## 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. xi'an Says:

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. xi'an Says:

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. arno Says:

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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