The Uniform Modulus of Continuity Of Iterated Brownian Motion 1

Transcription

1 The Uniform Modulus of Continuity Of Iterated Brownian Motion 1 By Davar Khoshnevisan Thomas M Lewis University of Utah & Furman University Summary Let X be a Brownian motion defined on the line with X0=0 and let Y be an independent Brownian motion defined on the nonnegative real numbers For all t 0, we define the iterated Brownian motion IBM, Z, by setting Z t = XYt In this paper we determine the exact uniform modulus of continuity of the process Z Keywords and Phrases Iterated Brownian motion, uniform modulus of continuity, the Ray Knight theorem Running Title Iterated Brownian Motion AMS Subject Classification F15; 60G17; 60J65 1 Research supported by NSF grant DMS

2 1 Introduction Let X 1, X 2 and Y be independent standard Brownian motions starting from 0 Define a Brownian motion, X, on the line as follows: Xt = X 1 t if t 0 X 2 t if t < 0 The iterated Brownian motion or iterated Wiener process, Z, is defined by setting Zt = XY t for all t 0 Iterated Brownian motions arise naturally in a variety of problems in probability and mathematical statistics For example, Funaki [F] used a modification of Z to give a probabilistic solution to the partial differential equation 4 u x 4 = 1 u 8 t with u0, x = u 0 x, while Deheuvels and Mason [DH] introduced iterated Brownian motions in their study of the Bahadur Kiefer process Recent attention has centered on the path properties of Z Burdzy [Bu1, Bu2] established the following local law of the iterated logarithm: 11 lim sup t 0 Zt t 1/4 ln ln1/t 3/4 = 25/4, as 33/4 Other notable results in this vein are: the functional form of 11 [HPS]; Chung s form of the law of the iterated logarithm [HPS] and [KL]; a characterization of the lower limits [S]; a characterization of the upper limits via a connection with the stable subordinator of index 1/4 [Be]; various global Strassen type theorems [CsCsFR1]; a study of the local time [CsCsFR2]; the Csörgő Révész modulus of non-differentibility [HS] Some of the techniques of the present paper have played an important rôle in the analysis of [HS] Viewing 11 as the modulus of continuity of Z at 0, Burdzy private communication has asked whether one can find the uniform modulus of continuity for Z The sole goal of this article is to answer this question To this end define for all δ 0, 1, Our theorem is the following: ωδ = sup 0 s,t 1 ψδ = δ 1/4 ln1/δ 3/4 sup Zs Zt 0 s t δ 1

3 Theorem 1 With probability one, ωδ lim δ 0 ψδ = 1 If we liberally estimate the modulus of Z by composing the modulus of X with that of Y, then, by the well known result of Paul Lévy see p 114 of [KS], we obtain: 12 lim sup ωδ/ψδ 2 1/4, δ 0 as Since 2 1/4 > 1, Theorem 1 shows that this simple argument yields the correct rate, but is not precise enough to anticipate the constant 1 This, of course, is not unexpected, since the method outlined above is rather crude What is surprising is that known general techniques, eg, metric entropy and majorizing measures, are not refined enough to prove the upper bound in Theorem 1 Indeed, some estimates of Burdzy [Bu1] can be used to show that there exists K > 0, such that for all t, ε > 0 and x > 1, P Zt + ε Zt ε 1/4 x K exp χx 4/3, where χ = 2 5/3 3 1 With this probability estimate and a direct imitation of Levy s uniform modulus argument for Brownian motion, one can at best obtain the following: lim sup δ 0 ωδ ψδ χ3/4, as Since χ 3/ > 1, even the upper bound which is usually the easier half of such results requires very new ideas Our innovation is an analysis of IBM through the excursions of Y from the fast points of X To this end, one of the key results used is the Ray Knight theorem see the proof of Lemma 21 2 Preliminaries Throughout, we shall use generic constants K, K 1 and K 2 ; these may change from line to line but not within the same line Given positive functions f and g, we will write ft gt as t a to mean lim t a ft/gt = 1 Next we define two classes of stopping times for the Brownian motion, Y For a > 0 and b R, let τ a = τa = inft > 0 : Y t = a}, σ b = σb = inft > 0 : Y t = b} 2

4 By Brownian scaling, a 2 τa and b 2 σb have the same distributions as τ1 and σ1, respectively We need the following well known estimate for the small ball probability of σ1 see, eg, p 80 of [KS]: as t 0, 21 P σ1 t Kt 1/2 exp 1 2t By the symmetry of Y, P σ1 t P τ1 t 2P σ1 t Therefore 21 implies the following: 22 K 1 t 1/2 exp 1 P τ1 t K 2 t 1/2 exp 1, 2t 2t for all t 1 and an appropriate choice of K 1 and K 2 Finally, by a theorem of Chung see p 221 of [C], as s, 23 P τ1 s K exp π2 8 s Fixing an R > 1, let D n =kr n : k Z} and r k,n = kr n, for all integers k, n 1 Let T 0,n = 0 and iteratively define for n, j 1, T j,n = inf s > Tj 1,n : Y s D n \ Y T j 1,n } }, T j,n = Tj,n T j 1,n For integers k, n 1, let us define the following sets of random indices: E k,n = j 1 : Y Tj 1,n = r k 1,n, Y T j,n = r k,n and T j,n 1 }, Ê k,n = j 1 : Y Tj,n = r k,n and T j,n 1 } We point out that the cardinality of E k,n denoted by #E k,n is the number of times Y upcrosses [r k 1,n, r k,n ] before time 1 Our next two lemmas are the most important of this section In our first calculation, we demonstrate that with high probability, Y upcrosses a significant number of the intervals [r k 1,n, r k,n ] a uniform number of times before time 1 3

5 Lemma 21 Fix 0 < θ < ζ < 1 Then there exists some K = Kθ, ζ, R > 0, such that for all integers k, n 1, P #E k,n R n1 ζ for some k R n1 2θ Kn 2 Proof Throughout, L x t will denote the local time in x of a standard Brownian motion For simplicity, set p n = R nθ and M n = R n1 2θ and define the following events: } Ω 0 = #E k,n R n1 ζ for some k M n Ω 1 = 4L r k,n 1 R nζ for all 0 k M n } It will suffice to demonstrate that P Ω c 1 and P Ω 0 Ω 1 are both bounded by Kn 2 Clearly, P Ω c 1 P σp n 1 + P Ω c 1, σp n 1 = I + II, notation being obvious Now, I is easily estimated: indeed, by the reflection principle, I = P max 0 s 1 Y s p n = P Y 1 p n Kp n, since Y 1 has a bounded density Next we estimate II Note that k M n implies that r k,n p 2 n Since σp n 1 and t L x t II P P inf 0 r k,n p 2 n inf 0 x p 2 n is nondecreasing, we obtain the bounds: L r k,n σp n 1 4 R nζ L x σp n 1 4 R nζ = P inf 0 x p n L x σ1 1 4 R nζ θ, where we have used scaling to obtain this last equality Since σ1 is the first time Y hits 1, it follows from the first Ray Knight theorem see Revuz and Yor [RY, Thm 22, Ch XI] that there exists a two dimensional Brownian motion, B, such that B0 = 0 and L 1 x σ1 = Bx 2 for all 0 x 1 Therefore II P inf Bx 2R nζ θ/2 1 p n x 1 4

6 This last probability can be estimated as follows: let e n =0, 1/n R 2 By the strong Markov property and a coupling argument, II is bounded above by P B1 p n n 1 + P inf Bx 2R nζ θ/2 B0 = e n 0 x p n Kn 2 + P sup Bx n 1 2R nζ θ/2 B0 = 0 0 x p n Kn 2 + P sup Bx 2n 1 B0 = 0 for all large n 0 x p n = Kn 2 + P sup Bx 2n 1 R nθ/2 0 x 1 Kn 2 + 2nR nθ/2 E Kn 2, sup Bx 0 x 1 by Chebychev s inequality and the fact that E sup 0 x 1 Bx < that In light of the above development, we are left to estimate P Ω 0 Ω 1 To do so, observe Ω0 Ω 1 } 0 k M n 2R n #E k,n 1/2L r } k,n 1 As a consequence of the estimates leading to 55 of [K], P Ω 0 Ω 1 is bounded above by, 1/2R K 1 n ln R 3n + P sup L x 1 16n 2 R 2n ln R 2 x To see this, take in the notation of [K], θ = 4, ζ = 1 and R = 1 You will see that K 1 above is the same as C 5 4, 1, 1 Since L 0 1 has the same distribution as Y 1, it has a bounded density Hence, we certainly have, P Ω 0 Ω 1 Kn 2, which is what we wished to show In our next lemma, we show that the embedded random walk, Y T j,n, j 0}, does not hit an excessive number of points in D n before time 1 Furthermore, we show that no point in D n gets hit too many times before time 1 Lemma 22 Let θ, ζ 0, 1 Then there exists some K = KR, θ, ζ > 0 such that for all integers n 1, 1 P τr nθ 1 KR nθ exp R 2nθ /2 2 P k Z #Êk,n > R n1+ζ } is dominated by K 1 R nθ exp R 2nθ /2 + K 2 exp R nζ + K 3 R n1+θ ζ/2 exp R nζ /2 Proof We obtain 1 by 22 and scaling Next we prove 2 5

7 Let j = jn =[R n2+ζ ] Observe that is dominated by P #Êk,n > R n1+ζ } k Z P τr nθ 1 + P T j,n 1 + KR n1+θ P ξ0, jn > R n1+ζ, where ξ0, denotes the local time at 0 of a one-dimensional simple symmetric random walk The first term has already been considered, and is given by 1 To estimate the second term, we will need the Laplace transform of τ a see, for example, p 67 of [RY]: 24 E exp λτ a = cosha 2λ 1 Observe T j,n = T 1,n + + T j,n can be written as the sum of jn independent and identically distributed random variables, where T 1,n has the same distribution as τr n Thus, by Chebyshev s inequality and 24, we obtain the following: jn P T j,n 1 e E exp T 1,n = e coshr n 2 jn Since coshx 1 + x 2 /2, ζ < 1 and ln1 + x x x 2 /2 for x 0, it follows that 25 P T j,n 1 K exp R nζ Finally, P ξ0, jn > R n1+ζ = P jn 1/2 ξ0, jn > R nζ/2 KR nζ/2 exp R nζ /2, for all n sufficiently large, where we have used Bernstein s inequality or Chernoff s form of the large deviation estimate for the simple symmetric random walk and the fact that ξ0, n has the same law as the absolute value of the simple symmetric random walk by time n see, eg, p 14 and p 95 of [R] Our next lemma provides a uniform upper bound on the spacings T i,n } when T i,n 1 Lemma 23 Let ζ > 0 Then there exist constants K l = K l R, ζ > 0 l = 1, 2, such that for all integers i, n 1, P T i,n n 2 R 2n for some T i,n 1 K 1 exp R nζ + K 2 R n2+ζ exp π 2 n 2 /8 6

8 Proof As in the proof of Lemma 22, let j = jn =[R n2+ζ ] Then the probability in question is dominated by 26 P T j,n 1 + jnp T 1,n n 2 R 2n Using 25 to estimate the first term, it remains to bound the second term By the strong Markov property and Brownian scaling, P T 1,n n 2 R 2n = P τr n n 2 R 2n = P τ1 n 2 The result follows from 23 and 26 3 The proof of the lower bound Define for all x 0, 1, F x = x ln1/x, Gx = x 2 ln1/x 1 Throughout, γ and α will denote real numbers satisfying: 1/4 < γ < 1/2 and 0 < α < 4 γ 1 Given γ and α, let θ and ζ be chosen so that 0 < θ < ζ < 1 and For convenience, let 2 α γ > ζ + 2θ 31 q = 2 2θ ζ α γ > 0 For all n 1 and k Z, let us define the following random variables and events: X k,n = Xr k,n Xr k 1,n A k,n = Xk,n αf R n }, B k,n = Tj,n γgr n for some j E k,n }, C k,n = Ak,n B k,n 7

9 Our goal is to show that at least one of the C k,n occurs for some index k with T k,n 1, almost surely and for all n sufficiently large To this end, our first goal is to establish the following: for all α, γ, θ and ζ satisfying 31, we have n=1 P k:t k,n 1} c Ck,n < For each k, n 1, let us define the set of random indices: E k,n =j 1 : Y Tj 1,n = r k 1,n, Y T j,n = r k,n } Thus E k,n is the set of indices j that correspond to upcrossings of the interval [r k 1,n, r k,n ] This is distinguished from E k,n in that we have no restrictions on time [R n1 2θ ], define the measurable event, B k,n = Tj,n > γgr n for the first [R n1 ζ ] elements of E k,n } For 0 k In words, this is the event that the time for each of the the first [R n1 ζ ] upcrossings of the interval [r k 1,n, r k,n ] by Y exceeds γgr n We observe that B c k,n #E k,n > R n1 ζ for all 0 k R n1 2θ} B k,n By the strong Markov property, the events A c k,n B k,n, 0 k R n1 2θ} are independent and P A c k,n B k,n = P A c 1,n B 1,n These considerations lead us to the estimate: c P Ck,n P #E k,n R n1 ζ for some k R n1 2θ 32 k:t k,n 1} + P K n 2 + P k R n1 2θ k R n1 2θ by Lemma 21 Since the events A k,n and B k,n are independent, 33 P k R n1 2θ A c k,n B k,n A c k,n B k,n, A c k,n B k,n, [R n1 2θ ] 1 P A 1,n P B1,n c exp [R n1 2θ ]P A 1,n P B c 1,n 8

10 By scaling and a standard calculation, we readily obtain: 34 P A 1,n = P X1 αr n/2 F R n Kn 1/2 R nα2 /2 By the strong Markov property and 22, P B1,n c = 1 1 P T 1,n γgr n [R n1 ζ ] Kn 1/2 R n/2γ [R n1 ζ ] Kn 1/2 R n 1 ζ 1/2γ, since, by definition, 2γ 1 > 1 ζ Recalling the definition of q > 0 from 31, we see from 33, 34 and 35 that P A c k,n B k,n exp Kn 1 R nq k R n1 2θ Hence, from 32 we obtain the estimate: c P Ck,n K 1 n 2 + exp K 2 n 1 R nq, which certainly sums k:t k,n 1} By the Borel Cantelli lemma there exists an an N 0 such that for all n N 0 there exists k = kn such that 36 X k,n αf R n and T k,n γgr n on a set of full measure We point out the elementary fact that G is invertible on 0, 1 and that F G 1 x 2 3/4 ψx as x 0 +, where G 1 is the inverse function to G Moreover the following asymptotic relationships are easily verified: F R n R 1/2 F R n+1 as n ; F R n+1 = F G 1 GR n+1 2 3/4 ψgr n+1 as n ; ψx γ 1/4 ψγx as x 0 + Given any ɛ > 0, from these considerations and 36 it follows that there almost surely exists an N 1 N 0 such that for all n N 1 there exists a k = kn for which X k,n 1 ε2 3/4 αγ 1/4 R 1/2 ψγgr n+1 and T k,n γgr n 9

11 Since X k,n = ZT k,n ZT k 1,n, then for any n N 1 we have ωγgr n 1 ε2 3/4 αγ 1/4 R 1/2 ψγgr n+1 For any δ > 0 small enough, there exists n N 1 such that γgr n+1 δ γgr n By monotonicity, for such a δ, ωδ ω γgr n+1 1 ε2 3/4 αγ 1/4 R 1/2 ψγgr n 1 ε2 3/4 αγ 1/4 R 1/2 ψδ Therefore, lim inf δ 0 ωδ ψδ 1 ε2 3/4 αγ 1/4 R 1/2, as Let ε 0 and R 1 to obtain the following: lim inf δ 0 ωδ ψδ 2 3/4 αγ 1/4, as Let θ, ζ 0, γ 1/2 and α 2 to obtain the promised lower bound 4 The proof of the upper bound Throughout let R > 1, θ, ζ, q 0, 1 We will continue to use the notation developed in Section 2, significantly the family of stopping times T i,n } First we will need some technical definitions Let I =[a, b] denote a generic closed interval with endpoints in D n and let I = b a denote its length For α 0, we say that I is α fast written I F α provided that Xb Xa αf I By a standard Brownian motion calculation, there exists some K = KR, α > 0 such that for any interval, I R 1 +, 41 P I F α Klog I 1 1/2 I α2 /2 as I 0 We will call a pair of stopping times T i,n, T l,n with j < l a crossing of I provided that either 10

12 i Y T i,n = a, Y T l,n = b and Y T k,n < b for all indices k satisfying j < k < l; or ii Y T i,n = b, Y T l,n = a and Y T k,n > a for all indices k satisfying j < k < l Corresponding to a crossing T i,n, T l,n of I, we define the crossing time of I as simply T l,n T i,n We will need the following collection of intervals: for n 1 and 1 j n let I j,n =[rk,n, r k+j,n ] : k R n1+θ + 1} Finally, for 0 α < 2 let 43 γ = γα = For future reference, we note that 1 4 α 2 + 2q + 2θ + 2ζ 44 q = α γ 2 ζ θ > 0 Let A n be the event that before time 1 at least one interval of the form [r k,n, r k+j,n ] with 1 j n which is in F α is crossed in time less than γgjr n by Y Likewise, for n 1 and 1 j n, let A j,n denote the event that at least one of the intervals from I j,n which is in F α is crossed in time less than γgjr n by one of the first R n1+ζ crossings of that interval Let l < k be integers and suppose that #Êl,n R n1+ζ and #Êk,n R n1+ζ Then the number of crossings of the interval [r l,n, r k,n ] by Y up to time 1 is bounded above by R n1+ζ Similarily, if τr nθ > 1, then Y t r k,n for all k R n1+θ + 1 and 0 t 1 Consequently, c A n τr nθ > 1} #Êk,n > R } n1+ζ k Z n A j,n j=1 Thus, by subadditivity, n P A n a n + P A j,n, j=1 where a n = P τr nθ 1 + P k Z #Êk,n > R n1+ζ } By Lemma 22, a n is summable; we are left to estimate P A j,n 11

13 Since X and Y are independent, by conditioning on the number of intervals in I j,n F α, then applying subadditivity and integrating, P A j,n is bounded above by the product of the following three terms: i the expected number of intervals in I j,n F α ; ii R n1+ζ ie, the number of observed crossings per interval; iii the probability that the time to cross an interval with length j/r n is faster than γgj/r n By 41, we see that the probability that a given interval in I j,n is in F α is bounded above by Kn 1/2 R nα2 /2 j α2 /2 Since the number of intervals in I j,n is asymptotically 2R n1+θ, it follows that the expected number of intervals in I j,n F α is bounded above by Kn 1/2 R n1+θ α2 /2 j α2 /2 This estimates the term described in i, and it remains to estimate the probability in iii; however, by 21, this probability is bounded above by Kn 1/2 R n/2γ j 1/2γ It follows that P A j,n KnR n2+θ+ζ α 2 /2 1/2γ j α2 /2+1/2γ By 44, Moreover, since q, γ, θ 0, 1, 44 yields: 2 + θ + ζ α γ = q α γ = 2 + q + γ + θ 5 Consequently, P A j,n KnR nq j 5, and n P A j,n Kn 7 R nq, j=1 which certainly sums in n, since q > 0 We have shown that n P A n < By the Borel-Cantelli lemma, it follows that 45 P A n, io = 0 12

14 Hence on a set of full measure and for n sufficiently large, we can make the following observation: let 0 i < j be integers and suppose that T j,n 1 Let a = Y T i,n Y T j,n and b = Y T i,n Y T j,n Whenever 46 b a n R n and Xb Xa αf b a, it too must be the case that T j,n T i,n γαgb a Indeed, by 46, the interval [a, b] determined by T i,n and T j,n is smaller in length than n/r n and is in F α ; thus by 45, T j,n T i,n must exceed some time to cross this interval, which, in turn, must exceed γαgb a Fix an integer m 1 and let α k = k 2/m for 0 k m+1 This determines m+1 intervals, [α 0, α 1 ] [α m, α m+1 ] Corresponding to each of the latter intervals, by 43, we get m + 1 values of γ: γα 0,, γα m Let } C m = 2 3/4 α1 max γ 1/4 α 0,, α m+1 γ 1/4 α m Let T i,n, T j,n, a and b be as in the previous paragraph By the usual modulus of continuity of X, α k F b a Xb Xa < α k+1 F b a, for some 0 k m Thus [a, b] is in F α k, for some α k Consequently, provided that b a n/r n, it follows that T j,n T i,n γα k Gb a and Xb Xa α k+1 F b a We recall that F G 1 x 2 3/4 ψx and ψγx γ 1/4 ψx as x 0 + Let ɛ > 0 be given Then for x 0 sufficiently small, F G 1 γx 1 + ɛ2 3/4 γ 1/4 ψx Thus, assuming that b a n/r n, we have shown: 47 Finally, let δ > 0 satisfy ZT j,n ZT i,n 1 + ɛc m ψt j,n T i,n n + 1 R n+1 2 δ logδ 1 n R n Let T i,n < T j,n 1 with T j,n T i,n δ Then, by the uniform modulus of continuity of Y, it follows that Y T j,n Y T i,n 2 δ logδ 1 n R n, 13

15 which shows that the interval [a, b] = [Y T i,n Y T j,n, Y T i,n Y T j,n ] is not too wide Thus we have shown that whenever T i,n < T j,n 1 with T j,n T i,n δ, then 48 ZT j,n ZT i,n 1 + ɛc m ψt j,n T i,n The above is the desired upper bound for the modulus of continuity of Z along the random subsequence given by T i,n ; i} We finish the proof of the upper bound and hence that of Theorem 1 by chaining deterministic times to the subsequence T i,n ; i} and showing that the slack in this chaining approximation is negligible Given 0 s < t 1 with t s δ, either i t s n 2 /R 2n, in which case we can use 12 ie, the inexact modulus of continuity of Z and conclude that for some K > 0, Zt Zs Kψn 2 /R 2n ; or ii n 2 /R n < t s δ In this case, using Lemma 23, for all n sufficiently large, T i,n n 2 /R 2n for all T i,n 1 It is a simple task to see that there exists i such that 1 n 2 R 2n T i,n 1 Use time reversal and the proof of Lemma 23, for instance Thus, we can find i and j, such that s T i,n T j,n t, T i,n s n 2 /R 2n and t T j,n n 2 /R 2n Thus, by the triangle inequality, 12, 47 and 48, it follows that Zt Zs Zt ZT i,n + Zs ZT j,n + + ZT i,n ZT j,n Kψn 2 /R 2n ɛc m ψ T j,n T i,n Kψn 2 /R 2n ɛc m ψ t s Kψn 2 /R 2n ɛc m ψ δ Therefore, ωδ Kψn 2 /R 2n ɛc m ψδ However, by 47, δ Kn 9 /R 2n as n This shows that ψn 2 /R 2n = o ψδ and from this observation, it follows that lim sup δ 0 + ωδ ψδ 1 + ɛc m as 14

16 We are done since since ɛ > 0 is arbitrary and, by 43 and the definitions of C m and α m, lim lim lim lim C α4 α 2 1/4 m = sup = 1 m ζ 0 θ 0 q 0 α [0, 2] 2 3/4 This finishes the proof of the upper bound and hence that of Theorem 1 Acknowledgements Part of this work was carried out while we were visiting the Center for the Mathematical Sciences at the University of Wisconsin Madison We wish to extend our thanks to Tom Kurtz and Jim Kuelbs for their hospitality, and to the National Science Foundation for their financial support Many thanks are due to Chris Burdzy for suggesting this problem, as well as his continued interest Our whole hearted thanks go to Zhan Shi for carefully reading and correcting the first draft of this paper References [Be] J Bertoin 1995 Iterated Brownian motion and stable 1/4 subordinator To appear in Stat Prob Lett [Bu1] K Burdzy 1993 Some path properties of iterated Brownian motion Sem Stoch Processes 1992, KL Chung, E Cinlar and MJ Sharpe, eds, Birkhauser 1993, pp [Bu2] K Burdzy 1993 Variation of iterated Brownian motion Measure valued Processes, Stochastic Partial Differential Equations and Interacting Particle Systems, Ed DA Dawson CRM Proceedings and Lecture Notes, 5, [C] KL Chung 1948 On the maximum of partial sums of sequences of independent random variables Trans Amer Math Soc, 64, pp [CsCsFR1] E Csáki, M Csörgo, A Földes and P Révész 1995 Global Strassen type theorems for iterated Brownian motion To appear in Stoch Process Appl [CsCsFR2] E Csáki, M Csörgo, A Földes and P Révész 1994 The local time of iterated Brownian motion Technical Report Series of the Laboratory for Research in Statatistic and Probability, 250 [F] T Funaki 1979 Probabilistic construction of the solution of some higher order parabolic differential equations, Proc Japan Acad 55, [HS] Y Hu and Z Shi 1994 The Csörgo Révész modulus of non-differentiability of iterated Brownian motion To appear in Stoch Process Appl 15

17 [HPS] Y Hu, D Pierre-Loti-Viaud and Z Shi 1994 Laws of the iterated logarithm for iterated Wiener processes To appear in J Theor Probab [K] D Khoshnevisan 1993 Exact rates of convergence to Brownian local time Ann Prob 22 3, [KL] D Khoshnevisan and T Lewis 1993 Chung s law of the iterated logarithm for iterated Brownian motion To appear in Ann Inst Henri Poincaré: Prob et Stat [KS] I Karatzas and S Shreve 1988 Brownian motion and Stochastic Calculus Springer Verlag, New York [R] P Révész 1990 Random Walk in Random and Non-Random Environments World Scientific, Singapore [RY] D Revuz and M Yor 1991 Continuous Martingales and Brownian Motion Springer Verlag, Berlin Heidelberg [S] Z Shi 1994 Lower limits of iterated Wiener process To appear in Stat Prob Lett D Khoshnevisan TM Lewis Department of Mathematics Department of Mathematics University of Utah Furman University Salt Lake City, UT Greenville, SC