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When running a Gibbs sampler (for $n=200$ Iterations) with two full conditionals, I get the output $\mathbf{x} = (x_1^{(n)},x_2^{(n)})_{n =1,...,200}$.

So $\mathbf{x}$ is the realizations of a Gibbs Markov chain, the so called Gibbs sequence. but are $(x_1^{(n)})_{n \in [1,...,200]}, (x_2^{(n)})_{n \in [1,...,100]}$ both realizations of a Markov chain too ?

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This two-block Gibbs sampler is the only generic case when sub-chains remain Markov chains per se, because $(X_1^{(n)})$ is generated via the kernel $$K(x_1,x_1^\prime)=\int f_2(x_2|x_1)f_1(x_1^\prime|x_2)\,\text{d}x_2$$ See our MCMC book for more details, but this is a case of interleaving property that also guarantees that the $X_1^{(n)}$'s are positively correlated with a correlation decreasing with the time difference and that Rao-Blackwellisation always reduce the variance of the resulting estimate.

Two Markov chains $(X^{(t)})$ and $(Y^{(t)})$ are said to be conjugate to each other with the interleaving property (or interleaved) if

  1. $X^{(t)}$ and $X^{(t+1)}$ are independent conditionally on $Y^{(t)}$;
  2. $Y^{(t-1)}$ and $Y^{(t)}$ are independent conditionally on $X^{(t)}$; and
  3. $(X^{(t)},Y^{(t-1)})$ and $(X^{(t)},Y^{(t)})$ are identically distributed under stationarity.
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  • $\begingroup$ so i build a hybrid sampler with 3 draws of the full conditionals and 1 Metropolis step. how can i check if the sub-chains are markov chains ? i used geweke-score for one "subchain" and then estimate the a-posteriori mean of the paramter. hope that this wasnt for free :((( $\endgroup$ – user2016445 Mar 27 '15 at 17:39
  • $\begingroup$ if you add a Metropolis, the Markov property does not necessarily hold for the subchains, as, for instance, using a joint move proposal on $(x_1,x_2)$. $\endgroup$ – Xi'an Mar 27 '15 at 17:45

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