341 Empirical entrance distributions

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As we pointed out earlier, splitting can be used to estimate expectations of more general functions of the sample paths than just the probability In particular,

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we argued in Section 321 that when all the particles have the same weight, the entrance distribution at any level does not depend on the choice of importance function, is the same as for the original chain, and can be estimated without bias by the empirical entrance distribution at that level, which we shall denote by N k This empirical distribution is already available at no extra cost when running the simulation More speci cally, recall that Nk particles are simulated in stage k, and Rk of i them hit Bk at the end of that stage Let { k , i = 1, , Nk } be the states of the i Nk chains at the end of stage k, and let Ik = {i : k Bk } be the subset of those states that have successfully hit Bk by their stopping time T Note that Ik has cardinality Rk We have N = k 1 Rk i ,

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(33)

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where x represents the Dirac mass at x k Proposition 1 For any measurable set C Bk , E[ N (C)] = k (C) This implies that for any measurable function , 1 i ( k ) E[E[ (X(Tk )) | Tk T ]] = E Rk

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342 Large-sample asymptotics

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We saw that the empirical entrance distribution N provides an unbiased estimate k of k , but what about the convergence (and speed of convergence) of N to k k when N The next proposition answers this question by providing a central limit theorem, which can be proved using the technology developed in [10] Proposition 2 Let : E R be a bounded and continuous function and 0 k n Then there is a constant vk ( ), which depends on and on the splitting implementation, such that 1 i N ( k ) E[ (X(Tk )) | Tk T ] N (0, vk ( )), Rk i I

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in distribution when N , where N (0, 2 ) is a normal random variable with mean 0 and variance 2 The result also extends to unbounded functions under appropriate uniform integrability conditions We also have a central limit theorem for the probability of reaching level k before T When k = n, this gives a central limit theorem for the estimator of , the probability of the rare event

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Proposition 3 For 0 k n, there is a constant Vk that depends on the splitting implementation, such that N p0 pk 1 p0 pk N (0, Vk )

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in distribution when N By combining these two propositions, we also obtain a central limit result for the unconditional average cost, when a cost is incurred when we hit Bk : p0 pk N i ( k ) E[ (X(min(T , Tk )))] N (0, vk ( )), p0 pk Rk i I

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in distribution when N , if we assume that (x) = 0 when x Bk By taking (x) equal to the indicator that x Bk , and vk ( ) = Vk , we recover the result of the second proposition An intuitive argument to justify these central limit theorems is that although the particles have dependent trajectories to a certain extent, the amount of dependence remains bounded, in some sense, when N The idea (roughly) is that the trajectories that start from the same initial state in stage 0 have some dependence, but those that start from different initial states are essentially independent (in xed splitting they are totally independent whereas in xed effort they are almost independent when N is large) When N while everything else is xed, the number of initial states giving rise to one or more successful trajectories eventually increases approximately linearly with N , while the average number of successful trajectories per successful initial state converges to a constant So the amount of independence increases (asymptotically) linearly with N , and this explains why these central limit theorems hold In what follows, we derive expressions for the asymptotic variance Vn , for selected splitting implementations Straightforward modi cations can provide expressions for Vk , for 0 k < n One may also rightfully argue that instead of normalizing by N in the central limit theorem, we should normalize by CN where CN = n Nk , the total number of particle levels simulated, which could k=0 be seen as the total amount of computation work if we assume that simulating one particle for one level represents one unit of work This makes sense if we assume that the expected work is the same at each level If CN /N C in probability as N , which is typically the case (in particular, CN /N = C = n + 1 exactly in the xed-effort implementations), then using Slutsty s lemma yields CN (p0 pn / 1) N (0, CVn ) in distribution as N So normalizing by CN instead of N only changes the variance by the constant factor C

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