Partial sums of typical multiplicative functions over short moving intervals

We prove that the $k$-th positive integer moment of partial sums of Steinhaus random multiplicative functions over the interval $(x, x+H]$ matches the corresponding Gaussian moment, as long as $H\ll x/(\log x)^{2k^2+2+o(1)}$ and $H$ tends to infinity with $x$. We show that properly normalized partial sums of typical multiplicative functions arising from realizations of random multiplicative functions have Gaussian limiting distribution in short moving intervals $(x, x+H]$ with $H\ll X/(\log X)^{W(X)}$ tending to infinity with $X$, where $x$ is uniformly chosen from $\{1,2,\dots, X\}$, and $W(X)$ tends to infinity with $X$ arbitrarily slowly. This makes some initial progress on a recent question of Harper.


Introduction
We are interested in the partial sums behavior of a family of completely multiplicative functions f supported on moving short intervals.Formally, for positive integers X, let [X] := {1, 2, . . ., X} and F X := {f : [X] → {|z| = 1}, f is completely multiplicative}.
For f ∈ F X , the function values f (n) for all n X are uniquely determined by (f (p)) p X .The Steinhaus random multiplicative function is defined by selecting f (p) uniformly at random from the complex unit circle and defining f (n) completely multiplicatively.One may view F X as the family of all Steinhaus random multiplicative functions.
Let H be another positive integer.We are interested in for a typical f ∈ F X+H , whether the random partial sums where x is uniformly chosen from [X], behave like a complex standard Gaussian.In this note, we provide a positive answer (Theorem 1.2) when H ≪ A X/ log A X holds for all A > 0. As we explain in §4, the answer is negative for H ≫ X exp(−(log log X) 1/2−ε ), but the question remains open between these two thresholds.
We formalize the question by explaining how to measure the elements in F X .Via complete multiplicativity of f ∈ F X , define on F X the product measure ν X := p X µ p , where for any given prime p, we let µ p denote the uniform distribution on the set {f (p)} = {|z| = 1}.For example, ν X (F X ) = 1.
Question 1.1 (Harper [14, open question (iv)]).What is the distribution of the normalized random sum defined in (1.1) (for most f ) as x is uniformly chosen from [X]?
1.1.Main results.In this note, we make some progress on Question 1.1.We use the notation d − → to denote convergence in distribution.
Theorem 1.2.Let integer X be large and W (X) tend to infinity arbitrarily slowly as X tends to infinity.Let H := H(X) ≪ X(log X) −W (X) and H → +∞ as X → +∞.Then, for almost all f ∈ F X+H , as X → +∞, where x is chosen uniformly from [X].
Here "almost all" means the total measure of such f is 1 − o X→+∞ (1) under ν X+H . 1 Also, CN (0, 1) denotes the standard complex normal distribution; a standard complex normal random variable Z (with mean 0 and variance 1) can be written as Z = X + iY , where X and Y are independent real normal random variables with mean 0 and variance 1/2.Recall that a real normal random variable W with mean 0 and variance σ 2 satisfies To prove Theorem 1.2, we establish moment statistics in several situations.We first show that the integer moments of random multiplicative functions f supported on suitable short intervals match the corresponding Gaussian moments.We write E f to mean "average over f ∈ F X with respect to ν X " (where F X is always clear from context).
with an implied constant depending only on k.
1 More precisely, there exist nonempty measurable sets (under ν X+H ) such that for every sequence of functions f X ∈ G X,H (X 1), the random variable on the left-hand side of (1.2) (with f = f X ) converges in distribution to CN (0, 1) as X → +∞.
Notice that k! is the 2k-th moment of the standard complex Gaussian distribution.Given an integer k 1, let E ′ (k) be the smallest real number r 0 such that for every ε > 0, we have Theorem 1.3 shows that E ′ (k) E(k). 2 The paper [5] studies the case k = 2, showing in particular that E ′ (2) 1.In the case that f is supported on {1, 2 . . ., x}, the 2k-th moments for general k were studied in [1,6,7,10,12,15,16] and it is known that the moments there do not match Gaussian moments: for instance, by [12,Theorem 1.1], there exists some constant c > 0 such that for all positive integers k c log x log log x (assuming x is large), , and an additional contribution of roughly k!H k from trivial solutions, it is plausible that one could improve the right-hand side in Theorem 1.
If true, this would suggest that E ′ (k) k − 1 and we believe this might be the true order.For a discussion of how one might improve on Theorem 1.3, see the beginning of §4.
By orthogonality, Theorem 1.3 is a statement about the Diophantine point count (1.4) The circle method, or modern versions thereof such as [8,17], might lead to an asymptotic for (1.4) uniformly over H ∈ [x 1−δ , x] for k = 2, unconditionally (cf.[17,Theorem 6]), or for k = 3, conditionally on standard number-theoretic hypotheses (cf.[27]).Alternatively, "multiplicative" harmonic analysis along the lines of [7,15,16] may in fact lead to an unconditional asymptotic over H ∈ [x 1−δ , x] for all k, with many main terms involving different powers of log x, log H. Nonetheless, for all k, we obtain an unconditional asymptotic for (1.4) uniformly over H ≪ x/(log x) Ck 2 , by replacing the complicated "off-diagonal" contribution to (1.4) with a larger but simpler quantity; see §2 for details.
Remark 1.5.An analog of (1.4) for polynomial values P (n i ) is studied in [21,28], and a similar flavor counting question to (1.4) is studied in [9] using the decoupling method.
2 After writing the paper, the authors learned that for H x/ exp(C k log x/ log log x), the Diophantine statement underlying Theorem 1.3 has essentially appeared before in the literature (see [4,proof of Theorem 34]).However, we handle a more delicate range of the form H x/(log x) Ck 2 .
After Theorem 1.3, our next step towards Theorem 1.2 is to establish concentration estimates for the moments of the random sums (1.1).We write E x to denote "expectation over x uniformly chosen from [X]" (where X is always clear from context).Theorem 1.6.Let X, k 1 be integers with X large.Suppose that H := H(X) → +∞ as X → +∞.There exists a large absolute constant A > 0 such that the following holds as long as H ≪ X(log X) −C k with C k = Ak Ak Ak .Let f ∈ F X+H ; then Furthermore, for any fixed positive integer ℓ < k, we have We prove Theorem 1.3 in §2, and then we prove Theorem 1.6 in §3.
Proof of Theorem 1.2, assuming Theorem 1.6.We use the notation A H (f, x) from (1.1).By Markov's inequality, Theorem 1.6 implies that there exists a set of the form We believe results similar to our theorems above should also hold in the (extended) Rademacher case, though we do not pursue that case in this paper.1.2.Notation.For any two functions f, g : R → R, we write f ≪ g, g ≫ f, g = Ω(f ) or f = O(g) if there exists a positive constant C such that |f | Cg, and we write f ≍ g or f = Θ(g) if f ≪ g and g ≫ f .We write O k to indicate that the implicit constant depends on k.We write o X→+∞ (g) to denote a quantity f such that f /g tends to zero as X tends to infinity.1.3.Acknowledgements.We thank Andrew Granville and the anonymous referee for many detailed comments that led us to significantly improve the results and presentation of our work.We thank and Adam Harper for helpful discussions and useful comments and corrections on earlier versions.We also thank Yuqiu Fu, Larry Guth, Kannan Soundararajan, Katharine Woo, and Liyang Yang for helpful discussions.Finally, we thank Peter Sarnak for introducing us (the authors) to each other during the "50 Years of Number Theory and Random Matrix Theory" Conference at IAS and making the collaboration possible.

Moments of random multiplicative functions in short intervals
In this section, we prove Theorem 1.3.For integers k, n 1, let τ k (n) denote the number of positive integer solutions (d 1 , . . ., d k ) to the equation [24, Theorem 1.29 and Corollary 1.36]) (2.1) As we mentioned before, when H x, Theorem 1.3 is implied by (1.3).From now on, we focus on the case H x. We split the proof into two cases: small H and large H.For small H, we illustrate the general strategy and carelessly use divisor bounds; for large H, we take advantage of bounds of Shiu [25] and Henriot [18] on mean values and correlations of multiplicative functions over short intervals, together with a decomposition idea.
Here we take ε to be a small absolute constant, e.g.ε = 1 100 .We begin with the following proposition.
Proposition 2.1.Let k, y, H 1 be integers.Suppose y is large and k log log y.Then N k (H; y), the number of integer tuples ).
Proof.The case k = 1 is trivial; one has N 1 (H; y) 2H/y.Suppose k 2. Whenever y|h 1 h 2 • • • h k = 0, there exists a factorization y = u 1 u 2 • • • u k where u i are positive integers such that u i |h i = 0 for all 1 i k. (Explicitly, one can take u 1 = gcd(h 1 , y) and if y H k .By the divisor bound (2.1), Proposition 2.1 follows.
Corollary 2.2.Let k, H, x 1 be integers.Suppose x is large and k log log x.
The 2k-th moment in Theorem 1.3 is H −k times the point count (1.4) for the Diophantine equation We will show that To this end, let N ′ k (x, H) denote the number of nontrivial solutions in (x, x + H] 2k with the further constraint that Then for any k 2, we have since for each (n 1 , . . ., n 2k ) ∈ (x, x + H] 2k , either (2.4) holds or there exists i A key observation 3 is that for nontrivial solutions to (2.3) with constraint (2.4), and if we write If x is large and k is fixed (or k log log x, say), then by the divisor bound (2.1), the quantity (2.6) is at most where in the last step we used Corollary 2.2.
By (2.5), it follows that x is large and k is fixed (or k log log x, say), then ), uniformly over k (log log x) 1/2−δ , say.However, in Case 2 below, our proof relies on a larger body of knowledge for which the k-dependence does not seem easy to work out; this is why we essentially keep k fixed in Theorem 1.3.
3 After writing the paper, the authors learned that this observation has appeared before in the literature (see [4, proof of Lemma 22]); however, we take the idea further, both in §2 and in §3.
Also useful to us will be the following immediate consequence of Shiu [25, Theorem 1].
Lemma 2.6.Let f ∈ M(A 1 , A 2 ) and β ∈ (0, 1).If x, y 2 are reals with x β y x, then We will apply the above results to f = τ k over intervals of the form [x, x + y] with y ≫ x 1/2k , say.Here τ k ∈ M(k, O k,ǫ (1)), by (2.1) and the fact that τ k (p) = k and for arbitrary integers m, n 1.
With the lemmas above in hand, we now build on the strategy from Case 1 to attack Case 2. As before, we let N ′ k (x, H) denote the number of nontrivial solutions (n 1 , . . ., n k , n k+1 , . . ., n 2k ) ∈ (x, x + H] 2k to (2.3) with constraint (2.4).Again, for such solutions we write h i = n i − n 2k ∈ [−H, H] \ {0}, and there exist positive integers u i H]; so u i H, and there exist signs As before, the quantity N ′ k (x, H) is at most (2.6).Upon splitting the range [H] for each u i into 1 + log 2 H ≪ 1 + log x dyadic intervals, we conclude that where we let n 2k := u 1 u 2 • • • u k and h i := ε i u i v i in the sum over u i , v i (for notational brevity), and where S(x, H) denotes the maximum of the quantity x + H. Now, for the rest of §2, fix a choice of ε 1 , . . ., ε k , U 1 , . . ., U k with S(x, H) = S( ε, U).
By symmetry, we may assume that U 1 2 • • U k .
We now bound S( ε, U ), assuming k 2. (For k = 1, we can directly note that N ′ 1 (x, H) = 0.) A key observation is that since By the sub-multiplicativity property (2.10), we have that S( ε, U) is at most (2.14)where u −i means that the factor u i is not included.But for each i 2 and u i ∈ [U i , 2U i ), Lemma 2.6 and (2.11) imply (since ( For the innermost sum, first note that ( (since H x 1−2εk −1 ); then by Lemma 2.6 and (2.11), we have (for any given u 2 , . . ., u k ) It follows that S( ε, U) is at most Plugging the above estimate into (2.13),we have (assuming k 2) in the given range of H. Then by using the first part of (2.7) (and noting that N 1 (x, H) = N ′ 1 (x, H) = 0) as before, we have (for arbitrary k 1) , which suffices for Theorem 1.3.

Proof of Theorem 1.6
In this section, we prove Theorem 1.6.Let rad k be the multiplicative function where [n 1 , . . ., n k ] denotes the least common multiple of n 1 , . . ., n k .In particular, for prime powers p ℓ we have rad k (p ℓ ) = p ⌈ℓ/k⌉ .(3.1) Recall that we use τ k (n) to denote the k-folder divisor function as defined in (2.6).We begin with the following sequence of lemmas.
Proof.Suppose that y|(x + t 1 ) . . .(x + t k ).Then there exist integers y 1 , . . ., y k 1 with For any given choice of y 1 , . . ., y k , t 1 , . . ., t k , the conditions y i |x + t i , when satisfiable, impose on x a congruence condition modulo [y 1 , . . ., y k ].It follows that for any given t 1 , . . ., t k , the number of values of x ∈ [X] with (x, t 1 , . . ., t k ) ∈ M k (X, H; y) is at most Remark 3.2.For a typical value of y X, Lemma 3.1 saves a factor of roughly y over the trivial bound H k X, even if H X 1−δ , say.Lemma 3.1 is close to optimal on average over y X, as one can prove by considering prime values of y, for instance.In some regimes, one can do better by other arguments: one can first fix a choice of y i (then select x and choose t i ≡ −x mod y i ) to get which beats Lemma 3.1 when H y and y/ rad k (y) is large, but not in general.
Lemma 3.3.Let k, y, X, H 1 be integers.Then B k (X, H; y), the set of integer tuples Proof.We write y = uv with u|(x + t 1 )(x + t 2 ) • • • (x + t k ) and v|h 1 h 2 respectively.This leads to the total bound .

If we allowed h
3 gives a relative saving of roughly y/H on average over y ≪ X; this follows from (the proof of) Lemma 3.5 below, whose proof requires the following lemma.Lemma 3.4.Let K, k 2 be integers.For integers i 1, let Then c i K K (ik) K .Furthermore, for all primes p and reals s > 1, we have Proof.The first part is clear, since c i The second part follows from the inequality which holds because we have ⌈j/k⌉ = i and j i whenever (i − 1)k < j ik.
It turns out that to prove the key Lemma 3.7 (below) for Theorem 1.6, we need a bound of the form (3.2).Lemma 3.5.Let k, X, H 1 be integers with X large and H X. There exists a positive integer ) ) (depending only on k) such that the following holds: Proof.The case k = 1 is clear by (2.11) (since rad 1 (y) = y), so suppose k 2 for the remainder of this proof.Let K := (2k) 2k •2k 2 k 4k+3 .Then τ 2k (y) 2k τ 2 (y)τ k (y) 2 τ K (y), since for all integers j 1 , j 2 1 we have τ j 1 (y)τ j 2 (y) τ j 1 j 2 (y) by [2, (3.2)].By Rankin's trick, the left-hand side of (3.2) is therefore at most H times By Lemma 3.4 and the multiplicativity of τ K and rad k , we find that for s > 1, we have where (2K/ log 2) K /e K , and e log 2 1).But then Therefore, the right-hand side of (3.3) is at most .
After plugging in s = 1 + 1/ log X and the bound c 1 K 2K , Lemma 3.5 follows.
We also need a simple but finicky combinatorial estimate.

We now know
Given integers x 1 , x 2 , H 1, let I j = (x j , x j + H] for j ∈ {1, 2}.We are now ready to estimate the size of the set (3.4) Lemma 3.7.Fix an integer k 1; let C k be as in Lemma 3.5.Let X, H be large integers with Proof.We roughly follow §2's proof of Theorem 1.3; however, the present situation is more complicated in some aspects, which we address using some new symmetry tricks.
First, let T ⋆ k (I 1 , I 2 ) be the subset of (3.4) satisfying the following conditions: In the notation of Lemma 3.6, applied with a = m and a = n (separately), we have #T In general, given an element n ∈ I 2k 1 × I 2k 2 of (3.4), let U be the set of integers u that violate at least one of the conditions (1)-( 4) above.Then n ∈ T ⋆ k (I 1 , I 2 ) if and only if U = ∅.This simple observation will help us estimate the size of (3.4).
Let N ⋆ k (I 1 , I 2 ) be the subset of (3.4) satisfying the following conditions: (1) n 2k / ∈ {n 1 , . . ., n k }. (This implies, but is not equivalent to, n 2k ∈ U.) Then (3.4) has size at least #T ⋆ k (I 1 , I 2 ) and we claim that (3.4) has size at most First note that for each element n of (3.4) lying outside of T ⋆ k (I 1 , I 2 ), there exist v ∈ U and (a, b, c) ∈ {m, n} × {0, k} × [k], with τ 2k (v) = max u∈U τ 2k (u), such that a b+c = v and a b+c / ∈ {a (k−b)+i : i ∈ [k]}; the existence of v with τ 2k (v) = max u∈U τ 2k (u) follows from the fact that U = ∅, and the existence of (a, b, c) then follows from the definition of U. The claim then follows from the definitions of N ⋆ k (I 1 , I 2 ) and N ⋆ k (I 2 , I 1 ), upon summing over all possibilities for a, b, c.
The right-hand side of (3.7) equals (k + τ 2k (n 2k )) 2k (k + 1) 2k τ 2k (n 2k ) 2k , so upon summing over (n 1 , . . ., n in the notation of Lemma 3.3.Therefore, summing (3.8) over x 2 ∈ [X] gives the inequality We next apply Lemma 3.3 to give an upper bound on |B k (X, H; y)|, which leads to Average over x 1 by using Lemma 3.5, to get (This is just for uniform notational convenience.)(a).We prove (1.5), a bound on the quantity where A H (f, x) is defined as in (1.1).By expanding the square, we can rewrite (3.11) as (3.12) The subtracted term in (3.12) can be computed by switching the summation: it equals We next focus on the first term in (3.12).We expand out the expression and switch the expectations to get that the first term in (3.12) is (b).We prove (1.6), a bound on the quantity (in the notation A H (f, x) from (1.1)) where 1 ℓ k − 1 and B H (x 1 , x 2 ) := E f A H (f, x 1 ) k A H (f, x 1 ) ℓ A H (f, x 2 ) k A H (f, x 2 ) ℓ .This is the same as counting solutions to where x 1 n i x 1 + H and x 2 m i x 2 + H for all 1 i k + ℓ.Suppose that x 1 x 2 .The left hand side in (3.16) is , while the right hand side in (3.16) is To make them equal, we must have which implies that (under the assumption Hk = o(x 2 )) x 2 x 1 x 2 + O(kH).

2 |
. (The upper bound on τ 2k (z) arises as follows: since z is the product of 2k − |S 1 | − |S 2 | elements u l of U, we have an upper bound 1
2.15)cf.theuseofLemma 2.6 and (2.11)in the proof of Lemma 2.7.By (2.15) (multiplied over 2 i k) and Lemma 2.8 (with V 1 = H/U 1 ), we conclude that the quantity (2.14) (and thus S( ε, U)) is at most (1))We estimate(3.13) by a combination of trivial bounds (based on the divisor bound (2.1)) and the moment estimate in Theorem 1.3.We split the sum E x E f |A H (f, x)| 2k into two ranges, and apply Theorem 1.3 and (3.10), to get that X• E x E f |A H (f, x)| 2k equals 1 x H(log X) 5Ck 2 E f |A H (f, x)| 2k + O (log X) −Ck 2 .Upon summing over both ranges of x above, it follows thatE x E f |A H (f, x)| 2k = k! + o X→+∞(1)in the given range of H (provided A is large enough that C k 10Ck 2 ).
.14) Now we use orthogonality and apply Lemma 3.7 to see that (3.14) is k! 2 + o X→+∞ (1) in the given range of H (if A is sufficiently large).Combining the above together, (1.5) follows.