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in PROBABILITY

CENTRAL LIMIT THEOREM FOR TRUNCATED HEAVY TAILED

BA-NACH VALUED RANDOM VECTORS

ARIJIT CHAKRABARTY1

Department of Mathematics, Indian Institute of Science, Bengaluru - 560012, India

email: arijit@math.iisc.ernet.in

SubmittedMay 31, 2010, accepted in final formAugust 19, 2010 AMS 2000 Subject classification: Primary 60F05; Secondary 60B12

Keywords: heavy tails, truncation, regular variation, central limit theorem, probability on Banach spaces

Abstract

In this paper the question of the extent to which truncated heavy tailed random vectors, taking values in a Banach space, retain the characteristic features of heavy tailed random vectors, is answered from the point of view of the central limit theorem.

1

Introduction

Situations where heavy-tailed distributions is a good fit, and at the same time there is a physical upper bound on the quantity of interest, are common in nature. Clearly, the natural model for phenomena like this is a truncated tailed distribution - a distribution that matches a heavy-tailed one till a specified limit and after that it decays significantly faster or simply vanishes. This leads to the general question: when can the upper bound be considered to be large enough so that the effect of truncating by that is negligible? The first attempt at answering this question, in finite dimensional spaces, was made in [Chakrabarty and Samorodnitsky, 2009]. In the current paper, the investigation started by that paper has been continued to achieve similar results in Banach spaces.

Suppose thatB is a separable Banach space and thatH,H1,H2, . . . areB-valued random variables in the domain of attraction of anα-stable random variableV with 0< α <2. This means that there are sequencesanandbnso that asn−→ ∞,

bn1 

n

X

j=1 Hjan

=⇒ V. (1.1)

We assume that the truncating threshold goes to infinity along with the sample size, and hence we essentially have a sequence of models. We denote both - the sample size and the number of the

1RESEARCH PARTLY SUPPORTED BY THE NSF GRANT “GRADUATE AND POSTDOCTORAL TRAINING IN PROBABIL-ITY AND ITS APPLICATIONS” AT CORNELL UNIVERSPROBABIL-ITY, AND THE CENTENARY POSTDOCTORAL FELLOWSHIP AT THE INDIAN INSTITUTE OF SCIENCE.

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model byn, and the truncating threshold in then-th model byMn. Thenth row of the triangular array will consist of observationsXn j, j=1, . . . ,n, which are assumed to be generated according to the following mechanism:

Xn j:=Hj1€kHjk ≤MnŠ+ Hj kHjk

(Mn+Lj)1€kHjk>MnŠ, (1.2)

j=1, . . . ,n,n=1, 2, . . .. Here(L,L1,L2, . . .)is a sequence of i.i.d. nonnegative random variables

independent of(H,H1,H2, . . .). For eachn=1, 2, . . . we view the observationXn j, j=1, . . . ,nas having power tails that are truncated at levelMn. The random variableLcan be thought of as to model that outside the ball of radiusMn, the tail “decays significantly faster or simply vanishes”.

Lis assumed to have finite second moment.

In[Chakrabarty and Samorodnitsky, 2009]two regimes depending on the growth rate ofMnand the tail of the random variableHwere introduced as follows: the tails in the model (1.2) are said to be

truncated softly if limn→∞nP kHk>Mn

=0 , truncated hard if limn→∞nP kHk>Mn

=∞. (1.3)

It was shown in that paper that as far as the central limit behavior of the row sum is concerned, observations with softly truncated tails behave like heavy tailed random variables, while obser-vations with hard truncated tails behave like light tailed random variables. In Theorem 2.1, the main result of this paper, we show that the result under hard truncation can be extended to Banach spaces, if the “small ball criterion” holds. Doing this is not straightforward because of the following reason. While in finite-dimensional spaces, convergence in law is equivalent to one-dimensional convergence of each linear functional in law to the linear functional evaluated at the limit, the same is not true in Banach spaces. In the latter spaces, one needs to check in addition some tight-ness conditions; see for example,[Ledoux and Talagrand, 1991] or[Araujo and Giné, 1980]for details.

Section 2 contains the results and their proofs. A couple of examples are studied in Section 3 -one where the hypothesis of Theorem 2.1 can be checked, and the other where the claim of that result does not hold. The examples serve the purpose of showing that there is a need for such a result, and that the result has some practical value.

2

A Central Limit Theorem for truncated heavy-tailed random

variables

The triangular array{Xn j: 1≤ jn} is as defined in (1.2). We would like to know if the row sumsSn, defined by

Sn:= n

X

j=1

Xn j, (2.1)

still converge in law after appropriate centering and scaling. Exactly same arguments as those in the proof of Theorem 2.1 in[Chakrabarty and Samorodnitsky, 2009] show that if the truncated heavy-tailed model is in the soft truncation regime as defined in (1.3), then

bn1(Snan) =⇒ V.

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the model is in the hard truncation regime,i.e., lim

n→∞nP(kHk>Mn) =∞. (2.2)

As mentioned earlier, easy-to-check criteria for satisfying the Central Limit Theorem on Banach spaces are not known. An example of the not-so-easy-to-check ones is Theorem 10.13, page 289 in

[Ledoux and Talagrand, 1991], known as the “small ball criterion”. The main result of this paper, Theorem 2.1, is an analogue of this theorem in the truncated setting under hard truncation. But before stating that, we need the following preliminary. It is known that (1.1) implies that there is a probability measureσon

S :={xB:kxk=1} such that as t−→ ∞,

P ‚

H

kHk ∈ ·

kHk>t

Œ

w

−→σ(·) (2.3)

weakly onS; see Corollary 6.20(b), page 151 in[Araujo and Giné, 1980].

Theorem 2.1. There is a Gaussian measureγon B such that

Bn1(SnESn)⇒γ (2.4)

if and only if the following hold:

1. (small ball criterion) For everyε >0 lim inf

n→∞ P(B −1

n kSnESnk< ε)>0 ,

2. supn1Bn1EkSnESnk<,

where

Bn:=

”

nMn2P(kHk>Mn)

—1/2

.

In that case, the characteristic function ofγis given by

ˆ

γ(f) =exp

‚

− 2 2−α

Z

S

f2(s)σ(ds)

Œ

,fB′. (2.5)

Here, Bis the dual of B, the space of linear functionals on B.

For the proof, we shall need the following one-dimensional lemma, which follows by exactly simi-lar arguments as those in Theorem 2.2 of[Chakrabarty and Samorodnitsky, 2009], and hence we omit the proof.

Lemma 2.1. For every f in B,

Bn−1(f(Sn)−E f(Sn))⇒N ‚

0, 2

2−α Z

S

f2(s)σ(ds)

Œ

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Proof of Theorem 2.1. First we prove the direct part,i.e., we assume that 1. and 2. hold. We first show that it suffices to check that{L(Zn)}is relatively compact where

Zn := Bn1 n

X

j=1 Yn j,

Yn j := Xn jXn j

and for everyn, Xn1,Xn2, . . . are i.i.d. copies of Xn1so that(Xn j : j ≥1)and(Xn j: j ≥1)are independent families. To see this, suppose that we have shown that{L(Zn)}is relatively compact. By Corollary 4.11, page 27 in[Araujo and Giné, 1980], it follows that the sequence{L(B−1

n Sn)} is relatively shift compact,i.e., there exists some sequence{vn} ⊂B such that {L(B−1

n Snvn)} is relatively compact. By Theorem 4.1 in de Acosta and Giné (1979), for relative compactness of {L[B−1

n (SnESn)]}, it suffices to check that lim

t→∞lim supn→∞ nE

kUnk1(kUnk>t)

=0 , (2.6)

where

Un:=Bn−1

H1(kHk ≤Mn) +

H

kHk(Mn+L)1(kHk>Mn))

. (2.7)

By (2.2), it follows thatBnMn. Thus, for fixedt>0 andnlarge enough,

nE

kUnk1(kUnk>t)

=nBn1P(kHk>Mn)

MnP(L>BntMn) +E

L1(L>BntMn) . SinceE L2<∞,

E

L1(L>BntMn)

E(L

2) BntMn and

MnP(L>BntMn)≤ Mn (BntMn)2

E L2=o(Bn1) asn−→ ∞. Thus, for all fixed t>0,

lim n→∞nE

kUnk1(kUnk>t)

=0 . (2.8)

This shows (2.6) and hence that{L[Bn1(SnESn)]}is relatively compact. In view of Lemma 2.1, this will complete the proof of the direct part.

First we record some properties of the random variables defined above, which shall be used in the proof. The hypotheses immediately imply that for allε >0

lim inf

n→∞ P(kZnk< ε)>0 (2.9)

and that

sup n≥1

EkZnk<∞. (2.10)

Let{Fk}be any sequence of increasing finite-dimensional subspaces so that

closure

[

k=1 Fk

!

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For any subspace F of B, denote byTF the canonical map fromBto the quotient space B/F. By Corollary 6.19 (page 151) in[Araujo and Giné, 1980], it follows that for everyk,TFk(H)is in the

domain of attraction of someα-stable law with the same scaling constant(bn)as that ofH, and that

lim k→∞supn1nP

€

kTFk(H)k>bn

Š

=0 . (2.12)

Clearly, for everyk, there isCk∈[0,∞)so that ast−→ ∞,

P(kTFk(H)k>t)∼CkP(kHk>t).

It follows by (2.12) that limk→∞Ck=0. Note that,

EkTFk(Xn1)k

2

= E”kTFk(H)k

21(kHk ≤M

n)

—

+E –

kTFk(H)k

2

kHk2 (Mn+L)21(kHk>Mn)

™

E”kTFk(H)k

21€

kTFk(H)k ≤Mn

Š—

+E –

kTFk(H)k

2

kHk2 1(kHk>Mn)

™

E(Mn+L)2.

By the Karamata theorem (Theorem B.1.5, page 363 in[de Haan and Ferreira, 2006]),

lim n→∞[M

2

nP(kHk>Mn)]−1E

€

kTFk(H)k21€kTFk(H)k ≤MnŠŠ= α 2−αCk.

By (2.3), it follows that asn−→ ∞,

E –

kTFk(H)k

2

kHk2 1(kHk>Mn)

™

P(kHk>Mn)

Z

S

kTFk(s)k

2σ(ds).

That (2.11) holds and the fact thatσis a finite measure implies that

lim k→∞

Z

S

kTFk(s)k

2σ(ds) =0 .

Thus, in view of the assumption thatE L2<∞, it follows that lim

k→∞lim supn→∞

[Mn2P(kHk>Mn)]−1EkTFk(Xn1)k

2=0 , (2.13)

which in turn implies that

lim k→∞lim supn→∞

[Mn2P(kHk>Mn)]−1EkTFk(Yn1)k

2=0 . (2.14)

Coming to the proof, in view of the criterion for relative compactness discussed in[Ledoux and Talagrand, 1991]

(page 40-41), it suffices to show that givenε >0, there is a finite dimensional subspaceF with lim sup

n→∞

P

kTF(Zn)k> ε

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Letǫ1,ǫ2, . . . be an i.i.d. sequence of Rademacher random variables, independent of(Xn,Xn,n≥ 1), and letdenote the conditional expectation given{Yn j}. It suffices to show that for allη >0,

lim

and that there is a numerical constantC>0 so that for everyδ >0, lim sup

whenever{Fk}is an increasing sequence of finite-dimensional subspaces satisfying (2.11). To establish (2.16), it suffices to check that

lim

β >0 is to be specified later. This is because fornlarge enough,

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asn−→ ∞. There isC∈(0,∞)so that

the equivalence in the second line following from Karamata’s theorem. This shows that

lim

and hence, showing (2.18) suffices for (2.16). Let

σn,F:=Bn−1 sup

By Theorem 4.7 in[Ledoux and Talagrand, 1991]on concentration of

Rademacher processes, with the median replaced by the expected value, as in page 292 of the same reference, it follows that

P

Thus all that needs to be shown is that given anyδ >0, there is a choice ofβdepending only on

δ, so that

which can be made as small as needed by (2.14). For the other part, note that by the contraction principle (Theorem 4.4 in[Ledoux and Talagrand, 1991]),

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For the proof of (2.17) we shall show that there is an universal constantC>0 so that whenever

F is a subspace satisfying

lim inf

it follows that

lim sup

The reason that this suffices is the following. Fix δ > 0 and a sequence of increasing finite-dimensional subspaces{Fk}satisfying (2.11). Note that for alln,k≥1,

P

By (2.16) and (2.9), it follows that

lim inf

By (2.20), (2.17) follows.

The proof of (2.20) uses an isoperimetric inequality; see Theorem 1.4 (page 26) in[Ledoux and Talagrand, 1991]. Let

In light of the isoperimetric inequality, by similar arguments as in page 291 of[Ledoux and Talagrand, 1991], it follows that fork,q≥1,

whereKis the universal constant in the isoperimetric inequality. Chooseq=2Kandkto be large enough (depending only onθ) so that

All that remains to be shown is

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Note that

max

jn kYn jk ≤maxjn kXn jk+maxjn k ˜

Xn jk and that

max

jn kXn jk ≤Mn+maxjn Lj.

SinceE L21<∞,{n−1/2maxjnLj}is a tight family. This shows (2.21) and thus establishes (2.20) withC=4q+1 and hence completes the proof of the direct part.

The converse is straightforward. For 1., note that if (2.4) holds, by the continuous mapping theorem,

lim n→∞P(B

−1

n kSnESnk ≤ε) =γ({xB:kxk ≤ε}),

the right hand side being positive because in a separable Banach space a centered Gaussian law puts positive mass on any ball with positive radius centered at origin, see the discussion on page 60-61 in [Ledoux and Talagrand, 1991]. For proving 2. we shall appeal to Theorem 4.2 in[de Acosta and Giné, 1979]. All that needs to be shown is

lim

t→∞lim supn→∞ nE

kξnk1(kξnk>t)

=0 , (2.22)

where

ξn:=UnE(Un) andUnis as defined in (2.7). Note that

EkUnk ≤ Bn−1

Mn+P(kHk>Mn)(Mn+E L)

→ 0 .

Fixt>0. Fornlarge enough so thatkE(Un)k<t/2, it follows that

nE

kξnk1(kξnk>t)

nE

kξnk1(kUnk>t/2)

nE

kUnk1(kUnk>t/2)

+nP(kUnk>t/2)EkUnk.

By (2.8), the first term goes to zero. For the second term, notice that fornlarge enough,

nP(kUnk>t/2)EkUnk ≤ nP(kHk>Mn)P 

L> t

2BnMn

‹ EkUnk = O€nP(kHk>Mn)Bn−2

Š

= o(1). This shows (2.22) and hence completes the proof.

Recall that a Banach spaceBis said to by of type 2 if there isC<∞so that for allN≥1 and zero mean independentB-valued random variablesX1, . . . ,XN,

E

n

X

j=1 Xj

2

C

N

X

j=1

EkXjk2.

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this is true. This is the statement of Theorem 10.5 (page 281) in[Ledoux and Talagrand, 1991]. We would like to mention at this point that while the assumption of type 2 is a rather restrictive one, this is a fairly large class. For example, every Hilbert space and Lp spaces for 2≤ p <∞ are Banach spaces of type 2. We show in the following result that (2.4) can be extended to these spaces.

Theorem 2.2. If B is of type 2 and the model with power law tails(1.2)is in the hard truncation regime, then there is a Gaussian measureγon B such that

Bn−1(SnESn)⇒γ

The characteristic function ofγis given by (2.5).

Proof. In view of Lemma 2.1 and using similar arguments as in the proof of Theorem 2.1, it suffices to prove that{L(Zn)}is relatively compact where the definition ofZn(andYn j) is exactly the same as in the proof of the latter theorem. Choose a sequence {Fk} of finite dimensional subspaces satisfying (2.11). Since B is of type 2, so isB/F for any closed subspace F, with the type 2 constant not larger than that ofB. Thus, there isC∈[0,∞)so that

EkTFk(Zn)k

2 C[M2

nP(kHk>Mn)]−1EkTFk(Yn1)k

2.

Using (2.14), it follows that limk→∞lim supn→∞EkTFk(Zn)k

2= 0 which shows (2.15) and thus

completes the proof.

3

Examples

In this section, we construct a couple of examples. In Example 1, the hypotheses of Theorem 2.1 can be verified. This helps to conclude that the result has some practical value. In Example 2, (2.4) does not hold, and hence there is a need for a result like Theorem 2.1 or Theorem 2.2.

Example 1.

Let{Tjk:j,k≥1}be i.i.d. R-valued symmetricα-stable (SαS) random variables with 0< α <2,

i.e., have the following characteristic function:

E[expiθT11] =e−|θ|α. For all j≥1, define theRN-valued random variableH

jas

Hj:=

X

k=1

akTjkek,

where(aj)is a sequence of non-negative numbers satisfying

X

j=1

aα/j 2<∞, (3.1)

andekis the element ofR

N

defined by

ek(n) =

1, k=n

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Recall thatP(|T11|>x) =O(xα); see Property 1.2.15, page 16 in[Samorodnitsky and Taqqu, 1994].

This ensures thatH1,H2, . . . are i.i.d. random variables taking values inc0, the space of sequences

limiting to zero, endowed with the sup norm. For that purpose, assuming thatP∞j=1j <∞would have been sufficient. However, we shall need (3.1) for other reasons. It is immediate thatH1, and

hence eachHj, is ac0valued symmetricα-stable random variable. This, in particular, means that

asn−→ ∞,

n−1

n

X

j=1

Hj=⇒H1. (3.3)

It is a well-known fact in finite dimensional spaces that the above implies

P(kH1k>x)∼C xα (3.4)

as x−→ ∞for someC∈(0,∞). However, since we could not find a reference for this on Banach spaces, we briefly sketch the argument for the sake of completeness. By Theorem 6.18, page 150 in

[Araujo and Giné, 1980]it follows that there is a measureµonB\{0}satisfyingµ(c D) =cαµ(D) for allc>0 andDB\ {0}, such that,

lim n→∞nP

€

n−1/αH1∈A Š

=µ(A) (3.5)

for allAB that is bounded away from the origin andµ(∂A) =0. It is also known that for all

δ >0, 0< µ({xB:kxk ≥δ})<∞. Set

A:={xB:kxk>1}. Clearly,µ(∂A) =0. Using (3.5) with thisAimplies that

C := limn→∞nP(kH1k > n1)exists, and is finite and positive. Let (xk) be any sequence of positive numbers going to infinity. Setnk :=⌊xkα⌋. Observe that x

α

kP(kH1k> xk)is sandwiched betweennkP(kH1k>(nk+1)1)and(n

k+1)P(kH1k>n 1

k ), and that both the bounds converge toC. Thus, (3.4) follows. The letterCwill be used to denote various such constants with possibly different definition throughout this section.

Let (Mn)be a sequence such that 1≪ Mnn1. Then, the truncation of H

j at level Mnwith

L≡0 is

Xn j:=

Hj

kHjk(kHjk ∧Mn). As before, define the row sum by

Sn:= n

X

j=1 Xn j.

We shall show that for this set up, the hypotheses of Theorem 2.1 can be verified by purely ele-mentary methods; the only sophisticated result that will be used is the contraction principle for finite dimensional spaces. All that needs to be shown is

lim inf n→∞ P(B

−1

n kSnk< ε)>0 for allε >0 , (3.6) and

sup n≥1

Bn1EkSnk<∞ (3.7)

where

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Note that in view of (3.4), this definition ofBndiffers from that in the statement of Theorem 2.1 by only a constant multiple in the limit.

The following is a sketch of how we plan to show (3.6). Define forK≥0,

SnK,1 := n

X

j=1

K

X

k=1 akekTjk

–

1(kHjk ≤Mn) +

Mn

kHjk1(kHjk>Mn)

™

,

SnK,2 := n

X

j=1

X

k=K+1 akekTjk

–

1(kHjk ≤Mn) +

Mn

kHjk1(kHjk>Mn)

™

.

We shall show that for allε >0,

sup K≥1

lim sup n→∞

P€Bn−1kSnK,1k> εŠ<1 , (3.8) and that

lim K→∞supn1B

−1

n EkS K,2

n k=0 . (3.9)

The reason that (3.8) and (3.9) suffice for (3.6) is the following. Fixε >0. Note that

Sn=SKn,1+S K,2

n ,

and hence it follows that

P(Bn−1kSnk> ε)≤P(Bn−1kS K,1

n k> ε/2) +P(B

−1

n kS K,2

n k> ε/2). Define

δ:=1−sup K≥1

lim sup n→∞

P€Bn1kSKn,1k> ε/2Š, which by (3.8) is positive. Using (3.9), chooseKto be large enough so that,

sup n≥1

Bn−1EkSnK,2k<εδ

2 .

Clearly, with this choice ofK, lim sup

n→∞

P€Bn1kSnK,1k> ε/2Š≤1−δ, and by the Markov inequality,

lim sup n→∞

P€Bn1kSnK,2k> ε/< δ. This shows (3.6).

Forn,K≥1, define

Un,K := n

X

j=1

K

X

k=1 akekTjk

–

1(ak|Tjk| ≤Mn) + Mn

ak|Tjk|

1(ak|Tjk|>Mn)

™

.

We start with showing thatSK,1

n is stochastically bounded byUn,K,i.e., for all x>0,

P”kSKn,1k>x—≤2P”kUn,Kk>x

—

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To that end, let(ǫjk:j,k≥1)be a family of i.i.d. Rademacher random variables, independent of

Using Theorem 4.4, page 95 in[Ledoux and Talagrand, 1991], it follows that

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whereGis a normal random variable with mean zero and varianceσ2. Thus, (3.8) will follows if

the following is shown: for allη >0,

Π∞k=1P(|G| ≤akα/2η)>0 . Clearly, it suffices to show that

X

k=1

P(|G|>akα/2η)<∞,

which immediately follows from the Markov inequality along with (3.1). Thus, (3.8) follows. For showing (3.9), note that for alln≥1 andK≥0,

Bn1EkSnK,2k ≤

X

k=K+1 E

Bn−1

n

X

j=1 akTjk

n

1(kHjk ≤Mn) + Mn kHjk

1(kHjk>Mn)

o

X

k=K+1 E1/2

– Bn1

n

X

j=1 akTjk

n

1(kHjk ≤Mn) +

Mn

kHjk1(kHjk>Mn)

o ™2

=

X

k=K+1

Bn−1n1/2E1/2 –

ak|T1k|

n

1(kH1k ≤Mn)

+ Mn kH1k

1(kH1k>Mn)

o™2

X

k=K+1

Bn1n1/2E1/2 –

ak|T1k|

n

1(ak|T1k| ≤Mn) (3.12)

+ Mn

ak|T1k|

1(ak|T1k|>Mn)

o ™2

C

X

k=K+1

aα/k 2 (3.13)

for some C < ∞ independent of n andK, where (3.11) has been used for (3.12), and (3.13) follows by Karamata Theorem, the estimation being similar to that leading to (2.13). This, in view of (3.1), shows (3.9). Thus, (3.6) follows. Also, using (3.13) forK=0, (3.7) follows. Thus, the hypotheses of Theorem 2.1 are satisfied.

Example 2

Fix 1< p <2. We first construct a bounded symmetric random variable X taking values in c0

(the space of sequences limiting to zero, equipped with the sup norm) so thatn−1/pPni=1Xidoes not converge to zero in probability, where X1,X2, . . . are i.i.d. copies ofX. Let(ǫj : j≥1)be a sequence of i.i.d. Rademacher random variables. We shall use the fact that there existsK∈(0,∞) so that

P

n

X

i=1 ǫi>t

!

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for all n ≥ 1 and t > 0 such that n1/2KtK−1n. This follows from (4.2) on page 90 in

[Ledoux and Talagrand, 1991]. Define

X:=

K is the constant in (3.14) and ej is as defined in (3.2). Clearly, X is a c0 valued symmetric

bounded random variable. LetX1,X2, . . . denote i.i.d. copies ofX. Note that forn≥1,

k=1Xk does not converge to zero in probability, it suffices to show that

To that aim, define

ln:=

š

exp€n2/pŠ,n≥1 , and note that

Π∞

Using (3.16), it follows that

n2/p−1al2

n =o(logln).

Thus, fornlarge enough it holds that

K n2/p−1al2

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and hence for such an,

What we have shown can be summed up as that fornlarge enough,

Π∞j=1P

Thus, (3.15) follows. FixxB\ {0}and define

Y :=X1(U=0) +xS1(U=1) (3.17)

whereS is a R-valued (symmetric) Cauchy random variable and U is a Bernoulli(1/2)random variable such thatX,S,Uare all independent. We start with showing thatY is in the domain of attraction of an 1-stable law onB. Let((Xi,Si,Ui):i≥1)denote i.i.d. copies of(X,S,U). SinceX has zero mean, by Theorem 9.21 in[Ledoux and Talagrand, 1991], it follows that

n−1

n

X

i=1

Xi−→P 0 .

We shall show by an application of the contraction principle (Theorem 4.4 in[Ledoux and Talagrand, 1991]) that

the inequality following by the contraction principle. This shows (3.18). By Theorem 3 on page 580 in[Feller, 1971], it follows that

for some Cauchy random variableZ. Thus, it is immediate that

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For a positive numberMn,

is the truncation ofYito the ball of radiusMn, as defined in (1.2) withLidentically equal to zero. Let similar to those leading to (3.18), it follows that foru>0,

P

Note that sinceX is bounded andMngoes to infinity, fornlarge enough,

Y1(kYk ≤Mn) =X1(U=0) +xS1(U=1)1(|S| ≤Mn/kxk).

Observing that if(ǫ1,ǫ2)are i.i.d. Rademacher random variables independent of(X,S,U), then

X1(U=0) +xS1(U=1)1(|S| ≤Mn/kxk) d

=ǫ1X1(U=0) +2|S|1(U=1)1(|S| ≤Mn/kxk), exactly same arguments as before will show that fornlarge enough andu>0,

P

The above can be summarized as that there existsN<∞so that

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Another application of the contraction principle shows that on the set{Nn>n/3},

PU

Nn

X

i=1 Xi

>u

!

≥ 1 2PU

n/3⌉

X

i=1 Xi

>u

!

= 1 2P

n/3⌉

X

i=1 Xi

>u

!

.

Thus, it follows that

P

n

X

i=1

Xi1(Ui=0)

>u

!

≥ 1 2P

n/3⌉

X

i=1 Xi

>u

!

P(Nn>n/3).

All the above calculations put together shows

P

n/3⌉

X

i=1 Xi

>u

!

=O P(kSnk>u)

uniformly inu. Sincen−1/pPni=1Xidoes not converge to zero in probability, it follows thatn−1/pSn does not converge to 0 in probability either.

The above calculations can be used to construct an example where (2.4) does not hold, in the following way. Fix 1< p <2 and a sequence (Mn)satisfying 1 ≪ Mnn2/p−1. DefineY by

(3.17). The argument that leads to (3.4) from (3.3), applied toY helps us conclude from (3.19) that P(kYk> x)∼ C x−1 as x −→ ∞, for someC ∈(0,∞). Since 2/p−1<1, it follows that

Mnn, which is a restatement of lim

n→∞nP(kYk>Mn) =∞.

Thus, the assumption of hard truncation is satisfied. Set L ≡0, Yni to be the truncation ofYi at levelMn,Snto be the row sum of the triangular array{Yni: 1≤in}and

Bn:=”nMn2P(kYk>Mn)—1/2. Thus,

B2n=O(nMn) =o

€ n2/pŠ.

This shows thatBn1Sndoes not converge weakly, for otherwise,n−1/pSnwould converge to zero in probability. Thus, (2.4) does not hold.

This is an example where the claim of Theorem 2.2 does not hold. The space c0 is not of Rademacher typepfor allp>1. Hence it was possible to construct a zero mean random variable with finitep-th moment, that does not satisfy the law of large numbers with raten1/p.

4

Acknowledgment

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References

[Araujo and Giné, 1980] Araujo, A. and Giné, E. (1980). The Central Limit Theorem for Real and Banach Valued Random Variables. Wiley, New York. MR0576407

[Chakrabarty and Samorodnitsky, 2009] Chakrabarty, A. and Samorodnitsky, G. (2009). Under-standing heavy tails in a bounded world or, is a truncated heavy tail heavy or not? Preprint available athttp://arxiv.org/pdf/1001.3218.

[de Acosta and Giné, 1979] de Acosta, A. and Giné, E. (1979). Convergence of moments and related functionals in the general central limit theorem in banach spaces. Z. Wahr. verw. Geb., 48:213–231. MR0534846

[de Haan and Ferreira, 2006] de Haan, L. and Ferreira, A. (2006). Extreme Value Theory: An Introduction. Springer, New York. MR2234156

[Feller, 1971] Feller, W. (1971). An Introduction to Probability Theory and its Applications, vol-ume 2. Wiley, New York, 2nd edition. MR0270403

[Ledoux and Talagrand, 1991] Ledoux, M. and Talagrand, M. (1991). Probability in Banach Spaces: Isoperimetry and Processes. Springer Verlag. MR1102015

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