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In the book, Markov Chains, the following theorem is mentioned:

Let $(X_n),n≥0$ be Markov$(λ,P)$. Then, conditional on $X_m=i,(X_{m+n})_{n≥0}$ is Markov$(\delta_i,P)$ and is independent of the random variables $X_0,\dots,X_m$.

This is proved by showing that for any event A determined by $X_0,\dots ,X_m$ we have: $P({X_m=i_m,...,X_{m+n}=i_{m+n}}\cap A|X_m=i)=\delta_{ii_m}*p_{i_m,i_{m+1}}...p_{i_{m+n−1},i_{m+n}}*P(A|X_m=i)$

I am wondering how this definition is related to the standard Markov property statement that $P(X_{n+1} | X_n, ..., X_0)=P(X_{n+1} | X_n)$ ?

Philip
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1 Answers1

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This is based on the following equivalence:

Consider three random variables or families of random variables $U$, $V$, and $W$, then $P(U\mid V,W)=P(U\mid V)$ if and only if $(U,W)$ is independent conditionally on $V$.

To see why in the discrete case (the general case being similar), note that the first condition reads $$P(U=u\mid V=v,W=w)=P(U=u\mid V=v),$$ for every $(u,v,w)$, that is, $$P(U=u,V=v,W=w)P(V=v)=P(U=u,V=v)P(V=v,W=w),$$ or, still equivalently, $$P(U=u,W=w\mid V=v)=P(U=u\mid V=v)P(W=w\mid V=v),$$ which is the second condition.

In the context of the question, use $U=(X_n)_{n\geqslant m}$, $V=X_m$ and $W=(X_n)_{n\leqslant m}$.

Did
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