Buraczewski et al. (2023) proved a functional limit theorem (FLT) and a law of the iterated logarithm (LIL) for a random Dirichlet series ∑k≥2(logk)αk1/2+sηk as s→0+, where α>−1/2 and η1,η2,… are independent identically distributed random variables with zero mean and finite variance. A FLT and a LIL are proved in a boundary case α=−1/2. The boundary case is more demanding technically than the case α>−1/2. A FLT and a LIL for ∑pηpp1/2+s as s→0+, where the sum is taken over the prime numbers, are stated as the conjectures.
Functional limit theoremlaw of the iterated logarithmrandom Dirichlet series60F1560F1760G50The present work was supported by the National Research Foundation of Ukraine (project 2023.03/0059 ‘Contribution to modern theory of random series’).Introduction
Recently there has been a surge of activity around limit theorems for random Dirichlet series and their zeros. Throughout the paper, by random Dirichlet series, we mean a random series parameterized by s>0:
Xα(s):=∑k≥2(logk)αk1/2+sηk,
where α∈R and η1, η2,… are independent copies of a random variable η with zero mean and finite variance, which live on a probability space (Ω,F,P). By Kolmogorov’s three series theorem, the series defining Xα(s) converges almost surely (a.s.) for each s>0. Furthermore, if α<−1/2, the series Xα(0+) converges a.s. by the same theorem. Thus, as far as limit theorems for Xα(s) as s→0+ are concerned, the case α<−1/2 is not interesting, hence excluded in the sequel.
The following functional limit theorem (FLT) and law of the iterated logarithm (LIL) were known in the case α>−1/2. We write ⟹ for weak convergence in a function space, and C(0,∞) and C[0,∞) for the spaces of real-valued continuous functions defined on (0,∞) and [0,∞), respectively. It is assumed that the spaces C(0,∞) and C[0,∞) are endowed with the topology of locally uniform convergence.
Assume thatE[η]=0,σ2=E[η2]∈(0,∞)and letα>−1/2. Then(s1/2+α∑k≥2(logk)αk1/2+stηk)t>0⟹(σ∫[0,∞)yαe−tydB(y))t>0,s→0+,onC(0,∞), where(B(y))y≥0is the standard Brownian motion.
Proposition 1 follows from Corollary 2.3 in [5]. In the cited article, the result was derived by a specialization of a FLT for Xα(s), with a complex-valued η, in the space of analytic functions.
For a family (xt) of real numbers denote by C((xt)) the set of its limit points.
Assume thatE[η]=0,σ2=E[η2]∈(0,∞)and letα>−1/2. Then, almost surely,C(((22ασ2Γ(1+2α)s1+2αloglog1/s)1/2∑k≥2(logk)αk1/2+sηk:s∈(0,e−1)))=[−1,1],where Γ is the Euler gamma function.
Proposition 2 can be found in Theorem 3.1 of [5]. A classical form of the LIL in terms of lim sup and lim inf was earlier obtained in Theorem 1.1 of [2] in a rather particular case P{η=±1}=1/2 and α=0. Nevertheless, we stress that the work [2] gave impetus to both [5] and the present paper.
Our purpose is to formulate and prove counterparts of Propositions 1 and 2 in a boundary case α=−1/2.
In the case α=0, real zeros of random Dirichlet series have been in focus of the recent papers [1, 3, 9] (it is assumed in [9] that the distribution of η is symmetric γ-stable for γ∈(0,2]). In the case α>−1/2, limit theorems for complex and real zeros of s↦Xα(s) were proved in [5]. Although we do not directly investigate zeros of s↦X−1/2(s) in the present paper, our LIL stated in Theorem 2 below entails that, a.s., s↦X−1/2(s) has infinitely many real zeros in any right neighborhood of 0.
Main results
We start by stating a FLT for X−1/2(s), properly scaled, as s→0+. If α>−1/2, the variance of Xα(s) exhibits a polynomial growth, whereas the growth of Var[X−1/2(s)] is logarithmic. This partly justifies the facts that the scaling of time in Proposition 1 is st, that is, standard, whereas the scaling of time in Theorem 1 is st.
Assume thatE[η]=0andσ2=E[η2]∈(0,∞). Then(1(log1/s)1/2∑k≥2(logk)−1/2k1/2+stηk)t≥0⟹(σB(t))t≥0,s→0+,onC[0,∞), where(B(t))t≥0is the standard Brownian motion.
Observe that the limit process in Theorem 1 is nowhere differentiable a.s., whereas the limit process in Proposition 1 is infinitely differentiable a.s. This distinction is a manifestation of intricacy of the boundary case α=−1/2.
We proceed with a LIL for X−1/2(s) as s→0+. The FLT given in Theorem 1 was used for guessing the LIL’s form, namely, the factor logloglog1/s in the normalization.
Since Var[X−1/2(s)]∼σ2log1/s as s→0+ (see the beginning of Section 3), we infer
loglogVar[X−1/2(s)]∼logloglog1/s,s→0+,
where, as usual, f(s)∼g(s) as s→0+ means that lims→0+(f(s)/g(s))=1. Thus, formulas (2) and (3) are indeed laws of the iterated logarithm.
One of the referees has kindly informed us that the results of Theorems 1 and 2 should hold with ∑pηpp1/2+s replacing ∑k≥2(logk)−1/2k1/2+sηk, where ∑p denotes the summation over the prime numbers. Based on this comment we formulate the following conjectures.
Assume thatE[η]=0andσ2=E[η2]∈(0,∞). Then(1(log1/s)1/2∑pηpp1/2+st)t≥0⟹(σB(t))t≥0,s→0+,onC[0,∞), where(B(t))t≥0is the standard Brownian motion.
At the moment we do not see a way to effectively estimate the difference ∑pηpp1/2+s−∑k≥2(logk)−1/2k1/2+sηk. Thus, Conjectures 1 and 2 do not seem to follow from Theorems 1 and 2. On the other hand, we think that both conjectures can be justified by a proper modification of the proofs of Theorems 1 and 2. Some of the modifications required are nonobvious and need a substantial technical work. As a consequence, proofs of these conjectures will be given elsewhere. Once Conjecture 2 has become a theorem, it provides a significant improvement over Theorem 1.3 in [6].
The remainder of the paper is structured as follows. Theorems 2 and 1 are proved in Sections 3 and 4, respectively. The reversed order of the proofs is necessitated by the fact that our proof of Theorem 1 uses some arguments and calculations from the proof of Theorem 2. At the first glance it looks plausible that an economical proof of the LIL may be based on a strong approximation by a Brownian motion of the standard random walk generated by η, that is, the random sequence (Tn)n≥0 defined by T0:=0, Tn:=η1+⋯+ηn for n∈N. In Section 5 we explain that this naive idea fails.
Proof of Theorem <xref rid="j_vmsta276_stat_004">2</xref>
Put
g(s):=E[(∑k≥2(logk)−1/2k1/2+sηk)2]=σ2∑k≥2(logk)−1k1+2s,s>0.
We start by showing that g(s)∼σ2log(1/s) as s→0+. By monotonicity,
∫2∞(logx)−1x1+2sdx≤∑k≥2(logk)−1k1+2s≤(log2)−121+2s+∫2∞(logx)−1x1+2sdx.
Plainly, lims→0+(log2)−121+2slog1/s=0. Changing the variable we infer
∫2∞(logx)−1x1+2sdx=∫2slog2∞e−yydy∼log1/s,s→0+.
It is convenient to split the presentation into two pieces. We proceed by proving one half of Theorem 2. In what follows we write log(3) for logloglog.
Under the assumptions of Theorem2,lim sups→0+(1log1/slog(3)1/s)1/2∑k≥2(logk)−1/2k1/2+sηk≤(2σ2)1/2a.s.andlim infs→0+(1log1/slog(3)1/s)1/2∑k≥2(logk)−1/2k1/2+sηk≥−(2σ2)1/2a.s.
Replacing ηk with ηk/σ we can work under the assumption that σ2=1. For s∈(0,e−e), put
f(s)=(12log1/slog(3)1/s)1/2.
Let M:(0,∞)→N0 denote a function satisfying lims→0+M(s)=+∞ and
lims→0+M(s)log1/s=0.
Here, as usual, N0:=N∪{0}.
In Lemma 1, we remove from the series in focus an initial fragment with a vanishing contribution. In all the lemmas given below we assume without further notice that the assumptions of Theorem 2 are in force.
The following limit relation holds:lims→0+f(s)∑k=2M(s)(logk)−1/2k1/2+sηk=0a.s.
Recall that T0=0 and Tn=η1+⋯+ηn for n∈N. According to the LIL for standard random walks,
|Tn|≤maxk≤n|Tk|=O((nloglogn)1/2),n→∞a.s.
Observe that
∑k=2M(s)(logk)−1/2k1/2+sηk=∫(3/2,M(s)]dT⌊x⌋(logx)1/2x1/2+s.
Integrating by parts we obtain
∫(3/2,M(s)](logx)−1/2x1/2+sdT⌊x⌋=(logM(s))−1/2TM(s)(M(s))1/2+s−(log3/2)−1/2η1(3/2)1/2+s+∫3/2M(s)((logx)−3/2/2+(1/2+s)(logx)−1/2)T⌊x⌋x3/2+sdx.
Relation (6) entails lims→0+(M(s))s=1. This in combination with (7) enables us to conclude that, as s→0+,
(logM(s))−1/2|TM(s)|(M(s))1/2+s∼(logM(s))−1/2|TM(s)|(M(s))1/2=O((logM(s))−1/2(loglogM(s))1/2)=o(1).
Since lims→0+f(s)=0, the latter ensures that
lims→0+f(s)(logM(s))−1/2|TM(s)|(M(s))1/2+s=0a.s.
To complete the proof, it is sufficient to show that
lims→0+f(s)∫3/2M(s)(logx)−1/2|T⌊x⌋|x3/2+sdx=0a.s.
To this end, write
∫3/2M(s)(logx)−1/2|T⌊x⌋|x3/2+sdx≤(maxk≤M(s)|Tk|)∫3/2M(s)(logx)−1/2x3/2+sdx=O((M(s)loglogM(s))1/2)O(1)=O((M(s)loglogM(s))1/2),s→0+a.s.,
having utilized (7) for the penultimate equality. Since (6) entails
lims→0+M(s)loglogM(s)log1/slog(3)1/s=0,
the claim follows. □
For k∈N, ρ>0 and s∈(0,e−e), define the event
Ak,ρ(s):={|ηk|>ρloglog1/s(k1+2slogkg(s)log(3)1/s)1/2}.
Next, we demonstrate that the second (remaining) part of the series also vanishes if the variables ηk are properly truncated.
For allρ>0,lims→0+∑k≥M(s)+1(logk)−1/2k1/2+s|ηk|1Ak,ρ(s)=0a.s.andlims→0+∑k≥M(s)+1(logk)−1/2k1/2+sE[|ηk|1Ak,ρ(s)]=0.
Put h(s):=(loglog1/s)(log(3)1/s)1/2. Observe that, for k≥3 and s≥0, k2slogk≥1. Hence, for s∈(0,e−e),
∑k≥M(s)+1(logk)−1/2k1/2+s|ηk|1Ak,ρ(s)≤∑k≥M(s)+1|ηk|k1/21{k−1/2|ηk|>ρ(g(s))1/2(h(s))−1}a.s.
The assumption E[η2]<∞ entails limk→∞k−1/2|ηk|=0 a.s. and thereupon supk≥1(k−1/2|ηk|)<∞ a.s. Since lims→0+((g(s))1/2(h(s))−1)=+∞, we infer
1{k−1/2|ηk|>ρ(g(s))1/2(h(s))−1}≤1{supk≥1(k−1/2|ηk|)>ρ(g(s))1/2(h(s))−1}=0
a.s. for small s. We have proved that the sum in (8) is equal to 0 a.s. for small enough s.
Relation (9) is justified as follows:
∑k≥M(s)+1(logk)−1/2k1/2+sE[|ηk|1Ak,ρ(s)]≤∑k≥M(s)+1k−1/2E[|η|1{ρ−1(g(s))−1/2h(s)|η|>k1/2}]≤E[|η|∑k=1⌊ρ−2(g(s))−1(h(s))2η2⌋k−1/2]≤2ρ−1E[η2](g(s))−1/2h(s)=2ρ−1(g(s))−1/2h(s)→0,s→0+.
The proof of Lemma 2 is complete. □
In what follows, (Ak,ρ(s))c denotes the complement of Ak,ρ(s), that is, for k∈N, ρ>0 and s∈(0,e−e),
(Ak,ρ(s))c:={|ηk|≤ρloglog1/s(k1+2slogkg(s)log(3)1/s)1/2}.
Our next result is related to the fragment of the series giving the principal contribution. However, this is a lighter version of what we really need, for the convergence here is only along a sequence.
Since (1−γ)(1+γ)2>1 whenever γ∈(0,(5−1)/2), ρ satisfying (10) does indeed exist.
Put f∗(s):=(2g(s)log(3)1/s)−1/2 for s∈(0,e−e). Since f∗(s)∼f(s) as s→0+, we can and do prove the result, with f∗ replacing f. Put
X(s)=f∗(s)∑k≥M(s)+1(logk)−1/2η˜k,ρ(s)k1/2+s,s∈(0,e−e).
Using ex≤1+x+(x2/2)e|x| for x∈R and E[η˜k,ρ(s)]=0 we deduce, for u∈R,
E[euX(s)]=∏k≥M(s)+1E[exp(uf∗(s)(logk)−1/2η˜k,ρ(s)k1/2+s)]≤∏k≥M(s)+1(1+u2(f∗(s))22(logk)−1k1+2s×E[(η˜k,ρ(s))2exp(|u|f∗(s)(logk)−1/2|η˜k,ρ(s)|k1/2+s)]).
The inequality
|η˜k,ρ(s)|≤|ηk|1(Ak,ρ(s))c+E[|ηk|1(Ak,ρ(s))c]≤2ρk1/2+sloglog1/s(logkg(s)log(3)1/s)1/2≤2ρk1/2+s(logkg(s)log(3)1/s)1/2a.s.,
which is valid for integer k≥2 and s∈(0,e−e), implies that
exp(|u|f∗(s)(logk)−1/2|η˜k,ρ(s)|k1/2+s)≤exp(2ρ|u|log(3)1/s)a.s.
Together with the inequalities E[(η˜k,ρ(s))2]≤1 and 1+x≤ex for x∈R this gives, for u∈R,
E[euX(s)]≤∏k≥M(s)+1exp(u2(f∗(s))22(logk)−1k1+2sexp(2ρ|u|log(3)1/s))≤exp(u24log(3)1/sexp(2ρ|u|log(3)1/s)).
An application of Markov’s inequality yields, for u≥0,
P{X(sn)>1+γ}≤e−(1+γ)uE[euX(sn)]≤exp(−(1+γ)u+u24log(3)1/snexp(2ρulog(3)1/sn)).
Putting u=2(1+γ)log(3)1/sn we conclude that
P{X(sn)>1+γ}≤exp(−(1+γ)2(2−exp(22(1+γ)ρ))log(3)1/sn)=1n(1−γ)(1+γ)2(2−exp(22(1+γ)ρ)).
Thus, in view of (10), ∑n≥1P{X(sn)>1+γ}<∞, and invoking the direct part of the Borel–Cantelli lemma completes the proof of Lemma 3. □
Now we present our final, and the most involved, preparatory result. It shows that the convergence along a sequence discussed in Lemma 3 can be lifted to the full convergence along the real numbers.
Let(sn)n∈Nbe as defined in Lemma3, whereγ∈(0,1/2), andM(s)=⌊log1/s/loglog1/s⌋fors∈(0,e−e). Fors∈[sn+1,sn], the following limit relation holds:limn→∞f(s)(∑k≥M(s)+1(logk)−1/2k1/2+sηk−∑k≥M(sn+1)+1(logk)−1/2k1/2+sn+1ηk)=0a.s.
Using the fact that M is a nonincreasing function for the arguments close to 0, write, for s∈[sn+1,sn],
∑k≥M(s)+1(logk)−1/2ηkk1/2+s−∑k≥M(sn+1)+1(logk)−1/2ηkk1/2+sn+1=∑k=M(s)+1M(sn+1)(logk)−1/2ηkk1/2+s+∑k≥M(sn+1)+1(logk)−1/2(1k1/2+s−1k1/2+sn+1)ηk=:In,1(s)+In,2(s).
Analysis ofIn,1(s). Actually, we shall prove that
limn→∞sups∈[sn+1,sn]|In,1(s)|=0a.s.
This relation is even more than we need because lims→0+f(s)=0. We obtain with the help of summation by parts
In,1(s)=(logM(sn+1))−1/2TM(sn+1)(M(sn+1))1/2+s−(log(M(s)+1))−1/2TM(s)(M(s)+1)1/2+s+∑k=M(s)+1M(sn+1)−1((logk)−1/2k1/2+s−(log(k+1))−1/2(k+1)1/2+s)Tk,
where, as before, Tn=η1+⋯+ηn for n∈N. Invoking formula (7) and limn→∞(M(sn+1))sn+1=1 we obtain
(logM(sn+1))−1/2|TM(sn+1)|(M(sn+1))1/2+s≤(logM(sn+1))−1/2|TM(sn+1)|(M(sn+1))1/2+sn+1∼(logM(sn+1))−1/2|TM(sn+1)|(M(sn+1))1/2=O((logM(sn+1))−1/2(loglogM(sn+1))1/2)→0,n→∞a.s.
By a similar argument we conclude that
limn→∞sups∈[sn+1,sn](log(M(s)+1))−1/2|TM(s)|(M(s)+1)1/2+s=0a.s.
Further, for s∈[sn+1,sn] and large n,
|∑k=M(s)+1M(sn+1)−1((logk)−1/2k1/2+s−(log(k+1))−1/2(k+1)1/2+s)Tk|≤∑k=M(s)+1M(sn+1)−1((logk)−1/2k1/2+s−(log(k+1))−1/2(k+1)1/2+s)|Tk|≤(supj≤M(sn+1)|Tj|)∑k=M(s)+1M(sn+1)−1((logk)−1/2k1/2+s−(log(k+1))−1/2(k+1)1/2+s)≤(supj≤M(sn+1)|Tj|)(logM(s))−1/2(M(s))1/2+s≤(supj≤M(sn+1)|Tj|)(logM(sn))−1/2(M(sn))1/2+sn+1=O((logM(sn+1))−1/2(loglogM(sn+1))1/2)→0,n→∞a.s.
We have used (7), limn→∞(M(sn+1)/M(sn))=1 and limn→∞(M(sn))sn+1=1 for the equality.
Analysis ofIn,2(s). One can show that
limn→∞sups≥sn+1∑k≥M(sn+1)+1(logk)−1/2(1k1/2+sn+1−1k1/2+s)|ηk|1{|ηk|>k1/2logn}=0a.s.
and
limn→∞sups≥sn+1∑k≥M(sn+1)+1(logk)−1/2(1k1/2+sn+1−1k1/2+s)×E[|ηk|1{|ηk|>k1/2logn}]=0.
For instance, relation (13) follows from supk≥1(k−1/2|ηk|)<∞ a.s. and
∑k≥M(sn+1)+1(logk)−1/2(1k1/2+sn+1−1k1/2+s)|ηk|1{|ηk|>k1/2logn}≤∑k≥M(sn+1)+1|ηk|k1/21{k−1/2|ηk|>logn}a.s.
The summands on the right-hand side are equal to 0 for large enough n. Here, we have used k2sn+1logk≥1 for k≥3 and n∈N. More details can be found in the proof of Lemma 2.
Put ηˆk(n):=ηk1{|ηk|≤k1/2logn}−E[ηk1{|ηk|≤k1/2logn}] for k,n∈N. For n∈N and small positive u, put
Yn∗(u):=∑k≥M(sn+1)+1(logk)−1/2ηˆk(n)k1/2+u.
In view of (13) and (14), it suffices to demonstrate that, for each s∈[sn+1,sn],
limn→∞f(sn)(Yn∗(s)−Yn∗(sn+1))=0a.s.
By a technical reason to be explained below, we shall show that, for each v∈[vn+1,vn],
limn→∞f(sn)(Yn(v)−Yn(vn+1))=0a.s.,
where Yn(v):=Yn∗(exp(−1/v)), vn:=1/(log1/sn)=exp(−n1−γ) for n∈N and v>0.
For j∈N0 and n∈N, put
Fj(n):={tj,m(n):=vn+1+2−jm(vn−vn+1):0≤m≤2j}.
Note that Fj(n)⊆Fj+1(n) and put F(n):=⋃j≥0Fj(n). The set F(n) is dense in the interval [vn+1,vn]. For any u∈[vn+1,vn], put
uj:=max{v∈Fj(n):v≤u}=vn+1+2−j(vn−vn+1)⌊2j(u−vn+1)vn−vn+1⌋.
Then limj→∞uj=u (we omit the dependence on n in the notation). An important observation is that either uj−1=uj or uj−1=uj−2−j(vn−vn+1). Consequently, uj=tj,m for some 0≤m≤2j, which implies that either uj−1=tj,m or uj−1=tj,m−1. Since Yn is a.s. continuous on [vn+1,vn], we obtain
|Yn(u)−Yn(vn+1)|=liml→∞|Yn(ul)−Yn(vn+1)|=liml→∞|∑j=1l(Yn(uj)−Yn(uj−1))+Yn(u0)−Yn(vn+1)|≤liml→∞∑j=0lmax1≤m≤2j|Yn(tj,m)−Yn(tj,m−1)|=∑j≥0max1≤m≤2j|Yn(tj,m)−Yn(tj,m−1)|.
Thus, our purpose is to prove that, for all ε>0 and sufficiently large n0∈N,
∑n≥n0P{∑j≥0max1≤m≤2jf(sn)|Yn(tj,m)−Yn(tj,m−1)|>ε}<∞.
Put aj:=(j+1)2−j/2 for j∈N0. In view of ∑j≥0aj<∞, it suffices to show that, for all ε>0,
∑n≥n0∑j≥0P{max1≤m≤2jf(sn)|Yn(tj,m)−Yn(tj,m−1)|>εaj}<∞.
Now we proceed similarly to the proof of Lemma 3 and refer to that proof regarding any missing details. Write, for u∈R and sufficiently large n,
E[exp(±u(Yn(tj,m)−Yn(tj,m−1)))]=E[exp(±u∑k≥M(sn+1)+11(klogk)1/2(1kexp(−1/tj,m)−1kexp(−1/tj,m−1))ηˆk(n))]≤∏k≥M(sn+1)+1(1+u221klogk(1kexp(−1/tj,m)−1kexp(−1/tj,m−1))2×E[(ηˆk(n))2exp(|u|(klogk)1/2(1kexp(−1/tj,m)−1kexp(−1/tj,m−1))|ηˆk(n)|)].
Now we prove that, for large n, all j∈N0 and integer m∈[0,2j],
A(j,n):=∑k≥M(sn+1)+11klogk(1kexp(−1/tj,m)−1kexp(−1/tj,m−1))2≤2−j(vn−vn+1)vn+12.
For fixed a,b>0, the function x↦x−1e−x(e−ax−e−bx)2 is decreasing on (1,∞). Indeed, x↦xe−x is decreasing on (1,∞), and
x↦e−min(a,b)xx−2(1−e−(max(a,b)−min(a,b))x)2
is decreasing on (0,∞) as the product of two positive decreasing functions. As far as monotonicity of the second function is concerned, observe that, up to a multiplicative constant, it is the Laplace–Stieltjes transform of the Lebesgue–Stieltjes convolution of the uniform distribution on [0,1] with itself. Using this argument with x=logk, a=exp(−1/tj,m) and b=exp(−1/tj,m−1) we conclude that the function of argument k under the sum defining A(j,n) is decreasing, whence
A(j,n)≤∫e∞1xlogx(1x2exp(−1/tj,m−1)+1x2exp(−1/tj,m)−2xexp(−1/tj,m−1)+exp(−1/tj,m))dx=∫2exp(−1/tj,m−1)∞e−xxdx+∫2exp(−1/tj,m)∞e−xxdx−2∫exp(−1/tj,m−1)+exp(−1/tj,m)∞e−xxdx=∫2exp(−1/tj,m−1)1e−xxdx+∫2exp(−1/tj,m)1e−xxdx−2∫exp(−1/tj,m−1)+exp(−1/tj,m)1e−xxdx.
Put I(y):=∫0yx−1(1−e−x)dx for y∈[0,1] and observe that
∫y1x−1e−xdx=−logy−I(1)+I(y),y∈(0,1].
The function x↦x−1(1−e−x) is decreasing on (0,∞) as the Laplace–Stieltjes transform of the uniform distribution on [0,1]. Hence, I is concave on [0,1] and thereupon, for a,b∈[0,1], I(2a)+I(2b)−2I(a+b)≤0. As a consequence,
A(j,n)≤−2log2+1tj,m−1+1tj,m+2log(e−1/tj,m−1+e−1/tj,m)≤tj,m−tj,m−1tj,m−1tj,m≤2−j(vn−vn+1)vn+12,
which proves the claim.
Next we work towards estimating
B(u,k,n):=exp(|u|(klogk)1/2(1kexp(−1/tj,m−1)−1kexp(−1/tj,m))|ηˆk(n)|)
for k≥M(sn+1)+1. Assume first that 2j≥vn+1−2(vn−vn+1). Observe that, for a>0 and 0<s<t,
0≤exp(−ae−1/s)−exp(−ae−1/t)≤aexp(−ae−1/s)e−1/t(1/s−1/t).
Using this inequality with a=logk, t=tj,m−1 and s=tj,m we obtain
0≤1kexp(−1/tj,m−1)−1kexp(−1/tj,m)≤logkkexp(−1/tj,m−1)e−1/tj,m(1tj,m−1−1tj,m).
In view of the inequality xe−ax≤(ea)−1 for x≥0, we conclude that the right-hand side does not exceed
1ee1/tj,m−1−1/tj,m(1tj,m−1−1tj,m)≤1eexp(2−j(vn−vn+1)vn+12)2−j(vn−vn+1)vn+12≤2−j(vn−vn+1)vn+12,
where the last inequality is justified by our present choice of j. Thus,
B(u,k,n)≤|u|2−jlogn(logM(sn+1))1/2vn−vn+1vn+12=:Cj(u,n)
whenever k≥M(sn+1)+1.
Assume now that 2j≤vn+1−2(vn−vn+1) for nonnegative integer j. Using a trivial estimate
1kexp(−1/tj,m−1)−1kexp(−1/tj,m)≤1
we infer
B(u,k,n)≤|u|logn(logM(sn+1))1/2=:Cj(u,n)
whenever k≥M(sn+1)+1.
Using now 1+x≤ex for x∈R we arrive at
E[exp(±u(Yn(tj,m)−Yn(tj,m−1)))]≤exp(u222−j(vn−vn+1)vn+12eCj(u,n)),u∈R.
By Markov’s inequality and the inequality eu|x|≤eux+e−ux for x∈R,
P{f(sn)|Yn(tj,m)−Yn(tj,m−1)|>εaj}≤exp(−uεajf(sn))E[exp(u|Yn(tj,m)−Yn(tj,m−1)|)]≤2exp(−uεajf(sn)+u222−j(vn−vn+1)vn+12eCj(u,n)).
Put
u=ε2j/2f(sn)vn+12vn−vn+1andkn:=1(f(sn))2vn+12vn−vn+1.
In view of
(logM(sn+1))−1/2∼nγ/2−1/2,(vn+1vn−vn+1)1/2∼(1−γ)−1/2nγ/2,n→∞,
and
limn→∞(log1/sn)vn+1=1,
we conclude that, for j satisfying 2j≥vn+1−2(vn−vn+1),
Cj(u,n)=|u|2−jlogn(logM(sn+1))1/2vn−vn+1vn+12=ε2−j/2lognf(sn)(logM(sn+1))1/2≤εlognf(sn)(logM(sn+1))1/2vn+1(vn−vn+1)1/2=ε21/2((log1/sn)vn+1)1/2(log(3)1/sn)1/2logn(logM(sn+1))1/2(vn+1vn−vn+1)1/2∼ε21/2nγ−1/2(logn)3/2→0,n→∞.
For nonnegative integer j satisfying 2j≤vn+1−2(vn−vn+1), Cj(u,n) admits the same upper bound:
Cj(u,n)=|u|logn(logM(sn+1))1/2=ε2j/2lognf(sn)(logM(sn+1))1/2vn+12vn−vn+1≤εlognf(sn)(logM(sn+1))1/2vn+1(vn−vn+1)1/2→0,n→∞.
Therefore, for sufficiently large n such that kn>ε−2log2 and eCj(u,n)≤3/2,
∑j≥0P{max1≤m≤2jf(sn)|Yn(tj,m)−Yn(tj,m−1)|>εaj}≤∑j≥02j2exp(−ε2(j+1)kn)exp(3ε2kn/4)=2exp(−ε2kn/4)1−2exp(−ε2kn).
Since kn∼2nγlogn as n→∞, (16) follows.
Now we comment on the reason behind the passage from Yn∗ to Yn via a logarithmic transformation. In order to have in the present context a successful approximation with the help of a dyadic partition of an interval [λn+1,λn], say, the endpoints should satisfy limn→∞(λn/λn+1)=1. This is the case for λn=vn and not the case for λn=sn. □
We are prepared to prove Proposition 3.
We only prove (4) because (5) follows from (4) by replacing ηk with −ηk. Recall our convention that σ2=1.
Choose any γ>0 sufficiently close to 0, put sn=exp(−exp(n1−γ)) for n∈N and select ρ=ρ(γ) such that (10) holds true. Let M(s)=⌊log1/s/loglog1/s⌋ for s∈(0,e−e). By Lemmas 2 and 3 and the fact that E[ηk1(Ak,ρ(s))c]=−E[ηk1Ak,ρ(s)],
lim supn→∞f(sn)∑k≥M(sn)+1(logk)−1/2k1/2+snηk≤1+γa.s.
Lemma 4 in combination with relation (18) secures
lim sups→0+f(s)∑k≥M(s)+1(logk)−1/2k1/2+sηk≤1+γa.s.
Finally, by Lemma 1,
lim sups→0+f(s)∑k≥2(logk)−1/2k1/2+sηk≤1+γa.s.,
which entails (4) because the left-hand side does not depend on γ. □
Under the assumptions of Theorem2,lim sups→0+(1log1/slog(3)1/s)1/2∑k≥2(logk)−1/2k1/2+sηk≥(2σ2)1/2a.s.andlim infs→0+(1log1/slog(3)1/s)1/2∑k≥2(logk)−1/2k1/2+sηk≤−(2σ2)1/2a.s.
Proposition 4 will be proved by using Lemmas 1 and 2, together with two additional lemmas. In the first of these, we show that an initial and a final fragments of the series in question do not contribute to the LIL.
As before, we assume without further notice that the assumptions of Theorem 2 hold true and that σ2=1. When proving Proposition 4 we use the sets Ak,ρ(s) and the corresponding variables η˜k,ρ(s) with ρ=1.
Fix anyγ>0and putsn:=exp(−exp(n1+γ))forn≥1. LetN1andN2be functions which take positive integer values, may depend on γ and satisfyN1(s)≥2for small positive s,lims→0+(loglogN1(s)/log1/s)=0,lims→0+(loglogN2(s)/log1/s)=1andlims→0+slogN2(s)=0. Thenlimn→∞f(sn)∑k=2N1(sn)(logk)−1/2k1/2+snη˜k,1(sn)=0a.s.andlimn→∞f(sn)∑k≥N2(sn)+1(logk)−1/2k1/2+snη˜k,1(sn)=0a.s.
As in the proof of Lemma 3, we obtain the result, with f∗ replacing f. For s>0 close to 0, put
Z1(s):=f∗(s)∑k=2N1(s)(logk)−1/2k1/2+sη˜k,1(s).
The reasoning used to derive both (21) and (22) is similar to the one applied in the proof of Lemma 3. Therefore, we give a proof of (21) and indicate the only minor change needed for a proof of (22).
Regarding (21), in view of the direct part of the Borel–Cantelli lemma, it is sufficient to demonstrate that, for all ε>0,
∑n≥1P{Z1(sn)>ε}<∞.
To this end, we obtain a counterpart of (12), for u∈R,
E[euZ1(s)]≤exp(u2(f∗(s))22∑k=2N1(s)(logk)−1k1+2sexp(2|u|log(3)1/s)).
For each fixed s>0, the function x→(logx)−1x−1−2s is decreasing on (1,∞). The assumption lims→0+(loglogN1(s)/log1/s)=0 entails lims→0+slogN1(s)=0. Hence,
∑k=2N1(s)(logk)−1k1+2s≤(log2)−121+2s+∫2N1(s)(logx)−1x1+2sdx=O(1)+∫(2log2)s2slogN1(s)y−1e−ydy=O(loglogN1(s)),s→0+.
Our choice of N1 guarantees that there exists an r>0 close to 0 and such that
∑k=2N1(s)(logk)−1k1+2s≤rg(s)
for small positive s, and also (1−δ)(1+γ)>1, where δ:=r(4ε2)−1exp(2ε−1) with ε appearing in (23). Then, for u∈R and small s>0,
E[euZ1(s)]≤exp(ru2g(s)(f∗(s))22exp(2|u|log(3)1/s))=exp(ru24log(3)1/sexp(2|u|log(3)1/s)).
Applying Markov’s inequality with u=(1/ε)log(3)1/sn yields, for large n,
P{Z1(sn)>ε}≤e−uεE[euZ1(sn)]≤exp(−(1−r(4ε2)−1e2ε−1)log(3)1/sn)=1n(1−δ)(1+γ),
thereby proving (23) and hence (21).
The proof of (22) is analogous to that of (21). The corresponding version of (24) reads
∑k≥N2(s)+1(logk)−1k1+2s≤∫N2(s)∞(logx)−1x1+2sdx=∫2slogN2(s)∞y−1e−ydy=o(1)
as s→0+. □
Lemma 6 treats the component of the series in question which gives a principal contribution to the LIL. Our proof is based on the converse part of the Borel–Cantelli lemma, which requires independence. The independence requirement complicates to some extent a selection of the essential component.
Fix sufficiently smallδ>0, pickγ>0satisfying(1+γ)(1−δ2/8)<1and let, as before,sn=exp(−exp(n1+γ))forn∈N. Thenlim supn→∞f(sn)∑k≥2(logk)−1/2k1/2+snη˜k,1(sn)≥1−δa.s.
Functions s↦loglogN1(s)/log1/s in Lemma 5 can tend to 0 as fast as we please. Observe that limn→∞(sn+1/sn)=∞. Thus, we can choose N1 satisfying
loglogN1(sn+1)log1/sn=loglogN1(sn+1)log1/sn+1log1/sn+1log1/sn→+∞,n→∞.
Let N2 be any function satisfying the assumptions of Lemma 5. Since
limn→∞loglogN2(sn)log1/sn=1,
we conclude that there exists n0∈N such that
loglogN1(sn+1)log1/sn≥loglogN2(sn)log1/snfor alln≥n0,
whence
N1(sn+1)≥N2(sn),n≥n0.
In view of Lemma 5, it is sufficient to check that
lim supn→∞Z2(sn)≥1−δa.s.,
where, for small s>0,
Z2(s):=f(s)∑k=N1(s)+1N2(s)(logk)−1/2k1/2+sη˜k,1(s).
We shall prove that there exists s‾>0 such that for all s∈(0,s‾),
P{Z2(s)>1−δ}≥3−1e−(1−δ2/8)log(3)1/s.
As a consequence,
∑n≥n1P{Z2(sn)>1−δ}≥3−1∑n≥n11n(1+γ)(1−δ2/8)=∞,
where n1≥n0 is chosen to ensure that sn<s‾ for n≥n1. In view of (26) the random variables Z2(sn1),Z2(sn1+1),… are independent. Hence, divergence of the series entails (27) by the converse part of the Borel–Cantelli lemma.
When proving (28) we use the event
Us:={1−δ<Z2(s)≤1}={(1−δ)V(s)<W(s)/(g(s))1/2≤V(s)},
where V(s)=(2log(3)1/s)1/2 and
W(s):=Z2(s)f(s)=∑k=N1(s)+1N2(s)(logk)−1/2k1/2+sη˜k,1(s).
For u∈R and small s>0, let Qs,u be a probability measure on (Ω,F) defined by
Qs,u(A)=E[euW(s)/(g(s))1/21A]E[euW(s)/(g(s))1/2],A∈F.
Then
E[eu(W(s)/(g(s))1/2−V(s))]Qs,u(Us)=e−uV(s)E[euW(s)/(g(s))1/21Us]≤P(Us)≤P{Z2(s)>1−δ}.
Now we show that by selecting an appropriate u=u(s)=O((log(3)1/s)1/2), the expectation on the left-hand side of (30) is bounded from below by e−(1−δ2/8)log(3)1/s for small positive s. Also, we shall prove that Qs,u(Us)≥1/3, thereby deriving (28).
We start by demonstrating that, with u=O((log(3)1/s)1/2),
E[euW(s)/(g(s))1/2]∼eu2/2+u2h(s),s→0+,
for some function h satisfying lims→0+h(s)=0. Put
ξk(s)=(logk)−1/2(g(s))1/2k1/2+sη˜k,1(s),k≥2,s∈(0,e−e).
As a consequence, for u∈R,
uW(s)/(g(s))1/2=∑k=N1(s)+1N2(s)uξk(s).
According to the second inequality in (11),
|uξk(s)|=O(1/loglog1/s)=o(1),s→0+a.s.
for each k≥2. Using
ex=1+x+x2/2+o(x2)andlog(1+x)=x+O(x2),x→0,
we obtain
E[euW(s)/(g(s))1/2]=∏k=N1(s)+1N2(s)E[exp(uξk(s))]=∏k=N1(s)+1N2(s)E[1+uξk(s)+u2ξk2(s)(1/2+o(1))]=exp∑k=N1(s)+1N2(s)log(1+u2E[ξk2(s)(1/2+o(1))])=exp(u2(1/2+o(1))∑k=N1(s)+1N2(s)E[ξk2(s)]+u4O(∑k=N1(s)+1N2(s)(E[ξk2(s)])2)).
Here, we have used the fact that the o(1) term can be chosen nonrandom. In view of (24) and (25),
∑k=N1(s)+1N2(s)(logk)−1k1+2s∼g(s),s→0+.
The latter, combined with uniformity in the integer k∈[N1(s)+1,N2(s)] of the limit relation
E[η˜k,12(s)]=E[ηk21(Ak,1(s))c]−(E[ηk1(Ak,1(s))c])2→1,s→0+,
results in
∑k=N1(s)+1N2(s)E[ξk2(s)]=1g(s)∑k=N1(s)+1N2(s)(logk)−1k1+2sE[η˜k,12(s)]→1,s→0+.
Finally,
u2∑k=N1(s)+1N2(s)(E[ξk2(s)])2=u2(g(s))2∑k=N1(s)+1N2(s)(logk)−2k2+4s(E[η˜k,12(s)])2≤u2(g(s))2∑k≥N1(s)+1(logk)−2k2=o(1),s→0+,
because both factors converge to 0 as s→0+. Now relations (33), (34) and (35) entail (31).
Formula (31), with u∈R fixed, reads lims→0+E[euW(s)/(g(s))1/2]=eu2/2, which implies a central limit theorem W(s)/(g(s))1/2→dN(0,1) as s→0+. Here, →d denotes convergence in distribution and N(0,1) denotes a random variable with the normal distribution with mean 0 and variance 1.
Passing to the proof of (28), put
u=u(s)=2(1−δ/2)(log(3)1/s)1/2.
Formula (31) entails
E[eu(W(s)/(g(s))1/2−V(s))]=e−(1−δ2/4)log(3)1/s+o(log(3)1/s)≥e−(1−δ2/8)log(3)1/s
for small s>0. Now we show that the Qs,u-distribution of
W(s)/(g(s))1/2−(u+2uh(s))
converges weakly as s→0+ to the P-distribution of N(0,1). To this end, we prove convergence of the moment generating functions. Let EQs,u denote the expectation with respect to the probability measure Qs,u. Invoking (31) we conclude that, for t∈R,
EQs,u[et(W(s)/(g(s))1/2−(u+2uh(s)))]∼E[e(t+u)W(s)/(g(s))1/2]E[euW(s)/(g(s))1/2]e−t(u+2uh(s))=exp((t+u)2/2+(t+u)2h(s)−u2/2−u2h(s)−t(u+2uh(s)))=exp((1/2+h(s))t2)→exp(t2/2)=E[exp(tN(0,1))],s→0+.
As a consequence of the weak convergence, we infer
lim sups→0+Qs,u{W(s)/(g(s))1/2≤(1−δ)V(s)}≤lims→0+Qs,u{W(s)/(g(s))1/2≤u+2uh(s)}=P{N(0,1)≤0}=1/2.
Further, the relation lims→0+(V(s)−(u+2uh(s)))=+∞ entails
lims→0+Qs,u{W(s)/(g(s))1/2≤V(s)}=1
and thereupon
Qs,u(Us)=Qs,u{W(s)/(g(s))1/2≤V(s)}−Qs,u{W(s)/(g(s))1/2≤(1−δ)V(s)}≥1/3
for small s>0. Now (28) follows from (30), (36) and (37). The proof of Lemma 6 is complete. □
It suffices to prove (19). To this end, pick sufficiently small δ>0 and γ>0, and let sn=exp(−exp(n1+γ)) for n≥1. We shall invoke a representation
f(sn)∑k≥2(logk)−1/2k1/2+snηk=f(sn)∑k=2⌊(log1/sn)1/2⌋(logk)−1/2k1/2+snηk−f(sn)∑k=2⌊(log1/sn)1/2⌋(logk)−1/2k1/2+sn(ηk1(Ak,1(sn))c−E[ηk1(Ak,1(sn))c])+f(sn)∑k≥⌊(log1/sn)1/2⌋+1(logk)−1/2k1/2+sn(ηk1Ak,1(sn)−E[ηk1Ak,1(sn)])+f(sn)∑k≥2(logk)−1/2k1/2+sn(ηk1(Ak,1(sn))c−E[ηk1(Ak,1(sn))c]).
Here, we have used that E[η]=0. The first three terms on the right-hand side converge to 0 a.s. as n→∞ by Lemma 1, formula (21) of Lemma 5, with N1(s)=M(s)=⌊(log1/s)1/2⌋, and Lemma 2, respectively. Lemma 6 ensures that
lim sups→0+f(s)∑k≥2(logk)−1/2k1/2+sηk≥lim supn→∞f(sn)∑k≥2(logk)−1/2k1/2+snηk≥1−δa.s.
Sending δ to 0+ we arrive at (19). □
A combination of Propositions 3 and 4 yields (2) and (3). To prove (1) we put H:={z∈C:Rez>0} and
X(z):=∑k≥2(logk)−1/2k1/2+zηk,z∈H,
and note that the so defined X is a random analytic function, see p. 247 in [5]. Hence, its restriction to positive arguments
s→X(s)=∑k≥2(logk)−1/2k−1/2−sηk
is a.s. continuous, and so is s↦(2σ2log1/slog(3)1/s)−1/2X(s) on (0,e−e). In view of (2) and (3), we obtain (1) with the help of the intermediate value theorem for continuous functions. □
Proof of Theorem <xref rid="j_vmsta276_stat_003">1</xref>
It is more convenient to prove the result in an equivalent form
(1s1/2∑k≥2(logk)−1/2k1/2+exp(−ts)ηk)t≥0⟹(σB(t))t≥0,s→+∞,
on C[0,∞). Without loss of generality we can and do assume that σ2=1.
In what follows we write X for X−1/2. We use a standard approach, which consists of two steps: (a) proving weak convergence of finite-dimensional distributions; (b) checking tightness.
(a) If t=0, then, for s>0, X(e−ts)=X(1)=∑k≥2(logk)−1/2k−3/2ηk, and s−1/2X(1) converges in probability to B(0)=0 as s→+∞.
Thus, it suffices to show that, for t1,t2∈(0,∞) (we do not need to consider t=0),
E[X(e−t1s)X(e−t2s)]∼min(t1,t2)s,s→+∞,
and check the Lindeberg–Feller condition: for all ε>0 and each fixed t>0,
lims→+∞1s∑k≥2E[((logk)−1/2k1/2+exp(−ts)ηk)21{(logk)−1/2|ηk|>εk1/2+exp(−ts)s1/2}]=0.Proof of (38). Using monotonicity to pass from the series to an integral we obtain
E[∑k≥2(logk)−1/2k1/2+exp(−t1s)ηk∑j≥2(logj)−1/2j1/2+exp(−t2s)ηj]=∑k≥2(logk)−1k1+exp(−t1s)+exp(−t2s)∼∫e∞(logy)−1y1+exp(−t1s)+exp(−t2s)dy=∫1∞e−(exp(−t1s)+exp(−t2s))yydy∼−log(e−t1s+e−t2s)∼min(t1,t2)s,s→+∞.Proof of (39). For each k≥2 and each s>0, (logk)−1/2k−1/2−exp(−ts)≤(log2)−1/2=:1/A. Hence, the expression under the limit on the left-hand side of (39) does not exceed
1s∑k≥2(logk)−1k1+2exp(−ts)E[η21{|η|>As1/2}].
As shown in the proof of (38), ∑k≥2(logk)−1k−1−2exp(−ts)∼ts as s→+∞. Further, E[η2]<∞ entails lims→+∞E[η21{|η|>As1/2}]=0. With these at hand, (39) follows.
(b) We have to prove tightness on C[0,T] (the set of continuous functions defined on [0,T]) for each T>0. Since (B(t))t∈[0,T] has the same distribution as T1/2(B(t))t∈[0,1], it is enough to investigate the case T=1 only.
Let M∗:(0,∞)→N0 denote a function satisfying lims→+∞M∗(s)=+∞ and M∗(s)=o(s) as s→+∞. We shall need a relation supk≤n|Tk|=O(n1/2) in probability as n→∞, which is a consequence of n−1/2maxk≤n|Tk|→d|N(0,1)| as n→∞. Repeating the proof of Lemma 1, with the aforementioned limit relation replacing (7), we infer
lims→+∞1s1/2supt∈[0,1]|∑k=2M∗(s)(logk)−1/2k1/2+exp(−ts)ηk|=0in probability.
From now on, we choose a particular M∗ defined as above, for instance, M∗(s)=⌊s1/2⌋ and put a(s):=(logM∗(s))1/2 for s≥0. Arguing as in the proof of Lemma 2 or in the analysis of In,2(s) in the proof of Lemma 4, we obtain
lims→+∞supt∈[0,1]∑k≥M∗(s)+1(logk)−1/2k1/2+exp(−ts)|ηk|1{|ηk|>k1/2a(s)}=0a.s.
and
lims→∞supt∈[0,1]∑k≥M∗(s)+1(logk)−1/2k1/2+exp(−ts)E[|ηk|1{|ηk|>k1/2a(s)}]=0.
Put ηk∗(s):=ηk1{|ηk|≤k1/2a(s)}−E[ηk1{|ηk|≤k1/2a(s)}] for k∈N and s>0, and then
X∗(t,s):=1s1/2∑k≥M∗(s)+1(logk)−1/2ηk∗(s)k1/2+exp(−ts),t≥0,s>0.
In view of (40), (41) and (42) it remains to prove tightness of the distributions of (X∗(t,s))t∈[0,1] for large s>0. According to formula (7.8) on p. 82 in [4], it is enough to show that, for all ε>0,
limi→∞lim sups→∞P{supu,v∈[0,1],|u−v|≤2−i|X∗(u,s)−X∗(v,s)|>ε}=0.
The proof of (43) follows closely the last part of the proof of Lemma 4. We use dyadic partitions of [0,1] by points tj,m∗:=2−jm for j∈N0 and m=0,1,…,2j. Similarly to the argument preceding formula (16) we infer
supu,v∈[0,1],|u−v|≤2−i|X∗(u,s)−X∗(v,s)|≤∑j≥imax1≤m≤2j|X∗(tj,m∗,s)−X∗(tj,m−1∗,s)|.
Thus, it suffices to prove that, for all ε>0,
limi→∞lim sups→+∞P{∑j≥imax1≤m≤2j|X∗(tj,m∗,s)−X∗(tj,m−1∗,s)|>εs1/2}=0.
Put aj∗:=2−j/2j2 for j∈N0. The last limit relation follows if we can show that, for all ε>0,
limi→∞lim sups→+∞∑j≥iP{max1≤m≤2j|X∗(tj,m∗)−X∗(tj,m−1∗)|>εaj∗s1/2}=0.
Denote by A∗(j,s) and B∗(u,k,s) the counterparts of A(j,n) and B(u,k,n) in the present situation. Then A∗(j,s)≤2−js and, if 2j≥s, B∗(u,k,s)≤|u|2−js=:Cj∗(u,s), if 2j≤s, B∗(j,k,s)≤|u|=:Cj∗(u,s). With these at hand we obtain
E[exp(±u(X∗(tj,m∗,s)−X∗(tj,m−1∗,s)))]≤exp(2−jsu22eCj∗(u,s)),u∈R,
and thereupon
P{|X∗(tj,m∗,s)−X∗(tj,m−1∗,s)|>εaj∗s1/2}≤exp(−uεaj∗s1/2)E[exp(u|X∗(tj,m∗,s)−X∗(tj,m−1∗,s)|)]≤2exp(−uε2−j/2j2s1/2+2−jsu22eCj∗(u,s)).
Put u=ε2j/2s−1/2. Then Cj∗(u,s)≤ε and further
∑j≥iP{max1≤m≤2j|X∗(tj,m∗,s)−X∗(tj,m−1∗,s)|>εaj∗s1/2}≤2exp(ε2eε/2)∑j≥i2je−ε2j2→0,i→∞.
The proof of Theorem 1 is complete.
A failure of approach based on a strong approximation
Our proof of Theorem 2 is quite technical. Naturally we wanted to work out a less technical argument. One promising possibility has been to exploit a strong approximation result for centered standard random walks with finite variance, see, for instance, Theorem 12.6 in [7].
There exists a standard Brownian motion(W(t))t≥0such thatT⌊t⌋−W(t)=o((tloglogt)1/2),t→∞a.s.
Implicit in the preceding proofs is the fact that the principal contribution to the LIL is made by the fragment of the original series which has the variance comparable to the variance of the full series. More precisely, one may reduce attention to the sum ∑k=N1(s)N2(s)(logk)−1/2k1/2+sηk, where N1 and N2 are positive integer-valued functions satisfying lims→0+(loglogN1(s)/log1/s)=0, lims→0+slogN2(s)=0 and
lims→0+(loglogN2(s)/log1/s)=1.
The reason is that
Var(∑k=N1(s)N2(s)(logk)−1/2k1/2+sηk)∼Var(∑k≥2(logk)−1/2k1/2+sηk)∼log1/s,s→0+.
We hoped it would be possible to work with Gaussian random variables given by ∫N1(s)N2(s)(logx)−1/2x1/2+sdW(x) in place of ∫N1(s)N2(s)(logx)−1/2x1/2+sdT⌊x⌋. Unfortunately, Lemma 7 does not seem to secure such a possibility. Indeed, after integrating by parts we intended to show that
lims→0+f(s)∫N1(s)N2(s)|T⌊x⌋−W(x)|(logx)1/2x3/2+sdx=0a.s.
In view of Lemma 7 the latter would be a consequence of
lims→0+f(s)∫N1(s)N2(s)(loglogx)1/2(logx)1/2x1+sdx=0.
Since
∫N1(s)N2(s)(loglogx)1/2(logx)1/2x1+sdx=1s1/2∫slogN1(s)slogN2(s)(log1/s+logz)1/2e−zz1/2dz≤(loglogN2(s))1/2s1/2∫0slogN2(s)e−zz1/2dz∼2(logN2(s)loglogN2(s))1/2,s→0+,
the validity of (45) required lims→0+(logN2(s)/log1/s)=0, which was incompatible with (44).
We note in passing that another strong approximation of (Tn)n≥0, this time by sums of independent Gaussian random variables, has a better error the big O of square root, see [8]. Revisiting the calculation above reveals that this approximation does not seem to help either.
Acknowledgement
We thank the three anonymous referees for their useful comments which helped to improve the presentation. In particular, we are indebted to one of the referees for providing a remark, which has led to the appearance of Conjectures 1 and 2.
ReferencesAymone, M.: Real zeros of random Dirichlet series. Electron. Commun. Probab.24, 1–8 (2019). MR4003128. https://doi.org/10.1214/19-ecp260Aymone, M., Frometa, S., Misturini, R.: Law of the iterated logarithm for a random Dirichlet series. Electron. Commun. Probab.25, 1–14 (2020). MR4137941. https://doi.org/10.3390/mca25010013Aymone, M., Frometa, S., Misturini, R.: How many real zeros does a random Dirichlet series have?Electron. J. Probab.29, 1–17 (2024). MR4688682. https://doi.org/10.1214/23-ejp1067Billingsley, P.: Convergence of Probability Measures, Second Edition. Wiley (1999). MR1700749. https://doi.org/10.1002/9780470316962Buraczewski, D., Dong, C., Iksanov, A., Marynych, A.: Limit theorems for random Dirichlet series. Stoch. Process. Appl.165, 246–274 (2023). MR4640264. https://doi.org/10.1016/j.spa.2023.08.007Geis, N., Hiary, G.: Counting sign changes of partial sums of random multiplicative functions. Q. J. Math.76, 19–45 (2025). MR4865832. https://doi.org/10.1093/qmath/haae061Kallenberg, O.: Foundations of Modern Probability. Springer (1997). MR1464694Major, P.: An improvement of Strassen’s invariance principle. Ann. Probab.7, 55–61 (1979). MR0515812Zhao, H., Huang, Y.: Random Dirichlet series with α-stable coefficients. Stat. Probab. Lett.208, 110047 (2024). MR4690584. https://doi.org/10.1016/j.spl.2024.110047