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where k = n=k+2 cj cj 1 /(2 (dk ck+1 )2 ) = O (1/dk ) We can now proj vide a convenient expression for k by observing that if we select k = 2dk k we obtain that ( , ) ( , ) = k dk + j zj k k ck+1 ck+1 2 dk dk j j 1
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Combining the terms in the previous expression into (86) and grouping the terms that do not depend on k directly yields ck+1 ck+1 (ck+1 1) 2dk 2 = ck+1 1 ck+1 1 2 2 dk 2 dk j2
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On the other hand, k k ck+1 ck+1 2 dk dk = k ck+1 2 dk j2 + j j 1 + k 2 dk k 2 dk j2
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COUNTING PROBLEMS
Therefore, we conclude that ( , ) ( , ) v ( , ) v ( , ) ( , ) = v ( , ) =
j =1 m j =1
j k dk
zj m j =1
k dk 1+
j zj + O j k dk
k 2 dk k 2 dk
j k 1+ dk k 2 dk
exp O
( , ) exp O v ( , )
This analysis yields the following result Theorem 2 Suppose that maxj n,i m cj , ri , then there exists a deterministic constant (0, ) such that: m,n Pr,c (Sn = 0) As a consequence, the estimator m,n is exponentially ef cient and since O d 2 operations are required to generate a copy of m,n and the corresponding IS algorithm is an FPRAS Proof The result follows by noting that the k remain uniformly bounded as n, m and since dk 1/k there must exist a constant (0, ) such that n n,m (r, c) = exp v (r, c) v (r, c) j2
j =1
The conclusion of the theorem then follows by noting that the kth increment requires O (ck m + n) operations to be generated (the term ck m comes from the conditional Poisson sampling and n arises in the computation of k ) The previous result can be extended to degree sequences that satisfy certain growth conditions [5] It is important to note that the selection of k = 2 k dk seems crucial in order to guarantee that the resultant importance estimator is exponentially ef cient On the other hand, the excellent numerical performance reported in [8] corresponds to the selection k = 1 The selection of k = 2 k dk can be motivated from another perspective Indeed, as has been discussed in previous chapters, note that the zero-variance change of measure, which corresponds to sampling the Sk conditional on the event Sn = 0, is Markovian and can be described by the transition kernel
Kk 1 ( , + z) =
m ck
I (z1 + + zm = ck )
u ( + z, ) , u ( , )
COUNTING PROBLEMS
where Sk = , = (ck+1 , , cn ), = (ck+2 , , cn ) and u ( , ) = P , (Sn k = 0) As a consequence, given that u ( , ) v ( , ) it seems natural to mimic the zero-variance change of measure directly using v ( ) by means of the Markov transition kernel Kk 1 ( , + z) = m ck
I (z1 + + zm = ck )
v ( + z, ) , w ( , )
where w ( , ) = E , (v ( + Xk+1 , )) is the normalizing constant that makes Kk 1 a well-de ned Markov transition kernel It turns out that generating increments according to Kk 1 corresponds to a strategy based on sampling the Xk by means of a conditional Poisson distribution with j (k) = j exp 2 k j j 1 + k j /dk (where Sk = ), therefore our selection of k , thereby motivating our choice of k directly The analysis of an IS strategy based on Kk 1 is studied in [5] A similar approximation procedure is also discussed brie y in [8] They mention that numerical experiments were also performed with the selection j (k) = j exp 2 k j and report very similar empirical performance of algorithms corresponding to this selection and that based on k = 1 Another example of a successful IS strategy for counting is given in [1] in the context of counting simple graphs This problem is equivalent to counting the number of symmetric binary tables with given margins and with zeros on the main diagonal It turns out that one can also pose this counting problem as a rare event estimation problem involving a suitably de ned random walk as in our previous example In [1] a change of measure is proposed that also can be shown to be exponentially ef cient An alternative approach, based on approximating the zero-variance change of measure (as previously discussed) is studied in [6]