A quadratic regression problem for two-state algebras with application to the Central Limit Theorem

0.24 MB·English

by  Marek Bozejko

A QUADRATIC REGRESSION PROBLEM FOR TWO-STATE
ALGEBRAS WITH APPLICATION TO THE CENTRAL LIMIT
THEOREM
9
0
0 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
2

n
a
J

6 Abstract. We extend the free version [8] of the Laha-Lukacs theorem to
2 probability spaces with two-states. We then use this result to generalize the
noncommutative centrallimittheoremofKargin[23]tothetwo-statesetting.

]
A
O 1. Introduction
. Both classical and free Meixner distributions first appeared in the theory of or-
h
t thogonalpolynomialsinthe worksofMeixner[30],Anshelevich[3], andSaitohand
a
Yoshida[34]. Morris[31]pointedouttherelevanceofclassicalMeixnerdistributions
m
forthetheoryofexponentialfamiliesinstatistics;Diaconis,KhareandSaloff-Coste
[
[18] gave an excellent overview of state of the art. Ismail and May [21] analyzed a

2 mathematically equivalent problem from the point of view of approximation oper-
v ators. Acounterpartof(someaspectsof)thistheoryforfreeMeixnerdistributions
6
appear in an unpublished manuscript by Bryc and Ismail [16] and in [15].
6
LahaandLukacs[28]characterizedallthe(classical)Meixnerdistributionsusing
2
0 aquadraticregressionpropertyandBoz˙ejkoandBryc[8]provedthecorresponding
. free version. Anshelevich[4] considereda Booleanversionofthis property showing
2
that in the Boolean theory Laha-Lukacs property characterizes only the Bernoulli
0
8 distributions.
0 AccordingtoExample3in[5]andProposition3.1ofFranz[19],Boolean,mono-
: tone, and free independence are all special cases of the c-freeness for algebras with
v
i two states. Our primary goalin this paper is to extend [8] and [4] to the two-state
X
setting under a weaker form of c-freeness, which we call (ϕψ)-freeness, and which
r shares with boolean and free independence a good descripti|on by cumulants.
a
As anapplicationofourmainresult,weprovethe centrallimittheoremundera
certaintypeof“weakdependence”whichincludesthesocalledsingletoncondition,
whoseimportancetocentrallimittheoremwaspointedoutinTheorem0ofBoz˙ejko
andSpeicher [11]; ourassumptions aremodeledonKargin[23]who weakenedfree-
ness assumption in the free central limit theorem. Our result addresses a question
of finding the “appropriate notions of independence or of weak dependence” for the
quantumcentrallimittheoremwhichwasraisedonpage11of[2]anddescribesthe
Date:Printed: January27,2009. File: noncom-clt-07-10.TEX.
2000 Mathematics Subject Classification. Primary: 46L53;Secondary: 60E05,05A18.
Key words and phrases. generalized two-state freeness, generalized freeMeixner distribution,
Laha-Lukacs theorem,noncommutative quadraticregression.
ResearchpartiallysupportedbytheTaftResearchCenter,KBNGrantNo1PO3A01330,and
NSFgrant#DMS-0504198.
1
2 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
limitlaw;ifoneisinterestedsolelyinconvergence,itcanbe deducedfromthe gen-
eraltheoryofthequantumcentrallimittheoremdevelopedbyAccardi,Hashimoto
and Obata [1]. Section 8.2 of Hora and Obata [20] discusses the role of singleton
condition and gives the central limit theorem under classical, free, boolean, and
monotone independence.
1.1. Atwo-statefreenesscondition. Let beaunital -algebrawithtwostates
ψ,ϕ: C. WeassumethatbothstatesfulAfilltheusuala∗ssumptionsofpositivity
A→
and normalization, and we assume tracial property ψ(ab) = ψ(ba) for ψ, but not
for ϕ.
A typical model of an algebra with two sates is a group algebra of a group
G = G , a free product of groups G . Here ϕ is the boolean product of the
i i i
∗
individual states (which was also called ”regularfree state”); the simplest example
is the free product of integers, G = Z, where G is a free group with arbitrary
i
number of generators, and ϕ is the Haagerup state, Φ(x) = rx, where x is the
| |
| |
length of word x G, 1 r 1, and state ψ is δ(0). For details see Bozejko
∈ − ≤ ≤
[6, 7].
A self-adjoint element X with moments that fulfill appropriate growthcon-
dition defines a pair µ,ν of∈prAobability measures on (R, ) such that
B
ϕ(Xk)= xkµ(dx) and ψ(Xk)= xkν(dx).
R R
Z Z
We will refer to measures µ,ν as the ϕ-law and the ψ-law of X, respectively.
With each set of a ,...,a and a pair of states (ϕ,ψ) we associate the
1 n
cumulants R = R , k = 1,2∈,.A.., which are the multilinear functions k C
k k,ϕ,ψ
A →
defined by
(1.1) ϕ(a a ...a )
1 2 n
n k−1 sr+1−1
= R (a ,a ,...,a )ϕ(a ...a ) ψ a .
k 1 s2 sk sk+1 n  j
Xk=11=s1<s2X<···<sk≤n rY=1 j=Ysr+1
 
We will use the notation
(1.2) r (a ,...,a ):=R (a ,...,a ).
n 1 n n,ψ,ψ 1 n
We remark that r are the free cumulants with respect to state ψ, as defined by
n
Speicher [35, 36]; see also [32]. For more general theory of cumulants, see [29].
Fix a , and consider the following formal power series
∈A
∞
(1.3) R(z) = R (a,...,a)zn 1,
n −
n=1
X
∞
(1.4) m(z) = znψ(an),
n=0
X
∞
(1.5) M(z) = znϕ(an).
n=0
X
By Theorem 5.1 of [9], Eqtn. (1.1) is equivalent to the following relation
(1.6) M(z)(1 zR(zm(z)))=1.
−
QUADRATIC REGRESSION AND CLT FOR TWO-STATE ALGEBRAS 3
Definition 1.1. We say that subalgebras , ,... are (ϕψ)-free if for every
1 2
A A |
choice of a ,...,a we have
1 n ∈ jAj
Rn(a1,...,anS)=0 except if all aj come from the same algebra.
Itisimportanttonotethat(ϕψ)-freenessisweakerthanc-freeness,asexplained
|
before Lemma 1.1. Thus we could have used the term weak c-freeness instead of
(ϕψ)-freeness.
|
When the algebras are (ψ ψ)-free, we will abbreviate this to ψ-free. From Ref.
|
[35] it follows that ψ-freeness coincides with the usual concept of freeness as intro-
duced by Voiculescu [37].
We will say that X,Y are (ϕψ)-free if the unital algebras C X and C Y are
| h i h i
(ϕψ)-free.
|
A related concept is the following.
Definition 1.2 (See Refs. [10] and [9]). We say that subalgebras , ,... are
1 2
A A
c-freeif for everychoiceofi =i = =i andeverychoiceofa suchthat
1 2 n j j
6 6 ···6 ∈A
ψ(a )=0 (thus a =1) we have
j j
6
n
(1.7) ϕ(a ...a )= ϕ(a ).
i1 in ik
k=1
Y
1.2. Properties of (ϕψ)-freeness. If , are (ϕψ)-free then for a ,b
1 2 1
| A A | ∈ A ∈
2
A
(1.8) ϕ(ab)=ϕ(a)ϕ(b).
For a ,a ,b we have
1 2 1 2
∈A ∈A
(1.9) ϕ(a ba )=ψ(b)ϕ(a a ) ψ(b)ϕ(a )ϕ(a )+ϕ(b)ϕ(a )ϕ(a ).
1 2 1 2 1 2 1 2
−
For a ,a ,b ,b we have
1 2 1 1 2 2
∈A ∈A
(1.10) ϕ(a b a b )=ϕ(a a )ψ(b )ψ(b ) ϕ(a )ψ(a )ϕ(b b )
1 1 2 2 1 2 1 2 1 2 1 2
−
+ϕ(a )ψ(a )ϕ(b )ϕ(b ) ϕ(a )ϕ(a )ϕ(b )ϕ(b ).
1 2 1 2 1 2 1 2
−
Formulas (1.8) (1.9) and are identical to formulas under c-freeness as given in
Lemma 2.1 of Ref. [9]. Together with formula (1.10) they imply that for a pair
of (ϕψ)-free algebras, (1.7) holds for n 4. One can check that if a,b are (ϕψ)-
| ≤ |
free and ψ(a) = ψ(b) = 0 but ψ(bab) = 0 then ϕ(ababa) = ϕ(a)3ϕ(b)2; thus the
6 6
concepts of c-freeness and of (ϕψ)-freeness are not equivalent. Nevertheless they
|
coincide for ψ-free algebras as noted in the following.
Lemma 1.1 (page 368 of Ref. [9]). Suppose , ,... are ψ-free. Then the
1 2
A A
algebras , ,... are (ϕψ)-free if and only if they are c-free.
1 2
A A |
(It would be interesting to characterize (ϕψ)-freeness without the freeness as-
|
sumption on ψ.)
We will also rely on the following fact.
Lemma 1.2 (Ref [9]). Given a noncommutative random variable X in a two-state
probability space, there exist a two-state algebra (which one can take as the algebra
ofnoncommutativepolynomialsC X,Y intwovariables)andtwonon-commutative
h i
random variables X,Y which are ψ-free, (ϕψ)-free, and both have the same ϕ-law
and ψ-law as X. |
e e
4 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
Proof. Theorem 1 of Ref. [10], see also Theorem 2.2 of Ref. [9], shows how to
extend both states to the free product of the original algebra so that the resulting
algebras are c-free and ψ-free. By Lemma 1.1, they are thus (ϕψ)-free. (cid:3)
|
2. A (ϕψ)-free quadratic regression problem
|
In this section we prove a two-state version of Theorem 3.2 of Ref. [8]. The
statement is fairly technical, but we found it useful for our proof of the central
limit theorem (Theorem 4.1 below).
Theorem 2.1. Suppose X,Y are self-adjoint (ϕψ)-free and
|
(2.1) ϕ(Xn)=ϕ(Yn), ψ(Xn)=ψ(Yn)
for all n. Furthermore, assume that ϕ(X) = 0, ϕ(X2) = 1. (This can always be
achieved by a shift and dilation, as long as ϕ(X2)=0.)
Let S=X+Y and suppose that there are a,c 6 R and b> 2 such that
∈ −
(2.2) ϕ (X Y)2Sn =cϕ (4I+2aS+bS2)Sn , n=0,1,2....
−
Then the ϕ-m(cid:0)oment gener(cid:1)ating fu(cid:0)nctions MS(z) := (cid:1) ∞ zkϕ(Sk) and mS(z) :=
k=0
∞ zkψ(Sk), which are defined as formal power series, are related as follows
k=0 P
P 2+b (2az+b)mS(z)
(2.3) MS(z)= − .
2+b (4z2+2az+b)mS(z)
−
Remark 2.1. Wewillapply(2.3)tothecasewhenmS(z)convergesforsmallenough
z , in the form as written. In general, the right hand side of (2.3) needs to be
| |
interpreted correctly. Recall that the composition p(q(z)) of two power series p,q
is well defined if q(z) has no constant term. Note that the formal power series
b+(4z2+2az+b)mS(z) has no constant term, so it can be composed with the
−
formalpowerseries ∞n=0 2n1+1zn,whichisaformalpowerexpansionofthefunction
1 . It is therefore natural to denote such a composition by
2 z P
−
1
.
2 ( b+(4z2+2az+b)mS(z))
− −
The right hand side of (2.3) is then interpreted as the product of this power series
with the formal power series 2+b (2az+b)mS(z).
−
Remark 2.2. Ourassumptionsonϕdonotallowustouseconditionalexpectations.
However,itisstillnaturaltoaskwhichpropertiesofconditionalexpectationswould
have implied assumptions of Theorem 2.1. To this end, we denote by ϕ( S) the
conditional expectation onto the commutative algebra generated by S. ·|
From equality of the laws (2.1) and (ϕψ)-freeness, one can deduce that
|
1
(2.4) ϕ(XSn)= ϕ Sn+1 , n=0,1,2....
2
(cid:0) (cid:1)
(See (2.7) below.) When the conditional expectation exists, this property follows
from ϕ(XS)= 1S. We can then derive (2.2) from the quadratic variance property
| 2
a b
(2.5) ϕ(X2 S) (ϕ(XS))2 =c I+ S+ S2 .
| − | 2 4
(cid:18) (cid:19)
QUADRATIC REGRESSION AND CLT FOR TWO-STATE ALGEBRAS 5
2.1. Proof of Theorem 2.1. We first remark that c = (2+b) 1. This follows
−
from (2.2) with n=0 since ϕ(X Y)2 =2 ϕ(XY) ϕ(YX)=2.
By definition, R (S,...,S) =±R (X,...,±X)+R (±Y,...,Y). From (2.1) we see
n n n
that R (X,...,X)=R (Y,...,Y). Thus
n n
(2.6) R (X Y,S,...,S)=R (X,S,...,S) R (Y,S,...,S)
n n n
− −
=R (X,...,X) R (Y,...,Y)=0
n n
−
for all n. By (1.1) this implies
(2.7) ϕ((X Y)Sn)=0.
−
Similarly, using multilinearity of R,
(2.8) R (X Y,X Y,S,...,S)
n
− −
=R (X,X Y,S,...,S) R (Y,X Y,S,...,S)
n n
− − −
=R (X,...,X)+R (Y,...,Y)=R (S,...,S)
n n n
for all n 2. Formula (1.1) therefore implies that
≥
ϕ (X Y)2Sn
−
n+(cid:0)2 (cid:1) k−1
= Rk(X Y,X Y,S...,S)ϕ(Sn−bk−1) ψ(Sbr+1−br−1)
− −
kX=21=b1<b2=2X<···<bk≤n+2 rY=1
n+2 k 1
−
+ Rk(X Y,S,...,S)ϕ(Sn−bk−1) ψ(Sbr+1−br−1).
−
Xk=11=b1<2<b2X<···<bk≤n+2 rY=1
By (2.6), the second sum vanishes. Using (2.8) we get
(2.9) ϕ (X Y)2Sn
−
=(cid:0)n+2 (cid:1) Rk(S,S,S...,S)ϕ(Sn−bk−1)k−1ψ(Sbr+1−br−1).
Xk=21=b1<b2=2X<···<bk≤n+2 rY=1
Comparing this with the decomposition for ϕ(Sn+2) we see that
ϕ (X Y)2Sn =ϕ(Sn+2)
−
(cid:0) n+2 (cid:1) Rk(S,S,...,S)ϕ(Sn−bk−1)k−1ψ(Sbr+1−br−1).
−
Xk=21=b1<2<b2X<···<bk≤n+2 rY=1
We now rewrite the last sum based on the value of m=b b , compare Ref. [8].
2 1
−
We have
ϕ (X Y)2Sn =ϕ(Sn+2)
−
n n+2
(cid:0) (cid:1)
ψ(Sm) Rk(S,S,...,S)ϕ(Sn−bk−1)
−
mX=1 Xk=21=b1<1+m=Xb2<···<bk≤n+2
k 1
− ψ(Sbr+1−br−1).
×
r=1
Y
6 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
Since b b 1=m, formula (2.8) gives
2 1
− −
n+2 k 1
−
Rk(S,S,...,S)ϕ(Sn−bk−1) ψ(Sbr+1−br−1)
Xk=21=b1<1+m=Xb2<···<bk≤n+2 rY=1
n+2
= Rk(X Y,X Y,S,...,S)ϕ(Sn−bk−1)ψ(Sm)
− −
kX=21=b1<1+m=Xb2<···<bk≤n+2
k 1
−
ψ(Sbr+1−br−1).
×
r=2
Y
Re-indexing the variables so that b =2 and inserting this into (2.9) we get
2
n
ϕ (X Y)2Sn =ϕ(Sn+2) ψ(Sm)ϕ((X Y)2Sn−m).
− − −
m=1
(cid:0) (cid:1) X
Thus from (2.2) we get
n
1
ϕ(Sn+2)= ψ(Sj) 4ϕ(Sn j)+2aϕ(Sn j+1)+bϕ(Sn j+2) .
− − −
2+b
j=0
X (cid:0) (cid:1)
A routine argument now relates the formal power series:
∞
MS(z)=1+z2 znϕ(Sn+2)
n=0
X
z2 ∞ n
=1+ zjψ(Sj)zn j 4ϕ(Sn j)+2aϕ(Sn j+1)+bϕ(Sn j+2)
− − − −
2+b
n=0j=0
XX (cid:0) (cid:1)
z2 ∞ ∞
=1+ zjψ(Sj) zn j 4ϕ(Sn j)+2aϕ(Sn j+1)+bϕ(Sn j+2)
− − − −
2+b
j=0 n=j
X X (cid:0) (cid:1)
=1+ mS(z) 4z2MS(z)+2az(MS(z) 1)+b(MS(z) 1) .
2+b − −
(cid:0) (cid:1)
3. The ϕ-law of X
In this section we are interested in one explicit case when Theorem 2.1 allows
us to determine the ϕ-law of X from the ψ-law of X. This case arises when X,Y
are ψ-free and (ϕψ)-free with compactly supported laws. Then the ϕ-law and the
ψ-law of X+Y a|re determined uniquely from the laws of X,Y by the generalized
convolution ⊛ which was introduced by Boz˙ejko and Speicher [10] and studied in
Refs. [9, 12, 13, 26, 27]. The generalized convolution is a binary operation on the
pairs of compactly supported probability measures (µ,ν). The analytic approach
from Theorem 5.2 in Ref. [9] is especially convenient for explicit calculations.
According to this result, the generalized convolution (µ ,ν )⊛(µ ,ν ) of pairs of
1 1 2 2
compactly supported probability measures is a pair (µ,ν) of compactly supported
probabilitymeasureswhichisdeterminedbythefollowingprocedure. Considerthe
Cauchy transforms
1 1
G (z)= µ (dx), g (z)= ν (dx), j =1,2.
j j j j
z x z x
Z − Z −
QUADRATIC REGRESSION AND CLT FOR TWO-STATE ALGEBRAS 7
Let k (z) be the inverse function of g (z) in a neighborhood of , and define
j j
∞
(3.1) r (z)=k (z) 1/z.
j j
−
On the second component the c-convolution acts as the free convolution [37], ν =
ν ⊞ν . Recallthatthefreeconvolutionνofmeasuresν ,ν istheuniqueprobability
1 2 1 2
measure with the Cauchy transform g(z) which solves the equation
1
g(z)= .
z r (g(z)) r (g(z))
1 2
− −
To define the action of the generalized convolution on the first component, let
R (z)=k (z) 1/G (k (z)).
j j j j
−
Thus
1
(3.2) G (z)= .
j
z R (g (z))
j j
−
The first component of the generalizedconvolutionis defined as the unique proba-
bility measure µ with the Cauchy transform
1
G(z)= .
z R (g(z)) R (g(z))
1 2
− −
We write
(µ,ν)=(µ ,ν )⊛(µ ,ν ).
1 1 2 2
We remark that
∞ ∞
r(z)= rkzk−1, R(z)= Rkzk−1
k=1 k=1
X X
are the generating functions for the ψ-free and (ϕψ)-free cumulants respectively,
|
see(1.3). Wealsonotethattheaboverelationscanbeinterpretedascombinatorial
relationsbetween ψ-moments and ϕ-moments; the assumption ofcompact support
allows us to determine the laws uniquely from moments.
3.1. The case of “constant conditional variance”.
Proposition 3.1. Suppose X,Y are ψ-free with the same compactly supported ψ-
law ν, and are (ϕψ)-free with the same ϕ-law. If (2.2) holds with a=b=0, then
the ϕ-law of X is|compactly supported and uniquely determined by ν.
Proof. The ψ-law of S is the free convolution ν⊞ν, so it is compactly supported.
ThereforemS(z)isgivenbyaseriesthatconvergesforsmallenough z . Then(2.3)
| |
reduces to
1
MS(z)= ,
1 2z2mS(z)
−
and MS(z) is also given by a convergent series. In particular, the ϕ-law of S is
compactly supported. So for z >0, the Cauchy transform is
ℑ
1 1
(3.3) GS(z)= MS(1/z)= .
z z 2gS(z)
−
ThusR (S,...,S)=0forallkexceptforR (S,S)=2. ThisshowsthatR (X,...,X)=
k 2 k
0 for all k except for R2(X,X)=1. Thus RX(z)=z and (1.6) gives
1
(3.4) MX(z)= .
1 z2mX(z)
−
8 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
This implies that ϕ-law of X has compact support, and its Cauchy transform is
uniquely determined by
1
(3.5) GX(z)= .
z gX(z)
−
(cid:3)
In particular, suppose ν is the semicircle law with mean zero and variance σ2,
so that gX(z) = z−√2zσ22−4σ2. Proposition 3.1 then shows that the ϕ-law of X has
Cauchy-Stieltjes transform
(σ2 1)z 1√z2 4σ2
(3.6) GX(z)= − 2 − 2 − .
1+(σ2 1) z2
−
This law plays the role of the “Gaussian limit” in Ref. [9].
3.2. The case of “linear conditional variance”. Suppose (2.2) holds with b=
0. Then (2.3) reduces to
1 azmS(z)
MS(z)= − .
1 (2z+a)zmS(z)
−
So again the Φ-law of S is compactly supported, if the ψ-law is, and the Cauchy
transform is
1 agS(z) 1
GS(z)= − =
z (2+az)gS(z) z RS(gS(z))
− −
with
2u
RS(u)= .
1 au
−
This shows that RX(z)= z and
1 az
−
1 agX(z)
GX(z)= − .
z (1+az)gX(z)
−
In particular, suppose that the ψ-law of X is Marchenko-Pastur with parameter
λ>0, so that
z+(1 λ) (z 1 λ)2 4λ
gX(z)= − − − − − .
2z
p
If a=1, then the ϕ-law of X is compactly supported, with Cauchy transform
1+λ z(1 2λ) (z 1 λ)2 4λ
GX(z)= − − − − − − .
2(1+z(1+λ)q z2(1 λ))
− −
Related laws appear in Eqtn. (17) of Ref. [17] and on page 380 in Ref. [9].
4. Central limit theorem for non-identical summands
The central limit theorem and the Poisson convergence theorem for sums of
(ϕψ)-free random variables that are also ψ-free appear in Theorems 4.3 and 4.4
|
of Ref. [9]. Recently Kargin [23] observed that in the free case one can dispense
with the assumption of identical laws and at the same time relax the freeness
assumption. A similar result in classical probability is due to Komlos [24] who
assumes a much weaker version of singleton condition (4.1) and has an inequality
in his condition (6) that substitutes for (4.3). Komlos’ conditions were motivated
by (classical) central limit theorem for the so called multiplicative systems. We
QUADRATIC REGRESSION AND CLT FOR TWO-STATE ALGEBRAS 9
also note that in classical probability Jakubowski and Kwapien´ [22] discovered a
beautiful connection between multiplicative systems and independent sequences.
No counterpart of this result is known in noncommutative setting; compare also
non-commutativep-orthogonalityandRemark2.4ofPisier[33],andworkofK¨ostler
and Speicher [25] on noncommutative versions of de Finetti’s theorem.
In this section we use Theorem 2.1 to deduce a two-state version of Kargin’s
result. TheconvergenceofmomentscanalsobeobtainedasacorollaryofTheorem
3 in Accardi Hashimoto and Obata [2], see also Theorem 3.3 in [1], Theorem 0 of
[11], and Section 8.2 in [20]. This theorem says that under the singleton condition
(4.1), in order to complete the proof of CLT, it suffices to control ergodic averages
of totally entangled pair partitions. The disentanglement can be achieved from
various conditions that include statistical conditions, such as the free case or the
generalized freeness given by conditions (4.2) and (4.3). This approach, as well
as classical CLT in Ref [24], suggests that one should seek a weaker version of
(4.3) that perhaps would be stated as an inequality. On the other hand, our proof
from Theorem 2.1 gives directly the formula for the Cauchy-Stieltjes transform of
the limit law which would require additional work if the techniques from [1] were
applied.
WealsonotethatWang[38]usesanalyticalmethodstostudylimittheoremsfor
additive c-convolutionwith measures of unbounded support. It is not obvious how
Kargin’sconditionAshouldbegeneralizedto this setting. Infact,ageneralization
of Theorem 2.1 to unbounded random variables would be interesting even in the
free case studied in [8].
Definition 4.1. We will say that a sequence of randomvariables X ,X ,... satis-
1 2
fies Kargin’s Condition A with respect to (ϕψ), if:
|
(i) For every k j ,...,j the following singleton conditions hold:
1 n
6∈{ }
(4.1) ϕ(X X ...X )=ϕ(X X X ...X )=...
k j1 jr j1 k j2 jr
=ϕ(X ...X X )=0.
j1 jr k
(4.2) ψ(X X ...X )=0.
k j1 jr
(In particular, ψ(X )=ϕ(X )=0.)
j j
(ii) For every k j ,...,j , and 0 p r,
1 r
6∈{ } ≤ ≤
(4.3) ϕ(X X ...X X X ...X )
k j1 jp k jp+1 jr
=ϕ(X2)ψ(X ...X )ϕ(X ...X ).
k j1 jp jp+1 jr
Weremarkthatconditions(4.1)and(4.3)areautomaticallysatisfiedifX ,X ,...
1 2
are ϕ-centered and (ϕψ)-free; clearly, condition (4.2) holds true if X ,X ,... are
1 2
|
ψ-centered and ψ-free but of course it is weaker and can hold also for classical
(commutative) independent random variables.
Theorem 4.1. Suppose that
(i) X ,X ,... satisfies Kargin’s Condition A with respect to (ϕψ);
1 2
|
(ii) All joint moments of order k are uniformly bounded
(4.4) sup ϕ(X ...X ) C < for k =1,2,....
| j1 jk |≤ k ∞
j1,...,jk≥1
10 MAREKBOZ˙EJKOANDWL ODZIMIERZBRYC
(iii) Sequences s2 :=ψ(X2) and S2 :=ϕ(X2) satisfy
j j j j
(4.5) (s2+ +s2)/n s and (S2+ +S2)/n S.
1 ··· n → 1 ··· n →
(iv) 0<s,S < .
(v) The ψ-mom∞ents of 1 n X converge to the corresponding mo-
√s2+ +s2 j=1 j
1 ··· n
ments of a compactly supported probability measure ν.
P
Thentheϕ-momentsof 1 n X convergestothemomentsoftheunique
√S2+ +S2 j=1 j
1 ··· n
compactlysupportedlawµwithCaucPhytransform (3.5),wheregX(z)= S ν(dx).
Sz sx
−
Combining Theorem 4.1 with Ref. [23] and formula (3.6) we get tRhe following
generalizationof Theorem 4.3 in Ref. [9].
Corollary 4.2. Suppose that
(i) X ,X ,... satisfies Kargin’s Condition A with respect to (ϕψ) and with
1 2
|
respect to (ψ ψ).
(ii) All moments|are uniformly bounded: (4.4) holds true, and sup ψ(Xk) <
n| n |
for k =1,2,....
(iii) ∞Sequence s2 := ψ(X2)= s2 and S2 :=ϕ(X2) satisfy (4.5) with 0<s,S <
j j j j j
.
∞
Then the ϕ-law of 1 n X converges to the law µ with the Cauchy-
√S2+ +S2 j=1 j
1 ··· n
Stieltjes transform (3.6) and σP=s/S.
Our proof of the central limit theorem is based on reduction to Laha-Lukacs
theoremwhich in classicalprobability was introducedin Section7.3.1 of Bryc [14].
4.1. Proof of Theorem 4.1. By Ref. [10] without loss of generality we may
assume that we have a two-state probability space with two copies of the original
sequence: (X ) and (Y ) each of them separately having the same ψ-moments
k k
and ϕ-moments as the original sequence, but such that the algebras X and Y
generated by (X ) and by (Y ), respectively, are ψ-free and (ϕψ)-free.A A
k k
Under this representation, the ψ-distribution of 1 | n (X +Y ) con-
√s2+ +s2 j=1 j j
vergestoν⊞ν. Ourgoalistoshowthattheϕ-distributi1on··o·f √nSP2+1 +S2 nj=1(Xj+
Yj) has the unique limit determined by the law with Cauchy-1St·i··eltjnesPtransform
(3.3). To do so, denote
n n
1 1
U = X , V = Y , S =U +V .
n j n j n n n
√n √n
j=1 j=1
X X
Denote
Z(jε) =XεjY1j−ε, ε=0,1.
Since the variablesdo not commute, we adopt a specialconventionfor the product
notation convention which relies on the order of the index set:
p
ϕ Zε(s) :=ϕ Zε(1)Zε(2) ...Zε(p) .
J(s) J(1) J(2) J(p)
!
sY=1 (cid:16) (cid:17)
Lemma 4.3. In the above setting, if X satisfies Kargin’s Condition A, then
j
X ,Y ,X ,Y ,... satisfies Kargin’s C{ond}ition A.
1 1 2 2
{ }
Proof. We first note the following.