Haar function construction of Brownian bridge and Brownian motion Wellner; 10/30/2008

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1 Haar function construction of Brownian bridge and Brownian motion Wellner; 1/3/28 Existence of Brownian motion and Brownian bridge as continuous processes on C[, 1] The aim of this subsection to convince you that both Brownian motion and Brownian bridge exist as continuous Gaussian processes on [, 1], and that we can then extend the definition of Brownian motion to [, ). Definition 1. Brownian motion (or standard Brownian motion, or a Wiener process) S is a Gaussian process with continuous sample functions and: (i) S() = ; (ii) E(S(t)) =, t 1; (iii) E{S(s)S(t)} = s t, s, t 1. Definition 2. A Brownian bridge process U is a Gaussian process with continuous sample functions and: (i) U() = U(1) = ; (ii) E(U(t)) =, t 1; (iii) E{U(s)U(t)} = s t st, s, t 1. Theorem 1. Brownian motion S and Brownian bridge U exist. Proof. We first construct a Brownian bridge process U. Let t t 1/2, h (t) h(t) 1 t 1/2 t 1, elsewhere. For n 1 let h nj (t) 2 n/2 h(2 n t j), j =,..., 2 n 1. (2) For example, h 1 (t) = 2 1/2 h(2t), h 11 (t) = 2 1/2 h(2t 1), while h 2 (t) = 2 1 h(4t), h 21 (t) = 2 1 h(4t 1), h 22 (t) = 2 1 h(4t 2), h 23 (t) = 2 1 h(4t 3). Note that h nj (t) 2 n/ The functions {h nj : j =,..., 2 n 1, n } are called the Schauder functions; they are integrals of the orthonormal (with respect to Lebesgue 1 (1)

2 measure on [, 1]) family of functions {g nj : j =,..., 2 n 1, n } called the Haar functions defined by Thus 1 and g (t) g(t) 21 [,1/2] (t) 1, g nj (t) 2 n/2 g (2 n t j), j =,..., 2 n 1, n 1. g 2 nj(t)dt = 1, 1 h nj (t) = g nj (t)g n j (t)dt = if n n, or j j, t (3) g nj (s)ds, t 1. (4) Furthermore, the family {g nj } 2n 1 j=, n {g( /2)} is complete: any f L 2(, 1) has an expansion in terms of the g s. In fact the Haar basis is the simplest wavelet basis of L 2 (, 1), and is the starting point for further developments in the area of wavelets. Now let {Z nj } 2n 1 j=, n be independent identically distributed N(, 1) random variables; if we wanted, we could construct all these random variables on the probability space ([, 1], B [,1], λ). Define For m > k where V n (t, ω) = U m (t, ω) = U m (t, ω) U k (t, ω) = V n (t, ω) 2 n 1 j= 2 n 1 j= Z nj (ω)h nj (t), m V n (t, ω). n= m n=k+1 V n (t, ω) m n=k+1 V n (t, ω) (5) Z nj (ω) h nj (t) 2 (n/2+1) max j 2 n 1 Z nj(ω) (6) since the h nj, j =,..., 2 n 1 are on disjoint t intervals. 2

3 Now P (Z nj > z) = 1 Φ(z) z 1 φ(z) for z > (by Mill s ratio ) so that P ( Z nj 2 n) = 2P ((Z nj 2 n) 2 (2 n) 1 e 2n. (7) 2π Hence ( P max Z nj 2 ) n 2 n P ( Z 2 n) 2n n 1/2 e 2n ; (8) j 2 n 1 2π since this is a term of a convergent series, by the Borel-Cantelli lemma max j 2 n 1 Z nj 2 n occurs infinitely often with probability zero; i.e. except on a null set, for all ω there is an N = N(ω) such that max j 2 n 1 X nj (ω) < 2 n for all n > N(ω). Hence sup U m (t) U k (t) t 1 m n=k+1 2 n/2 n 1/2 (9) for all k, m N N(ω). Thus U m (t, ω) converges uniformly as m with probability one to the (necessarily continuous) function U(t, ω) V n (t, ω). (1) n= Define U on the exceptional set. Then U is continuous for all ω. Now {U(t) : t 1} is clearly a Gaussian process since it is the sum of Gaussian processes. We now show that U is in fact a Brownian bridge: by formal calculation (it remains only to justify the interchange of summation 3

4 and expectation), { E{U(s)U(t)} = E V n (s) = = = = = n= } V m (t) m= E{V n (s)v n (t)} n= E n= n= { 2 n 1 j= 2 n 1 s j= 2 n 1 n= j= 1 = s t st 1 Z nj s g nj dλ t 2 n 1 t g nj dλ Z nk 1 [,s] g nj dλ g nj dλ 1 1 [,s] (u)1 [,t] (u)du st k= g nk dλ } 1 [,t] g nj dλ + st st where the next to last equality follows from Parseval s identity. Thus U is Brownian bridge. Now let Z be one additional N(, 1) random variable independent of all the others used in the construction, and define S(t) U(t) + tz = Then S is also Gaussian with mean and V n (t) + tz. (11) n= Cov[S(s), S(t)] = Cov[U(s) + sz, U(t) + tz] = Cov[U(s), U(t)] + stv ar(z) = s t st + st = s t. Thus S is Brownian motion. Since U has continuous sample paths, so does S. The following figures illustrate the construction given in the theorem. 4

5 Figure 1: The Schauder function h. Here is the Mathematica code that produced the plots for the case m = 8. Needs["Histograms "] ndist = NormalDistribution[, 1] m = 8 Z = Table[RandomReal[ndist, 2^n], {n,, m}] Histogram[Flatten[Z]] h[t_] := t /; <= t <= 1/2 h[t_] := 1 - t /; 1/2 < t <= 1 h[t_] := /; t < t > 1 h1[t_, n_, j_] := 2^(-n/2)*h[2^n *t - j] V[t_] := Z[[1, 1]]*h[t] V[t_, n_] := Sum[Z[[n + 1, j + 1]]*h1[t, n, j], {j,, 2^n - 1}] U[t_, m_] := Sum[V[t, n], {n, 1, m}] + V[t] P1 = Plot[h[t], {t,, 1}] P2 = Plot[h1[t, 1, ], {t,, 1}] P3 = Plot[h1[t, 1, 1], {t,, 1}] Show[P2, P3] P4 = Plot[h1[t, 2, ], {t,, 1}, PlotRange -> {,.5}] P5 = Plot[h1[t, 2, 1], {t,, 1}, PlotRange -> {,.5}] P6 = Plot[h1[t, 2, 2], {t,, 1}, PlotRange -> {,.5}] 5

6 Figure 2: The Schauder functions h 1 and h 11. P7 = Plot[h1[t, 2, 3], {t,, 1}, PlotRange -> {,.5}] Show[P4, P5, P6, P7] Plot[V[t], {t,, 1}] Plot[V[t, 1], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.,., 1.], Plot[V[t, 2], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.1,.,.9], Plot[V[t, 3], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.2,.,.8], Plot[V[t, 4], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.3,.,.7], Plot[V[t, 5], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.4,.,.6], Plot[V[t, 6], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.5,.,.5], Plot[V[t, 7], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.6,.,.4], Plot[V[t, 8], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.7,.,.3], Plot[U[t, 8], {t,, 1}, PlotStyle -> {Thickness[1/2], RGBColor[.,.5,.5], 6

7 Figure 3: The Schauder functions h 2, h 21, h 22, h Figure 4: A sample path of the random function V (t). 7

8 Figure 5: A sample path of the random function V 1 (t) Figure 6: A sample path of the random function V 2 (t). 8

9 Figure 7: A sample path of the random function V 3 (t) Figure 8: A sample path of the random function V 4 (t). 9

10 Figure 9: A sample path of the random function V 5 (t) Figure 1: A sample path of the random function V 6 (t). 1

11 Figure 11: A sample path of the random function V 7 (t) Figure 12: A sample path of the random function V 8 (t). 11

12 Figure 13: A sample path of the random function U 8 (t) Figure 14: A sample path of the random function S 8 (t). 12