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.)

3 Responses to “Markov chains 101”

  1. 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.

  2. 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)….

  3. 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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