# Average case tractability of additive random fields with Korobov kernels

We investigate average case tractability of approximation of additive random fields with marginal random processes corresponding to the Korobov kernels for the non-homogeneous case. We use the absolute error criterion (ABS) or the normalized error criterion (NOR). We show that the problem is always polynomially tractable for ABS or NOR, and give sufficient and necessary conditions for strong polynomial tractability for ABS or NOR.

## Authors

• 14 publications
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## 1. Introduction

Let be a sequence of independent random processes, where . Suppose that every random element has zero mean and a covariance function . Let be the covariance operator with the covariance kernel for every . Then for any and , we have

 KXjf(t)=∫[0,1]KXj(t,s)f(s)ds.

We consider the random field

 (1.1) Yd(t):=d∑j=1Xj(tj),  t=(t1,⋯,td)∈[0,1]d,

with the following zero-mean covariance function

 (1.2) KYd(t,s)=d∑j=1KXj(tj,sj),

and covariance operator

 (1.3) KYdf(t)=∫[0,1]dKYd(t,s)f(s)ds,

where This random field is called an additive random field. There are many papers which investigated this random field, see [1, 3, 4, 5].

In this paper, we investigate the average case approximation of , as a random element of the space equipped with inner product and norm , by a finite rank random field.

The th minimal 2-average case error, for , is defined by

Here is the class of all linear algorithms with rank defined by

 AYdn:={n∑m=1(Yd,φm)2,dψm:φm,ψm∈L2([0,1]d)}.

The so-called initial average case error is given by

 eYd(0):=(E∥Yd∥22,d)1/2.

We use either the absolute error criterion (ABS) or the normalized error criterion (NOR). For any , we study information complexity of approximation of the random fields defined by

 nYd,Z(ε):=min{n∈N:eYd(n)≤εCRId},

where ,

 CRId:={1, for Z=ABS,eYd(0),for Z=NOR.

Let . First we consider average case tractability of . Various notions of tractability have been discussed for multivariate problems. We recall some of the basic tractability notions (see [7, 8, 9, 10]).

For , say is

strongly polynomially tractable (SPT) iff there exist non-negative numbers and such that for all ,

 nYd,Z(ε)≤C(ε−1)p;

The infimum of satisfying the above inequality is called the exponent of strong polynomial tractability and is denoted by .

polynomially tractable (PT) iff there exist non-negative numbers and such that for all ,

 nYd,Z(ε)≤Cdq(ε−1)p.

quasi-polynomially tractable (QPT) iff there exist two constants such that for all ,

 nYd,Z(ε)≤Cexp(t(1+lnε−1)(1+lnd));

uniformly weakly tractable (UWT) iff for all ,

 limε−1+d→∞lnnYd,Z(ε)(ε−1)s+dt=0;

weakly tractable (WT) iff

 limε−1+d→∞lnnYd,Z(ε)ε−1+d=0.

This paper is devoted to studying average case tractability of the additive random field under ABS and NOR. For additive random fields similar problems were investigated in [2, 5, 6, 11] in various settings for the homogeneous case and in [3, 4] for the non-homogeneous case. Here, the homogeneous case means that approximated additive random fields are constructed (in a special way) from copies of one marginal process, while the non-homogeneous case means that the random fields are composed of a whole sequence of marginal random processes with generally different covariance functions. Specifically, the authors in [3] obtained the growth of for arbitrary fixed and for the non-homogeneous case and gave application to the additive random fields with marginal random processes corresponding to the Korobov kernels.

It should be noted, however, that all these works deal only with NOR. In this paper, we consider average case tractability of the problem of the additive random fields with marginal random processes corresponding to the Korobov kernels under ABS and NOR. We shall show that the problem is always polynomially tractable for ABS or NOR. Obviously, PT implies all QPT, UWT, WT. We also give sufficient and necessary conditions for which is SPT for ABS or NOR.

The paper is organized as follows. In Section 2 we give preliminaries about the additive random fields with marginal random processes corresponding to the Korobov kernels and introduce main results, i.e., Theorems 2.1-2.3. Section 3 is devoted to proving Theorems 2.1-2.3.

## 2. Preliminaries and main results

Let us consider the sequence of additive random fields , , , defined by (1.1). Let be the sequence of eigenpairs of the covariance operator of . Under the additive structure (1.2

), the eigenvalues

, are generally unknown or not easily depend on . However, under the following condition, we can explicitly describe the eigenvalues .

For every there exist such that for all . We denote by

the eigenvalues corresponding to eigenvector 1. Let

and be the non-increasing sequence of the remaining eigenvalues and the corresponding sequence of eigenvectors of , respectively. It is known that the family of eigenvectors

 1∪{¯ψXjk(tj):k∈N,j=1,⋯,d},t=(t1,⋯,td)∈[0,1]d,

is an orthogonal system in for every , see [3]. Hence the identical 1 is an eigenvector of with the eigenvalue , and the pairs , for all and , are the remaining eigenpairs of .

Let and be the non-increasing sequence of the eigenvalues and the corresponding sequence of eigenvectors of defined by (1.3). Then the average case information complexity can also be described in terms of eigenvalues of by

 (2.1) nYd,Z(ε):=min{n∈N:∞∑j=n+1λd,j≤ε2CRI2d},

where

 CRId={1, for Z=ABS,(∑∞j=1λd,j)1/2,for Z=NOR,

see [7].

Particularly, we study additive random fields with marginal random processes corresponding to the Korobov kernels. Let , be a zero-mean random field with the following covariance function

 Kα,β,σ(x,y)=α+2β∞∑k=1k−σcos(2πk(x−y)),x,y∈[0,1].

Here , and . Let be the covariance operator with kernel of , and for any and ,

 KBα,β,σf(x):=∫[0,1]Kα,β,σ(x,y)f(y)dy.

The eigenpairs of the covariance operator are known, see [7]. The identical 1 is an eigenvector of with the eigenvalue . The other eigenpairs of have the following form:

 ¯λBα,β,σ2k−1=¯λBα,β,σ2k=βkσ,¯ψBα,β,σ2k−1(x)=e−i2πkx,¯ψBα,β,σ2k(x)=ei2πkx,

for any .

Suppose that is a sequence of independent zero-mean random fields with covariance functions , respectively. Let , , be the sequence of zero-mean random fields with the covariance functions

 KBd(x,y)=d∑j=1Kαj,βj,σj(xj,yj),

where , , and the parameters , for all , and .

Let be the covariance operator of . We have

 KBdf(x)=∫[0,1]dd∑j=1Kαj,βj,σj(xj,yj)f(yj)dy,

for any and . Then the identical 1 is an eigenvector of with the eigenvalue

 (2.2) ¯λBd0=d∑j=1αj

and the remaining eigenvalues and eigenvectors are

 (2.3) ¯λBd2k−1=¯λBd2k=βjkσj,  k∈N, j=1,…,d,

and

 ¯ψBd2k−1(xj)=e−i2πkxj,¯ψBd2k(xj)=ei2πkxj,  x∈[0,1]d, k∈N, j=1,…,d,

respectively. Let be the sequence of non-increasing rearrangement of the eigenvalues of . Then we have

 (2.4) ∞∑j=1λd,j=d∑j=1αj+2d∑j=1∞∑k=1βjkσj=d∑j=1αj+2d∑j=1βjζ(σj),

and for any and ,

 (2.5) ∞∑j=1λτd,j=(d∑j=1αj)τ+2d∑j=1∞∑k=1βτjkτσj=(d∑j=1αj)τ+2d∑j=1βτjζ(τσj),

where , is the Riemann zeta-function.

In the sequel we always assume that the sequences satisfy

 (2.6) αj≥0,    1≥β1≥β2≥⋯>0,  1<σ1≤σ2≤⋯.

In this paper, we consider the tractability of the problem

 B={Bd:L2([0,1]d)→L2([0,1]d)}

under ABS and NOR, where the sequences satisfy (2.6). Our main results can be formulated as follows.

###### Theorem 2.1.

Let the sequences satisfy (2.6). Then the problem

(i) is always PT for ABS or NOR;

(ii) is SPT for ABS iff

 (2.7) A∗:=lim––––d→∞ln1βdlnd>1.

The exponent of SPT is

 (2.8) pstr−avg=max{2A∗−1,2σ1−1}.

In order to investigate the strong polynomial tractability of the problem for NOR, we consider two cases.

###### Theorem 2.2.

Let satisfy (2.6) and

 0≤αj≤cβj,for allj∈N,

where is a constant. Then for NOR the problem is SPT iff

 A∗:=lim––––d→∞ln1βdlnd>1.

The exponent of SPT is

 pstr−avg=max{2A∗−1,2σ1−1}.
###### Theorem 2.3.

Let satisfy (2.6). Further assume that

 0

and there exists a constant such that for every ,

 d∑j=1αj≤cdαd.

Then the problem is SPT for NOR iff

 (2.9) B∗:=lim––––d→∞lnαdβdlnd>0.

The exponent of SPT is

 (2.10) pstr−avg=max{2B∗,2σ1−1}.

In particular, if for , then the problem is SPT for NOR iff

 A∗:=lim––––d→∞ln1βdlnd>0.
###### Remark 2.4.

In [3], Khartov and Zani investigated for arbitrary fixed and of the above problem with the parameters

 (2.11) βj∼cj−s,αj/βj→r,σj→+∞,j→∞,

where , , and . They obtained the following results.

(1) For and either , or , . Then

 supd∈NnBd,NOR(ε)<∞,for% everyε∈(0,1).

(2) For , and ,

 nBd,NOR(ε)∼2Q(ε)⋅d,d→∞,for every ε∈(0,ε0),

where

 Q(ε)=(1−(ε/ε0)2)11−s,  ε0=(1+r/2)−1/2.

(3) For , and , then

 lnnBd,NOR(ε)=(1−(ε/ε0)2)⋅lnd+o(lnd),d→∞,ε∈(0,ε0).

We remark that in some sense the above results give the tractability results of the problem for NOR under the condition (2.11). Specifically, the above result (1) corresponds to SPT of for NOR, and results (2) and (3) relate to PT of for NOR, but more explicitly. Comparing with [3], we obtain the tractability of the problem with general parameters satisfying (2.6) for ABS and NOR. Also we get the exponent of SPT for ABS and NOR.

###### Remark 2.5.

For the above problem , let be the additive random fields with marginal random processes corresponding to the Korobov covariance functions

with parameters satisfying

 \mathaccent869αj=0,  \mathaccent869βj=βj,  \mathaccent869σj=σj,  j∈N.

Let and be the sequences of non-increasing rearrangement of the eigenvalues of the covariance operators and , respectively. Then for some ,

 λd,j0=d∑j=1αj,

and

 \mathaccent869λd,j=λd,j  if  j

From the above equalities we obtain that

 λd,j≥\mathaccent869λd,j≥λd,j+1,   j∈N.

It follows from (2.1) and the inequality that for ,

 nBd,ABS(ε)≥n\mathaccent869Bd,ABS(ε).

On the other hand, we note that for ,

 ∞∑j=n0+1\mathaccent869λd,j≤ε2,

where . This gives that

 ∞∑j=n0+2λd,j≤∞∑j=n0+1\mathaccent869λd,j≤ε2,

which means that

 nBd,ABS(ε)≤n0+1=n\mathaccent869Bd,ABS(ε)+1.

Hence we have for ,

 n\mathaccent869Bd,ABS(ε)≤nBd,ABS(ε)≤n\mathaccent869Bd,ABS(ε)+1.

It follows that the problems

 Bd:L2([0,1]d)→L2([0,1]d)and\mathaccent869Bd:L2([0,1]d)→L2([0,1]d),d∈N,

have the same tractability for ABS and the same exponent of SPT for ABS if the problem is SPT for ABS.

###### Remark 2.6.

Let the sequences satisfy (2.6). If the problem is SPT (or PT) for ABS, then it is also SPT (or PT) for NOR.

Indeed, it follows from (2.4) that

 ∞∑j=1λd,j=d∑j=1αj+2d∑j=1βjζ(σj)≥2β1>0.

By (2.1) we have

 nBd,ABS(ε(2β1)1/2)=min{n∈N:∞∑j=n+1λd,j≤2β1ε2≤ε2∞∑j=1λd,j},

which means

 nBd,NOR(ε)≤nBd,ABS(ε(2β1)1/2).

Therefore if the problem is SPT (or PT) for ABS, then it is also SPT (or PT) for NOR.

## 3. Proofs of Theorems 2.1-2.3

Proof of Theorem 2.1.

(i) From Remark 2.6, it is sufficient to show that is PT for ABS. By Remark 2.5 we get that is PT for ABS iff is PT for ABS. Hence it suffices to prove that is PT for ABS.

We note

 (∞∑j=1\mathaccent869λτd,j)1τ=(2d∑j=1βτjζ(τσj))1τ,

and

 (3.1) 1≤ζ(τσj)≤ζ(τσ1)<+∞,forτ∈(1σ1,1).

It follows that for ,

 (∞∑j=1\mathaccent869λτd,j)1τ≤(2ζ(τσ1))1τ(d∑j=1βτj)1τ≤(2ζ(τσ1))1τd1τ,

where in the last inequality we used for all . This forces

 supd∈N(∞∑j=1\mathaccent869λτd,j)1τd−1τ≤(2ζ(τσ1))1τ<+∞.

Due to [7, Theorem 6.1] we obtain that is PT for ABS. Therefore the problem is always PT for ABS or NOR.

(ii) From Remark 2.5, it is sufficient to prove that the problem

 \mathaccent869B={\mathaccent869Bd:L2([0,1]d)→L2([0,1]d)},

is SPT for ABS iff (2.7) holds, and that the exponent of satisfies (2.8).

Assume that (2.7) holds, i.e.,

 A∗=lim––––d→∞ln1βdlnd>1.

We want to prove that is SPT for ABS. Indeed, by (3.1) we have for any ,

 (∞∑j=1\mathaccent869λτd,j)1τ ≤(2ζ(τσ1))1τ(d∑j=1βτj)1τ (3.2) ≤(2ζ(τσ1))1τ(∞∑j=1βτj)1τ.

Next, we shall prove that for any ,

 (∞∑j=1βτj)1τ<+∞.

Indeed, for such , there exists a for which

 τ=1A∗(1−δ).

Since

 A∗=lim––––d→∞ln1βdlnd>1,

there exists a such that

 ln1βdlnd≥(1−δ/2)A∗ford>d0,

which means

 βd≤d−(1−δ/2)A∗  for  d>d0.

It follows that

 ∞∑j=1βτj≤d0∑j=1βτj+∞∑d0+1j−1−δ/21−δ≤d0+∞∑j=1j−1−δ/21−δ<+∞,

for any . Hence we obtain

 (3.3) supd∈N(∞∑j=1\mathaccent869λτd,j)1τ<+∞,

for any . We note that is equivalent to

 2τ1−τ>2σ1−1and2τ1−τ>2A∗−1,

due to the monotonicity of the function

 φ(x)=x1−x, x∈(0,1).

It follows from [7, Theorem 6.1] that if (2.7) holds, then is SPT for ABS, and the exponent of SPT satisfies

 (3.4) pstr−avg≤inf{2τ1−τ ∣∣ τ satisfies (???)}≤max{2A∗−1,2σ1−1}.

On the other hand, assume that is SPT for ABS. Then there exist positive and such that

 +∞>C :=supd∈N(∞∑j=⌈C1⌉\mathaccent869λτd,j)1τ=supd∈N(∞∑j=1\mathaccent869λτd,j−⌈C1⌉−1∑j=1\mathaccent869λτd,j)1τ =supd∈N(2d∑j=1βτjζ(τσj)−⌈C1⌉−1∑j=1\mathaccent869λτd,j)1τ ≥supd∈N(2d∑j=1βτj−(⌈C1⌉−1)βτ1)1τ (3.5) ≥supd∈N(2dβτd−⌈C1⌉+1)1τ,

where we used

 1≥β1≥⋯>0,   \mathaccent869λd,j≤β1≤1,  ζ(τσj)≥1

in the above inequalities. Obviously, by (3) we have

 τ>1σ1.

It follows from (3) that

 dβτd≤C2,

where , which yields that

 ln1βdlnd≥1τ−lnC2τlnd,

Letting we get

 A∗=lim––––d→∞ln1βdlnd≥1τ>1.

which means

 τ≥1A∗.

Hence if is SPT for ABS, then we have

 A