arXiv:0804.4258v2 [math.PR] 24 Feb 2011 rcs,ifiiedvsblt,Leypoes Peres–Solo distribu process, L´evy of symmetrization number, divisibility, (P.V.) Vijayaraghavan infinite process, 09 o.3,N.1 250–274 1, No. DOI: 37, Vol. Probability2009, of Annals The amn,PttadYr[ Yor and Petit Carmona, where ihiiilcondition initial with
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2. Necessary and sufficient conditions for infinite divisibility. Suppose that (N ,Y ),t 0 is a bivariate L´evy process such that N is a Poisson { t t ≥ } { t} process with parameter a> 0 and Yt is a Poisson process with parameter b> 0. It then follows easily that {(N },Y ) has no Gaussian part, no drift, { t t } and a L´evy measure ν(N,Y ) concentrated on the set (1, 0), (0, 1), (1, 1) , consisting of three points (e.g., [21], Proposition 11.10).{ Denote } u := ν ( (1, 0) ), v := ν ( (0, 1) ), w := ν ( (1, 1) ). (N,Y ) { } (N,Y ) { } (N,Y ) { } Then u, v, w 0, u + w = a and v + w = b. Let ≥ u v w p := , q := , r := , u + v + w u + v + w u + v + w so that p, q, r [0, 1], p + q + r = 1, p + r> 0 and q + r> 0. These give the ∈ normalized L´evy measure on the three points. The two processes Nt and Y are independent if and only if r =0. If r =1, then N = Y for{ all t}with { t} t t probability one. So we call r the dependence parameter of (Nt,Yt) . The law µ in (1.2) depends on c, u, v and w. But it will turn out{ (Proposition} 2.1) that µ depends only on c, q and r. Thus, for c> 1 denote ∞− (2.1) µ := c−Ns− dY . c,q,r L s Z0 If r =1, then
∞− ∞− ∞ −Ns− −Ns− −j c c dYs = c dNs = c = , 0 0 c 1 Z Z Xj=0 − which is degenerate to a constant. So, from now on, we assume that p+q> 0 in addition to the above-mentioned conditions p + r> 0 and q + r> 0. That is, p,q,r < 1. In this section we are interested in whether µc,q,r is infinitely sym divisible or not. It is also of interest whether the symmetrization (µc,q,r) PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 7
sym of µc,q,r is infinitely divisible or not. Recall that the symmetrization µ of a distribution µ is defined to be the distribution with characteristic function µ(z) 2. Infinite divisibility of µ implies that of µsym, but the converse is not |true,| as is mentioned in the Introduction. b We define ρq,r in the following way: If q> 0, denote by σq a geometric k distribution with parameter 1 q, that is, σq( k ) = (1 q)q for k = 0, 1,... and denote − { } − (2.2) ρ = ρ := (1 + r/q)σ (r/q)δ , q,r q − 0 so that ρq,r is a probability distribution concentrated on N0 with (2.3) ρ ( 0 )=(1+ r/q)(1 q) (r/q)= p; q,r { } − − if q = 0, let ρ0,r be a Bernoulli distribution with parameter r (0, 1), that is, ∈ (2.4) ρ ( 1 ) = 1 ρ ( 0 )= r. 0,r { } − 0,r { } Proposition 2.1. We have (2.5) µ (z)= ρ (z)µ (c−1z), z R. c,q,r q,r c,q,r ∈ −1 In particular, µc,q,r is c -decomposable and determined by c, q and r. b b b Proof. As is explained in the Introduction, we have only to show that (Y )= ρ , where T is the time of the first jump of N , that is, the time L T q,r { t} of the first jump of (Nt,Yt) with size in (1, 0), (1, 1) . Let Si be the size of the ith jump of {(N ,Y ) }. Then we have{ for k 1 } { t t } ≥ Y = k [S = = S = (0, 1),S = (1, 1)] T ⇐⇒ 1 · · · k−1 k or [S = = S = (0, 1),S = (1, 0)], 1 · · · k k+1 as well as Y = 0 S = (1, 0). T ⇐⇒ 1 Since
P [Si = (1, 0)] = p, P [Si = (0, 1)] = q, P [Si = (1, 1)] = r, it follows that P (Y =0)= p and, for k 1, P (Y = k)= qk−1r + qkp. From T ≥ T this it follows easily that (YT )= ρq,r for q > 0, while it is a Bernoulli distribution with parameterLr for q = 0.
We can now formulate criteria when ρq,r and µc,q,r and their symmetriza- tions are infinitely divisible. As is seen in (1.4), infinite divisibility of ρq,r 8 A. LINDNER AND K. SATO
sym implies that of µc,q,r. Similarly, infinite divisibility of (ρq,r) implies that sym of (µc,q,r) . The converse of these two implications is by no means clear, as we know Niedbalska-Rajba’s example mentioned in the Introduction. How- ever the following theorem will say that the converse is true for µc,q,r and sym sym ρq,r and for (µc,q,r) and (ρq,r) . Thus infinite divisibility of µc,q,r does sym not depend on c. Another remarkable consequence is that (µc,q,r) can be sym infinitely divisible without µc,q,r being infinitely divisible and that (ρq,r) can be infinitely divisible without ρq,r being infinitely divisible.
Theorem 2.2. Assume that the parameters p, q, r satisfy p,q,r< 1. Let c> 1. For µc,q,r defined in (2.1) and ρq,r in (2.2) and (2.4), the following hold true: (a) If p = 0, then ρq,r and µc,q,r are infinitely divisible. (b) If p> 0 and q> 0, then the following conditions are equivalent: (i) µc,q,r is infinitely divisible. (ii) ρq,r is infinitely divisible. (iii) r pq. ≤ (c) If p> 0, q> 0 and r>pq, then the following conditions are equiva- lent: sym (i) (µc,q,r) is infinitely divisible. sym (ii) (ρq,r) is infinitely divisible. (iii) p qr. ≤ sym sym (d) If q = 0, then none of ρq,r, µc,q,r, (ρq,r) and (µc,q,r) is infinitely divisible.
In the proof, we will first settle the question of infinite divisibility of ρq,r sym and (ρq,r) .
Lemma 2.3. Assume q > 0 and let ρ = ρq,r. Then the following hold true: (a) If r pq, or if p = 0, then ρ is infinitely divisible. (b) If r>pq≤ and p> 0, then ρ is not infinitely divisible. (c) Assume that r>pq and p> 0. Then ρsym is infinitely divisible if and only if (2.6) p qr. ≤
We remark that if 0 α 1, then (1 α)σq + αδ0 is infinitely divisible, since convex combinations≤ of≤ two geometric− distributions are infinitely divis- ible (see pages 379–380 in Steutel and van Harn [22]), and the Dirac measure δ0 is a limit of geometric distributions. Assertions (a) and (b) show to what extent this fact can be generalized to negative α. PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 9
Proof of Lemma 2.3. Since σ (z) = (1 q)/(1 qeiz), we have q − − p + reiz (2.7) ρ(z)= b , z R. 1 qeiz ∈ − k−1 (a) If p =0, then ρ( 0b)=0 and ρ( k ) = (1 q)q for k = 1, 2,... and thus ρ is a geometric distribution{ } translated{ } by− 1, hence infinitely divisible. So assume that r pq. Then p> 0, recalling that p + r > 0. Since p = (1 q)/(1 + r/p), it≤ follows from (2.7) that − r r ρ(z)=exp log(1 q) log 1+ + log 1+ eiz log(1 qeiz) . − − p p − − Henceb ∞ qk r k (2.8) ρ(z)=exp (eikz 1) 1 . " − k − −pq # Xk=1 Recall that r/(bpq) 1. It follows that ρ is infinitely divisible with L´evy ≤−1 k k measure νρ( k )= k q (1 ( r/(pq)) ), k = 1, 2,..., and drift 0. (b) Now assume{ } that r>pq− −and p> 0. By Katti’s criterion ([12] or Corol- ∞ lary 51.2 of [21]), a distribution n=0 pnδn with p0 > 0 is infinitely divisible if and only if there are qn 0, n = 1, 2,..., such that ≥ P n npn = kqkpn−k, n = 1, 2,.... Xk=1 In fact, the equations above determine qn, n = 1, 2,..., successively in a ∞ unique way; infinite divisibility of n=0 pnδn is equivalent to nonnegativity of all qn. Now let pn = ρ( n ). The first two equations are p1 = q1p0 and { } P 2p2 = q1p1 + 2q2p0. Hence q1 = p1/p0 > 0, but 2 2p2 q1p1 (1 + r/q)(1 q)q q2 = − = 2 − [1 q (r/q)(1 + q)] < 0, 2p0 2p − − since r>pq. This shows that ρ is not infinitely divisible. (c) Assume again that r>pq and p> 0. From (2.7) it can be seen that ρ will have a real zero if and only if p = r. In that case, ρ 2 will also have a real zero, and hence ρsym cannot be infinite divisible in| | agreement with theb fact that (2.6) is violated for p = r. So in the followingb we assume that p = r. From (2.7) we have 6 log( ρ(z) 2) = log(p2 + 2pr cos z + r2) log(1 2q cos z + q2). | | − − Write b 2pr 2q p2 + r2 A = , B = , C = . p2 + r2 1+ q2 1+ q2 10 A. LINDNER AND K. SATO
Then 0
2 ixz (2.12) log( ρ(z) )= (e 1 ixz1(−1,1)(x))ν(dx), | | R − − Z b PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 11 where ν is the symmetric signed measure ∞ (2.13) ν = Em(δm + δ−m). m=1 X Let F = r/(pq). Then F > 1. A simple calculation then shows that A B if and only if F 1 q2(F 2 F ), which is equivalent to 1 q2F , that is,≤ (2.6). − ≤ − ≤ Now, if (2.6) holds, then A B and hence Dk 0 for all k, which implies E 0 for all m and ρsym is≤ infinitely divisible≥ with the L´evy–Khintchine m ≥ representation (2.12) with (2.13). If (2.6) does not hold, then A>B, Dk < 0 for all even k, and Em < 0 for all even m, which implies, by (2.12) and (2.13), that ρsym is not infinitely divisible (see Exercise 12.3 of [21]).
Proof of Theorem 2.2. Write µ = µc,q,r and ρ = ρq,r. (a) Suppose p = 0. Then ρ is infinitely divisible by Lemma 2.3, and hence so is µ by (2.5). (b) Suppose that p,q > 0. Under these conditions, the equivalence of (ii) and (iii) follows from Lemma 2.3. Further, (ii) implies (i) by (2.5), so that it remains to show that (i) implies (iii). For that, suppose that r>pq, and in order to show that µ is not infinitely divisible, we will distinguish three cases: p = r, p > r and p < r. The first case is easy, but in the second and third cases we have to use rather involved arguments resorting to different conditions that guarantee noninfinite-divisibility. Case 1: Suppose that p = r. Then ρ will have a real zero as argued in the proof of Lemma 2.3(c). By (2.5), µ will also have a real zero, so that µ cannot be infinitely divisible. b Case 2: Suppose that p > r. Then ρbcan be expressed as in (2.8) with the same derivation. Together with (2.5) and (1.4) this implies ∞ ∞ b −n 1 (2.14) µ(z)=exp (eimc z 1) (qm ( r/p)m) , z R. − m − − ∈ "n=0 m=1 # X X Absoluteb convergence of this double series follows from c> 1, q < 1 and r/p < 1. Define the real numbers a , m N, and the signed measure ν by m ∈ ∞ ∞ 1 m m a := (q ( r/p) ) and ν := a δ −n . m m − − m c m nX=0 mX=1 It follows that µ in (2.14) has the same form as the L´evy–Khintchine rep- resentation with the signed measure ν in place of a L´evy measure, so that infinite divisibilityb of µ is equivalent to the signed measure ν having nega- tive part 0; see Exercise 12.3 in [21]. Thus, to show that µ is not infinitely divisible, we will show that there is a point x such that ν( x ) < 0. Since { } k r/p>q, it follows that am < 0 if m is even and that am > 0 if m is odd. If c 12 A. LINDNER AND K. SATO
−n is irrational for all k N, then the points c m with n N0 and m N are distinct, which implies∈ ν( c−nm ) < 0 for all even m and∈ µ is not infinitely∈ divisible. { } Suppose that ck is rational for some k N. Let k be the smallest such ∈ 0 k and write ck0 = α/β with α, β N such that α and β have no common ∈ t divisor. Let f be the largest t N0 such that 2 divides β. Let m be even. Denote ∈ ′ G := (n′,m′) N N : c−n m′ = m, m′ odd , m { ∈ 0 × } ′ H := (n′,m′) N N : c−n m′ = m, m′ even . m { ∈ 0 × } Then
(2.15) ν( m )= am′ am + am′ . { } ′ ′ ≤ ′ ′ (n ,m )X∈Gm∪Hm (n ,mX)∈Gm We claim that the set Gm contains at most one element. To show this, ′ ′ ′ n′ ′ ′ let (n ,m ) Gm. Then n =0 and c = m /m and thus n = lk0 for some l N. Then∈m′/m = (α/β)l6 and hence βl divides m and m/βl is odd. Thus m∈= 2lf m′′ with some odd integer m′′. It follows that lf is determined by m. Hence l is determined by m and c. Hence n′ is determined by m and c, which shows that Gm contains at most one element. It also follows that f 1 whenever G = ∅ for some even m. If there is some even m such ≥ m 6 that Gm = ∅, then ν( m ) < 0 by (2.15), and we are done. So suppose from { } jf now on that Gm = ∅ for every even m. Let mj = 2 for j = 1, 2,.... The 6 ′ ′ argument above shows that the unique element (nj,mj) in Gmj is given by ′ ′ jk0 nj = jk0 and mj = c mj. Noting that 0
1 mj mj 1 mj amj = (q (r/p) ) (r/p) mj − ≤−2mj and
1 m′ m′ 3 m′ 3 2m a ′ = (q j + (r/p) j ) (r/p) j < (r/p) j . mj ′ ′ mj ≤ 2mj 4mj Thus
1 m m ′ j j ν( mj ) amj + am (r/p) ( 1 + (3/2)(r/p) ) < 0 { } ≤ j ≤ 2mj − for large enough j, showing that µ is not infinitely divisible under the con- ditions of Case 2. Case 3: Suppose that p < r and, by way of contradiction, assume that µ is infinitely divisible. Denote by L (θ)= e−θxµ(dx), θ 0, the Laplace µ R ≥ transform of µ. Then L (θ)= e−ϕ(θ) where ϕ has a completely monotone µ R PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 13 derivative ψ(θ) on (0, ), that is, ( 1)nψ(n)(θ) 0 on (0, ) for n = 0, 1,... (see Feller [7], page 450).∞ By (2.5)− and (2.7) we≥ have ∞ ∞ p + rfn(θ) (2.16) ϕ(θ)= log Lµ(θ)= log , − − 1 qfn(θ) Xn=0 − −θ −n −n where f0(θ)= e and fn(θ) = exp( c θ)= f0(c θ), n = 1, 2,.... Con- − d vergence of the summation in (2.16) is easily established. Since ψ = dθ ϕ −1 −1 d −1 is completely monotone, so is θ c ψ(c θ)= dθ (ϕ(c θ)). Consider the function 7→ p ξ(θ)= , θ (0, ). p + re−θ ∈ ∞
Then 1/(1 qf0(θ)) ξ(θ) is the difference of two completely monotone functions, because− − d d p + rf (θ) 1 (ϕ(θ) ϕ(c−1θ)) = log 0 = ξ(θ). dθ − dθ − 1 qf (θ) 1 qf (θ) − − 0 − 0 ∞ k −kθ Since 1/(1 qf0(θ)) = k=0 q e is completely monotone, ξ(θ) is itself the difference of− two completely monotone functions. Applying Bernstein’s theo- rem, there must exist aP signed measure σ on [0, ) such that e−θx σ (dx) < ∞ [0,∞) | | and ξ(θ)= e−θxσ(dx) for all θ (0, ). However, introducing the ∞ [0,∞) ∈ ∞ R signed measure τ := ∞ ( r/p)kδ , we have R k=0 − k ∞ P r k ξ(θ)= e−θ = e−θxτ(dx), −p [0,∞) Xk=0 Z −θ0x −θ0x if θ>θ0 := log(r/p). Thus e σ(dx) and e τ(dx) have a common Laplace transform ξ(θ0 + θ), θ> 0. Now from the uniqueness theorem in Laplace transform theory (page 430 of Feller [7]) combined with the Hahn–Jordan decomposition of signed measures, it follows that e−θ0xσ(dx)= e−θ0xτ(dx), that is, σ = τ. But e−θx τ (dx)= for 0 <θ θ , contradicting the [0,∞) | | ∞ ≤ 0 property of σ. This finishes the proof of (b). R (c) Suppose that p,q> 0 and that r>pq. The equivalence of (ii) and (iii) then follows from Lemma 2.3, and (ii) implies (i) by (2.5), so that it remains to show that (i) implies (iii). If p = r, then ρ 2 and hence µ 2 have real zeros as shown in the proof of Lemma 2.3(c) and| | µsym is not| infinitely| divisible. Hence we can assume that p = r. With A, Bb, Dk and Em bas in the proof of Lemma 2.3(c), it follows from6 µ(z) 2 = ∞ ρ(c−nz) 2 and (2.10) that | | n=0 | | ∞ ∞ Q (2.17) log( µ(z) 2) = 2 b E (cos(bmc−nz) 1). | | m − n=0 m=1 X X b 14 A. LINDNER AND K. SATO
Since ∞ ∞ 2 E cos(mc−nz) 1 | m|| − | nX=0 mX=1 ∞ ∞ k k = D cos((k 2l)c−nz) 1 | k| l | − − | nX=0 Xk=1 Xl=0 ∞ ∞ ∞ ∞ D 2k(kc−nz)2 z2 c−2n k(Ak + Bk) < , ≤ | k| ≤ ∞ n=0 n=0 X Xk=1 X Xk=1 we can consider the right-hand side of (2.17) as an integral with respect to a signed measure. Thus
2 ixz (2.18) log( µ(z) )= (e 1 ixz1(−1,1)(x))ν(dx), | | R − − Z where ν is the symmetricb signed measure e ∞ ∞ (2.19) e ν = Em(δmc−n + δ−mc−n ). n=0 m=1 X X Now suppose that p >e qr. As observed in the proof of Lemma 2.3(c), this is equivalent to A>B. In order to show that µ is not infinitely divisible, we use Exercise 12.3 of [21] again. We need to show that ν has a nontrivial negative part. Recall that Em > 0 for all odd m and Em < 0 for all even m. k N If c is irrational for all k , then ν( m )= Em < 0 fore even m. Hence, k ∈ { } suppose that c is rational for some k N. We first estimate Em. Since m+2h 2m+2h, it follows from (2.9) and∈ (2.11) that h ≤ e ∞ 1 2Am (2.20) E 2Am+2h , m N. | m|≤ m + 2h ≤ m(1 A2) ∈ Xh=0 − Choose γ (0, 1) such that A/γ < 1, and choose α N such that (α + ∈ ∈ 1/2)/(α + 1) γ. By Stirling’s formula, there exists a constant d1 > 0 such that for every≥m N, ∈ m + 2αm m + 2αm 1/2 (m + 2αm)m+2αm d αm ≥ 1 (m + αm)αm (m + αm)m+αm(αm)αm d 1 (2γ)m+2αm. ≥ (αm)1/2
Since Dk < 0 for every even k, we conclude m + 2αm E D | m| ≥ | m+2αm| αm PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 15
(2.21) d 1 (Aγ)m+2αm(1 (B/A)m+2αm) ≥ (αm)1/2(m + 2αm) − for every even m 2. For even m define G , H , k and f as in the proof of ≥ m m 0 (b)—Case 2. If Gm = ∅ for some even m, then ν( m ) Em < 0 similarly to (2.15). So suppose that G = ∅ for all even m { 2.} As≤ seen in the proof m 6 ≥ of (b), this implies that Gm consists of a single elemente and that f 1. Let m = m = 2jf with j N, then the unique element (n′ ,m′ ) in G ≥satisfies j ∈ j j mj ′ jk0 ′ mj/mj = c . Recall that mj is odd by the definition of Gm. For large j, ′ we then have mj/2 >mj + 2αmj , and from (2.20) and (2.21) it follows that there exists some constant d2 > 0 such that
E ′ mj ′ ′ mj /2 d2 mj(A/γ) 0 as j , Emj ≤ → →∞ | | q so that ν( mj ) Em + E ′ < Em /2 < 0 for large j, finishing the proof { } ≤ j mj j of (c). (d) Suppose q = 0. By (2.4), ρ = ρ0,r is Bernoulli distributed with pa- ∞ −n rameter r. Further, µ = µc,0,r is the distribution of n=0 c Un, where Un,n N is an i.i.d. sequence with distribution ρ. The support of µ is then a{ subset∈ of} [0, c/(c 1)]. It follows that also ρsym andPµsym have bounded support. Moreover− none of them is degenerate to a Dirac measure. Hence they are not infinitely divisible.
sym sym Example 2.4. (a) Let p = q> 0. Then ρq,r, µc,q,r, (ρq,r) and (µc,q,r) will all be infinitely divisible if r [0, 3 2√2], and none of them is infinitely divisible if r> 3 2√2 0.17157.∈ Recall− that r is the dependence parameter. (b) Let 2p = q>− 0. Then≈ ρ and µ will be infinitely divisible for r q,r c,q,r ∈ [0, (13 3√17)/4] and fail to be infinitely divisible for r> (13 3√17)/4 − sym sym − ≈ 0.15767. On the other hand, (ρq,r) and (µc,q,r) are infinitely divisible if and only if r [0, (13 3√17)/4] [1/2, 1). ∈ − ∪ 3. Continuous-singularity and absolute continuity. We continue to study the distribution ∞− µ = c−Ns− dY c,q,r L s Z0 defined by a process (Nt,Yt),t 0 and a constant c> 1 in Section 2. The parameters p, q and {r with p +≥q +} r = 1 are assumed to satisfy p,q,r < 1 and p, q, r 0 throughout this section (see the first paragraph of Section ≥ 2). In this section continuity properties of µc,q,r are considered. Since µc,q,r is c−1-decomposable and nondegenerate, it is either continuous-singular or 16 A. LINDNER AND K. SATO absolutely continuous, as Wolfe’s theorem [25] says. So our problem is to specify the continuous-singular case and the absolutely continuous case. To get complete criteria for the two cases is a difficult problem, far from being achieved. We use two classes of numbers, namely Pisot–Vijayaraghavan (P.V.) num- bers (sometimes called Pisot numbers) and Peres–Solomyak (P.S.) numbers. A number c> 1 is called a P.V. number if there exists a polynomial F (x) with integer coefficients with leading coefficient 1 such that c is a simple root of F (x) and all other roots have a modulus of less than 1. Every positive integer greater than 1 is a P.V. number, but also (1 + √5)/2 and the unique real root of x3 x 1 = 0 are nontrivial examples. There exist countably infinitely many− P.V.− numbers which are not integers. See Peres, Schlag and Solomyak [19] for related information. On the other hand, fol- lowing Watanabe [23], we call c−1 a P.S. number if c> 1 and if there are p0 (1/2, 1) and k N such that the kth power of the characteristic func- tion∈ of the distribution∈ of ∞ c−nU , where U is Bernoulli i.i.d. with n=0 n { n} P [Un = 0] = 1 P [Un =1]= p0, is integrable. Watanabe [23] points out that the paper [20]− of Peres andP Solomyak contains the proof that the set of P.S. numbers in the interval (0, 1) has Lebesgue measure 1. However, according to [23], an explicit example of a P.S. number is not known so far. As follows from the results of [23], the set of P.V. numbers and the set of reciprocals of P.S. numbers are disjoint.
Theorem 3.1. Assume that c is a P.V. number and that q> 0. Then µc,q,r is continuous-singular.
Recall that the assumption q > 0 merely excludes the case of infinite Bernoulli convolutions.
Proof of Theorem 3.1. Write µ = µc,q,r. The following proof of continu- ous-singularity of µ is based on an idea of Erd˝os [5]. It is enough to show that it is not absolutely continuous. Thus, by virtue of the Riemann–Lebesgue theorem, it is enough to find a sequence z such that k →∞ lim sup µ(zk) > 0. k→∞ | | By the definition of a P.V. number,b there is a polynomial F (x)= xN + a xN−1 + + a x + a such that a , . . ., a Z, F (c)=0, and the N−1 · · · 1 0 N−1 0 ∈ totality α1,...,αN of roots of F (x) satisfies α1 = c and αj < 1 for 2 { } k | | ≤ j N. Choose zk = 2πc . Now we divide the proof into three cases: (Case 1)≤p> 0 and r pq; (Case 2) p = 0; (Case 3) p> 0 and r>pq. Recall that q> 0 is always≤ assumed. PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 17
Case 1: As in the proofs of Theorem 2.2(b)—Case 2, we have ∞ ∞ −n (3.1) µ(z)=exp (eimc z 1)a , z R, − m ∈ "n=0 m=1 # X X with b (3.2) a = m−1qm(1 ( r/(pq))m) 0. m − − ≥ The double series above is absolutely convergent. We then have ∞ ∞ (3.3) µ(z) = exp (1 cos(mc−nz))a . | | − − m " n=0 m=1 # X X Thus b ∞ ∞ ∞ µ(z ) = exp (1 cos(2πmck−n))a = exp (S +R )a | k | − − m − m m m " n=0 m=1 # " m=1 # X X X withb k ∞ S = (1 cos(2πmck−n)),R = (1 cos(2πmck−n)). m − m − n=0 X n=Xk+1 Now k k N S = (1 cos(2πmcn)) = 1 cos 2πm α n , m − − j n=0 n=0 !! X X Xj=2 since cn = N α n N α n and N α n is an integer. The latter is j=1 j − j=2 j j=1 j a consequence of the symmetric function theorem in algebra (e.g., Lang P P PN n [15], Section IV.6), implying that j=1 αj , as a symmetric function of α1,...,αN , can be expressed as a polynomial with integer coefficients in the elementary symmetric functions, whichP are integer valued themselves since F has integer coefficients with leading coefficient 1. Choose 0 <δ< 1 such that α < δ for j = 2,...,N. Then, with some constants C ,C ,C , | j| 1 2 3 k N 2 k N S C m α n C m2 α 2n m ≤ 1 j ≤ 2 | j| n=0 ! n=0 X Xj=2 X Xj=2 k C m2 δ2n C m2/(1 δ2). ≤ 3 ≤ 3 − n=0 X Further, we have ∞ R C (mc−n)2 = C m2/(c2 1). m ≤ 1 1 − n=1 X 18 A. LINDNER AND K. SATO
Hence, it follows that ∞ 2 C3 C1 µ(zk) exp amm 2 + 2 . | |≥ "− 1 δ c 1 # mX=1 − − ∞ 2 This shows thatb limsupk→∞ µ(zk) > 0, since m=1 amm < . Case 2: We have | | ∞ ∞ P b qm ρq,r(z) = exp (cos mz 1) | | " − m # mX=1 by the remark at theb beginning of the proof of Lemma 2.3(a). Hence the situation is the same as in Case 1. Case 3: Recall the proof of Lemma 2.3(c). We have ∞ ρ (z) 2 = exp 2 E (cos mz 1) | q,r | m − " m=1 # X with Em of (2.11).b Hence ∞ ∞ ∞ µ(z) 2 = ρ (c−nz) 2 exp 2 E+(1 cos(mc−nz)) , | | | q,r | ≥ − m − n=0 " n=0 m=1 # Y X X wherebE+ = max Eb , 0 . We have ∞ E+m2 < , since m { m } m=1 m ∞ ∞ ∞ ∞ P m + 2h E+m2 m2 D m ≤ | m+2h| h m=1 m=1 X X Xh=0 ∞ k = m2 D | k| (k m)/2 mX=1 k−mXeven ≥0 − ∞ ⌊(k−1)/2⌋ ∞ k k k2 D (Ak + Bk) < , ≤ | k| l ≤ 2 ∞ Xk=1 Xl=0 Xk=1 ⌊(k−1)/2⌋ k k−1 −1 −k k k noting that l=0 l 2 and Dk k 2 (A + B ) with 0 < ≤ | |≤ 2 A< 1 and 0
Recall that Lebesgue almost all c (1, ) are the reciprocals of P.S. num- bers. ∈ ∞
Proof of Theorem 3.2. Let µ = µc,q,r. Let p0 (1/2, 1) and k N as in the definition of a P.S. number. The following proof∈ was suggested∈ by an argument of Watanabe [23]. Let K := k log(2p0 1) /2, which is positive. Then (2.4) of [23] tells us that | − | ∞ ∞ (3.4) exp α (cos(c−nu) 1) du < whenever α K. − ∞ ≥ −∞ ( n=0 ) Z X Under the condition that p> 0, q> 0 and r pq, we have (3.3) with am of ∞ ≤ (3.2). Let α0 = m=1 am. Then it follows from Jensen’s inequality that ∞ ∞ ∞ ∞ P 1 −n µ(z) dz = 2 exp am α0 (cos(mc z) 1) dz −∞ | | 0 "α0 − !# Z Z mX=1 nX=0 b ∞ ∞ ∞ 1 −n 2 am exp α0 (cos(mc z) 1) dz ≤ 0 "α0 − !# Z mX=1 nX=0 ∞ ∞ 2 a ∞ = m exp α (cos(c−nu) 1) du. α m 0 − 0 m=1 ! 0 n=0 ! X Z X −1 m The last integral is finite whenever α0 K by (3.4). We have am m q for m odd, and it follows that α tends≥ to as q 1. Thus there is≥ε = ε(c) 0 ∞ ↑ such that α0 K for all q 1 ε. Hence µ has bounded continuous density whenever p>≥0, r pq and≥ q− 1 ε. The case when p =0 and q 1 ε ≤ −≥1 m− ≥ − follows similarly, with am = m q in the above calculations. To see the second half of the theorem, suppose that q C p C r with ≥ 1 ≥ 2 C1,C2 > 0. Then p> 0 since p + r> 0, and hence q> 0. Thus p r −1 p C p −1 1 1 −1 q = 1+ + 1+ + 1 1+ + . q q ≥ q C q ≥ C C 2 1 2 Hence, q 1 ε(c) if C and C are large enough. We also have ≥ − 1 2 r C 1 1 1 1+ + . pq ≤ C C C 2 1 2 Hence r/(pq) 1 if C1 is fixed and C2 is large. Thus there are C1 and C2 such that q ≤1 ε(c), p> 0 and r/(pq) 1 whenever q C p C r. ≥ − ≤ ≥ 1 ≥ 2 Now we use the entropy H(ρ) of a discrete probability measure ρ on R. Here discrete means that ρ is concentrated on a countable set. We define H(ρ) := ρ( a ) log ρ( a ), − { } { } aX∈C 20 A. LINDNER AND K. SATO where C is the carrier (the set of points with positive mass) of ρ.
Theorem 3.3. Assume that q> 0. Then 1 1 1 q + r 1 (3.5) H(ρ ) = (q + r) log + log log + p log , q,r 1 q 1 q q − q p − − where p log(1/p) is understood to be zero for p = 0. The following are true: (a) The Hausdorff dimension of µc,q,r is estimated as H(ρ ) (3.6) dim(µ ) q,r . c,q,r ≤ log c
(b) For each c> 1, there exists a constant C3 = C3(c) > 0 such that µc,q,r is continuous-singular whenever max q, r C p. { }≤ 3 (c) Fix q and r. Then there exists a constant C4 = C4(q, r) > 0 such that µ is continuous-singular whenever c C . c,q,r ≥ 4
The estimate (3.6) is meaningful only when H(ρq,r)/ log c< 1, as the Hausdorff dimension of any measure on the line is less than or equal to 1. In this case (3.6) not only tells the continuous-singularity of µc,q,r, but also gives finer information on a set of full measure for µc,q,r.
Proof of Theorem 3.3. Recall that ρq,r is defined by (2.2). The ge- ometric distribution σq has entropy q (3.7) H(σ )= log(1 q) log q q − − − 1 q − and H(ρq,r) is readily calculated as H(ρ )=(1+ r/q)[H(σ ) + (1 q) log(1 q) q log(1 + r/q)] p log p, q,r q − − − − which shows (3.5). (a) Applying the remarkable Theorem 2.2 of Watanabe [23] on c−1-decompo- sable distributions, we obtain (3.6). (b) Notice that, since µ is continuous-singular or absolutely continuous, it must be continuous-singular if its Hausdorff dimension is less than 1. Suppose that max q, r C3p with C3 > 0. Then p> 0 and hence p =(1+ −1 { }≤−1 q/p + r/p) (1 + 2C3) , which tends to 1 as C3 0. Hence q 0 and r 0 as C ≥0. Thus → → → 3 → 1 q 1 1 1 H(ρ ) = (q + r) log + log + log + p log 0 q,r 1 q 1 q q q + r p → − − and hence sup H(ρ ):max q, r C p 0, C 0. { q,r { }≤ 3 }→ 3 → PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 21
This shows (b). (c) For given q, r, take any C4 > exp H(ρq,r). Then the assertion follows from the estimate (3.6).
It is known that the distributions of some L´evy processes have time evolu- tion, that is, change their qualitative properties as time passes (see Chapters 5 and 10 of Sato [21] and Watanabe [24]). It was Watanabe [23] who showed that the distributions of some semi-self-decomposable processes have time evolution in continuous-singularity and absolute continuity. It is of interest that the L´evy process Zt,t 0 determined by the distribution µc,q,r when it is infinitely divisible{ gives≥ an} explicit example of time evolution of this sort as the following theorem shows. Note that (Z ) = (µ )t∗ for t 0. L t c,q,r ≥ Theorem 3.4. Assume that either p> 0, q> 0 and r pq or p = 0. ≤ Write ρ = ρq,r and µ = µc,q,r. Then the following are true: (a) There are t1 = t1(c, q, r) and t2 = t2(c, q, r) with 0
Lemma 3.5. If ρ is a distribution on Z with finite absolute moment of order 1+ ε for some ε> 0, then its entropy H(ρ) is finite.
Proof. Let ρ = ∞ p δ . Then ∞ m 1+εp < . Hence m=−∞ m m m=−∞ | | m ∞ there is a constant C > 0 such that p C m −1−ε. The function f(x)= P m P x log(1/x) is increasing for 0 x e−1. Hence,≤ | | ≤ ≤ H(ρ) p log(1/p )+ C m −1−ε log((C m −1−ε)−1) < ≤ m m | | | | ∞ |mX|≤m0 |mX|>m0 with an appropriate choice of m0. 22 A. LINDNER AND K. SATO
Proof of Theorem 3.4. Observe that under our assumption µ is in- finitely divisible (see Theorem 2.2), so that µt∗ is definable for all t 0. If p> 0 and r pq, it follows from (2.8) that ρ is a compound Poisson distribu-≥ tion, concentrated≤ on N , with finite second moment, since ∞ m2a < 0 m=1 m ∞ for am = νρ( m ). If p =0, then ρ is a geometric distribution shifted by 1 (see the proof{ of} Lemma 2.3). In both cases, H(ρ) < byP Lemma 3.5. The property that µ is c−1-decomposable is preserved to∞ convolution powers, since (1.3) implies
µt∗(z)= ρt∗(z)µt∗(c−1z) for any t 0. Thus we have ≥ c c c ∞ ∞ ∞ t∗ t∗ −n 0 imc−nz (3.9) µ (z)= ρ (c z)=exp itγp z + t (e 1)am , − ! nY=0 nX=0 mX=1 c0 c 0 ∞ −n where γp =0 for p> 0 and γp = n=0 c = c/(c 1) for p =0. If p> 0, q> 0 and r =0, then ρ is geometric. If p = 0, then the− result is the same as in the case of ρ being a geometricP distribution, since shifts do not change entropy, Hausdorff dimension and continuity properties. Hence, from now on, we exclude the case p = 0. Let us prove (d) first. We obtain the properties of H(ρt) from Proposition 5.1 of Watanabe [23] or Exercise 29.24 of Sato [21]. Applying Theorem 2.2 of [23] to the law µt∗, we get (3.8). (a) It follows from (d) that µt∗ is continuous-singular for all sufficiently t∗ small t> 0. Let t1 be the supremum of t> 0 for which µ is continuous- ′ t∗ singular. Then 0
In the rest of this section we consider the case where Nt and Yt are independent, that is, the case where r = 0. The following{ theorem} is{ largely} a repetition of Theorems 3.1–3.3 in this case but, since the L´evy measure of µc,q,0 is increasing with respect to q, we obtain stronger statements. PROPERTIES OF STATIONARY LAW OF GENERALIZED O–U PROCESSES 23
Theorem 3.6. Under the condition that 0
Proof. Let us begin with the proof of (d). The estimate (3.10) follows from (3.6) of Theorem 3.3. The expression (3.11) is exactly (3.7), since ρq,0 = σq. (e) and (f) These come from (d), as a distribution with Hausdorff dimen- sion < 1 cannot be absolutely continuous. We get the sufficient condition (3.12), since 1 1 1 1 H(ρ )= (1 q) log + q log log 2 q,0 1 q − 1 q q ≤ 1 q − − − by strict concavity of the function log x. (a) Recall that µc,q,0 has L´evy measure ∞ ∞ qm ν = δ −n . c,q,0 m c m n=0 m=1 X X 24 A. LINDNER AND K. SATO
′ Hence, if q Example 3.7. (a) In the case c = e, µe,q,0 is continuous-singular if q 1 log 2 0.30685. ≤ − ≈ This follows from (3.12) in Theorem 3.6. (b) In the case q = 1/2, µc,1/2,0 is continuous-singular if c> 4 since H(ρ1/2,0)= 2 log 2, as (f) of Theorem 3.6 says. In the independent case (0 Proposition 3.8. If 0 Proof. Write ρ = ρq,0. Since ρ equals geometric distribution σq with parameter p, the distribution ρt∗, t> 0, is a negative binomial distribution with parameters t and p, that is, t∗ t t k ρ ( k )= − p ( q) , k N0. { } k − ∈ To estimate H(ρt∗) from above, observe that tptqk/k ρt∗( k ) tqk for 0 ∞ k 2 where we used k=1 kq = q/p and ∞ ∞ k P 1 1 1 qk log k qk = log , ≤ n p p Xk=1 Xk=1 nX=1 compare Gradshteyn and Ryzhik [10], Formula 1.513.6. Recalling that pt p since t 1 and this can be further estimated to ≥ ≤ 2 1 q 1 q 1 H(ρt∗) t log + log + log . ≤ p p p t p2 q Together with (q/p) log(1/q) = (q/p) log(1 + p/q) 1, this gives (3.13). ≤ Acknowledgments. 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