\in \cal{S}$, and $n\geq0$, where $\cal{S}$ is the state space of $\{X_n \}$.

My question is: can we get $$ P(X_{n+1}=j|X_0\neq x_0,\ldots,X_{n-1}\neq x_{n-1},X_n=i)=P(X_{n+1}=j|X_n=i) $$ or $$ P(X_{n+1}=j|X_0\neq x_0,\ldots,X_k\neq x_k,X_{k+1}=x_{k+1},\ldots X_{n-1}= x_{n-1},X_n=i)=P(X_{n+1}=j|X_n=i)? $$

Xiangdong Meng
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1 Answers1

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Yes. The Markov property states that for any $k\geq 1$, $V\subseteq S^k$ and $x_k,x_{k+1},...,x_n\in S$ we have:

$\mathbb{P}(X_{k+1}=x_{k+1},...,X_n=x_n|X_k=x_k,(X_0,...,X_{k-1})\in V)=\mathbb{P}(X_0=x_k,X_1=x_{k+1},...,X_{n-k}=x_n)$

In particular it follows that we will not change the left hand side if we just change the set $V$. And if we take $V=S^k$ the left hand side will be exactly $\mathbb{P}(X_{k+1}=x_{k+1},...,X_n=x_n|X_k=x_k)$.

Mark
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