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To simulate from the posterior distribution $p(\theta|Y)$ where $\theta = (\mu,\lambda_1,\lambda_2)$, I run a Gibbs sampler to draw approximately random values from $p(\theta|Y)$. This Gibbs sampler returns as output $$\lbrace \mu^{(n)}, \lambda_1^{(n)}, \lambda_2^{(n)} \rbrace_{n=1}^{N}$$ (after burn-in).

If interested in the parameter $\lambda_1$, to estimate this parameter $\lambda_1$, I use the statistic : $$\frac{1}{N}\sum_{i=1}^{N} \lambda_1^{(i)}$$ This is the naive Monte Carlo estimator that approximates the expectation of $\lambda_1$ by the strong law of large numbers (the sample has to be iid to make this work).

My question: does the Gibbs sampler produce an iid sample from the posterior $p(\theta|Y)$ ?

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  • $\begingroup$ Gibbs sampling (and MCMC more generally) produces dependent values, so no, they're not iid. $\endgroup$
    – Glen_b
    Commented Apr 1, 2015 at 23:37

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The validations for both the convergence of the Gibbs Markov chain and of the empirical average based on this Markov chain are (both) called ergodic theorems.

For a Harris positive chain $(X_n)$, with invariant distribution $\pi$, an atom $\alpha$ is ergodic if $$ \lim_{n\rightarrow\infty}\; |K^n(\alpha,\alpha)-\pi(\alpha)|=0 \;. $$ In the countable case, the existence of an ergodic atom is, in fact, sufficient to establish convergence according to the total variation norm, $$ \|\mu_1-\mu_2\|_{TV} = \sup_A \; |\mu_1(A)-\mu_2(A)| . $$ Proposition 6.48: If $(X_n)$ is Harris positive on $\mathcal{X}$ and denumerable, and if there exists an ergodic atom $\alpha\subset \mathcal{X}$, then, for every $x \in \mathcal{X} $, $$ \lim_{n \rightarrow \infty} \; \|K^n(x,\cdot)-\pi\|_{TV} = 0 \;. $$ (extract from Monte Carlo Statistical Methods (2004), p.231)

and

Proposition 6.63: If $(X_n)$ has a $\sigma$-finite invariant measure $\pi$, the following two statements are equivalent:

  1. If $f,g\in L^1(\pi)$ with $\int g(x) d\pi(x)\neq 0$, then$$\lim_{N \rightarrow \infty} \; \frac{\sum_{i=1}^N f(X_i)}{\sum_{i=1}^N g(X_i)} ={\int f(x) \text{d}\pi(x) \over \int g(x) \text{d}\pi(x)} \;.$$
  2. The Markov chain $(X_n)$ is Harris recurrent.

(extract from Monte Carlo Statistical Methods (2004), p.241)

None of those results requires i.i.d. sampling.

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  • $\begingroup$ so that this sample mean of the single parameter $\lambda_1$ converges to the expectation value (a posterori mean - bayes estimator in this case) follows directly from the ergodic theorem for the gibbs markovchain ? i am a little bit confused because in your book you write the ergodic theorem for the whole chain $\mathbf{X} = (X_1,X_2,X_3)$ but i am only interested in a single one. But follows this fact too ? $\endgroup$
    – user2016445
    Commented Mar 28, 2015 at 20:07
  • $\begingroup$ If you pick $f(\lbrace \mu^{(n)}, \lambda_1^{(n)}, \lambda_2^{(n)} \rbrace)=\lambda_1^{(n)}$, this is a particular function of the chain. $\endgroup$
    – Xi'an
    Commented Mar 28, 2015 at 20:13
  • $\begingroup$ not at all: the subchains like $\{\lambda_1^{(n)}\}_n$ are more complex objects than the chain itself and most often not Markov chains per se, but only hidden Markov chains. $\endgroup$
    – Xi'an
    Commented Mar 28, 2015 at 20:21
  • $\begingroup$ what means hidden markov chain in this context ? because in my thesis (that i ve already handed in :( ) i wrote that the subchains are markov chains from this 3-stage gibbs-sampler (that in general holds for the 2-stage case) ! so is there a opportunity to save a well deserved grade :) ? $\endgroup$
    – user2016445
    Commented Mar 28, 2015 at 20:26
  • $\begingroup$ In general, the sub-chains of a three-stage Gibbs sampler are not Markov chains themselves. This is why we distinguish the two-stage case in our book, as the only one that produces sub-Markov-chains that are again Markov. What I mean by hidden Markov is tautological: $\{\lambda_1^{(n)}\}_n$ is a part of a Markov chain... $\endgroup$
    – Xi'an
    Commented Mar 28, 2015 at 21:25

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