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

Jaspar Wiart, Christiane Lemieux and Gracia Y. Dong

Abstract

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

Vi=x(b-1)bi+1

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

x=k=1xkbkandy=k=1ykbk

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

hi=k=1ixkbk=k=1iykbk,

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

bVi-Vi-1=bxbi-xbi-1=0.

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

xbi-1-xy+hi-1(x+y-hi-1-1bi-1)
=hi-1+rxi-1bi-1-(hi-1+rxi-1)(hi-1+ryi-1)+hi-1(hi-1+rxi-1+ryi-1-1bi-1)
=hi-1+rxi-1bi-1-rxi-1ryi-1-hi-1bi-1hi-1+rxi-1bi-1-rxi-1bi-1-hi-1bi-1=0

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:

b(xy-xbi+1-ki(x+y-ki-1bi))-(ki(x+y-ki-1bi)-hi-1(x+y-hi-1-1bi-1))
=(b+1)(xy-xbi-ki(x+y-ki-1bi))-(xy-xbi-1-hi-1(x+y-hi-1-1bi-1))
=(b+1)(rxiryi-rxibi)-(rxi-1ryi-1-rxi-1bi-1)
=brxiryi-rxibi-xi2b2i-xi(rxi+ryi)bi+xib2i-1.

Multiply by bi to get

bi+1rxiryi-rxi-xi2bi-xirxi-xiryi+xibi-1,

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

(bi+1rxi-xi)ryi-rxi-xi2bi-xirxi+xibi-1(bi+1rxi-xi)xi+1bi+1-rxi-xi2bi-xirxi+xibi-1
=xibi+1(b2-(b+1)xi-1)
xibi+1(b2-(b+1)(b-1)-1)=0.

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

xibi-1-xi2bi-xiryi+(bi+1ryi-xi-1)rxixibi-1-xi2bi-xiryi+(bi+1ryi-xi-1)xibi+1
=xibi+1(b2-(b+1)xi-1))
xibi+1(b2-(b+1)(b-1)-1))=0.

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

bi+1rxiryi-rxi-xi2bi-xirxi-xiryi+xibi-1=xibi-1-xi2bi-xiryi+(bi+1ryi-xi-1)rxi
xibi-1-xi2bi-xiryi
xi(1bi-1-b-1bi-1bi)=0.

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

(A.1)bhi+1(x+y-hi+1-1bi+1)-(b+1)ki(x+y-ki-1bi)+hi-1(x+y-hi-1-1bi-1)

is greater than or equal to zero. Using the identities

hi+1=hi-1+xibi+xi+1bi+1,ki=hi-1+xibi,x=hi-1+xibi+xi+1bi+1+rxi+1,y=hi-1+xibi+xi+1bi+1+ryi+1,

write

hi+1(x+y-hi+1-1bi+1)=(hi-1+xibi+xi+1bi+1)(hi-1+xibi+xi+1bi+1+rxi+1+ryi+1-1bi+1)
=hi-12+xi2b2i+xi+12b2i+2+2hi-1xibi+2hi-1xi+1bi+1+2xixi+1b2i+1+hi-1(rxi+1+ryi+1)
+xi(rxi+1+ryi+1)bi+xi+1(rxi+1+ryi+1)bi+1-hi-1bi+1-xib2i+1-xi+1b2i+2,

and

ki(x+y-ki-1bi)=(hi-1+xibi)(hi-1+xibi+2xi+1bi+1+rxi+1+ryi+1-1bi)
=hi-12+xi2b2i+2hi-1xibi+2hi-1xi+1bi+1+2xixi+1b2i+1
+hi-1(rxi+1+ryi+1)+xi(rxi+1+ryi+1)bi-hi-1bi-xib2i,

and

hi-1(x+y-hi-1-1bi-1)=hi-1(hi-1+2xibi+2xi+1bi+1+rxi+1+ryi+1-1bi-1)
=hi-12+2hi-1xibi+2hi-1xi+1bi+1+hi-1(rxi+1+ryi+1)-hi-1bi-1.

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

bVi+1-Vi=0(hi-12+2hi-1xibi+2hi-1xi+1bi+1+hi-1(rxi+1+ryi+1))
-(xi2b2i+2xixi+1b2i+1+xi(rxi+1+ryi+1)bi)+b(xi+12b2i+2+xi+1(rxi+1+ryi+1)bi+1)
-hi-1(bbi+1-b+1bi+1bi-1)-xi(bb2i+1-b+1b2i)-bxi+1b2i+2
=xi+12b2i+1+xib2i-1-xi2b2i-2xixi+1b2i+1-xi+1b2i+1+(xi+1-xi)(rxi+1+ryi+1)bi.

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

xi+12+b2xi-bxi2-2xixi+1-xi+1+bi+1(xi+1-xi)(rxi+1+ryi+1)

is non-negative.

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

xi+12+b2xi-bxi2-2xixi+1-xi+1+bi+1(xi+1-xi)(rxi+1+ryi+1)
xi+12+b2xi-bxi2-2xixi+1-xi+1
xi+12+bxi(xi+1+1)-bxixi+1-2xixi+1-xi+1
=xi+12+bxi-2xixi+1-xi+1
xi+12+(xi+1+1)xi-2xixi+1-xi+1
xi+12-xixi+1-xi+1+xi
xi+1(xi+1-xi-1)
0.

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

xi+12+b2xi-bxi2-2xixi+1-xi+1+bi+1(xi+1-xi)(rxi+1+ryi+1)
xi+12+b2xi-bxi2-2xixi+1+xi+1-2xi
=(xi+1-xi)2+b2xi-(b+1)xi2-2xi+xi+1
(xi+1-xi)2+xi(b2-(b+1)(b-1)-2)+xi+1
=(xi+1-xi)2+xi+1-xi
=(xi+1-xi)(xi+1-xi+1)
=(xi-xi+1)(xi-xi+1-1)0.

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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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