Large deviations principles for lacunary sums Nina Gantert joint - - PowerPoint PPT Presentation

Large deviations principles for lacunary sums

Nina Gantert joint work, soon to be finished with Christoph Aistleitner, Zakhar Kabluchko, Joscha Prochno, Kavita Ramanan

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Thanks to: Aicke Hinrichs, Joscha Prochno, Christoph Th¨ ale, Elisabeth Werner for organizing “New Perspectives and Computational Challenges in High Dimensions” MFO, February 2020

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What is a lacunary sum?

First: What is a lacunary sum? Let Ω = [0, 1], P = Lebesgue measure, take an increasing sequence (ak)k∈N of positive integers satifying the Hadamard gap condition ak+1 ak > q for some q > 1. Then, Sn(ω) =

n

• k=1

cos(2πakω) is called lacunary (trigonometric) sum.

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What is a lacunary sum?

Question

Does (Sn) share the properties of a sum of i.i.d. random variables?

Remark

Sn =

n

• k=1

Xk with Xk = cos(2πakω) Xj and Xk are not independent if j = k. X1, X2, . . . are identically distributed Xj and Xk are uncorrelated if j = k.

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Law of large numbers

Law of large numbers:

Theorem (Hermann Weyl)

Sn n → 0 P − a.s. for n → ∞ . (1) Indeed this is true for all ω ∈ [0, 1] \ Q.

Remark

If ak = qk for some q ∈ {2, 3, 4, . . .}, (1) follows from the ergodic theorem: ω = (ω1, ω2, ω3, . . .) where the ωi are i.i.d. and uniform on {0, 1, . . . , q − 1}. qkω mod 1 = T kω where T is the shift transformation: Tω = (ω2, ω3, ω4, . . .). Hence Sn(ω) =

n

• k=1

f (T kω) with f (x) = cos(2πx).

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Law of large numbers

Remark

The last argument is still true if we replace x → cos(2πx) by another 1-periodic function f .

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Central limit theorem

Central limit theorem:

Theorem (Salem, Zygmund)

P

• 1
• n/2

n

• k=1

cos(2πakω) ≤ t

• →

1 √ 2π

t

• −∞

exp

• −u2

2

• du

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Central limit theorem

Question

Does the CLT still hold if we replace x → cos(2πx) by another 1-periodic function f ? Answer: in general, no!

Example (Erd¨

• s, Fortet)

Take f (x) = cos(2πx) + cos(4πx) and ak = 2k − 1, then P

• 1
• n/2

n

• k=1

cos(2πakω) ≤ t

• → F(t)

where F is not the distribution function of a Gaussian law.

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Central limit theorem

Theorem (Kac)

If f is BV and 1

0 f (x)dx = 0 and ak = 2k, the CLT holds with

σ2 =

1

• f (x)2dx + 2

∞

• k=1

1

• f (x)f (2kx)dk .

.

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Law of the iterated logarithm

Law of the iterated logarithm:

Theorem (Erd¨

• s, Gal)

lim sup

n→∞ n

• k=1

cos(2πakω) √n log log n = 1 P -a.s..

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Our question: large deviations?

Our question: large deviations? Let x > 0. Then, P Sn n ≥ x

• → 0 for n → ∞ .

Exponential rate of decay?

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Our question: large deviations?

Theorem

If (ak) obeys the large gap condition ak+1/ak → ∞, then the random variables Sn/n satisfy a large deviation principle with rate function I, where I is given by

• I(x) = − lim

n→∞

1 n log P Sn n ≥ x

• for x > 0, where

Sn =

n

• k=1

cos(2πakUk) with U1, U2, . . . i.i.d. and uniform

• n [0, 1].

Hence, the rate function is the same as for a sum of i.i.d random variables.

• I is convex,

I(1) = I(−1) = +∞.

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Our question: large deviations?

Remark

Sn =

n

• k=1

cos(2πakU1),

• Sn =

n

• k=1

cos(2πakUk) . To show the statement, it suffices to show that for all θ ∈ R lim

n→∞

1 n log E

• eθSn

= lim

n→∞

1 n log E

• eθ

Sn

= Λ(θ) (2) and that θ → Λ(θ) is differentiable.

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Our question: large deviations?

The proof of (2) is simpler in the case when mk := ak+1/ak ∈ N, ∀k. Again, in this case ω can be written as a sequence ω = (ω1, ω2, ω3, . . .) where the ωk are independent, with uniform distribution on {0, 1, . . . , mk − 1}. More precisely, define the filtration F1 ⊂ F2 ⊂ . . ., where Fk+1 := σ

• Jk+1,i, i = 0, . . . , ak+1 − 1
• ,

k ∈ N0, where Jk+1,i :=

• i

ak+1 , i + 1 ak+1

• ,

i = 0, . . . , ak+1 − 1. Let Yk := E [Xk|Fk+1] Then Yk, k ∈ N0 are independent (Yk is a function of ωk!) Xk − Yk∞ ≤ 2π ak ak+1 . (3)

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Our question: large deviations?

Reason for (3): max

ω∈Jk+1,i

|Xk(ω) − Yk(ω)| = max

ω∈Jk+1,i

• Xk(ω) − ak+1
• Jk+1,i

Xk(y) dy

• =

max

ω∈Jk+1,i

• Xk(ω) − Xk(y0)
• ≤

max

ω∈Jk+1,i

2πak

• ω − y0
• ≤ 2π ak

ak+1 , where y0 ∈ Jk+1,i comes from the mean value theorem. Hence, for Sn := n

k=1 Xk and

Sn := n

k=1 Yk,

Sn − Sn∞ ≤ 2π

n

• k=1

ak ak+1 = o(n), n → ∞, because by assumption ak/ak+1 → 0 as k → ∞. This suffices to show (2).

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Our question: large deviations?

Theorem

If ak = qk for some q ∈ {2, 3, 4, . . .}, the random variables Sn/n satisfy a large deviation principle with rate function Iq, Iq = I, Iq(x) → I(x) for q → ∞, for all x ∈ (−1, 1).

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Our question: large deviations?

Theorem

There exists a sequence of positive integers (ak)k∈N satisfying ak+1/ak ≥ q for some q > 1 and all k ∈ N, for which Sn/n does not satisfy a large deviation principle. More precisely, for this sequence (ak)k∈N there exists ¯ x0 ∈ (0, 1) such that for all x0 ∈ (0, ¯ x0), 0 < lim inf

n→∞ −1

n log P(Sn > nx0) < lim sup

n→∞ −1

n log P(Sn > nx0) < ∞.

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Our question: large deviations?

Theorem

There is a sequence (ak) such that ak+1/ak → 2 and the random variables Sn/n satisfy a large deviation principle with rate function I = I2. Hence, large deviations are sensitive to the properties (not only the growth) of the sequence (ak)!

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Our question: large deviations?

Remark (Sublacunary case)

Assume ak → ∞, 1

k log ak → 0, i.e. the Hadamard gap condition is not

• satisfied. Then

1 n log P Sn n ≥ 1 − ε

• → 0 .

Proof There is δ > 0 such that 2πakω ≤ δ ⇒ cos(2πakω) ≥ 1 − ε . But P [2πakω ≤ δ for k ∈ {1, 2, . . . n}] = P [2πanω ≤ δ] = δ 2πan . Hence 1 n log P Sn n ≥ 1 − ε

• ≥ 1

n log δ 2πan → 0 .

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Our question: large deviations?

Thanks for your attention!

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