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Licensed Unlicensed Requires Authentication Published by De Gruyter January 10, 2021

On the dependence structure and quality of scrambled (t, m, s)-nets

Jaspar Wiart, Christiane Lemieux and Gracia Y. Dong


In this paper we develop a framework to study the dependence structure of scrambled (t,m,s)-nets. It relies on values denoted by Cb(𝒌;Pn), which are related to how many distinct pairs of points from Pn lie in the same elementary 𝒌-interval in base b. These values quantify the equidistribution properties of Pn in a more informative way than the parameter t. They also play a key role in determining if a scrambled set P~n is negative lower orthant dependent (NLOD). Indeed, this property holds if and only if Cb(𝒌;Pn)1 for all 𝒌s, which in turn implies that a scrambled digital (t,m,s)-net in base b is NLOD if and only if t=0. Through numerical examples we demonstrate that these Cb(𝒌;Pn) values are a powerful tool to compare the quality of different (t,m,s)-nets, and to enhance our understanding of how scrambling can improve the quality of deterministic point sets.

Funding source: Natural Sciences and Engineering Research Council of Canada

Award Identifier / Grant number: 238959

Funding source: Austrian Science Fund

Award Identifier / Grant number: F5506-N26

Award Identifier / Grant number: F5509-N26

Funding statement: The authors wish to acknowledge the support of the Natural Science and Engineering Research Council (NSERC) of Canada for its financial support via grant #238959. The first author is also partially supported by the Austrian Science Fund (FWF): Projects F5506-N26 and F5509-N26, which are parts of the Special Research Program “Quasi-Monte Carlo Methods: Theory and Applications”.

A Appendix

Proof of Lemma 2.5.

When either x or y is 1, from Lemma 2.4 we know that


and thus bVi-Vi-1=0 in this case. So for the remainder of the proof, we assume x,y[0,1). Let


be the base-b digital expansion of x and y chosen so that only finitely many digits are non-zero. Recall that ki=bimin(x,y)b-i for i0. When γb(x,y)1, then for i{1,,γb(x,y)} we have


and k0=0. We also define rxi=x-hi, and ryi=y-hi for i0. Without loss of generality we assume xy. There are four cases.

Case 1: γb(x,y)<i-1. In this case


Case 2: γb(x,y)=i-1. In this case, bVi-Vi-1 becomes


because ryi-11bi-1.

Case 3: γb(x,y)=i. We use the calculation in Case 2 and the identities rxi-1=xibi+rxi and ryi-1=xibi+ryi to simplify bVi-Vi-1:


Multiply by bi to get


which will be shown to be non-negative. Note that by assumption x<y and since their base b expansions differ for the first time at the (i+1)st digit, we always have xi+1<yi+1.

Case 3a: xixi+1<yi+1. The assumption implies 0bi+1rxi-xi and xi+1bi+1ryi. We estimate


Case 3b: (xi+1<xi<yi+1). The assumption implies rxixibi+1 and (bi+1ryi-xi-1)0. We estimate


Case 3c: (xi+1<yi+1xi). The assumption implies 0(bi+1ryi-xi-1). We estimate


Case 4: γb(x,y)>i. In this case we need to show that


is greater than or equal to zero. Using the identities








Now substituting into (A.1) and simplifying, we get


By multiplying the above by b2i+1 we see that to finish the proof we need to show that


is non-negative.

Case 4a: xi<xi+1. We have


Case 4b: xi+1xi. Since rxi+1,ryi+11bi+1, we have



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Received: 2020-07-28
Revised: 2020-11-08
Accepted: 2020-12-18
Published Online: 2021-01-10
Published in Print: 2021-03-01

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