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On the asymptotic normality of persistent Betti numbers Johannes T. N. Krebs †‡ Wolfgang Polonik § November 8, 2018 Abstract Persistent Betti numbers are a major tool in persistent homology, a subfield of topological data analysis. Many tools in persistent homology rely on the properties of persistent Betti numbers considered as a two-dimensional stochastic process (r, s) 7! n -1/2 (β r,s q (K(Sn)) - E β r,s q (K(Sn)) ). So far, pointwise limit theorems have been established in different set-ups. In particular, the pointwise asymptotic normality of (persistent) Betti numbers has been established for stationary Poisson processes and binomial processes with constant intensity function, see Yogeshwaran et al. [2017] and Hiraoka et al. [2018]. In this contribution, we generalize the existing results to broader class of underlying Poisson and binomial processes with smooth intensity functions and prove the asymptotic normality of the finite-dimensional distributions of the above mentioned empirical process. Keywords: Limit theorems; Persistent Betti numbers; Point processes; Topological data analysis; Random geometric complexes; Weak convergence. MSC 2010: Primary: 60G55; 60D05; Secondary: 60F05. In this manuscript we address an important question in topological data analysis (TDA), namely, the study of the weak convergence of the empirical process of persistent Betti numbers (r, s) 7! n -1/2 β r,s q (K(n 1/d S n )) - E h β r,s q (K(n 1/d S n )) i⌘ , where S n is either a binomial or a Poisson process with intensity function n on the unit cube [0, 1] d and q =0,...,d-1. is a smooth intensity function which is not necessarily constant. So far, there exist results on the pointwise asymptotic normality for Betti numbers in the case of a homogeneous Poisson process or a binomial process, see Yogeshwaran et al. [2017]. In the case of a homogenous Poisson process this result was extended to persistent Betti numbers by Hiraoka et al. [2018]. We will use this latter result to derive the pointwise asymptotic normality for the general case. The recent contributions Owada et al. [2018] and Owada and Thomas [2018] also study the limiting behavior of Betti numbers. TDA is a comparably young field that has emerged from several contributions in algebraic topology and compu- tational geometry. Milestone contributions which helped to popularize TDA in its early days are Edelsbrunner et al. [2000], Zomorodian and Carlsson [2005] and Carlsson [2009]. TDA consists of various techniques which aim at under- standing the topology of a d-dimensional manifold based on an approximating point cloud. In practice, we can think of a probability distribution, whose topological properties are of interest, and a sample from this distribution. The various This research was partially supported by the German Research Foundation (DFG), Grant Number KR-4977/1. Department of Statistics, University of California, Davis, CA, 95616, USA, email: [email protected] Corresponding author § Department of Statistics, University of California, Davis, CA, 95616, USA, email: [email protected] 2
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Page 1: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

On the asymptotic normality of persistent Betti numbers⇤

Johannes T. N. Krebs† ‡ Wolfgang Polonik§

November 8, 2018

Abstract

Persistent Betti numbers are a major tool in persistent homology, a subfield of topological data analysis. Many tools

in persistent homology rely on the properties of persistent Betti numbers considered as a two-dimensional stochastic

process (r, s) 7! n�1/2(�r,sq (K(Sn)) � E

⇥�r,sq (K(Sn))

⇤). So far, pointwise limit theorems have been established in

different set-ups. In particular, the pointwise asymptotic normality of (persistent) Betti numbers has been established

for stationary Poisson processes and binomial processes with constant intensity function, see Yogeshwaran et al. [2017]

and Hiraoka et al. [2018].

In this contribution, we generalize the existing results to broader class of underlying Poisson and binomial processes

with smooth intensity functions and prove the asymptotic normality of the finite-dimensional distributions of the above

mentioned empirical process.

Keywords: Limit theorems; Persistent Betti numbers; Point processes; Topological data analysis; Random geometric

complexes; Weak convergence.

MSC 2010: Primary: 60G55; 60D05; Secondary: 60F05.

In this manuscript we address an important question in topological data analysis (TDA), namely, the study of the weakconvergence of the empirical process of persistent Betti numbers

(r, s) 7! n�1/2⇣

�r,s

q

(K(n1/dSn

))� Eh

�r,s

q

(K(n1/dSn

))

i⌘

,

where Sn

is either a binomial or a Poisson process with intensity function n on the unit cube [0, 1]d and q = 0, . . . , d�1. is a smooth intensity function which is not necessarily constant.

So far, there exist results on the pointwise asymptotic normality for Betti numbers in the case of a homogeneousPoisson process or a binomial process, see Yogeshwaran et al. [2017]. In the case of a homogenous Poisson processthis result was extended to persistent Betti numbers by Hiraoka et al. [2018]. We will use this latter result to derivethe pointwise asymptotic normality for the general case. The recent contributions Owada et al. [2018] and Owada andThomas [2018] also study the limiting behavior of Betti numbers.

TDA is a comparably young field that has emerged from several contributions in algebraic topology and compu-tational geometry. Milestone contributions which helped to popularize TDA in its early days are Edelsbrunner et al.[2000], Zomorodian and Carlsson [2005] and Carlsson [2009]. TDA consists of various techniques which aim at under-standing the topology of a d-dimensional manifold based on an approximating point cloud. In practice, we can think ofa probability distribution, whose topological properties are of interest, and a sample from this distribution. The various

⇤This research was partially supported by the German Research Foundation (DFG), Grant Number KR-4977/1.†Department of Statistics, University of California, Davis, CA, 95616, USA, email: [email protected]‡Corresponding author§Department of Statistics, University of California, Davis, CA, 95616, USA, email: [email protected]

2

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methods of TDA have been successfully implemented in applied sciences such as biology (Yao et al. [2009]), mate-rial sciences (Lee et al. [2017]) or chemistry (Nakamura et al. [2015]). From the mathematical statistician’s point ofview, a particular interest deserves the application of TDA to time series, see, e.g., the pioneering works of Severskyet al. [2016], Umeda [2017] and the contributions of Gidea et al. to financial time series (Gidea and Katz [2018], Gidea[2017], Gidea et al. [2018]).

This present contribution falls into the area of persistent homology which is one of the major tools in TDA. We canonly give a short introduction to this topic here, a more detailed introduction which offers insights to the basic concepts,ideas and applications of persistent homology can be found in Chazal and Michel [2017], Oudot [2015] and Wasserman[2018].

The basic ingredient for the study of persistent Betti numbers is a realization of a point cloud in Rd (a sample of apoint process) and simplicial complexes built from this point cloud according to a rule which describes the neighborhoodrelation between points. The two most frequent simplicial complex models are the Rips-Vietoris and the Cech complex.When considered as geometric structures, simplicial complexes are characterized by the number of their q-dimensionalholes, most notably connected components, loops and cavities (0, 1 and 2 dimensional features). These holes areprecisely defined with a tool from algebraic topology, the so-called homology. The qth homology of a simplicial complexis determined by a quotient space. Its dimension is the so-called qth Betti number. Intuitively, the qth Betti number countsthe number of q-dimensional holes in the simplicial complex.

For a given simplicial complex model, we can construct an increasing sequence of simplicial complexes which isindexed by one parameter that can be understood as time, a so-called filtration. Given a filtration on a finite time interval,we can consider the evolution of the qth homology groups, i.e., of the dynamical behavior of the Betti numbers. As theunderlying simple point process (e.g., a Poisson process on the Euclidean space) is random, these Betti numbers are alsorandom and we consider a stochastic process.

From the applied point of view, the mere knowledge of the evolution of the Betti numbers is often not enough,especially when considering objects obtained from persistence diagrams, such as persistent landscapes. In this contextthe more general concept of persistent Betti numbers is the appropriate tool.

The remainder of this manuscript is organized as follows. In Section 1 we introduce the detailed framework andnotation. Section 2 offers a short overview of related results which are important and needed in our present study.Section 3 contains the main results of this manuscript which are in detail derived in Section 4 and in Appendix A.

1 Definitions and notation

Given a finite subset P of the Euclidean space Rd the Cech filtration C(P ) = (Cr

(P ) : r � 0) and the Rips-Vietorisfiltration R(P ) = (R

r

(P ) : r � 0) are defined as

Cr

(P ) = {finite � ⇢ P,[

x2�

B(x, r) 6= ;},

Rr

(P ) = {finite � ⇢ P, diam(�) r},

here B(x, r) = {y 2 Rd

: kx � yk r} and diam is the diameter of a measurable set. Throughout this article, weconsider either the Cech or the Rips-Vietoris filtration and write K(P ) for the underlying filtration. Furthermore, if werefer to a generic simplicial complex, we write K for simplicity.

Consider a filtration K(P ) and a time r � 0. Write Hq

(Kr

(P )) for the homology of the simplicial complex Kr

(P )

w.r.t. to the base field F2 = {0, 1}. Write Zq

(Kr

(P )) for the qth cycle group of the simplicial complex Kr

(P ) andB

q

(Kr

(P )) for the qth boundary group. Let 0 q d� 1. Then (r, s)-persistent Betti number (see Edelsbrunner et al.

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[2000]) of a simplicial complex K(P ) is defined by

�r,s

q

(K(P )) = dim

Zq

(Kr

(P ))

Zq

(Kr

(P )) \Bq

(Ks

(P ))

= dimZq

(Kr

(P ))� dimZq

(Kr

(P )) \Bq

(Ks

(P )).

The persistent Betti number is closely related to the persistence diagram of the underlying point cloud P , see Hiraokaet al. [2018] for further details. Visually, �r,s

q

(K(P )) counts the number of generators of the persistence diagram bornbefore time r and are still alive at time s, see Figure 1. The Betti number �r

q

(K(P )) is then defined as �r,r

q

(K(P )),r � 0; this means �r

q

(K(P )) = dimZq

(Kr

(P ))� dimBq

(Kr

(P )).The persistent Betti numbers are translation invariant, i.e., �r,s

q

(K(P + v)) = �r,s

q

(K(P )) for each v 2 Rd. Forthat reason the add-one cost function D0�

r,s

q

(K(P )) = �r,s

q

(K(P [ {0})) � �r,s

q

(K(P )) is an important tool in ouranalysis. Moreover, if A ✓ Rd, we write K

q

(P, r,A) for the q-simplices in Kr

(P ) with at least one edge in A and wewrite K

q

(P, r) for the entire set of q-simplices in Kr

(P ).

death

birth

s

r

1

Figure 1: The measure induced by the persistent Betti number �r,s

q

(K(P )) are the points in the gray-shaded rectangle.The point on the dashed red line is not included in �r,s

q

(K(P )) whereas the point on the solid red line is.

A density function is blocked if =

P

m

d

i=1 ai 1{Ai

}, where ai

> 0 and the Ai

are subcubes forming a partitionof [0, 1]d as follows: each edge of the cube [0, 1]d is partitioned into m intervals of the same length and the products ofthese intervals form the cubes A

i

with volume m�d for i = 1, . . . ,md.The density function satisfies on [0, 1]d the following smoothness criterion: (1) is continuous such that 0 <

inf sup < 1 and (2) for each " > 0 there is a blocked density function "

such that sup | � "

| ".E.g., a continuously differentiable intensity function is admissible. In the case of the Rips-Vietoris or Cech filtrationa continuous (continuously differentiable) intensity implies that also the persistence diagram of the underlying pointprocess admits a continuous (continuously differentiable) density (w.r.t. the Lebesgue measure) by the work of Chazaland Divol [2018].

We conclude this section with the introduction of further notation used throughout this article. For y, z 2 Zd, wewrite y � z (resp. y � z) if y precedes (resp. succeeds) z in the lexicographic ordering on Zd and write y � z (resp.y ⌫ z) if either y � z (resp. y � z) or y = z. Set Q(z) = (�1/2, 1/2]d + z for z 2 Zd. If f : R ! R, write f(t�)

(resp. f(t+)) for the limit of f from the left (resp. the right) at t if this limit exists. We let ) denote convergence indistribution of a sequence of random variables or weak convergence of a stochastic process.

In our analysis, the definition of the radius of weak stabilization will be useful. To this end, consider a point processP on Rd without accumulation points and let Q be a finite subset of Rd centered around a point z, i.e., Q = Q\B(z, L)

for some L � 0. (E.g., {0} is centered around 0 and the set Q(z) is centered around z for each z 2 Rd.) Write for shortK

r,a

= Kr

(P \B(z, a)) and K0r,a

= Kr

((P [Q) \B(z, a)) for a, r � 0.Let � = {(r, s) : 0 r s < 1} and fix a subset �c

= {(r, s) : a r s b} for some 0 a b < 1.

4

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Define the radius of weak stabilization of (r, s) by

⇢(r,s)(P,Q) = inf{R > 0 : dimZq

(K0r,a

)� dimZq

(Kr,a

) = const. 8a � R

and dimZq

(K0r,a

) \Bq

(K0s,a

)� dimZq

(Kr,a

) \Bq

(Ks,a

) = const. 8a � R}(1.1)

and define the radius of weak stabilization of �c by

⇢(P,Q) = sup

(r,s)2�c

⇢(r,s)(P,Q). (1.2)

If Q = {0}, we simply write ⇢(r,s)(P ) (resp. ⇢(P )). One can use similar ideas as in Lemma 5.3 in Hiraoka et al. [2018]to show that ⇢(r,s)(P,Q) and ⇢(P,Q) are a.s. finite; we do this in Lemma A.4 in the Appendix.

2 Related results

Below we quote results which are closely related to our study. The techniques employed to obtain these results are toolsfrom geometric probability, which studies geometric quantities deduced from simple point processes in the Euclideanspace. A classical result of Steele [1988] proves the convergence of the total length of the minimum spanning tree builtfrom an i.i.d. sample of n points in the unit cube. There are several generalizations of this work, for notable contributionssee McGivney and Yukich [1999], Penrose and Yukich [2003], Yukich [2000] and the monograph of Penrose [2003].

A different type of contribution equally important is Penrose and Yukich [2001] which studies the asymptotic nor-mality of functionals built on the Poisson and binomial process. We will heavily use the ideas given therein to obtainlimit expressions for covariance function of the finite-dimensional distributions of the persistent Betti numbers. Forcompleteness, we mention that the study of Gaussian limits is not limited to the total mass functional (as, e.g., in Pen-rose and Yukich [2001] and Penrose and Yukich [2003]) but can also be extended to random point measures obtainedfrom the points of a marked point process, see, e.g., Baryshnikov and Yukich [2005], Penrose [2007] and Blaszczyszynet al. [2016].

Recently, the limiting expression for the expectation of persistent Betti numbers was obtained.

Proposition 2.1 (Divol and Polonik [2018], Lemma 9). Let 0 < r s < 1. Let Sn

be either a Poisson or ann-binomial process with intensity n. Let X 0 be independent of S

n

with density . Then

lim

n!1n�1E

h

�r,s

q

(K(n1/dSn

))

i

= Eh

ˆbq

(r(X 0)

1/d, s(X 0)

1/d)

i

,

where ˆbq

(r, s) is the limit of n�1E⇥

�r,s

q

(K((n1/dX⇤n

))

for a homogeneous binomial process X⇤n

on [0, 1]d.

In another recent article Goel et al. [2018] give convergence results of Betti numbers addressing both a.s. conver-gence and convergence in the mean.

So far normality results for (persistent) Betti numbers exist only in a pointwise sense and are rather direct conse-quences of Theorem 2.1 and 3.1 given in Penrose and Yukich [2001]. We quote them here in a sense which makes themmore in-line with our framework. For this we need the notion of the interval of co-existence I

d

(P) defined by the criticalradius for percolation of the occupied and the critical radius of percolation of the vacant component of a Poisson processP with unit intensity on Rd. We refer to Yogeshwaran et al. [2017] for the exact definition.

Proposition 2.2 (Pointwise normality of (persistent) Betti numbers).

(i) [Hiraoka et al. [2018] Theorem 5.2] Let P be a homogeneous point process with unit intensity on Rd and letP(n) = P \ [0, n1/d

]

d. Let 0 r s < 1. Then there is a �2(r, s) 2 R+ such that

n�1/2(�r,s

q

(K(P(n)))� E⇥

�r,s

q

(K(P(n)))⇤

) ) N(0,�2(r, s)).

5

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(ii) [Yogeshwaran et al. [2017] Theorem 4.7] Let K(·) be the Cech filtration and let 0 r < 1 such that r /2 Id

(P).Let X

n

be a binomial process of length n and intensity n on [0, 1]d for each n 2 N+. Then there is a 0 < ⌧2(r) �2

(r, r) such thatn�1/2

�r

q

(K(n1/dXn

))� Eh

�r

q

(K(n1/dXn

))

i⌘

) N(0, ⌧2(r)).

First, we remark that the above statements in their original version are also valid for more general domains Wn

✓ Rd

which are not necessarily rectangular domains such as [0, n1/d]

d. Furthermore, we remark that Hiraoka et al. [2018]prove their theorem for a general class of filtrations which contains among others the Cech and the Rips-Vietoris fil-tration. Moreover, Theorem 4.7 of Yogeshwaran et al. [2017] also contains a version of (ii) for Betti numbers of thehomogeneous Poisson process, this is however contained in the result (i). Finally, we remark that Yogeshwaran et al.[2017] point out that the condition r /2 I

d

(P) is likely to be superfluous. As a by-product of our results, we show that,in fact, the condition can be removed.

3 Main results

This section contains the main results, the convergence of finite-dimensional distributions to a normal distribution. LetP,P0 be independent and homogeneous Poisson processes with unit intensity on Rd. Set B

n

= [�2

�1n1/d, 2�1n1/d]

d.By the stabilizing property of the persistent Betti numbers, there are random variables �r,s

0 (1) 2 Z and N0 2 N suchthat

�r,s

q

(P \Bn

)� �r,s

q

(((P \Q(0)) [ (P0 \Q(0))) \Bn

) ⌘ �

r,s

0 (1) for all n � N0,

see Lemma 4.6. Set �((u, v), (r, s)) = E [E [�

u,v

0 (1)|F0]E [�

r,s

0 (1)|F0]].

Theorem 3.1. Let Pn

be a Poisson process with intensity n on [0, 1]d. Let ` 2 N+ and (ri

, si

) 2 � for i = 1, . . . , `.Then

0

B

B

@

n�1/2�

�r1,s1q

(K(n1/dPn

))� E⇥

�r1,s1q

(K(n1/dPn

))

⇤�

...n�1/2

�r`,s`q

(K(n1/dPn

))� E⇥

�r`,s`q

(K(n1/dPn

))

⇤�

1

C

C

A

) N(0,⌃),

where the covariance matrix ⌃ is given by

⌃(i, j) = Eh

�((X)

1/d((r

i

, si

), (rj

, sj

)))

i

(1 i, j `).

Moreover, let P⌧

be a homogeneous Poisson process with intensity ⌧ on Rd for each ⌧ 2 R+. Define for 0 r s < 1

↵(r, s) :=

Z

[0,1]dE⇥

D0�r,s

q

(K(P(x)))

(x)dx. (3.1)

Theorem 3.2. Let Xn

be a binomial process with intensity n. Let ` 2 N+ and (ri

, si

) 2 � for i = 1, . . . , `. Then

0

B

B

@

n�1/2�

�r1,s1q

(K(n1/dXn

))� E⇥

�r1,s1q

(K(n1/dXn

))

⇤�

...n�1/2

�r`,s`q

(K(n1/dXn

))� E⇥

�r`,s`q

(K(n1/dXn

))

⇤�

1

C

C

A

) N(0, ˜⌃),

where the covariance matrix ˜

⌃ is given by

˜

⌃(i, j) = Eh

�((X)

1/d((r

i

, si

), (rj

, sj

)))

i

� ↵(ri

, si

)↵(rj

, sj

) (1 i, j `).

6

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4 Technical results

This section consists of two parts. First, we derive the asymptotic normality of the finite-dimensional distributions ofthe persistent Betti numbers obtained from Poisson processes. Then, we repeat these considerations for an underlyingbinomial process.

The next result is crucial for the upcoming proofs, the so-called geometric lemma enables us to obtain upper boundson moments. The result for Betti numbers is well-known (to topologists), we quote here a generalized version due toHiraoka et al. [2018] (Lemma 2.11).

Lemma 4.1 (Geometric lemma). Let X ✓ Y be two finite point sets of Rd. Then

��r,s

q

(K(Y))� �r,s

q

(K(X))�

� q+1X

j=q

|Kj

(Y, s) \Kj

(X, s)|.

4.1 The Poisson case

In the first step, we consider blocked density functions =

P

m

d

i=1 ai 1{Ai

}.

Proposition 4.2. Let be a blocked density and let Pn

be a Poisson process with intensity n. Let X be a randomvariable with density . Then for each pair (r, s) 2 �, with �2

(r, s) from Proposition 2.2 (i),

n�1/2⇣

�r,s

q

(K(n1/dPn

))� Eh

�r,s

q

(K(n1/dPn

))

i⌘

) N⇣

0,Eh

�2((X)

1/d(r, s))

i⌘

, n ! 1.

Proof. Define the filtration

�K(n1/dP

n

, r) =

m

d[

i=1

K(n1/d(P

n

\Ai

), r). (4.1)

The proof consists of two steps. First, we prove the desired convergence for the reduced filtration�K. Second, we

demonstrate that the Betti numbers obtained from�K and K are sufficiently close as n ! 1.

Note that

�r,s

q

(

�K(n1/dP

n

)) =

m

dX

i=1

�r,s

q

(K(n1/d(P

n

\Ai

))),

as�K is the union is of disjoint complexes. Moreover, �r,s

q

(K(n1/d(P

n

\ Ai

))) and �r,s

q

(K(n1/d(P

n

\ Aj

))) areindependent if i 6= j as we are dealing with a Poisson process. We can use the geometric properties of the filtration toobtain the following equality

�r,s

q

(K(n1/d(P

n

\Ai

))) = �ra

1/di ,sa

1/di

q

(K((nm�dai

)

1/dm(Pn

\Ai

))),

where (nm�dai

)

1/dm(Pn

\ Ai

)) is a homogeneous Poisson process with unit intensity on [0, (nm�dai

)

1/d]

d. Thus,

the distribution of this persistent Betti number equals the distribution of �ra

1/di ,sa

1/di

q

(K(P0 \ Bn,i

)), where Bn,i

=

[�2

�1(nm�da

i

)

1/d, 2�1(nm�da

i

)

1/d]

d and where P0 is a stationary Poisson process on Rd with unit intensity. SetB

k

= [�2

�1k1/d, 2�1k1/d]d for k 2 N. Then

n�1/2(�u,v

q

(K(P0 \Bn

))� E⇥

�u,v

q

(K(P0 \Bn

))

) ) N(0,�2(u, v)) (n ! 1)

by Theorem 5.1 in Hiraoka et al. [2018] for all (u, v) 2 �. In particular, one can show that for (u, v) 2 �

7

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Lemma 4.3.

sup

n2NE⇥

|�u,v

q

(K(P0 \Bbnm�daic))� �u,v

q

(K(P0 \Bn,i

))|2⇤

< 1.

Proof. The distribution of P0 \ Bn,i

equals that of (Z1, . . . , ZMn), where Zi

are uniformly i.i.d. on Bn,i

and Mn

isPoisson distributed with mean nm�da

i

. Set Cn,i

= Bn,i

\Bbnm�daic. Hence, using the geometric lemma we obtain

E⇥

|�u,v

q

(K(P0 \Bbnm�daic))� �u,v

q

(K(P0 \Bn,i

))|2⇤

E

2

6

4

0

@

MnX

k=1

1�

Zk

2 Bn,i

\Bbnm�daic

MnX

j=1

1{|Zk

� Zj

| 2v}�

q

1

A

23

7

5

X

`2NP(M

n

= `)

`

X

k1,k2=1

E

2

41{Zk1 2 C

n,i

} 1{Zk2 2 C

n,i

}E

2

4

`

X

j=1

1{|Zk1 � Z

j

| 2v}�

2q��

Zk1 , Zk2

3

5

3

5

1/2

⇥ E

2

41{Zk1 2 C

n,i

} 1{Zk2 2 C

n,i

}E

2

4

`

X

j=1

1{|Zk2 � Z

j

| 2v}�

2q��

Zk1 , Zk2

3

5

3

5

1/2

,

where we use the Cauchy–Schwarz inequality in the derivation of the last inequality.Next, use that E

h

|P

`

j=1 1{|Zk1 � Zj

| 2s} |2q|Zk1 , Zk2

i

is dominated by 2 plus the expectation of a binomialrandom variable with parameters ` � 2 and a success probability proportional to n�1. Moreover, P(Z

k

2 Cn,i

) =

O(n�1). Thus, up to a multiplicative constant the last equation is bounded above by

X

`2NP(M

n

= `)

`

X

k1,k2=1

E [1{Zk1 2 C

n,i

} 1{Zk2 2 C

n,i

}] (2 + `n�1+ (`n�1

)

2q)

c1X

`2NP(M

n

= `)(`n�1+ (`n�1

)

2q+2) c2 < 1

for some c1, c2 2 R+ and for all n 2 N.

We obtain that

(nm�dai

)

�1/2(�

ra

1/di ,sa

1/di

q

(K(P0 \Bn,i

))� E[�ra

1/di ,sa

1/di

q

(K(P0 \Bn,i

))]) ) N(0,�2(ra

1/di

, sa1/di

)).

Consequently, using independence

n�1/2(�r,s

q

(

�K(n1/dP

n

))� E[�r,s

q

(

�K(n1/dP

n

))]) ) N

0,

Z

[0,1]d�2

((x)1/d(r, s))(x)dx

!

.

This proves the first step. Next, we show that for each " > 0,

lim sup

n!1n�1

Var

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

". (4.2)

which implies the claim. Let z1, . . . , zkn 2 Zd be the points with the property that Q(zj

) = (�1/2, 1/2]d + zj

intersects with [0, n1/d]

d. The zj

are ordered lexicographically, i.e., zj�1 � z

j

. We have kn

/n ! 1. Let P0n

be anotherPoisson process with intensity function n, independent of P

n

. Define, for each zj

, a new Poisson process with intensity(n�1/d · ) on the cube [0, n1/d

]

d by

P00n

(zj

) =

n

(n1/dPn

) \Q(zj

)

o

[n

(n1/dP0n

) \Q(zj

)

o

.

8

Page 8: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

Set Gn,0 = {;,⌦} and let G

n,j

be the smallest �-field such that the number of Poisson points of n1/dPn

in [j

`=1Q(z`

)

is measurable. Then

E

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

Gn,j�1

= E

�r,s

q

(K(P00n

(zj

))� �r,s

q

(

�K(P00

n

(zj

)))

Gn,j

.

In particular,

n�1Var

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

= n�1Var

2

4

knX

j=1

E

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

Gn,j

� E

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

Gn,j�1

3

5

= n�1Var

2

4

knX

j=1

E

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))� �r,s

q

(K(P00n

(zj

))� �r,s

q

(

�K(P00

n

(zj

)))

Gn,j

3

5

n�1knX

j=1

E"

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))� �r,s

q

(K(P00n

(zj

)) + �r,s

q

(

�K(P00

n

(zj

)))

2#

.

Again, we can insert terms ±�r,s

q

(K(n1/dPn

[ (n1/dP0n

\Q(zj

)))) and ±�r,s

q

(

�K(n1/dP

n

[ (n1/dP0n

\Q(zj

)))) on theright-hand side of the last inequality. Then, it suffices to consider

n�1knX

j=1

Eh

�r,s

q

(K(n1/dPn

[ (n1/dP0n

\Q(zj

))))� �r,s

q

(K(n1/dPn

))

+ �r,s

q

(

�K(n1/dP

n

))� �r,s

q

(

�K(n1/dP

n

[ (n1/dP0n

\Q(zj

))))

2i

,

(4.3)

the other summand works in the same fashion. We apply the tightness results from the Appendix A to both filtrations K

and�K, note that the application to

�K is possible because it is the union of disjoint complexes. We write ⇢K resp. ⇢ �

Kfor

the stabilization radius from 1.1. Then using Proposition A.5, for each � > 0, there is an L > 0 such that

sup

n,j

P(⇢K(n1/dPn

, n1/dP0n

\Q(zj

)) > L) _ sup

n,j

P�

⇢ �K(n1/dP

n

, n1/dP0n

\Q(zj

)) > L�

�.

Moreover,

sup

n,j

Eh

�r,s

q

(K(n1/dPn

[ (n1/dP0n

\Q(zj

))))� �r,s

q

(K(n1/dPn

))

4i

< 1,

sup

n,j

Eh

�r,s

q

(

�K(n1/dP

n

[ (n1/dP0n

\Q(zj

))))� �r,s

q

(

�K(n1/dP

n

))

4i

< 1.

(4.4)

Indeed, by Lemma 4.1 (similar considerations also apply to�K)

|�r,s

q

(K(n1/dPn

[ (n1/dP0n

\Q(zj

))))� �r,s

q

(K(n1/dPn

))|

X

`=q,q+1

|K`

(n1/dPn

[ (n1/dP0n

\Q(zj

)), s) \K`

(n1/dPn

, s)|.

So, it remains to show that E⇥

|K`

(n1/dPn

[ (n1/dP0n

\Q(zj

)), s) \K`

(n1/dPn

, s)|4⇤

is uniformly bounded for ` =

q, q + 1, n 2 N and j = 1, . . . , kn

.

9

Page 9: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

Note that we can write Pn

as (X1, . . . , XNn) and P0n

as (Y1, . . . , YMn) where Nn

and Mn

are independently Poissondistributed with parameter n and the X

i

and Yi

are i.i.d. with density . Then the first line in (4.4) is bounded above by(times a constant)

E

2

4

MnX

k=1

1{Yk

2 Q(zj

)}�

NnX

i=1

1{|Xi

� Yk

| 2r}�

q

!43

5

+ E

2

4

MnX

k=1

1{Yk

2 Q(zj

)}�

MnX

i=1

1{|Yi

� Yk

| 2r}�

q

!43

5 .

(4.5)

Again, we only consider the second term in (4.5), the first term works similarly. Using the conditional structure of thePoisson process and the Holder inequality, this term is bounded above by

X

`2NP(M

n

= `)

`

X

k1,k2=1

E"

1{Yk1 2 Q(z

j

)} 1{Yk2 2 Q(z

j

)}�

`

X

i=1

1{|Yi

� Yk1 | 2r}

4q#1/2

⇥ E"

1{Yk1 2 Q(z

j

)} 1{Yk2 2 Q(z

j

)}�

`

X

i=1

1{|Yi

� Yk2 | 2r}

4q#1/2

CX

`2NP(M

n

= `)

`

X

k1,k2=1

E [1{Yk1 2 Q(z

j

)} 1{Yk2 2 Q(z

j

)}] (1 + `n�1+ (`n�1

)

4q), (4.6)

where we use that the conditional expectation of the inner summands when conditioned on Yk1 and Y

k2 is dominated by2 plus are the expectation of a binomial random variable of length ` � 2 and a success probability proportional to n�1.Moreover, the pth moment of a Poisson distribution with parameter � is a polynomial in � of degree p. Consequently,(4.6) is uniformly bounded in n and j. In particular, this proves that the expectation in (4.4) is uniformly bounded.

Thus, (4.3) reduces to

n�1knX

j=1

Eh

�r,s

q

(K([n1/dPn

[ (n1/dP0n

\Q(zj

))] \B(zj

, L)))� �r,s

q

(K(n1/dPn

\B(zj

, L)))

+ �r,s

q

(

�K(n1/dP

n

\B(zj

, L)))� �r,s

q

(

�K([n1/dP

n

[ (n1/dP0n

\Q(zj

))] \B(zj

, L)))�

2i

.

(4.7)

First notice that the inner expectation can only be non-zero if B(zj

, L) intersects with n1/d([m

d

i=1@Ai

)

(2s). The totalnumber of such intersections is O(n(d�1)/dn�1

) = O(n�1/d). Hence, as k

n

/n ! 1, it is sufficient to show that theexpectations in (4.7) are uniformly bounded. This follows however along the same lines as (4.4). Consequently, (4.2) issatisfied and the proof is complete.

Proposition 4.4. Let Pn

be a Poisson process with intensity n, where is a general density which satisfies: for all" > 0 there is a blocked density

"

such that sup | � "

| ". Then there are coupled Poisson processes Qn

withintensity function n

"

such that

sup

n2Nn�1

Var

h

�r,s

q

(K(n1/dPn

))� �r,s

q

(K(n1/dQn

))

i

= O(").

Proof. Let " > 0 be arbitrary but fixed. The proof is divided in two main steps. Some preparation is necessary first. Let¯M ⇠ Poi(sup) and choose m 2 N such that P( ¯M � m)(1 + mp

) ", where p is as in (4.13) below. Let P,P0 be

10

Page 10: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

independent Poisson processes with unit intensity on Rd ⇥ [0,1). Set

P(n) = {x | 9 0 t (n�1/dx) and (x, t) 2 P}, PS

= {x | 9 0 t sup+ " and (x, t) 2 P},

P0(n) = {x | 9 0 t (n�1/dx) and (x, t) 2 P0} and (P0

)

S

= {x | 9 0 t sup+ " and (x, t) 2 P0}.

Then L(P(n)) = L(P0(n)) = L(n1/dP

n

) and PS is a homogeneous Poisson process with intensity sup + " on Rd

such that P(n) ✓ PS for all n 2 N+. In the same way, P0(n) ✓ (P0

)

S for all n 2 N.Let L 2 N be such that P(⇢(PS [ (P0

)

S

) � L) (1 + m4(q+1))

�1". By assumption there is a blocked densityfunction

"

such that sup |� "

| ((1 +m4(q+1))(Ld _ 1))

�1". In particular, k� "

k1 ".Furthermore define,

Q(n) = {x | 9 0 t "

(n�1/dx) and (x, t) 2 P}.

Then Q(n) is a Poisson process with intensity "

(n�1/d · ) and Q(n) ✓ PS because "

+ ".Moreover, there is a coupling (X 0, Y 0

) – independent of P and P0 – such that X 0= (X 0

k

: k 2 N+) are i.i.d. withdensity and Y 0

= (Y 0k

: k 2 N+) are i.i.d. with density "

such that P(X 0k

6= Y 0k

) 2". Moreover, let N 0n

⇠ Poi(n).Set P0

n

= (X 01, . . . , X

0N

0n) and Q0

n

= (Y 01 , . . . , Y

0N

0n). Then L(P(n)) = L(n1/dP0

n

) and L(Q(n)) = L(n1/dQ0n

).We use a similar representation of the persistent Betti numbers as in the proof of Proposition 4.2 and consider

n�1Var

�r,s

q

(K(P(n))� �r,s

q

(K(Q(n))⇤

. (4.8)

Let zn,1, . . . , zn,kn 2 Zd be the points such that Q(z

n,j

) = (�1/2, 1/2]d + zn,j

intersects with [0, n1/d]

d. The zn,j

areordered lexicographically, i.e., z

n,j�1 � zn,j

. Observe that kn

/n ! 1.Define for each z

n,j

a new Poisson processes on [0, n1/d]

d by

P00n

(zn,j

) = {P(n) \Q(zn,j

)} [n

(n1/dP0n

) \Q(zn,j

)

o

,

Q00n

(zn,j

) = {Q(n) \Q(zn,j

)} [n

(n1/dQ0n

) \Q(zn,j

)

o

,

the corresponding intensity functions are also (n�1/d · ) and "

(n�1/d · ). Set Gn,0 = {;,⌦} and let G

n,j

be thesmallest �-field such that the number of Poisson points of P(n) and Q(n) in [j

`=1Q(zn,`

) is measurable. Then, (4.8) canbe expanded in an MDS as

n�1knX

j=1

Eh

E⇥

�r,s

q

(K(P(n)))� �r,s

q

(K(P00n

(zj

)))� [�r,s

q

(K(Q(n)))� �r,s

q

(K(Q00n

(zj

)))]|Gn,j

⇤2i

. (4.9)

Next, insert ±�r,s

q

(K((P(n) [ P00n

(zn,j

)))) and ±�r,s

q

(K((Q(n) [ Q00n

(zn,j

)))) in (4.9). This reduces the problem to

n�1knX

j=1

Eh

�r,s

q

(K(P(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(P(n)))

� [�r,s

q

(K(Q(n) [ (n1/dQ0n

\Q(zn,j

))))� �r,s

q

(K(Q(n)))]�

2i(4.10)

and

n�1knX

j=1

Eh

�r,s

q

(K(P(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(P00n

(zn,j

)))

� [�r,s

q

(K(Q(n) [ (n1/dQ0n

\Q(zn,j

))))� �r,s

q

(K(Q00n

(zn,j

)))]

2i

.

11

Page 11: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

Clearly, it suffices to consider the summand in (4.10) and to show that it is bounded by " times a constant which isindependent of n and the choice of ". This then completes the preparations.

We begin with the first step and show that we can replace the process n1/dQ0n

\Q(zn,j

) with n1/dP0n

\Q(zn,j

) inthe third persistent Betti number in (4.10) at cost of O("). Observe that

��r,s

q

(K(Q(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(Q(n) [ (n1/dQ0n

\Q(zn,j

))))

X

i=q,q+1

|Ki

(Q(n) [ (n1/d(P0

n

[ Q0n

) \Q(zn,j

)), s) \Ki

(Q(n) [ (n1/dP0n

\Q(zn,j

)), s)|

+

X

i=q,q+1

|Ki

(Q(n) [ (n1/d(P0

n

[ Q0n

) \Q(zn,j

)), s) \Ki

(Q(n) [ (n1/dQ0n

\Q(zn,j

)), s)|.(4.11)

Again, it suffices to consider the first summand and argue similarly as in the proof of Proposition 4.2. We can thinkof the Poisson process Q(n) as n1/d

(Y1, . . . , YNn), where Nn

⇠ Poi(n) and (Yi

: i 2 N+) is an i.i.d. sequence withcommon density

"

. Using the independence between Q(n) and P0n

,Q0n

, an application of the last inequality then yields,for instance, for the term in the first line of (4.11) if i = q + 1

Eh

|Kq+1(Q(n) [ (n1/d

(P0n

[ Q0n

) \Q(zn,j

)), s) \Kq+1(Q(n) [ (n1/dP0

n

\Q(zn,j

)), s)|2i

Eh⇣

N

0n

X

`=1

1{Y 0`

6= X 0`

} 1n

n1/dY 0`

2 Q(zn,j

)

o⇣

NnX

k=1

1n

|Yk

� Y 0`

| 2sn�1/do

+

N

0n

X

k=1

1n

|X 0k

� Y 0`

| 2sn�1/do

+

N

0n

X

k=1

1n

|Y 0k

� Y 0`

| 2sn�1/do⌘

q+1⌘2i

Conditional on the event {Nn

= m,N 0n

= m0}, 1 `1, `2 m0 and p = 2(q + 1)

E"

m

X

k=1

1n

|Yk

� Y 0`1| 2sn�1/d

o

p

X 0`1, Y 0

`1, X 0

`2, Y 0

`2

#

C⇣

1 +

m

n+

⇣m

n

p

,

E

2

4

m

0X

k=1

1n

|X 0k

� Y 0`1| 2sn�1/d

o

p

X 0`1, Y 0

`1, X 0

`2, Y 0

`2

3

5 C

1 +

m0

n+

m0

n

p

,

E

2

4

m

0X

k=1

1n

|Y 0k

� Y 0`1| 2sn�1/d

o

p

X 0`1, Y 0

`1, X 0

`2, Y 0

`2

3

5 C

1 +

m0

n+

m0

n

p

,

for a constant C 2 R+, using the properties of the binomial distribution. Moreover,

n�1knX

j=1

m

0X

`1,`2=1

Eh

1�

Y 0`1

6= X 0`1

1n

n1/dY 0`1

2 Q(zj

)

o

1�

Y 0`2

6= X 0`2

1n

n1/dY 0`2

2 Q(zj

)

oi

C"

m0

n+

m0

n

◆2!

.

Similar bounds are valid for the remaining terms in (4.11). Combining these arguments, we obtain that (4.10) is boundedabove by " times a constant (which is independent of n and the choice of "). This finishes the first step.

12

Page 12: On the asymptotic normality of persistent Betti numbersanson.ucdavis.edu/~polonik/Normality of Persistent Betti Numbers.pdfOn the asymptotic normality of persistent Betti numbers⇤

In the second step, it remains to study (4.10) after n1/dQ0n

has been replaced with n1/dP0n

, i.e.,

n�1knX

j=1

Eh

�r,s

q

(K(P(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(P(n)))

� [�r,s

q

(K(Q(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(Q(n)))]�

2i

.

(4.12)

Choose a 1 j kn

and consider the expectation in (4.12). We model the Poisson process n1/dP0n

\ Q(zn,j

)

as (Zn,j,1, . . . , Zn,j,Mn) where M

n

is Poi(�) with � 2 [inf , sup]. First we focus on the tail of this expectation.Arguing similarly as above, we have for M

n

= k

E

�r,s

q

(K(P(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(P(n)))�

2 ��

Mn

= k

C(1 + kp) (4.13)

for some p 2 N. The expectation which involves Q(n) instead of P(n) admits the same upper bound. As the Poissonparameter � of M

n

is uniformly bounded by sup, we have that supn2N P(M

n

> m)(1+mp

) " for the above choicem 2 N+. Hence, we can neglect extreme values of M

n

and and can consider (4.12) conditional on the event Mn

m.We have

Eh

�r,s

q

(K(P(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(P(n)))

� [�r,s

q

(K(Q(n) [ (n1/dP0n

\Q(zn,j

))))� �r,s

q

(K(Q(n)))]�

2��

Mn

mi

m

m

X

`=0

P(Mn

= `)

`

X

k=1

Eh

D0�r,s

q

(P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)

�D0�r,s

q

(Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)

2��

Mn

= `i

.

(4.14)

We use the stabilizing property of the point processes to bound above the expectation in (4.14). Let ⇢ be given as in(1.1), then the expectation in (4.14) is at most

Eh

D0�r,s

q

((P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) \B(0, L))

�D0�r,s

q

((Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) \B(0, L))�

2��

Mn

= `i

(4.15)

+ Eh

D0�r,s

q

(P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)

�D0�r,s

q

(Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)

4��

Mn

= `i1/2

⇥ P({⇢(P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) � L} or

{⇢(Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) � L})1/2.

(4.16)

First, we consider the stabilization radii in the probability in (4.16) and show that the corresponding measures are tight.Using the subset principle from Proposition A.5 and the fact that the Z

n,j,i

are i.i.d., we obtain

P({⇢(P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) � L}|Mn

= `)

P({⇢(PS [ {Zn,j,1, . . . , Zn,j,l�1}� Z

n,j,l

) � L}|Mn

= `)

P({⇢(PS [ {Zn,j,1, . . . , Zn,j,l�1}� Z 0

n

) � L}|Mn

= `), (4.17)

where Z 0n

is an independent random variable. Next, we use that Zn,j,1, . . . , Zn,j,Mn

L= P0

(n) \ Q(zn,j

). So, (4.17) is

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dominated byP(⇢(PS [ (P0

)

S � Z 0n

) � L) = P(⇢(PS [ (P0)

S

) � L) (1 +m4(q+1))

�1",

where the first equality follows from the stationarity of PS and (P0)

Sand the second is due to the above choice ofL. A similar argument applies to P(⇢(Q(n) [ {Z

n,j,1, . . . , Zn,j,k�1} � Zn,j,k

) � L) P(⇢(PS [ (P0)

S

) � L) (1 + m4(q+1)

)

�1". Note, that these upper bounds hold uniformly in n and j. Consider the fourth moment in (4.16),elementary calculations show that it is uniformly bounded in j and in n. Indeed, using the geometric lemma, we obtainfor k ` m

E

D0�r,s

q

(P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)

4|M

n

= `

C1Eh

|B(0, 2s) \ PS |4(q+1)+m4(q+1)

i

C2(1 +m4(q+1)),

which is uniformly bounded in n and j for the above choice of m 2 N+ and certain C1, C2 2 R+.In the same way, E[|D0�

r,s

q

(Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

)|4 |Mn

= `] C2(1 +m4(q+1)). This shows

that given m 2 N+, (4.16) is bounded byp" times a constant uniformly in n and j.

It remains to consider (4.15) given the choices of m 2 N+ and L 2 R+. Here we can use that P(n) and Q(n) agreewith a high probability on bounded sets. (4.15) is at most (times a constant)

Eh

D0�r,s

q

((P(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) \B(0, L))�

4

+

D0�r,s

q

((Q(n) [ {Zn,j,1, . . . , Zn,j,k�1}� Z

n,j,k

) \B(0, L))�

4��

Mn

= `i1/2

⇥ P(P(n) \B(Zn,j,k

, L) 6= Q(n) \B(Zn,j,k

, L))1/2.

Again the fourth moments are uniformly bounded by 1+m4(q+1) times a constant. Moreover, the probability is at mostZ

[0,1]d(y)

Z

B(n�1/dy,n

�1/dL)

n|(z)� "

(z)|dzdy CLd

sup |� "

| (1 +m4(q+1))

�1".

This shows that given m 2 N+, also (4.15) is bounded byp" times a constant uniformly in n and j and the proof is

complete.

Proposition 4.5. Let P be a stationary Poisson process with unit intensity on Rd and let Bn

= [�2

�1n1/d, 2�1n1/d]

d.Then for each two pairs (r, s), (u, v) 2 �

lim

n!1n�1

Cov

�u,v

q

(K(P \Bn

)),�r,s

q

(K(P \Bn

))

= �((u, v), (r, s)).

Proof. The proof heavily makes use of the ideas given in the proof of Theorem 3.1 in Penrose and Yukich [2001]. Forz 2 Zd, denote by F

z

the �-algebra generated by the Poisson points of P in the set [y�z

Q(y). So Fz

is the smallest�-algebra such that P is measurable in any bounded Borel subset of [

y�z

Q(y). Let B0n

= {z1, . . . , zkn} be the set oflattice points z

j

2 Zd which satisfy Q(zj

) \Bn

6= ;. Note that kn

/n ! 1.Set G

j

= Fzj for j = 1, . . . , k

n

and G0 = {;,⌦}. Then (Gj

)

knj=0 is a filtration. Let P0 be a Poisson process with unit

intensity on Rd which is independent of P. Set P00(z) = (P \Q(z)) [ (P0 \Q(z)) for z 2 Rd. Then

�r,s

q

(K(P \Bn

))� E⇥

�r,s

q

(K(P \Bn

))

=

knX

j=1

E⇥

�r,s

q

(K(P \Bn

))� �r,s

q

(K(P00(z

j

) \Bn

))|Fzj

,

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the same equality is satisfied if (r, s) is replaced with (u, v). Moreover, by the probabilistic nature of the Poisson process

E⇥

�r,s

q

(K(P \Bn

))� �r,s

q

(K(P00(z

i

) \Bn

))|Fzj

= 0

if i > j. Consequently, we obtain

n�1Cov

�u,v

q

(K(P \Bn

)),�r,s

q

(K(P \Bn

))

= n�1Ehn

knX

j=1

E⇥

�u,v

q

(K(P \Bn

))� �u,v

q

(K(P00(z

j

) \Bn

))|Fzj

o

⇥n

knX

j=1

E⇥

�r,s

q

(K(P \Bn

))� �r,s

q

(K(P00(z

j

) \Bn

))|Fzj

oi

= n�1knX

j=1

Eh

E⇥

�u,v

q

(K(P \Bn

))� �u,v

q

(K(P00(z

j

) \Bn

))|Fzj

⇥ E⇥

�r,s

q

(K(P \Bn

))� �r,s

q

(K(P00(z

j

) \Bn

))|Fzj

i

.

(4.18)

First, we show that the difference �

r,s

z

(Bn

) = �r,s

q

(K(P \Bn

))� �r,s

q

(K(P00(z) \B

n

)) stabilizes.

Lemma 4.6. For each (r, s) 2 � and for each z 2 Zd, there is a random variable �r,s

z

(1) and an N0 = N0(z, (r, s)) 2N such that �r,s

z

(Bn

) ⌘ �

r,s

z

(1) a.s. for all n � N0.

Proof. Let z 2 Zd. Then

r,s

z

(Bn

)

= dimZq

(Kr

(P \Bn

))� dimZq

(Kr

(P00(z) \B

n

)) (4.19)

��

dimBq

(Ks

(P \Bn

)) \ Zq

(Kr

(P \Bn

))� dimBq

(Ks

(P00(z) \B

n

)) \ Zq

(Kr

(P00(z) \B

n

))

, (4.20)

where P00(z)\B

n

= {(P\Q(z))\Bn

}[{P0\Q(z)\Bn

}. For simplicity, define K0u,n

= Ku

((P[(P0\Q(z)))\Bn

)

and Ku,n

= Ku

(P\Bn

) for u � 0. First, consider the difference in (4.19) and add the terms ⌥ dimZq

(K0r,n

). If m � n,then

Zq

(K0r,n

)

.

Zq

(Kr,n

) ,! Zq

(K0r,m

)

.

Zq

(Kr,m

).

with the principle from Lemma A.1. In particular, dimZq

(K0r,n

)/Zq

(Kr,n

) dimZq

(K0r,m

)/Zq

(Kr,m

). Moreover,

dimZq

(K0r,n

)

.

Zq

(Kr,n

) Kq

(((P \Q(z)(2s)) [ (P0 \Q(z))) \Bn

, r,P0 \Q(z))

and the latter becomes independent of n once Q(z)(2s) is fully contained in Bn

. This boundedness property combinedwith the fact that the rank is integer-valued and the increasing property, show that there is a (random) n1 2 N such that

dimZq

(Kr

(P \Bn

))� dimZq

(Kr

((P [ (P0 \Q(z))) \Bn

))

is constant for all n � n1. Similarly, one can show the existence of an n2 2 N such that

dimZq

(Kr

((P [ (P0 \Q(z))) \Bn

))� dimZq

(Kr

(P00(z) \B

n

))

is constant for all n � n2. Second, consider the difference in (4.20). We add ± dimBq

(K0s,n

) \Zq

(K0r,n

) this time andproceed similarly as before. With similar arguments as those following Condition A.2 (see in particular Lemma A.4) in

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the appendix, one can show that there is an n3 2 N such that

dimBq

(Ks

(P \Bn

)) \ Zq

(Kr

(P \Bn

))

� dimBq

(Ks

((P [ (P0 \Q(z))) \Bn

)) \ Zq

(Kr

((P [ (P0 \Q(z))) \Bn

))

is constant for all n � n3. In the same way, there is an n4 2 N such that

dimBq

(Ks

((P [ (P0 \Q(z))) \Bn

)) \ Zq

(Kr

((P [ (P0 \Q(z))) \Bn

))

� dimBq

(Ks

(P00(z) \B

n

)) \ Zq

(Kr

(P00(z) \B

n

))

is constant for all n � n4. Combining these results, there is some n0 2 N such that for all n � n0

r,s

z

(Bn

) ⌘ �

r,s

z

(Bn0) ⌘ �

r,s

z

(1).

We define Gr,s

z

(Bn

) = E [�

r,s

z

(Bn

)|Fz

] and Gr,s

z

= E [�

r,s

z

(1)|Fz

] as well as

� = �((u, v), (r, s)) = E [Gu,v

0 Gr,s

0 ] .

Note that �((u, v), (r, s)) = �((r, s), (u, v)) and that the family {(Gu,v

z

, Gr,s

z

) : z 2 Zd} is stationary.We show in the second step that

n�1knX

j=1

Gu,v

zjGr,s

zj! � in L1

(P). (4.21)

This is a consequence of the pointwise ergodic theorem. Indeed, let e1 = (1, 0, . . . , 0) 2 Zd. Then the averages of{Gu,v

e1Gr,s

e1, Gu,v

2e1Gr,s

2e1 , . . . , Gu,v

ne1Gr,s

ne1} converge to � a.s. in L1

(P). Thus, we can partition the set B0n

into Jn

one-dimensional intervals I

k

of a length length `n

each. Here Jn

is O(n(d�1)/d) and `

n

is O(n1/d). We mean by one-

dimensional interval maximal subsets of Bn

of the type (Z \ [a, b]) ⇥ {z2} ⇥ . . . ⇥ {zd

} with a, b, z2, . . . , zd in Z. SoB0

n

= [Jnk=1Ik. Let " > 0 be arbitrary, then (using the translation invariance due to stationarity) there is an integer N1

such that for all n � N1 and all k = 1, . . . , Jn

`�1n

X

z2Ik

Gr,s

z

Gu,v

z

� ��

L

1 ".

Consequently, for a certain constant c which is independent of n it is also true that�

n�1P

kn

j=1 Gr,s

zjGu,v

zj� ��

L

1 c".

This proves (4.21). Finally, we consider (4.18) and show that

n�1knX

j=1

Gu,v

zj(B

n

)Gr,s

zj(B

n

) ! � in L1(P). (4.22)

Therefore, we demonstrate first that Gu,v

0 (Bn

)Gr,s

0 (Bn

)�Gr,s

0 Gu,v

0 ! 0 in L1(P). We have

E [|Gr,s

0 (Bn

)Gu,v

0 (Bn

)�Gr,s

0 Gu,v

0 |] E [|Gr,s

0 (Bn

)�Gr,s

0 ||Gu,v

0 (Bn

)�Gu,v

0 |]

+ E [|Gr,s

0 ||Gu,v

0 (Bn

)�Gu,v

0 |] + E [|Gr,s

0 (Bn

)�Gr,s

0 ||Gu,v

0 |] .

But, E⇥

|Gr,s

0 (Bn

)�Gr,s

0 |2⇤

E⇥

|�r,s

0 (Bn

)��

r,s

0 (1)|2⇤

, an analog relation holds for (u, v). Furthermore, as in the

16

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proof of Lemma 4.1 in Yogeshwaran et al. [2017] we find using the geometric lemma that

sup

n2N,z2Zd

E⇥

|�r,s

z

(Bn

)|4⇤

+ E⇥

|�u,v

z

(Bn

)|4⇤

< 1. (4.23)

Using Lemma 4.6 and uniform integrability arguments, we obtain that Gu,v

0 (Bn

)Gr,s

0 (Bn

)�Gr,s

0 Gu,v

0 ! 0 in L1(P).

Next, we show that n�1P

kn

j=1 Gu,v

zj(B

n

)Gr,s

zj(B

n

) ! � in L1. Due to the result from the last paragraph, we onlyneed to show n�1

P

kn

j=1 Gu,v

zj(B

n

)Gr,s

zj(B

n

)�Gu,v

zjGr,s

zj! 0 in L1.

Let " > 0 be arbitrary but fix. Choose L > 0 such that

P(⇢(r_u,s_v)(P,P0 \Q(0)) � L) ". (4.24)

By the bounded moments condition from (4.23), it is clearly sufficient to consider only those z 2 B0n

\ Bn

with adistance of at least L to @B

n

for the convergence result, write B00n

for this set. Then

sup

z2B

00n

E [|Gu,v

z

(Bn

)Gr,s

z

(Bn

)�Gu,v

z

Gr,s

z

|] (4.25)

sup

z2B

00n

E [|Gu,v

z

(Bn

)�Gu,v

z

||Gr,s

z

(Bn

)|+ |Gu,v

z

||Gr,s

z

(Bn

)�Gr,s

z

|] .

Again, by the bounded moments condition from (4.23) and the properties of the conditional expectation, it is sufficientto consider sup

z2B

00nE⇥

|�r,s

z

(Bn

)��

r,s

z

|2⇤

and the same supremum for (u, v).

sup

z2B

00n

E⇥

|�r,s

z

(Bn

)��

r,s

z

|2⇤

sup

z2B

00n

Eh

1{N0(z, (r, s)) � n}⇥�

�r,s

q

(K([P \Bn

� z] \B(0, L)))

� �r,s

q

(K([P00(z) \B

n

� z] \B(0, L)))�⇣

�r,s

q

(K([P \BN0 � z] \B(0, L)))

� �r,s

q

(K([P00(z) \B

N0 � z] \B(0, L)))⌘

2i

(4.26)

+ C P({⇢(r,s)(P \Bk

� z,P0 \Q(z) \Bk

� z) � L, for k = n or k = N0} or

{⇢(r,s)((P \Q(z)) [ (P0 \Q(z)) \Bk

� z,P \Q(z) \Bk

� z) � L, for k = n or k = N0})1/2.(4.27)

Note that (4.26) is zero because we only consider the case where B(z, L) ✓ Bn

✓ BN0 . The square-root of the

probability in (4.27) is bounded by 2

p" because of (4.24).

Proposition 4.7. Let Pn

be a Poisson process with intensity n, where is a general density which satisfies: for all" > 0 there is a blocked density

"

such that sup | � "

| ". Then the finite-dimensional distributions converge to anormal distribution. In particular, for two pairs (u, v), (r, s) 2 � the covariance function satisfies

n�1Cov

�u,v

q

(K(n1/dPn

)),�r,s

q

(K(n1/dPn

))

! Eh

�((X)

1/d(u, v),(X)

1/d(r, s))

i

, n ! 1,

where X is distributed with density and �((u, v), (r, s)) is as in Lemma 4.5.

Proof. Note that once the limiting covariance expression is demonstrated, the claim concerning the finite-dimensionaldistributions is an immediate consequence of Theorem 2.3 in McLeish [1974]: for any finite linear combination ofpersistent Betti numbers criteria (a) and (b) in this theorem are satisfied, this follows from the one-dimensional resultsfrom Propositions 4.2 and 4.4. The convergence in criterion (c) of this very theorem is then guaranteed if the limit of thecovariance exists.

To prove the existence of this limit, we proceed as follows: in the first step, we show similar as in the proof of

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Proposition 4.2 the convergence for blocked intensity functions. In the second step, we proceed as in the proof ofProposition 4.4 and demonstrate that the covariance function of a general intensity can be approximated arbitrarilyclosely by that of a blocked intensity function.

We begin with the first step. Consider two pairs (u, v), (r, s) 2 �. Assume once more that =

P

m

d

i=1 ai 1{Ai

} andconsider the filtration from (4.1). Observe that due to the probabilistic structure of the Poisson process

Cov

�u,v

q

(K(n1/d(P

n

\Ai

))),�r,s

q

(K(n1/d(P

n

\Aj

)))

= 0

if i 6= j. Thus,

Cov

�u,v

q

(

�K(n1/dP

n

)),�r,s

q

(

�K(n1/dP

n

))

=

m

dX

i=1

Cov

�ua

1/di ,va

1/di

q

(K(P \Bn,i

)),�ra

1/di ,sa

1/di

q

(K(P \Bn,i

))

,

where P is a stationary Poisson process with unit intensity and Bn,i

= [�2

�1(nm�da

i

)

1/d, 2�1(nm�da

i

)

1/d]

d. Next,we verify the following two properties: for general (u, v), (r, s) 2 �

n�1Cov

�u,v

q

(K(P \Bn

)),�r,s

q

(K(P \Bn

))

! �((u, v), (r, s)). (4.28)

And

sup

n2N

Cov

�u,v

q

(K(P \Bn,i

)),�r,s

q

(K(P \Bn,i

))

� Cov

�u,v

q

(K(P \B[nm�dai])),�

r,s

q

(K(P \B[nm�dai]))

< 1.

(4.29)

Equation 4.29 is a direct consequence of Lemma 4.3. (4.28), however, follows directly from Lemma 4.5.All in all, we have shown that

n�1Cov

�u,v

q

(

�K(n1/dP

n

)),�r,s

q

(

�K(n1/dP

n

))

!m

dX

i=1

�(a1/di

(u, v), a1/di

(r, s))(m�dai

)

=

Z

[0,1]d�((x)1/d(u, v),(x)1/d(r, s))(x)dx.

(4.30)

We complete the first step of the proof when showing that we can replace�K with the regular filtration K in (4.30). This

follows however immediately from (4.2), where we show that for each pair (r, s) 2 �

n�1Var

�r,s

q

(K(n1/dPn

))� �r,s

q

(

�K(n1/dP

n

))

! 0, n ! 1, (4.31)

and the fact that the limit of n�1Var

�r,s

q

(K(n1/dPn

))

exists by Proposition 4.2 for blocked intensity functions.This finishes the first step and we come to the second step which is the approximation of a regular intensity function

in the covariance function by a blocked intensity function "

. Again, this follows from a previous result. By Proposition4.4, we find for each " > 0 and each pair (r, s) 2 � a blocked intensity function

"

and a coupling of Poisson processes

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Pn

and Qn

with intensities and "

such that

sup

n2Nn�1

Var

h

�r,s

q

(K(n1/dPn

))� �r,s

q

(K(n1/dQn

))

i

= O(").

This implies that for two pairs (u, v), (r, s) 2 �

sup

n2Nn�1

Cov

�r,s

q

(K(n1/dPn

)),�u,v

q

(K(n1/dPn

))

� Cov

�u,v

q

(K(n1/dQn

)),�r,s

q

(K(n1/dQn

))

= O(").

Thus, the proof is complete.

4.2 The binomial case

We apply the classical de-Poissonization trick to obtain the asymptotic normality in the binomical case. To this end, let(U

m,n

: m 2 N) be a sequence of binomial processes for each n 2 N+ such that Um,n

= (Yn,1, . . . , Yn,m

) for an i.i.d.sequence (Y

n,i

: i 2 N+) with common density . We study how

Rr,s

m,n

:

= �r,s

q

(K(n1/dUm+1,n))� �r,s

q

(K(n1/dUm,n

)) ⌘ D0�r,s

q

(K(n1/d(U

m,n

� Yn,m+1))) (4.32)

can be approximated by ↵(r, s) from (3.1) for n large and m close to n. For this we need some coupled random variables.

Lemma 4.8. Let " > 0 and K > 0. Let X 0 be an independent r.v. with density . Let h be a real-valued function suchthat h(n) ! 1 and h(n)/n ! 0 as n ! 1. There is an n0 2 N such that for all n � n0 and m 2 [n�h(n), n+h(n)]

there is a random variable Wm,n

which has the same law as Um,n

and there is a Cox process Hn

with intensity measure(X 0

) on Rd such that

P(n1/d(W

m,n

�X 0) \B(0,K) 6= H

n

\B(0,K)) ". (4.33)

Proof. We can proceed as in the proof of Lemma 3.1 in Penrose and Yukich [2003]. Let K > 0 and " > 0. Let P be ahomogeneous Poisson process with unit intensity on Rd ⇥ [0,1) independent of X 0. Suppose we are given n.

Set P(n) = {x | 9(x, t) 2 P : t n(x)}. Then P(n) is a Poisson process on [0, 1]d with intensity n consistingof N(n) ⇠ Poi(n) points with common density . Discard (N(n) �m)

+ points of P(n) and add (m �N(n))+ i.i.d.points with density to P(n) to obtain W

m,n

. Then Wm,n

has the same distribution as Um,n

Moreover, define Pn

= {(x, t) 2 P : t n(X 0)}. Given X 0

= x, Pn is a homogeneous Poisson process onRd ⇥ [0, n(x)] with intensity 1. Also define H

n

= {n1/d(x�X 0

) : (x, t) 2 Pn} which is a Cox process with intensity(X 0

) on Rd.Let x be a Lebesgue point of , note that almost every point (w.r.t. the Lebesgue measure) has this property as

2 L1. Consider X 0= x and proceed as in the proof of Lemma 3.1 in Penrose and Yukich [2003] to show that

lim

n!1P(P(n) \B(x,Kn�1/d

) 6= (n�1/dHn

+ x) \B(x,Kn�1/d)|X 0

= x) = 0.

Consequently, using dominated convergence, we find that

P(P(n) \B(X 0,Kn�1/d) 6= (n�1/dH

n

+X 0) \B(X 0,Kn�1/d

)) ! 0. (4.34)

In particular, there is an n1 2 N such that the probability in (4.34) is at most "/2. Moreover, for all x 2 [0, 1]d

P(P(n) \B(x,Kn�1/d) 6= W

m,n

\B(x,Kn�1/d)) 2E [|N(n)�m|]n�1 C(n�1

(n1/2+ h(n))), (4.35)

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where the constant is independent of m and also independent of x as sup < 1. Let n2 be such that the left-hand sidein (4.35) is at most "/2. Set n0 = n1 _ n2, then (4.33) holds for all n � n0 and for all m 2 [n� h(n), n+ h(n)].

Lemma 4.9. Let (r, s) 2 �. Let h be a real-valued function such that h(n) ! 1 and h(n)/n ! 0 as n ! 1. Then

lim

n!1sup

n�h(n)mn+h(n)|E⇥

Rr,s

m,n

� ↵(r, s)| = 0.

Proof. We use the coupled random variables from Lemma 4.8 and rewrite ↵(r, s) as E⇥

D0�r,s

q

(K((Hn

\B(0, wn

)))

,

where wn

is ⇢(r,s)(Hn

) from (1.1). Note that

|E⇥

Rr,s

m,n

� ↵(r, s)| Eh

|D0�r,s

q

(K((n1/d(W

m,n

�X 0))))�D0�

r,s

q

(K(Hn

\B(0, wn

)))|i

. (4.36)

Let " > 0. In the first step, we show that

sup

m2[n�h(n),n+h(n)]P(|D0�

r,s

q

(K((n1/d(W

m,n

�X 0))))�D0�

r,s

q

(K((Hn

\B(0, wn

))))| > ") ", (4.37)

n ! 1. The claim follows then, using moment bounds and that the Betti numbers are integer valued. Therefore, definevm,n

= ⇢(r,s)(n1/d

(Wm,n

� X 0). The family of probability measures {P

vm,n : n 2 N,m 2 [n � h(n), n + h(n)]} istight by Proposition A.5. In particular, there is an `0 2 R+ such that sup

n2N,m2[n�h(n),n+h(n)] P(vm,n

� `0) "/3.Moreover, using the tightness of a single probability measure on R and the fact that L(w

n

) = L(w1) for all n 2 N+,there is an `1 2 R+ such that P(w

n

> `1) "/3. Set L = `0 _ `1. Then

P(|D0�r,s

q

(K(n1/d(W

m,n

�X 0)))�D0�

r,s

q

(K((Hn

\B(0, wn

))))| > ")

P(vm,n

� L) + P(w1 � L)

+ P(|D0�r,s

q

(K((n1/d(W

m,n

�X 0) \B(0, L))))�D0�

r,s

q

(K((Hn

\B(0, L))))| > ").

Now by Lemma 4.8, there is an n0 such that the last probability is at most "/3 for all m 2 [n� h(n), n+ h(n)] and forall n � n0. This shows (4.37).

Moreover, as in Lemma 4.1 in Yogeshwaran et al. [2017] both

sup

n2N+,

m2[n�h(n),n+h(n)]

Eh

|D0�r,s

q

(K((n1/d(W

m,n

�X 0))))|4

i

< 1 and sup

n2NE⇥

|D0�r,s

q

(K((Hn

\B(0, wn

))))|4⇤

< 1.

Together with (4.37) this yields that (4.36) vanishes uniformly in m 2 [n�h(n), n+h(n)] and the proof is complete.

In the same manner, it is true that

Lemma 4.10. Let (r, s) 2 �. Let h(n) ! 1 and h(n)/n ! 0 as n ! 1. Then for Rr,s

m,n

from (4.32)

lim

n!1sup

n�h(n)m<m

0n+h(n)

Eh

Rr,s

m,n

Rr,s

m

0,n

i

� ↵(r, s)2�

= 0.

Proof. In the first part of the proof we construct a suitable coupling between certain random variables similar as it isused in the proof of Proposition 3.1 in Penrose and Yukich [2003]. Let n 2 N and n � h(n) m < m0 n + h(n).Let P,Q be independent Poisson processes on Rd ⇥ [0,1) with unit intensity and let X 0, Y 0 be i.i.d. with density andindependent of P and Q.

Define P(n) = {x | 9t such that (x, t) 2 P : t n(x)}. Let N(n) be the number of points of P(n) and choosean ordering of these points at random and denote by Z1, . . . , Z

N(n). Also set ZN(n)+1 = U1, Z

N(n)+2 = U2 and so

20

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on, where Ui

are i.i.d. with density . Set Wm,n

= (Z1, . . . , Zm

) and Wm

0,n

= (Z1, . . . , Zm

, X 0, Zm+1, . . . , Zm

0�1).Clearly, (U

m,n

, Um

0,n

) have the same law as (Wm,n

,Wm

0,n

).Moreover define F

X

0= {x 2 Rd

: d(x,X 0) d(x, Y 0

)} and FY

0= Rd \ F

X

0 . Set Pn

X

0 = {(x, t) 2 P \ (FX

0 ⇥[0, n(X 0

)])} and Qn

X

0 = {(x, t) 2 Q \ (FY

0 ⇥ [0, n(X 0)])}. Given X 0

= x, the point process Pn

X

0 [ Qn

X

0 is ahomogeneous Poisson process of intensity 1 on Rd ⇥ [0, n(x)].

Set HX

0

n

= {n1/d(x�X 0

) | 9t such that (x, t) 2 Pn

X

0 [ Qn

X

0}. Then HX

0

n

is a Cox process with intensity (X 0).

In the same way, define Pn

Y

0 = {(x, t) 2 P \ (FY

0 ⇥ [0, n(Y 0)])} and Qn

Y

0 = {(x, t) 2 Q \ (FX

0 ⇥ [0, n(Y 0)])}.

Set HY

0

n

= {n1/d(x� Y 0

) | 9t such that (x, t) 2 Pn

Y

0 [ Qn

Y

0}.Again, let " > 0 and K > 0 be arbitrary. Then there is an n0 2 N such that for all n � n0 and for all m 2

[n� h(n), n+ h(n)]

P(n1/d(W

m,n

�X 0) \B(0,K) 6= HX

0

n

\B(0,K)) ",

P(n1/d(W

m

0,n

� Y 0) \B(0,K) 6= HY

0

n

\B(0,K)) ".(4.38)

Here, we need that limn!1 P(B(X 0,Kn�1/d

) 6✓ FX

0) = 0 and that lim

n!1 P(B(Y 0,Kn�1/d) 6✓ F

Y

0) = 0,

however, this is evident. We only verify the second probability because it involves the additional point X 0. GivenY 0

= y such that B(y,Kn�1/d) ✓ F

Y

0 , we have

P(Wm

0,n

\B(y,Kn�1/d) 6= Pn

Y

0 \B(y,Kn�1/d) |Y 0

= y)

Eh

(Wm

0,n

\ Pn

Y

0) \B(y,Kn�1/d) |Y 0

= yi

+ Eh

(Pn

Y

0 \Wm

0,n

) \B(y,Kn�1/d) |Y 0

= yi

2n�1(E [|m0 �N(n)|] + 1) C(n�1

(n1/2+ h(n))),

where the constant is independent of m and also independent of y as sup < 1. Moreover, we find as in the proof ofLemma 4.8 that

lim

n!1P((Pn

Y

0 [Qn

Y

0) \B(Y 0,Kn�1/d) 6= (n�1/dHY

0

n

+ Y 0) \B(Y 0,Kn�1/d

)) = 0,

we omit further details for the sake of brevity. This proves both statements in (4.38).Furthermore, define ⇣X

0

n

= D0�r,s

q

(K((HX

0

n

\ B(0, wX

0

n

)))) and ⇣Y0

n

= D0�r,s

q

(K((HY

0

n

\ B(0, wY

0

n

)))), whereagain wX

0

n

= ⇢(r,s)(HX

0

n

) and wY

0

n

= ⇢(r,s)(HY

0

n

). Note that for each n these two ⇣X0

n

and ⇣Y0

n

are independent becauseHX

0

n

and HY

0

n

are independent. This finishes the first part of the proof.In the second part, we show the claim. Since the Betti numbers are integer-valued, we have for " sufficiently small

P(|D0�r,s

q

(K(n1/d(W

m,n

�X 0)))D0�

r,s

q

(K(n1/d(W

m

0,n

� Y 0)))� ⇣X

0

n

⇣Y0

n

| � 1)

P(|D0�r,s

q

(K(n1/d(W

m,n

�X 0)))� ⇣X

0

n

| > ") + P(|D0�r,s

q

(K(n1/d(W

m,n

� Y 0)))� ⇣Y

0

n

| > ").(4.39)

We can again apply the tightness criterion from Proposition A.5 to see that both probabilities on the right-hand sidevanish uniformly in m 2 [n� h(n), n+ h(n)]. Hence, also the probability on the left-hand side vanishes uniformly inm 2 [n� h(n), n+ h(n)]. Next, as m < m0 and due to independence

|Eh

Rr,s

m,n

Rr,s

m

0,n

i

� ↵(r, s)2| Eh

|D0�r,s

q

(K(n1/d(W

m,n

�X 0)))D0�

r,s

q

(K(n1/d(W

m

0,n

� Y 0)))� ⇣X

0

n

⇣Y0

n

|i

.

Note that the right-hand side is uniformly integrable (using the well-known moment bounds for persistent Betti numbers)and vanishes in probability by (4.39). Hence, it also vanishes in L1. This completes the proof.

21

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Proposition 4.11. Let be a general smooth density function and let (r, s) 2 �. Then

n�1/2⇣

�r,s

q

(K(n1/dXn

))� Eh

�r,s

q

(K(n1/dXn

))

i⌘

) N⇣

0,Eh

�2((X)

1/d(r, s))

i

� ↵(r, s)2⌘

, n ! 1,

where X is a random variable with density . Moreover, the finite-dimensional distributions converge to a normaldistribution. In particular, the covariance function of the binomial process satisfies

lim

n!1n�1

Cov

�u,v

q

(K(n1/dXn

)),�r,s

q

(K(n1/dXn

))

= Eh

�((X)

1/d(u, v),(X)

1/d(r, s))

i

� ↵(u, v)↵(r, s).

Proof. First we prove the statement concerning the asymptotic normality and we show the limit of the covariance in thesecond step. We can assume throughout the proof that a Poisson process P

n

of intensity n is given by (X1, . . . , XNn),where the X

i

are i.i.d. on [0, 1] with density and Nn

⇠ Poi(n).The first part of the proof works similar as the proof of Theorem 2.1 in Penrose and Yukich [2001]. We show that

for (r, s) 2 � and Pn

and Xn

= (X1, . . . , Xn

)

n�1E

�r,s

q

(K(n1/dPn

))� �r,s

q

(K(n1/dXn

))� ↵(r, s)(Nn

� n)⌘2�

! 0, n ! 1. (4.40)

Set h(n) = n3/4. We use the definitions from above to rewrite (4.40) as

n�1E

�r,s

q

(K(n1/dUNn,n))� �r,s

q

(K(n1/dUn,n

))� ↵(r, s)(Nn

� n)⌘2�

n�1dn+h(n)eX

m=n

P(Nn

= m)E✓

m�1X

k=n

(Rr,s

k,n

� ↵(r, s))

◆2�

+ n�1n

X

m=bn�h(n)c

P(Nn

= m)E✓

n�1X

k=m

(Rr,s

k,n

� ↵(r, s))

◆2�(4.41)

+ n�1E

�r,s

q

(K(n1/dUNn,n))� �r,s

q

(K(n1/dUn,n

))� ↵(r, s)(Nn

� n)⌘21{|N

n

� n| > h(n)}�

. (4.42)

Consider the first term in (4.41). Note that supn,m2N E

(Rr,s

m,n

)

2⇤

< 1. Using the results of Lemmas 4.9 and 4.10, itis straightforward to show that this term is vanishing. Indeed, let " > 0. Then for m 2 [n, dn+ h(n)e]

E✓

m�1X

k=n

(Rr,s

k,n

� ↵(r, s))

◆2�

"(m� n)2 + c(m� n).

A similar bound holds also if m 2 [bn � h(n)c, n]. Consequently, (4.41) is at most n�1E⇥

"(Nn

� n)2 + c(Nn

� n)⇤

which is bounded by " times a constant for n large enough. Furthermore, one can show that also (4.42) vanishes usingthe moment bounds on Rr,s

m,n

and the exponential decay of P(|Nn

� n| > n3/4). This shows (4.40) and also implies

n�1�

Eh

�r,s

q

(K(n1/dPn

))� �r,s

q

(K(n1/dXn

))

i

! 0. (4.43)

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Combining (4.40) and (4.43) with the results from Propositions 4.2 and 4.4, we find that

n�1/2n

�r,s

q

(K(n1/dXn

)) + ↵(r, s)(Nn

� n)� Eh

�r,s

q

(K(n1/dXn

))

io

) N⇣

0,Eh

�2((X)

1/d(r, s))

i⌘

, n ! 1.(4.44)

Furthermore, as �r,s

q

(K(n1/dXn

)) and Nn

are independent and as n�1/2(N

n

� n) ) N(0, 1) (n ! 1)

n�1/2n

�r,s

q

(K(n1/dXn

))� Eh

�r,s

q

(K(n1/dXn

))

io

) N⇣

0,Eh

�2((X)

1/d(r, s))

i

� ↵(r, s)2⌘

, n ! 1.

This proves the first statement.The second statement is now immediate; with the result from (4.40)

lim

n!1n�1

Cov

�r,s

q

(n1/dXn

),�u,v

q

(n1/dXn

)

= lim

n!1n�1

Cov

�r,s

q

(n1/dPn

)� ↵(r, s)(Nn

� n),�u,v

q

(n1/dPn

)� ↵(u, v)(Nn

� n)⌘

.(4.45)

Consequently, it remains to compute the limit of n�1Cov

Nn

,�r,s

q

(n1/dPn

)

. We make again use of the representationof n1/dP

n

as UNn,n and obtain

n�1Cov

Nn

,�r,s

q

(UNn,n)

= n�1X

k2NP(N

n

= k)(k � n)E⇥

�r,s

q

(Uk,n

)

= n�1X

k2N,|n�k|n

3/4

P(Nn

= k)(k � n)⇣

k�1X

m=n

E [Rm,n

] + E⇥

�r,s

q

(Un,n

)

+ n�1X

k2N,|n�k|>n

3/4

P(Nn

= k)(k � n)E⇥

�r,s

q

(Uk,n

)

.

(4.46)

Note that the second summand in (4.46) vanishes as the Poisson distribution decays exponentially whereas the persistentBetti number admits a polynomial bound. The first summand in (4.46) equals

↵(r, s)n�1X

k2N,|n�k|n

3/4

P(Nn

= k)(k � n)

+ n�1X

k2N,|n�k|n

3/4

P(Nn

= k)(k � n)⇣

k�1X

m=n

E [Rm,n

]� ↵(r, s)⌘

(4.47)

The second summand in (4.47) converges to 0 because

lim

n!1sup

m2N,|n�m|n

3/4

|E [Rm,n

]� ↵(r, s)| = 0.

The first summand in (4.47) converges to ↵(r, s). This proves that limn!1 n�1

Cov

Nn

,�r,s

q

(n1/dPn

)

= ↵(r, s). Inparticular, combining (4.45) with this result yields the desired covariance expression and the proof is complete.

23

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A Appendix

Throughout this appendix we consider a filtration that satisfies K(1) to K(3) from Hiraoka et al. [2018], e.g., the Cech orthe Rips-Vietoris filtration. The following principle will be important.

Lemma A.1. Let K and K0 be two simplicial complexes. Then Cq

(K) \ Cq

(K0) = C

q

(K \K0). Moreover, Z

q

(K) \Zq

(K0) = Z

q

(K \K0) and B

q

(K) \Bq

(K0) ◆ B

q

(K \K0).

Proof. We consider the first claim concerning the spaces Cq

. The inclusion ”◆” is clear and we only prove ”✓”. Thisinclusion can be deduced from the fact that C

q

is a free module over F2 generated by the corresponding q-simplices inthe filtration. We can write c 2 C

q

(K \ K0) as

P

i

ai

�i

, where �i

2 Kq

, ai

2 F2 and asP

j

bj

�j

where �j

2 K 0q

,bj

2 F2. Hence,P

i

ai

�i

�P

j

bj

�j

= 0. If �i

2 K \K0, the coefficient ai

is zero, as this basis element cannot occurin the filtration K0. The same holds in the other direction, if �

j

2 K0 \K, bj

is zero.The amendment Z

q

(K)\Zq

(K0) = Z

q

(K\K0) follows immediately. Again the inclusion ”◆” is clear and we only

prove ”✓”. If c 2 Zq

(K)\Zq

(K0), then by the above c 2 C

q

(K\K0) and by assumption @c = 0. Thus, c 2 Z

q

(K\K0)

as desired. The inclusion concerning the boundary groups is immediate.

We remark that the inclusion Bq

(K) \ Bq

(K0) ✓ B

q

(K \K0) is not true in general. For instance if the sum of all

basis elements in Cq+1(K) and in C

q+1(K0) form two disjoint connected components with the same boundary (”two

arcs”), then we have Bq

(K \K0) = {0}, but B

q

(K) \Bq

(K0) contains a nontrivial element.

However, we can derive some different implications given certain conditions. We assume

Condition A.2. P ✓ U are two simple point clouds on Rd without accumulation points and Q ✓ V are finite subsetsof Rd centered around a z 2 Rd such that Q = Q \B(z, L) and V = V \B(z, L). Moreover, P \Q = ; = U \ V .

Moreover, use the following abbreviations

Ks,a

= Ks

(P \B(z, a)), K0s,a

= Ks

((P [Q) \B(z, a)),

˜Ks,a

= Ks

(U \B(z, a)), ˜K0s,a

= Ks

((U [ V ) \B(z, a)).

Set a⇤ = a⇤(s) = µ(s) +L, where µ(s) is the upper bound on the diameter of a simplex in the filtration at time s whichis guaranteed by the assumptions of Hiraoka et al. [2018] on the filtration and L is an upper bound on the diameter ofthe sets Q and V .

Consider two points a⇤ a1 a2 such that the difference C0(Ks,a2 \Ks,a1) contains exactly one additional pointfrom P and write

Cq+1(K

0s,a2

\K0s,a1

) = h�1, . . . ,�n

i.

W.l.o.g. we can assume that the simplices are already in the right order, i.e.,

Bq

(K0s,a2

) = Bq

(K0s,a1

)� h@�1, . . . , @�i

i, (A.1)

such that @�j

6= 0 mod Bq

(K0s,a1

)� h@�1, . . . , @�j�1i for j = 1, . . . , i and @�j

= 0 mod Bq

(K0s,a1

)� h@�1, . . . , @�i

ifor j = i+1, . . . , n. As a1 is sufficiently large, we have that each of the simplices �

j

is also contained in Ks,a2 . Hence,

as Bq

(Ks,a

) is a subspace of Bq

(K0s,a

), we have that

Bq

(Ks,a2) = B

q

(Ks,a1)� h@�1, . . . , @�i

i � h@�jl : for certain indices i+ 1 j

l

ni. (A.2)

This shows in particular that the map a 7! dimBq

K0s,a

/Bq

Ks,a

is non increasing if a � a⇤. Clearly, the same is true,for the map a 7! dimB

q

˜K0s,a

/Bq

˜Ks,a

.

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Consider the situation where a unique cycle which goes through points in Q is generated when passing from a1 toa2 adding exactly one new point. More precisely,

Lemma A.3. If i < n and if there is a unique 0 6= c 2 Bq

(K0s,a1

)� h@�1, . . . , @�i

i such that for some i+ 1 ` n

@�`

+ c = 0 and c \Q 6= ;, (A.3)

then dimBq

(K0s,a2

)/Bq

(Ks,a2) < dimB

q

(K0s,a1

)/Bq

(Ks,a1).

Proof. Note that @�`

cannot close another cycle in Bq

(K0s,a1

)� h@�1, . . . , @�i

i as this would contradict the minimalityof the added set h@�1, . . . , @�i

i or the fact that c 6= 0. In particular, adding the chain @�`

to the smaller boundary spacecannot close a cycle there, i.e., @�

`

6= 0 mod Bq

(Ks,a1) � h@�1, . . . , @�i

i. This proves that in the assumed situationdimB

q

(K0s,a2

)/Bq

(Ks,a2) < dimB

q

(K0s,a1

)/Bq

(Ks,a1).

In the following we are interested in the probability measures induced by ⇢(r,s)(P,Q) and ⇢(P,Q) from (1.1) and(1.2). We write P

⇢(r,s)(P,Q) and P⇢(P,Q) for these measures.

Lemma A.4. The quantities ⇢(r,s)(P,Q) and ⇢(P,Q) from (1.1) and (1.2) are well-defined.

Proof. It is sufficient to consider ⇢(r,s)(P,Q). One can use the geometric lemma to show that the nonnegative mappings

a 7! dim

Zq

(K0r,a

)

Zq

(Kr,a

)

and a 7! dim

Zq

(K0r,a

) \Bq

(K0s,a

)

Zq

(Kr,a

) \Bq

(Ks,a

)

(A.4)

are bounded above. Moreover, the mapping Zq

(K0r,a1

)/Zq

(Kr,a1) ,! Z

q

(K0r,a2

)/Zq

(Kr,a2) is injective for all 0

a1 a2, this follows from Lemma A.1. Thus, there is an a⇤1 2 R+ such that the first mapping in (A.4), which is integer-valued, is constant for all a � a⇤1. Next, we show that the second mapping in (A.4) becomes also constant as a ! 1.To this end, we first show that a 7! dimB

q

(K0s,a

)/Bq

(Ks,a

) is constant for all a � a⇤2 for a certain a⇤2 2 R+. Thisfollows however from the non-increasing property (see paragraph right after (A.2)) and the boundedness from belowof this mapping. Now, we can return to the second mapping in (A.4), which is also integer-valued. First, assume thatthis map increases at an a � a⇤1 _ a⇤2. So, there is a q-simplex �⇤ such that �⇤ 6= 0 mod B

q

(K0s,a�) and �⇤ 6= 0 mod

Zq

(K0r,a�) but either �⇤

= 0 mod Bq

(Ks,a�) or �⇤

= 0 mod Zq

(Kr,a�). This is a contradiction.

Second, if the map decreases at an a � a⇤1_a⇤2, there is a q-simplex �⇤ such that �⇤ 6= 0 mod Bq

(Ks,a�) and �⇤ 6= 0

mod Zq

(Kr,a�) but either �⇤

= 0 mod Bq

(K0s,a�) or �⇤

= 0 mod Zq

(K0r,a�). Again, this is a contradiction.

Recall �c

= {(r, s) 2 � : s w} for some w 2 R+. We write a⇤(w) = µ(w) + L, where as before µ(w) is anupper bound on the diameter of simplices in �

c and L is an upper bound on the diameter of V from Condition A.2.

Proposition A.5 (Tightness of differences of persistent Betti numbers). Assume Condition A.2. Let (r, s), (u, v) 2 �

c

such that r u, s v. Then ⇢(r,s)(P,Q) ⇢(u,v)(U, V ) _ a⇤ a.s., where a⇤ = a⇤(w).Moreover, let I be an index set. Then

(1) Let (Pi

: i 2 I) be a family of Poisson processes on Rd with intensity functions i

such that supi2I

supi

< 1.If

i

j

for i, j 2 I , then P(⇢(Pi

) � L) P(⇢(Pj

) � L). In particular, the family {P⇢(Pi) : i 2 I} is tight.

(2) Let be a probability density on [0, 1]d such that sup < 1 and let Pn

,P0n

be independent Poisson processeswith intensity n for n 2 N+. Then the family {P

⇢(n1/dPn,n1/dP0

n\Q(z)) : n 2 N+, z 2 Rd} is tight.

(3) Let Xn

be a binomial process on [0, 1]d obtained from an i.i.d. sequence (Xk

: k 2 N+) with common density .Let X 0 be an independent r.v. with density . Write Q

m,n

for the point process n1/d(X

m

� X 0) for m 2 J

n

=

[n � h(n), n + h(n)], where the function h satisfies h(n) ! 1 and h(n)/n ! 0 as n ! 1. Then the family{P

⇢(Qm,n) : n 2 N+,m 2 Jn

} is tight.

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Proof. We begin with the subset principle and consider a pair (r, s) 2 �

c. If P ✓ U and Q ✓ V , where Q and V

are centered around z, then ⇢(r,s)(P,Q) ⇢(r,s)(U, V ) + a⇤ a.s. Indeed, as the definition of ⇢(r,s) is meaningful byLemma A.4, there is an a.s. finite r.v. ⇢ = ⇢(r,s)(U, V ) such that for all a � ⇢

dimZq

(

˜K0r,a

)� dimZq

(

˜Kr,a

) ⌘ dimZq

(

˜K0r,⇢

)� dimZq

(

˜K0r,⇢

)

(A.5)

is constant. In particular, P⇢

is a tight probability measure. Using the inclusions P ✓ U and Q ✓ V , we consider forgeneral a � a0 > 0 the sequence of injective mappings

Zq

(K0r,a

0)

Zq

(Kr,a

0)

f

,!Zq

(K0r,a

)

Zq

(Kr,a

)

g

,!Zq

(

˜K0r,a

)

Zq

(

˜Kr,a

)

h

,!Zq

(

˜K0r,⇢

)

Zq

(

˜Kr,⇢

)

= Qf

. (A.6)

The injectivity of f and g follows from Lemma A.1 and so does the injectivity of h if a ⇢. If a > ⇢, then again

Zq

(

˜K0r,⇢

)

Zq

(

˜Kr,⇢

)

h

0

,!Zq

(

˜K0r,a

)

Zq

(

˜Kr,a

)

.

Together with (A.5) this yields that h0 is actually a bijection with inverse h. Note that we can apply this last argument(i.e., injectivity plus equal ranks) because the base ring F2 is a field and, thus, possesses the invariant basis property.

Next, assume there is a a > ⇢ such that

dimQ1 = dim

Zq

(K0r,⇢

)

Zq

(Kr,⇢

)

< dim

Zq

(K0r,a

)

Zq

(Kr,a

)

= dimQ2.

Then there is a [c] 2 Q2 which is not contained in Q1. However, this [c] is already contained in Qf

. Hence c 2Zq

(

˜K0r,⇢

) \ Zq

(K0r,a

). Consequently by Lemma A.1, c 2 Zq

(K0r,⇢

); this contradicts [c] /2 Q1. Thus, the mappinga 7! dimZ

q

(K0r,a

)/Zq

(Kr,a

) is constant for all a � ⇢.We come to the stabilization of the intersection of the B

q

and the Zq

. First consider the space Bq

only; assume thata 7! dimB

q

(K0s,a

)/Bq

(Ks,a

) stabilizes after ⇢ _ a⇤. We show that this is a contradiction. It suffices to take two pointsa⇤ a1 < a2 such that C0(Ks,a2 \Ks,a1) contains exactly one point from P . W.l.o.g. we can use a representation asin (A.1) and (A.2), i.e.,

Bq

(K0s,a2

) = Bq

(K0s,a1

)� h@�1, . . . , @�i

i and Bq

(Ks,a2) = B

q

(Ks,a1)� h@�1, . . . , @�j

i.

Clearly, dimBq

(K0s,a2

)/Bq

(Ks,a2) < dimB

q

(K0s,a1

)/Bq

(Ks,a1) if and only if j > i. In this situation there is at least

one chain @�`

for i+ 1 ` j which meets the requirements of Lemma A.3. Now, consider the simplicial complexesfrom U as we pass from a1 to a2. We can also find a minimal representation

Bq

(

˜K0s,a2

) = Bq

(

˜K0s,a1

)� h@�1, . . . , @�u

i � h@�1, . . . , @�v

i,

where v i and where �1, . . . , �u

are further chains which originate from (q + 1)-simplices and which contain pointsfrom U \P . By assumption, during this process, the points from the difference U \P cannot generate a cycle which con-tains points from V , otherwise we could apply Lemma A.3 (after replacing K by ˜K) and dimB

q

(

˜K0s,a2

)/Bq

(

˜Ks,a2) <

Bq

(

˜K0s,a1

)/Bq

(

˜Ks,a1) due to monotonicity and this would contradict ⇢ a⇤. However, this implies that adding @�

`

todimB

q

(

˜K0s,a2

) generates a cycle which cannot exist in the set

Bq

(

˜Ks,a2) = B

q

(

˜Ks,a1)� h@�1, . . . , @�u

i � h@�1, . . . , @�v

i.

26

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Thus, @�`

6= 0 mod Bq

(

˜Ks,a1)� h@�1, . . . , @�u

i � h@�1, . . . , @�v

i. Hence,

dimBq

(

˜K0s,a2

)/Bq

(

˜Ks,a2) < B

q

(

˜K0s,a1

)/Bq

(

˜Ks,a1)

which is a contradiction to ⇢ a⇤. So, a 7! dimBq

(K0s,a

)/Bq

(Ks,a

) is constant for all a � ⇢ _ a⇤.Finally, consider the map

a 7! dim

Zq

(K0r,a

) \Bq

(K0s,a

)

Zq

(Kr,a

) \Bq

(Ks,a

)

= dimZq

(K0r,a

) \Bq

(K0s,a

)� dimZq

(Kr,a

) \Bq

(Ks,a

). (A.7)

First, assume that the map in (A.7) increases at an a � ⇢ _ a⇤. Hence, there is a q-simplex �⇤ such that �⇤ 6= 0 modB

q

(K0s,a�) and �⇤ 6= 0 mod Z

q

(K0r,a�) but either �⇤

= 0 mod Bq

(Ks,a�) or �⇤

= 0 mod Zq

(Kr,a�). This is a

contradiction.Second, if the map decreases for some a � ⇢ _ a⇤, there is a q-simplex �⇤ such that �⇤ 6= 0 mod B

q

(Ks,a�) and

�⇤ 6= 0 mod Zq

(Kr,a�) but either �⇤

= 0 mod Bq

(K0s,a�) or �⇤

= 0 mod Zq

(K0r,a�). Again, this is a contradiction.

In the next step, we show that ⇢ = ⇢(r,s)(U, V ) ⇢0 = ⇢(u,v)(U, V ). Again

Zq

(

˜Kr,a

)

Zq

(

˜Kr,a

)

,! Zq

(

˜Ku,a

)

Zq

(

˜Ku,a

)

. (A.8)

Assume now that dimZq

(

˜K0r,⇢

0)/Zq

(

˜Kr,⇢

0) < dimZ

q

(

˜K0r,a

)/Zq

(

˜Kr,a

) for some a � ⇢0. Then there is a

[c] 2Zq

(

˜K0r,a

)

Zq

(

˜Kr,a

)

-

Zq

(

˜K0r,⇢

0)

Zq

(

˜Kr,⇢

0)

.

Then [c] is also contained in Zq

(

˜K0u,a

)/Zq

(

˜Ku,a

) by (A.8). Consequently, as above and by the definition of ⇢0, [c] 2Zq

(

˜K0u,⇢

0)/Zq

(

˜Ku,⇢

0). Hence, c 2 Z

q

(

˜K0u,⇢

0) \ Zq

(

˜K0r,a

). By Lemma A.1 c 2 Zq

(

˜K0r,⇢

0). This is a contradiction to theassumption.

We consider the second difference which comes from the intersection of Bq

with Zq

. Again, we show first that themap a 7! dimB

q

(

˜K0s,a

)/Bq

(

˜Ks,a

) is constant for a � ⇢0 _ a⇤. This works as before, one shows if a1 < a2 and if

dim

Bq

(

˜K0s,a2

)

Bq

(

˜Ks,a2)

< dim

Bq

(

˜K0s,a1

)

Bq

(

˜Ks,a1)

,

then also

dim

Bq

(

˜K0u,a2

)

Bq

(

˜Ku,a2)

< dim

Bq

(

˜K0u,a1

)

Bq

(

˜Ku,a1)

.

Thus, the map

a 7! dim

Zq

(

˜K0r,a

) \Bq

(

˜K0s,a

)

Zq

(

˜Kr,a

) \Bq

(

˜Ks,a

)

is also constant for a � ⇢0_a⇤, we omit the details here. So ⇢ ⇢0_a⇤ and the proof of the subset principle is complete.Proof of (1). Let (r, s) 2 �

c. We begin with the first claim concerning the measures derived from Poisson processes.Let a homogeneous Poisson process P on Rd ⇥ [0,1) with unit intensity. Then w.l.o.g. we can assume that the Poissonprocesses P

i

are given as projections by

Pi

= {x | 9 0 t i

(x) such that (x, t) 2 P}, i 2 I. (A.9)

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Moreover, define the homogeneous Poisson process on Rd with intensity sup

i2I

sup

x2Rd i

(x)

PS

= {x | 9 0 t sup

i2I

sup

x2Rd

i

(x) such that (x, t) 2 P}. (A.10)

In combination with the above, we obtain ⇢(r,s)(Pi

) ⇢(r,s)(PS

) ⇢(w,w)(PS

). In particular, this yields thetightness of the family of probability measures given by {P

⇢(Pi) : i 2 I}. So the first part of the proof is complete.Proof of (2). Let again (r, s) 2 �

c. and begin as follows: let P,P0 be independent Poisson processes with unitintensity on Rd ⇥ [0,1). Set

P(n) = {x | 9 0 t n (n�1(n�1/d ·)) such that (x, t) 2 P},

P0(n) = {x | 9 0 t n (n�1(n�1/d ·)) such that (x, t) 2 P0}

for n 2 N+. Then L(n1/dPn

) = L(P(n)) and L(n1/dP0n

) = L(P0(n)). Also define the stationary and independent

Poisson processes with intensity sup on Rd by

PS

= {x | 9 0 t sup such that (x, t) 2 P},

(P0)

S

= {x | 9 0 t sup such that (x, t) 2 P0}.

Then P(n) ✓ PS and P0(n) ✓ (P0

)

S for all n 2 N+. Consequently, using the above subset principle, we have for eachz 2 Rd with corresponding box Q(z) = (�1/2, 1/2]d + z

⇢(r,s)(n1/dP

n

, n1/dP0n

\Q(z))L= ⇢(r,s)(P(n),P

0(n) \Q(z)) ⇢(r,s)(P

S , (P0)

S \Q(z))

L= ⇢(r,s)(P

S , (P0)

S \Q(0)),

where the last equality is due to the stationarity of the Poisson processes PS and (P0)

S . Hence, it suffices to considerthe case for Q = Q(0)

We can model the Poisson processes (P0)

S\Q by an array of uniformly distributed random variables (Z1, . . . , ZM

) ✓Q, where E [M ] < 1 as the intensity function is essentially bounded and the cube Q is compact.

We can use this construction to extend the difference in the persistent Betti numbers as

�r,s

q

(K([PS [ ((P0)

S \Q)] \B(0, a)))� �r,s

q

(K(PS \B(0, a)))

=

M

X

k=1

�r,s

q

(K([PS [ (

k

[

i=1

Zi

)] \B(0, a)))� �r,s

q

(K([PS [ (

k�1[

i=1

Zi

)] \B(0, a))),

where we use that each set {Zi

} is centered around 0. Consequently, for each L � 0 and each `0 2 N+

P(⇢(r,s)(PS , (P0)

S \Q) � L) P⇣

sup

1kM

⇢(r,s)(PS [ {Z1, . . . , Zk�1}, {Zk

}) � L⌘

`0X

`=0

P(M = `)

`

X

k=1

P(⇢(r,s)(PS [ {Z1, . . . , Zk�1}, {Zk

}) � L) + P(M > `0).

We can choose `0 such that the last probability is small. Consider the first double sum, it is at most

`0X

`=0

P(M = `) ` P(⇢(r,s)(PS [ {Z1, . . . , Z`0�1}, {Z`0}) � L)

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`0X

`=0

P(M = `) ` P(⇢(w,w)(PS [ {Z1, . . . , Z`0�1}, {Z`0}) � L),

this inequality follows from the subset principle applied to ⇢(r,s) and from the fact that the Zi

are i.i.d.Again, ⇢(w,w)(P

S [ {Z1, . . . , Z`0�1}, {Z`0}) is a.s. finite. Moreover,P

`0

`=0 P(M = `) ` E [M ] < 1. Conse-quently, for all " > 0, there is a L > 0 such that sup

n2N+,z2Rd,(r,s)2�c P(⇢(r,s)(n1/dP

n

, n1/dP0n

\ Q(z)) � L) ".This completes the second claim.

Proof of (3). We come to the last claim and use the following coupled processes. Again, let (r, s) 2 �

c. P is aPoisson process with unit intensity on Rd ⇥ [0,1) and set = sup. Set

P(n, i) = {x | 9(i� 1) < t (i� 1)+ (·/n1/d) such that (x, t) 2 P},

PS

i

= {x | 9(i� 1) < t i such that (x, t) 2 P}.

Then P(n, i) ✓ PS

i

for all n 2 N and (P (n, i) : i 2 N+) and (PS

i

: i 2 N+) are independent families of randomvariables. Denote by N

n,i

the number of Poisson points of P (n, i), so Nn,i

⇠ Poi(n).The process n1/dX

m

is coupled to a process Um,n

as follows. The first m � (m � Nn,1)

+ points are taken fromP(n, 1). If N

n,1 < m, then take additional m � Nn,1 � (m � N

n,1 � Nn,2)

+ points from Pn,2 and so forth. This

procedure stops a.s. and yields Um,n

. We have that L(n1/dXm

) = L(Um,n

) and Um,n

✓ Um+1,n. Moreover, set

Z 0n

= n1/dX 0. Then, using again the monotonicity property of ⇢, we obtain

P(⇢(r,s)(Um,n

� Z 0n

) � L) P(⇢(Un+h(n),n � Z 0

n

) � L)

P({⇢(r,s)(P(n, 1) [ P(n, 2)� Z 0n

) � L} \ {Nn,1 +N

n,2 � n+ h(n)})

+ P(Nn,1 +N

n,2 < n+ h(n)).(A.11)

for all m 2 [n� h(n), n+ h(n)] and L > 0. As Nn,1 and N

n,2 are independent, Nn,1 +N

n,2 ⇠ Poi(2n) and

P(Nn,1 +N

n,2 n+ h(n)) exp

� (2n� (n+ h(n)))2

4n� (n+ h(n))

exp

�n

6

,

if n � 264 (with the relation P(X �� x) exp(�x2(�+ x)�1

)). This shows that the second probability in (A.11)is negligible. The first probability in (A.11) is at most

P({⇢(r,s)(PS

1 [ PS

2 � Z 0n

) � L}) P({⇢(r,s)(PS

1 [ PS

2 ) � L}) P({⇢(w,w)(PS

1 [ PS

2 ) � L}),

which is independent of n. So the proof of the last statement is complete.

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