Krzysztof Burdzy University of Washington. = X(Y (t)), t 0}

VARIATION OF ITERATED BROWNIAN MOTION Krzysztof Burdzy University of Washington 1. Introduction and main results. Suppose that X 1, X 2 and Y are independent standard Brownian motions starting from 0 and let { X 1 t) if t 0, 1) Xt) = X 2 t) if t < 0. We will consider the process 2) {Zt) df = XY t)), t 0} which we will call iterated Brownian motion or simply IBM. It can be proved that Z uniquely determines X and Y see Burdzy 1992) for a precise statement). A Law of Iterated Logarithm for IBM is also proved in Burdzy 1992). We consider IBM to be a process of independent interest but there exists an intriguing relationship between this process strictly speaking its modification) and squared Laplacian which was discovered by Funaki 1979). So far, the probabilistic approach to bi-harmonic functions is much less successful than the probabilistic treatment of harmonic functions. Krylov 1960) and Hochberg 1978) attacked the problem using a signed finitely additive measure with infinite variation. M adrecki 1992) and M adrecki and Rybaczuk 1992) have a genuine probabilistic approach but their processes take values in an exotic space. Both models are used to define stochastic integrals for processes with 4-th order scaling properties. Higher order variations of the process play an important role in M adrecki and Rybaczuk s construction of the stochastic integral. It is no surprise that the 4-th variation of their process is a deterministic linear function. The quadratic variation of their process is, in a suitable sense, a Brownian motion. See 3.16) in Hochberg s paper for a result with similar intuitive content. In this paper, we study higher order variations of IBM with view towards possible applications to the construction of the stochastic integral with respect to IBM. We prove that the 4-th variation of IBM is a deterministic linear function. This clearly means that the quadratic variation is infinite although we do not prove this). We show that, in a weak sense, the signed quadratic variation of IBM is distributed like Brownian motion. Suppose that Λ = {s = t 0 t 1 t n = t} is a partition of [s, t]. The mesh of the partition Λ is defined as Λ = df max 1 k n t k t k 1. Supported in part by NSF grant DMS 91-00244 and AMS Centennial Research Fellowship 1 Typeset by AMS-TEX

2 KRZYSZTOF BURDZY Theorem 1. i) Fix some 0 s < t. The following limit exists in L p for every p <. n 3) lim Zt k ) Zt k 1 )) 4 = 3t s). Λ 0 k=1 ii) Suppose in addition that t k t k 1 = t s)/n for every k. Then n 4) lim Zt k ) Zt k 1 )) 3 = 0 in L p, p <. Λ 0 k=1 Theorem 2. Suppose that t 0 = 0 and t k t k 1 = 1/n for k 1. Let m V n t m ) = Zt k ) Zt k 1 )) 2 sgnzt k ) Zt k 1 )). k=1 Extend V n continuously to [0, ) by linear interpolation on each interval [t k 1, t k ]. The processes {V n s), s 0} converge in distribution as n to a Brownian motion {Bs), s 0} with variance VarBs) = 3s. Remarks. i) The assumption that t k t k 1 = t j t j 1 for all j and k is imposed for convenience in Theorem 1 ii) and Theorem 2. The assumption seems to be unnecessary but it makes the calculations somewhat more manageable. ii) A heuristic argument suggests that for a fixed s > 0, the sequence {V n s)} n 1 has no subsequences converging in probability. iii) The models considered by Funaki 1979) and M adrecki and Rybaczuk 1992) involve complex numbers. It might be worth having a look at the complex version of IBM. Suppose that Y is a standard one-dimensional Brownian motion, X is a two-sided complex i.e., two-dimensional but written in complex notation) Brownian motion and Zt) = XY t)). Let V n t m ) = m k=1 Zt k) Zt k 1 )) 2, in the notation of Theorem 2. Then Theorem 2 holds for this complex analogue of quadratic variation. The limiting process B for V n s is a complex i.e., two-dimensional) Brownian motion with the quadratic variation 3 times as large as the standard one. This result may be proved just like Theorem 2 by using the method of moments. The proof of Theorem 2 is based on estimates of moments of V n s. The estimates are quite delicate and it would take enormous amount of space to write them down in all detail. We will carefully examine one crucial estimate and indicate how this can be generalized to other moments. We would like to thank Ron Pyke for simple proofs of Lemmas 1 and 2 below. 2. Proofs. Throughout the paper, c will stand for a strictly positive and finite constant which may change the value from line to line. We will need the following standard estimate. Let a > 0. 5) a x 2 1 2πt exp x 2 /2t)dx = x t/2π exp x 2 /2t) a t/2π exp a 2 /2t) + x= x=a a + a t/2π exp x 2 /2t)dx x/a) t/2π exp x 2 /2t)dx = a t/2π exp a 2 /2t) + t/a) t/2π exp a 2 /2t) = a + t/a) t/2π exp a 2 /2t).

VARIATION OF ITERATED BROWNIAN MOTION 3 The next estimate may be derived in an analogous way using integration by parts. 6) a x 4 1 2πt exp x 2 /2t)dx a 3 + ta + t 2 /a) t/2π exp a 2 /2t). Lemma 1. Suppose that for every integer) k 1, k 2, the k-th moment of a random variable R is the same as that of a normal random variable U with mean 0 and variance σ 2. Then R and U have the same distribution. The point of the lemma is that we do not assume that the variances of R and U are identical. The lemma would follow immediately from known results see Durrett 1991) Theorem 3.9)) if we added this assumption. Proof. The 2k-th moment µ 2k of U is equal to σ 2k 2k 1)!!. Thus lim sup k µ 1/2k 2k /2k = lim supσ 2k 2k 1)!!) 1/2k /2k = 0. k Durrett 1991) shows in the proof of Theorem 3.9) that this implies that the characteristic function ϕ U has the following series expansion valid on the whole real line. ϕ U t) = 1 + k=1 t k k! ϕk) U 0). The characteristic function ϕ R of R is represented by an analogous series. For every k 2, the k-th moment of R is the same as that of U so ϕ k) R 0) = ϕk) U 0) and it follows that the series for ϕ R and ϕ U may differ by at most one term. Hence ϕ R t) = ϕ U t) + at 2 and ϕ R t)/t 2 = ϕ U t)/t 2 + a for all t 0. Since ϕ U t) 1 and ϕ R t) 1, 0 = lim t ϕ R t)/t 2 = lim t ϕ U t)/t 2 + a = a. It follows that a = 0 and, therefore, U and R have identical characteristic functions. Lemma 2. Let f σ x) denote the centered normal density with standard deviation σ and let ψx, σ) = f σ 0) f σ x). For all k, n 0 there exists c = ck, n) < such that x n ψx, ρ 1 )... ψx, ρ k )f σ x)dx cρ 3 1... ρ 3 for all ρ 1,..., ρ k 0 c does not depend on ρ j s or σ). Proof. Since 1 e y y for all y 0, ψx, ρ) = 1 1 exp x 2 /2ρ 2 )) 1 2πρ 2 x 2 2π ρ 3. k σn+2k

4 KRZYSZTOF BURDZY Let ξ denote a standard normal random variable. Then the integral in the statement of the lemma equals E σξ n k j=1 ψσξ, ρ j ) E σξ n+2k 2 2π) k k j=1 ρ 3 j = ck, n)σ n+2k k j=1 ρ 3 j. Proof of Theorem 1 i). We will only prove the convergence in L p for p = 2. The general case may be treated in an analogous way. Recall that Λ = {s = t 0 t 1 t n = t} is a partition of [s, t]. Let i t = df t i t i 1 and i Z = df Zt i ) Zt i 1 ). We have 7) [ n 2 [ n i Z) 4 3t s)] = i Z) 4 3 i t) i=1 = i=1 ] 2 n i Z) 4 3 i t) j Z) 4 3 j t). i,j=1 It will suffice to prove that the expectation of the above random variable goes to 0 as Λ goes to 0. Fix some α 0, 1) and suppose that i j. Fix some numbers u i 1 < u i < u j 1 < u j. We will compute E i Z) 4 3 i t) j Z) 4 3 j t) given A 1 = A 1 u i 1, u i, u j 1, u j ) df = {Y t i 1 ) = u i 1, Y t i ) = u i, Y t j 1 ) = u j 1, Y t j ) = u j }. Given this condition, the processes {Xu i +t) Xu i ), t 0} and {Xu i t) Xu i ), t 0} are independent standard Brownian motions. Given A 1, the random variable i Z) 4 3 i t is defined in terms of the first process and j Z) 4 3 j t is defined in terms of the second one. Since EXs 1 ) Xs 2 )) 4 = 3s 1 s 2 ) 2, it follows that E i Z) 4 3 i t) j Z) 4 3 j t) A 1 ) 8) = E i Z) 4 3 i t A 1 )E j Z) 4 3 j t A 1 ) = EXu i ) Xu i 1 )) 4 3 i t A 1 )EXu j ) Xu j 1 )) 4 3 j t A 1 ) = 3u i u i 1 ) 2 3 i t)3u j u j 1 ) 2 3 j t). The same argument works for any u i 1, u i, u j 1 and u j such that the interval with endpoints u i 1 and u i is disjoint from the interval with endpoints u j 1 and u j. Suppose that r > 2 Λ α/2. Let A 2 = A 2 r) df = { Y t i ) Y t j 1 ) = r, Y t i 1 ) Y t i ) < i t α/2, Y t j ) Y t j 1 ) < j t α/2 }. The increments Y t i 1 ) Y t i ) and Y t j ) Y t j 1 ) are independent given A 2. If the event A 2 occurs then the interval with endpoints Y t i 1 ) and Y t i ) is disjoint from the interval

VARIATION OF ITERATED BROWNIAN MOTION 5 with endpoints Y t j 1 ) and Y t j ). Hence we may integrate over suitable u i 1, u i, u j 1 and u j in 8) to obtain 9) Since E i Z) 4 3 i t) j Z) 4 3 j t) A 2 ) = i t α/2 3u 2 1 3 i t) i t α/2 2π i t exp u2 /2 i t)du j t α/2 5) implies that for some β > 0 and small i t, 10) i t α/2 3v 2 1 3 j t) j t α/2 2π j t exp v2 /2 j t)dv. 3u 2 1 3 i t) 2π i t exp u2 /2 i t)du = 0, 3u 2 1 3 i t) i t α/2 2π i t exp u2 /2 i t)du = 2 3u 2 1 3 i t) i t α/2 2π i t exp u2 /2 i t)du 6u 2 1 i t α/2 2π i t exp u2 /2 i t)du 6 i t α/2 + i t/ i t α/2 ) i t/2π exp i t α /2 i t) exp i t β ). This and 9) show that for small Λ 11) E i Z) 4 3 i t) j Z) 4 3 j t) A 2 ) exp i t β j t β ). Let A 3 = A 3 Λ ) df = { Y t i ) Y t j 1 ) > 2 Λ α/2, Y t i 1 ) Y t i ) < i t α/2, Y t j ) Y t j 1 ) < j t α/2 }. It follows from 11) that 12) E i Z) 4 3 i t) j Z) 4 3 j t)1 A3 ) E i Z) 4 3 i t) j Z) 4 3 j t) A 3 ) exp i t β j t β ). Since EX 8 t = ct 4, 13) E i Z) 8 + 9 i t) 2 Y t i 1 ) Y t i ) = r) = cr 4 + 9 i t) 2. An argument similar to that in 10) except that we would use 6) rather than 5)) gives for small i t and some η > 0 14) E i Z) 8 + 9 i t) 2 )1 {Y ti 1 ) Y t i )> α/2 i } ) = 2 cu 4 + 9 i t) 2 1 ) i t α/2 2π i t exp u2 /2 i t)du exp it η ).

6 KRZYSZTOF BURDZY Let A 4 = A 4 Λ ) df = { Y t i ) Y t j 1 ) > 2 Λ α/2 } Then 14) yields [ { Y t i 1 ) Y t i ) > i t α/2 } { Y t j ) Y t j 1 ) > j t α/2 } ]. 15) E i Z) 4 3 i t) j Z) 4 3 j t)1 A4 ) E i Z) 4 3 i t) 2 1 A4 ) 1/2 E j Z) 4 3 j t) 2 1 A4 ) 1/2 E2 i Z) 8 + 9 i t) 2 )1 A4 ) 1/2 E2 j Z) 8 + 9 j t) 2 )1 A4 ) 1/2 2 exp i t η /2 j t η /2). We define A 5 = A 5 Λ ) to be { Y t i ) Y t j 1 ) > 2 Λ α/2 }. Combining 12) and 15) yields for small i t and j t 16) E i Z) 4 3 i t) j Z) 4 3 j t)1 A5 ) exp i t β j t β ) + 2 exp i t η /2 j t η /2) 3 exp 2 i t β j t η ). Choose α, γ 0, 1) such that δ df = α/2 γ/2 > 0. Let A 6 = { Y t i ) Y t j 1 ) 2 Λ α/2 } and suppose that t i t j 1 > Λ γ. Taking the expectation on both sides of 13) gives It is easy to see that By the independence of increments of Y, Hence E i Z) 8 + 9 i t) 2 ) = c i t 2. P A 6 ) c Λ α/2 / Λ γ/2 = c Λ δ. E i Z) 8 + 9 i t) 2 )1 A6 ) = c i t 2 P A 6 ) = c i t 2 Λ δ, E j Z) 8 + 9 j t) 2 )1 A6 ) = c j t 2 Λ δ. 17) E i Z) 4 3 i t) j Z) 4 3 j t)1 A6 ) E i Z) 4 3 i t) 2 1 A6 ) 1/2 E j Z) 4 3 j t) 2 1 A6 ) 1/2 E2 i Z) 8 + 9 i t) 2 )1 A6 ) 1/2 E2 j Z) 8 + 9 j t) 2 )1 A6 ) 1/2 2c i t 2 Λ δ ) 1/2 c j t 2 Λ δ ) 1/2 = 2c i t j t Λ δ. A similar application of the Schwarz inequality gives 18) E i Z) 4 3 i t) j Z) 4 3 j t) c i t j t

VARIATION OF ITERATED BROWNIAN MOTION 7 for any i and j. We conclude from 16) and 17) that for sufficiently small Λ 19) n E i Z) 4 3 i t) j Z) 4 3 j t) i,j=1 t i t j 1 > Λ γ n 3 exp 2 i t β j t η ) + 2c i t j t Λ δ ) i,j=1 t i t j 1 > Λ γ c Λ δ. As for the remaining terms, we use the estimate 18). n E i Z) i,j=1 t i t j 1 Λ γ This and 19) show that 4 3 i t) j Z) 4 3 j t) n c i t j t c Λ i,j=1 t i t j 1 Λ γ γ. 20) n i,j=1 E i Z) 4 3 i t) j Z) 4 3 j t) 0 as Λ 0. This completes the proof of 3) in the case p = 2. We can prove in a similar way that 21) lim for any p <. p <. n Λ 0 j 1,...,j p =1 E p jk Z) 4 3 jk t) = 0 k=1 This can be used to show that the limit in 3) exists in L p for every Proof of Theorem 2. The proof will be based on the method of moments, i.e., we will show that the moments of V converge to the moments of B. Recall that t 0 = 0 and t k t k 1 = 1/n for k 1. Fix some 0 s 1 < s 2. Let Θ = Θn) be the set of all k such that s 1 t k 1 < t k s 2. The set Θn) is non-empty for sufficiently large n. Recall that k t = t k t k 1, k Z = Zt k ) Zt k 1 ), k Y = Y t k ) Y t k 1 ) and let ± k Z2 = Zt k ) Zt k 1 )) 2 sgnzt k ) Zt k 1 )). We start with some estimates needed for computing the moments of the increments of V n. For every s, the distribution of Xs) is normal so EX 2j s) = 2j 1)!! s j Durrett 1991), Excercise 3.18). By conditioning on the value of k Y we obtain for some d j > 0, 22) E k Z) 2j = = EX 2j 1 s) 2π k t exp s2 /2 k t)ds 2j 1)!! s j 1 2π k t exp s2 /2 k t)ds = d j k t) j/2.

8 KRZYSZTOF BURDZY Hence, E k Z) 2j < for all j <. The main contribution in our moment estimates will come from the expectations of the form 23) E ± k 1 Z 2 ) 2... ± k m Z 2 ) ). 2 Suppose that m = 2. We have ± j Z2 ) 2 ± k Z2 ) 2 = j Z 4 k Z 4 [ ] 2 = j Z) 4 3 j t) 9 j t k t + 2 3 j Z 4 k t. j Θ It is easy to check that d 2 = 3 in 22). Thus and, therefore, E E3 j Z 4 k t = 9 j t k t [ ] 2 ± j Z2 ) 2 ± k Z2 ) 2 = E j Z) 4 3 j t) + j Θ 9 j t k t. The expectation on the right hand side goes to 0 as n goes to, by 20). Hence 24) lim E ± j Z2 ) 2 ± k Z2 ) 2 = lim 9 j t k t = 9s 2 s 1 ) 2. In order to estimate the expectations in 23) for m 3, we use induction. We will treat only the case m = 3. i, By 21), 22) and 24), ± i Z2 ) 2 ± j Z2 ) 2 ± k Z2 ) 2 = i, [ ] 3 = j Z) 4 3 j t) + j Θ 3 i, 9 i Z 4 j t k t + 3 i Z 4 j Z 4 k Z 4 i, i, [ 3 lim E j Z) 4 3 j t)] = 0, j Θ 27 i t j t k t 3 i Z 4 j Z 4 k t. [ lim E 3 ] [ 9 i Z 4 j t k t = lim E 3 ] 27 i t j t k t = 81s 2 s 1 ) 3, i, i,

VARIATION OF ITERATED BROWNIAN MOTION 9 [ lim E 3 ] [ 3 i Z 4 j Z 4 k t = lim s 2 s 1 )E 3 ] 3 i Z 4 j Z 4 = 81s 2 s 1 ) 3. i, i,j Θ Hence, lim E i, ± i Z2 ) 2 ± j Z2 ) 2 ± k Z2 ) 2 = lim E i, 27 i t j t k t = 27s 2 s 1 ) 3. In the same way, using induction, we may prove that ) 25) lim E ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 = 3 m s 2 s 1 ) m. Suppose that q 1,..., q m 1 are integers and at least one of them is strictly greater than 1. By Hölder s inequality and 22), E[ ± k 1 Z 2 ) 2q 1 ± k 2 Z 2 ) 2q 2... ± k m Z 2 ) 2q m ] and, therefore, 26) lim E ± k 1 Z 2 ) 2mq 1 ) 1/m... E ± k m Z 2 ) 2mq m ) 1/m d 2mq1 k1 t) mq 1 ) 1/m... d 2mqm km t) mq m ) 1/m c1/n) q 1+ +q m, ± k 1 Z 2 ) 2q 1 ± k 2 Z 2 ) 2q 2... ± k m Z 2 ) 2q m The absolute value of the difference between ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 and k 1,...,k m distinct ) ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 is bounded by a finite sum of the expressions of the form ± k 1 Z 2 ) 2q 1 ± k 2 Z 2 ) 2q 2... ± k i Z 2 ) 2q i k 1,...,k i Θ lim nm c1/n) q 1+ +q m = 0. where at least one of the q j s is greater than 1. It follows from 25) and 26) that 27) lim E ) ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 = 3 m s 2 s 1 ) m. k 1,...,k m distinct Next we tackle the expectations of the form ) E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) q m,

10 KRZYSZTOF BURDZY where q 1,..., q m 1 are integers but they are not necessarily even. We will tacitly assume that the sum is taken over indices which are pairwise distinct the other terms appear in sums with different exponents q k ). We will illustrate the method by analyzing in detail only one sum, namely, E k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 ± k 1 Z 2 ± k 3 Z 2 ). Note that the indices in the last sum are ordered the sum with unordered indices may be obtained by adding a finite number of sums with ordered indices. Let A 1 = A 1 u k1 1, u k1, u k2 1, u k2, u k3 1, u k3, u k4 1, u k4 ) df = {Y t ki 1) = u ki 1, Y t ki ) = u ki, i = 1, 2, 3, 4}. Let A 2 denote the event that there exists a number a such that for each i, the interval with endpoints u ki 1 and u ki is either contained in a, ) or in, a) and that each half-line contains at least one of these intervals. Suppose for a moment that u ki 1 and u ki are such that A 2 holds, for example, the intervals corresponding to i = 1, 2 are in a, ) and the other two are contained in the other half-line. The same argument that leads to 8) gives in the present case 28) E ± k 1 Z 2 ± k 3 Z 2 A 1 ) = E ± k 1 Z 2 A 1 )E ± k 3 Z 2 A 1 ). Both conditional expectations on the right hand side are equal to zero since the random variables have symmetric conditional) distributions. If the event A 2 is realized in some other way, the conditional expectation on the left hand side of 28) may be factored in some other way such that at least one conditional expectation on the right hand side is equal to 0 because of symmetry of the involved distribution. Hence, 29) E ± k 1 Z 2 ± k 3 Z 2 1 A2 ) = 0. For arbitrary values of u i s, 30) E ± k 1 Z 2 ± k 3 Z 2 A 1 ) E ± k 1 Z 2 ) 4 A 1 )) 1/2 E ± k 2 Z 2 ) 2 A 1 )) 1/2 E ± k 3 Z 2 ) 4 A 1 )) 1/2 E ± k 4 Z 2 ) 2 A 1 )) 1/2 = EX 8 u k1 1 u k1 )) 1/2 EX 4 u k2 1 u k2 )) 1/2 EX 8 u k3 1 u k3 )) 1/2 EX 4 u k4 1 u k4 )) 1/2 c u k1 1 u k1 2 u k2 1 u k2 u k3 1 u k3 2 u k4 1 u k4. Note that E ± k 1 Z 2 ± k 3 Z 2 A 1 u k1 1, u k1, u k2 1, u k2, u k3 1, u k3, u k4 1, u k4 )) 31) = E ± k 1 Z 2 ± k 3 Z 2 A 1 u k1 1, u k1, u k2, u k2 1, u k3 1, u k3, u k4 1, u k4 ))

VARIATION OF ITERATED BROWNIAN MOTION 11 because when we exchange the roles of u k2 1 and u k2, we change, in a sense, the sign of ± k 2 Z 2. For the same reason we have E ± k 1 Z 2 ± k 3 Z 2 A 1 u k1 1, u k1, u k2 1, u k2, u k3 1, u k3, u k4 1, u k4 )) 32) = E ± k 1 Z 2 ± k 3 Z 2 A 1 u k1 1, u k1, u k2 1, u k2, u k3 1, u k3, u k4, u k4 1)). Let ρ = ρr 1, r 2, r 3, r 4 ) = r 1 + r 2 + r 3 + r 4. We claim that E ± k 1 Z 2 ± k 3 Z 2 ) [ r 1 2 r 2 r 3 2 r 4 P Y t k1 ) Y t k1 1) dr 1 )P Y t k2 ) Y t k2 1) dr 2 ) ] P Y t k3 ) Y t k3 1) dr 3 )P Y t k4 ) Y t k4 1) dr 4 ) [ 1 r5 ρ1 r6 ρ P Y t k2 1) Y t k1 ) dr 5 )P Y t k3 1) Y t k2 ) dr 6 ) ] P Y t k2 1) Y t k1 ) dr 5 + r 2 ))P Y t k3 1) Y t k2 ) dr 6 r 2 )) [ 33) P Y t k4 1) Y t k3 ) dr 7 ) P Y t k4 1) Y t k3 ) dr 7 + r 4 ))]. 1 r7 ρ The terms in the first pair of large square brackets are justified by 30). The terms in the second and third pair of large square brackets come from 31)-32). The presence of indicator functions follows from 29). We will now estimate 1 r5 ρ1 r6 ρ[p Y t k2 1) Y t k1 ) dr 5 )P Y t k3 1) Y t k2 ) dr 6 ) 34) P Y t k2 1) Y t k1 ) dr 5 + r 2 ))P Y t k3 1) Y t k2 ) dr 6 r 2 ))]. Recall ψx, σ) from Lemma 2. Since t k2 1 t k1 = k 2 1 k 1 )/n, the standard deviation of Y t k2 1) Y t k1 ) is equal to k 2 1 k 1 )/n) 1/2. Hence P Y t k2 1) Y t k1 ) dr 5 ) P Y t k2 1) Y t k1 ) dr 5 + r 2 )) 35) ψr 5, k 2 1 k 1 )/n) 1/2 ) + ψr 5 + r 2, k 2 1 k 1 )/n) 1/2 ))dr 5 2ψ2ρ, k 2 1 k 1 )/n) 1/2 )dr 5 provided r 5 ρ. Similarly, 36) P Y t k3 1) Y t k2 ) dr 6 ) P Y t k3 1) Y t k2 ) dr 6 r 2 )) 2ψ2ρ, k 3 1 k 2 )/n) 1/2 )dr 6 assuming r 6 ρ. We will assume temporarily that k j k j 1 > 1 for all j in order to avoid normal variables with zero variance and in order to be able to use Lemma 2. We will get rid of this assumption later.

12 KRZYSZTOF BURDZY For any reals a, a, b and b, ab a + a)b + b) a b + b a + a b. This, 35) and 36) imply P Y t k2 1) Y t k1 ) dr 5 )P Y t k3 1) Y t k2 ) dr 6 ) P Y t k2 1) Y t k1 ) dr 5 + r 2 ))P Y t k3 1) Y t k2 ) dr 6 r 2 )) 2ψ2ρ, k 2 1 k 1 )/n) 1/2 )dr 5 P Y t k3 1) Y t k2 ) dr 6 ) + 2ψ2ρ, k 3 1 k 2 )/n) 1/2 )dr 6 P Y t k2 1) Y t k1 ) dr 5 ) + 2ψ2ρ, k 2 1 k 1 )/n) 1/2 )dr 5 2ψ2ρ, k 3 1 k 2 )/n) 1/2 )dr 6. The integral in 34) is, therefore, bounded by 1 r5 ρ1 r6 ρ 37) [2ψ2ρ, k 2 1 k 1 )/n) 1/2 )dr 5 P Y t k3 1) Y t k2 ) dr 6 ) + 2ψ2ρ, k 3 1 k 2 )/n) 1/2 )dr 6 P Y t k2 1) Y t k1 ) dr 5 ) + 2ψ2ρ, k 2 1 k 1 )/n) 1/2 )dr 5 2ψ2ρ, k 3 1 k 2 )/n) 1/2 )dr 6 ] cψ2ρ, k 2 1 k 1 )/n) 1/2 )ρ 2 k 3 1 k 2 )/n) 1/2 + cψ2ρ, k 3 1 k 2 )/n) 1/2 )ρ 2 k 2 1 k 1 )/n) 1/2 + cψ2ρ, k 2 1 k 1 )/n) 1/2 )ρ 2 ψ2ρ, k 3 1 k 2 )/n) 1/2 ). We can prove in the same way that 1 r7 ρ[p Y t k4 1) Y t k3 ) dr 7 ) P Y t k4 1) Y t k3 ) dr 7 + r 4 ))] 38) cψ2ρ, k 4 1 k 3 )/n) 1/2 )ρ. We now substitute this estimate and 37) into 33) to obtain E ± k 1 Z 2 ± k 3 Z 2 ) r 1 2 r 2 r 3 2 r 4 P Y t k1 ) Y t k1 1) dr 1 )P Y t k2 ) Y t k2 1) dr 2 ) P Y t k3 ) Y t k3 1) dr 3 )P Y t k4 ) Y t k4 1) dr 4 ) [ cψ2ρ, k2 1 k 1 )/n) 1/2 )ρ 2 k 3 1 k 2 )/n) 1/2 39) + cψ2ρ, k 3 1 k 2 )/n) 1/2 )ρ 2 k 2 1 k 1 )/n) 1/2 + cψ2ρ, k 2 1 k 1 )/n) 1/2 )ρ 2 ψ2ρ, k 3 1 k 2 )/n) 1/2 ) ] cψ2ρ, k 4 1 k 3 )/n) 1/2 )ρ.

VARIATION OF ITERATED BROWNIAN MOTION 13 By multiplying out the expression in brackets on the right hand side we obtain three terms under the integral sign. The first one is equal to r 1 2 r 2 r 3 2 r 4 cψ2ρ, k 2 1 k 1 )/n) 1/2 )ρ 2 k 3 1 k 2 )/n) 1/2 cψ2ρ, k 4 1 k 3 )/n) 1/2 )ρp Y t k1 ) Y t k1 1) dr 1 )P Y t k2 ) Y t k2 1) dr 2 ) P Y t k3 ) Y t k3 1) dr 3 )P Y t k4 ) Y t k4 1) dr 4 ) = r 1 2 r 2 r 3 2 r 4 cψ2 r 1 + r 2 + r 3 + r 4 ), k 2 1 k 1 )/n) 1/2 ) 40) r 1 + r 2 + r 3 + r 4 ) 2 k 3 1 k 2 )/n) 1/2 cψ2 r 1 + r 2 + r 3 + r 4 ), k 4 1 k 3 )/n) 1/2 ) r 1 + r 2 + r 3 + r 4 ) P Y t k1 ) Y t k1 1) dr 1 )P Y t k2 ) Y t k2 1) dr 2 ) P Y t k3 ) Y t k3 1) dr 3 )P Y t k4 ) Y t k4 1) dr 4 ). Note that 4 ψ2 r 1 + r 2 + r 3 + r 4 ), k 2 1 k 1 )/n) 1/2 ) ψ8 r j, k 2 1 k 1 )/n) 1/2 ) j=1 and the standard deviation of Y t kj ) Y t kj 1) is equal to n 1/2. This and Lemma 2 can be used to show that 40) is bounded by ck 2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2. The other terms in 39) may be treated in a similar way so 41) E ± k 1 Z 2 ± k 3 Z 2 ) ck 2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2 + ck 3 1 k 2 )/n) 3/2 k 2 1 k 1 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2 + ck 2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 3/2 k 4 1 k 3 )/n) 3/2 n 15/2. Now we discuss our temporary assumption that k j k j 1 > 1. If k 4 k 3 = 1 then the last term in large square brackets in 33) should be replaced by 1. If k 4 k 3 = 2 then the effect of the same term on our estimate is that of a multiplicative constant see, e.g., 38)). It follows that the terms corresponding to k j k j 1 = 1 contribute to our sums as much as those corresponding to k j k j 1 = 2 up to a multiplicative constant). Having this in

14 KRZYSZTOF BURDZY mind, we may write E k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 ) ± k 1 Z 2 ± k 3 Z 2 k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 k j k j 1 >1 [ ck2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2 42) + ck 3 1 k 2 )/n) 3/2 k 2 1 k 1 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2 + ck 2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 3/2 k 4 1 k 3 )/n) 3/2 n 15/2]. We have 43) k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 k j k j 1 >1 ck 2 1 k 1 )/n) 3/2 k 3 1 k 2 )/n) 1/2 k 4 1 k 3 )/n) 3/2 n 13/2 cn 3 n j 3/2 1 k 1 =1 j 1 =1 cn 3 nn 1/2 = cn 3/2. n j 2 =1 j 1/2 2 j 3 =1 Similar bounds hold for other terms in 42) so that E and, therefore, k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 lim E j 3/2 3 ± k 1 Z 2 ± k 3 Z 2 ) cn 3/2 + cn 3/2 + cn 2 k 1,...,k 4 Θ k 1 <k 2 <k 3 <k 4 ± k 1 Z 2 ± k 3 Z 2 ) = 0. Now we will explain how this result may be generalized. Suppose that q j is either equal to 1 or 2 for j = 1,..., m. The expectation E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) q m ) may be bounded as in 41) by a product of factors corresponding to ± k j Z 2 ) q j. If q j = 1 and j > 1 then the factor is of the form and when q j = 2, j > 1, then the factor is ck j 1 k j 1 )/n) 3/2 n 2 ck j 1 k j 1 )/n) 1/2 n 3/2.

VARIATION OF ITERATED BROWNIAN MOTION 15 For j = 1, the factor is n 1 or n 1/2, depending on whether q j = 2 or 1. Summing as in 43) shows that each factor corresponding to ± k j Z 2 ) q j contributes cn 1/2 to the sum of expectations provided j > 1. The contribution from the first factor is either c or cn 1/2. Hence 44) E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) m) q cn m+2)/2. If m > 3 then 45) lim E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) q m) = 0. Suppose that some q j s are greater than 2 and at least one of them is odd we need this assumption to prove 29)). The only part of the proof that will be affected by this change in the assumptions is that the powers of r j s in 33) will increase. If every q j is equal to 2p j + 1 or 2p j + 2 for some integer p j 0 then instead of 44) we will have 46) E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) m) q cn p 1 p m + m+2)/2. Next we will analyse the fourth moment of V n s. We have 47) ) 4 ± k Z2 = 3 k Θ j,k distinct ± j Z2 ) 2 ± k Z2 ) 2 + k Θ + c ± j Z2 ) 3 ± k Z2 + c + c j k i,j,k,m Θ i<j<k<m j,k,m Θ m j k<m ± k Z2 ) 4 ± i Z2 ± j Z2 ± k Z2 ± mz 2. ± j Z2 ) 2 ± k Z2 ± mz 2 It follows from 26), 27), 45) and 46) that lim E ) 3 ± j Z2 ) 2 ± k Z2 ) 2 = 27s 2 s 1 ) 2, j,k distinct ) lim E k Θ ± k Z2 ) 4 = 0, lim E c ± j Z2 ) 3 ± k Z2 + c j k + c i,j,k,m Θ i<j<k<m j,k,m Θ m j k<m ± i Z2 ± j Z2 ± k Z2 ± mz 2 ) = 0. ± j Z2 ) 2 ± k Z2 ± mz 2

16 KRZYSZTOF BURDZY Hence 48) lim E k Θ ± k Z2 ) 4 = 27s 2 s 1 ) 2. More generally, suppose that m > 1 is an integer. Then ) 2m ± k Z2 = 2m 1)!! k Θ + + 2q 1 + +q j )=2m q 1... q j 2 q 1 + +q j =2m k 1,...,k m distinct cq 1,..., q j ) cq 1,..., q j ) ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 k 1,...,k j Θ ± k 1 Z 2 ) 2q 1 ± k 2 Z 2 ) 2q 2... ± k j Z 2 ) 2q j ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k j Z 2 ) q j, where q i s are positive integers and in the last sum, at least two q i s are odd. By 27) lim E 2m 1)!! We have lim E k 1,...,k m distinct 2m=q 2q 1 + +q j )=2m q 1... q j 2 + q 1 + +q j =2m cq 1,..., q j ) cq 1,..., q j ) by 26), 45) and 46). We conclude that 49) lim E for m > 1. Note that k Θ 50) lim E ± k 1 Z 2 ) 2... ± k m Z 2 ) 2 ) = 2m 1)!!3 m s 2 s 1 ) m. k 1,...,k j Θ ± k 1 Z 2 ) 2q 1 ± k 2 Z 2 ) 2q 2... ± k j Z 2 ) 2q j ) ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k j Z 2 ) q j = 0 ± k Z2 ) 2m = 2m 1)!!3 m s 2 s 1 ) m k Θ ± k Z2 ) 2m+1 = lim 0 = 0 since the random variables under the expectation have symmetric distributions. We see from 49)-50) that the moments of Ṽns 2 ) Ṽns 1 ) = df k Θ ± k Z2 except possibly the second moment) converge as n goes to infinity to the moments of the normal distribution with mean 0 and variance 3s 2 s 1 ). The difference Ṽns 2 ) Ṽns 1 ) V n s 2 ) V n s 1 ))

VARIATION OF ITERATED BROWNIAN MOTION 17 is bounded by ± j Z2 + ± k Z2 for appropriate j and k. Since lim E ± j Z2 + ± k Z2 p = 0 for every p < and every choice of j and k, an application of Minkowski s inequality shows that the moments of V n s 2 ) V n s 1 ) have the same limits as those of Ṽns 2 ) Ṽns 1 ). It is perhaps appropriate to explain why we have not proved the convergence of the second moments of V n s 2 ) V n s 1 ). In order to do it we would have to have very accurate estimates of ) E ± j Z2 ± k Z2 which cannot be found using our method. A simple modification of the proof of Theorem 4.5.5 of Chung 1974) shows that every subsequence of {V n s 2 ) V n s 1 )} n 1 has a further subsequence which converges in distribution to a random variable which has the same moments with possible exception of variance) as the centered normal with variance 3s 2 s 1 ). Lemma 1 implies that there is only one distribution which has the same moments of order greater than 2 as N0, 3s 2 s 1 )) and so {V n s 2 ) V n s 1 )} n 1 converges in distribution to N0, 3s 2 s 1 )). We will indicate how one can prove that the finite-dimensional distributions of V n also converge to those of Brownian motion Bs) with variance 3s. In order to prove that a pair of random variables has a two-dimensional normal distribution it suffices to show that all linear combinations of the random variables are normal. One can show this by finding the moments of all linear combinations. Let us fix some 0 s 1 < s 2 u 1 < u 2 and let Θs) and Θu) be the obvious analogues of Θ. Let a and b be arbitrary real numbers. In order to find lim E a ± k Z2 + b ) m ± k Z2 we might calculate k Θs) [ lim E a ) m1 ± k Z2 b k Θs) k Θu) k Θu) ± k Z2 ) m2 ] for various values of m 1 and m 2. Doing this would require estimates of E ± k 1 Z 2 ) q 1 ± k 2 Z 2 ) q 2... ± k m Z 2 ) q m s) k 1,...,k i Θu) ± k 1 Z 2 ) p 1 ± k 2 Z 2 ) p 2... ± k i Z 2 ) p i It can be shown that these expectations converge to the desired limits. The method of proof is a routine adaptation of the one used in the case of the one-dimensional distributions of V n. We omit the details as they are tedious. The case of m-dimensional distributions, m > 2, can be dealt with in the same way. Since the the finite-dimensional distributions of V n converge as n to those of B it remains to check that the distributions of V n are tight. It will suffice to show that there exists c < such that 51) EV n s 2 ) V n s 1 )) 4 cs 2 s 1 ) 2 ).

18 KRZYSZTOF BURDZY for all s 1, s 2 and all n see Billingsley 1968), 12.51)). It follows easily from the scaling properties of Brownian motions X and Y that Z has the 4-th order scaling properties while V n s have the Brownian scaling properties. In other words, the distribution of { n/mv n s m/n), s 0} is the same as that of {V m s), s 0}. It follows that 51) holds for all n if it holds for a single n. We will show that 51) holds for n = 1. Suppose that s 1 = 0. Then we want to show that EV 1 s 2 )) 4 cs 2 2. This is true for s 2 = 1 and some c since V 1 has all moments. It is easy to extend it to all s 2 [0, 1] since V 1 is by definition) a linear function on [0, 1]. Suppose that s 2 is an integer greater than 1. Then EV 1 s 2 )) 4 = E s 2 V s2 s 2 /s 2 )) 4 = s 2 2EV s2 1)) 4 by the scaling property of V n s. We have proved in 48) that so lim sup EV s2 1)) 4 < c < s 2 52) EV 1 s 2 )) 4 cs 2 2 for all integer s 2. It is elementary to extend the inequality to non-integer s 2 using the fact that V 1 is a linear function between integers. Now we use another invariance property of V n s, namely, that of translation invariance of their increments. The distribution of {V 1 s) V 1 0), s 0} is the same as that of {V 1 s + a) V 1 a), s 0} provided a > 0 is an integer. This and 52) imply that EV 1 s 2 ) V 1 s 1 )) 4 cs 2 s 1 ) 2 for integer numbers s 1 and all s 2 > s 1. Extending the result to all s 1 is easy. This concludes the proof of the weak convergence of V n s to B. Proof of Theorem 1 ii). We will prove the theorem only for s = 0 and t = 1. The estimate 46) obtained in the proof of Theorem 2 will be used to prove 4). We will apply the estimate with s 1 = 0 and s 2 = t recall the notation form the previous proof). We want to show that lim E n ) p j Z) 3 = 0. j=1 It will suffice to show that 53) lim E k1 Z) 3q 1 k2 Z) 3q 2... km Z) 3q m) = 0 for integer q j 1 such that q 1 + + q m = p. The sum extends over distinct k j s. If all q j s are even then 53) follows from 26). Assume that at least one of q j s is odd. Let us adopt the following convention for integer m ± k Z2 ) m+1/2 = k Z) 2m+1.

VARIATION OF ITERATED BROWNIAN MOTION 19 One can verify that 46) remains true if some of q j s are not integers but have the form k + 1/2 for some integer k. Such q j s should be treated as odd integers for the purpose of the decomposition q j = 2p j + 1 used in 46). We obtain from 46) 54) E k1 Z) 3q 1 k2 Z) 3q 2... km Z) 3q m ) cn r 1 r m + m+2)/2 where 3q j /2 = 2r j + 2 if 3q j /2 is an even integer and 3q j /2 = 2r j + 1 otherwise. If m > 2 then 54) implies 53). If m = 2 and p is large then r 1 + r 2 > 0 and again 54) implies 53). References 1. P. Billingsley, Convergence of Probability Measures, Wiley, New York, 1968. 2. K. Burdzy, Some path properties of iterated Brownian motion, Seminar on Stochastic Processes 1992 E. Çinlar, K.L. Chung and M. Sharpe, eds.), Birkhäuser, Boston, 1993, pp. 67 87. 3. R. Durrett, Probability: Theory and Examples, Wadsworth, Belmont, CA, 1991. 4. T. Funaki, Probabilistic construction of the solution of some higher order parabolic differential equations, Proc. Japan Acad. 55 1979), 176 179. 5. K.J. Hochberg, A signed measure on path space related to Wiener measure, Ann. Probab. 6 1978), 433 458. 6. V. Yu. Krylov, Some properties of the distribution corresponding to the equation u/ t = 1) q+1 2q u/ x 2q, Soviet Math. Dokl. 1 1960), 760 763. 7. A. M adrecki, The 4-a-stable motions, construction, properties and applications, preprint) 1992). 8. A. M adrecki and M. Rybaczuk, New Feynman-Kac type formula, Reports Math. Phys 1992), to appear). Department of Mathematics, GN-50, Seattle, WA 98195