Bull. Sci. math. 125, 6-7 (2001) 431–456 2001 Éditions scientifiques et médicales Elsevier SAS. All rights reserved S0007-4497(01)01095-8/FLA
MALLIAVIN CALCULUS AND MARTINGALE EXPANSION *
BY
NAKAHIRO YOSHIDA Graduate School of Mathematical Sciences, University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153, Japan
ABSTRACT. – We proved the validity of the asymptotic expansion for the distribution of a martingale with jumps. A sufficient condition is presented in terms of the decay of certain integrations of Fourier type. In order to estimate such Fourier type integrals, we use the Malliavin calculus and show that it becomes a key to our program. 2001 Éditions scientifiques et médicales Elsevier SAS
1. Introduction
Asymptotic expansion is now admitted to be fundaments of the higher-order asymptotic statistical theory beyond the first-order theory; cf. Akahira and Takeuchi [1], Bhattacharya and Ghosh [2], Ghosh [5], Pfanzagl [14,15], Taniguchi [20]. This is also the case when we consider statistical problems for semimartingales. In spite of the importance, the asymptotic expansion of martingales is rather a new topic discussed in the latest mathematical statistics. Mykland [11] is a prominent work which presented, for continuous martingales (Mn,t :0 t Tn), an asymptotic [ ] 2 expansion of the expectation E g(Mn,Tn ) for a class of C -functions g. Consecutively, in [12,13], he presented expansions for martingales with discrete time parameter and for martingales with jumps, in a class of C2- functions g. In certain problems, for instance, problems of confidence intervals, sta- tisticians need asymptotic expansion without the C2-regularity condition.
* This work was in part supported by the Research Fund for Young Scientists of the Ministry of Science, Education and Culture, and by Cooperative Research Program of the Institute of Statistical Mathematics. 432 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456
Generally speaking, it is possible to remove such a regularity condition if we may assume a condition which ensures the regularity of the distribu- tion of Mn,Tn . It is well known that the so-called Cramér condition serves as such a condition for independent observations. In this article, we will consider asymptotic expansion for martingales with jumps. In order to validate the expansion for martingales it will be necessary to examine the order of the decay of certain Fourier integrals. For this purpose, we will take advantage of the Malliavin calculus. In fact, by means of the associated integration-by-parts setting, it is possible to [ ] prove the validity of the expansion of E g(Mn,Tn ) for any measurable function g. It seems to be the only handy method which universally applies to functionals appearing in statistics for stochastic processes with continuous-time parameter. In the previous paper [26], continuous martingales were treated, however the same methodology can be applied to general martingales with jumps. The aim of this article is to clarify the idea and to carry out this program for martingales admitting jumps. For functionals of ε-Markov processes with mixing property, the lo- cal approach provides an efficient way to the asymptotic expansion (cf. Götze and Hipp [6,7], Kusuoka and Yoshida [10], Yoshida [28] and Sakamoto and Yoshida [18,19]). However, the present martingale ap- proach (global approach) still has advantages of being widely applied: a class of diffusion processes and estimation of a diffusion coefficient ([26]), the M-estimator over Wiener space (Sakamoto and Yoshida [17]), and the second order expansions in the small σ -theory ([22–25], Der- moune and Kutoyants [4]). Besides, in order to illustrate another merit, we will in Section 3 give a simple example of a long-range-dependent- energy martingale to which the geometric-mixing approach cannot apply but our martingale method is still applicable. The organization of the present article is as follows. First, we will summarize the Malliavin calculus for jump processes in Section 2. The main result will be stated in Section 3 and the proof of it will be presented in Section 4. Since it is not yet clear whether the embedding technique for martingales preserves the smoothness of functionals in the sense of the Malliavin calculus, we do not use the embedding technique as Mykland did (of course, in his case, such smoothness is not necessary because of the smoothness of functions g). The method here is based on the Fourier analysis rather classical. N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 433
2. Malliavin calculus for jump processes
In this section, we give a review of the Malliavin calculus for jump processes formulated by Bichteler, Gravereaux and Jacod [3] among many other possible formulations. Any formulation is possible to use if a sufficiently higher order integration-by-parts formula in differential form or in difference form is available in it. Let (Θ, B,Π)be a probability space. The following formulation of the Malliavin calculus was adopted by Bichteler et al. [3]. L D L ∈ p DEFINITION.–A linear operator on ( ) p>1 L (Π) into p p>1 L (Π) is called a Malliavin operator if: (1) B is generated by D(L). 2 n n (2) For f ∈ C↑(R ), n ∈ N, and F ∈ D(L) , f ◦ F ∈ D(L). (3) For any F,G∈ D(L), EΠ [F LG]=EΠ [GLF ]. (4) For F ∈ D(L), L(F 2) 2F LF . Namely, the bilinear operator Γ on D(L) × D(L) associated with L by Γ(F,G)= L(F G) − F LG − GLF is nonnegative definite. 1 n n 2 n (5) For F = (F ,...,F ) ∈ D(L) , n ∈ N, and f ∈ C↑(R ), n n i 1 i j L(f ◦ F)= ∂if ◦ F LF + ∂i∂j f ◦ FΓ F ,F . i=1 2 i,j=1
Let (L, D(L)) be a Malliavin operator. We define F 2 by Dp 1/2 F 2 = F + LF + Γ (F, F ) Dp p p p 2 for p 2. Denote by D the completion of D(L) with respect to · 2 . p Dp 2 Then spaces (D , · 2 ) become Banach spaces with natural inclusions p Dp between them. Let 2 = 2 D∞− Dp. p2 There exists an integration-by-parts setting (IBPS). See Theorem 8- 18 of [3], p. 107, for details. We shall rewrite the integration-by- parts formula for a partially nondegenerate situation. Let ∆ = det σF , i,j i,j = d = i j σF (σF )i,j=1,whereσF Γ(F ,F ) and σ[i,i ] is the (i, i )-cofactor 2 d of σF . Assume that F ∈ D∞−(R ). Also we suppose that a trunction functional ψ satisfies the condition that ∆ = 0 ⇒ ∆−1ψ = 0 a.s., 434 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456
· −1 = i,j ∈ 2 −1 ∈ 2 hence ∆ ∆ ψ ψ a.s. If σF D∞− and ∆ ψ D∞−, then the 2 d integration-by-parts formula holds, i.e., for f ∈ C↑(R ), Π = Π J F (1) E ∂if(F)ψ E f(F) i ψ , where the operator J F → −1 ∈ 2 → i : ψ: Θ R such that ∆ ψ D∞− Lp(Π) p>1 is defined by d − − J F =− 1 L i + 1 i i ψ 2∆ ψσ[i,i ] F Γ ∆ ψσ[i,i ],F . i =1 ∈ [ ] [ ] For k N,defineSk F and Sk ψ as follows: [ ]:={ i,j = } ∈ 2 d S1 F σF : i, j 1,...,d if F D∞−(R ); [ ]:={ i,j L i [ ] i = } ∈ 2 d Sk F σF , F ,Γ(Sk−1 F ,F ): i, j 1,...,d if F D∞−(R ) [ ]⊂ 2 and Sk−1 F D∞−; [ ; ]:={ −1 } = −1 = S1 ψ F ∆ ψ if ∆ 0 implies ∆ ψ 0; [ ; ]:={ −1 [ ; ] −1 [ ; ] i = } Sk ψ F ∆ Sk−1 ψ F ,∆ Γ(Sk−1 ψ F ,F ): i 1,...,d [ ; ]⊂ 2 = −1 [ ; ]∪ if Sk−1 ψ F D∞− and if ∆ 0 implies ∆ Sk−1 ψ F −1 [ ; ] ={ } ∆ Γ(Sk−1 ψ F ,F) 0 . Moreover, we prepare the following notation: ∆ := ∆F . [ ; ]:= [ ]∪ [ ; ] ∈ 2 d S1 ψ F S1 F S1 ψ F if F D∞−(R ); [ ; ]:= [ ; ]∪ [ ]∪ [ ; ] ∈ 2 d [ ] Sk ψ F Sk−1 ψ F Sk F Sk ψ F if F D∞−(R ), Sk−1 F ⊂ 2 [ ; ]⊂ 2 = −1 [ ; ]∪ D∞− and Sk−1 ψ F D∞−,andif∆ 0 implies ∆ Sk−1 ψ F −1 [ ; ] ={ } ={ : ∆ Γ(Sk−1 ψ F ,F) 0 . Here we denoted Γ(A,B) Γ(a,b) a ∈ A,b ∈ B} for function sets A and B. 2 2 PROPOSITION 1. – Suppose that F ∈ D∞−.IfSk[ψ; F ]⊂D∞−,then k+1 d for f ∈ C↑ (R ), EΠ ∂ ∂ ···∂ f(F)ψ = EΠ f(F)J F ···J F J F ψ . i1 i2 ik ik i2 i1 2 Proof. Ð From assumption that Sk[ψ; F ]⊂D∞−, it follows that ∆−1J F ···J F (ψ) (1 j k − 1) are well defined, and that ∆−1J F ··· ij i1 ij J F (ψ) ∈ L{S [ψ; F ]} ⊗ Alg (S [F ]). The assertion of the propo- i1 j+1 R j+1 sition can be proved by this fact and by induction with the aid of the integration-by-parts formula (1). ✷ N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 435
3. Main result
We consider a sequence of d-dimensional random vectors {Xn} with stochastic expansion: = + Xn Mn,Tn rnNn, where Mn,Tn and Nn are d-dimensional random vectors, both of which n F n n = are defined on a probability space (Ω , ,P ),andMn (Mn)t∈[0,Tn] F n is a d-dimensional local martingale with respect to a filtration ( t )t∈[0,Tn] n F n n = c over (Ω , ,P ). We assume that Mn,0 0 and denote by Mn the d continuous part of Mn and by Mn the purely discontinuous part of Mn. {rn} is a sequence of positive numbers satisfying limn→∞ rn = 0. Here we use the parameter n ∈ N, however it is clearly possible to consider any directed set in place of N for our results. The optional quadratic covariation process of the local martingale [ ]=[ ]= [ i j ] d Mn is denoted by Mn Mn,Mn ( Mn,Mn )i,j=1, and the pre- dictable covariation process is denoted by Mn=Mn,Mn= i j d = ∗[ ]+ ∗ = ( Mn,Mn )i,j=1. Moreover we put Ξn α Mn β Mn and Ξn ∗ Ξn − I ,whereI is the identity matrix. Though we will take α = 1/3 and β∗ = 2/3 later as Mykland found it, we will leave them for a while to = −1 see why those numbers should be chosen in this way. Put ξn rn Ξn,Tn . For semimartingale X, µX denotes the integer-valued random measure d on R+ × R , and it is defined by X ; = µ (ω dt,dx) 1{∆Xs =0}δ(s,∆Xs)(dt,dx). s>0 It is well known that there exists a unique random measure ν = νX which X compensates µ . The random measure νn denotes the compensator Mn corresponding to µn = µ . In this section, we assume that each (Ωn, F n,Pn) has a Malliavin Ln n n,2 = −2| |4 ∗ operator with corresponding Γ and D∞−.Letκn rn x µn,Tn = −2| |4 ∗ Sn ={ } and let λn rn x νn,Tn .Put −1 Mn,Tn ,Nn . Given a nonnegative n Sn Sn ⊂ n,2 random variable sn on Ω , 0 is defined only when −1 D∞− by
Sn = n 0 sn,σMn,Tn ,Γ (Mn,Tn ,Nn), σNn ,Mn,Tn ,ξn,Nn,κn,λn . Sn Sn ⊂ n,2 Sn ={Ln }∪Sn ∪ The set 1 is defined only when 0 D∞− by 1 Xn 0 n Sn ∈ Sn Sn ⊂ n,2 Γ ( 0 ,Xn).Fork N, k 2, k is defined only when k−1 D∞− by Sn = Sn ∪ n Sn k k−1 Γ ( k−1,Xn). 436 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456
We will assume that [A1] For a random vector (Z,ξ,η) taking values in Rd × (Rd ⊗ Rd ) × Rd , →d →∞ (Mn,Tn ,ξn,Nn) (Z,ξ,η) as n . n S · n [A2] The set l is bounded with respect to Lp(P ) uniformly in n ∈ N for any p>1. We denote by φ the d-dimensional standard normal density. De- d → d ⊗ d d → d = jk d = fine A : R R R and B : R R by A(x) (A (x))j,k=1 [ | = ] = j d = [ | = ] P ξ Z x and B(x) (B (x))j=1 P η Z x ; moreover, let d d 1 jk j pn(x) = φ(x) + rn ∂j ∂k A (x)φ(x) − rn ∂j B (x)φ(x) . 2 j,k=1 j=1 Remark 1. – Under the assumptions, it is easy to show the existence of the derivatives of Ajk and Bj , and the integrability of the each term on the right-hand side of the above definition of pn. Let E(M, γ ) be the set of measurable functions f : Rd → R satisfying that |f(x)| M(1 +|x|γ ). The following is our main result:
THEOREM 1. – Let Yn denote either Xn or Mn,Tn . Suppose that conditions [A1] and [A2] for some l>2d + 4. Moreover, suppose that n[ −p] ∞ n[ ] supn∈N P sn < for any p>1, and that limn→∞ P det σYn sn = 1. Then, for any M>0, γ>0, a1 >(3 + d)/(l − d) and a2 < 1,there exist a constant C and a sequence ε = o(r ) such that n n − n = + a1 n[ ]a2 + P f(Xn) f(x)pn(x) dx Crn P det σYn Xt = θZt + et , where Z = (Zt )t∈N and e = (et )t∈N are independent and e is an i.i.d. sequence with E[et ]=0andVar[et ]=1. Given observations ˆ (Xt ,Zt )t=1,...,T , the least square estimator θT for θ is given by √ √ T ˆ i=1 Zt et / T T(θT − θ)= . T 2 t=1 Zt /T In the present situation, we can naturally embed the processes (Xt ,Zt )t=1,...,T into the continuous-time processes (X[t],Z[t])t∈R+ ,andto simplify our argument, for a while, let us confine our attention to the mar- T = T tingale which forms the principal part: the martingale M (Mt )t∈R+ defined by t T 1 M = √ Z[s+1] dws , t T 0 = where w (wt )t∈R+ is a Wiener process independent of Z,andthe F F = [ ∈ ] reference filtration ( t )t∈R+ is given by t σ Zn (n Z+), ws (s t) . [ ]= Let (Zn)n∈Z+ be a stationary Gaussian sequence satisfying E Zn 0and [ ]= T = · 2 Var Zn 1. Then M 0 Z[s+1] ds/T and hence, for the Hermite polynomial H2 of degree 2, T = −1 −1 ξT rT T H2(Zt ) t=1 for T ∈ Z+. Suppose that the autocovariance function −q r(n):= E[Z0Zn]∼|n| L(n) as n →∞for some constant q (0 T −1 →d AT H2(Zt ) X(1) t=1 as T →∞, T ∈ N,ifq<1/2. Take rT = AT /T. Then we see that T + E exp iuMT ivξT ) T 1 T = − 2 −1 2 + −1 E exp u T Z[t+1] dt ivAT H2(Zt ) 2 t=1 0 1 → E exp − u2 + ivX(1) ; 2 T →d thus (MT ,ξT ) (Z ,X(1)),whereZ is independent of X(1).From our theorem, it is possible to obtain the asymptotic expansion of the T ∼ −q distribution of MT . The order of the second term is rT T L(T );since the conditional expectation of ξ given Z vanishes, what we obtained is a Berry–Esseen type bound, i.e., for any q 4. Preliminaries and proofs Suppose that we are given a one-dimensional random variable (trun- n cation functional) ψn : Ω →[0, 1] for each n ∈ N. We will use the d d α multi-index: for x = (x1,...,xd ) ∈ R and α = (α1,...,αd ) ∈ Z+, x = α1 ··· αd α = α1 ··· αd = = x1 xd ; moreover, ∂x ∂1 ∂d with ∂i ∂/∂xi , i 1,...,d. d For α = (α1,...,αd ) ∈ Z+,let · ˆα = n iu Xn α gn (u) P e ψnXn . α d → We define gn : R R by 1 d gα(x) = e−iu·xgˆα(u) du, x ∈ Rd , n 2π n Rd ˆα 0 if gn is integrable. The function gn is referred to as the local density of Xn. It is easy to see that under regularity conditions, α = α 0 gn (x) x gn(x) d d for x ∈ R and α ∈ Z+. n d d Define cn ∈ R by cn = P [ψn]; moreover, for α ∈ Z+ and x ∈ R ,let α = α 0 hn(x) x hn(x),where 1 d d 0 = + jk − j hn(x) cnφ(x) rn ∂j ∂k A (x)φ(x) rn ∂j B (x)φ(x) . 2 j,k=1 j=1 Define Aα(u, x) and Bα(u, x) by: d jk j k α j α Aα(u, x) = A (x) (iu) (iu) x + (iu) ∂kx j,k=1 k α α + (iu) ∂j x + ∂j ∂kx , and d j j α α Bα(u, x) = B (x) (iu) x + ∂j x , j=1 respectively. Set 1 hˆα(u) = eiu·x c xα + r A (u, x) + r B (u, x) φ(x)dx. n n 2 n α n α Rd 440 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 d d Denote by S the set of all d × d-symmetric matrices. For α ∈ Z+, d d u, z ∈ R and r ∈ S ,definePα(u,z,r)and Qα(u,z,r)by 1 P (u,z,r)= exp −iu · z − r(u, u) (−i∂ )α α 2 u 1 × exp iu · z + r(u, u) , 2 and 1 r(u,u) Qα(u,z,r)= e 2 Pα(u,z,r), respectively. Clearly, 1 (−i∂ )α exp iu · z + r(u, u) = eiu·zQ (u,z,r). u 2 α ˆα In the sequel, we often denote Mn,Tn simply by Mn. The integral gn (u) ˆ α of Fourier type will be approximated by hn(u). The gap between them is decomposed into three parts as follows: ˆα − ˆα = α + α + α gn (u) hn(u) Jn (u) Kn (u) Ln(u), where α n α iu·Xn J (u) = P ψnX e n α − · − P n ψ (r N )α βQ u, M , Ξ eiu Xn n β n n β n n,Tn 0βα 1 iu·Z − rnE e Aα(u, Z) , 2 α − · Kα(u) = P n ψ (r N )α β Q u, M , Ξ eiu Xn n n β n n β n n,Tn 0βα · · − n iu Mn − iu Z P ψnQα u, Mn, Ξn,Tn e rnE e Bα(u, Z) and · α = n iu Mn Ln(u) P ψnQα u, Mn, Ξn,Tn e n α − 1 |u|2 − P ψn(−i∂u) e 2 . d d d For β ∈ Z+, u ∈ R and r ∈ S ,let − 1 1 | | = 2 r(u,u) β 2 r(u,u) = β Sβ (u, r) e ∂u e i Pβ (u, 0,r). N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 441 LEMMA 1. – |β| (1) There exist constants cj (independent of u, r) such that |β| | | |β|| |j | |(j+|β|)/2 Sβ (u, r) cj u r j=0 even = odd = and that codd 0 and ceven 0. (2) For u, z ∈ Rd , r ∈ Sd , α | | − P (u,z,r)= (−i) β zα βS (u, r). α β β 0βα Proof. Ð It is easy to show (1). For (2), we may use the multi-index Leibniz rule: 1 (−i∂ )α exp iu · z + r(u, u) u 2 |α| α α−β iu·z β 1 r(u,u) = (−i) (∂ ) e · (∂ ) e 2 . ✷ β u u 0βα Let q>1andρ>0. Before applying the Malliavin calculus, we will clarify what kind conditions should be verified to obtain the asymptotic expansion. [C1] −2 | |4 ∗ +| |4 ∗ ∞ sup rn x µn,Tn x νn,Tn q < . n∈N p [C2] (i) ψn → 1asn →∞. (ii) For any p1,p2,p3 ∈ Z+, n p1 p2 p3 sup P ψn|Mn| |ξn| |Nn| < ∞. n∈N (iii) Assumption [A1] is satisfied. = 2ρ | | { } [C3] For bn rn , Ξn,Tn bn a.s. on the set ψn > 0 . [C4] There exists a constant A independent of n ∈ N such that on {ψn > 0}, −ρ d rn sup ∆Mn,s A a.s. sTn and 442 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 −2ρ d,i d,i 2 rn sup ∆ Mn ,Mn s A a.s. sTn for all i = 1,...,d. d [C5]l For any α ∈ Z+, · | |l n iu Xn α ∞ sup sup u P e ψnXn < . n∈N −ρ u: |u|rn d [C6]j For any α ∈ Z+, · | |j n iu Xn α ∞ sup sup u P e ψnAn,u < , n∈N −ρ u:1|u|rn where 1 α − Aα = (r N )α β n,u ( +|u|2)r β n n 1 n 0βα × Qβ (u, Mn, 0) − Qβ (u, Mn,rnξn) . d [C7]k For any α ∈ Z+, · | |k iu Xn α ∞ sup sup u E e ψnBn,u < , n∈N −ρ u:1|u|rn where 1 α − Bα = (r N )α βQ (u, M ,r ξ ) n,u +| | β n n β n n n (1 u )rn 0βα −iu·rnNn − Qα(u, Mn,rnξn) e . We will denote by C a generic constant independent of n and u.Also we write F(n,u) G(n, u) if there exists a constant C independent of n and u such that F(n,u) CG(n,u) for any n and u.Forafixed 1 ={ ∈ d ; | | −ρ } 0 ={ ∈ d ; positive number ρ,letΛn u R 1 u rn and Λn u R | | −ρ } = d ∈ d u rn .Putek (δk,i)i=1 R :thekth vector of the standard basis of Rd . We assume [C3]. α = →∞ LEMMA 2. – If [C2] is satisfied, then Jn (u) o(rn)(n ) for d each u ∈ R . Moreover, if [C6]j holds, then −1 α −(j−2) r J (u) 1 1 (u) C(1 +|u|) n n Λn N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 443 for any u ∈ Rd . Under [C2] and [C6] ,ifj>d+ 2,then j −1 α r J (u) 1 0 (u) du → 0 n n Λn Rd as n →∞. Proof. Ð Put α − j α (u) = ψ (r N )α β 1,n n β n n β:0βα |β||α|−1 iu·Xn × Qβ (u, Mn, 0) − Qβ (u, Mn,rnξn) e and · α = − iu Xn j2,n(u) ψn Qα(u, Mn, 0) Qα(u, Mn,rnξn) e . −1 n[| α |] = ∈ d Clearly, [C2] implies that rn P j1,n(u) o(1) for each u R .On α the other hand, since Qα(u, z, 0) = z (multi-index), − − α | | − r 1j α (u) = ψ r 1 Mα − (−i) j Mα j n 2,n n n n j n j:0jα 1 rnξn(u,u) iu·Xn × Sj (u, rnξn) e 2 e · − 1 = iu Xn 1 α − α 2 rnξn(u,u) e ψnrn Mn Mn e d α − · 1 − − α ek k · 2 rnξn(u,u) ( i)Mn rnξn u e ek k=1 − − α α ek1 ek2 − (−1)Mn ek + ek k ,k :1k k d 1 2 1 2 1 2 d d 1 × k1k2 + 2 k1l1 k2l2 l1 l2 2 rnξn(u,u) rnξn rn ξn ξn u u e l =1 l =1 1 2 α |j| α−j 1 r ξ (u,u) − (−i) M S (u, r ξ ) e 2 n n . j n j n n j: jα |j|3 d Write α = (α1,...,αd ) ∈ Z+. It follows from [C2] and the implied uniform integrability that 444 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 −1 n α rn P j2,n(u) 1 d iu·Z α α−lk k· → E e − Z E[ξ|Z](u, u) + iαkZ E ξ |Z · u 2 k=1 α − − + Zα l1 l2 E ξ k1k2 |Z ek + ek k ,k :1k k d 1 2 1 2 1 2 1 d = E eiu·Z iuj iuk Ajk(Z)Zα 2 j,k=1 d d + j kj α | + α | k1k2 2 iu A (Z)∂k(x ) x=Z ∂k1 ∂k2 x x=ZA (Z) j,k=1 k1,k2=1 1 = E eiu·ZA (u, Z) . 2 α Thus we have obtained the first assertion. With [C6]j , taking limit, we have iu·Z −(j−2) E e Aα(u, Z) (1 +|u|) , and then the second and the third assertions are easy consequences. ✷ α = →∞ LEMMA 3. – If [C2] is satisfied, then Kn (u) o(rn)(n ) for d each u ∈ R . Moreover, if [C7]k holds, then −1 α −(k−1) r K (u) 1 1 (u) C(1 +|u|) n n Λn d for any u ∈ R . Under [C2] and [C7]k ,ifk>d+ 1,then −1 α r K (u) 1 0 (u) du → 0 n n Λn Rd as n →∞. Proof. Ð Let · · α = iu Xn − iu Mn k1,n ψnQα(u, Mn,rnξn) e e α − · + ψ (r N )α β Q (u, M ,r ξ ) eiu Xn n β n n β n n n β:0βα |β|=|α|−1 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 445 and α − · kα = ψ (r N )α β Q (u, M ,r ξ ) eiu Xn . 2,n n β n n β n n n β:0βα |β||α|−2 −1 n[ α ]→ Clearly, [C2] implies that rn P k2,n(u) 0andthat − · α − · r 1P n kα (u) → E Zα eiu Z iu · η + ηα β Zβ eiu Z n 1,n β β:0βα |β|=|α|−1 as n →∞.Since · α − E eiu Z Zα(iu) · η + Zβ ηα β β β:0βα |β|=|α|−1 · α − = E eiu Z Zα(iu) · B(Z) + Zβ B(Z)α β β β:0βα |β|=|α|−1 d = iu·Z α · + α j E e Z (iu) B(Z) ∂j x x=ZB(Z) j=1 iu·Z = E e Bα(u, Z) , it holds that −1 α = −1 n α + α − iu·Z → rn Kn (u) rn P k1,n k2,n E e Bα(u, Z) 0 as n →∞. This shows the first assertion. What rest easily follow from [C7]k. ✷ Though the following lemma is more or less well known, we will state it here for later convenience. LEMMA 4. – Suppose that | |2 ∗ ∞ x µn,Tn 1 < . Then the following inequalities hold true: d d = ⊗2 { } − ˆ ⊗2 = ⊗2 × (1) 0 ∆ Mn ,Mn s Rd x νn( s , dx) (xn,s ) Rd x { } ˆ = ·; { } = νn( s , dx),wherexn,s Rd xνn( s , dx) 0. | d d |2 =| ⊗2 { } |2 | |4 { } (2) ∆ Mn ,Mn s Rd x νn( s , dx) Rd x νn( s , dx). 446 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 ∗ = 1 ∗ = 2 LEMMA 5. – Let α 3 and β 3 . Suppose that conditions [C1], [C2], [C3] and [C4] are satisfied. If p>2q(2q − 1)−1, then for some α constants Cp and Cα (independent of p), it holds that α α 3 2 4 L (u) 1 0 (u) C |u| + 1 ψ − 1 + C r |u| + 1 n Λn p n p α n for any n ∈ N and u ∈ Rd . Moreover, α α,d −ρ(3+d) d 2−ρ(4+d) L (u) 1 0 (u) du C r ψ − 1 + C r n Λn p n n p α n Rd for any n ∈ N. Proof. Ð Step 1. Let 1 Y = iu · M + Ξ (u, u). n,t n,t 2 n,t Define a stopping time τn by τn = inf t; sup Ξn,t (u, u) bn , u: |u|=1 = 2ρ = where bn rn .DefineΦi , i 1,...,5, as follows: = n Yn,− · · Φ1 P ψn e (iu Mn)τn , (note that on the right-hand side, the first dot stands for the stochastic integration and the second one for the inner product) 1 ∗ ⊗2 Φ = α − 1 P n ψ eYn,s− ∆Md (u, u) 2 2 n n,s sτn 1 ∗ + β P n ψ eYn,s− · Md (u, u) , n n τn 2 1 ∗ 1 ⊗3 =− − 3 n Yn,s− d Φ3 α i P ψn e ∆Mn,s (u,u,u) , 2 3 sτ n 1 ∗ Φ =− i3β P n ψ eYn,s− ∆Md (u)∆ Md (u, u) , 4 2 n n,s n s sτn and n Φ5 = P ψn R2,n(u) + R3,n(u) + R4,n(u) . Here Ri,n(u), i = 2, 3, 4, are defined by: N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 447 1 R (u) = eYn,s− (∆Y )2 − P (u), 2,n 2 n,s 2,n sTn 1 1 ⊗3 R (u) = eYn,s− (∆Y )3 − eYn,s− i3 ∆Md (u,u,u) 3,n 6 n,s 6 n,s sTn sTn and 1 1 Yn,s− 4 3 v∆Yn,s R4,n(u) = e (∆Yn,s) (1 − v) e ds, 6 s Tn 0 where 1 ⊗2 1 ∗ ⊗3 = Yn,s− d − d P2,n(u) e ∆Mn,s (iu, iu) α ∆Mn,s (iu, iu, iu) sT 2 2 n 1 − i3β∗∆Md ⊗ ∆ Md (u; u, u) . 2 n,s n s ∗ ∗ Note that if α = 1/3andβ = 2/3, we have Φ3 = 0and 1 ⊗ n Yn,s− d 2 Yn,− d Φ2 =− P ψn e ∆M (u, u) − e · M (u, u) . 3 n,s n τn sτn 0 We then have a decomposition of Ln(u): 5 0 = (2) Ln(u) Φi . i=1 In fact, by using Itô’s formula, we see that a.s. on {ψn > 0}, Y − 1 |u|2 (3) e n,Tn − e 2 = eYn,− · M (iu) n,τn 1 ∗ ⊗2 ∗ + eYn,− · α ∆Md (u, u) + β Md ,Md (u, u) n,s τn n n τn 2 s· Yn,s− ∆Yn,s · + e e − 1 − iu ∆Yn,s , sτn where Mn(u) denotes the 1-form Mn ·u. In view of the choice of numbers α∗ and β∗, we see 448 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 1 ⊗2 0 = + + n Yn,s− d Ln(u) Φ1 Φ2 P ψn e ∆Mn,s (u, u) 2 sτ n n ∆Yn,s · + P ψn e − 1 − iu ∆Yn,s sτn 1 ⊗2 = Φ + Φ + P n ψ eYn,s− ∆Md (u, u) 1 2 2 n n,s sτn n Yn,s− 2 + P ψn e (∆Yn,s ) /2! sτ n n Yn,s− 3 n + P ψn e (∆Yn,s ) /3! + P ψnR4,n(u) sτ n = + + n − 3 ∗ d ⊗3 Φ1 Φ2 P ψn i /2 α ∆Mn,s (u,u,u) sτ n 1 + n − 3 ∗ d ⊗ d d ; P ψn i β ∆Mn,s ∆ Mn ,Mn s(u u, u) 2 sτ n n n Yn,s− 3 n + P ψnR2,n(u) + P ψn e (∆Yn,s ) /3! + P [ψnR4,n] sτ n = Φ + Φ + P n ψ R (u) 1 2 n 3,n n n + Φ4 + P ψnR2,n(u) + P ψnR4,n(u) 5 = Φi, i=1 which is the desired decomposition (2). 0 Differentiating Ln by u, we obtain 5 α = − α Ln (u) ( i∂u) Φi . i=1 Step 2: By Burkholder–Davis–Gundy’s inequality, Hölder’s inequality and Doob’s inequality, we have · ∗ | |∗ | | (b Mn)τn r C b Tn p Mn,Tn q N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 449 for Rd -valued (bounded) predictable processes b and p,q,r > 1 with 1/p + 1/q = 1/r. Here dot means both the inner product and the stochastic integration. Use the above inequality for b = U(u,Mn, Ξn), where α iu·z+ 1 r(u,u) U(u,z,r)= (−i∂u) e 2 iu , · we see that U(u,Mn, Ξn)− Mn,τn∧· is a uniformly integrable martingale, and hence, the marginal mean vanishes and for p,p > 1 (1/p + 1/p = 1), α n (−i∂ ) Φ = P ψ U(u,M , Ξ )− · M u 1 n n n n,τn n = P (ψ − 1)U(u, M , Ξ )− · M n n n n,τn − · C ψn 1 p U(u,Mn, Ξn)− Mn,τn p 1 | |2 − α | |+ 2 bn u ψn 1 pCp ( u 1) e . Next we will consider Φ2.Fork = 1,...,d,let M = k d k d − k d k d 1 Mn , Mn Mn , Mn . Mτn ∈ Mn Then 1 , the set of (element-wise) uniformly integrable martin- d,τn 2 −[ d,τn d,τn ]∈Mn gales. Indeed, from [9] I.4.50(b), (Mn ) Mn ,Mn ;sim- d,τn 2 − d,τn ∈Mn = ilarly from [9] I.4.2., (Mn ) Mn .Forj,k 1,...,d and l ∈ Z+, we have the estimate: j l ∗ Mn,− · M1 τn p j l j l 1/2 Mn,− · M1, Mn,− · M1 τn p j l ∗ [M M ]1/2 Mn,− 1, 1 T Tn p1 n p2 j l ∗ 4 1/2 4 1/2 M − |x| ∗ µ + |x| ∗ ν . n, Tn p1 n,Tn p2/2 n,Tn p2/2 Here p ,p1 > 1, 2 Yn,s− into account, we see that for p (p >p/(p− 1), in particular, p>2q/(2q − 1)), 1 2 α 2 bn|u| (−i∂) Φ2 ψn − 1 p |u| + 1 e 2 ∈ 0 for u Λn. Put M = d · d,τn 2 ∆ Mn Mn . Here the integration on the right-hand side reads as a tri-linear form valued process. Then 1 ∗ (−∂ )αΦ =− i3β P n ψ (−∂ )α eYn,− · M (u; u, u) . u 4 2 n u 2,τn Since ∗ 1 |M | | |2 ∗ 2 2,τn p x µn,τn p ∈ 0 d d { } for u Λn,and∆ Mn ,Mn (3/2)Ξn is bounded on ψn > 0 a.s. because of [C4], we obtain for p,p1 > 1 (1/p + 1/p1 < 1), 1 2 1 α 3 bn|u| 2 2 (−∂u) Φ4 ψn − 1 p 1 +|u| e 2 1 + |x| ∗ µn,τ n p1 ∈ 0 for u Λn. Step 3: It follows from definition that 2 1 1 ∗ ⊗4 (4) R = eYn,s− α ∆Md (u,u,u,u) 2,n 2 2 n,s sTn 2 1 ⊗ + β∗ ∆ Md ,Md 2(u, u; u, u) 2 n n s 1 ⊗ + α∗β∗ ∆Md 2 ⊗ ∆ Md ,Md (u, u; u, u) . 2 n,s n n s Differentiating (4) and using [C1] and [C2] (moment), it is not difficult to obtain 1 | |2 n − α 2 +| |4 2 bn u ρ | | (5) P ψn( ∂u) R2,n(u) rn 1 u e P1,α rn u , where P1,α is a polynomial. X j := j d,j Yjk := 1 ∗ j k d,j d,k Zjk := With iu ∆Mn,s , 2 α u u ∆Mn,s ∆Mn,s and 1 ∗ j k d,j d,k 2 β u u ∆ Mn ,Mn s , R3,n(u) is expressed as N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 451 1 3 R (u) = eYn,s− X j + Yjk + Zjk 3,n 6 sTn j j,k j,k − X j X kX l . j k l This is a third order homogenous polynomial of X j , Yjk, Zjk without the terms X j X kX l . Differentiating this expression and taking [C1], [C4] and Lemma 4 into account, we obtain the estimate: 1 | |2 n − α 2 bn u | |4 + 2 ρ | | (6) P ψn( ∂u) R3,n(u) e u 1 rn P2,α rn u , where P2,α is a polynomial independent of n and u. In a similar way, it is easy to show (7) P n ψ (−∂ )αR (u) n u 4,n 1 1 r2 1 +|u|4 P rρ |u| exp α∗ + β∗ A2r2ρ |u|2 + b |u|2 n 3,α n 2 n 2 n for some polynomial P3,α. d Thus from (5), (6) and (7), we see that for α ∈ Z+, there exists a polynomial Pα independent of n and u such that 1 1 (−∂ )αΦ r2 |u|4 + 1 P rρ |u| exp A2|u|2r2ρ + b |u|2 . u 5 n α n 2 n 2 n α Step 4: Finally, we will estimate Ln . From above Steps, it follows that α α for some constant Cp and C , 5 α α L (u) 1 0 (u) (−i∂ ) Φ 1 0 (u) n Λn u i Λn i=1 α | |3 + − + α 2 | |4 + Cp u 1 ψn 1 p C rn u 1 for any n ∈ N and u ∈ Rd . Consequently, α L (u) 1 0 (u) du n Λn Rd α | |3 + − + 2 | |4 + Cp u 1 du ψn 1 p rn u 1 du 0 0 Λn Λn ∼ α −ρ(3+d) − + α −ρ(4+d)+2 Cp (d)rn ψn 1 p C (d)rn , which completes the proof of Lemma 5. ✷ 452 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 LEMMA 6. – Suppose conditions [C1]–[C7]k are satisfied. Suppose that j>d+ 2, k>d+ 1, l>2d + 4 and p>2q(2q − 1)−1. Then, for −1 −1 d ρ satisfying (l − d) <ρ<(4 + d) and for each α ∈ Z+, there exists α,d a constant Cp such that α − α α,d −ρ(3+d) − + sup gn (x) hn(x) Cp rn ψn 1 p o(rn). x∈Rd Proof. Ð It follows from the Fourier inversion formula that α − α (8) sup gn (x) hn(x) x∈Rd 1 d = −iu·x ˆα − ˆα sup e gn (u) hn(u) du x∈Rd 2π Rd 1 d gˆα(u) + hˆα(u) du 2π n n d 0 R −Λn 1 d + gˆα(u) − hˆα(u) du. 2π n n 0 Λn By Lemmas 2, 3 and 5, one has ˆα − ˆα α + α + α gn (u) hn(u) du Jn (u) Kn (u) Ln(u) du 0 0 Λn Λn + α,d −ρ(3+d) − + α,d 2−ρ(4+d) o(rn) Cp rn ψn 1 p C rn = + α,d −ρ(3+d) − o(rn) Cp rn ψn 1 p. On the other hand, condition [C5]l yields ∞ C (9) gˆα(u) du rd−1 dr ∼ rρ(l−d) = o(r ). n (1 + r)l n n d − 0 −ρ R Λn rn We then obtain the result from (8), (9), and a similar estimate for the ˆα ✷ integration of hn . This completes the proof. PROPOSITION 2. – Let M,γ > 0. Under the same assumptions as Lemma 6, there exist constants C1 = C1(M,γ,d,p), C2 = C2(M,γ,d) and a sequence εn with εn = o(rn) such that N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 453 ! n − [ ] +| |γ + −ρ(3+d) − P f(Xn) Pn f M sup 1 Xn p C1rn ψn 1 p n∈N + C2 ψn − 1 1 + εn ∈ ∈ E = − [ ]= for any n N and any f (M, γ ),wherep p/(p 1) and Pn f Rd f(x)pn(x) dx. n n n = −1 n Proof. Ð Define a measure Q on Ω by dQ cn ψn dP .By condition [C5], we see that the characteristic function n − · Q [ ] = 1 n iu Xn Ch Xn (u) cn P e ψn n n of the distribution L{Xn | Q } is du-integrable, and hence, L{Xn | Q } −1 0 n[ ]= has a density which is nothing but cn gn. In particular, P f(Xn)ψn 0 Rd f(x)gn(x) dx.Letd be the minimum even number greater than d, γ the minimum even number greater or equal to γ ,andk = d + γ . Clearly, for f ∈ E(M, γ ), 0 − 0 f(x) gn(x) hn(x) dx Rd −1 +| |d · α − α M 1 x dx Cd ,γ sup gn (x) hn(x) . | | x∈Rd Rd α: α k |α|:even By using this inequality and applying Lemma 6, we have P n f(X ) − P [f ] n n n n P f(Xn)ψn − P f(Xn) + 0 − 0 f(x)gn(x) dx f(x)hn(x) dx Rd Rd + 0 − f(x)hn(x) dx f(x)pn(x) dx d d R R +| |γ − M sup 1 Xn p ψn 1 p n∈N + C(M,γ,d,p)r−ρ(3+d) ψ − 1 + o(r ) n n p n γ +|1 − cn| M 1 +|x| φ(x)dx, Rd from which one has the result. ✷ 454 N. YOSHIDA / Bull. Sci. math. 125 (2001) 431–456 Proof of Theorem 1. Ð Since a1 >(d+ 3)/(l − d) and l>2d + 4, we can take ρ so that 1/(l − d)<ρ min{a1/(d + 3), 1/(d + 4)}.Let a = 1 − 2ρ and b = 2a. Take a smooth function ϕ : R →[0, 1] satisfying = | | 1 = | | that ϕ(x) 1if x 2 ,andϕ(x) 0if x 1. 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