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 1/20

Thanks to: Aicke Hinrichs, Joscha Prochno, Christoph Th¨ ale, Elisabeth Werner for organizing “New Perspectives and Computational Challenges in High Dimensions” MFO, February 2020 2/20

What is a lacunary sum? First: What is a lacunary sum? Let Ω = [0 , 1] , P = Lebesgue measure, take an increasing sequence ( a k ) k ∈ N of positive integers satifying the Hadamard gap condition a k +1 > q for some q > 1 . a k Then, n � S n ( ω ) = cos(2 π a k ω ) k =1 is called lacunary (trigonometric) sum. 3/20

What is a lacunary sum? Question Does ( S n ) share the properties of a sum of i.i.d. random variables? Remark n � S n = X k with X k = cos(2 π a k ω ) k =1 X j and X k are not independent if j � = k. X 1 , X 2 , . . . are identically distributed X j and X k are uncorrelated if j � = k. 4/20

Law of large numbers Law of large numbers: Theorem (Hermann Weyl) S n n → 0 P − a.s. for n → ∞ . (1) Indeed this is true for all ω ∈ [0 , 1] \ Q . Remark If a k = q k for some q ∈ { 2 , 3 , 4 , . . . } , (1) follows from the ergodic theorem: ω = ( ω 1 , ω 2 , ω 3 , . . . ) where the ω i are i.i.d. and uniform on { 0 , 1 , . . . , q − 1 } . q k ω mod 1 = T k ω where T is the shift transformation: T ω = ( ω 2 , ω 3 , ω 4 , . . . ) . Hence n � f ( T k ω ) S n ( ω ) = k =1 with f ( x ) = cos(2 π x ) . 5/20

Law of large numbers Remark The last argument is still true if we replace x → cos(2 π x ) by another 1 -periodic function f . 6/20

Central limit theorem Central limit theorem: Theorem (Salem, Zygmund) � � � t � � n � − u 2 1 1 √ P � cos(2 π a k ω ) ≤ t → exp du 2 n / 2 2 π k =1 −∞ 7/20

Central limit theorem Question Does the CLT still hold if we replace x → cos(2 π x ) by another 1 -periodic function f ? Answer: in general, no! Example (Erd¨ os, Fortet) Take f ( x ) = cos(2 π x ) + cos(4 π x ) and a k = 2 k − 1, then � � n � 1 � cos(2 π a k ω ) ≤ t → F ( t ) P n / 2 k =1 where F is not the distribution function of a Gaussian law. 8/20

Central limit theorem Theorem (Kac) � 1 0 f ( x ) dx = 0 and a k = 2 k , the CLT holds with If f is BV and � 1 � 1 � ∞ σ 2 = f ( x ) 2 dx + 2 f ( x ) f (2 k x ) dk . k =1 0 0 . 9/20

Law of the iterated logarithm Law of the iterated logarithm: Theorem (Erd¨ os, Gal) � n cos(2 π a k ω ) k =1 lim sup √ n log log n = 1 P -a.s. . n →∞ 10/20

Our question: large deviations? Our question: large deviations? Let x > 0. Then, � S n � n ≥ x → 0 for n → ∞ . P Exponential rate of decay? 11/20

Our question: large deviations? Theorem If ( a k ) obeys the large gap condition a k +1 / a k → ∞ , then the random variables S n / n satisfy a large deviation principle with rate function � I, where � I is given by � � � 1 S n � I ( x ) = − lim n log P n ≥ x n →∞ � n for x > 0 , where � S n = cos(2 π a k U k ) with U 1 , U 2 , . . . i.i.d. and uniform k =1 on [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) = + ∞ . 12/20

Our question: large deviations? Remark n n � � � S n = cos(2 π a k U 1 ) , S n = cos(2 π a k U k ) . k =1 k =1 To show the statement, it suffices to show that for all θ ∈ R � e θ S n � � S n � 1 1 e θ � lim n log E = lim n log E = Λ( θ ) (2) n →∞ n →∞ and that θ → Λ( θ ) is differentiable. 13/20

Our question: large deviations? The proof of (2) is simpler in the case when m k := a k +1 / a k ∈ N , ∀ k . Again, in this case ω can be written as a sequence ω = ( ω 1 , ω 2 , ω 3 , . . . ) where the ω k are independent, with uniform distribution on { 0 , 1 , . . . , m k − 1 } . More precisely, define the filtration F 1 ⊂ F 2 ⊂ . . . , where � � F k +1 := σ J k +1 , i , i = 0 , . . . , a k +1 − 1 , k ∈ N 0 , where � � , i + 1 i J k +1 , i := , i = 0 , . . . , a k +1 − 1 . a k +1 a k +1 Let Y k := E [ X k | F k +1 ] Then Y k , k ∈ N 0 are independent ( Y k is a function of ω k !) a k � X k − Y k � ∞ ≤ 2 π . (3) a k +1 14/20

Our question: large deviations? Reason for (3): � � � � � � � max | X k ( ω ) − Y k ( ω ) | = max � X k ( ω ) − a k +1 X k ( y ) dy � ω ∈ J k +1 , i ω ∈ J k +1 , i J k +1 , i � � � X k ( ω ) − X k ( y 0 ) � = max ω ∈ J k +1 , i � � � ω − y 0 � ≤ max 2 π a k ω ∈ J k +1 , i ≤ 2 π a k , a k +1 where y 0 ∈ J k +1 , i comes from the mean value theorem. Hence, for S n := � n S n := � n k =1 X k and � k =1 Y k , n � a k � S n − � S n � ∞ ≤ 2 π = o ( n ) , n → ∞ , a k +1 k =1 because by assumption a k / a k +1 → 0 as k → ∞ . This suffices to show (2). 15/20

Our question: large deviations? Theorem If a k = q k for some q ∈ { 2 , 3 , 4 , . . . } , the random variables S n / n satisfy a large deviation principle with rate function I q , I q � = � I, I q ( x ) → � I ( x ) for q → ∞ , for all x ∈ ( − 1 , 1) . 16/20

Our question: large deviations? Theorem There exists a sequence of positive integers ( a k ) k ∈ N satisfying a k +1 / a k ≥ q for some q > 1 and all k ∈ N , for which S n / n does not satisfy a large deviation principle. More precisely, for this sequence ( a k ) k ∈ N there exists ¯ x 0 ∈ (0 , 1) such that for all x 0 ∈ (0 , ¯ x 0 ) , n →∞ − 1 n →∞ − 1 0 < lim inf n log P ( S n > nx 0 ) < lim sup n log P ( S n > nx 0 ) < ∞ . 17/20

Our question: large deviations? Theorem There is a sequence ( a k ) such that a k +1 / a k → 2 and the random variables S n / n satisfy a large deviation principle with rate function � I � = I 2 . Hence, large deviations are sensitive to the properties (not only the growth) of the sequence ( a k )! 18/20

Our question: large deviations? Remark (Sublacunary case) Assume a k → ∞ , 1 k log a k → 0 , i.e. the Hadamard gap condition is not satisfied. Then � S n � 1 n log P n ≥ 1 − ε → 0 . Proof There is δ > 0 such that 2 π a k ω ≤ δ ⇒ cos(2 π a k ω ) ≥ 1 − ε . But δ P [2 π a k ω ≤ δ for k ∈ { 1 , 2 , . . . n } ] = P [2 π a n ω ≤ δ ] = . 2 π a n Hence � S n � 1 ≥ 1 δ n log P n ≥ 1 − ε n log → 0 . 2 π a n 19/20

Our question: large deviations? Thanks for your attention! 20/20