The Contraction Method on C([0, 1]) and Donsker s Theorem

Transcription

1 The Contraction Method on C([0, 1]) and Donsker s Theorem Henning Sulzbach J. W. Goethe-Universität Frankfurt a. M. YEP VII Probability, random trees and algorithms Eindhoven, March 12, 2010 joint work with Ralph Neininger

2 Quickselect / Find Task: Given a list of n different elements S, FIND the element of rank k.

3 Quickselect / Find Task: Given a list of n different elements S, FIND the element of rank k. Algorithm: Choose one element x at random (pivot),

4 Quickselect / Find Task: Given a list of n different elements S, FIND the element of rank k. Algorithm: Choose one element x at random (pivot), Comparing all other elements with x gives sublists S, S >,

5 Quickselect / Find Task: Given a list of n different elements S, FIND the element of rank k. Algorithm: Choose one element x at random (pivot), Comparing all other elements with x gives sublists S, S >, If I n = S k, search recursively in S for the element of rank k, otherwise search recursively ins > for element of rank k I n.

6 Quickselect / Find Task: Given a list of n different elements S, FIND the element of rank k. Algorithm: Choose one element x at random (pivot), Comparing all other elements with x gives sublists S, S >, If I n = S k, search recursively in S for the element of rank k, otherwise search recursively ins > for element of rank k I n. Stop, if the considered set is of size 1.

7 Quickselect / Find For simplicity k = 1. X n = number of key comparisons, I n = size of left sublist

8 Quickselect / Find For simplicity k = 1. X n = number of key comparisons, I n = size of left sublist This leads to X n d = XIn + n 1, where (X j ), I n are independent, I n is distributed uniformly on {0,..., n 1}.

9 Quickselect / Find For simplicity k = 1. X n = number of key comparisons, I n = size of left sublist This leads to X n d = XIn + n 1, where (X j ), I n are independent, I n is distributed uniformly on {0,..., n 1}. Scaling gives. Y n := X n n d = I n n Y I n + n 1 n

10 Recursive equation Y n := X n n d = I n n Y I n + n 1 n.

11 Recursive equation Y n := X n n A possible limit Y should satisfy for ind. U, Y with U uniform on [0, 1]. d = I n n Y I n + n 1 n. Y d = UY + 1, (1)

12 Recursive equation Y n := X n n A possible limit Y should satisfy for ind. U, Y with U uniform on [0, 1]. d = I n n Y I n + n 1 n. Y d = UY + 1, (1) Observe: µ M(R) satisfies (1), iff F (µ) = µ for F : M(R) M(R), F (µ) = L(UY + 1) with U, Y ind. and L(Y ) = µ.

13 Contraction

14 Contraction For µ, ν M 1 (R) let l 1 (µ, ν) = inf E[ X Y ] X,Y :L(X )=µ,l(y )=ν

15 Contraction For µ, ν M 1 (R) let l 1 (µ, ν) = inf E[ X Y ] X,Y :L(X )=µ,l(y )=ν l 1 is a complete metric on M 1 (R) and l 1 (µ n, µ) 0 µ n w µ.

16 Contraction For µ, ν M 1 (R) let l 1 (µ, ν) = inf E[ X Y ] X,Y :L(X )=µ,l(y )=ν l 1 is a complete metric on M 1 (R) and l 1 (µ n, µ) 0 µ n w µ. Show: F is a contraction according to l 1 in M 1 (R).

17 Contraction Proof: Let X, Y s.t. L(X ) = µ, L(Y ) = ν and E[ X Y ] d(µ, ν) + ε.

18 Contraction Proof: Let X, Y s.t. L(X ) = µ, L(Y ) = ν and E[ X Y ] d(µ, ν) + ε. Then l 1 (F (µ), F (ν)) E[ UX + 1 (UY + 1) ] = EUE[ X Y ] EU(l 1 (µ, ν) + ε).

19 Contraction Proof: Let X, Y s.t. L(X ) = µ, L(Y ) = ν and E[ X Y ] d(µ, ν) + ε. Then l 1 (F (µ), F (ν)) E[ UX + 1 (UY + 1) ] = EUE[ X Y ] EU(l 1 (µ, ν) + ε). This gives l 1 (F (µ), F (ν)) EUl 1 (µ, ν)

20 The stochastic fixed-point equation has a unique solution in M 1 (R) Y d = UY + 1

21 The stochastic fixed-point equation Y = d UY + 1 has a unique solution in M 1 (R) and it is easy to show l 1 (Y n, Y ) 0 for L(Y ) = µ.

22 Process version Consider all 1 k n uniformly in the same algorithm X n (k, ) ( Rescaling gives a the r.v. Y k ) n n = X n(k) n points k n Fit Y n to a stepfunction in D([0, 1]) Grübel, Rösler ( 95) prove defined on lattice Y n d Y in (D([0, 1]), d Sk )

23 Donsker s Theorem Let X 1, X 2,... be iid r.v. with EX 1 = 0, EX 2 1 = 1 and E X 1 2+ε < for some ε > 0.

24 Donsker s Theorem Let X 1, X 2,... be iid r.v. with EX 1 = 0, EX 2 1 = 1 and E X 1 2+ε < for some ε > 0. S n = (S n t ) t [0,1] with St n = 1 nt X k + (nt nt )X n nt +1, t [0, 1] k=1

25 Donsker s Theorem Let X 1, X 2,... be iid r.v. with EX 1 = 0, EX 2 1 = 1 and E X 1 2+ε < for some ε > 0. S n = (S n t ) t [0,1] with St n = 1 nt X k + (nt nt )X n nt +1, t [0, 1] k=1 S S^n

26 Donsker s Theorem

27 Donsker s Theorem St n = 1 nt X k + (nt nt )X n nt +1, t [0, 1] k=1 Theorem (Donsker, 1951) S n Motion. d B in (C([0, 1]), sup ), where B is a standard Brownian

28 Decomposition of Brownian Motion

29 In space: Decomposition of Brownian Motion

30 In space: Decomposition of Brownian Motion B B'

31 In space: Decomposition of Brownian Motion (B+B')/Sqrt(2) B B'

32 Decomposition of Brownian Motion It holds ( d Bt + B ) (B t ) t = t 2 for two independent standard Brownian Motions B, B. t (2)

33 Decomposition of Brownian Motion It holds ( d Bt + B ) (B t ) t = t 2 for two independent standard Brownian Motions B, B. In general, there is no spacian decomposition of S n. Equation (2) is satisfied by any gaussian process. t (2)

34 In time: Decomposition of Brownian Motion

35 In time: Decomposition of Brownian Motion Concatenation B B'

36 Decomposition of Brownian Motion ( ) d 1 ( ) 1 (B t ) t = 2 1{t 1/2} B 2t + 1 {t>1/2} B {t>1/2} B 2t 1 t = 1 ϕ 2 (B) + 1 ψ 2 (B ), (3) 2 2

37 Decomposition of Brownian Motion ( ) d 1 ( ) 1 (B t ) t = 2 1{t 1/2} B 2t + 1 {t>1/2} B {t>1/2} B 2t 1 t = 1 ϕ 2 (B) + 1 ψ 2 (B ), (3) 2 2 with (continuous and linear) functions ϕ β, ψ β : C([0, 1]) C([0, 1]) ϕ β (f )(t) = 1 {t 1/β} f (βt) + 1 {t>1/β} f (1), ψ β (f )(t) = 1 {t 1/β} f (0) + 1 {t>1/β} f ( βt 1 β 1 ).

38 Decomposition of Brownian Motion ( ) d 1 ( ) 1 (B t ) t = 2 1{t 1/2} B 2t + 1 {t>1/2} B {t>1/2} B 2t 1 t = 1 ϕ 2 (B) + 1 ψ 2 (B ), (3) 2 2 with (continuous and linear) functions ϕ β, ψ β : C([0, 1]) C([0, 1]) ϕ β (f )(t) = 1 {t 1/β} f (βt) + 1 {t>1/β} f (1), ψ β (f )(t) = 1 {t 1/β} f (0) + 1 {t>1/β} f ( βt 1 β 1 ).

39 Theorem (Uniqueness - Characterization of Brownian Motion) Let X = (X t ) t [0,1] be a continuous real-valued stochastic process satisfying (3). Then there exists a constant σ 0 such that σx is a standard Brownian Motion.

40 Time-decomposition of the random walk

41 Time-decomposition of the random walk Concatenation S_ S_

42 Time-decomposition of the random walk It holds: S n d n/2 ( = ϕ n S n/2 ) + n n/2 n/2 n ψ n n/2 (Ŝ n/2 ), (4)

43 Time-decomposition of the random walk It holds: Now S n d n/2 ( = ϕ n S n/2 ) + n n/2 n/2 n ψ n n/2 (Ŝ n/2 ), (4) n/2 n 1 2, n/2 n 1 2. suggests weak convergence S n B.

44 The Zolotarev metric on C([0, 1])

45 The Zolotarev metric on C([0, 1]) Let M(C([0, 1])) be the set of probability measures on C([0, 1]). For µ, ν M(C([0, 1])) and s > 0 define the Zolotarev distance of µ and ν by ζ s (µ, ν) = sup f F s E[f (X ) f (Y )], with L(X ) = µ, L(Y ) = ν

46 The Zolotarev metric on C([0, 1]) Let M(C([0, 1])) be the set of probability measures on C([0, 1]). For µ, ν M(C([0, 1])) and s > 0 define the Zolotarev distance of µ and ν by ζ s (µ, ν) = sup f F s E[f (X ) f (Y )], with L(X ) = µ, L(Y ) = ν and for s = m + α with 0 < α 1 and m := s 1 N 0 F s := {f C m (C([0, 1]), R) : D m f (x) D m f (y) x y α, x, y C(

47 The Zolotarev metric on C([0, 1]) Let M(C([0, 1])) be the set of probability measures on C([0, 1]). For µ, ν M(C([0, 1])) and s > 0 define the Zolotarev distance of µ and ν by ζ s (µ, ν) = sup f F s E[f (X ) f (Y )], with L(X ) = µ, L(Y ) = ν and for s = m + α with 0 < α 1 and m := s 1 N 0 F s := {f C m (C([0, 1]), R) : D m f (x) D m f (y) x y α, x, y C( Set ζ s (X, Y ) = ζ s (L(X ), L(Y )).

48 Properties of the Zolotarev distance ζ s

49 Properties of the Zolotarev distance ζ s It holds ζ s (X, Y ) <, if E X s, E Y s <, E[g(X,..., X )] = E[g(Y,..., Y )] for all k m and multilinear, bounded functions g : C([0, 1]) k R. In the following assume finiteness of the considered ζ s -distances

50 Lemma Properties of the Zolotarev distance ζ s ζ s is (s,+) - ideal, i.e. ζ s (ϕ(x ), ϕ(y )) ϕ s ζ s (X, Y ) for any continuous and linear function ϕ : C([0, 1]) C([0, 1]) with ϕ = sup ϕ(f ). f C([0,1]), f =1

51 Properties of the Zolotarev distance ζ s Lemma ζ s is (s,+) - ideal, i.e. ζ s (ϕ(x ), ϕ(y )) ϕ s ζ s (X, Y ) for any continuous and linear function ϕ : C([0, 1]) C([0, 1]) with Furthermore ϕ = sup ϕ(f ). f C([0,1]), f =1 ζ s (X 1 + X 2, Y 1 + Y 2 ) ζ s (X 1, Y 1 ) + ζ s (X 2, Y 2 ) for (X 1, Y 1 ) and (X 2, Y 2 ) independent. Proof: Zolotarev ( 76)

52 Proof of Donsker s Theorem (Sketch) We are in the case 2 < s < 3 and have to consider E[f (X, X )] for continuous, bilinear functions f : C([0, 1]) 2 R. This is done by controlling the covariance function.

53 Proof of Donsker s Theorem (Sketch) We are in the case 2 < s < 3 and have to consider E[f (X, X )] for continuous, bilinear functions f : C([0, 1]) 2 R. This is done by controlling the covariance function. Since S n and B do not share their covariance function (of course E[Ss n, St n ] min(s, t)) we also consider the process B n defined by B n t = B nt n + (nt nt ) ( B nt +1 n B nt n ). B

54 Proof of Donsker s Theorem (Sketch) An contraction argument shows for δ < ε/2. ζ 2+ε (S n, B n ) = O(n δ )

55 Proof of Donsker s Theorem (Sketch) An contraction argument shows for δ < ε/2. Together with ζ 2+ε (S n, B n ) = O(n δ ) B n B 0 a.s.

56 Proof of Donsker s Theorem (Sketch) An contraction argument shows for δ < ε/2. Together with ζ 2+ε (S n, B n ) = O(n δ ) B n B 0 a.s. Donsker s Theorem follows with the following properties of Zolotarev s distance.

57 Properties of ζ s in C([0, 1])

58 Properties of ζ s in C([0, 1]) Theorem Let B = C([0, 1]) and 0 < s 3, i.e, m {0, 1, 2}. Let (X n ) n 1, X be random variables in (C([0, 1]), ) where, for each n 1, X n is piecewise linear with intervals of length at least r n. If ) ζ s (X n, X ) = o (log m 1rn, as n, then X n X in distribution.

59 Properties of ζ s in C([0, 1]) Theorem Let B = C([0, 1]) and 0 < s 3, i.e, m {0, 1, 2}. Let (X n ) n 1, X be random variables in (C([0, 1]), ) where, for each n 1, X n is piecewise linear with intervals of length at least r n. If ) ζ s (X n, X ) = o (log m 1rn, as n, then X n X in distribution. Proof: This basically follows from a result of Barbour from the context of Stein s method.

60 Properties of ζ s in C([0, 1]) Theorem Suppose 0 < s 3, i.e, m {0, 1, 2}. Let (X n ) n 1, (Y n ) n 1, Z be random variables in (C([0, 1]), ) where, for each n 1, X n, Y n are piecewise linear with intervals of length at least r n. If Y n Z in distribution and ) ζ s (X n, Y n ) = o (log m 1rn, as n, then X n Z in distribution.