# A Note on the Gaps in the Support of Discretely Infinitely Divisible Laws

Authors: Anthony G. Pakes; S. Satheesh
Journal: ISRN Probability and Statistics
 (2013)

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## Abstract

We discuss the nature of gaps in the support of a discretely infinitely divisible distribution from the angle of compound Poisson laws/processes. The discussion is extended to infinitely divisible distributions on the nonnegative real line.


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## Body

## 1. Introduction

We begin with the following definitions in whichN denotes the set of positive integers and Z+ the set of nonnegative integers.Definition 1.
AZ+-valued random variable (r.v) X is infinitely divisible (ID) if for every n∈N there exists a r.v Xn such that
(1)X=Xn,1+⋯+Xn,n,wheretheXn,i’sareindependentcopiesofXn.Definition 2.
AZ+-valued r.v X is discretely infinitely divisible (DID) if for every n∈N there exists a Z+-valued r.v Xn such that (1) is true.The distinction between the notions of infinite divisibility (ID) of discrete distributions in general and discrete infinite divisibility (DID) is often not made and this sometimes causes confusion. For example, the geometric distribution on{1,2,…} is ID but not DID whereas that on Z+={0,1,2,…} is DID and thus is ID. Blurring these concepts may have lead to Remark 9 in Bose et al. [1] asserting that if a Z+-valued ID distribution assigns a positive probability to the integer 1, then its support cannot have any gaps. Satheesh [2] gave the following simple example to show that this is not always true.Example 3.
Consider the r.vX with probability generating function (PGF) P1(s)=ps/(1-(1-p)sk), k>1 integer. Obviously, P{X=1}>0, but its support has gaps as the atoms are separated by k-1 integers.In the sequelsupp(X) or supp(F) denotes the support of X or that of its distribution function (d.f.) F. To emphasize what now is obvious, if X is ID and ℓX=min{j∈Z+:P{X=j}>0} is its left extremity and if k∈N, then k(X-ℓX) is DID and its support has gaps if k≥2. Also, in terms of Definition 1, ℓXn=ℓX/n, and hence supp(Xn)=supp(X) only if X is DID. To avoid such cases, unless otherwise stated, we assume that supp(X-ℓX) is aperiodic; that is, its greatest common divisor is unity. This natural restriction implies the following definition.Definition 4.
The supportsupp(X) of the ID discrete r.v. X on Z+ contains gaps if it is a proper subset of ℓX+Z+. (Note that supp(X) is necessarily an infinite set).The distinction between Definitions1 and 2 is made clear in Grandell [3, page 26; a result due to Kallenberg], Satheesh [2], and Steutel and van Harn [4, page 23]. Implications of the r.v X being DID include the following.(i)
ℓX=0.(ii)
LetFn be the nth convolution root of F; that is, Fn is the d.f. of Xn in (1). Then supp(X)=supp(Xn) or in terms of their d.f.’s supp(F) = supp(Fn) for all n∈N.(iii)
LetP and Pn denote the PGF’s of X and Xn, respectively. Then P(s)=[Pn(s)]n, |s|≤1, for every n∈N, that is, for all n∈N; the nth root of a DID PGF is again a PGF.(iv)
X is compound Poisson; that is, P(s)=exp{-λ[1-Q(s)]}, where Q is a PGF and the rate parameter λ>0.We conclude this review by noting that Satheesh [2] modified Remark 9 in Bose et al. [1] as follows.Theorem 5.
LetX be a Z+-valued ID r.v with P{X=0}>0. Then supp(X) cannot have any gaps if and only if P{X=1}>0.The purpose of this note is to complement the discussion on the gaps in the support of DID laws in Satheesh [2] from the angle of compound Poisson laws/processes. Some general considerations about the supports of positive ID distributions are discussed in Section 3.


## 2. Compound Poisson Processes and DID

If we think in terms of compound Poisson laws/processes, the picture is clearer. Let{N(t),t≥0} be a Poisson process with rate parameter λ>0 and let {Yi,i≥1} be a sequence of independent copies of a generic jump random variable Y. Then, X(t)=∑i=1N(t)‍Yi describes a compound Poisson process, that is, a subordinator with zero drift and Lévy measure which assigns mass λP(Y=j) to {j}. In keeping with the usual convention, we assume that P(Y=0)=0.Suppose that the support of the jump distribution (ofY) is {yi,y2,…,ym} with all yi’s positive integers and m≤∞. From (iv) above one may identify the correspondence that Pt(s)=exp{-λt[1-Q(s)]} is the PGF of X(t), and we set X=X(1). It is clear that supp(X(t))={n1y1+n2y2+⋯:ni∈Z+} is independent of t. The following examples illustrate possible behaviors of gaps in supp(X).Example 6.
Consider the case whenm=2, y1=1, and y2=3. Here supp(X(t)=Z+; there are no gaps. Although this is obvious, it can be seen better by considering the semigroup elements generated by successively increasing values of the Poisson driving process. So consider the unordered pairwise partitions (n1,n2) of n=0,1,…, respectively. The possible values of X(t) corresponding to the first few values of n are shown in Table 1.Table 1
N
(
t
)
=
n
=
n
1
+
n
2
0
1
2
3
4
5
⋯
X
(
t
)
=
n
1
y
1
+
n
2
y
2
0
1, 3
2, 4, 6
3, 5, 7, 9
4, 6, 8, 10, 12
5, 7, 9, 11, 13, 15
⋯Example 7.
Next consider the case whenm=2, y1=2, and y2=3. As before we now have the possible outcomes in Table 2.
Thus,supp(X(t))={0,2,…} where, here and below, the dots denote all integers larger than the last one written. Clearly, the support omits {1}, a gap.Table 2
N
(
t
)
=
n
=
n
1
+
n
2
0
1
2
3
4
⋯
X
(
t
)
=
n
1
y
1
+
n
2
y
2
0
2, 3
4, 5, 6
6, 7, 8, 9
8, 9, 10, 11, 12
⋯Example 8.
If we consider the case whenm=2, y1=3 and y2=7 then see Table 3.
Thus,supp(X(t))={0,3,6,7,9,10,12,…}, which omits {1,2,4,5,8,11}, so several gaps.Table 3
N
(
t
)
=
n
=
n
1
+
n
2
0
1
2
3
4
⋯
X
(
t
)
=
n
1
y
1
+
n
2
y
2
0
3, 7
6, 10, 14
9, 21, 13, 17
12, 28, 20, 16, 24
⋯Example 9.
If we consider the case whenm=2, y1=4, and y2=9, then (see Table 4).
Thussupp(X(t))={0,4,8,9,12,13,16,17,18,20,21,22,24,…} and there are many gaps.Table 4
N
(
t
)
=
n
=
n
1
+
n
2
0
1
2
3
4
⋯
X
(
t
)
=
n
1
y
1
+
n
2
y
2
0
4, 9
8, 13, 18
12, 17, 22, 27
16, 21, 26, 31, 36
⋯It is important to notice that the widths of successive gaps vary haphazardly but they do tend to become narrower as they approach the infinite region of the support containing all larger integers; that is,supp(X(t)) contains a translate of Z+. In the last example, the gap lengths are (4, 3, 2) and it illustrates that the gap lengths decrease. In general, we have the following result.Theorem 10.
(i)
The support of the compound Poisson processX(t) is ∪{n1y1+⋯+nmym; n1,…,nm∈Z+}; the additive semigroup generated by the yi’s. That is, supp({X(t)})=s.g.(supp(Yi)).
(ii)
Given2≤n∈N the support relation supp(Fn)=supp(F) holds if and only if X is DID, and then it holds for all n.Pakes et al.  [5] alludes to (i) and it is an immediate consequence of remarks preceding Example 6. Part (ii) follows from Implication (iv) because the supports of the components in (1) coincide since they have the same jump distribution; only their rate parameters differ.We end this section with the following formal extension. Suppose now that the support ofYi comprises a set of positive integer multiples of ν∈N. Thus, supp(Yi) has an arithmetic support with span ν; that is, ν is the greatest common divisor of supp(Yi). A number theoretic lemma asserts that there exists an integer j′ such that jν∈s.g.(supp(Yi)) if j≥j′. This same lemma is used to prove that, in a Markov chain, diagonal n-step transition probabilities of an aperiodic state are eventually all positive; see Seneta [6, Lemma A.3]. In this context we say that supp(X) has no gaps if P{X=jν}>0 for all j∈Z+. The following result is now clear, part (i) of which appears in Steutel and van Harn [4, page 53] with a different proof and part (ii) is a formal extension of Theorem 5 in Satheesh [2].Theorem 11.
Suppose thatX is DID with d.f.  F and that supp(Yi) is arithmetic with span ν. Then(i)
supp(F) has at most finitely many gaps.(ii)
supp(F) has no gaps if and if only F({ν})>0.One point of this extension is that if the distribution ofX is gap-free in the sense just defined, then so in an obvious sense is the distribution of k+X for any k∈N. With this extended notion of “gap-free,” Example 3 ceases to be a contradiction of Remark 9 in Bose et al. [1]. Note that the term “gap” is not defined in this reference. However, even with our broader notion, taking X as in Examples 7–9, the distribution of 1+X provides counterexamples to Remark 9.


## 3. Some General Considerations

LetX≥0 be a r.v with d.f. F and Laplace-Stieltjes transform ϕ(θ)=E(e-θX). If X is also ID, then ϕ has the canonical form
(2)ϕ(θ)=E(e-θX)=exp[-ℓθ-∫0∞‍(1-e-xθ)M(dx)],
where ℓ≥0 is a constant and M is a (Lévy) measure with support supp(M)⊂[0,∞) and satisfying ∫0∞‍(1∧x)M(dx)<∞.As above, letFn denote the n’th convolution root of F and ln=sup{x≥0:Fn(x)=0} its left extremity, n∈N. The following result about left extremities generalizes Lemma 1.1 in Seneta and Vere-Jones [7]. Their result is specific to ID distributions and they prove it using an ϵ-δ type of analytic argument. The proof below differs.Lemma 12.
(i)
LetX≥0 be as above. The left extremity ℓX of X is given by
(3)e-ℓX=limθ→∞[ϕ(θ)]1/θ.
(ii)
IfX is also ID with Laplace-Stieltjes transform (2), then ℓn=ℓ/n, n∈N.Proof.
(i)
Forθ>0 the θ-norm of Z=exp(-X) is 𝒩Z(θ)=(EZθ)1/θ=(ϕ(θ))1/θ. A well-known consequence of Jensen’s inequality is that 𝒩Z(θ) is nondecreasing in θ (named Lyapunov’s inequality by some authors). Hence, the limit asserted in (i) exists, and it equals 𝒩Z(∞)=a.s.supZ=e-ℓ; see Loève [8, page 162].
(ii)
Referring to (2) and defining the Lévy tail mass function M¯(v)=M(v,∞), we have
(4)θ-1∫0∞‍(1-e-θx)M(dx)=∫0∞‍∫0x‍e-θvdvM(dx)θ-1∫0∞‍(1-e-θx)M(dx)=∫0∞‍e-θvM¯(v)dv,
and the monotone convergence theorem implies that the right-hand side converges to zero asθ→∞. The assertion follows from (2).It follows thatℓX=ℓ, thus explaining our notation for the so-called drift parameter in (2). We see also that the Fn have a common support only if ℓ=0. We assume this condition in what follows and it is equivalent to the inclusion 0∈supp(F). These equivalent conditions thus generalize Definition 2. In addition, it follows from (2) and monotone convergence that Fn(0)=exp(-n-1M¯(0+)). This is zero if and only if M is an infinite measure in which case ℓM=0; that is, M charges all intervals (0,ϵ) for each ϵ>0. The following lemma is a restatement of a known result.Lemma 13.
Ifℓ=0, then supp(Fn)=[0,∞) if and only if ℓM=0.The direct assertion is proved by Tucker [9] as his Case (i). The “only if” assertion is obvious because if ℓM>0; then supp(Fn) omits at least the interval (0,ℓM), a gap. Lemma 13 follows from the general result asserting that if {X(t):t≥0} denotes the Lévy process determined by M, that is, X(1)=dX, then supp(X(t))={0}∪s.g.(supp(M))¯. See Steutel and van Harn [4, pages 110 and 111] for an essentially analytic proof. Lemma 13 is intuitively obvious from the fact that the sample paths of {X(t)} increase by jumps whose sizes and rates are controlled by M. So if ℓM=0, then the Lévy process can hit any subinterval of (0,∞) with positive probability. This follows in essence because jumps can be arbitrarily small, and hence no interval can be “jumped across” by almost all sample paths.It is worth noting that ifM is an infinite measure and hence that supp(Fn)=[0,∞), then all Fn are continuous and they can be either absolutely continuous or singular continuous (or have both as components). This follows from small changes to constructions such as in Orey [10].So gaps are possible in the general case only for compound Poisson distributions. The following example complements Examples7, 8, and 9 by showing that if M charges an interval, no matter how small, then there are only finitely many gaps, possibly none.Example 14.
Choose numbers0<δ<c<∞ and suppose that supp(M)=[c,c+δ]. Thus,
(5)s.g.(supp(M))=∪k≥1[ck,(c+δ)k].
This set has gaps but only finitely many because, for all sufficiently large integersk, we have δk>c; that is, (c+δ)k>c(k+1), meaning that the gaps 𝒢k=((c+δ)k,c(k+1))=∅. In addition, the lengths |𝒢k|=c-δk of nonempty gaps decrease as k grows.More generally, ifsupp(M) has positive Lebesgue measure, then s.g.(supp(M))¯ contains an interval. It thus follows from Example 14 that s.g.(supp(M))¯ contains a semi-infinite interval and hence so does supp(Fn) for all n. Finally, discrete analogues can be constructed by choosing c and c+δ above to be rational and then supposing that M assigns positive measure to each rational member of [c,c+δ].



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