Abstract

We introduce the fractional mixed fractional Brownian sheet and investigate the small ball behavior of its sup-norm statistic by establishing a general result on the small ball probability of the sum of two not necessarily independent joint Gaussian random vectors. Then, we state general conditions and characterize the sufficiency part of the lower classes of some statistics of the above process by an integral test. Finally, when we consider the sup-norm statistic, the necessity part is given by a second integral test.

1. Introduction

Let {𝐵𝐻𝑖(𝑠),𝑠0} be a fractional Brownian motion (FBM) with index 0<𝐻𝑖<1,𝑖, and {𝐵𝐻𝑗,𝐻𝑘(𝑠,𝑠),𝑠0,𝑠0} a fractional Brownian sheet (FBS) with index 0<𝐻𝑗,𝐻𝑘<1,𝑗,𝑘. We refer to [1] for further information about the FBM and the FBS. Denote by 𝜆1 and 𝜆2 two real numbers such that 𝜆1𝜆20.

Define a fractional mixed fractional Gaussian process by a suitable combination of some appropriate fractional Gaussian processes. In the sequel, we consider the following three examples.

Example 1.1. The fractional mixed fractional Brownian motion (FMFBM) is defined by 𝑋𝑤1,𝑤2,𝑠=𝜆1𝑠𝐻2𝐵𝐻1𝑤1+𝜆2𝑠𝐻1𝐵𝐻2𝑤2,(1.1) where 𝐵𝐻1 and 𝐵𝐻2 are independent FBM with 𝐻1𝐻2.

Example 1.2. The fractional mixed fractional Brownian motion and fractional Brownian sheet (FMFBMFBS) are defined by 𝑋𝑤1,𝑤2,𝑤3,𝑠=𝜆1𝑠𝐻2+𝐻3𝐵𝐻1𝑤1+𝜆2𝑠𝐻1𝐵𝐻2,𝐻3𝑤2,𝑤3,(1.2) where the FBM 𝐵𝐻1 and the FBS 𝐵𝐻2,𝐻3 are independent.

Example 1.3. The fractional mixed fractional Brownian sheet (FMFBS) is defined by 𝑋𝑤1,𝑤2,𝑤3,𝑤4,𝑠=𝜆1𝑠𝐻3+𝐻4𝐵𝐻1,𝐻2𝑤1,𝑤2+𝜆2𝑠𝐻1+𝐻2𝐵𝐻3,𝐻4𝑤3,𝑤4,(1.3) where 𝐵𝐻1,𝐻2 and 𝐵𝐻3,𝐻4 are independent FBS with (𝐻1,𝐻2)(𝐻3,𝐻4).

The motivation supporting this paper is threefold.

(i)The first goal of the FMFBS deals with the potential applications. Since the FMFBM, the FMFBMFBS, and the FMFBS can be analyzed based on the large bodies of knowledge on FBM and FBS, it can be used in the same fields, that is, natural time series in economics, fluctuations in solid, hydrology, and, more recently, by new problems in mathematical finance, telecommunication networks, and the environment (see [24]).(ii)A second application deals with the small ball probability problem of the sum of two not necessarily joint centered Gaussian random vectors 𝑋 and 𝑌 in a separable Banach space 𝐸 with norm (see [5]). The small ball behavior of the FMFBS under the uniform norm can be investigated as a special case of the small ball probability problem of the sum of two centered Gaussian random vectors, having a log-type small ball factor (see [6]).(iii)Last but not least, this article extends El-Nouty's results [69] and consequently answers some new questions. Recall first two definitions of the Lévy classes, stated in [10]. Let {𝑍(𝑡),𝑡0} be a stochastic process defined on the basic probability space (Ω,𝒜).

Deffinition 1.4. The function 𝑓(𝑡),𝑡0, belongs to the lower-lower class of the process 𝑍,(𝑓LLC(𝑍)), if, for almost all 𝜔Ω, there exists 𝑡0=𝑡0(𝜔) such that 𝑍(𝑡)𝑓(𝑡) for every 𝑡>𝑡0.

Deffinition 1.5. The function 𝑓(𝑡),𝑡0, belongs to the lower-upper class of the process 𝑍,(𝑓LUC(𝑍)), if, for almost all 𝜔Ω, there exists a sequence 0<𝑡1=𝑡1(𝜔)<𝑡2=𝑡2(𝜔)< with 𝑡𝑛+, as 𝑛+, such that 𝑍(𝑡𝑛)𝑓(𝑡𝑛),𝑛.

In the spirit of [69, 11], the main aim of this paper is to characterize the lower classes of the uniform norm of the FMFBS for any 0<𝐻1,𝐻2,𝐻3,𝐻4<1. More precisely, we want to compare the influence of two FBSs and to measure the weight of a log-type small ball factor versus another one.

2. Main Results

Our first result is given in the following theorem.

Theorem 2.1. Let 𝑋 and 𝑌 be any two joint Gaussian random vectors in a separable Banach space with norm . Assume that there exist 𝐶𝑋1 and 𝐶𝑌1 such that one has, for any 𝜖>0 small enough, 𝐶𝑋𝜖1/𝛼log(1/𝜖)𝛽1log𝑋𝜖𝐶𝑋,𝐶𝑌𝜖1/𝛼log(1/𝜖)𝛽1log𝑌𝜖𝐶𝑌,(2.1) with 0<𝛼,𝛼<+,0𝛽,𝛽<+ and (𝛼,𝛽)(𝛼,𝛽).
If (𝛼<𝛼) or (𝛼=𝛼 and 𝛽>𝛽), then there exists 𝐾𝑋𝐶𝑋 depending on 𝐶𝑋 only such that one has, for any 𝜖>0 small enough, 𝐾𝑋𝜖1/𝛼log(1/𝜖)𝛽1log𝑋+𝑌𝜖𝐾𝑋.(2.2)

Since the study of the lower classes of the FMFBM (resp., FMFBMFBS) under the uniform norm was investigated in [8] (resp., [9]), we focus our attention to the FMFBS. Set 𝑌(𝑡)=sup0𝑠𝑡sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝑋𝑤1,𝑤2,𝑤3,𝑤4||,𝑠,𝑡0.(2.3) Note first that, by the scaling property, we have, for any 𝜖>0, 𝑌(𝑡)𝜖𝑡𝐻1+𝐻2+𝐻3+𝐻4=sup0𝑠1sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝑋𝑤1,𝑤2,𝑤3,𝑤4||,𝑠𝜖=𝑌(1)𝜖=𝜙(𝜖),(2.4) where 𝜙 is named the small ball function and 𝛾=𝐻1+𝐻2+𝐻3+𝐻4 the scaling factor.

Recall that the small ball behavior of the FBS under the uniform norm was studied in [12, 13].

Set 𝛼=min(𝐻1,𝐻2,𝐻3,𝐻4), which is in ]0,1[. We introduce the number 𝛽 taking its values in {0,1+1/𝛼}. As a direct consequence of Theorem 2.1, we have the following corollary.

Corollary 2.2. There is a constant 𝐾0,0<𝐾01, depending on 𝐻1,𝐻2,𝐻3,𝐻4,𝜆1 and 𝜆2 only, such that one has, for any 𝜖>0 small enough, explog(1/𝜖)𝛽𝐾0𝜖1/𝛼𝐾𝜙(𝜖)exp0log(1/𝜖)𝛽𝜖1/𝛼.(2.5)

Recall that we suppose (𝐻1,𝐻2)(𝐻3,𝐻4). In the sequel, there is no loss of generality to assume also that 𝐻1𝐻2,𝐻3𝐻4.(2.6) Thus when (𝐻1=𝐻2,𝐻3<𝐻4 and 𝐻1𝐻3), (𝐻1=𝐻2,𝐻3=𝐻4 and 𝐻1<𝐻3), or (𝐻1<𝐻2,𝐻3=𝐻4 and 𝐻3𝐻1), we emphasize that 𝛽=1+1/𝛼, that is, we have a log-type small ball factor.

Note first that the minimum 𝛼 plays a key role. This is not really surprising. Indeed, this phenomenon was already observed in [8, 9].

It appears that the sufficiency part of the lower classes of 𝑌 can be stated in a general framework. Roughly speaking, we follow the same lines as those of [6, 7].

Let {𝑌0(𝑡),𝑡0} be a real-valued statistic of the two independent FBSs, 𝐵𝐻1,𝐻2 and 𝐵𝐻3,𝐻4, such that 𝑌0(𝑡) is a nondecreasing function of 𝑡0.

The following notation is needed. If 𝕂 is a Hausdorff compact space, we denote, by 𝐶(𝕂), the space of all continuous functions from 𝕂 to equipped with the classical sup-norm. Let 𝕏=𝐶([0,1]2)×𝐶([0,1]2) be the product space equipped with the product topology. Denote, by L (𝐵𝐻1,𝐻2,𝐵𝐻3,𝐻4), the Gaussian measure associated to 𝐵𝐻1,𝐻2 and 𝐵𝐻3,𝐻4 and defined on 𝔹, the Borel 𝜎-field of 𝕏.

We assume that 𝑌0 satisfies the following three conditions:

(i)The scaling condition. There exists 𝛾0>0 such that 𝑌0(𝑡)𝜖𝑡𝛾0𝑌=0(1)𝜖=𝜙(𝜖).(2.7)(ii)The convexity condition. There exists a convex and 𝔹-measurable function 𝑔: (𝕏,L(𝐵𝐻1,𝐻2,𝐵𝐻3,𝐻4)) such that, for any 𝑡0,𝑌0(𝑡)=𝑔(𝐵𝐻1,𝐻2(𝑠1𝑡,𝑠2𝑡), 𝐵𝐻3,𝐻4(𝑠3𝑡,𝑠4𝑡);0𝑠1,𝑠2,𝑠3,𝑠41), and 𝑌0(𝑡)<+, with probability 1.(iii)The log-type small ball condition. There exist 𝛼0]0,𝛾0],𝛽0 and a constant 𝐾,0<𝐾1, depending on 𝐻1,𝐻2,𝐻3,𝐻4,𝛾0,𝛼0 and 𝛽0 only such that we have, for any 𝜖>0 small enough, explog(1/𝜖)𝛽0𝐾𝜖1/𝛼0𝐾𝜙(𝜖)explog(1/𝜖)𝛽0𝜖1/𝛼0.(2.8)

Note that these conditions generalize those of [6, 7]. The small ball function still plays a key role. The convexity of the function 𝜓 defined by 𝜓(𝜖)=log𝜙(𝜖),0<𝜖<1, is ensured by (ii) (see [14, 15]).

Our second result is given in the following theorem.

Theorem 2.3. Let 𝑓(𝑡) be a positive nondecreasing function of 𝑡0. Assume that there exists 𝑚>0 such that (𝑓(𝑡)/𝑡𝛾0𝛼0)(log𝑡𝛾0/𝑓(𝑡))𝛽0𝛼0𝑚.
If 𝑓(𝑡)𝑡𝛾0isboundedand0+𝑓(𝑡)1/𝛼0𝑡(𝛾0/𝛼0)1𝑡log𝛾0𝑓(𝑡)𝛽0𝜙𝑓(𝑡)𝑡𝛾0𝑑𝑡<+,(2.9) then one has 𝑌𝑓LLC0.(2.10)

The sup-norm statistic 𝑌 clearly satisfies the three above conditions with 𝛾0=𝛾=𝐻1+𝐻2+𝐻3+𝐻4,𝛼0=𝛼=min(𝐻1,𝐻2,𝐻3,𝐻4)=min(𝐻1,𝐻3),𝛽0=𝛽{0,1+1/𝛼}, and 𝐾=𝐾0. Now, we characterize the necessity part of the lower classes of the FMFBMFBS. Our main result is stated in the following theorem.

Theorem 2.4. Let 𝑓(𝑡) be a positive nondecreasing function of 𝑡0 such that 𝑓(𝑡)/𝑡𝛾 is a nonincreasing function of 𝑡>0.
If 𝑓LLC(𝑌),(2.11) then we have lim𝑡+𝑓(𝑡)𝑡𝛾=0,0+𝑓(𝑡)1/𝛾𝜙𝑓(𝑡)𝑡𝛾𝑑𝑡<+.(2.12)

First, we can notice that Theorem 2.3 involves 𝛾0,𝛼0 and 𝛽0. If 𝛽0=0, Theorem 2.3 looks like [7, Theorem 1] or else like [6, Theorem 1.1]. Theorem 2.4 has the same form as the necessity part of [11, Theorem 1.2] or as the theorems obtained in [69]. In his previous works on the study of the lower classes, the author showed that the methodology in [11] led to two integral tests, these tests are actually identical when 𝛾=𝛼 and 𝛽=0. Here, 𝛼=min(𝐻1,𝐻3)<𝛾=𝐻1+𝐻2+𝐻3+𝐻4. This is why the integral tests of Theorems 2.3 and 2.4 have different forms. Moreover, since 𝛼<𝛾, we must assume, as in [69], that 𝑓(𝑡)/𝑡𝛾 is not only bounded, but also a nonincreasing function of 𝑡>0. This last assumption will play a key role in some proofs. Finally, although they are two different integral tests, Theorems 2.3 and 2.4 are sharp. Indeed, set, if 𝛽=0, 𝑓(𝑡)=𝜆𝑡𝛾(loglog𝑡)𝛼,𝑡3,𝜆>0,(2.13) or else (i.e., 𝛽=1+(1/𝛼)) 𝑓(𝑡)=𝑡𝛾(𝜆logloglog𝑡)1+𝛼(loglog𝑡)𝛼,𝑡16,𝜆>0.(2.14) If 𝜆 is small enough, then Theorem 2.3 yields 𝑓LLC(𝑌), or else if 𝜆 is large enough, then 𝑓LUC(𝑌) by applying Theorem 2.4.

In Section 3, we prove Theorem 2.1. The proof of Theorem 2.4 is postponed to Sections 4 and 5. In the latter, we establish some key small ball estimates. Note also that these estimates can be of independent interest. The proofs, which are modifications of those of [6, 7], will be consequently omitted, in particular, the proof of Theorem 2.3.

3. Proof of Theorem 2.1

Recall, first, a Gaussian correlation inequality stated in [5].

Theorem A. Let 𝜇 be a centered Gaussian measure on a separable Banach space 𝐸. Then for any 0<𝜆<1, and any symmetric convex sets 𝐴 and 𝐵 in 𝐸, 𝜆𝜇(𝐴𝐵)𝜇2𝐴+1𝜆2𝐵𝜇(𝜆𝐴)𝜇1𝜆21/2𝐵.(3.1) In particular, 𝜇(𝐴𝐵)𝜇(𝜆𝐴)𝜇1𝜆21/2𝐵.(3.2)

Rougly speaking, the proof follows the same lines as those in [15] and will be split into two parts: the lower bound and the upper one.

Part I. The Lower Bound
Theorem A implies, for any 0<𝛿<1 and 0<𝜆<1, 𝑋+𝑌𝜖𝑋(1𝛿)𝜖,𝑌𝛿𝜖𝑋𝜆(1𝛿)𝜖𝑌1𝜆21/2.𝛿𝜖(3.3)

Then we get, by using (2.1), log𝑋+𝑌𝜖𝐶𝑋log1/𝜆(1𝛿)𝜖𝛽𝜆(1𝛿)𝜖1/𝛼𝐶𝑌log1/1𝜆21/2𝛿𝜖𝛽1𝜆21/2𝛿𝜖1/𝛼.(3.4) Hence, since (𝛼<𝛼) or (𝛼=𝛼 and 𝛽>𝛽), there exists 𝐶𝑋𝐶𝑋 depending on 𝐶𝑋 only such that we have, for any 𝜖>0 small enough, 𝜖1/𝛼log(1/𝜖)𝛽𝐶log𝑋+𝑌𝜖𝑋𝜆(1𝛿)1/𝛼,(3.5) and the lower bound follows by taking 𝛿0 and 𝜆1.

Part II. The Upper Bound
A new use of Theorem A implies, for any 0<𝛿<1 and 0<𝜆<1, 𝜖𝑋𝜖(1𝛿)𝜆𝑋+𝑌𝜆,𝑌𝛿𝜖(1𝛿)𝜆𝑋+𝑌𝜖𝑌1𝜆21/2𝛿𝜖.(1𝛿)𝜆(3.6)

Then combining (2.1) with the fact that (𝛼<𝛼) or (𝛼=𝛼 and 𝛽>𝛽), there exists 𝐶𝑋𝐶𝑋 depending on 𝐶𝑋 only such that we have, for any 𝜖>0 small enough, 𝜖1/𝛼log(1/𝜖)𝛽log𝑋+𝑌𝜖𝜆(1𝛿)1/𝛼𝐶𝑋,(3.7) and the upper bound follows by taking 𝛿0 and 𝜆1.

By choosing 𝐾𝑋=max(𝐶𝑋,𝐶𝑋), we complete the proof of Theorem 2.1.

Remark 3.1. When 𝑋 and 𝑌 are independent, there is a simple proof without using the correlation inequality in the spirit of [5]. A direct proof of Corollary 2.2 can also be done as in [9].

4. Proof of Theorem 2.4, Part I

To simplify the reading of our paper, we introduce the following notation. Set 𝑎𝑡=𝑓(𝑡)/𝑡𝛾 and 𝑏𝑡=𝜙(𝑎𝑡).

Suppose here that, with probability 1, 𝑓(𝑡)𝑌(𝑡) for all 𝑡 large enough. We want to prove that lim𝑡+𝑎𝑡=0 and 0𝑎𝑡1/𝛾𝑏𝑡(𝑑𝑡/𝑡)<+.

In the sequel, there is no loss of generality to assume that 𝑓(𝑡) is a continuous function of 𝑡0.

Lemma 4.1. One has lim𝑡+𝑎𝑡=0.(4.1)

To prove Theorem 2.4, we will show that 𝑓LUC(𝑌) when 0𝑎𝑡1/𝛾𝑏𝑡(𝑑𝑡/𝑡)=+ and lim𝑡+𝑎𝑡=0.

Following the same lines as those in [11], our aim is to construct a suitable subset 𝐽 of such that we have the following property for an appropriate family of sets (𝐸𝑖)𝑖𝐽 in a basic probability space: given 𝜖>0, there exist a number 𝐾 and an integer 𝑝 such that 𝑛𝐽,𝑛𝑝𝑚𝐽,𝑚>𝑛𝐸𝑛𝐸𝑚𝐸𝑛𝐾+(1+𝜖)𝑚𝐽,𝑚>𝑛𝐸𝑚.(4.2)

Lemma 4.2. When 0𝑎𝑡1/𝛾𝑏𝑡(𝑑𝑡/𝑡)=+ and lim𝑡+𝑎𝑡=0, we can find a sequence {𝑡𝑛,𝑛1} with the two following properties: 𝑡𝑛+1𝑡𝑛1+𝑎𝑡1/𝛾𝑛,𝑛=1𝑏𝑡𝑛=+.(4.3)

Remark 4.3. The condition “𝑓(𝑡)/𝑡𝛾 is a nonincreasing function of 𝑡>0” is essential to prove Lemma 4.2 (see [7, page 373]).

To continue the construction of the set 𝐽, we need the following definition and notation.

Definition 4.4. Consider the interval 𝐴𝑘=[2𝑘,2𝑘+1[,𝑘. If 𝑎𝑡1/𝛾𝑖𝐴𝑘,𝑖, then one notes 𝑢(𝑖)=𝑘.

Next, set 𝐼𝑘={𝑖,𝑢(𝑖)=𝑘} which is finite by Lemma 4.1 and 𝑁𝑘𝐾=exp0𝛾log2𝛽𝑘𝛽2𝛾(𝑘1)/𝛼,(4.4) where 𝐾0 was defined in Corollary 2.2.

Notation
(i)𝑈𝑚,𝑘={𝑖,𝑖𝐼𝑘𝐼,𝑖<𝑚,card𝑘[𝑖,𝑚[𝑁𝑘},𝑚,𝑘;(ii)𝑘0=inf{𝑛,2𝛾𝑛/𝛼2𝛾/𝛼/(𝐾0(2𝛾/𝛼1)(𝛾log2)𝛽)+22𝛾/𝛼/𝐾20(2𝛾/𝛼1)}, (𝑘0 depends on 𝛾,𝛼,𝛽 and 𝐾0 only);(iii)𝑉𝑚=𝑘𝑈𝑚,𝑘, where 𝑚 is fixed, 𝑢(𝑚)=𝑘1, and 𝑘𝑘1+𝑘0;(iv)𝑊=𝑚1𝑉𝑚.

Now, we can define the set 𝐽 as follows: 𝐽=𝑊.(4.5) Since it is assumed that 𝑓(𝑡)/𝑡𝛾 is a nonincreasing function of 𝑡>0 (being a particular case of the condition “𝑓(𝑡)/𝑡𝛾 is bounded”),𝑖<𝑚𝑘=𝑢(𝑖)𝑢(𝑚)=𝑘1.(4.6) Moreover, since 𝑘01, we get 𝑘<𝑘1+𝑘0. Hence 𝑉𝑚 is always an empty set (by construction). Thus we obtain 𝐽=.

We have, by Lemma 4.2, 𝑛𝐽𝑏𝑡𝑛=+.(4.7)

Lemma 4.5. 𝑛𝐽,𝑚𝐽,𝑛<𝑚, such that card (𝐼𝑢(𝑚)[𝑛,𝑚])>exp(𝐾02𝑢(𝑚)1), one has 𝑡𝑚𝑡𝑛𝐾expexp042min(𝑢(𝑛),𝑢(𝑚)).(4.8)

Proof. Set 𝑘=𝑢(𝑛),𝑘1=𝑢(𝑚) and 𝐺=𝐼𝑘1[𝑛,𝑚]={𝑖1,𝑖2,...,𝑖𝑧}, where 𝑛𝑖1<𝑖2<<𝑖𝑧𝑚. We have 𝑡𝑚𝑡𝑛=𝑡𝑚𝑡𝑖𝑧𝑡𝑖𝑧𝑡𝑖𝑧1𝑡𝑖1𝑡𝑛.(4.9) Note that, when 𝑖𝐼𝑘1, we have 𝑡𝑖+1𝑡𝑖(1+𝑎𝑡1/𝛾𝑖)𝑡𝑖(1+2𝑘11). Moreover, since card(𝐺)>exp(𝐾02𝑘11) by hypothesis, (4.9) implies 𝑡𝑚𝑡𝑛𝐾expexp02𝑘11log1+2𝑘11𝐾expexp042𝑘1,(4.10) when 𝑛 hence 𝑘1 are large enough.
Thus, since 𝑘𝑘1, (4.10) implies (4.8).
The proof of Lemma 4.5 is now complete.

5. Proof of Theorem 2.4, Part II

Consider, now, the events 𝐸𝑛={𝑌(𝑡𝑛)<𝑓(𝑡𝑛)}. We have directly (𝐸𝑛)=𝑏𝑡𝑛, and, by (4.7), 𝑛𝐽𝑏𝑡𝑛=+. To prove (4.2), we remark that, given 𝑛𝐽,𝐽 can be rewritten as follows: 𝐽=𝐽(𝑘𝐽𝑘)𝐽, where 𝐽={𝑚𝐽,𝑡𝑛𝑡𝑚2𝑡𝑛},𝐽𝑘={𝑚𝐽𝐼𝑘,𝑡𝑚>2𝑡𝑛,card(𝐼𝑘[𝑛,𝑚])exp(𝐾02𝑘1)}, and 𝐽=𝐽(𝐽(𝑘𝐽𝑘)).

Our first key small ball estimate is given in the following lemma.

Lemma 5.1. Consider 0<𝑡<𝑢, and 𝜃,𝜈>0. Then one has 𝑌(𝑡)𝜃𝑡𝛾𝐾𝑌(𝑢)𝜈exp5𝑌(𝑡)𝜃𝑡𝛾𝐾exp5(𝑢𝑡)𝜈1/𝛾,(5.1) where 𝐾5 depends on 𝐻1,𝐻2,𝐻3,𝐻4,𝜆1, and 𝜆2 only.

Proof. Set 𝐹1={𝑌(𝑡)𝜃𝑡𝛾} and 𝐹2={𝑌(𝑢)𝜈}. We have 𝐹1𝐹2𝐹=1sup0𝑠𝑢sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝑋𝑤1,𝑤2,𝑤3,𝑤4||𝐹,𝑠𝜈1sup𝑡𝑠𝑢sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝑋𝑤1,𝑤2,𝑤3,𝑤4||𝐹,𝑠𝜈1sup𝑡𝑠𝑢sup0𝑤1,𝑤2𝑠||𝜆1𝑠𝐻3+𝐻4𝐵𝐻1,𝐻2𝑤1,𝑤2||.𝜈(5.2) Denote, by [𝑥], the integer part of a real 𝑥. Let 𝛿>0. We consider the sequence 𝑡𝑘,𝑘{0,,𝑛}, where 𝑡0=𝑡,𝑡𝑘+1=𝑡𝑘+𝛿, and 𝑛=[(𝑢𝑡)/𝛿]. Consider also the rectangles 𝑅𝑗=[𝑡𝑗,𝑡𝑗+1]×[𝑡𝑗,𝑡𝑗+1], where 𝑗{0,,𝑛1}. Their area is (𝑡𝑗+1𝑡𝑗)2=𝛿2. Let 𝐺𝑗 be the event defined by 𝐺𝑗=𝐹1sup𝑡𝑠𝑡𝑗sup0𝑤1,𝑤2𝑠||𝜆1𝑠𝐻3+𝐻4𝐵𝐻1,𝐻2𝑤1,𝑤2||.𝜈(5.3) We have 𝐹1𝐹2𝐺𝑗.
Moreover, we have also 𝐺𝑗+1𝐺𝑗𝑍𝑗,4𝜈(5.4) where 𝑍𝑗=𝜆1𝑡𝐻3+𝐻4𝑗+1𝐵𝐻1,𝐻2𝑡𝑗+1,𝑡𝑗+1𝐵𝐻1,𝐻2𝑡𝑗,𝑡𝑗+1𝐵𝐻1,𝐻2𝑡𝑗+1,𝑡𝑗+𝐵𝐻1,𝐻2𝑡𝑗,𝑡𝑗.(5.5) Before rewritting 𝑍𝑗, we recall the integral representation of 𝐵𝐻𝑘,𝐻𝑙,𝑘,𝑙, given by 𝐵𝐻𝑘,𝐻𝑙𝑠𝑘,𝑠𝑙=𝑠𝑘𝑠𝑙𝑔𝐻𝑘𝑠𝑘,𝑢𝑘𝑔𝐻𝑙𝑠𝑙,𝑢𝑙𝑊𝑑𝑢𝑘,𝑢𝑙,(5.6) where 𝑊(𝑢𝑘,𝑢𝑙),𝑢𝑘,𝑢𝑙, is a standard Brownian sheet, 𝑔𝐻(𝑠,𝑢)=𝑘12𝐻(max(𝑠𝑢,0)𝐻1/2max(𝑢,0)𝐻1/2), and 𝑘2𝐻 is a normalizing constant.
Hence 𝑍𝑗 can be rewritten by (5.6) as follows: 𝑍𝑗=𝑍𝑗,1+𝑍𝑗,2, where 𝑍𝑗,1=𝜆1𝑘12𝐻1𝑘12𝐻2𝑡𝐻3+𝐻4𝑗+1𝑡𝑗+1𝑡𝑗𝑡𝑗+1𝑡𝑗𝑡𝑗+1𝑢1𝐻11/2𝑡𝑗+1𝑢2𝐻21/2𝑊𝑑𝑢1,𝑢2.(5.7) Note also that 𝑍𝑗,1 and 𝑍𝑗,2 are independent.
Since (|𝑍𝑗,1+𝑥|4𝜈) is maximum at 𝑥=0 and 𝑍𝑗,1 and 𝐺𝑗 are independent, we have 𝐺𝑗+1𝐺𝑗||𝑍𝑗,1||.4𝜈(5.8) The integral representation of 𝑍𝑗,1 implies that 𝔼(𝑍𝑗,1)=0 and Var𝑍𝑗,1=𝜆21𝑘22𝐻1𝑘22𝐻24𝐻1𝐻2𝛿2(𝐻1+𝐻2)𝑡2(𝐻3+𝐻4)𝑗+1𝜆21𝑘22𝐻1𝑘22𝐻24𝐻1𝐻2𝛿2𝛾=𝐿2𝛿2𝛾.(5.9) Denote, by Φ, the distribution function of the absolute value of a standard Gaussian random variable. Then we obtain 𝑍𝑗+1𝑍𝑗Φ4𝜈𝐿𝛿𝛾,(5.10) and therefore, (𝐹1𝐹2)(𝐹1)Φ(4𝜈/𝐿𝛿𝛾)𝑛.
Choosing 𝛿=𝜈1/𝛾, we get 𝐾5=logΦ(2/𝐿). Lemma 5.1 is proved.

Lemma 5.2. 𝑚𝐽(𝐸𝑛𝐸𝑚)𝐾𝑏𝑡𝑛 and 𝑚(𝑘𝐽𝑘)(𝐸𝑛𝐸𝑚)𝐾𝑏𝑡𝑛, where 𝐾 and 𝐾 are numbers.

Proof. Setting 𝑢=𝑡𝑚,𝑡=𝑡𝑛,𝜃=𝑎𝑡𝑛 and 𝜈=𝑓(𝑡𝑚), Lemma 5.1 implies 𝐸𝑛𝐸𝑚𝐾exp5𝑏𝑡𝑛𝐾exp5𝑡𝑚𝑡𝑛𝑓𝑡𝑚1/𝛾.(5.11) Consider, first, the case when 𝑚𝐽.
Lemma 4.2 implies that, for all 𝑖𝑛, we have 𝑡𝑖+1𝑡𝑖𝑡𝑖𝑎𝑡1/𝛾𝑖=𝑓(𝑡𝑖)1/𝛾𝑓(𝑡𝑛)1/𝛾. Then we can establish 𝑡𝑚𝑡𝑛𝑡(𝑚𝑛)𝑓𝑛1/𝛾𝑡,𝑓𝑚𝑡𝑓𝑛𝑡𝑚𝑡𝑛𝛾2𝛾𝑓𝑡𝑛.(5.12) Combining (5.11) with (5.12), we get 𝐸𝑛𝐸𝑚𝐾exp5𝑏𝑡𝑛𝐾exp5(𝑚𝑛)2,(5.13) which is the first part of Lemma 5.2.
Consider, now, the case 𝑚𝐽𝑘.
Combining (5.11) with the definition of 𝐽𝑘, we have 𝐸𝑛𝐸𝑚𝐾exp5𝑏𝑡𝑛𝐾exp52𝑎𝑡𝑚1/𝛾.(5.14) Since 𝑢(𝑚)=𝑘, we get 𝐸𝑛𝐸𝑚𝐾exp5𝑏𝑡𝑛exp𝐾52𝑘1,(5.15) and consequently, by noting that card𝐽𝑘card(𝐼𝑘[𝑛,𝑚])exp(𝐾02𝑘1) and by assuming 𝐾0<𝐾5, we have 𝑚𝐽𝑘𝐸𝑛𝐸𝑚𝐾exp5𝑏𝑡𝑛𝐾exp0𝐾52𝑘1.(5.16) Lemma 5.2 is, therefore, proved.

To deal with the set 𝐽, we first state a standard large deviation result and a technical lemma (see [6]).

Lemma A. Let 𝑋={𝑋(𝑠1,𝑠2),(𝑠1,𝑠2)[0,1]2} be a separable real-valued centered Gaussian process such that 𝑋(0,0)=0 with probability 1 and satisfying, for any [𝑠1,𝑠1+1]×[𝑠2,𝑠2+2][0,1]2,𝑠𝔼𝑋1,𝑠1+1×𝑠2,𝑠2+221/2𝜅1,2𝑐𝜅𝛼11𝛼22,𝛼1>0,𝛼2>0,(5.17) where 𝑋𝑠1,𝑡1×𝑠2,𝑡2=[𝑠1,𝑡1]×[𝑠2,𝑡2]𝑋𝑑𝑢1,𝑢2.(5.18) Then one has, for 𝑐𝜅1𝛿>1,sup(𝑠1,𝑠2)[0,1]2||𝑋𝑠1,𝑠2||1𝛿𝐶𝑐exp𝐶𝜅1𝛿2,(5.19) where 𝐶 is a positive constant independent of 𝑐𝜅 and 𝛿.

Lemma B. One has, for 𝜖1>𝜖/2, where 𝜖 is small enough, exp𝐾3|𝜖1𝜖|log1/𝜖𝛽𝜖1+1/𝛼𝜙𝜖1𝐾𝜙(𝜖)exp3|𝜖1𝜖|log1/𝜖𝛽𝜖1+1/𝛼,(5.20) where 𝐾3>0.

Building on Lemmas A and B, one can establish our last key small ball estimate in the following result.

Lemma 5.3. Let 𝜆 be a real number such that 1/2<𝜆<1. Set 𝐻𝑟=min1max1,𝐻2,𝐻3,𝐻43,(1𝜆)𝛼3.(5.21) Then one has, for 𝑢2𝑡,𝑌(𝑡)𝜃𝑡𝛾,𝑌(𝑢)𝜈𝑢𝛾2𝑡𝜙(𝜃)𝜙(𝜈)exp𝑢𝑟𝐾3log(1/𝜃)𝛽𝜃1+(1/𝛼)+log(1/𝜈)𝛽𝜈1+(1/𝛼)1+3𝐶12,2𝐶exp12,24𝜆21𝐾2𝐻1,2𝑢𝑡𝑟+1𝐶34,2𝐶exp34,24𝜆22𝐾2𝐻3,2𝑢𝑡𝑟1+3𝐶12,1𝐶exp12,14𝜆21𝐾2𝐻1,1𝑢𝑡𝑟+1𝐶34,1𝐶exp34,14𝜆22𝐾2𝐻3,1𝑢𝑡𝑟,(5.22) where 𝐾𝐻1,1,𝐾𝐻1,2>0 depend on 𝐻1(𝐻1𝐻2) only, 𝐾𝐻3,1,𝐾𝐻3,2>0 depend on 𝐻3(𝐻3𝐻4) only, 𝐾3>0 is defined as in Lemma B, and 𝐶12,1,𝐶12,2,𝐶34,1,𝐶34,2>0 are defined as in Lemma A.

Proof. Set 𝑄=(𝑌(𝑡)𝜃𝑡𝛾,𝑌(𝑢)𝜈𝑢𝛾).
Set 𝑣=𝑢𝑡. If 𝑡=𝑜(𝑢), then 𝑡=𝑜(𝑣) and 𝑣=𝑜(𝑢).
Based on (5.6), 𝐵𝐻1,𝐻2 and 𝐵𝐻3,𝐻4 can be split as follows: 𝐵𝐻1,𝐻2=𝐵𝐻1,𝐻2,1+𝐵𝐻1,𝐻2,2,𝐵𝐻3,𝐻4=𝐵𝐻3,𝐻4,1+𝐵𝐻3,𝐻4,2,(5.23) where we have, for (𝑖,𝑗){(1,2),(3,4)}, 𝐵𝐻𝑖,𝐻𝑗,1𝑤𝑖,𝑤𝑗=|𝑥𝑖|𝑣𝑤𝑗𝑔𝐻𝑖𝑤𝑖,𝑥𝑖𝑔𝐻𝑗𝑤𝑗,𝑥𝑗𝑊𝑑𝑥𝑖,𝑥𝑗.(5.24) Note that 𝐵𝐻1,𝐻2,1 and 𝐵𝐻1,𝐻2,2 are independent as 𝐵𝐻3,𝐻4,1 and 𝐵𝐻3,𝐻4,2.
Equation (5.23) implies that the FMFBS 𝑋 can be rewritten as follows: 𝑋=𝑋1+𝑋2, where 𝑋𝑖𝑤1,𝑤2,𝑤3,𝑤4,𝑠=𝜆1𝑠𝐻3+𝐻4𝐵𝐻1,𝐻2,𝑖𝑤1,𝑤2+𝜆2𝑠𝐻1+𝐻2𝐵𝐻3,𝐻4,𝑖𝑤3,𝑤4.(5.25) Set 𝑌𝑖(𝑡)=sup0𝑠𝑡sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝑋𝑖𝑤1,𝑤2,𝑤3,𝑤4||,𝑠,𝑡0,𝑖{1,2}.(5.26) Then, given 𝛿>0, we have (see [11]) 𝑌𝑄𝜙(𝜃+2𝛿)𝜙(𝜈+2𝛿)+32(𝑡)>𝛿𝑡𝛾𝑌+31(𝑢)>𝛿𝑢𝛾.(5.27) Equation (5.20) implies 𝜙(𝜃+2𝛿)𝜙(𝜃)exp2𝛿𝐾3log(1/𝜃)𝛽𝜃1+(1/𝛼),(5.28) and, consequently, 𝜙(𝜃+2𝛿)𝜙(𝜈+2𝛿)𝜙(𝜃)𝜙(𝜈)exp2𝛿𝐾3log(1/𝜃)𝛽𝜃1+(1/𝛼)+log(1/𝜈)𝛽𝜈1+(1/𝛼).(5.29) If we choose 𝛿=(𝑡/𝑢)𝑟, then we get the first term of the RHS of Lemma 5.3.
Next, we want to obtain an upper bound of 𝑌2(𝑡)>𝛿𝑡𝛾=sup0𝑠1sup0𝑤1,𝑤2,𝑤3,𝑤4𝑠||𝜆1𝑠𝐻3+𝐻4𝐿𝐻1,𝐻2,2𝑤1,𝑤2+𝜆2𝑠𝐻1+𝐻2𝐿𝐻3,𝐻4,2𝑤3,𝑤4||,>𝛿(5.30) where we have, for (𝑖,𝑗){(1,2),(3,4)}, 𝐿𝐻𝑖,𝐻𝑗,2𝑤𝑖,𝑤𝑗=|𝑥𝑖|𝑣/𝑡𝑤𝑗𝑔𝐻𝑖𝑤𝑖,𝑥𝑖𝑔𝐻𝑗𝑤𝑗,𝑥𝑗𝑊𝑑𝑥𝑖,𝑥𝑗.(5.31) We can show, by standard computations, that 𝑌2(𝑡)>𝛿𝑡𝛾sup0𝑤1,𝑤21||𝐿𝐻1,𝐻2,2𝑤1,𝑤2||>𝛿2||𝜆1||+sup0𝑤3,𝑤41||𝐿𝐻3,𝐻4,2𝑤3,𝑤4||>𝛿2||𝜆2||.(5.32) Denote, by 𝜎𝐻, the covariance function of a FBM 𝐵𝐻. Set 𝜎𝐻,2, the covariance function of the process {𝐵𝐻,2(𝑤),0𝑤1}, defined by 𝐵𝐻,2(𝑤)=|𝑥|𝑣/𝑡𝑔𝐻𝑊(𝑤,𝑥)(𝑑𝑥),(5.33) where 𝑊(𝑥),𝑥, is a Wiener process.
Since 𝔼𝐿𝐻1,𝐻2,2𝑤1,𝑤2𝐿𝐻1,𝐻2,2𝑤1,𝑤2=𝜎𝐻1,2𝑤1,𝑤1×𝜎𝐻2𝑤2,𝑤2,(5.34) we have, for any [𝑤1,𝑤1+1]×[𝑤2,𝑤2+2][0,1]2,𝔼𝐿𝐻1,𝐻2,2𝑤1,𝑤1+1×𝑤2,𝑤2+22=𝔼[𝑤1,𝑤1+1]×[𝑤2,𝑤2+2]𝐿𝐻1,𝐻2,2𝑑𝑥1,𝑥2×[𝑤1,𝑤1+1]×[𝑤2,𝑤2+2]𝐿𝐻1,𝐻2,2𝑑𝑥1,𝑥2𝑤1+1𝑤1𝑤1+1𝑤1||𝜎𝐻1,2𝑥1,𝑥1||𝑑𝑥1𝑑𝑥1×𝑤2+2𝑤2𝑤2+2𝑤2||𝜎𝐻2𝑥2,𝑥2||𝑑𝑥2𝑑𝑥2=I×II.(5.35) Consider II first. We get, by the inequality of Cauchy-Schwarz, II𝑤2+2𝑤2𝑤2+2𝑤2𝑥𝐻22𝑥𝐻22𝑑𝑥2𝑑𝑥222.(5.36) Consider I now.
A straight computation implies that there exists 𝐾𝐻1,2>0 depending on 𝐻1 such that 𝔼𝐵𝐻1,2𝑥12𝐾2𝐻1,2𝑥21𝑣/𝑡2𝐻12,(5.37) and, consequently, by the inequality of Cauchy-Schwarz, ||𝜎𝐻1,2𝑥1,𝑥1||𝐾2𝐻1,2𝑥1𝑥1(𝑣/𝑡)2𝐻12.(5.38) So we get 𝐼𝐾2𝐻1,2(𝑣/𝑡)2𝐻1221.(5.39) Hence, combining (5.35) with (5.36) and (5.39), we have 𝔼𝐿𝐻1,𝐻2,2𝑤1,𝑤1+1×𝑤2,𝑤2+22𝐾2𝐻1,2(𝑣/𝑡)2𝐻122122.(5.40) An application of Lemma A with 𝛼1=𝛼2=1, 𝑐𝜅=𝐾𝐻1,2(𝑣/𝑡)𝐻11, and 𝑐𝜅1𝛿>1 implies that sup0𝑤1,𝑤21||𝐿𝐻1,𝐻2,2𝑤1,𝑤2||>𝛿2||𝜆1||1𝐶12,2𝐶exp12,2𝐾2𝐻1,2𝑣2/𝑡2𝐻11𝛿24𝜆21.(5.41) Similarly, we can establish sup0𝑤3,𝑤41||𝐿𝐻3,𝐻4,2𝑤3,𝑤4||>𝛿2||𝜆2||1𝐶34,2𝐶exp34,2𝐾2𝐻3,2𝑣2/𝑡2𝐻31𝛿24𝜆22.(5.42) Set 𝛿=(𝑡/𝑢)𝑟. Recall that 𝑣2=𝑢𝑡 and 𝑟(1max(𝐻1,𝐻2,𝐻3,𝐻4))/3. Combining (5.32) with (5.41) and (5.42), we get 𝑌2(𝑡)>𝛿𝑡𝛾1𝐶12,2𝐶exp12,24𝜆21𝐾2𝐻1,2𝑢𝑡𝑟+1𝐶34,2𝐶exp34,24𝜆22𝐾2𝐻3,2𝑢𝑡𝑟,(5.43) that is the second term of the RHS of Lemma 5.3.
Finally, we can establish a similar result for (𝑌1(𝑢)>𝛿𝑢𝛾), that is, 𝑌1(𝑢)>𝛿𝑢𝛾1𝐶12,1𝐶exp12,14𝜆21𝐾2𝐻1,1𝑢𝑡𝑟+1𝐶34,1𝐶exp34,14𝜆22𝐾2𝐻3,1𝑢𝑡𝑟,(5.44) which achieves the proof of Lemma 5.3.

Finally, we state the last technical lemma.

Lemma 5.4. There exists an integer 𝑝 such that if 𝑛>sup𝑠𝑝(sup𝐼𝑠), then for 𝑚𝐽,𝑚>𝑛, given 𝜖>0, one has (𝐸𝑛𝐸𝑚)(1+𝜖)𝑏𝑡𝑛𝑏𝑡𝑚.

Proof. Let 𝑢(𝑛)=𝑘 and 𝑢(𝑚)=𝑘1. We have, by Lemma 4.5, 𝑡𝑚𝑡𝑛𝐾expexp042min(𝑘,𝑘1).(5.45) Let 𝑝. Then 𝑘>𝑝 and 𝑘1>𝑝. Thus we have min(𝑘,𝑘1)>𝑝.
Set 𝑡=𝑡𝑛,𝑢=𝑡𝑚,𝜃=𝑎𝑡𝑛 and 𝜈=𝑎𝑡𝑚. Note that log(1/𝜃)1/𝜃2(𝑘+1)𝛼,log(1/𝜈)1/𝜈2(𝑘1+1)𝛼, and 1/𝑏𝑡𝑛𝑏𝑡𝑚=exp(𝜓(𝜃)+𝜓(𝜈)).
By using Lemma 5.3 and letting 𝑝+, we end the proof of Lemma 5.4.

Lemmas 5.2 and 5.4 yield that (4.2) holds. Combining Borel-Cantelli's second lemma with (4.2) and (4.7), we show that, given 𝜖>0, 𝑛𝐽𝐸𝑛1+2𝐾𝜖11+2𝜖𝑛𝐽𝐸𝑛=𝑛𝐽𝑌𝑡𝑛𝑡𝑓𝑛,(5.46) and, consequently, 𝑓LUC(𝑌). The proof of Theorem 2.4 is now complete.

Acknowledgment

The author thanks the referee for the insightful comments.