Yet another proof of the Nualart-Peccati criterion

by Ivan Nourdin

∗†

Université Nancy 1

This version: December15th, 2011

Abstract: In[14],NualartandPeccatishowedthat,surprisingly,theconvergence indistribution

of a normalized sequence of multiple Wiener-Itô integrals towards a standard Gaussian law is

equivalent to convergence of just the fourth moment to 3. In [3], this result is extended to a

sequence of multiple Wigner integrals, in the context of freeBrownian motion. The goal of the

present paper isto oer an elementary, unifyingproof of these two results. The only advanced,

neededtoolisthe productformulafor multipleintegrals. Apartfromthisformula,therestofthe

proofonly relies on softcombinatorial arguments.

Keywords: Brownian motion; free Brownian motion; multiple Wiener-Itô integrals; multiple

Wignerintegrals; Nualart-Peccaticriterion; productformula.

1 Introduction

Thefollowingsurprisingresult,provedin[14 ],showsthattheconvergenceindistributionofa

nor-malized sequenceof multiple Wiener-Itôintegrals towards a standard Gaussian lawis equivalent

to convergence ofjust the fourthmoment to 3.

Theorem 1.1 (Nualart-Peccati) Fix an integer

p > 2

. Let

{B(t)}

t∈[0,T ]

be a classical Brown-ian motion,and let

(F

n

)

n>1

be a sequence of multiple integrals of the form

F

n

=

Z

[0,T ]

p

f

n

(t

1

, . . . , t

p

)dB(t

1

) . . . dB(t

p

),

(1.1) where each

f

n

∈ L

2

_{([0, T ]}

p

_;

_R)

issymmetric(itisnotarestrictiveassumption). Supposemoreover

that

E[F

2

n

]

→ 1

as

n

→ ∞

. Then, as

n

→ ∞

, the following two assertions are equivalent: (i) The sequence

(F

n

)

converges in distribution to

B(1)

∼ N(0, 1)

;

(ii)

E[F

4

n

]

→ E[B(1)

4

] = 3

.

In[14],theoriginalproofof

(ii)

⇒ (i)

reliesontoolsfromBrownianstochasticanalysis. Precisely, using thesymmetryof

f

n

,onecan rewrite

F

n

as

F

n

= p!

Z

T

0

dB(t

1

)

Z

t

1

0

dB(t

2

) . . .

Z

t

p−1

0

dB(t

p

)f

n

(t

1

, . . . , t

p

),

∗

Institut Élie Cartan, Université Henri Poincaré, BP 70239, 54506 Vandoeuvre-lès-Nancy, France,

inourdin@gmail.com

†

supported in part by the two following (french) ANR grants: `Exploration des Chemins Rugueux'

[ANR-09-BLAN-0114]and `Malliavin, Steinand Stochastic EquationswithIrregular Coecients'

andthenmakeuseoftheDambis-Dubins-Schwarztheoremtotransformitinto

F

n

= β

(n)

hF

n

i

,where

β

(n)

isa classicalBrownian motionand

hF

n

i = p!

2

Z

T

0

dt

1

Z

t

1

0

dB(t

2

) . . .

Z

t

p−1

0

dB(t

p

)f

n

(t

1

, . . . , t

p

)



2

.

(1.2)

Therefore,to getthat

(i)

holds true, itisnowenough to prove that

(ii)

implies

hF

n

i

L

2

→ 1

,which isexactlywhat Nualartand Peccati didin[14 ].

Since the publication of [14 ], several researchers have been interested inunderstanding more

deeplywhyTheorem 1.1holds. Letus mentionsome worksinthis direction:

1. In[13 ],NualartandOrtiz-Latorregave anotherproofofTheorem 1.1usingexclusively the

toolsof Malliavin calculus. The mainingredient oftheir proofis theidentity

δD =

−L

,where

δ

,

D

and

L

arebasic operators inMalliavincalculus.

2. Based on the ideas developed in[7 ],thefollowing boundis shownin[8 , Theorem 3.6](see

also [11 ]): if

E[F

2

n

] = 1

,then

sup

A∈B(

R)







P [F

n

∈ A] −

1

√

2π

Z

A

e

−u

2

/2

du









6

2

r

p

_{− 1}

3p

p

|E[F

4

n

]

− 3|.

(1.3)

Of course,with (1.3) in hand, itis totally straightforward to obtain Theorem 1.1 asa corollary.

However, the proof of (1.3), albeit not that dicult, requires the knowledge of both Malliavin

calculus andStein's method.

3. ByusingthetoolsofMalliavincalculus, PeccatiandIcomputedin[9]anewexpressionfor

thecumulants of

F

n

,in termsof thecontractions of thekernels

f

n

. Asan immediate byproduct ofthis formula, we areableto recover Theorem1.1, see[9, Theorem5.8] for thedetails. Seealso

[5] for anextension inthe multivariatesetting.

4. In[6 ],Theorem1.1isextendedto thecasewhere, insteadof

B(1)

∼ N(0, 1)

inthelimit,a centeredchi-squarerandomvariable,say

Z

,isconsidered. Moreprecisely,itisprovedinthislatter reference that an adequably normalized sequence

F

n

of the form (1.1) converges in distribution towards

Z

ifand only if

E[F

4

n

]

− 12E[F

n

3

]

→ E[Z

4

]

− 12E[Z

3

]

. Hereagain, theproof isbasedon theuseof thebasicoperatorsof Malliavin calculus.

5. Thefollowingresult,provedin[3],istheexactanalogueofTheorem1.1,butinthesituation

where theclassical Brownian motion

B

isreplacedbyits free counterpart

S

.

Theorem 1.2 (Kemp-Nourdin-Peccati-Speicher) Fix an integer

p > 2

. Let

{S(t)}

t∈[0,T ]

be a free Brownian motion,and let

(F

n

)

n>1

be a sequence of multiple integrals of the form

F

n

=

Z

[0,T ]

p

f

n

(t

1

, . . . , t

p

)dS(t

1

) . . . dS(t

p

),

where each

f

n

∈ L

2

_{([0, T ]}

p

_;

_R)

ismirrorsymmetric(thatis,satises

f

n

(t

1

, . . . , t

p

) = f

n

(t

p

, . . . , t

1

)

for all

t

1

, . . . , t

p

∈ [0, 1]

). Suppose moreover that

E[F

2

(i) For all

k > 3

,

E[F

k

n

]

→ E[S(1)

k

]

; (ii)

E[F

4

n

]

→ E[S(1)

4

] = 2

.

TheproofofTheorem1.2containedin[3]isbasedontheuseofcombinatorial featuresrelatedto

thefreeprobabilityrealm, including non-crossingpairing and partitions.

Thus, there isalready several proofsof Theorem 1.1. Each of them hasits own interest,

be-causeitallowstounderstandmoredeeplyaparticularaspectofthisbeautifulresult. Ontheother

hand, all these proofs require at some point to deal with sophisticated tools, such as stochastic

Brownian analysis, Malliavin calculus orStein's method.

The goalofthispaperistooer anelementary,unifyingproofof both Theorems 1.1and1.2.

Asanticipated,theonlyadvancedresultwewillneedistheproductformulaformultipleintegrals,

thatis, theexplicitexpression for theproductoftwo multiples integrals oforder

p

and

q

,say,as alinearcombinationofmultipleintegrals oforderlessor equalto

p + q

. Apartfromthis formula, therest ofthe proof only relieson `soft'combinatorial arguments.

Thelevelofourpaperis(hopefully)availabletoanygoodstudent. Fromouropinionhowever,

its interestis not only to provide a new, simple proofof a knownresult. Itis indeed noteworthy

thatthe numberof required tools hasbeen reduced to its maximum(the productformula being

essentially the only one we need), so that our approach might represent a valuable strategy to

follow in order to generalize Theorem(s) 1.1 (and 1.2) in other situations. For instance, let us

mentionthatthetwoworks[10,2]haveindeedfollowedour lineofreasoning, andsuccessfully

ex-tendedTheorem1.2inthecasewherethelimitisthefreePoissondistributionandthe(so-called)

tetilla lawrespectively.

The rest of the paperis organized asfollows. Section 2 deals with some preliminary results.

Section3containsourproofofTheorem1.2,whereasSection4isdevotedtotheproofofTheorem

1.1.

2 Preliminaries

2.1 Multiple integrals with respect to classical Brownian motion

Inthissection,our mainreferenceisNualart'sbook[12]. Tosimplifytheexposition, withoutloss

of generality we xthe timehorizon tobe

T = 1

. Let

{B(t)}

t∈[0,1]

be a classical Brownian motion, that is, a stochastic process dened on a probabilityspace

(Ω, F , P )

,startingfrom 0,withindependent increments, and suchthat

B(t)

−

B(s)

isa centered Gaussian randomvariablewithvariance

t

− s

for all

t > s

. For a given real-valued kernel

f

belonging to

L

2

_{([0, 1]}

p

₎

, let us quickly sketch out the

con-struction ofthe multiple Wiener-Itôintegral of

f

with respectto

B

,written

I

p

(f ) =

Z

[0,1]

p

D

p

_{⊂ [0, 1]}

p

bethe collection ofall diagonals, i.e.

D

p

=

_{(t

1

, . . . , t

p

)

∈ [0, 1]

p

: t

i

= t

j

for some

i

6= j}.

(2.5) Asarststep, when

f

hastheform ofacharacteristic function

f = 1

A

,with

A = [u

1

, v

1

]

× . . . ×

[u

p

, v

p

]

⊂ [0, 1]

p

such that

A

∩ D

p

₌

_∅

,the

p

th multipleintegral of

f

is dened by

I

p

(f ) = (B(v

1

)

− B(u

1

)) . . . (B(v

p

)

− B(u

p

)).

Then, this denition is extended by linearity to simple functions of the form

f =

P

k

i=1

α

i

1

A

i

,

where

A

i

= [u

i

1

, v

1

i

]

× . . . × [u

i

p

, v

i

p

]

are disjoint

p

-dimensional rectangles as above which do not meet thediagonals. Simplecomputationsshow that

E[I

p

(f )] = 0

(2.6)

I

p

(f ) = I

p

( e

f )

(2.7)

E[I

p

(g)I

p

(f )] = p!

heg, e

f

i

L

2

_([0,1]

p

₎

.

(2.8)

Here,

f

e

∈ L

2

_{([0, 1]}

p

₎

denotesthesymmetrizationof

f

,thatis,thesymmetricfunctioncanonically associatedto

f

,given by

e

f (t

1

, . . . , t

p

) =

1

p!

X

π∈S

p

f (t

_π(1)

, . . . , t

_π(p)

).

(2.9) Since each

f

∈ L

2

_{([0, 1]}

p

₎

can be approximated in

L

2

-norm by simple functions, we can nally

extend thedenition of(2.4) to all

f

∈ L

2

_{([0, 1]}

p

₎

. Notethat, byconstruction, (2.6)-(2.8) isstill

trueinthis general setting. Then, one easily seesthat, inaddition,

E[I

p

(f )I

q

(g)] = 0

for any

p

6= q

,

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

_).

(2.10)

Before being inposition to state theproduct formula for two multiple integrals, we need to

introduce thefollowing quantity.

Denition 2.1 For symmetric functions

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

₎

, thecontractions

f

_⊗

r

g

∈ L

2

([0, 1]

p+q−2r

)

(

0 6 r 6 min(p, q)

) are the (notnecessarily symmetric) functionsgiven by

f

_⊗

r

g(t

1

, . . . , t

p+q−2r

) :=

Z

[0,1]

r

f (t

1

, . . . , t

p−r

, s

1

, . . . , s

r

)g(t

p−r+1

, . . . , t

p+q−2r

, s

1

, . . . , s

r

)ds

1

. . . ds

r

.

By convention, we set

f

⊗

0

g = f

⊗ g

, the tensor product of

f

and

g

.

The symmetrization of

f

⊗

r

g

is written

f e

⊗

r

g

. Observe that

f

⊗

p

g = f e

⊗

p

g =

hf, gi

L

2

_([0,1]

p

₎

whenever

p = q

. Also,using Cauchy-Schwarz inequality, itisimmediate to prove that

for all

r = 0, . . . , min(p, q)

. (Itisactually an equalityfor

r = 0

.) Moreover, a simple application of the triangleinequality leadsto

kf e

⊗

r

g

k

L

2

_([0,1]

p+q−2r

₎

6

kf ⊗

r

g

k

L

2

_([0,1]

p+q−2r

₎

.

Wecan nowstate theproduct formula,which isthemain ingredient ofour proof ofTheorem

1.1. Bytakingthe expectationin(2.11),observethat we recoverboth (2.8)and (2.10).

Theorem 2.2 For symmetric functions

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

₎

,we have

I

p

(f )I

q

(g) =

min(p,q)

_X

r=0

r!



p

r



q

r



I

p+q−2r

(f e

⊗

r

g).

(2.11)

2.2 Multiple integrals with respect to free Brownian motion

In this section, our main references are:

(i)

the monograph [4] by Nica and Speicher for the generalities about freeprobability;

(ii)

thepaper[1 ] byBianeandSpeicherforthefreestochastic analysis. We referthe reader to themfor anyunexplained notionor result.

Let

{S(t)}

t∈[0,1]

be a free Brownian motion, that is, a stochastic process dened on a non-commutative probability space

(A , E)

,starting from 0,withfreelyindependent increments, and such that

S(t)

− S(s)

isacentered semicircularrandom variablewithvariance

t

− s

for all

t > s

. We maythink offreeBrownian motionas`innite-dimensional matrix-valuedBrownian motion'.

For more details about the construction andfeatures of

S

,see [1 , Section1.1] and thereferences therein.

When

f

∈ L

2

_{([0, 1]}

p

₎

isreal-valued, wewrite

f

∗

toindicatethefunctionof

L

2

_{([0, 1]}

p

₎

givenby

f

∗

(t

1

, . . . , t

p

) = f (t

p

, . . . , t

1

)

. (Hence, to say that

f

n

is mirror-symmetric in Theorem 1.2 means that

f

n

= f

∗

n

.) We quickly sketch out theconstruction ofthe multiple Wigner integral of

f

with respect to

S

. Let

D

p

_{⊂ [0, 1]}

p

be the collection of all diagonals, see (2.5). For a characteristic

function

f = 1

A

,where

A

⊂ [0, 1]

p

hastheform

A = [u

1

, v

1

]

× . . . × [u

p

, v

p

]

with

A

∩ D

p

₌

_∅

,the

p

th multiple Wignerintegralof

f

,written

I

p

(f ) =

Z

[0,1]

p

f (t

1

, . . . , t

p

)dS(t

1

) . . . dS(t

p

),

isdened by

I

p

(f ) = (S(v

1

)

− S(u

1

)) . . . (S(v

p

)

− S(u

p

)).

Then, asinthe previous sectionwe extend this denition by linearityto simple functionsof the

form

f =

P

k

i=1

α

i

1

A

i

,

where

A

i

= [u

i

1

, v

i

1

]

× . . . × [u

i

p

, v

i

p

]

aredisjoint

p

-dimensional rectangles as above which donot meet thediagonals. Simplecomputationsshow that

E[I

p

(f )] = 0

(2.12)

E[I

p

(f )I

p

(g)] =

hf, g

∗

i

L

2

_([0,1]

p

₎

.

(2.13)

By approximation, the denition of

I

p

(f )

is extended to all

f

∈ L

2

_{([0, 1]}

p

₎

, and (2.12)-(2.13)

continue to holdtrue inthis more general setting. Itturns out that

E[I

p

(f )I

q

(g)] = 0

for

p

6= q

,

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

_).

(2.14)

Before giving the product formula inthe free context,we need to introduce theanalogue for

Denition 2.3 For functions

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

₎

,the contractions

f

_ g

r

∈ L

2

([0, 1]

p+q−2r

)

(

0 6 r 6 min(p, q)

)

are the functionsgiven by

f

_ g(t

r

1

, . . . , t

p+q−2r

) :=

Z

[0,1]

r

f (t

1

, . . . , t

p−r

, s

1

, . . . , s

r

)g(s

r

, . . . , s

1

, t

p−r+1

, . . . , t

p+q−2r

)ds

1

. . . ds

r

.

By convention, we set

f

0

_ g = f

⊗ g

, the tensorproduct of

f

and

g

. Observethat

f

p

_ g =

hf, g

∗

_i

L

2

_([0,1]

p

₎

whenever

p = q

. Also,using Cauchy-Schwarz, itis immedi-ate to prove that

kf

r

_ g

k

L

2

_([0,1]

p+q−2r

₎

6

kfk

_L

2

_([0,1]

p

₎

kgk

_L

2

_([0,1]

q

₎

for all

r = 0, . . . , min(p, q)

. (It isactually an equalityfor

r = 0

.)

We can now state the product formula in the free context, which turns out to be simpler

compared to the classical case(Theorem 2.2).

Theorem 2.4 For functions

f

∈ L

2

_{([0, 1]}

p

₎

and

g

∈ L

2

_{([0, 1]}

q

₎

,we have

I

p

(f )I

q

(g) =

min(p,q)

_X

r=0

I

p+q−2r

(f

_ g).

r

(2.15) 3 Proof of Theorem 1.2

Let the notation and assumptions of Theorem 1.2 prevail. Without loss of generality, we may

assume that

E[F

2

n

] = 1

for all

n

(instead of

E[F

2

n

]

→ 1

as

n

→ ∞

). Moreover, because

f

n

= f

∗

n

, observe that

kf

n

k

2

L

2

_([0,1]

p

₎

= E[F

n

2

] = 1

.

Itistrivialthat

(i)

implies

(ii)

. Conversely,assumethat

(ii)

isinorder, andletusprove that

(i)

holds. Fixaninteger

k > 3

. Iterative applicationsof theproductformula(2.15) leads to

F

_n

k

= I

p

(f

n

)

k

=

X

(r

1

,...,r

_k−1

)∈A

k

I

kp−2r

1

−...−2r

_k−1

f

n

r

1

_ . . .

r

_ f

k−1

n

,

(3.16) where

A

k

=



(r

1

, . . . , r

k−1

)

∈ {0, 1, . . . , p}

k−1

: r

2

6

2p

− 2r

1

, r

3

6

3p

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

(k

− 1)p − 2r

1

− . . . − 2r

k−2

.

Inordertosimplifytheexposition, notethatwehaveremovedthebracketsinthewritingof

f

n

r

1

_

. . .

r

k−1

_ f

n

. Weusetheimplicitconventionthatthesequantitiesarealwaysdenediterativelyfrom thelefttotheright. Forinstance,

f

n

r

1

_ f

n

_ f

r

2

n

_ f

r

3

n

actuallystandsfor

((f

n

r

1

_ f

n

)

_ f

r

2

n

)

_ f

r

3

n

. By takingthe expectationin(3.16),wededucethat

E[F

_n

k

] =

X

(r

1

,...,r

_k−1

)∈B

k

f

n

r

1

with

B

k

=



(r

1

, . . . , r

k−1

)

∈ A

k

: 2r

1

+ . . . + 2r

k−1

= kp

. We decompose

B

k

as

C

k

∪ E

k

, with

C

k

= B

k

∩ {0, p}

k−1

and

E

k

= B

k

\ C

k

. We thenhave,for all

k > 3

,

E[F

_n

k

] =

X

(r

1

,...,r

_k−1

)∈C

k

f

n

_ . . .

r

1

r

k−1

_ f

n

+

X

(r

1

,...,r

_k−1

)∈E

k

f

n

_ . . .

r

1

r

k−1

_ f

n

.

(3.18)

Lemmas 3.2 and 3.4 imply together thatthe rst sum in (3.18) is equal to

E[S(1)

k

_]

. Moreover,

by Lemma 3.1 and because

(ii)

is in order, we have that

kf

n

r

_ f

n

k

L

2

_([0,1]

2

p−2r

₎

→ 0

for all

r = 1, . . . , p

− 1

. Hence, thesecondsumin(3.18) mustconverge tozerobyLemma3.5. Thus,

(i)

isin order, and the proofof the theoremis concluded.

2

Lemma 3.1 We have

E[F

4

n

] = 2 +

P

p−1

r=1

kf

n

_ f

r

n

k

2

_L

2

_([0,1]

2

p−2r

₎

. Proof. The product formula (2.15) yields

F

2

n

=

P

p

r=0

I

2p−2r

(f

n

_ f

r

n

).

Using (2.13)-(2.14), we infer

E[F

_n

4

] =

_kf

n

⊗ f

n

k

2

L

2

_([0,1]

2p

₎

+

kf

n

k

2

L

2

_([0,1]

p

₎

2

+

p−1

X

r=1

hf

n

_ f

r

n

, (f

n

_ f

r

n

)

∗

i

L

2

_([0,1]

2

_p−2r

₎

= 2

kf

n

k

4

_L

2

_([0,1]

p

₎

+

p−1

X

r=1

kf

n

_ f

r

n

k

2

_L

2

_([0,1]

2

p−2r

₎

= 2 +

p−1

X

r=1

kf

n

_ f

r

n

k

2

_L

2

_([0,1]

2

p−2r

₎

,

since

kf

n

k

2

L

2

_([0,1]

p

₎

= 1

and

f

n

_ f

r

n

(t

1

, . . . , t

2p−2r

)

=

Z

[0,1]

r

f

n

(t

1

, . . . , t

p−r

, s

1

, . . . , s

r

)f

n

(s

r

, . . . , s

1

, t

p−r+1

, . . . , t

2p−2r

)ds

1

. . . ds

r

=

Z

[0,1]

r

f

n

(s

r

, . . . , s

1

, t

p−r

, . . . , t

1

)f

n

(t

2p−2r

, . . . , t

p−r+1

, s

1

, . . . , s

r

)ds

1

. . . ds

r

= f

n

_ f

r

n

(t

2p−2r

, . . . , t

1

) = (f

n

_ f

r

n

)

∗

(t

1

, . . . , t

2p−2r

).

2

Lemma 3.2 For all

k > 3

, the cardinality of

C

k

coincides with

E[S(1)

k

_]

.

Proof. Bydividing all the

r

i

'sby

p

,one getthat

C

k

bij.

≡

C

e

k

:=



(r

1

, . . . , r

k−1

)

∈ {0, 1}

k−1

: r

2

6

2

− 2r

1

, r

3

6

3

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

k

− 1 − 2r

1

− . . . − 2r

k−2

, 2r

1

+ . . . + 2r

k−1

= k

.

Ontheotherhand,considertherepresentation

S(1) = I

1

(1

[0,1]

)

. Asabove,iterativeapplications of the productformula(2.15) leads to

e

A

k

=



(r

1

, . . . , r

k−1

)

∈ {0, 1}

k−1

: r

2

6

2

− 2r

1

, r

3

6

3

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

k

− 1 − 2r

1

− . . . − 2r

k−2

.

By takingtheexpectation, we deducethat

E[S(1)

k

] =

X

(r

1

,...,r

k−1

)∈ e

C

k

1

[0,1]

r

1

_ . . .

r

_ 1

k−1

[0,1]

=

X

(r

1

,...,r

k−1

)∈ e

C

k

1 = # e

C

k

= #C

k

.

2

Remark 3.3 When

k

is even, it is well-known that

E[S(1)

k

_]

is given by

Cat

k/2

, the Catalan number of order

k/2

. There is manycombinatorial ways to dene this number. One of them is to see it at the number of paths inthe lattice

Z

2

which start at

(0, 0)

, end at

(k, 0)

, make steps of the form

(1, 1)

or

(1,

−1)

,and neverliesbelow the

x

-axis, i.e., all their pointsareof the form

(i, j)

with

j > 0

.

Letthe notation oftheproof ofLemma 3.2 prevail. Set

s

i

= 1

− 2r

i

. Then

e

C

k

bij.

≡



(s

1

, . . . , s

k−1

)

∈ {−1, 1}

k−1

: 1 + s

1

>

1

2

(1

− s

2

), 1 + s

1

+ s

2

>

1

2

(1

− s

3

),

. . . , 1 + s

1

+ . . . + s

k−2

>

1

2

(1

− s

k−1

), 1 + s

1

+ . . . + s

k−1

= 0



.

It turnsout thatthe set ofconditions







s

j

∈ {−1, 1},

j = 1, . . . , k

− 1

1 + s

1

+ . . . + s

j

>

1

₂

(1

− s

j+1

),

j = 1, . . . , k

− 2

1 + s

1

+ . . . + s

k−1

= 0,

(3.19) isequivalent to







s

j

∈ {−1, 1},

j = 1, . . . , k

− 1

1 + s

1

+ . . . + s

j

>

0,

j = 1, . . . , k

− 2

1 + s

1

+ . . . + s

k−1

= 0.

(3.20)

Indeed, it isclear that(3.19) implies(3.20). Conversely, suppose that(3.20) is inorder, and let

j

∈ {1, . . . , k − 2}

. Because

1

2

(1

− s

j+1

) 6 1

,one has that

1 + s

1

+ . . . + s

j

>

1

2

(1

− s

j+1

)

when

1 + s

1

+ . . . + s

j

>

1

. If

1 + s

1

+ . . . + s

j

= 0

then, because

1 + s

1

+ . . . + s

j+1

>

0

(even if

j = k

_{− 2}

),one has

s

j+1

= 1

,implying inturn

1 + s

1

+ . . . + s

j

>

1

2

(1

− s

j+1

) = 0

. Thus

e

C

k

bij.

≡

n

(s

1

, . . . , s

k−1

)

∈ {−1, 1}

k−1

: 1 + s

1

>

0, 1 + s

1

+ s

2

>

0,

. . . , 1 + s

1

+ . . . + s

k−2

>

0, 1 + s

1

+ . . . + s

k−1

= 0



,

and we recoverthe resultof Lemma 3.2when

k

is even. (The casewhere

k

isoddis trivial.)

Lemma 3.4 We have

f

n

r

1

Proof. Itis evident,using the identities

f

n

0

_ f

n

= f

n

⊗ f

n

and

f

n

p

_ f

n

=

Z

[0,1]

p

f

n

(t

1

, . . . , t

p

)f

n

(t

p

, . . . , t

1

)dt

1

. . . dt

p

=

kf

n

k

2

L

2

_([0,1]

p

₎

= 1.

2

Lemma 3.5 As

n

→ ∞

, assume that

kf

n

r

_ f

n

k

L

2

_([0,1]

2

p−2r

₎

→ 0

for all

r = 1, . . . , p

− 1

. Then, as

n

→ ∞

we have

f

n

r

1

_ . . .

r

_ f

k−1

n

→ 0

for all

k > 3

and all

(r

1

, . . . , r

k−1

)

∈ E

k

.

Proof. Fix

(r

1

, . . . , r

k−1

)

∈ E

k

, and let

j

∈ {1, . . . , k − 1}

be the smallest integer such that

r

j

∈ {1, . . . , p − 1}

. Recallthat

f

n

0

_ f

n

= f

n

⊗ f

n

. Then



f

n

_ . . .

r

1

r

k−1

_ f

n





=



f

n

_ . . .

r

1

r

_j−1

_ f

n

r

j

_ f

n

r

j+1

_ . . .

r

_ f

k−1

n





=



(f

n

⊗ . . . ⊗ f

n

)

r

j

_ f

n

r

j+1

_ . . .

r

k−1

_ f

n





(using

f

n

p

_ f

n

= 1

)

6

_k(f

_n

_{⊗ . . . ⊗ f}

_n

)

_{⊗ (f}

n

r

j

_ f

n

)

k

L

2

_([0,1]

q

₎

kf

_n

k

k−j−1

L

2

_([0,1]

p

₎

(byCauchy-Schwarz, for a certain

q

)

=

_kf

n

r

j

_ f

n

k

(because

kf

n

k

2

L

2

_([0,1]

p

₎

= 1

)

−→ 0

as

n

→ ∞.

2

4 Proof of Theorem 1.1

WefollowthesamerouteasintheproofofTheorem1.2,thatis,weutilizethemethodofmoments.

(Itis well-knownthat the

N (0, 1)

lawis uniquely determined byits moments.) Let thenotation andassumptionsofTheorem1.1prevail. Withoutlossofgenerality,wemayassumethat

E[F

2

n

] =

1

forall

n

(insteadof

E[F

2

n

]

→ 1

as

n

→ ∞

). Moreover,observethat

p!

kf

n

k

2

L

2

_([0,1]

p

₎

= E[F

n

2

] = 1

. Fixan integer

k > 3

. Iterative applications oftheproductformula(2.11) leads to

F

_n

k

= I

p

(f

n

)

k

=

X

(r

1

,...,r

_k−1

)∈A

k

I

kp−2r

1

−...−2r

_k−1

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

(4.21)

×

k−1

_Y

j=1

r

j

!



p

r

j



jp

_{− 2r}

1

− . . . − 2r

j−1

r

j



,

where

A

k

=



(r

1

, . . . , r

k−1

)

∈ {0, 1, . . . , p}

k−1

: r

2

6

2p

− 2r

1

, r

3

6

3p

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

(k

− 1)p − 2r

1

− . . . − 2r

k−2

.

In order to simplify the exposition, notethat we have removed all thebrackets in thewriting of

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

. We usethe implicit convention thatthese quantities arealways dened itera-tivelyfromthelefttotheright. Forinstance,

f

n

⊗

e

r

1

f

n

⊗

e

r

2

f

n

⊗

e

r

3

f

n

standsfor

((f

n

⊗

e

r

E[F

_n

k

] =

X

(r

1

,...,r

_k−1

)∈B

k

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

×

k−1

_Y

j=1

r

j

!



p

r

j



jp

_{− 2r}

1

− . . . − 2r

j−1

r

j



,

(4.22) with

B

k

=



(r

1

, . . . , r

k−1

)

∈ A

k

: 2r

1

+ . . . + 2r

k−1

= kp

. Combining (4.22) with the crude bound(consequenceof Cauchy-Schwarz)

kf

n

⊗

e

r

f

n

k

L

2

_([0,1]

2

p−2r

₎

6

kf

n

k

2

L

2

_([0,1]

p

₎

= 1/p! 6 1,

we havethat

E[F

k

n

] 6 #B

k

,thatis, for every

k

the

k

th moment of

F

n

isuniformlybounded. Assume that

(i)

is inorder. Because of the uniform boundedness of the moments, standard arguments implies that

E[F

4

n

]

→ E[B(1)

4

]

. Conversely, assume that

(ii)

is in order and let us prove that, for all

k > 1

,

E[F

_n

k

]

_{→ E[B(1)}

k

]

as

n

→ ∞

. (4.23)

The cases

k = 1

and

k = 2

being immediate, assume that

k > 3

is given. We decompose

B

k

as

C

k

∪ E

k

,with

C

k

= B

k

∩ {0, p}

k−1

and

E

k

= B

k

\ C

k

. We have

E[F

_n

k

] =

X

(r

1

,...,r

k−1

)∈C

k

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

×

k−1

_Y

j=1

r

j

!



jp

_{− 2r}

1

− . . . − 2r

j−1

r

j



(4.24)

+

X

(r

1

,...,r

k−1

)∈E

k

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

×

k−1

_Y

j=1

r

j

!



p

r

j



jp

− 2r

1

− . . . − 2r

j−1

r

j



.

By Lemma 4.1 together with assumption

(ii)

, we have that

kf

n

⊗

r

f

n

k

L

2

_([0,1]

2

p−2r

₎

(as well as

kf

n

⊗

e

r

f

n

k

L

2

_([0,1]

2

p−2r

₎

) tendsto zerofor any

r = 1, . . . , p

− 1

. Lemmas4.2 and4.3imply together thatthe rst sumin(4.24) converges to

E[B(1)

k

_]

,whereas thesecond sum converges to zeroby

Lemma 4.4. Thus,(4.23) is inorder, andtheproof ofthetheorem isconcluded.

2

Lemma 4.1 We have

E[F

_n

4

] = 3+

p−1

X

r=1



p

r



2

"

(p!)

2

_kf

n

⊗

r

f

n

k

2

_L

2

_([0,1]

2

_p−2r

₎

+ (r!)

2



p

r



2

(2p

_{− 2r)!kf}

n

⊗

e

r

f

n

k

2

_L

2

_([0,1]

2

_p−2r

₎

#

.

Proof (following[14]). Let

π

∈ S

2p

. If

r

∈ {0, . . . , p}

denotes thecardinalityof

{π(1), . . . , π(p)} ∩

{1, . . . , p}

thenit isreadily checked that

r

is alsothe cardinalityof

{π(p + 1), . . . , π(2p)} ∩ {p +

Moreover, for any xed

r

∈ {0, . . . , p}

, there are

p

r

2

(p!)

2

permutations

π

∈ S

2p

such that

#

_{{π(1), . . . , π(p)} ∩ {1, . . . , p} = r}

. (Indeed, sucha permutationiscompletely determined bythe choiceof:

(a) r

distinct elements

x

1

, . . . , x

r

of

{1, . . . , p}

;

(b) p

− r

distinct elements

x

r+1

, . . . , x

p

of

{p + 1, . . . , 2p}

;

(c)

a bijection between

{1, . . . , p}

and

{x

1

, . . . , x

p

}

;

(d)

a bijection between

{p + 1, . . . , 2p}

and

{1, . . . , 2p} \ {x

1

, . . . , x

p

}

.) Now, recallfrom(2.9) thatthesymmetrizationof

f

n

⊗ f

n

is given by

f

n

⊗f

e

n

(t

1

, . . . , t

2p

) =

1

(2p)!

X

π∈S

2p

f

n

(t

π(1)

, . . . , t

π(p)

)f

n

(t

π(p+1)

, . . . , t

π(2p)

).

Therefore,

kf

n

⊗f

e

n

k

2

L

2

_([0,1]

2p

₎

=

1

(2p)!

2

X

π,π

0

_∈S

_2p

Z

[0,1]

2p

f

n

(t

π(1)

, . . . , t

π(p)

)f

n

(t

π(p+1)

, . . . , t

π(2p)

)

×f

n

(t

π

0

₍₁₎

, . . . , t

_π

0

_(p)

)f

n

(t

π

0

_(p+1)

, . . . , t

_π

0

_(2p)

)dt

1

. . . dt

2p

=

1

(2p)!

X

π∈S

2p

Z

[0,1]

2p

f

n

(t

1

, . . . , t

p

)f

n

(t

p+1

, . . . , t

2p

)

×f

n

(t

π(1)

, . . . , t

π(p)

)f

n

(t

π(p+1)

, . . . , t

π(2p)

)dt

1

. . . dt

2p

=

1

(2p)!

p

X

r=0

X

π∈S

2

p

{π(1),...,π(p)}∩{1,...,p}=r

Z

[0,1]

2

p

f

n

(t

1

, . . . , t

p

)f

n

(t

p+1

, . . . , t

2p

)

×f

n

(t

π(1)

, . . . , t

π(p)

)f

n

(t

π(p+1)

, . . . , t

π(2p)

)dt

1

. . . dt

2p

.

Hence, using(4.25), we deducethat

(2p)!

kf

n

⊗f

e

n

k

2

_L

2

_([0,1]

2p

₎

= 2(p!)

2

kf

n

k

4

_L

2

_([0,1]

p

₎

+ (p!)

2

p−1

X

r=1



p

r



2

kf

n

⊗

r

f

n

k

2

_L

2

_([0,1]

2

p−2r

₎

= 2 + (p!)

2

p−1

X

r=1



p

r



2

kf

n

⊗

r

f

n

k

2

_L

2

_([0,1]

2

p−2r

₎

.

(4.26)

Theproductformula(2.11)leadsto

F

2

n

=

P

p

r=0

r!

p

r

2

I

2p−2r

(f

n

⊗

e

r

f

n

).

Using(2.8)-(2.10),weinfer

E[F

_n

4

] =

p

X

r=0

(r!)

2



p

r



4

(2p

− 2r)!kf

n

⊗

e

r

f

n

k

2

_L

2

_([0,1]

2

p−2r

₎

= (2p)!

_kf

n

⊗f

e

n

k

2

_L

2

_([0,1]

2

p

₎

+ 1 +

p−1

X

r=1

(r!)

2



p

r



4

(2p

_{− 2r)!kf}

n

⊗

e

r

f

n

k

2

_L

2

_([0,1]

2

_p−2r

₎

.

By inserting (4.26)in theprevious identity, we getthedesiredresult.

2

Lemma 4.2 As

n

→ ∞

, assume that

Then,for all

k > 3

and all

(r

1

, . . . , r

k−1

)

∈ C

k

, we have

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

→

k−1

_Y

j=1

j−2r

1

/p−...−2r

_j−1

/p

r

j

/p

(r

j

)!

jp−2r

1

−...−2r

_r

_j

j−1

as

n

→ ∞.

Proof. In all the proof, for sake of conciseness we write

f

e

⊗d

n

instead of

d

times

z

}|

{

f

n

⊗ . . . e

e

⊗f

n

. (Here, 

d

times justmeansthat

f

n

appears

d

timesintheexpression.) Itisreadilychecked that

f

e

⊗d

n

= g

f

n

⊗d

sothat, accordingto (2.9),

f

_n

⊗d

e

⊗

p

f

n

(t

1

, . . . , t

dp−p

) =

1

(dp)!

X

π∈S

dp

Z

[0,1]

p

f

n

(t

π(1)

, . . . , t

π(d)

) . . . f

n

(t

π(dp−p+1)

, . . . , t

π(dp)

)

×f

n

(t

dp−p+1

, . . . , t

dp

)dt

dp−d+1

. . . dt

dp

.

Let

π

∈ S

dp

. When

{π(jp − p + 1), . . . , π(jp)} 6= {dp − p + 1, . . . , dp}

for all

j = 1, . . . , d

, it is readily checked, using (4.27) aswellasCauchy-Schwarz, thatthefunction

(t

1

, . . . , t

dp−p

)

7→

Z

[0,1]

p

f

n

(t

π(1)

, . . . , t

π(d)

) . . . f

n

(t

π(dp−p+1)

, . . . , t

π(dp)

)

×f

n

(t

dp−p+1

, . . . , t

dp

)dt

dp−d+1

. . . dt

dp

tendsto zeroin

L

2

_{([0, 1]}

dp−p

₎

. Let

A

dp

bethesetof permutations

π

∈ S

dp

for which thereexists (atleastone)

j

∈ {1, . . . , d}

suchthat

{π(jp − p + 1), . . . , π(jp)} = {dp − p + 1, . . . , dp}

. Wethen have

f

_n

⊗d

e

_⊗

p

f

n

(t

1

, . . . , t

dp−p

)

≈

1

(dp)!

X

π∈A

dp

Z

[0,1]

p

f

n

(t

π(1)

, . . . , t

π(d)

) . . . f

n

(t

π(dp−p+1)

, . . . , t

π(dp)

)

×f

n

(t

dp−p+1

, . . . , t

dp

)dt

dp−d+1

. . . dt

dp

,

where, here and in the rest of the proof, we use the notation

h

n

≈ g

n

(for

h

n

and

g

n

two functions of, say,

q

arguments) to mean that

h

n

− g

n

tends to zero in

L

2

_{([0, 1]}

q

₎

. Because a

permutation

π

of

A

dp

is completely characterized by the choice ofthesmallest index

j

for which

{π(jp−p+1), . . . , π(jp)} = {dp−p+1, . . . , dp}

aswellastwopermutations

τ

∈ S

p

and

σ

∈ S

pd−p

, and usingmoreoverthat

f

n

⊗

p

f

n

=

kf

n

k

2

L

2

_([0,1]

p

₎

=

_p!

1

and that

f

n

issymmetric, we deducethat

f

_n

⊗d

e

⊗

p

f

n

(t

1

, . . . , t

dp−p

)

≈

d

(dp)!

X

σ∈S

dp−p

f

n

(t

σ(1)

, . . . , t

σ(d)

) . . . f

n

(t

σ(dp−2p+1)

, . . . , t

σ(dp−p)

)

≈

d

p!

dp

_p

^

f

n

⊗(d−1)

(t

1

, . . . , t

dp−p

) =

d

p!

dp

_p

f

e

⊗(d−1)

n

(t

1

, . . . , t

dp−p

).

(4.28)

Becausetheright-handside of(4.28) is asymmetric function,we eventually getthat

withtheconventionthat

f

e

⊗0

n

= 1

. Ontheotherhand,wehave

f

e

⊗d

n

⊗

e

0

f

n

= f

n

⊗d

e

⊗f

e

n

= f

n

⊗(d+1)

e

by the very denition of

f

e

⊗d

n

. We can summarize these two last identities by writing that, for any

r

_{∈ {0, p}}

,

f

_n

⊗d

e

⊗

e

r

f

n

≈

d

r/p

r!

dp

_r

f

e

⊗(d+1−2r/p)

n

.

(4.29)

Now,let

k > 3

and

(r

1

, . . . , r

k−1

)

∈ C

k

. Thanksto(4.29),wehave

f

n

⊗

e

r

1

f

n

=

(

1

r1/p

)

(r

1

)!

(

_r1

p

)

f

e

⊗(2−2r

1

/p)

n

,

f

n

⊗

e

r

1

f

n

⊗

e

r

2

f

n

≈

1

r

1

/p

2−2r

1

/p

r

2

/p

(r

1

)!

_r

p

1

(r

2

)!

2p−2r

_r

2

1

f

⊗(3−2r

e

1

/p−2r

2

/p)

n

,

and soon. Iterating this procedureleads eventually to

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

≈

k−1

_Y

j=1

j−2r

1

/p−...−2r

j−1

/p

r

j

/p

(r

j

)!

jp−2r

1

−...−2r

_r

_j

j−1

,

(4.30)

which isexactlythe desiredformula. The proofof the lemmais done.

2

Lemma 4.3 For all

k > 3

, we have

E[B(1)

k

] =

X

(r

1

,...,r

_k−1

)∈C

k

k−1

_Y

j=1



j

− 2r

1

/p

− . . . − 2r

j−1

/p

r

j

/p



.

Proof. Theidentity is clear when

k

is an oddinteger, because

C

k

=

∅

inthis case. Assume now that

k

iseven. Considertherepresentation

B(1) = I

1

(1

[0,1]

)

. Iterativeapplicationsoftheproduct formula (2.11) leadsto

B(1)

k

= I

1

(1

[0,1]

)

k

=

X

(r

1

,...,r

_k−1

)∈ e

A

k

I

k−2r

1

−...−2r

k−1

1

[0,1]

⊗

e

r

1

. . . e

⊗

r

_k−1

1

[0,1]

×

k−1

_Y

j=1



j

− 2r

1

− . . . − 2r

j−1

r

j



,

where

e

A

k

=



(r

1

, . . . , r

k−1

)

∈ {0, 1}

k−1

: r

2

6

2

− 2r

1

, r

3

6

3

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

k

− 1 − 2r

1

− . . . − 2r

k−2

.

By takingtheexpectation, we deducethat

e

C

k

=



(r

1

, . . . , r

k−1

)

∈ {0, 1}

k−1

: r

2

6

2

− 2r

1

, r

3

6

3

− 2r

1

− 2r

2

, . . . ,

r

k−1

6

k

− 1 − 2r

1

− . . . − 2r

k−2

, 2r

1

+ . . . + 2r

k−1

= k

.

Itisreadilycheckedthat

1

[0,1]

⊗

e

r

1

. . . e

⊗

r

_k−1

1

[0,1]

= 1

[0,1]

⊗

r

1

. . .

⊗

r

_k−1

1

[0,1]

= 1

forall

(r

1

, . . . , r

k−1

)

∈

e

C

k

. Hence

E[B(1)

k

] =

X

(r

1

,...,r

k−1

)∈ e

C

k

k−1

_Y

j=1



j

_{− 2r}

1

− . . . − 2r

j−1

r

j



=

X

(r

1

,...,r

k−1

)∈C

k

k−1

_Y

j=1



j

− 2r

1

/p

− . . . − 2r

j−1

/p

r

j

/p



,

which isthedesiredconclusion.

2

Lemma 4.4 As

n

→ ∞

, assume that

kf

n

⊗

e

r

f

n

k

L

2

([0,1]

2

p−2r

)

→ 0

for all

r = 1, . . . , p

− 1

. Then, as

n

→ ∞

we have

f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n

→ 0

for all

k > 3

and all

(r

1

, . . . , r

k−1

)

∈ E

k

.

Proof. Fix

k > 3

and

(r

1

, . . . , r

k−1

)

∈ E

k

, and let

j

∈ {1, . . . , k − 1}

be the smallest integer such that

r

j

∈ {1, . . . , p − 1}

. Asin theproof of Lemma 4.2, when

h

n

and

g

n

arefunctions of

q

argumentslet us write

h

n

≈ g

n

to indicate that

h

n

− g

n

tendsto zero in

L

2

_{([0, 1]}

q

₎

. Recall from (4.29) that

f

e

⊗d

n

⊗

e

p

f

n

≈

_p!

₍

d

dp

p

)

f

n

⊗(d−1)

e

. Then



f

n

⊗

e

r

1

. . . e

⊗

r

_k−1

f

n





=



f

n

⊗

e

r

1

. . . e

⊗

r

j−1

f

n

⊗

e

r

j

f

n

⊗

e

r

j+1

. . . e

⊗

r

k−1

f

n





≈

c



(f

n

⊗ . . . e

e

⊗f

n

) e

⊗

r

j

f

n

⊗

e

r

j+1

. . . e

⊗

r

k−1

f

n





(forsome constant

c > 0

independent of

n

)

6

c

_k(f

n

⊗ . . . e

e

⊗f

n

) e

⊗(f

n

⊗

e

r

j

f

n

)

k

L

2

([0,1]

q

)

kf

n

k

k−j−1

L

2

_([0,1]

p

₎

(byCauchy-Schwarz, fora certain

q

)

6

c

_kf

n

⊗

e

r

j

f

n

k

(because

kf

n

k

2

L

2

_([0,1]

p

₎

=

_p!

1

6

1

)

−→ 0

as

n

→ ∞.

2

Acknowledgement. Ithank oneanonymousreferee for his/herthoroughreading andinsightful

comments.

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