# Monte Carlo Methods: [closed]

Can someone explain to me the following statement from “Introducing Monte Carlo methods with R!” By Robert Christian.

“If the exploration mechanism has enough energy to reach as far as the boundaries of the support of the target f, the method will eventually uncover the complexity of the target.”

• His name is Christian Robert, not Robert Christian, and he's on here with username @Xi'an. – Thomas Lumley Mar 4 at 6:03

The study of independent Metropolis-Hastings algorithms is certainly interesting, but their practical implementation is more problematic in that they are delicate to use in complex settings because the construction of the proposal is complicated—if we are using simulation, it is often because deriving estimates like MLEs is difficult—and because the choice of the proposal is highly influential on the performance of the algorithm. Rather than building a proposal from scratch or suggesting a non-parametric approximation based on a preliminary run—because it is unlikely to work for moderate to high dimensions—it is therefore more realistic to gather information about the target stepwise, that is, by exploring the neighborhood of the current value of the chain. If the exploration mechanism has enough energy to reach as far as the boundaries of the support of the target $$f$$, the method will eventually uncover the complexity of the target. (This is fundamentally the same intuition at work in the simulated annealing algorithm of Section 2.3.3 and the stochastic gradient method of Section 2.3.2.)
The ability to reach out to every part of the support of the target $$\pi(\cdot)$$ crucially depends on the proposal distribution $$q(\cdot\,;\cdot)$$ used in the Metropolis-Hastings algorithm: if the possible moves allowed by $$q(\cdot\,;\cdot)$$ are too limited with regard to the support of $$\pi(\cdot)$$, the Markov chain will remain stuck in a subset of the support of $$\pi(\cdot)$$ and hence produce a simulation of $$\pi(\cdot)$$ restricted to this subset. On the opposite, if the possible moves allowed by $$q(\cdot\,;\cdot)$$ are wide-ranging, again with regard to the support of $$\pi(\cdot)$$, the Markov chain will have a positive probability to reach everywhere in the support of $$\pi(\cdot)$$ (if not necessarily in one step).