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Suppose $X_i$ is a Markov chain, which is defined by

$$p(X_1,X_2,\ldots,X_n) = p(X_1)p(X_1\mid X_2)p(X_2\mid X_3)\ldots p(X_{n-1}\mid X_n)$$

For arbitrary subchain $X_{\alpha_k}$, where $1\le \alpha_1 < \alpha_2 < \cdots < \alpha_m \le n$, is it a Markov chain?

Michael Hardy
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1 Answers1

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Yes. The Markov property can be stated as $X_3$ being conditionally independent of $X_1$ given $X_2$, or $p(X_1,X_3|X_2)=p(X_1|X_2)p(X_3|X_2)$. More generally, you can apply this property multiple times to get, for some $m$,

$$ p(X_1,X_2,\dots,X_{m-1},X_{m+1},\dots,X_n|X_m)=p(X_1,X_2,\dots,X_{m-1}|X_m)p(X_{m+1},\dots,X_n|X_m). $$

Now take $1\le \alpha <\beta<\gamma\le n$. Using above property you can write

$$ p(X_\alpha,X_\gamma|X_\beta) = \sum_{X_i:i\in\{\alpha+1,\dots,\beta-1,\beta+1,\dots\gamma-1\}} p(X_\alpha,X_{\alpha+1},\dots,X_{\beta-1},X_{\beta+1},\dots,X_\gamma|X_\beta)\\ = \sum_{X_i:i\in\{\alpha+1,\dots,\beta-1,\beta+1,\dots\gamma-1\}} p(X_\alpha,X_{\alpha+1},\dots,X_{\beta-1}|X_\beta)p(X_{\beta+1},\dots,X_\gamma|X_\beta)\\ = \sum_{X_i:i\in\{\alpha+1,\dots,\beta-1\}}\sum_{X_j:j\in\{\beta+1,\dots\gamma-1\}} p(X_\alpha,X_{\alpha+1},\dots,X_{\beta-1}|X_\beta)p(X_{\beta+1},\dots,X_\gamma|X_\beta)\\ = \left(\sum_{X_i:i\in\{\alpha+1,\dots,\beta-1\}} p(X_\alpha,X_{\alpha+1},\dots,X_{\beta-1}|X_\beta)\right)\left(\sum_{X_j:j\in\{\beta+1,\dots\gamma-1\}}p(X_{\beta+1},\dots,X_\gamma|X_\beta)\right)\\ =p(X_\alpha|X_\beta)p(X_\gamma|X_\beta) $$

Bernhard
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