On trigonometric sums with random frequencies

We prove that if I k are disjoint blocks of positive integers and n k are independent random variables with uniform distribution on I k , then N −


Introduction
Salem and Zygmund [7] proved that if (n k ) is a sequence of positive integers satisfying the Hadamard gap condition then the sequence sin 2πn k x, k ≥ 1 obeys the central limit theorem, i.e.
with respect the the probability space (0, 1) equipped with Borel sets and Lebesgue measure.Here the exponential growth condition (1.1) can be weakened, but as Erdős Theorem 1.Let I 1 , I 2 , . . .be disjoint blocks of consecutive positive integers with size d and let n 1 , n 2 , . . .be a sequence of independent random variables on a probability space (Ω, A, P) such that n k is uniformly distributed over over the probability space ((0, 1), B, λ), where B is the Borel σ-algebra in (0, 1), λ is the Lebesgue measure, and N (0, g) denotes the distribution with characteristic function 1 0 e −g(x)t 2 /2 dx.
Here g ≥ 0 and N (0, g) is the distribution of √ gζ, where ζ is a standard normal random variable on (0, 1), independent of g.Clearly, N (0, g) is a variance mixture of zero mean Gaussian distributions.Note that N k=1 λ k (x) = E( N k=1 sin 2πn k x) is the averaged version of N k=1 sin 2πn k x, a nonrandom trigonometric sum and Theorem 1 states that the fluctuations of the random trigonometric sum N k=1 sin 2πn k x around its nonrandom average part always have a mixed normal limit distribution.If ∪ ∞ k=1 |I k | = N, i.e. there are no gaps between the blocks I k , then n k=1 λ k (x) = O(1) for any fixed x and thus (1.3) holds without the λ k (x), yielding the result of Bobkov and Götze [2].Letting ∆ k denote the number of integers between I k and I k+1 (the "gaps"), we will see that the CLT (1.3) also holds with λ k (x) = 0 if ∆ k is nondecreasing and ∆ k = O(k γ ) for some γ < 1/4.If ∆ k grows at least exponentially, then so does the sequence (A k ), where A k denotes the smallest integer of I k .Now and from the CLT of Salem and Zygmund [7] it follows that the limit distribution of where ) and the convolution of these two mixed Gaussian laws is N (0, 1/2), which is exactly the limit distribution of N −1/2 N k=1 sin 2πn k x by the theorem of Salem and Zygmund, since (n k ) grows exponentially.Thus the pure Gaussian limit distribution of N −1/2 N k=1 sin 2πn k x is obtained as the combination of two mixed Gaussian distributions N (0, g) with g in (1.4) and N (0, g * ) with g * in (1.6).
It is worth noting that for any fixed x ∈ (0, 1), sin 2πn k x − λ k (x) are independent, uniformly bounded mean zero random variables on (Ω, A, P) and by elementary calculations.Thus by the law of the iterated logarithm we have for any fixed x ∈ (0, 1) with P-probability 1 (1.7)By Fubini's theorem, with P-probability 1 relation (1.7) holds for almost every x ∈ (0, 1) with respect to Lebesgue measure, yielding the LIL corresponding to (1.3).Actually, the previous argument also shows that for any fixed x ∈ (0, 1) we have (1.3) over the probability space (Ω, A, P), with N (0, g) replaced by N (0, g(x)).However, Fubini's theorem does not work for distributional results and thus we cannot interchange the role of x ∈ (0, 1) and ω ∈ Ω and we will need an elaborate argument in Section 2 to prove Theorem 1. Formula (1.4) shows that for any 0 < x < 1 we have lim x→∞ g(x) = 1/2 and thus for large d the sequence sin 2πn k x − λ k (x) nearly satisfies the ordinary CLT and LIL with limit distribution N (0, 1/2) and limsup = 1/2, just as lacunary trigonometric series with exponential gaps.Formally, this is not surprising since for large d the expected gaps E(n k+1 − n k ) in our sequence are large.As the pictures of g for d = 3 and d = 10 below show, however, the near CLT and LIL actually hold for relatively small values of d such as d = 10.Thus the reason of the near CLT and LIL is not solely large gaps in the the sequence (n k ) but the random fluctuations of the sequence (n k ) as well.
The analogue of Theorem 1 is valid also in the case when n 1 , n 2 , . . .have continuous uniform distribution over the intervals I 1 , I 2 , . ... To formulate the result, define the probability measure µ on the Borel sets of R by Theorem 2. Let n 1 , n 2 , . . .be a sequence of independent random variables on a probability space (Ω, A, P) such that n k has continuous uniform distribution on the interval with respect to the probability space (R, B, µ), where the characteristic function of F is (1.9)

Proofs
We will give the proof of Theorem 2, where the calculations are slightly simpler.Let and By A k+1 − A k ≥ B + 2 and the fact that (see e.g.Hartman [5]) it follows that for every fixed ω ∈ Ω the functions ϕ k are orthogonal over L 2 µ (R) and thus elementary algebra shows that the and Thus From (2.10),A k+1 − A k ≥ B + 2 and elementary trigonometric identities it follows that the functions cos(2A k + B)x are orthogonal in L 2 µ (R) and thus the Rademacher-Menshov convergence theorem implies that ∞ k=1 k −1 cos(2A k + B)x converges µalmost everywhere.Consequently, the Kronecker lemma implies and thus Since for fixed x ϕ 2 k (x) − Eϕ 2 k (x), k = 1, 2, . . .are independent, uniformly bounded, zero mean random variables, the strong law of large numbers yields and thus we conclude that for µ-a.e.x we have P-almost surely (2.12) computation of the sum.Thus we can assume ∆ k ↑ ∞, and then also A k+1 − A k ↑ ∞.
Recalling that the A k are integers, let us break the sum N k=1 e 2πiA k x into subsums

.18)
Clearly Z N,r consists of M r consecutive terms of N k=1 e 2πiA k x for some M r ≥ 0 and thus in the case M r ≥ 1 we have for some integer P r ≥ 0, except when rx is an integer, where C is an absolute constant and t denotes the distance of t from the nearest integer.From a well known result in Diophantine approximation theory (see e.

e 2πiA k x ≤ C 2 r≤C 1 1 / 4 ,
g. Kuipers and Niederreiter[6], Definition 3.3.on p. 121 and Exercise 3.5 on page 130), for every ε > 0 and almost all x in the sense of Lebesgue measure we have nx ≥ cn −(1+ε) for some constant c = c(x) > 0 and all n ≥ 1.This shows that Z N,r = O(r 1+ε ) a.e. and since by (2.16) the largest r actually occurring in breaking N k=1 e 2πiA k x into a sum of Z N,r 's is at most C 1 N γ , we have N k=1 upon choosing ε small enough.