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where r! = r1 ! rm !, c! = c1 ! cn ! and (r, c) = 2 d2
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Approximation (83) can be rigorously justi ed under our assumptions of bounded row and column sums [2] We record these observations in the following result See [5, 12, 20] for additional extensions Theorem 1 Assume that supm,n 1 maxj n,i m {cj , ri } < Then as d , M(r, c) = d! exp( (r, c)) (1 + o(1)) r!c!
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Now we return to showing the exponential ef ciency of the estimator m,n as d The idea is to show that there exists a constant 1 (0, ) such that m,n 1 Pr,c (Sn = 0) For 0 k n 1, given Sk = and = (ck+1 , , cn ) with j = cj +k , let us de ne
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and set v ( , ) = dk ! exp ( ( , )) m ! ! 1
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where dk = 1 + + m Now let = Sk+1 , = (ck+2 , , cn ) and note that ( , ) ( , ) Kk ( , ) v ( , ) ( , ) Kk ( , ) = = v ( , ) v ( , ) v ( , ) v ( , ) Kk ( , ) Kk ( , ) (dk ck+1 )! !ck+1 ! m exp ( ( , ) ( , )) dk ! ! ck+1
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We write = z = (z1 , , zm ) (with z1 + + zm = ck+1 ) and note that Kk ( , ) (dk ck )! !ck ! m dk ! ! ck Kk ( , )
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(84)
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1 + j k /dk
dk 1
1 dk
ck+1
ck+1 !
1 1 (ck+1 1) dk
COUNTING PROBLEMS
ck+1 1
Observe that 1
j =1
j dk
= exp
ck+1 (ck+1 1) +O 2dk
1 2 dk
and therefore, the expression in (84) is bounded by wk ( ) ck+1 ! k ck+1 (ck+1 1) ck+1 + +O ck+1 exp dk 2dk dk 1 2 dk (85)
The next proposition provides an estimate for wk ( ) Proposition 1 There exists a constant 2 (0, ) such that: ck+1 ck+1 m m j2 d ck+1 exp 2 k 1 + k j2 exp wk ( ) = k 2 2 2 ck+1 ! 2 dk j =1 dk dk j =1
Proof We give a sketch of the proof of this result Let J1 , , Jck+1 be iid random variables such that P (J = j ) = j 1 + j k dk 1
2 where = m=1 j 1 + j k /dk = dk (1 + k m=1 j2 /dk ) Let Ii,j = 1 when j j Ji = Jj and set N = i<j Ii,j Then, we have that
wk ( ) =
1 ck+1 !
ck+1 P (N = 0) =
1 ck+1 !
ck+1 exp E(N ) + O
1 2 dk
This Poisson-based approximation, which can be rigorously justi ed using the inclusion exclusion principle, is the only missing piece in the proof of this proposition Note that ck+1 E(N ) = 2
m j =1
j2 1 + j k /dk
2 dk 1 + k
m 2 2 j =1 j /dk
ck+1 2 ck+1
d2 j =1 k
k 2 dk
On the other hand, we have that ck+1 Therefore, c dk k+1 1 + k wk ( ) = 2 ck+1 ! dk k c = dk k+1 1 + 2 dk ck+1 j2
m j =1
j2
exp O k 2 dk
m j =1
ck+1 exp 2
d2 j =1 k
COUNTING PROBLEMS
Using the proposition above together with the upper bound in (85), we obtain the following estimate of the right-hand side of (84): j k zj ( , ) ( , ) ck+1 (ck+1 1) exp 1+ v( , ) v ( , ) dk 2dk j =1 m m j2 k ck+1 k 2 +O exp 2 2 2 2 dk j =1 j dk dk j =1
(86)
exp ( ( , ) ( , )) Finally, to bound the difference ( , ) ( , ), note that it equals ( , ) ( , ) = 2 k + k j zj 2 k ck+1 2 k