Markov chains 101

An interesting conundrum from Arnaud Guillin: can you find an MCMC setting where there are two Markov chains $(x_n)$ and $(y_n)$ such that the pair $(x_n,y_n)$ is not a Markov chain. The standard setting is the opposite when a pair is Markov but at least one of the components is not… Data augmentation being an exception. (Arnaud has a normal example available that is unrelated with MCMC samplers.)

This entry was posted on November 19, 2009 at 2:04 pm and is filed under Statistics with tags Markov chains, MCMC, simulation. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Markov chains 101”

xi'an Says:
November 19, 2009 at 11:13 pm

Antoine Dreyer also has a simple alternative: take two iid sequences $(Y_n)$ and $(\epsilon_n)$ and define

$X_{n+1}=X_n+Y_{n-1}+\epsilon_n$

then both $(X_n)$ and $(Y_n)$ are Markov chains, but $(X_n,Y_n)$ is not a (first order) Markov chain.

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xi'an Says:
November 19, 2009 at 5:53 pm

Nice, Randal!

I was thinking of a coupling strategy on a finite set, with $X_n$ and $Y_n$ merging when entering a certain atom $\alpha$ together. Marginaly, they are both Markov while jointly, they are not: $(X_n,Y_n)$ must remember the last time it entered $\alpha$ together (or not)….

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arno Says:
November 19, 2009 at 5:24 pm

randie star m’en a donne un autre…: je te mets son mail

“Prends un noyau Q ayant densité q par rapport disons à Lebesgue. Tu construis $X_n\sim q(X_{n-1},\cdot)$ . Tu poses $X=(X_n)_n$ . Tu poses $Y_{2k}=X_{2k}$ et tu consideres

$Y_{2k+1}|X \sim q(Y_{2k-2},\cdot)q(\cdot,Y_{2k})/q2(Y_{2k-2},Y_{2k})$

c’est à dire la loi conditionnelle à la valeur precedente et suivante.

On a clairement que $X\text{ et }Y$ sont marginalement des chaines de Markov. Mais si tu ecris la loi jointe de $Z=(X,Y)$ , tu vas voir que ca ne s’écrit pas en produit de fonctions de $Z_{k-1}$ et $Z_k$ , ca suffit pour dire que ce n’est pas une chaine de Markov. “

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