# Asymptotic Distributions for Power Variations of the Solution to the Spatially Colored Stochastic Heat Equation

Authors: Wensheng Wang; Xiaoying Chang; Wang Liao
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

Letuα,d=uα,dt,x,t∈0,T,x∈ℝd be the solution to the stochastic heat equations (SHEs) with spatially colored noise. We study the realized power variations for the process uα,d, in time, having infinite quadratic variation and dimension-dependent Gaussian asymptotic distributions. We use the underlying explicit kernels and spectral/harmonic analysis, yielding temporal central limit theorems for SHEs with spatially colored noise. This work builds on the recent works on delicate analysis of variations of general Gaussian processes and SHEs driven by space-time white noise.


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## Body

## 1. Introduction

Throughout this work, we will consider the followingd-dimensional stochastic heat equation (SHE):(1)∂∂tuα,dt,x=ε2∂2∂x2uα,dt,x+σuα,dt,xW˙α,d,t∈+,x∈ℝd,uα,d0,x=wx,x∈ℝd,with ε>0 and Gaussian space-time colored noise Wα,d. The noise Wα,d is assumed to have a particular covariance structure (see [1]):(2)EWα,dt,AWα,ds,B=t∧s∫A∫Bfα,dx−ydxdy,t,s∈s+,A,B∈ℬbℝd,where(3)fα,dx=cα,dx−d+α,0<α<d,with cα,d=2d−απd/2Γd−α/2/Γα/2. The initial condition, wx, is taken to be bounded and ρ-Hölder continuous. We will also assume σ to be Lipschitz continuous, and there exists c0≥0 such that σx−σy≤c0x−y and σx≤c01+x. Stochastic PDEs (SPDEs) such as (1) have been studied in [1–6] and others.It is known (see [1, 7–10]) that (1) admits a unique mild solution if and only if d<2+α, and this mild solution is interpreted as the solution of the following integral equation:(4)uα,dt,x=∫ℝdGt,x−yuα,d0,ydy+∫0t∫ℝdGt−s,x−yσuα,ds,yWα,dds,dy,for t∈ℝ+,x∈ℝd, where the above integral is a Wiener integral with respect to the noise Wα,d (see, e.g., [2] for the definition) and G is the Green kernel of the heat equation given by(5)Gt,x=2πεt−1/2e−x2/2εt,ift>0,x∈ℝd,0,ift≤0,x∈ℝd.Bezdek [11] investigated weak convergence of probability measures corresponding to the solution of (1) in d=1. He showed that probability measures corresponding to uα,1 weakly converge to those corresponding to the solution to the SHE with white noise when α↑1, that is, the solution of (1) converges in the appropriate sense to the solution of the same equation, but with white noise W instead of colored noise Wα,1 as α↑1. By that, we mean the solution to(6)∂∂tut,x=ε2∂2∂x2ut,x+σut,xW˙,t∈ℝ+,x∈ℝ,u0,x=wx,x∈ℝ,where W denotes white noise. SPDEs such as (6) have been studied in [1, 2, 7, 10, 12, 13] and others.Among others, Tudor and Xiao [14] investigated the exact uniform and local moduli of continuity and Chung-type laws of the iterated logarithm of the process uα,d in time. In fact, they investigated these path properties for a more wide class, namely, the solution to the linear SHE driven by a fractional noise in time with correlated spatial structure. Swanson [13] showed that the solutions of the SHEs in (6) with ε=σ=1, in time, have infinite quadratic variation and are not semimartingales and also investigated central limit theorems (CLTs) for modifications of the quadratic variations of the solutions of the SHEs with white noise. Pospíšil and Tribe [12] investigated the quartic variations of the solutions of the SHEs in (6) with ε=σ=1, in time, having Gaussian asymptotic distributions. Inspired by Swanson [13] and Pospíšil and Tribe [12], in this work, we show that the realized power variations of the solutions of the SHEs in (1) with colored noise, in time, have infinite quadratic variation and Gaussian asymptotic distributions.Forp>0, the p-power variation of a process X, with respect to a subdivision πn=0=tn,0<tn,1<…<tn,n=1 of 0,1, is defined to be the sum(7)∑j=1nXtn,j−Xtn,j−1p.For simplicity, consider from now on the case wheretn,j=j/n, for n∈ℕ and j∈1,…,n. In this work, we wish to point out some interesting phenomena when X is the solution to a SHE with colored noise. In fact, we will also drop the absolute value (when p is odd). More precisely, we will consider(8)∑j=1nΔXjp,where ΔXj=ΔXj/n denotes the increment Xj/n−Xj−1/n.The analysis of the asymptotic behavior of quantities of type (8) is motivated, for instance, by the study of the exact rates of convergence of some approximation schemes of scalar stochastic differential equations driven by a Brownian motion (BM) B (see, e.g., [15–17]), besides, of course, the traditional applications of quadratic variations to parameter estimation problems.Now, let us recall some known results concerning thep-power variations (for p∈ℕ+), which are today more or less classical. First, assume that B is the standard BM. Let μp denote the p-moment of a standard Gaussian random variable following an N0,1 law, that is, μ2p−1=0 and μ2p=2p−1!!=2p!/p!2p for all p∈ℕ+. By the scaling property of the BM and using the CLT, it is immediate that (see, e.g., [17]), as n⟶∞:(9)1n∑j=1nnp/2ΔBjp−μp⟶ℒN0,μ2p−μp2.Assume thatH≠1/2, that is, the case where the fractional Brownian motion (FBM) BH has no independent increments anymore. Then, (9) has been extended by Corcuera et al. [15], Nourdin [17], Dobrushin and Major [18], Taqqu [19], Breuer and Major [20], Giraitis and Surgailis [21], Wang [22], and Wang and Wang [23]. Swanson [13] extended (9) to modifications of the quadratic variation of the solutions of SHE driven by space-time white noise. Motivated by (9), in this work, we show that (9) with different mean and variance also holds for the solution to SHE with colored noise.Our proofs are based on the method of Swanson [13]. We make use of the product moments of various orders of the normal correlation surface of two variates in Pearson and Young [24] to establish exact convergence rates of variances of the realized power variation of the process u with respect to time. This work builds on the recent works on delicate analysis of variations of general Gaussian processes and SHEs driven by space-time white noise.


## 2. Results

In order to state our results, we first introduce some notations. LetXα,dt=uα,dt,x, where x∈ℝ is fixed. We consider discrete Riemann sums over a uniformly spaced time partition tj=jΔt, where Δt=n−1. Let ΔXα,d;j=Xα,dtj−Xα,dtj−1 and σα,d;j2=EΔXα,d;j2. For any p∈ℕ+ and n∈ℕ+, we define(10)VpnXα,dt=∑j=1ntΔXα,d;jp.Here and in the sequel,⌊a⌋ denotes an integer satisfying a−1<⌊a⌋≤a for a∈ℝ+.Letθ=θα,d=d−α/2. For j∈ℕ+, let aj=aα,d,j=2j1−θ−j−11−θ−j+11−θ. For real number r≥1, define br=bα,d,r=∑j=1∞ajr. It follows from (44) below that br is a positive and finite constant depending on α,d, and r. For any p∈ℕ+, we put(11)κα,d,p=Kα,dpμ2p−μp2+p!p!2p−1∑u=1p/222ub2up/2−u!p/2−u!2u!,ifpis even,Kα,dpμ2p−p!p!2p−2∑u=0p/222ub2u+1p/2−u!p/2−u!2u+1!,ifpis odd,where(12)Kα,d=Kα,dθ,ε=Γθ2d+α/2πd/2Γd/21−θ2εd/4,where Γs=∫0∞xs−1e−xdx, s>0, is the gamma function.We will first show the exact convergence rate of variance for the realized power variation of the processXα,d.Theorem 1.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for each fixed t>0 and any p∈ℕ+,(13)n−1+p1−θVarVpnXα,dt⟶κα,d,pt,as n tends to infinity.By (13), we have the following convergence in probability for the realized power variation of the process Xα,d.Corollary 2.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for each fixed t>0 and any p∈ℕ+,(14)n−1+p1−θV2pnXα,dt⟶Kα,dpμ2pt,in L2 and in probability as n tends to infinity.Remark 3.
SinceV2pnXα,dt is monotone, (14) implies that n−1+p1−θV2pnXα,dt⟶Kα,dpμ2pt uniform convergence in probability in the time interval 0,T with some T>0. Moreover, (14) implies that the process Xα,d has infinite quadratic variation.Example 4.
Ifα↑1 and d=2, the 4-th variation, namely, p=2 in (14), the corresponding constant of the right-hand side of (14) is equal to 3/επ.
The CLT for the realized power variation of the processXα,d is as follows.Theorem 2.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for any p∈ℕ+,(15)Xα,dt,1n∑j=1ntnp1−θ/2ΔXα,d;jp−Kα,dp/2μp⟶ℒXα,dt,κα,d,p1/2Bt,as n tends to infinity, where B=Bt,t∈0,T is a BM independent of the process Xα,d, and the convergence is in the space D0,T2 equipped with the Skorokhod topology.Remark 6.
Comparing (15) and (9), we have that the realized power variations of the process Xα,d for α+1≤d<α+2 share similar Gaussian asymptotic properties with those of BM.
Throughout this paper, positive and finite constants are numbered asc2,1, c2,2, … or c3,1, c3,2, ….


## 3. Proofs

### 3.1. Preliminaries

We need the following product moment of various orders of the normal correlation surface of two variates, which are equations (9) and (12) in Pearson and Young [24].Lemma 7.
Suppose thatξ,η∼N0,σ12ρρσ22, where ρ=σ1σ2−1Eξη. Then,(16)Eξpηp=p!p!2pσ1pσ2p∑j=1p/22ρ2jp/2−j!p/2−j!2j!,ifpis even,ρp!p!2p−1σ1pσ2p∑j=0p/22ρ2jp/2−j!p/2−j!2j+1!,ifpis odd.We also derive some needed estimates on the covariance function and the variance function of increments ofXα,d.Lemma 8.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for all s,t∈0,T,(17)EXα,dtXα,ds=Kα,dt+s1−θ−t−s1−θ,(18)c2,1t−s1−θ≤EXα,dt−Xα,ds2≤c2,2t−s1−θ,and(19)EXα,dt−Xα,ds2−Kα,dt−s1−θ≤c2,3sθ+1t−s2,where Kα,d is given in (12).Proof.
By Proposition 2.3 of Tudor [10], one has that (17) holds with(20)Kα,d=2π−d1−θ∫ℝddξξαe−ξ2/2.
By the following integral formula (see Corollary on page 23 in [25]):(21)∫ℝdf∑i=1dxi2dx1…dxd=πd/2Γd/2∫0∞yd/2−1fydy,the constant Kα,d becomes(22)Kα,d=Γθ2d+α/2πd/2Γd/21−θ.
This is (12) and yields (17).
Equation (18) is cited from Theorem 2.2 in Tudor [10]. It remains to show (19). To show (19), we define the following pinned string process in time Uα,dt,t≥0 by(23)Uα,dt=∫0t∫ℝdGt−u,x−yWα,ddu,dy+∫−∞0∫ℝdGt−u,x−y−G−u,x−yWα,ddu,dy.
Note thatUα,d0=0 and Uα,dt can be expressed as(24)Uα,dt=∫ℝ∫ℝdGt−u+,x−y−G−u+,x−yWα,ddu,dy.
In the above,a+=maxa,0. Now for every t≥0, one has the following decomposition:(25)Xα,dt=Uα,dt−Yα,dt,where(26)Yα,dt=∫R∫RdGt−u,x−yI0>u−G−u+,x−yWα,ddu,dy.
Following the same lines as the Proof of Theorem 1 of Tudor and Xiao [14], for any 0≤s<t,(27)EUα,dt−Uα,ds2=Kα,dt−s1−θ.
Denote byμdξ=ξ−αdξ the tempered non-negative measure on ℝd. Let ℱφ denote the Fourier transform of the function u⟼φu and f be the Riesz kernel defined in (3). Then, for any φ,ψ∈Sℝd (see, e.g., [10, 14]),(28)∫ℝd∫ℝdφxfx−yψydxdy=2π−d∫ℝdℱφξℱψξ¯μdξ.
It follows from (28) that for any 0≤s<t,(29)EYα,dt−Yα,ds2=E∫ℝ∫ℝdGt−u,x−yI0>u−Gs−u,x−yI0>uWα,ddu,dy2=∫ℝdu∫ℝd∫ℝdGt−u,x−yI0>u−Gs−u,x−yI0>u×Gt−u,x−y′I0>u−Gs−u,x−y′I0>ufy−y′dydy′=2π−d∫Rdu∫ℝdμdξℱGt−u,x−⋅I0>u−Gs−u,x−⋅I0>uξ×ℱGt−u,x−⋅I0>u−Gs−u,x−⋅I0>uξ¯=2π−d∫ℝdμdξ∫ℝe−t−uεξ2/2I0>u−e−s−uεξ2/2I0>u2du=2π−d∫ℝdμdξ∫−∞0e−t−uεξ2/2−e−s−uεξ2/22du.
Since1−e−x≤2x for all x≥0, one has for all 0≤s<t and ξ∈ℝ,(30)e−tεξ2/2−e−sεξ2/2=e−sεξ2/21−e−t−sεξ2/2≤t−sεξ2e−sεξ2/2.
Thus, by (21), for any 0<s<t,(31)∫ℝdμdξ∫−∞0e−t−uεξ2/2−e−s−uεξ2/22du=∫ℝdμdξ∫−∞0euεξ2e−tεξ2/2−e−sεξ2/22du=∫ℝd1ξα+2e−tεξ2/2−e−sεξ2/22dξ≤t−s2∫ℝd1ξα−2e−sεξ2dξ=πd/2Γd/2t−s2sθ+1∫0∞xθe−xdx≤c2,4sθ+1t−s2.
Combining (29) and (31), one has(32)EYα,dt−Yα,ds2≤c2,5t−s2sθ+1.
It follows from the argument of (29) that(33)EXα,dt−Xα,ds2−EUα,dt−Uα,ds2=EYα,dt−Yα,ds2.
This, together with (27) and (32), yields (19). The proof of Lemma 8 is completed.


### 3.2. Proof of Theorem 1

Proof.
of Theorem 1. It is sufficient to prove (13) for the even p case since the odd p case can be proved similarly. For 1≤i<j≤nt, define ρα,d;ij=σα,d;iσα,d;j−1EΔXα,d;iΔXα,d;j. Note that for a random variable ξ following an N0,σ2 law,(34)Eξp=μpσp,∀p∈ℕ+.
By (16) and (34), one has(35)VarVpnXα,dt=E∑j=1ntΔXα,d;jp−μpσα,d;jp2=∑j=1ntEΔXα,d;jp−μpσα,d;jp2+2∑i=1nt∑j=i+1ntEΔXα,d;ip−μpσα,d;ipΔXα,d;jp−μpσα,d;jp=∑j=1ntEΔXα,d;j2p−μp2σα,d;j2p+2∑i=1nt∑j=i+1ntEΔXα,d;ipΔXα,d;jp−μp2σα,d;ipσα,d;jp=μ2p−μp2∑j=1ntσα,d;j2p+p!p!2p−1∑u=1p/222up/2−u!p/2−u!2u!∑i=1nt∑j=i+1ntσα,d;ipσα,d;jpρα,d;ij2u.
It follows from (18) that(36)σα,d;j2≤c2,6n−1+θfor all1≤j≤nt.
By (19), (36), and Lagrange mean value theorem, it holds that for any real number r>0 and 1<j≤nt,(37)σα,d;jr−Kα,dn−1+θr/2≤c2,7σα,d;jr−2+Kα,dn−1+θr−2/2σα,d;j2−Kα,dn−1+θ≤c2,8n−2+−1+θr−2/2tj−1−θ+1.
Note that sinceα+1≤d<α+2, one has 1/2≤θ<1. Thus,(38)1n∑j=2nttj−1−θ+1/2⟶∫0tx−θ+1/2dx=21−θt−θ+1/2.
It follows from (37) (with r=2p) and (38) that(39)n−1+p1−θ∑j=1ntσα,d;j2p−Kα,dn−1+θp⟶0.
Hence,(40)n−1+p1−θ∑j=1ntσα,d;j2p=n−1+p1−θ∑j=1ntσα,d;j2p−Kα,dn−1+θp+n−1+p1−θKα,dn−1+θpnt⟶Kα,dpt.
It follows from (17) that(41)EΔXα,d;iΔXα,d;j=Kα,dn−1+θj+i1−θ−j−i1−θ−j+i−11−θ+j−i+11−θ−j+i−11−θ+j−i−11−θ+j+i−21−θ−j−i1−θ,which simplifies to(42)EΔXα,d;iΔXα,d;j=−Kα,dn−1+θaj+i−1+n−1+θaj−i,where aj=2j1−θ−j−11−θ−j+11−θ. Thus, by binomial expansion, for every 1≤u≤p/2 and 1≤i<j≤nt,(43)σα,d;ipσα,d;jpρα,d;ij2u=σα,d;ip−2uσα,d;jp−2uEΔXα,d;iΔXα,d;j2u=Kα,d2uσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1+n−1+θaj−i2u=Kα,d2u∑v=02u2uvσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1vn−1+θaj−i2u−v.
If we writeak=fk−1−fk, where fx=x+11−θ−x1−θ, then for each k≥2, the Lagrange mean value theorem gives ak=f′k−ζ1=θ1−θk−ζ1+ζ2−θ−1 for some ζ1,ζ2∈0,1. This yields that for all k∈ℕ+,(44)0<ak≤c2,9kθ+1,and hence for any r≥1,(45)∑k=1Makr⟶br,with some br=br>0 as M⟶∞.
Note that sincej+i−1≥j+i/2, one has(46)n−1+θaj+i−1≤c2,10n21ti+tjθ+1.
Note that (44) gives n−1+θaj−i≤c2,11n−1+θ and n−1+θaj+i−1≤c2,12n−1+θ for all 1≤i<j≤nt. Thus, by (36) and (46), for every 1≤u≤p/2 and 1≤v≤2u,(47)n−1+p1−θ∑i=1nt∑j=i+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1vn−1+θaj−i2u−v≤c2,13n−θ∑i=1nt∑j=i+1ntn−1+θaj+i−1≤c2,14n−2−θ∑i=1nt∑j=i+1nt1ti+tjθ+1,which tends to zero as n⟶∞ since ∫0t∫0tx+y−θ+1dxdy<∞.
We now consider the termv=0 in (43). Let BH=BHt,t∈ℝ+ be a FBM with index H∈0,1, which is a centered Gaussian process with EBHt−BHs2=s−t2H for s,t∈ℝ+. Then, for H0=1−θ/2,(48)EBH0j+1n−BH0jnBH0i+1n−BH0in=−122j−in1−θ−j−i−1n1−θ−j−i+1n1−θ=−12n−1+θaj−i.
Thus,(49)n−1+θ∑i=1nt∑j=i+1ntaj−i=n−1+θ∑i=1nt−1∑j=i+1ntaj−i=−2∑i=1nt−1∑j=i+1ntEBH0j+1n−BH0jnBH0i+1n−BH0in=−2∑i=1nt−1EBH0nt+1n−BH0i+1nBH0i+1n−BH0in=−∑i=1nt−1−nt−in1−θ+nt+1−in1−θ−1n1−θ=−ntn1−θ+1n1−θ+ntn−1+θ.
This yields(50)n−θ∑i=1nt∑j=i+1ntn−1+θaj−i⟶t.
By (36) and (44), for every 1≤u≤p/2 and any M>0,(51)n−1+p1−θ∑i=1nt∑j=i+M+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u≤c2,15M−θ+12u−1n−θ∑i=1nt∑j=i+M+1ntn−1+θaj−i≤c2,16M−θ+12u−1n−θ∑i=1nt∑j=i+1ntn−1+θaj−i.
This, together with (45), yields(52)n−1+p1−θ∑i=1nt∑j=i+M+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u≤c2,17M−θ+12u−1t,which tends to zero by letting M⟶∞.
By (37) (with r=p−2u), (36), and (48), for every 1≤u≤p/2,(53)n−1+p1−θ∑i=2nt∑j=i+1ntσα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2un−1+θaj−i2u≤c2,18n−1−2θ∑i=2nt1ti−1θ+1∑j=i+1ntn−1+θaj−i=−2c2,18n−1−2θ∑i=2nt1ti−1θ+1−nt−in1−θ+nt+1−in1−θ−1n1−θ≤c2,19n−θ∑i=2nt−nt−in1−θ+nt+1−in1−θ+c2,20n−2−θ∑i=2nt1ti−1θ+1≤c2,21n−θ1n1−θ+nt−1n1−θ+c2,22n−3/2−θ/2∑i=2ntti−1−θ+1/2,which tends to zero as n⟶∞ since ∫0tx−θ+1/2dx<∞. Hence, for every 1≤u≤p/2,(54)n−1+p1−θ∑i=2nt∑j=i+1ntσα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2un−1+θaj−i2u⟶0.
Similarly, for every1≤u≤p/2,(55)n−1+p1−θ∑i=2nt∑j=i+1ntKα,dn−1+θp−2u/2×σα,d;jp−2u−Kα,dn−1+θp−2u/2n−1+θaj−i2u⟶0.
For every1≤u≤p/2 and any M>0,(56)n−1+p1−θ∑i=2nt∑j=i+1i+MKα,dn−1+θp−2un−1+θaj−i2u=Kα,dp−2unt−1n∑j=1Maj2u⟶Kα,dp−2ub2ut,as n⟶∞ and M⟶∞.
Note that for every1≤u≤p/2 and 1≤i<j≤nt,(57)σα,d;ip−2uσα,d;jp−2u=σα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2u+Kα,dn−1+θp−2u/2σα,d;jp−2u−Kα,dn−1+θp−2u/2+Kα,dn−1+θp−2u.
Hence, by (54)–(57), for every 1≤u≤p/2,(58)n−1+p1−θ∑i=2nt∑j=i+1i+Mσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u⟶Kα,dp−2ub2ut,as n⟶∞ and M⟶∞. It follows from (36) that(59)n−1+p1−θ∑j=21+Mσα,d;ip−2uσα,d;jp−2un−1+θaj−12u⟶0.
This, together with (43), (47), and (58), yields for every 1≤u≤p/2,(60)n−1+p1−θ∑i=1nt∑j=i+1ntσα,d;ipσα,d;jpρα,d;ij2u⟶Kα,dpb2ut.
Therefore, by (35), (40), and (60), one has(61)n−1+p1−θVarVpnXα,dt⟶Kα,dpμ2p−μp2+p!p!2p−1∑u=1p/222ub2up/2−u!p/2−u!2u!t=κα,d,pt.
This proves (13). The Proof of Theorem 1 is completed.Proof.
of Corollary2. Write(62)n−1+p1−θ/2VpnXα,dt−Kα,dp/2μpt=n−1+p1−θ/2VpnXα,dt−EVpnXα,dt+μpn−1+p1−θ/2∑j=1ntσα,d;jp−Kα,dn−1+θp/2+Kα,dp/2μpntn−t.
Obviously, the third term of (62) tends to zero as n⟶∞. It follows from (37) (with r=p) and (38) that the second term of (62) tends to zero as n⟶∞. Thus, by (13),(63)En−1+p1−θ/2VpnXα,dt−Kα,dp/2μpt2⟶0.
This proves (14).


### 3.3. Proof of Theorem 2

The following lemma is needed to prove Theorem2.Lemma 9.
LetF1,…,F4 be normal random variables with mean zero, EFj2=1 and ρij=EFiFj. Put ξj=Fjp−EFjp. Then, for any p∈ℕ+,(64)E∏j=14ξj≤c3,1ρ12ρ34+11−ρ122maxi≤2<jρij,whenever ρ12<1. Moreover,(65)E∏j=14ξj≤c3,2max2≤j≤4ρ1j.Furthermore, there existsε>0 such that(66)E∏j=14ξj≤c3,3max1≤i≠j≤4ρij2,whenever ρij<ε for all 1≤i≠j≤4.Proof.
Following the same lines as the proof of Lemma 3.3 in Swanson [13] with hjFj=ξj, 1≤j≤4, we get Lemma 9 immediately.Proposition 10.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Fix r∈ℕ+. Put(67)WrnXα,dt=n−1/2+r1−θ/2∑i=1ntΔXα,d;ir−μrσα,d;ir.Then, for all0≤s<t and all n∈ℕ+,(68)EWrnXα,dt−WrnXα,ds4≤c3,4nt−nsn2.The sequenceWrnXα,d is therefore relatively compact in the Skorokhod space Dℝ0,∞.Proof.
We follow the method of Proposition 3.5 in Swanson [13] to prove (68). Let S=j∈ℕ+4:⌊ns⌋+1≤j1≤…≤j4≤nt. For j∈S and k∈1,2,3, define hk=jk+1−jk and let Sk=j∈S:hk=maxh1,h2,h3. Define N=nt−⌊ns⌋+1, and for i∈0,1,…,N, let Ski=j∈Sk:maxh1,h2,h3=i. Further define Tkℓ=Tki,ℓ=j∈Ski:minh1,h2,h3=ℓ and Vkv=Vki,ℓ,v=j∈Tkℓ:medh1,h2,h3=v, where “med” denotes the median function. For j∈S, define(69)Uα,d;j=∏k=14ΔXα,d;jkr−μrσα,d;jkr.
Observe that(70)EWrnXα,dt−WrnXα,ds4=n−2+2r1−θE∑i=nt+1ntΔXα,d;ir−μrσα,d;ir4≤4!n−2+2r1−θ∑j∈SEUα,d;j≤4!n−2+2r1−θ∑k=13∑j∈SkEUα,d;j,and that(71)∑j∈SkEUα,d;j=∑i=0N∑j∈SkiEUα,d;j=∑i=0N∑ℓ=0iθ∑j∈TkℓEUα,d;j+∑i=0N∑ℓ=iθ+1i∑j∈TkℓEUα,d;j=∑i=0N∑ℓ=0iθ∑v=ℓi∑j∈VkvEUα,d;j+∑i=0N∑ℓ=iθ+1i∑v=ℓi∑j∈VkvEUα,d;j.
LetFα,d;k=σα,d;jk−1ΔXα,d;jk and(72)ξα,d;k=Fα,d;kr−EFα,d;kr=σα,d;jk−rΔXα,d;jkr−μrσα,d;jkr.
Then,(73)EUα,d;j=∏k=14σα,d;jkrE∏k=14ξα,d;k.
By (42) and (44), for all k≠l∈ℕ+,(74)EΔXα,d;kΔXα,d;l≤c3,5n−1+θk−lθ+1.
It follows from (36) and (74) that(75)ρα,d;kl=EFα,d;kFα,d,l=σα,d;jk−1σα,d;jl−1EΔXα,d;jkΔXα,d;jl≤c3,6jk−jlθ+1.
Suppose0≤ℓ≤iθ. Fix v and let j∈Vkv be arbitrary. If k=1, then i=maxh1,h2,h3=h1=j2−j1. If k=3, then i=maxh1,h2,h3=h3=j4−j3. In either case, by (65), (36), (73), and (75), one has(76)EUα,d;j≤c3,7n−2r1−θiθ+1≤c3,71ℓvθ+1+1iθ+1n−2r1−θ.
Ifk=2, then i=maxh1,h2,h3=h2=j3−j2 and ℓv=h3h1=j4−j3j2−j1. Hence, by (64), (36), (73), and (75),(77)EUα,d;j≤c3,81ℓvθ+1+1iθ+1n−2r1−θ.
Now choosek′≠k such that hk′=ℓ. With k′ given, j is determined by jk. Since there are two possibilities for k′ and N+1 possibilities for jk, Vkv≤2N+1. Therefore,(78)∑ℓ=0iθ∑v=ℓi∑j∈VkvEUα,d;j≤c3,9N+1∑ℓ=0iθ∑v=ℓi1ℓvθ+1+1iθ+1n−2r1−θ≤c3,10N+1∑ℓ=0iθ1ℓθ+1+1iθn−2r1−θ≤c3,11N+1n−2r1−θ.
For the second summation, supposeiθ+1≤ℓ≤i. In this case, if j∈Tkℓ, then ℓ=minh1,h2,h3, so that by (66), (36), (73), and (75),(79)EUα,d;j≤c3,12n−2r1−θℓ2θ+1.
Since∑v=ℓiVkv≤2N+1i and 1/2≤θ<1, one has(80)∑ℓ=iθ+1i∑v=ℓi∑j∈VkvEUα,d;j≤c3,13N+1i∑iθ+1in−2r1−θℓ2θ+1≤c3,14N+1i∫iθ∞1x2θ+1dxn−2r1−θ≤c3,15N+1n−2r1−θ.
Thus, using (70), (71), (78), and (80), one has(81)n−2+2r1−θE∑j=ns+1ntΔXα,d;jr−μrσα,d;jr4≤c3,16∑i=0NN+1n−2=c3,16nt−nsn2,which is (68).
To show that a sequence of càdlàg processesXn is relatively compact, it suffices to show that for each T>1, there exist constants β>0, C>0, and q>1 such that(82)RXnt,h=EXnt+h−XntβXnt−Xnt−hβ≤Chq,for all n∈ℕ, all t∈0,T, and all h∈0,t (see, e.g., Theorem 3.8.8 in [26]). Taking β=2 and using (68) together with Hölder inequality gives(83)RWrnXα,dt,h≤c3,17nt+nh−ntnnt−nt−nhn.
Ifnh<1/2, then the right-hand side of this inequality is zero. Assume nh≥1/2. Then,(84)nt+nh−ntn≤nh+1n≤3h.
The other factor is similarly bounded, so thatRWrnXα,dt,h≤c3,18h2.Proposition 11.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for any 0≤s<t and r∈ℕ+,(85)WrnXα,dt−WrnXα,ds⟶ℒκα,d,r1/2t−s1/2N,as n⟶∞, where N is a standard normal random variable.Proof.
Letnjj=1∞ be any sequence of natural numbers. We will prove that there exists a subsequence njm such that WrnjmXα,dt−WrnjmXα,ds converges in law to the given random variable.
For eachm∈ℕ+, choose njm∈nj such that njm>njm−1 and njm≥m2/θt−s−1. Let b=bm=njmt−s/m. For 0≤k≤m, define uk=njms+kb, so that(86)WrnjmXα,dt−WrnjmXα,ds=njm−1/2+r1−θ/2∑i=njms+1njmtΔXα,d;ir−μrσα,d;ir=njm−1/2+r1−θ/2∑k=1m∑i=uk−1+1ukΔXα,d;ir−μrσα,d;ir.
Let us now introduce the filtration(87)ℱt=σWα,dA:A⊂0,t×ℝd,λA<∞,where λ denotes Lebesgue measure on ℝd+1. Let τk=njm−1uk−1. For each pair i,k such that uk−1<i≤uk, define(88)ξα,d;ik=ΔXα,d;i−EΔXα,d;i|ℱτk.
Note thatξα,d;ik is ℱτk+1-measurable and independent of ℱτk. Recall that(89)Xα,dt=∫0t∫ℝdGt−x,yWα,ddx,dy.
Also, given constants0≤τ≤s≤t, one has(90)EXα,dt|ℱτ=∫0τ∫ℝdGt−x,yWα,ddx,dy.
It follows from (89) and (90) that(91)Xα,dt+τk−EXα,dt+τk|ℱτk=∫τkt+τk∫ℝdGt+τk−x,yWα,ddx,dy.
This yields thatξα,d;ik has the same law as ΔXα,d;i−uk−1.
Now defineσα,d;ik2=Eξα,d;ik2=σα,d;i−uk−12 and(92)ζα,d;mk=∑i=uk−1+1ukξα,d;ikr−μrσα,d;ikr,so that ζα,d;mk, 1≤k≤m, are independent and(93)WrnjmXα,dt−WrnjmXα,ds=njm−1/2+r1−θ/2∑k=1mζα,d;mk+εα,d;m,where(94)εα,d;m=njm−1/2+r1−θ/2∑k=1m∑i=uk−1+1ukΔXα,d;ir−μrσα,d;ir−ξα,d;ikr−μrσα,d;ikr.
Sinceξα,d;ik and ΔXα,d;i−ξα,d;ik=EΔXα,d;i|ℱτk are independent, one has(95)σα,d;i2=EΔXα,d;i2=Eξα,d;ik2+EΔXα,d;i−ξα,d;ik2=σα,d;i−uk−12+EΔXα,d;i−ξα,d;ik2.
This, together with (19), gives(96)EΔXα,d;i−ξα,d;ik2=σα,d;i2−σα,d;i−uk−12≤c3,19njm−1+θi−uk−1θ+1.
Thus, sinceΔXα,d;i−ξα,d;ik is Gaussian, by (34) and (96),(97)EΔXα,d;i−ξα,d;ik4≤c3,20njm−2+2θi−uk−12θ+2.
Note that (34) and (36) give EΔXα,d;i4r−4≤c3,21σα,d;i4r−4≤c3,22njm−1+θ2r−2 and Eξα,d;ik4r−4≤c3,23σα,d;i−uk−14r−4≤c3,24njm−1+θ2r−2. By Lagrange mean value theorem,(98)ΔXα,d;ir−ξα,d;ikr≤c3,25ΔXα,d;ir−1+ξα,d;ikr−1ΔXα,d;i−ξα,d;ik.
Thus, by (97) and Hölder inequality,(99)EΔXα,d;ir−ξα,d;ikr2≤c3,26EΔXα,d;i4r−4+Eξα,d;ik4r−41/2EΔXα,d;i−ξα,d;ik41/2≤c3,27njm−r1−θi−uk−1θ+1.
Similarly, by (96) and Lagrange mean value theorem,(100)σα,d;ir−σα,d;ikr≤c3,28σα,d;ir−2+σα,d;ikr−2σα,d;i2−σα,d;ik2≤c3,29njm−r1−θ/2i−uk−1θ+1.
Therefore, by (99), (100), and Hölder inequality,(101)Eεα,d;m≤njm−1/2+r1−θ/2∑k=1m∑j=uk−1+1ukEΔXα,d;ir−ξα,d;ikr21/2+μrσα,d;jr−σα,d;jkr≤c3,30njm−1/2∑k=1m∑i=uk−1+1uki−uk−1−θ+1/2=c3,31njm−1/2∑k=1m∑i=1uk−uk−1i−θ+1/2.
Sinceuk−uk−1≤b, this gives(102)Eεα,d;m≤c3,32njm−1/2mb1−θ/2=c3,32mθ+1/2njm−θ/2t−s1−θ/2.
But sincenjm was chosen so that njm≥m2/θt−s−1, one has Eεα,d;m≤c3,33m−1−θ/2t−s1/2 and ϵα,d;m⟶0 in L1 and in probability. Therefore, by (93), one needs only to show that(103)njm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒκα,d,r1/2t−s1/2Nin order to complete the proof.
For this, we will use the Lindeberg–Feller theorem (see, e.g., Theorem 2.4.5 in [27]), which states the following: for each m, let ζα,d;mk,1≤k≤m, be independent random variables with Eζα,d;mk=0. Suppose(a)
njm−1+r1−θ∑k=1mEζα,d;mk2⟶ν2.(b)
For allε>0, limm⟶∞njm−1+r1−θ∑k=1mEζα,d;mk22Injm+r1−θ/2×ζα,d;mk>ε−1/2⟶0.
Then,njm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒνN as n⟶∞.
To verify these conditions, recall thatξα,d;ik and ΔXα,d;i−uk−1 have the same law, so that(104)Eζα,d;mk4=njm−2+2r1−θE∑i=1uk−uk−1ΔXα,d;ir−μrσα,d;ir4.
Hence, by (68),(105)njm−2+2r1−θEζα,d;mk4≤c3,34uk−uk−12njm−2.
Jensen inequality now givesm−1+r1−θ∑k=1mEζα,d;mk2≤c3,35mbnjm−1=c3,35t−s, so that by passing to a subsequence, one may assume that (a) holds for some ν≥0.
For (b), letε>0 be arbitrary. Then,(106)njm−1+r1−θ∑k=1mEζα,d;mk2Injm−1/2+r1−θ/2ζα,d;mk>ε≤ε−2njm−2+2r1−θ∑k=1mEζα,d;mk4≤c3,36ε−2mb2njm−2=c3,36ε−2m−1t−s2,which tends to zero as m⟶∞.
It therefore follows thatnjm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒνN as n⟶∞ and it remains only to show that ν=κα,d,r1/2t−s1/2. For this, observe that the continuous mapping theorem implies that WrmXα,dt−WrmXα,ds2⟶ℒν2N2. By the Skorokhod representation theorem, one may assume that the convergence is a.s. By Proposition 10, the family WrmXα,dt−WrmXα,ds2 is uniformly integrable. Hence, WrmXα,dt−WrmXα,ds2⟶ν2N2 in L1, which implies EWrmXα,dt−WrmXα,ds2⟶ν2. But by Theorem 1, EWrmXα,dt−WrmXα,ds2⟶κα,d,rt−s, so ν=κα,d,r1/2t−s1/2 and the proof is complete.Proof.
of Theorem 2It is sufficient to prove (15) for the even p case since the odd p case can be proved similarly. Let njj=1∞ be any sequence of natural numbers. By Proposition 10, the sequence Xα,d,WpnjXα,d is relatively compact. Therefore, there exists a subsequence njk and a càdlàg process Yα,d such that Xα,d,WpnjkXα,d⟶ℒXα,d,Yα,d. Fix 0<s1<s2<…<sℓ<s<t. With notation as in Proposition 11, let(107)ζα,d;njk=njk−1/2+p1−θ/2∑i=njks+2njktξα,d;ikp−μpσα,d;ikp,and define(108)ηα,d;njk=WpnjkXα,dt−WpnjkXα,ds−ζα,d;njk.
As in the proof of Proposition 11,ηα,d;njk⟶0 in probability. It therefore follows that(109)WpnjkXα,ds1,…,WpnjkXα,dsℓ,ζα,d;njk⟶ℒYα,ds1,…,Yα,dsℓ,Yα,dt−Yα,ds.
Note thatℱ⌊njks⌋+1njk−1 and ζα,d;njk are independent. Hence, WpnjkXα,ds1,…,WpnjkXα,dsℓ, …,WpnjkXα,dsℓ and ζα,d;njk are independent, which implies that Yα,dt−Yα,ds and Yα,ds1,…,Yα,dsℓ are independent. This yields that the process Yα,d has independent increments.
By Proposition11, the increment Yα,dt−Yα,ds is normally distributed with mean zero and variance κα,d,pt−s. Also, Xα,d0=0 since WpnXα,d0=0 for all n. Hence, Yα,d is equal in law to κα,d,p1/2B, where B is a standard BM. It remains only to show that Xα,d and B are independent.
Fix0<s1<s2<…<sℓ≤T. Let Zα,d=Xα,ds1,…,Xα,dsℓT and Σα,d=EZα,dZα,dT. It is easy to see that Σα,d is invertible. Hence, one may define the vectors vα,d;j∈ℝℓ by vα,d;j=EZα,dΔXα,d;j, and wα,d;j=Σα,d−1vα,d;j. Let ξα,d;j=ΔXα,d;j−wα,d;jTZα,d, so that ξα,d;j and Zα,d are independent.
Define(110)W˜pnXα,dt=n−1/2+p1−θ/2∑j=1ntξα,d;jp−μpσα,d;jp.
Then,(111)WpnXα,dt−W˜pnXα,dt≤n−1/2+p1−θ/2∑j=1ntΔXα,d;jp−ξα,d;jp.
By (34), binomial expansion, and Hölder inequality,(112)Esup0≤t≤TWpnXα,dt−W˜pnXα,dt≤c3,37n−1/2+p1−θ/2∑ν=1p∑j=1ntEΔXα,d;j2p−2ν1/2Ewα,d;jTZα,d2ν1/2≤c3,38∑ν=1pn−1/2+ν1−θ/2∑j=1ntEwα,d;jTZα,d2ν1/2≤c3,39max1≤i≤ℓ∑ν=1pn−1/2+ν1−θ/2∑j=1ntEXα,dsiΔXα,d;jν.
Note that by (36) and Hölder inequality, one has EXα,dsiΔXα,d;j≤c3,40σα,d;j≤c3,41n−1−θ/2 for all 1≤i≤ℓ and 1≤j≤nt, and note that by (17) and Lagrange mean value theorem, for any 1≤i≤ℓ and 1≤j≤nt,(113)EXα,dsiΔXα,d;j=Kα,dsi+tj1−θ−si+tj−11−θ−si−tj1−θ+si−tj−11−θ=Kα,d1−θnsi+j−ζ1n−θ+si−j−ζ2n−θ≤2Kα,d1−θnsi−j−ζ2n−θ,where ζ1,ζ2∈0,1. Then, for any 1≤i≤ℓ and 1≤ν≤2p,(114)n−1/2+ν1−θ/2∑j=1ntEXα,dsiΔXα,d;jν≤c3,42n1/2−ν1+θ/21n∑j=1ntsi−j−ζ2n−θ,which tends to zero as n⟶∞ since ∫0Tsi−x−θdx<∞. Thus,(115)Zα,d,W˜pnXα,ds1,…,W˜pnXα,dsd⟶ℒZα,d,κα,d,p1/2Bs1,…,κα,d,p1/2Bsℓ.
SinceZα,d and W˜pnXα,d are independent, this gives that Xα,d and B are independent.
We now can complete the proof. Note that by (37) and (38),(116)max0≤t≤T1n∑j=1ntnp1−θ/2ΔXα,d;jp−Kα,dp/2μp−WpnXα,dt≤μpn−1/2+p1−θ/2∑j=1ntσα,d;jp−Kα,dn−1+θp/2⟶0.
This finishes the proof.


## 3.1. Preliminaries

We need the following product moment of various orders of the normal correlation surface of two variates, which are equations (9) and (12) in Pearson and Young [24].Lemma 7.
Suppose thatξ,η∼N0,σ12ρρσ22, where ρ=σ1σ2−1Eξη. Then,(16)Eξpηp=p!p!2pσ1pσ2p∑j=1p/22ρ2jp/2−j!p/2−j!2j!,ifpis even,ρp!p!2p−1σ1pσ2p∑j=0p/22ρ2jp/2−j!p/2−j!2j+1!,ifpis odd.We also derive some needed estimates on the covariance function and the variance function of increments ofXα,d.Lemma 8.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for all s,t∈0,T,(17)EXα,dtXα,ds=Kα,dt+s1−θ−t−s1−θ,(18)c2,1t−s1−θ≤EXα,dt−Xα,ds2≤c2,2t−s1−θ,and(19)EXα,dt−Xα,ds2−Kα,dt−s1−θ≤c2,3sθ+1t−s2,where Kα,d is given in (12).Proof.
By Proposition 2.3 of Tudor [10], one has that (17) holds with(20)Kα,d=2π−d1−θ∫ℝddξξαe−ξ2/2.
By the following integral formula (see Corollary on page 23 in [25]):(21)∫ℝdf∑i=1dxi2dx1…dxd=πd/2Γd/2∫0∞yd/2−1fydy,the constant Kα,d becomes(22)Kα,d=Γθ2d+α/2πd/2Γd/21−θ.
This is (12) and yields (17).
Equation (18) is cited from Theorem 2.2 in Tudor [10]. It remains to show (19). To show (19), we define the following pinned string process in time Uα,dt,t≥0 by(23)Uα,dt=∫0t∫ℝdGt−u,x−yWα,ddu,dy+∫−∞0∫ℝdGt−u,x−y−G−u,x−yWα,ddu,dy.
Note thatUα,d0=0 and Uα,dt can be expressed as(24)Uα,dt=∫ℝ∫ℝdGt−u+,x−y−G−u+,x−yWα,ddu,dy.
In the above,a+=maxa,0. Now for every t≥0, one has the following decomposition:(25)Xα,dt=Uα,dt−Yα,dt,where(26)Yα,dt=∫R∫RdGt−u,x−yI0>u−G−u+,x−yWα,ddu,dy.
Following the same lines as the Proof of Theorem 1 of Tudor and Xiao [14], for any 0≤s<t,(27)EUα,dt−Uα,ds2=Kα,dt−s1−θ.
Denote byμdξ=ξ−αdξ the tempered non-negative measure on ℝd. Let ℱφ denote the Fourier transform of the function u⟼φu and f be the Riesz kernel defined in (3). Then, for any φ,ψ∈Sℝd (see, e.g., [10, 14]),(28)∫ℝd∫ℝdφxfx−yψydxdy=2π−d∫ℝdℱφξℱψξ¯μdξ.
It follows from (28) that for any 0≤s<t,(29)EYα,dt−Yα,ds2=E∫ℝ∫ℝdGt−u,x−yI0>u−Gs−u,x−yI0>uWα,ddu,dy2=∫ℝdu∫ℝd∫ℝdGt−u,x−yI0>u−Gs−u,x−yI0>u×Gt−u,x−y′I0>u−Gs−u,x−y′I0>ufy−y′dydy′=2π−d∫Rdu∫ℝdμdξℱGt−u,x−⋅I0>u−Gs−u,x−⋅I0>uξ×ℱGt−u,x−⋅I0>u−Gs−u,x−⋅I0>uξ¯=2π−d∫ℝdμdξ∫ℝe−t−uεξ2/2I0>u−e−s−uεξ2/2I0>u2du=2π−d∫ℝdμdξ∫−∞0e−t−uεξ2/2−e−s−uεξ2/22du.
Since1−e−x≤2x for all x≥0, one has for all 0≤s<t and ξ∈ℝ,(30)e−tεξ2/2−e−sεξ2/2=e−sεξ2/21−e−t−sεξ2/2≤t−sεξ2e−sεξ2/2.
Thus, by (21), for any 0<s<t,(31)∫ℝdμdξ∫−∞0e−t−uεξ2/2−e−s−uεξ2/22du=∫ℝdμdξ∫−∞0euεξ2e−tεξ2/2−e−sεξ2/22du=∫ℝd1ξα+2e−tεξ2/2−e−sεξ2/22dξ≤t−s2∫ℝd1ξα−2e−sεξ2dξ=πd/2Γd/2t−s2sθ+1∫0∞xθe−xdx≤c2,4sθ+1t−s2.
Combining (29) and (31), one has(32)EYα,dt−Yα,ds2≤c2,5t−s2sθ+1.
It follows from the argument of (29) that(33)EXα,dt−Xα,ds2−EUα,dt−Uα,ds2=EYα,dt−Yα,ds2.
This, together with (27) and (32), yields (19). The proof of Lemma 8 is completed.


## 3.2. Proof of Theorem 1

Proof.
of Theorem 1. It is sufficient to prove (13) for the even p case since the odd p case can be proved similarly. For 1≤i<j≤nt, define ρα,d;ij=σα,d;iσα,d;j−1EΔXα,d;iΔXα,d;j. Note that for a random variable ξ following an N0,σ2 law,(34)Eξp=μpσp,∀p∈ℕ+.
By (16) and (34), one has(35)VarVpnXα,dt=E∑j=1ntΔXα,d;jp−μpσα,d;jp2=∑j=1ntEΔXα,d;jp−μpσα,d;jp2+2∑i=1nt∑j=i+1ntEΔXα,d;ip−μpσα,d;ipΔXα,d;jp−μpσα,d;jp=∑j=1ntEΔXα,d;j2p−μp2σα,d;j2p+2∑i=1nt∑j=i+1ntEΔXα,d;ipΔXα,d;jp−μp2σα,d;ipσα,d;jp=μ2p−μp2∑j=1ntσα,d;j2p+p!p!2p−1∑u=1p/222up/2−u!p/2−u!2u!∑i=1nt∑j=i+1ntσα,d;ipσα,d;jpρα,d;ij2u.
It follows from (18) that(36)σα,d;j2≤c2,6n−1+θfor all1≤j≤nt.
By (19), (36), and Lagrange mean value theorem, it holds that for any real number r>0 and 1<j≤nt,(37)σα,d;jr−Kα,dn−1+θr/2≤c2,7σα,d;jr−2+Kα,dn−1+θr−2/2σα,d;j2−Kα,dn−1+θ≤c2,8n−2+−1+θr−2/2tj−1−θ+1.
Note that sinceα+1≤d<α+2, one has 1/2≤θ<1. Thus,(38)1n∑j=2nttj−1−θ+1/2⟶∫0tx−θ+1/2dx=21−θt−θ+1/2.
It follows from (37) (with r=2p) and (38) that(39)n−1+p1−θ∑j=1ntσα,d;j2p−Kα,dn−1+θp⟶0.
Hence,(40)n−1+p1−θ∑j=1ntσα,d;j2p=n−1+p1−θ∑j=1ntσα,d;j2p−Kα,dn−1+θp+n−1+p1−θKα,dn−1+θpnt⟶Kα,dpt.
It follows from (17) that(41)EΔXα,d;iΔXα,d;j=Kα,dn−1+θj+i1−θ−j−i1−θ−j+i−11−θ+j−i+11−θ−j+i−11−θ+j−i−11−θ+j+i−21−θ−j−i1−θ,which simplifies to(42)EΔXα,d;iΔXα,d;j=−Kα,dn−1+θaj+i−1+n−1+θaj−i,where aj=2j1−θ−j−11−θ−j+11−θ. Thus, by binomial expansion, for every 1≤u≤p/2 and 1≤i<j≤nt,(43)σα,d;ipσα,d;jpρα,d;ij2u=σα,d;ip−2uσα,d;jp−2uEΔXα,d;iΔXα,d;j2u=Kα,d2uσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1+n−1+θaj−i2u=Kα,d2u∑v=02u2uvσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1vn−1+θaj−i2u−v.
If we writeak=fk−1−fk, where fx=x+11−θ−x1−θ, then for each k≥2, the Lagrange mean value theorem gives ak=f′k−ζ1=θ1−θk−ζ1+ζ2−θ−1 for some ζ1,ζ2∈0,1. This yields that for all k∈ℕ+,(44)0<ak≤c2,9kθ+1,and hence for any r≥1,(45)∑k=1Makr⟶br,with some br=br>0 as M⟶∞.
Note that sincej+i−1≥j+i/2, one has(46)n−1+θaj+i−1≤c2,10n21ti+tjθ+1.
Note that (44) gives n−1+θaj−i≤c2,11n−1+θ and n−1+θaj+i−1≤c2,12n−1+θ for all 1≤i<j≤nt. Thus, by (36) and (46), for every 1≤u≤p/2 and 1≤v≤2u,(47)n−1+p1−θ∑i=1nt∑j=i+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj+i−1vn−1+θaj−i2u−v≤c2,13n−θ∑i=1nt∑j=i+1ntn−1+θaj+i−1≤c2,14n−2−θ∑i=1nt∑j=i+1nt1ti+tjθ+1,which tends to zero as n⟶∞ since ∫0t∫0tx+y−θ+1dxdy<∞.
We now consider the termv=0 in (43). Let BH=BHt,t∈ℝ+ be a FBM with index H∈0,1, which is a centered Gaussian process with EBHt−BHs2=s−t2H for s,t∈ℝ+. Then, for H0=1−θ/2,(48)EBH0j+1n−BH0jnBH0i+1n−BH0in=−122j−in1−θ−j−i−1n1−θ−j−i+1n1−θ=−12n−1+θaj−i.
Thus,(49)n−1+θ∑i=1nt∑j=i+1ntaj−i=n−1+θ∑i=1nt−1∑j=i+1ntaj−i=−2∑i=1nt−1∑j=i+1ntEBH0j+1n−BH0jnBH0i+1n−BH0in=−2∑i=1nt−1EBH0nt+1n−BH0i+1nBH0i+1n−BH0in=−∑i=1nt−1−nt−in1−θ+nt+1−in1−θ−1n1−θ=−ntn1−θ+1n1−θ+ntn−1+θ.
This yields(50)n−θ∑i=1nt∑j=i+1ntn−1+θaj−i⟶t.
By (36) and (44), for every 1≤u≤p/2 and any M>0,(51)n−1+p1−θ∑i=1nt∑j=i+M+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u≤c2,15M−θ+12u−1n−θ∑i=1nt∑j=i+M+1ntn−1+θaj−i≤c2,16M−θ+12u−1n−θ∑i=1nt∑j=i+1ntn−1+θaj−i.
This, together with (45), yields(52)n−1+p1−θ∑i=1nt∑j=i+M+1ntσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u≤c2,17M−θ+12u−1t,which tends to zero by letting M⟶∞.
By (37) (with r=p−2u), (36), and (48), for every 1≤u≤p/2,(53)n−1+p1−θ∑i=2nt∑j=i+1ntσα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2un−1+θaj−i2u≤c2,18n−1−2θ∑i=2nt1ti−1θ+1∑j=i+1ntn−1+θaj−i=−2c2,18n−1−2θ∑i=2nt1ti−1θ+1−nt−in1−θ+nt+1−in1−θ−1n1−θ≤c2,19n−θ∑i=2nt−nt−in1−θ+nt+1−in1−θ+c2,20n−2−θ∑i=2nt1ti−1θ+1≤c2,21n−θ1n1−θ+nt−1n1−θ+c2,22n−3/2−θ/2∑i=2ntti−1−θ+1/2,which tends to zero as n⟶∞ since ∫0tx−θ+1/2dx<∞. Hence, for every 1≤u≤p/2,(54)n−1+p1−θ∑i=2nt∑j=i+1ntσα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2un−1+θaj−i2u⟶0.
Similarly, for every1≤u≤p/2,(55)n−1+p1−θ∑i=2nt∑j=i+1ntKα,dn−1+θp−2u/2×σα,d;jp−2u−Kα,dn−1+θp−2u/2n−1+θaj−i2u⟶0.
For every1≤u≤p/2 and any M>0,(56)n−1+p1−θ∑i=2nt∑j=i+1i+MKα,dn−1+θp−2un−1+θaj−i2u=Kα,dp−2unt−1n∑j=1Maj2u⟶Kα,dp−2ub2ut,as n⟶∞ and M⟶∞.
Note that for every1≤u≤p/2 and 1≤i<j≤nt,(57)σα,d;ip−2uσα,d;jp−2u=σα,d;ip−2u−Kα,dn−1+θp−2u/2σα,d;jp−2u+Kα,dn−1+θp−2u/2σα,d;jp−2u−Kα,dn−1+θp−2u/2+Kα,dn−1+θp−2u.
Hence, by (54)–(57), for every 1≤u≤p/2,(58)n−1+p1−θ∑i=2nt∑j=i+1i+Mσα,d;ip−2uσα,d;jp−2un−1+θaj−i2u⟶Kα,dp−2ub2ut,as n⟶∞ and M⟶∞. It follows from (36) that(59)n−1+p1−θ∑j=21+Mσα,d;ip−2uσα,d;jp−2un−1+θaj−12u⟶0.
This, together with (43), (47), and (58), yields for every 1≤u≤p/2,(60)n−1+p1−θ∑i=1nt∑j=i+1ntσα,d;ipσα,d;jpρα,d;ij2u⟶Kα,dpb2ut.
Therefore, by (35), (40), and (60), one has(61)n−1+p1−θVarVpnXα,dt⟶Kα,dpμ2p−μp2+p!p!2p−1∑u=1p/222ub2up/2−u!p/2−u!2u!t=κα,d,pt.
This proves (13). The Proof of Theorem 1 is completed.Proof.
of Corollary2. Write(62)n−1+p1−θ/2VpnXα,dt−Kα,dp/2μpt=n−1+p1−θ/2VpnXα,dt−EVpnXα,dt+μpn−1+p1−θ/2∑j=1ntσα,d;jp−Kα,dn−1+θp/2+Kα,dp/2μpntn−t.
Obviously, the third term of (62) tends to zero as n⟶∞. It follows from (37) (with r=p) and (38) that the second term of (62) tends to zero as n⟶∞. Thus, by (13),(63)En−1+p1−θ/2VpnXα,dt−Kα,dp/2μpt2⟶0.
This proves (14).


## 3.3. Proof of Theorem 2

The following lemma is needed to prove Theorem2.Lemma 9.
LetF1,…,F4 be normal random variables with mean zero, EFj2=1 and ρij=EFiFj. Put ξj=Fjp−EFjp. Then, for any p∈ℕ+,(64)E∏j=14ξj≤c3,1ρ12ρ34+11−ρ122maxi≤2<jρij,whenever ρ12<1. Moreover,(65)E∏j=14ξj≤c3,2max2≤j≤4ρ1j.Furthermore, there existsε>0 such that(66)E∏j=14ξj≤c3,3max1≤i≠j≤4ρij2,whenever ρij<ε for all 1≤i≠j≤4.Proof.
Following the same lines as the proof of Lemma 3.3 in Swanson [13] with hjFj=ξj, 1≤j≤4, we get Lemma 9 immediately.Proposition 10.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Fix r∈ℕ+. Put(67)WrnXα,dt=n−1/2+r1−θ/2∑i=1ntΔXα,d;ir−μrσα,d;ir.Then, for all0≤s<t and all n∈ℕ+,(68)EWrnXα,dt−WrnXα,ds4≤c3,4nt−nsn2.The sequenceWrnXα,d is therefore relatively compact in the Skorokhod space Dℝ0,∞.Proof.
We follow the method of Proposition 3.5 in Swanson [13] to prove (68). Let S=j∈ℕ+4:⌊ns⌋+1≤j1≤…≤j4≤nt. For j∈S and k∈1,2,3, define hk=jk+1−jk and let Sk=j∈S:hk=maxh1,h2,h3. Define N=nt−⌊ns⌋+1, and for i∈0,1,…,N, let Ski=j∈Sk:maxh1,h2,h3=i. Further define Tkℓ=Tki,ℓ=j∈Ski:minh1,h2,h3=ℓ and Vkv=Vki,ℓ,v=j∈Tkℓ:medh1,h2,h3=v, where “med” denotes the median function. For j∈S, define(69)Uα,d;j=∏k=14ΔXα,d;jkr−μrσα,d;jkr.
Observe that(70)EWrnXα,dt−WrnXα,ds4=n−2+2r1−θE∑i=nt+1ntΔXα,d;ir−μrσα,d;ir4≤4!n−2+2r1−θ∑j∈SEUα,d;j≤4!n−2+2r1−θ∑k=13∑j∈SkEUα,d;j,and that(71)∑j∈SkEUα,d;j=∑i=0N∑j∈SkiEUα,d;j=∑i=0N∑ℓ=0iθ∑j∈TkℓEUα,d;j+∑i=0N∑ℓ=iθ+1i∑j∈TkℓEUα,d;j=∑i=0N∑ℓ=0iθ∑v=ℓi∑j∈VkvEUα,d;j+∑i=0N∑ℓ=iθ+1i∑v=ℓi∑j∈VkvEUα,d;j.
LetFα,d;k=σα,d;jk−1ΔXα,d;jk and(72)ξα,d;k=Fα,d;kr−EFα,d;kr=σα,d;jk−rΔXα,d;jkr−μrσα,d;jkr.
Then,(73)EUα,d;j=∏k=14σα,d;jkrE∏k=14ξα,d;k.
By (42) and (44), for all k≠l∈ℕ+,(74)EΔXα,d;kΔXα,d;l≤c3,5n−1+θk−lθ+1.
It follows from (36) and (74) that(75)ρα,d;kl=EFα,d;kFα,d,l=σα,d;jk−1σα,d;jl−1EΔXα,d;jkΔXα,d;jl≤c3,6jk−jlθ+1.
Suppose0≤ℓ≤iθ. Fix v and let j∈Vkv be arbitrary. If k=1, then i=maxh1,h2,h3=h1=j2−j1. If k=3, then i=maxh1,h2,h3=h3=j4−j3. In either case, by (65), (36), (73), and (75), one has(76)EUα,d;j≤c3,7n−2r1−θiθ+1≤c3,71ℓvθ+1+1iθ+1n−2r1−θ.
Ifk=2, then i=maxh1,h2,h3=h2=j3−j2 and ℓv=h3h1=j4−j3j2−j1. Hence, by (64), (36), (73), and (75),(77)EUα,d;j≤c3,81ℓvθ+1+1iθ+1n−2r1−θ.
Now choosek′≠k such that hk′=ℓ. With k′ given, j is determined by jk. Since there are two possibilities for k′ and N+1 possibilities for jk, Vkv≤2N+1. Therefore,(78)∑ℓ=0iθ∑v=ℓi∑j∈VkvEUα,d;j≤c3,9N+1∑ℓ=0iθ∑v=ℓi1ℓvθ+1+1iθ+1n−2r1−θ≤c3,10N+1∑ℓ=0iθ1ℓθ+1+1iθn−2r1−θ≤c3,11N+1n−2r1−θ.
For the second summation, supposeiθ+1≤ℓ≤i. In this case, if j∈Tkℓ, then ℓ=minh1,h2,h3, so that by (66), (36), (73), and (75),(79)EUα,d;j≤c3,12n−2r1−θℓ2θ+1.
Since∑v=ℓiVkv≤2N+1i and 1/2≤θ<1, one has(80)∑ℓ=iθ+1i∑v=ℓi∑j∈VkvEUα,d;j≤c3,13N+1i∑iθ+1in−2r1−θℓ2θ+1≤c3,14N+1i∫iθ∞1x2θ+1dxn−2r1−θ≤c3,15N+1n−2r1−θ.
Thus, using (70), (71), (78), and (80), one has(81)n−2+2r1−θE∑j=ns+1ntΔXα,d;jr−μrσα,d;jr4≤c3,16∑i=0NN+1n−2=c3,16nt−nsn2,which is (68).
To show that a sequence of càdlàg processesXn is relatively compact, it suffices to show that for each T>1, there exist constants β>0, C>0, and q>1 such that(82)RXnt,h=EXnt+h−XntβXnt−Xnt−hβ≤Chq,for all n∈ℕ, all t∈0,T, and all h∈0,t (see, e.g., Theorem 3.8.8 in [26]). Taking β=2 and using (68) together with Hölder inequality gives(83)RWrnXα,dt,h≤c3,17nt+nh−ntnnt−nt−nhn.
Ifnh<1/2, then the right-hand side of this inequality is zero. Assume nh≥1/2. Then,(84)nt+nh−ntn≤nh+1n≤3h.
The other factor is similarly bounded, so thatRWrnXα,dt,h≤c3,18h2.Proposition 11.
Fixε>0 and x∈ℝd and assume α+1≤d<α+2. Assume that w=0 and σ=1 in (1). Then, for any 0≤s<t and r∈ℕ+,(85)WrnXα,dt−WrnXα,ds⟶ℒκα,d,r1/2t−s1/2N,as n⟶∞, where N is a standard normal random variable.Proof.
Letnjj=1∞ be any sequence of natural numbers. We will prove that there exists a subsequence njm such that WrnjmXα,dt−WrnjmXα,ds converges in law to the given random variable.
For eachm∈ℕ+, choose njm∈nj such that njm>njm−1 and njm≥m2/θt−s−1. Let b=bm=njmt−s/m. For 0≤k≤m, define uk=njms+kb, so that(86)WrnjmXα,dt−WrnjmXα,ds=njm−1/2+r1−θ/2∑i=njms+1njmtΔXα,d;ir−μrσα,d;ir=njm−1/2+r1−θ/2∑k=1m∑i=uk−1+1ukΔXα,d;ir−μrσα,d;ir.
Let us now introduce the filtration(87)ℱt=σWα,dA:A⊂0,t×ℝd,λA<∞,where λ denotes Lebesgue measure on ℝd+1. Let τk=njm−1uk−1. For each pair i,k such that uk−1<i≤uk, define(88)ξα,d;ik=ΔXα,d;i−EΔXα,d;i|ℱτk.
Note thatξα,d;ik is ℱτk+1-measurable and independent of ℱτk. Recall that(89)Xα,dt=∫0t∫ℝdGt−x,yWα,ddx,dy.
Also, given constants0≤τ≤s≤t, one has(90)EXα,dt|ℱτ=∫0τ∫ℝdGt−x,yWα,ddx,dy.
It follows from (89) and (90) that(91)Xα,dt+τk−EXα,dt+τk|ℱτk=∫τkt+τk∫ℝdGt+τk−x,yWα,ddx,dy.
This yields thatξα,d;ik has the same law as ΔXα,d;i−uk−1.
Now defineσα,d;ik2=Eξα,d;ik2=σα,d;i−uk−12 and(92)ζα,d;mk=∑i=uk−1+1ukξα,d;ikr−μrσα,d;ikr,so that ζα,d;mk, 1≤k≤m, are independent and(93)WrnjmXα,dt−WrnjmXα,ds=njm−1/2+r1−θ/2∑k=1mζα,d;mk+εα,d;m,where(94)εα,d;m=njm−1/2+r1−θ/2∑k=1m∑i=uk−1+1ukΔXα,d;ir−μrσα,d;ir−ξα,d;ikr−μrσα,d;ikr.
Sinceξα,d;ik and ΔXα,d;i−ξα,d;ik=EΔXα,d;i|ℱτk are independent, one has(95)σα,d;i2=EΔXα,d;i2=Eξα,d;ik2+EΔXα,d;i−ξα,d;ik2=σα,d;i−uk−12+EΔXα,d;i−ξα,d;ik2.
This, together with (19), gives(96)EΔXα,d;i−ξα,d;ik2=σα,d;i2−σα,d;i−uk−12≤c3,19njm−1+θi−uk−1θ+1.
Thus, sinceΔXα,d;i−ξα,d;ik is Gaussian, by (34) and (96),(97)EΔXα,d;i−ξα,d;ik4≤c3,20njm−2+2θi−uk−12θ+2.
Note that (34) and (36) give EΔXα,d;i4r−4≤c3,21σα,d;i4r−4≤c3,22njm−1+θ2r−2 and Eξα,d;ik4r−4≤c3,23σα,d;i−uk−14r−4≤c3,24njm−1+θ2r−2. By Lagrange mean value theorem,(98)ΔXα,d;ir−ξα,d;ikr≤c3,25ΔXα,d;ir−1+ξα,d;ikr−1ΔXα,d;i−ξα,d;ik.
Thus, by (97) and Hölder inequality,(99)EΔXα,d;ir−ξα,d;ikr2≤c3,26EΔXα,d;i4r−4+Eξα,d;ik4r−41/2EΔXα,d;i−ξα,d;ik41/2≤c3,27njm−r1−θi−uk−1θ+1.
Similarly, by (96) and Lagrange mean value theorem,(100)σα,d;ir−σα,d;ikr≤c3,28σα,d;ir−2+σα,d;ikr−2σα,d;i2−σα,d;ik2≤c3,29njm−r1−θ/2i−uk−1θ+1.
Therefore, by (99), (100), and Hölder inequality,(101)Eεα,d;m≤njm−1/2+r1−θ/2∑k=1m∑j=uk−1+1ukEΔXα,d;ir−ξα,d;ikr21/2+μrσα,d;jr−σα,d;jkr≤c3,30njm−1/2∑k=1m∑i=uk−1+1uki−uk−1−θ+1/2=c3,31njm−1/2∑k=1m∑i=1uk−uk−1i−θ+1/2.
Sinceuk−uk−1≤b, this gives(102)Eεα,d;m≤c3,32njm−1/2mb1−θ/2=c3,32mθ+1/2njm−θ/2t−s1−θ/2.
But sincenjm was chosen so that njm≥m2/θt−s−1, one has Eεα,d;m≤c3,33m−1−θ/2t−s1/2 and ϵα,d;m⟶0 in L1 and in probability. Therefore, by (93), one needs only to show that(103)njm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒκα,d,r1/2t−s1/2Nin order to complete the proof.
For this, we will use the Lindeberg–Feller theorem (see, e.g., Theorem 2.4.5 in [27]), which states the following: for each m, let ζα,d;mk,1≤k≤m, be independent random variables with Eζα,d;mk=0. Suppose(a)
njm−1+r1−θ∑k=1mEζα,d;mk2⟶ν2.(b)
For allε>0, limm⟶∞njm−1+r1−θ∑k=1mEζα,d;mk22Injm+r1−θ/2×ζα,d;mk>ε−1/2⟶0.
Then,njm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒνN as n⟶∞.
To verify these conditions, recall thatξα,d;ik and ΔXα,d;i−uk−1 have the same law, so that(104)Eζα,d;mk4=njm−2+2r1−θE∑i=1uk−uk−1ΔXα,d;ir−μrσα,d;ir4.
Hence, by (68),(105)njm−2+2r1−θEζα,d;mk4≤c3,34uk−uk−12njm−2.
Jensen inequality now givesm−1+r1−θ∑k=1mEζα,d;mk2≤c3,35mbnjm−1=c3,35t−s, so that by passing to a subsequence, one may assume that (a) holds for some ν≥0.
For (b), letε>0 be arbitrary. Then,(106)njm−1+r1−θ∑k=1mEζα,d;mk2Injm−1/2+r1−θ/2ζα,d;mk>ε≤ε−2njm−2+2r1−θ∑k=1mEζα,d;mk4≤c3,36ε−2mb2njm−2=c3,36ε−2m−1t−s2,which tends to zero as m⟶∞.
It therefore follows thatnjm−1/2+r1−θ/2∑k=1mζα,d;mk⟶ℒνN as n⟶∞ and it remains only to show that ν=κα,d,r1/2t−s1/2. For this, observe that the continuous mapping theorem implies that WrmXα,dt−WrmXα,ds2⟶ℒν2N2. By the Skorokhod representation theorem, one may assume that the convergence is a.s. By Proposition 10, the family WrmXα,dt−WrmXα,ds2 is uniformly integrable. Hence, WrmXα,dt−WrmXα,ds2⟶ν2N2 in L1, which implies EWrmXα,dt−WrmXα,ds2⟶ν2. But by Theorem 1, EWrmXα,dt−WrmXα,ds2⟶κα,d,rt−s, so ν=κα,d,r1/2t−s1/2 and the proof is complete.Proof.
of Theorem 2It is sufficient to prove (15) for the even p case since the odd p case can be proved similarly. Let njj=1∞ be any sequence of natural numbers. By Proposition 10, the sequence Xα,d,WpnjXα,d is relatively compact. Therefore, there exists a subsequence njk and a càdlàg process Yα,d such that Xα,d,WpnjkXα,d⟶ℒXα,d,Yα,d. Fix 0<s1<s2<…<sℓ<s<t. With notation as in Proposition 11, let(107)ζα,d;njk=njk−1/2+p1−θ/2∑i=njks+2njktξα,d;ikp−μpσα,d;ikp,and define(108)ηα,d;njk=WpnjkXα,dt−WpnjkXα,ds−ζα,d;njk.
As in the proof of Proposition 11,ηα,d;njk⟶0 in probability. It therefore follows that(109)WpnjkXα,ds1,…,WpnjkXα,dsℓ,ζα,d;njk⟶ℒYα,ds1,…,Yα,dsℓ,Yα,dt−Yα,ds.
Note thatℱ⌊njks⌋+1njk−1 and ζα,d;njk are independent. Hence, WpnjkXα,ds1,…,WpnjkXα,dsℓ, …,WpnjkXα,dsℓ and ζα,d;njk are independent, which implies that Yα,dt−Yα,ds and Yα,ds1,…,Yα,dsℓ are independent. This yields that the process Yα,d has independent increments.
By Proposition11, the increment Yα,dt−Yα,ds is normally distributed with mean zero and variance κα,d,pt−s. Also, Xα,d0=0 since WpnXα,d0=0 for all n. Hence, Yα,d is equal in law to κα,d,p1/2B, where B is a standard BM. It remains only to show that Xα,d and B are independent.
Fix0<s1<s2<…<sℓ≤T. Let Zα,d=Xα,ds1,…,Xα,dsℓT and Σα,d=EZα,dZα,dT. It is easy to see that Σα,d is invertible. Hence, one may define the vectors vα,d;j∈ℝℓ by vα,d;j=EZα,dΔXα,d;j, and wα,d;j=Σα,d−1vα,d;j. Let ξα,d;j=ΔXα,d;j−wα,d;jTZα,d, so that ξα,d;j and Zα,d are independent.
Define(110)W˜pnXα,dt=n−1/2+p1−θ/2∑j=1ntξα,d;jp−μpσα,d;jp.
Then,(111)WpnXα,dt−W˜pnXα,dt≤n−1/2+p1−θ/2∑j=1ntΔXα,d;jp−ξα,d;jp.
By (34), binomial expansion, and Hölder inequality,(112)Esup0≤t≤TWpnXα,dt−W˜pnXα,dt≤c3,37n−1/2+p1−θ/2∑ν=1p∑j=1ntEΔXα,d;j2p−2ν1/2Ewα,d;jTZα,d2ν1/2≤c3,38∑ν=1pn−1/2+ν1−θ/2∑j=1ntEwα,d;jTZα,d2ν1/2≤c3,39max1≤i≤ℓ∑ν=1pn−1/2+ν1−θ/2∑j=1ntEXα,dsiΔXα,d;jν.
Note that by (36) and Hölder inequality, one has EXα,dsiΔXα,d;j≤c3,40σα,d;j≤c3,41n−1−θ/2 for all 1≤i≤ℓ and 1≤j≤nt, and note that by (17) and Lagrange mean value theorem, for any 1≤i≤ℓ and 1≤j≤nt,(113)EXα,dsiΔXα,d;j=Kα,dsi+tj1−θ−si+tj−11−θ−si−tj1−θ+si−tj−11−θ=Kα,d1−θnsi+j−ζ1n−θ+si−j−ζ2n−θ≤2Kα,d1−θnsi−j−ζ2n−θ,where ζ1,ζ2∈0,1. Then, for any 1≤i≤ℓ and 1≤ν≤2p,(114)n−1/2+ν1−θ/2∑j=1ntEXα,dsiΔXα,d;jν≤c3,42n1/2−ν1+θ/21n∑j=1ntsi−j−ζ2n−θ,which tends to zero as n⟶∞ since ∫0Tsi−x−θdx<∞. Thus,(115)Zα,d,W˜pnXα,ds1,…,W˜pnXα,dsd⟶ℒZα,d,κα,d,p1/2Bs1,…,κα,d,p1/2Bsℓ.
SinceZα,d and W˜pnXα,d are independent, this gives that Xα,d and B are independent.
We now can complete the proof. Note that by (37) and (38),(116)max0≤t≤T1n∑j=1ntnp1−θ/2ΔXα,d;jp−Kα,dp/2μp−WpnXα,dt≤μpn−1/2+p1−θ/2∑j=1ntσα,d;jp−Kα,dn−1+θp/2⟶0.
This finishes the proof.



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*Source: 8208934-2021-10-12.xml*