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Convergence of Sequential Monte Carlo Methods. Dan Crisan, Arnaud Doucet. Problem Statement. X: signal, Y: observation process X satisfies and evolves according to the following equation, Y satisfies. Bayes ’ recursion. Prediction Updating. A Sequential Monte Carlo Methods.
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Convergence of Sequential Monte Carlo Methods Dan Crisan, Arnaud Doucet
Problem Statement • X: signal, Y: observation process • X satisfies and evolves according to the following equation, • Y satisfies
Bayes’ recursion • Prediction • Updating
A Sequential Monte Carlo Methods • Empirical measure • Transition kernel • Importance distribution • : abs. continuous with respect to • : strictly positive Radon Nykodym derivative • Then is also continuous w.r.t. and
Algorithm • Step 1:Sequential importance sampling • sample: • evaluate normalized importance weights and let
Step 2: Selection step • multiply/discard particles with high/low importance weights to obtain N particles let assoc.empirical measure • Step 3: MCMC step • sample ,where K is a Markov kernel of invariant distribution and let
Convergence Study • denote • convergence to 0 of average mean square error under quite general conditions • Then prove (almost sure) convergence of toward under more restrictive conditions
Bounds for mean square errors • Assumptions • 1.-A Importance distribution and weights • is assumed abs.continuous with respect to for all is a bounded function in argument define
There exists a constant s. t. for all there exists with s.t. • There exists s. t. and a constant s.t.
First Assumption ensures that • Importance function is chosen so that the corresponding importance weights are bounded above. • Sampling kernel and importance weights depend “ continuously” on the measure variable. • Second assumption ensures that • Selection scheme does not introduce too strong a “discrepancy”.
Lemma 1 • Let us assume that for any then after step 1, for any • Lemma 2 • Let us assume that for any then for any
Lemma 3 • Let us assume that for any then after step 2, for any • Lemma 4 • Let us assume that for any then for any
Theorem 1 • For all , there exists independent of s.t. for any