proces… · Gaussian approximations for weighted empirical processes under dependence Weidong Liu...

Gaussian approximations for weighted empirical

processes under dependence

Weidong Liu∗

April 24, 2018

Abstract

Let Xn be a function of i.i.d random variables and define the weighted empirical

process Fq(s, t) = q(s)∑[t]

k=1(I{Xk≤s} − F (s)), where F (s) is the distribution of Xn

and q(s) is a weighted function. We show that Fq(s, t) can be approximated by

q(s)K(s, t), where K(s, t) is a two-parameters Gaussian process.

Keywords. Empirical processes; Gaussian approximation; linear process; GARCH

process; short-range dependence; weak convergence.

1 Introduction and main results

Let X1, X2, · · · be a sequence of random variables with common distribution F (x). De-

fine the weighted empirical process Fq(s, t) = q(s)∑[t]

k=1(I{Xk≤s} − F (s)), where q(s) is a

weighted function. In this paper, we shall consider the Gaussion approximation for Fq(s, t).

This subject has been extensively explored by many authors when X1, X2, · · · are i.i.d. ran-

dom variables and q(s) ≡ 1. We refer the reader to Csorgo and Revesz (1981) for deep

∗Department of Mathematics, Institute of Natural Sciences and MOE-LSC, Shanghai Jiao Tong Uni-

versity. Research supported by NSFC, Grants No.11201298, No.11322107 and No.11431006, the Pro-

gram for Professor of Special Appointment (Eastern Scholar) at Shanghai Institutions of Higher Learning,

Shanghai Pujiang Program, Shanghai Shuguang Program and 973 Program (2015CB856004). Email:

[email protected].

1

results on this issue. Various generalizations have also been made without independent

assumption and most of them were focused on mixing sequences or some special processes.

For example, assuming q(s) ≡ 1 and {Xn} is a sequence of strong mixing and uniform

random variables with certain mixing rates, Berkes and Philipp (1977) proved

sup0≤s≤1

sup0≤t≤n

|Fq(s, t)−K(s, t)| = O(n1/2/(log n)λ) a.s. (1.1)

for some λ > 0, where K(s, t) is a Gaussian process with mean zero and covariance

function min{t, t′}Γ(s, s′) defined in (1.10). Based on the method developed by Berkes

and Philipp (1977), Berkes and Horvath (2001) proved (1.1) for GARCH process and

q(s) ≡ 1. For a general function q(s), Shao and Yu (1996) obtained some results concerning

the weak convergence Fq(·, n)/√n ⇒ q(·)B(·), where B(·) is a Gaussian process with

covariance function Γ(s, s′). Their results extent the Chibisov-O’Reilly Theorem to the

case of dependent random variables. For other work on this direction, we refer to Mehra

and Rao (1975), Csorgo and Yu (1996), Wu (2008) and the references therein.

There seems few studies on the a.s. Gaussian approximation for the weighted empirical

process under dependence assumptions. It is reasonable to conjecture that

sups∈R

sup0≤t≤n

|Fq(s, t)− q(s)K(s, t)| = O(n1/2/(log n)λ) a.s. (1.2)

for some λ > 0, if we impose some suitable conditions on the underlying process and the

weighted function q(s). (1.2) is sufficient to obtained the weak convergence results and

the functional law of iterated logarithm for the weighted empirical process. It also help us

to understand the classical empirical process∑n

k=1(I{Xk≤s} − F (s)) at the two extremes

s → ±∞. For the dependence assumption in this paper, we assume that Xn admits a

functional form of i.i.d. sequence. That is,

Xn = g(· · · , εn−1, εn) (1.3)

where {εn;n ∈ Z} are i.i.d. random variables and g is a measurable function such that

Xn is well-defined. The class of processes under the framework (1.3) is huge; see Wiener

(1958), Priestley (1988), Tong (1990) and Wu (2005) among others. We also use the

physical dependence measure introduced in Wu (2005) to character the dependence of

{Xn}. Let {ε′n;n ∈ Z} be an i.i.d. copy of {εn;n ∈ Z} and X′n = g(· · · , ε−1, ε

′0, ε1, · · · , εn).

2

Assume X0 ∈ Lp, p > 0, and let

θn,p = ‖Xn −X′

n‖p. (1.4)

Wu (2005) called θn,p the physical dependence measure. It will be used throughout this

paper. We also require the weighted funciton satisfies

|q(s)| ≤ C(1 + |s|)q for some q ≥ 0. (1.5)

To state the main results, we need some other conditions. Let FX1|ξ0(x) = E(I{X1≤x}|F0)

be the conditional distribution of X1 given F0 = (· · · , ε−1, ε0). Suppose that for some

0 < C <∞, 0 < v ≤ 1 and any x, y ∈ R,∣∣∣FX1|ξ0(x)− FX1|ξ0(y)∣∣∣ ≤ C|x− y|v a.s. (1.6)

Condition 1. Let q = 0, (1.6) hold and

P(|Xn −X

′

n| ≥ n−θ/v)

= O(n−θ) for some θ > 2. (1.7)

Condition 2. Let q > 0 and (1.6) hold. Assume that E|X1|2q+δ <∞ for some δ > 0 and

θn,2q = ‖Xn −X′n‖2q = O(ρn) for some 0 < ρ < 1.

Let Fε0(x) be the distribution function of ε0 and assume that

Fε0(x) is Lipschitz continuous on R. (1.8)

Suppose that Gn = G(· · · , εn−1, εn), where G is a measurable function.

Condition 3. Let q > 0 and (1.8) hold. Suppose that Xn = a0εn +Gn−1 and

E|ε0|2qI{|ε0|≤x} is Lipschitz continuous on R. (1.9)

(i) Let E|Gn|2q+2 <∞ and Θn,2q+2 = O(n−θ) for some θ > 0. (ii) Let E|Gn|2q+2 <∞ and

Θn,2q+2 = O((log n)−θ) for some θ > 3.

Condition 4. Let q > 0 (1.8) hold. Suppose that Xn =∑∞

j=0 ajεn−j and (1.9) holds. (i)

Let E|ε0|max(2q,2)+δ < ∞ for some δ > 0 and∑∞

j=n |aj| = O(n−θ) for some θ > 0. (ii) Let

E|ε0|max(2q,2) <∞ and∑∞

j=n |aj| = O((log n)−θ) for some θ > 3.

3

Remark 1. (1.9) is satisfied if supx |x|2qfε0(x) = O(1), which in turn will be implied by

the assumption that E|ε0|2q <∞ and fε0(x) is continuous on R, where fε0(x) is the density

of ε0.

Let Y k(x) = I{Xk≤x} − F (x) and

Γ(s, s′) = E(Y 1(s)Y 1(s

′)) +

∞∑n=2

E(Y 1(s)Y n(s′)) +

∞∑n=2

E(Y n(s)Y 1(s′)). (1.10)

Theorem 1 (i) below concerns the Gaussian approximation for the classical empirical

process (i.e. q(s) ≡ 1), while Theorem 1 (ii) considers the general q(s).

Theorem 1. (i). Suppose Condition 1 holds and q(s) ≡ 1. There exists a Kiefer process

K(s, t) with covariance function min{t, t′}Γ(s, s′) such that for some λ > 0,

sups∈R

sup0≤t≤n

|Fq(s, t)−K(s, t)| = O(n1/2(log n)−λ) a.s. (1.11)

(ii). Assume that Condition 2 or Condition 3 (i) or Condition 4 (i) holds. Then for some

λ > 0,

sups∈R

sup0≤t≤n

|Fq(s, t)− q(s)K(s, t)| = O(n1/2(log n)−λ) a.s. (1.12)

The following weak invariance principle can be obtained from the proof of Theorem 1.

Theorem 2. (i). Suppose that Condition 3 (ii) holds and E|G1|2q+2(log+ |G1|)p < ∞for some p > max{(2q + 2)/(2q), 2}. Then n−1/2Fq(s, n) ⇒ q(·)B(·). (ii). Suppose

that Condition 4 (ii) holds and E|ε0|max(2q,2)(log+ |ε0|)p < ∞ for some p > 2. Then the

conclusion in (i) still holds.

Remark 2. We give some remarks on Conditions 1-4. Due to q(s) ≡ 1, the moment

in Condition 1 is very weak. (1.7) holds for many time series such as AR(1) models,

Nonlinear AR models and GARCH models; see Section 2 for more details. The moment

condition in Condition 2 is also weak for q > 0. An example can be found in Wu (2008)

which implies that it cannot be replaced by the weaker one (say)

E|X1|2q log−1(1 + |X1|)[log log(10 + |X1|)]−λ <∞

4

for some λ > 0. The dependence assumptions in Condition 3 is much weaker than θn,2q =

O(ρn) in Condition 2 by strengthening the moment condition. For the linear process in

Condition 4, the moment condition can be further weakened. It is almost necessary when

q ≥ 1. The condition∑∞

j=n |aj| = O((log n)−θ) is also almost necessary for the short

memory process.

Remark 3. It is clearly that our proof is still available if (1.8) is replaced by the weaker

one:

|Fε0(x)− Fε0(y)| ≤ C|x− y|θ for some C > 0 and 0 < θ ≤ 1. (1.13)

In this case, we should impose some stronger conditions on θn,2q+2 or an. We will not seek

the possible bounds for them here. A careful check of the proofs of Theorems 1 and 2 indi-

cates that analogous results also holds for the two-sided process Xn = g(· · · , εj−1, εj, εj+1, · · ·)since our main tool is the m-dependence approximation. We only need to replace E(I{Xk≤x}|Fk−m,k)by E(I{Xk≤x}|Fk−m,k+m) in the proofs and FX1|ξ0(·) in (1.6) by E(I{X1≤·}|F0 ∪ F1,∞).

Remark 4. In a recent paper by Berkes et al. (2008), it was shown that (1.11) holds

under q(s) ≡ 1,

|F (x)− F (y)| ≤ C|x− y|v (1.14)

and if there exist some n-dependence random variables Xk,n such that for some θ > 4,

P(|Xk −Xk,n| ≥ n−θ/v) ≤ n−θ. (1.15)

Their result and Theorem 1(i) have different ranges of applicability. Generally, (1.7) is

weaker than (1.15). Let Xn =∑

j≥0 ajεn−j. Then (1.7) is reduced to P(|anε0| ≥ n−θ/v|) ≤n−θ for some θ > 2, while (1.15) becomes P(|

∑i≥n aiε−i| ≥ n−θ/v|) ≤ n−θ for some

θ > 4. For other examples, see Corollary 3. On the other hand, (1.6) is stronger than

(1.14). There are many time series satisfying (1.6) and (1.14) simultaneously; see Section

2. It should be noted that the approximation method in this paper is different with that of

Berkes et al. (2008). The latter paper used I{Xk,m≤x} to approximate I{Xk≤x} and we use

E(I{Xk≤x}|Fk−m,k).

The rest of the paper is structured as follows. Applications of the main results are given

in Section 2. In Section 3.1, we prove some auxiliary lemmas. The proofs of theorems are

5

given in Section 3.2. Throughout, we let C, C(·) denote positive constants and may be

different in every place. For two real sequences {an} and {bn}, write an = O(bn) if there

exists a constant C such that |an| ≤ C|bn| holds for large n, an = o(bn) if limn→∞ an/bn = 0

and an � bn if C1bn ≤ an ≤ C2bn.

2 Examples

In this section, some examples are provided to motivate the application of Theorem 1. For

the sake of briefness, we only consider the case q(s) ≡ 1.

2.1 Linear process

Let Xn =∑∞

i=0 aiεn−i with {an} satisfying |an| = O(ρn) for some 0 < ρ < 1.

Corollary 1. Let q(s) ≡ 1 and (1.13) hold. Suppose that |an| = O(ρn) for some 0 < ρ < 1

and E(log+ |ε0|)p <∞ for some p > 2. Then (1.11) holds.

2.2 Nonlinear AR model

Define the nonlinear autoregressive model by

Xn = f(Xn−1) + εn, n ∈ Z, (2.1)

where |f(x)− f(y)| ≤ ρ|x− y|, 0 < ρ < 1. Special cases of (2.1) include TAR model (Tong

1990) and the exponential autoregressive model (Haggan and Ozaki 1981).

Corollary 2. Let q(s) ≡ 1 and (1.13) hold. Suppose that E(log+ |ε0|)p < ∞ for some

p > 2. Then (1.11) holds.

2.3 GARCH model

We only consider GARCH (1,1) process and similar results can be proved for GARCH(p, q)

processes and augmented GARCH processes. Let Xk satisfy the following equations:

Xk = σkεk, (2.2)

6

and

σ2k = δ + βσ2

k−1 + αX2k−1, (2.3)

where δ > 0, β and α are nonnegative constants. The GARCH process was introduced

by Bollerslev (1986). (2.2) and (2.3) admit a unique stationary solution if and only if

E log(β + αε20) < 0; see Nelson (1990). The solution can be written as

Xk = δ∞∑i=1

εk

i−1∏j=1

(β + αε2k−j).

Corollary 3. Let q(s) ≡ 1, (1.13) hold and E(log+ |ε0|)p < ∞ for some p > 2. Suppose

that E log(β + αε20) < 0 and E| log(β + αε20)|p <∞ for some p > 6. Then (1.11) holds.

Remark 5. To get 1.11) for GARCH (1,1) process, Berkes and Horvath (2001) requires

E| log(β + αε20)|p <∞ for some p > 2 + 16/v.

3 Proof

Theorems 1 and 2 will be proven by a series of lemmas. In Lemma 1, we approximate

Fq(s, t) by the weighted empirical process of m-dependent random variables. Then by

Lemmas 5 and 6, it suffices to prove the theorems on the case of q(s) ≡ 1. Finally, we

complete the proofs by the method of Berkes and Philipp (1977).

Define Pk(X) = E[X|Fk]−E[X|Fk−1]. Let Yn(x) = I{Xk≤x}, Yn,m(x) = E[I{Xn≤x}|Fn−m,n],

Zn,m(x) = Yn(x)−Yn,m(x), Sn(x) =∑n

k=1 Yk(x)−nF (x). For any function g(x), we denote

g(x, y) = g(x)− g(y).

3.1 Auxiliary lemmas

Lemma 1. Let m = [nα] and 0 < α < 1.

(i). Let q = 0 and Condition 1 hold.

(ii). Suppose one of Conditions 2, 3 (ii), or 4 (ii) holds. Let E|X1|2q(log+ |X1|)p <∞ for

some p > 2.

Then under (i) or (ii), we have

P(

supx∈R

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Zk,m(x)∣∣∣ ≥ n1/2(log n)−λ

)= O((log n)−β) (3.1)

7

for some λ > 0 and β > 1

Proof. We first prove the lemma under Condition 1 and q = 0. Define xi = F−1(i/n) for

0 ≤ i ≤ n, and let bn = min{|x1|, xn−1}. Then we have

E supx≤x1,x≥xn−1

max1≤i≤n

∣∣∣ i∑k=1

Zk,m(x)∣∣∣ ≤ nP(|X0| ≥ |x1| ∧ xn−1) ≤ 1. (3.2)

In view of (3.2), it suffices to show that

P(

sup|x|≤bn

max1≤i≤n

∣∣∣ i∑k=1

Zk,m(x)∣∣∣ ≥ n1/2(log n)−λ

)= O((log n)−β). (3.3)

Write

Zk,m(x) = Zk,m(x)− E(Zk,m(x)|Fk−1) + E(Zk,m(x)|Fk−1) =: Z1,k + Z2,k.

Define Y ?k (x) and X?

k by replacing (· · · , εk−m−1εk−m) with (· · · , ε′k−m−1ε′

k−m) in Yk(x) and

Xk respectively. It is easy to see that Zk,m(x) = E(Yk(x)− Y ?k (x)|Fk). Noting that

|I{X≤x} − I{Y≤x}| ≤ I{|X−x|≤|X−Y |} + I{|Y−x|≤|X−Y |} (3.4)

for any random variables X, Y , we can see that

|Zk,m(x)| ≤ E(I{|Xk−x|≤|Xk−X?k |}|Fk) + E(I{|X?

k−x|≤|Xk−X?k |}|Fk).

Since F is a Lipschitz continuous function with order v, it follows that for any s > 1,

P1 := P(

supx

n∑k=1

(log n)sE(Z2

1,k(x)|Fk−1) ≥ n)

≤ C(log n)sE(supx

E[Z21,m(x)|F0])

≤ Cn−2/5(log n)−1−ε + (log n)sP(|X1 −X?

1 |v ≥ n−2/5)

≤ Cn−1/5 + (log n)s∞∑k=m

P(|Xk −X

′

k|v ≥ k−1.5)

≤ C(log n)−β. (3.5)

Setting the event A ={

supx∑n

k=1(log n)sE(Z2

1,k(x)|Fk−1) ≤ n}

and using Freedman’s

inequality (cf. Freedman (1975)), we have for λ < s/2,

n−1∑j=1

P(

max1≤i≤n

∣∣∣(1 + |xj|)qi∑

k=1

Zk,1(xj)∣∣∣ ≥ √n/(log n)λ,A

)8

≤ 2n exp(− 4−1(log n)s−2λ

)+ 2ne−4

−1n1/2/(logn)λ ≤ C(log n)−β. (3.6)

We now deal with Z2,k(x). Clearly, Zk,m(x) can be decomposed as

Zk,m(x) =∞∑

j=m−k

{E(I{Xk≤x}|F−j−1,k)− E(I{Xk≤x}|F−j,k)} =:∞∑

j=m−k

Rk,j(x).

By this equation, we obtain that∑n

k=1 Z2,k(x) =∑∞

j=m−n∑n

k=1∨(m−j) E(Rk,j(x)|Fk−1) =:∑∞j=m−n Un,j(x). Let λ1 > 1 + λ and set

U(q)n,j (x) = Un,j(x)I{|Un,j(x)|≤

√n(logn)−λ1 (1+|x|)−q}, U

(q)

n,j(x) = U(q)n,j (x)− E(U

(q)n,j (x)|F−j,∞).

Then using the inequality in (3.4) and recall θ in Condition 1, we have

P(

supx

max1≤i≤n

∣∣∣ ∞∑j=m−i

(Ui,j(x)− U (0)

i,j (x))∣∣∣ ≥ √n(log n)−λ

)≤ Cn−1(log n)λ1+λ

∞∑j=m−n

E{ n∑k=1∨(m−j)

supx

E(|Rk,j(x)||Fk−1)}2

≤ Cn−1(log n)λ1+λ∞∑

j=m−n

E{ n∑k=1∨(m−j)

bk+j

}2

+ Cn−1(log n)λ1+λ∞∑

j=m−n

{ n∑k=1∨(m−j)

E1/2(I{|Xk−X∗k,{−j}|v≥b2k+j})}2

≤ C(log n)−β, (3.7)

where bn = n−θ/2 and X∗k,{−j} is defined by replacing ε−j with ε∗−j in Xk. Set the event

B = {supx∈R

∞∑j=m−i

E(|U (0)

i,j (x)|2|F−j,∞) ≥ n/(log n)λ1+λ}.

From the arguments in (3.7), we see that P(B) = O((log n)−β). Hence Freedman’s inequal-

ity, (3.7) and the fact λ1 − λ > 1 imply that

P(

max1≤j≤n−1

max1≤i≤n

∣∣∣ i∑k=1

Z2,k(xj)∣∣∣ ≥ n1/2(log n)−λ

)≤ P

(max

1≤j≤n−1max1≤i≤n

∣∣∣ i∑k=1

Z2,k(xj)∣∣∣ ≥ n1/2(log n)−λ,Bc

)+O((log n)−β)

9

≤ n2 max1≤j≤n−1

max1≤i≤n

P(∣∣∣ i∑

k=1

Z2,k(xj)∣∣∣ ≥ n1/2(log n)−λ,Bc

)+O((log n)−β)

≤ Cn2(

exp(− 4−1(log n)λ1−λ

)+O((log n)−β)

≤ C(log n)−β. (3.8)

By the monotonicity of the indicator function, we have for xj ≤ x ≤ xj+1,

|Zk,m(xj)− Zk,m(x)| ≤ Zk,m(xj, xj+1)− EZk,m(xj, xj+1)

+2Yk,m(xj, xj+1)− 2EYk,m(xj, xj+1) + 2n−1,

and thus

P(

max1≤j≤n−2

supx∈[xj ,xj+1]

max1≤i≤n

∣∣∣ i∑k=1

{Zk,m(xj)− Zk,m(x)}∣∣∣ ≥ √n/(log n)λ

)≤ P

(max

1≤j≤n−2

∣∣∣ n∑k=1

{Zk,m(xj, xj+1)− EZk,m(xj, xj+1)}∣∣∣ ≥ √n/(log n)λ

)+P(

max1≤j≤n−2

∣∣∣ n∑k=1

{Yk,m(xj, xj+1)− EYk,m(xj, xj+1)}∣∣∣ ≥ √n/(log n)λ

)=: A1 + A2. (3.9)

Proceeding the same proofs as in (3.6) and (3.8), we have A1 = O((log n)−β). For A2, due

to the m-dependence, we have

A2 ≤ n2 exp(− n

nm(log n)2λ[F (xj+1)− F (xj)]

)+ n2 exp

(− n

(log n)λm

)= O((log n)−β).

This together with (3.6), (3.8) and (3.9) proves (3.1).

We next prove the lemma under Condition 2. For q > 0, we have

E sup|x|≥n1/q

|x|q max1≤i≤n

∣∣∣ i∑k=1

Zk,m(x)∣∣∣ ≤ 1.

Let −n1/q = y1 < · · · < ytn = n1/q, where tn � n1/q+max(q,5)/v and |yi− yi+1| = n−max(q,5)/v.

As the proof above, by Condition 2 and Holder’s inequality, we have for τ close to 1 enough,

P(

supx

max1≤i≤n

∣∣∣(1 + |x|)q∞∑

j=m−i

(Ui,j(x)− U (q)

i,j (x))∣∣∣ ≥ √n/(log n)λ

)10


j=m−n

E{ n∑k=1∨(m−j)

supx

E1−τ (|Rk,j(x)||Fk−1)Eτ (|Xk|q/τ |Fk−1)}2


j=m−n

{ n∑k=1∨(m−j)

‖ supx

E(|Rk,j(x)||Fk−1)‖1−τ2

}2


j=m−n

E{ n∑k=1∨(m−j)

bk+j

}2

+ Cn−1(log n)λ1+λ∞∑

j=m−n

{ n∑k=1∨(m−j)

P(1−τ)/2(|Xk −X∗k,{−j}|v ≥ b

2/(1−τ)k+j

)}2

≤ C(log n)−β,

where in the third inequality we used (3.4) and bk = k−2. Hence it follows from Freedman’s

inequality that

P(

maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Z2,k(yk)∣∣∣ ≥ √n/(log n)λ

)= O((log n)−β). (3.10)

For λ1 > 1 + λ, set

Z1,k(x) = (1 + |x|)qZ1,k(x)I{(1+|x|)q |Z1,k(x)|≤n1/2(logn)−λ1},

Z1,k(x) = Z1,k(x)− E(Z1,k(x)|Fk−1).

Then

P0,q := P(

supx≥nδ

max1≤i≤n

|i∑

k=1

[(1 + |x|)qZ1,k(x)− Z1,k(x)]| ≥ n1/2(log n)−λ)

≤ (log n)λ1+λE{

sup|x|≥nδ

(1 + |x|)2qZ2

1,k(x))}

≤ C(log n)λ1+λ−pE|X1|2q(log |X1|)p. (3.11)

Similarly as in (3.5), we have for sufficiently small δ > 0 and λ1 − p < −β,

P1,q := P(

supx

n∑k=1

(log n)λ1(1 + |x|)2qE(Z2

1,k(x)|Fk−1) ≥ n)

≤ Cn2δq(log n)λ1E{

sup|x|≤nδ

E(Z2

1,k(x)|F0)}

+C(log n)λ1E{

sup|x|≥nδ

(1 + |x|)2qE(Z2

1,k(x)|F0)}

11

≤ C(log n)−β + (log n)sn2δqP(|X1 −X?

1 |v ≥ n−2δq(log n)−β−λ1)

+C(log n)λ1−pE|X1|2q(log |X1|)p

≤ C(log n)−β. (3.12)

Therefore by (3.16), (3.12) and Freedman’s inequality,

P(

maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Z1,k(yk)∣∣∣ ≥ √n/(log n)λ

)= O((log n)−β). (3.13)

Recall that FX1|ξ0(x) a Lipschitz continuous function with order v. Elementary calculations

show that for j ≤ tn,

supx∈[yj ,yj+1]

|(1 + |yj|)qZ1,k(yj)− (1 + |x|)qZ1,k(x)|

≤ C(1 + |yj|)q{n−1 +

[Yk(yj, yj+1) + E(Yk(yj, yj+1)|Fk−m,k)

]}≤ C(1 + |yj|)q

{n−1 +

[Yk(yj, yj+1)− E(Yk(yj, yj+1)|Fk−m,k)

]+2E(Yk(yj, yj+1)|Fk−m,k)− 2E(Yk(yj, yj+1)) + 2Cn−q

}≤ C(1 + |yj|)q(Yk,1 + 2Yk,2) + C, (3.14)

where

Yk,1(j) = Yk(yj, yj+1)− E(Yk(yj, yj+1)|Fk−m,k)

Yk,2(j) = E(Yk(yj, yj+1)|Fk−m,k)− 2E(Yk(yj, yj+1).

For notation briefness, we use Yk,j (Yk,j,m) to denote Yk(yj, yj+1) (Yk,m(yj, yj+1)). Thus, by

Freedman’s inequality,

P2 := P(

maxj

supx∈[yj ,yj+1]

max1≤i≤n

∣∣∣ i∑k=1

{(1 + |yj|)qZ1,k(yj)− (1 + |x|)qZ1,k(x)}∣∣∣ ≥ n1/2/(log n)λ

)≤ P

(maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Yk,1(j)∣∣∣ ≥ n1/2/(log n)λ

)+P(

maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Yk,2(j)∣∣∣ ≥ n1/2/(log n)λ

). (3.15)

It follows from (3.10) and (3.13) that

P(

maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Yk,1(j)∣∣∣ ≥ n1/2/(log n)λ

)≤ C(log n)−β.

12

Note that

Yk,2(j) = Yk,2(j)− E(Yk,2(j)|Fk−1) + E(Yk,2(j)|Fk−1) =: Yk,3(j) + E(Yk,2(j)|Fk−1),

Yk,3(j), 1 ≤ k ≤ n are martingale differences and |E(Yk,2(j)|Fk−1)| ≤ Cn−2. Also we have

maxj≤tn

(1 + |yj|)2qn∑k=1

E(Y2k,2(j)|Fk−1) ≤ C a.s.

So by Freedman’s inequality,

P(

maxj:|yj |≤n1/(4q)

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Yk,3(j)∣∣∣ ≥ n1/2/(log n)λ

)≤ C(log n)−β.

As the proof above, for λ1 > 1 + λ, set

Zk(j) = (1 + |yj|)qYk,3(j)I{(1+|yj |)q |Yk,3(j)|≤n1/2(logn)−λ1},

Zk(j) = Zk(j)− E(Zk(j)|Fk−1).

It follows that

P(

supj:|yj |≥n1/(4q)

max1≤i≤n

|i∑

k=1

[(1 + |yj|)qYk,3(j)− Z1,k(yj)]| ≥ n1/2(log n)−λ)

≤ (log n)λ1+λE{

supj:|yj |≥n1/(4q)

(1 + |yj|)2qY2k,3(j)

}≤ C(log n)λ1+λ−pE|X1|2q(log |X1|)p. (3.16)

By Freedman’s inequality again,

P(

maxj:|yj |≥n1/(4q)

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Z1,k(yj)∣∣∣ ≥ n1/2/(log n)λ

)≤ C(log n)−β.

Then we have

P(

maxj

max1≤i≤n

∣∣∣(1 + |yj|)qi∑

k=1

Yk,3(j)∣∣∣ ≥ n1/2/(log n)λ

)≤ C(log n)−β.

Next we prove the lemma under Condition 3 (ii). DefineG?k = G(F ′k−m−1, εk−m, · · · , εk−1).

By (3.4), E(Z2

1,k(x)|Fk−1) ≤ Fε0(x+ |Gk −G?k| −Gk)− Fε0(x− |Gk −G?

k| −Gk). For any

13

random variables X and Y independent of ε0, set T(i)x (X, Y ) = (1 + |x|)2q|Fε0(x + X) −

Fε0(x+ Y )|i for i = 1, 2. We have by (1.9) that

supxT (i)x (X, Y )I{max(|X|, |Y |) ≤ |x|/2}

≤ C∣∣∣E(|ε0|2qI{x+Y≤ε0≤x+X} + |ε0|2qI{x+X≤ε0≤x+Y }|X, Y )

∣∣∣i ≤ C|X − Y |i,

supxT (i)x (X, Y )I{max(|X|, |Y |) ≥ |x|/2} ≤ C|X − Y |i max(|X|2q, |Y |2q, 1).

Now let X = |Gk −G?k| −Gk, Y = −|Gk −G?

k| −Gk. In virtue of Holder’s inequality and

Condition 3 (ii), it is easily seen that P1,q ≤ C(log n)λ1E[supx(1 + |x|)2qE[Z21,m(x)|F0]] ≤

C(log n)λ1Θm,2q+2. Observing that

E[

supx

(1 + |x|)2q|Rk,j(x)|2]≤ E

[supx

(1 + |x|)2q|Fε0(x−G1)− Fε0(x−G∗1)|2]

= E supxT (2)x (Gk, G

∗k) ≤ Cθ2k+j,2q+2,

it is readily seen that

∞∑j=m−n

E{ n∑k=1∨(m−j)

supx

(1 + |x|)qE[|Rk,j(x)||Fk−1]}2

≤ CnΘ2m,2q+2. (3.17)

By the bound of P1,q, (3.17) and proceeding similar arguments as those under Condition

2, we can get (3.1).

It remains to show the lemma under Condition 4 (ii). Set ξk =∑∞

j=1 akεk−j, ξk,j =∑∞j=1 akεk−j − ak+jε−j and ξ∗k = ξk,j + ak+jε

′−j. Hence by the independence between

{ε−j, ε′−j} and ξk,j,

E supxT (2)x (ξk, ξ

∗k)I{|ξk,j| ≥ x/4} ≤ a2k+jE(ε−j − ε

′

−j)2E|ξk,j|2p,


∗k)I{|ak+jε−j| ≥ x/8} ≤ C|ak+j|max(2,2q),

and


∗k)I{|ξk+j|+ |ak+jε−j|+ |ak+jε

′

−j| ≤ x/2}

≤ CqE sup|x|≥1

(|x|−q

∫ x−ξ∗k

x−ξk|y|2qdFε0(y)

)2+ Ca2k+jE|ε0 − ε

′

0|2 ≤ Ca2k+jE|ε0 − ε′

0|2.

14

Note that E[Z21,m(x)|F0]] ≤ |Fε0(x − ξ2,j) − Fε0(x − ξ?1,m − ξ1,m − ξ2,m)|, where ξ1,m =∑∞

j=m ajε1−j, ξ2,m =∑m−1

j=1 ajε1−j, ξ?1,m =

∑∞j=m ajε

′1−j. Using the independence between

ξ?1,m + ξ1,m and ξ2,j, we can show that P1,q ≤ C(log n)λ1∑∞

j=m |aj|. ♦

Remark 6. Under (1.8), we can see that for |x| ≥ 1 and any 0 ≤ t ≤ 2,

T (2)x (X, Y )I{max(|X|, |Y |) ≥ |x|/2} ≤ C|X − Y |2−t max(|X|2qI{|X|≥x/2}, |Y |2qI{|Y |≥x/2}),

T (2)x (X, Y )I{max(|X|, |Y |) ≤ |x|/2} ≤ Cx−q|X − Y |2.

These two inequalities will be used to prove Lemmas 2 and 6.

Set Sn(x) =∑n

k=1 Yk(x)−nF (x), Sn,m(x) =∑n

k=1 Yk,m(x)−nF (x), and σ2 = σ2(x, y) =

EY 2

0(x, y) + 2∑∞

k=1 E[Y 0(x, y)Y k(x, y)], where Y k(x, y) = Yk(x, y)− F (x, y).

Lemma 2. (i) Suppose the conditions in Theorem 1.1 hold. Then there exists a positive

number cθ such that

supx,y

∣∣∣E[Sn,m(x)Sn,m(y)]− nΓ(x, y)

∣∣∣ ≤ C(nm−cθ + n1−θ

), (3.18)

E[ n∑k=1

Y k(x, y)]2

= nσ2(x, y) +O(n1−cθ |F (x)− F (y)|cθ), (3.19)

E[ n∑k=1

Y k,m(x, y)]2

= nσ2(x, y) +O(n1−cθ |F (x)− F (y)|cθ). (3.20)

(ii). Suppose that Condition 2 or 3 (i) or 4 (i) holds. Then for x < y ≤ ∞ and some

τ > 0,

max(1, |y − x|−τ ) supn≥1

n−1E( n∑k=1

Yk,m(x, y)− nF (x, y))2

= o(x−2q−τ ) as x→∞.

(iii). Suppose that Condition 3 (ii) or 4 (ii) holds. For x < y ≤ ∞,

(log(max(e, |y − x|−1)))θ supn≥1

n−1E( n∑k=1

Yk,m(x, y)− nF (x, y))2

= o(x−2qφ(x)) as x→∞,

where φ(x) = (E|G1|2q+2I{|G1|≥x/8})2q/(2q+2) + x−q if Condition 3 (ii) holds, and φ(x) =

E|G1|max(2q,2)I{|G1| ≥ x/8}+ x−q + E|ε0|max(2q,2)I{|ε0| ≥ x} if Condition 4 (ii) holds.

15

where Y k(x) = Yk(x) − F (x), k ∈ Z. Clearly, Y k(x) =∑k

j=−∞PjY k(x) and PjY k(x) =

E(Fε0(x−Gk)−Fε0(x−G∗k,{j})|Fj). Hence supx ‖PjY k(x)‖2 ≤ Cθk−j,2. Since E[Y 0(x)Y k(y)] =∑0j=−∞ E

[PjY 0(x)PjY k(y)

], Θn,2 = O(n−θ), we have

∣∣∣E[Sn(x)Sn(y)]− nΓ(x, y)

∣∣∣ ≤ n

∞∑k=n

(|E(Y k(x)Y 0(y))|+ |E(Y k(y)Y 0(x))|

)+

n∑k=1

k(|E(Y k(x)Y 0(y))|+ |E(Y k(y)Y 0(x))|

)≤ CnΘn,2 +

0∑j=−∞

θ−j

n∑k=1

kθk−j

≤ Cn1−θ, (3.24)

which together with (3.21) proves (3.18). Note that ‖PjY k(x, y)‖2 ≤ min(θn−j,2, |F (x) −F (y)|1/2). It is easy to see that

∑∞j=n min(θj,2, |F (x) − F (y)|1/2) ≤ Cn−cθ |F (x) − F (y)|cθ

for some cθ > 0. Since

|E(Y 0(x, y)Y n(x, y))| ≤ Cn∑

j=−∞

min(θ−j,2, |F (x)− F (y)|1/2) min(θn−j,2, |F (x)− F (y)|1/2),

we can see that (3.19) holds by similar arguments as in (3.24). (3.20) now follows from

(3.19) and (3.21).

Next we prove (ii) under Condition 3 (i) and the proofs of the others are similar. We

shall show that there is a positive number 0 < τ < 1 such that∑∞

k=n θτk,2q+2 = O(n−θ1) for

some θ1 > 0. Set N1 = {k ≥ n : θk,2q+2 ≤ k−2}, N2 = {k ≥ n : θk,2q+2 > k−2}. Clearly, for

1/2 < τ < 1, we have∑

k∈N1θτk,2q+2 ≤ Cn−(2τ−1). Moreover, since

∑∞i=n θi,2q+2 = O(n−θ),

it follows that for τ close to 1 enough,

∞∑i=n

2i+1−1∑k=2i,k∈N2

θτk,2q+2 ≤∞∑i=n

2i+1−1∑k=2i,k∈N2

k−2(τ−1)θk,2q+2 ≤ C∞∑i=n

22(1−τ−θ)i ≤ C22(1−τ−θ)n.

Hence we get∑∞

k=n θτk,2q+2 = O(n−θ1). Using the equation PjY k,m(x) = E[F (x − Gk) −

F (x−G∗k,{j})|Fk−m,j] and setting Gk(x, y) = Fε0(y−Gk)−Fε0(x−Gk) and G∗k,{j}(x, y) =

Fε0(y−G∗k,{j})−Fε0(x−G∗k,{j}), then PjY k,m(x, y) = E[Gk(x, y)−G∗k,{j}(x, y)|Fk−m,j]. By

Remark 5, we have

‖|x|q+τ/2|Fε0(x−Gk)− Fε0(x−G∗k)|‖22 ≤ CE|Gk −G∗k|2−τ |Gk|2q+τ . (3.25)

17

On the other hand, by Holder’s inequality,

‖|x|q+τ/2|PjY k,m(x, y)|‖22 ≤ ‖|x|q+τ/2Yk(x, y)‖22 ≤ E|Xk|q+τ/2I{x≤Xk≤y} ≤ C|x− y|1/2.(3.26)

Combining (3.25) and (3.26) we obtain that for some τ > 0, ‖|x|q+τ/2PjY k,m(x, y)‖22 ≤C min(E|Gk − G∗k|2−τ |Gk|2q+τ , |y − x|1/2) ≤ C min(θ2−τk+j,2q+2, |y − x|1/2). Thus (ii) can be

proved by similar inequalities as (3.23) and (3.24). Analogously, we can prove (iii). ♦

Lemma 3. For every z > 3m and x, y ∈ R, we have

P(

max1≤j≤n

∣∣∣ j∑k=1

Y k,m(x, y)∣∣∣ ≥ z

)≤ C exp

(− C z2

nm−1E(∑m

k=1 Y k,m(x, y))2

)+ C exp

(− C z

m

).

Proof. Note that Y k,m(x, y), 1 ≤ k ≤ n, are m-dependent random variables. Write

n∑k=1

Y k,m(x, y) =

[n/m]∑i=1

im∑k=(i−1)m+1

Y k,m(x, y) +n∑

k=[n/m]m+1

Y k,m(x, y).

The lemma now follows from Bernstein’s inequality. ♦Recall φ(x) defined in Lemma 2. Let φ

′(x) = 1/φ(x) if Condition 3 (ii) or Condition 4

(ii) is satisfied; φ′(x) = xτ if Condition 2 or 3 (i) or 4 (i) holds (recall τ in Lemma 2 (ii)).

Let Q > 2q and m = [nγ] with 0 < γ < 1/2− q/Q. Let Y k,m(x) = Yk,m(x)− F (x).

Lemma 4. Let q > 0. For 1 ≤ j ≤ n1/Q and λ ≥ 0, we have

P(

supj≤x≤j+1

max1≤i≤n

|x|q∣∣∣ i∑k=1

Y k,m(x)∣∣∣ ≥ √n/(log n)λ

)≤ C exp

(− Cφ

′(j)

(log n)2λ

)+ C exp

(− Cn(1/2−q/Q−γ)/2

).

Proof. For j ≤ x ≤ j+ 1, write x = j+∑∞

k=1 σk2−k = j+

∑lk=1 σk2

−k + σ2−l, where l will

be specialized later, σk = 0, 1 and 0 ≤ σ ≤ 1. It can be shown that for j ≤ x ≤ j + 1,

max1≤i≤n

∣∣∣|x|q i∑t=1

Y t,m(j, x)∣∣∣ ≤ 2

l∑k=0

max1≤i≤n

∣∣∣jq i∑t=1

Y t,m(ak,j,1, ak,j,2)∣∣∣+ n3/22−l,

where ak,j,1 = j + ak2−k, ak,j,2 = j + (ak + 1)2−k, 0 ≤ ak ≤ 2k. Hence Lemmas 2 and 3

yield that for 1 < α < (θ − 1)/2 (recall θ in Lemma 2(iii)),

P(

max1≤i≤n

∣∣∣jq i∑t=1

Y t,m(ak,j,1, ak,j,2)∣∣∣ ≥ k−αn1/2(log n)−λ

)≤ exp

(− C k−2αnφ

′(j)

nk−θ(log n)2λ

)+ exp

(− Ck

−αn1/2−q/Q

m(log n)λ

).

18

Taking l = 2[log n/ log 2], we have

P(

supj≤x≤j+1

max1≤t≤n

∣∣∣(1 + |x|)qi∑t=1

Y t,m(j, x)∣∣∣ ≥ n1/2(log n)−λ

)≤

l∑k=0

2k exp(− C k−2αnφ

′(j)

nk−θ(log n)2λ

)+ Cn2 exp

(− C n1/2−q/Q

m(log n)λ+α

)≤ C exp

(− Cφ

′(j)

(log n)2λ

)+ C exp

(− Cn(1/2−q/Q−γ)/2

).

The proof is complete by noting that

P(

max1≤i≤n

|j|q∣∣∣ i∑k=1

Y k,m(j)∣∣∣ ≥ √n/(log n)λ

)≤ C exp

(− Cφ

′(j)

(log n)2λ

)+ C exp

(− Cn(1/2−q/Q−γ)/2

).

.

Remark 7. Note that E(∑m

t=1 Y t,m(x, y))2≤ Cma(x, y), where a(x, y) = |F (y)−F (x)|cθ

for some cθ > 0 if conditions in Theorem 1 hold; a(x, y) = (log |F (x) − F (y)|−1)−θ if

conditions in Theorem 2 hold. For every y ≤ x ≤ y + δ,

F (x) = F (y) + F (y, y + δ)l∑

k=1

σk2−k + F (y, y + δ)σ2−l.

Hence

max1≤i≤n

∣∣∣ i∑t=1

Y t,m(x, y)∣∣∣ ≤ 2

l∑k=0

max1≤i≤n

∣∣∣ i∑t=1

Y t,m(bk,1, bk,2)∣∣∣+ n2−l,

where bk,1 = F−1(ak,1), bk,2 = F−1(ak,2), ak,1 = F (y) + F (y, y + δ)ak2−k, ak,2 = F (y) +

F (y, y + δ)(ak + 1)2−k, 0 ≤ ak ≤ 2k. Under the conditions of Theorem 1 or Theorem 2,

we have E(∑m

t=1 Y t,m(bk,1, bk,2))2≤ Cma(y, y + δ)τk−(1−τ)θ for any 0 < τ < 1. Then by

letting q = 0 in the proof of Lemma 4, we have for some 0 < τ ≤ 1/3 and any α ∈ R,

P(

supy≤x≤y+δ

max1≤i≤n

∣∣∣ i∑k=1

Y k,m(x, y)∣∣∣ ≥ √n/(log n)α

)≤ C exp

(− Ca−τ (y, y + δ)

(log n)2α

)+ C exp

(− Cn(1/2−γ)/2

). (3.27)

19

Lemma 5. Let m = [nγ]. Under Condition 2 or 3 (i) or 4 (i), we have for any δ > 0,

there exist some λ > 0, β > 1 and γ > 0 small enough such that

P(

sup|x|≥(logn)δ

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k,m(x)∣∣∣ ≥ n1/2(log n)−λ

)= O((log n)−β). (3.28)

Proof. Since there exists some Q > 2q such that E|X0|Q <∞, we have

E[

sup|x|≥n1/Q

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k,m(x)∣∣∣] ≤ Cn1/2(log n)−λ−β.

By Lemma 4, it is readily seen that

P(

sup(logn)δ≤|x|≤n1/Q

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k,m(x)∣∣∣ ≥ n1/2(log n)−λ

)= O((log n)−β). (3.29)

Hence the lemma is proved. ♦

Lemma 6. If conditions in Theorem 1.2 hold, then for any ε > 0,

limz→∞

supn≥1

P(

sup|x|≥z

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k,m(x)∣∣∣ ≥ εn1/2

)= 0. (3.30)

Proof. We first show that

P(

sup|x|≥n1/Q

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k(x)∣∣∣ ≥ εn1/2

)= o(1). (3.31)

Let m = −1 in the proof of Lemma 1 (under Condition 2) and recall conditions in Theorem

1.2. We have P0,q = o(1) and

P(

sup|x|≥n1/Q

n∑k=1

(log n)s(1 + |x|)2qE(Z2

1,k(x)|Fk−1) ≥ n)

≤ C(log n)s−pE|X1|2q(log |X1|)p = o(1).

Note that E[|Rk,j(x)||Fk−1] = E[Fε0(x − Gk) − Fε0(x − G∗k,{−j−1})|F−j−1,k−1]. It follows

Remark 6 that∞∑

j=m−n

E{ n∑k=1∨(m−j)

sup|x|≥n1/Q

(1 + |x|)qE[|Rk,j(x)||Fk−1]}2

≤ Cn(log n)−p.

Thus (3.31) follows from the proof of Lemma 1. Now, by (3.31), we can get

P(

sup|x|≥n1/Q

max1≤i≤n

∣∣∣(1 + |x|)qi∑

k=1

Y k,m(x)∣∣∣ ≥ εn1/2

)= o(1).

The proof now is completed by Lemma 4. ♦

20

3.2 Proof of theorems

Proof of Theorem 1. In order to prove Theorem 1, by Lemmas 1 and 5, it is enough to show

(1.11) under one of those conditions in Theorem 1.1. As in Berkes and Philipp (1977), we

use the classical Bernstein blocking method. Define blocks of integers H1, I1, H2, I2, · · ·by requiring that Hk contains hk and Ik contains ik consecutive integers and that there are

no gaps between consecutive blocks, where

hi = |Hi| = [aia−1 exp(ia)], ik = |Ii| = [aia−1 exp(ia/4)]

for some a ∈ (1/2, 1). Let Nm =∑m

i=1 card(Hi

⋃Ii) ∼ exp(ma). Clearly, for each n there

exists a unique mn such that Nmn ≤ n < Nmn+1 and we have mn ∼ (log n)1/a. One can

easily show that 0 ≤ mn ≤ (log n)1/a, n ≥ 4. Define

ξj(x) =∑k∈Hj

Yk,m(x) and ηj(x) =∑k∈Ij

Yk,m(x)

for j ≥ 1. In the following lemmas, we always assume that the conditions in Theorem 1.1

hold.

Note that supx∈R

∣∣∣∑nk=1 ηk(x)

∣∣∣ ≤ naena/4 = o(en

a/2/naλ) a.s. So it suffices to show that

max1≤t≤mn

supx∈R

∣∣∣ t∑j=1

ξj(x)−K(x, t)∣∣∣ = o(n1/2/(log n)λ) a.s.

Let qn = [log n/ log 4] and d = dn = 2qn ≤ n1/2. We divide [0, 1] into dn small intervals.

Let ti = tn,i = (i−1)/dn, 1 ≤ i ≤ dn. Let si = inf{x : F (x) ≥ ti} and define dn-dimensional

vector

Xn = (ξn(si), 1 ≤ i ≤ dn).

We will approximate Xn by an normal vector with the same covariance matrix. In order

to do this, we split the interval Hn into Hn,1, In,1, · · ·Hn,pn , In,pn small intervals, where

|Hn,i| = ena/4 and |In,i| = en

a/16 for 1 ≤ i ≤ pn. The last intervals Hn,pn and In,pn may be

incomplete. Set

ξn,i(x) =∑j∈Hn,i

ξj(x), ηn,i(x) =∑j∈In,i

ξj(x)

ξn,i = (ξn,i(sj), 1 ≤ j ≤ dn), ηn,i = (ηn,i(sj), 1 ≤ j ≤ dn).

21

By Bernstein’s inequality, it is easy to see that

P(∣∣∣ pn∑

i=1

ηn,i

∣∣∣ ≥ e7na/16)≤ ne−n

a/32.

The following lemma come from Einmahl and Mason (1997), Fact 2.2. The constants

c1, c2 are specialized in Zaitsev (1987).

Lemma 7. Let Xj = (X(1)j , · · · , X(d)

j ), 1 ≤ j ≤ n, be independent mean zero random

vectors satisfying max1≤j≤n |Xj| ≤ M . If the underlying probability space is rich enough;

one can construct independent normally distributed mean zero random vectors V1, · · · , Vnwith Cov(Xi) = Cov(Vi), 1 ≤ i ≤ n, such that

P(∣∣∣ n∑

i=1

(Xi − Vi)∣∣∣ ≥ x

)≤ c2 exp

(− c1x

M

),

where c1 = c3d−2, c2 = c4d

2, c3, c4 are universal constants.

Since |ξn,i| ≤ n1/2ena/4, we can construct a sequence of independent normal vectors

Gn,i) with Cov(Gn,i) = Cov(ξn,i) =: Σn,i, 1 ≤ i ≤ pn, such that

P(∣∣∣ pn∑

i=1

(ξn,i −Gn,i)∣∣∣ ≥ en

a/3)≤ c4n exp

(− c3n−1en

a/12).

By Lemma 2, we have |Σn,i − ena/4Γd| ≤ ne(1−τ)n

a/4. Hence it holds that∣∣∣E exp(i⟨u, p−1/2n e−n

a/8

pn∑i=1

Gn,i

⟩)− exp

(− 1

2i〈u,Γdu〉

)∣∣∣ ≤ C|u|2ne−τna/4. (3.32)

By Lemma 2.2 in Berkes and Philipp (1979), it follows that

ΨPL

(p−1/2n e−n

a/8

pn∑i=1

Gn,i, N(0,Γd))

≤ 16d

T+ T d

∫|u|≤T

∣∣∣E exp(i⟨u, p−1/2n e−n

a/8

pn∑i=1

Gn,i

⟩)− exp

(− 1

2i〈u,Γdu〉

)∣∣∣du+P(|N(0,Γd)| ≥ T/2

)≤ 16n1/2e−n

1/5

+ Cen4/5−τna/4 + de−CT

2/d

≤ Ce−n1/5

.

22

where T = en1/5

. Hence we can define independent N(0,Γd) random vectors ηn, n ≥ 1,

such that

P(∣∣∣p−1/2n e−n

a/8

pn∑i=1

Gn,i − ηn∣∣∣ ≥ Ce−n

1/5)≤ Ce−n

1/5

.

So we have ∣∣∣ pn∑i=1

Gn,i − p1/2n |Hn,1|1/2ηn)∣∣∣ ≤ |Hn|1/2e−n

1/5

a.s.

Note that |p1/2n |Hn,1|1/2 − |Hn|1/2| ≤ |Hn|−1/2ena/4. It is easy to see that

∣∣∣|Hn|1/2ηn −

p1/2n |Hn,1|1/2ηn

∣∣∣ ≤ |Hn|1/2e−n1/5

a.s. and∣∣∣|Hn|1/2ηn− (|Hn|+ |In|)1/2ηn

∣∣∣ ≤ |Hn|1/2e−n1/5

a.s.

In view of the above arguments, we can get

|Xn − (|Hn|+ |In|)1/2ηn| ≤ |Hn|1/2e−n1/5

a.s.

Write ηn = (ηn(sj), 1 ≤ j ≤ dn). We can assume that there exists a Kiefer process K(s, t)

with covariance function min{t, t′}Γ(s, s′) such that

[K(sj, tk+1)−K(sj, tk)]/(tk+1 − tk)1/2 = ηn(sj), 1 ≤ j ≤ dn.

To complete the rest proof, we only need a geometrical picture argument as in Berkes and

Philipp (1977), page 136, and the follow lemma which can be obtained by (3.27).

Lemma 8. Under the conditions of Theorem 1.1, we have for some ε > 0,

P(

max1≤i≤dn

supsi≤x≤si+1

∣∣∣ ∑j∈Hn

Y j,m(si, x)∣∣∣ ≥ en

a/2/n(1−a)/2+ε)

= O(n−2).

and

P(

supx∈R

maxNn≤k≤Nn+1

∣∣∣ k∑j=Nn

Y j,m(x)∣∣∣ ≥ en

a/2/nε)

= O(n−2).

Proof of Theorem 2. By Lemma 6, we only need to show that n−1/2Fq(·, n)⇒ B(·) under

q(s) ≡ 1. The finite-dimensional convergence easily follows from Theorem 1(i) in Hannan

(1973) and the assumptions Θ0,2q+2 <∞ or∑

j≥0 |aj| <∞. So we only need to check the

tightness of n−1/2Fq(·, n), which follows from (3.27) by letting α = 0. ♦

23

3.3 Proofs of corollaries

Proof of Corollary 1. It is easy to see that Xn =∑

i≥0 aiεn−i is well defined if E(log+ |ε0|) <∞. Also (1.7) holds and the proof is complete. ♦

Proof of Corollary 2. It is easy to see that Xn admits a unique stationary solution if

E(log+ |ε0|) <∞. We have |Xn −X′n| ≤ Cρn|ε0 − ε

′0| and hence (1.7) holds. ♦

Proof of Corollary 2. Set ηj = (β + αε2−j). We can see that

P(|Xn −X

′

n| ≥ n−θ/v)≤ 2P

( n−1∏j=1

ηj−n|δ∞∑i=1

ε1

i−1∏j=1

ηj−1| ≥ n−θ/v/2)

≤ 2P( n−1∏j=1

ηj−n ≥ ρn)

+ 2P(|δ∞∑i=1

ε1

i−1∏j=1

ηj−1| ≥ ρ−n).

Recall that log η1 < 0. Then

P( n−1∏j=1

ηj−n ≥ ρn)

≤ P( n−1∑j=1

(log(ηj−n)− E log(ηj−n)) ≥ −nE log(ηj−n) + n log ρ)

= O(n−θ)

for some θ > 2. Following from a similar proof of Lemma 4 in Hormann (2008), we can

get

P(|δ∞∑i=1

ε1

i−1∏j=1

ηj−1| ≥ ρ−n)

= O(n−θ).

The proof is complete. ♦REFERENCES

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