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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. I would like to prove that

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

Could you confirm if my below understanding is correct?

Proof We have $$ \begin{align} & \mathbb P [X_{n+1} = x_{n+1} | X_n = x_n, \ldots, X_m=x_m] \\ = & \sum_{x_{m-1}, \ldots, x_0 \in S} \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_0=x_0] \mathbb P[X_{m-1} = x_{m-1}, \ldots, X_0=x_0 | X_n = x_n, \ldots, X_m=x_m] \\ = & \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n] \sum_{x_{m-1}, \ldots, x_0 \in S} \mathbb P[X_{m-1} = x_{m-1}, \ldots, X_0=x_0 | X_n = x_n, \ldots, X_m=x_m] \\ =& \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n]. \end{align} $$

Above,

  • the first equality follows from the law of total conditional probability,
  • the second one follows from Markov property, and
  • the third one follows from the fact that $\sum_n \mathbb P[A_n|B] = 1$ if $(A_n)$ is a collection of pairwise disjoints events whose union is the whole sample space.
Akira
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1 Answers1

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A typo in the theorem: $X_n = x_1$ should be $X_n = x_n$. Your proof is correct.

Another proof by double expectation formula: $$ \begin{align} P[X_{n+1} = x_{n+1} | X_n = x_n, ..., X_m = x_m] &= E\left[P\left[X_{n+1} = x_{n+1} | X_n = x_n, ..., X_m = x_m\right] | X_{m-1}, ..., X_0\right] \\ &= EP[X_{n+1} = x_{n+1} | X_n = x_n] \\ &= P[X_{n+1} = x_{n+1} | X_n = x_n] \end{align} $$

Chia
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