On the law of large numbers for free identically distributed random variables 1
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1 General Mathematics Vol. 5, No. 4 (2007), On the law of large numbers for free identically distributed random variables Bogdan Gheorghe Munteanu Abstract A version of law of large numbers for free identically distributed random variables is considering at this work. It shown that lim t µ(x : x > t) = 0 t is a sufficient and necessary condition for the weak law of large numbers for the sequence X, X 2,..., free random variables Mathematical Subject Classification: 45L54, 60F05, 47C5 Key words: Law of large numbers, free random variables, free convolution Introduction Analytic theory of free additive convolution is useful in frame of this article. The calculation of free additive convolution is based on an analoque of the Fourier transform first introduced by Voiculescu [6]. We need the version of Received 4 July, 2007 Accepted for publication (in revised form) 4 December,
2 56 Bogdan Gheorghe Munteanu this apparatus which is suitable for the convolution of arbitrary probability measures []. First, some notation. Let C denote the complex field, C + and C the upper and lower half plane. We consider Γ α = {z = x + iy : y > 0 and x < αy}, Γ α,β = {z Γ α : y > β}, α,β > 0. where α and β are positive numbers. Given a probability measure µ on R, its Cauchy transform G µ : C + C is defined as G µ (z) := µ dx z x = E( (z X) ) The reciprocal Cauchy transform is defined by F µ : C + C +, F µ (z) = /G µ (z). 2 Proof of the main result First, we would like to formulate the theorem in terms of free convolutions rather than random variables. To do this, observe that given a self-adjoint random variable X affiliated with some algebras A and a scalar λ > 0, we have µ λ X = D λ µ X where D λ is the dilation of a measure µ defined by D λ µ(a) = µ(λ A), (A R measurable). Theorem 2.. Let µ be a probability measure R. The following conditions are equivalent:
3 On the law of large numbers (i) There exist real constants M,M 2,... R such that lim ν n = δ 0, n where ν n = D /n µ... D /n µ δ Mn and δ 0 Dirac distribution; }{{} n ori (ii) The measure µ satisfies lim t µ (x : x > t) = 0. Moreover, if (ii) is t +n satisfied, the constants M n in (i) can be chosen to be M n = tdµ(t). For the proof of this theorem, we establish some preparatory lemmas. The foolowing result is related by Bercovici and Voiculescu (993) in Proposition 4.5. to (993,[7]). Lemma 2.. ([7]) Let µ be a probability measure on R. Given a truncated cone Γ α,β. Then exists a truncated cone Γ α,β n such that F µ(γ α,β ) Γ α,β. Proof. Fix a number α (0, ) such that γ = tan α, and choose β > 0 so large that (2.) F µ (u) u sin γ u, for all Iu > β and β > β. The relation (2.) is possible beacause α ug µ (u) = z u dµ(t), thus Fµ(u) u, forall u Γ z t u α. R We note the disk D u = {w w u < sin γ u }. We observe that F µ (u) D u if u Γ α,β Γ α, while the implication D u Γ α,β if u Γ α,β is justify in figure, where u Γ α,β, while Iu > Iu In the sequel we will use the following notation. If y 0, we denote = [ y,y] and = (, y) (y, + ). Proposition 2.. ([8]) Let {µ n } n= be a sequence of probability measures on R. The following assertions are equivalent (a) The sequence {µ n } n= converges weakly to a probability measure µ;
4 58 Bogdan Gheorghe Munteanu Figure : Graphic justfy of implication D u Γ α,β, u Γ α (b) There exist α,β > 0 such that the sequence (ψ νn ) n=, ψ, ψ Γ α,β and ψ νn (u) = 0( u ) if u, u Γ α,β ; (c) There exist α,β > 0 such that the functions ψ µn are defined on Γ α,β for every n, lim ψ µn (iy) exists for every y > β n uniformly in n as y. and ψ µn (iy) = o(y) Lemma 2.2. Let µ be a probability measure on R satisfying condition (i) of Theorem 2.. Then lim (IF µ(iy) y) = 0. Proof. By Proposition 2.,(ii) weak convergence of ν 0 to δ 0 implies that there exist α 0, β 0 > 0 such that ψ νn (u) 0 for u Γ α0,β 0. However, ψ νn (u) = nψ D/n µ(u) M n = n n ψ µ(nu) M n = ψ µ (nu) M n, which implies (2.2) lim n Iψ νn (u) = lim n Iψ µ (nu) = 0,
5 On the law of large numbers for all u Γ α0,β 0. By Lemma 2., exists α 2,α 3,β 2,β 3 > 0 such that α 0 > α 2 and β 0 < β 2 such that F µ has an inverse on Γ α2,β 2 and Γ α0,β 0 Γ α2,β 2 F µ (Γ α3,β 3 ),(α 2 > α 3 and β 3 > β 2 ). Therefore for any for all u Γ α3,β 3 it follows that (2.3) F µ (u) = u ψ µ (F µ (u)). Indeed, in condition ψ µ (u) = F (u) u replace u F µ (u) and we obtain eager relation. In particular, defining α,β such that α 0 > α > α 2 and β 0 < β < β 2 then the relation (2.2) holds u Γ α,β ( because Γ α0,β 0 Γ α,β ). Since nγ α,β = Γ α0,β 0, it is immediate now that n= (2.4) lim u ; u Γ α,β Iψ µ (u) = 0. Since Fµ(u) u u, u Γ α, we conclude that F µ (iy) iy, and hence F µ (iy). Then lim Iψ µ (F µ (iy)) = 0 (in relation (2.4) replace u by F µ (iy)). If in (2.3) we replace u by iy, we obtain IF µ (iy) y = Iψ µ (F µ (iy)), which concludes the proof. Lemma 2.3. Let σ be a finite positive measure on R such that lim yσ( ) = 0. Then the following hold (i) lim y y (ii) lim + +y y (iii) lim y k +y y y 2 +t 2 dσ(t) = 0; dσ(t) = 0; k+ dσ(t) = 0,k > 0. Proof. Consider the function y y, t [ y, y] y f y (t) = 2 +t 2 0, otherwise.
6 60 Bogdan Gheorghe Munteanu Then lim f y (t) = 0, for all t R and for all y > 0 we have f y (t) = y y y 2 + t 2 y2 y 2 + y L (σ). This condition is the condition of the Lebesque convergence theorem (2000,[4], + Theoreme 3.8., page 76), which implies that lim f y (t) dσ(t) = 0. The function g(t) = The result now follows because (2.5) y y But y y y y 2 +t 2 is increasing for t [0, y] and y >. y 2 + t 2 dσ(t) = y y f y (t) dσ(t) + y y y y 2 + t 2 dσ(t). dσ(t) y 2 +t 2 2 yσ( y ) if to count on inequality y. y+ 2 In condition (2.5) through on limit when y, lim y y y 2 + t 2 dσ(t) = lim yσ( y ) = 0. To prove (ii) we integrate by parts and we count on σ ( ) = σ (R) σ( ) = σ ([ y,y]) = 2y, we then have σ( t ) = 2 σ(t). }{{} t We have y y dσ(t) = 2 dσ(t) = 2 tσ(t) y 0 σ(t) dt 0 0 = 2yσ(y) 2 y 0 σ(t) dt = [ σ( )]σ(y) + = yσ( ) + y 0 y 0 σ( t ) dt. σ( t ) dt y
7 On the law of large numbers... 6 Then dσ(t) = y σ() + y σ( t ) dt. 0 It is clear that y σ() = o() if y. Select ǫ > 0 and choose N > 0 large enough such that tσ( t ) < ǫ, t N. Then, as y, y σ( t ) dt = 0 N 0 Nσ(R) ǫ + o(). σ( t ) dt + y N + y N ǫ t dt We will use the notation M y = t dµ(t) if y Z. tσ( t ) t Lemma 2.4. Let µ be a probability measure on R such that lim yµ( ) = 0. Then F µ (iy) = iy M y + o() as y. Proof. We will prove the estimate G µ (iy) = iy My y 2 + y 2 o() as y. We can estimate separately the real and the imaginary parts of G µ (iy). For the real part we have thus G µ (z) = RG µ (iy) = = = M y y 2 + dµ(t) z t = G µ(iy) = t y 2 + t 2 dµ(t) = ty 2 t 3 y 2 (y 2 + t 2 ) dµ(t) + dµ(t) iy t = t y 2 + t dµ(t) + 2 t 3 y 2 (y 2 + t 2 ) dµ(t) + t 3 y 2 (y 2 + t 2 ) dµ(t) + dt iy + t y 2 t 2 dµ(t), t y 2 + t 2 dµ(t) t y 2 + t 2 dµ(t). t y 2 + t 2 dµ(t)
8 62 Bogdan Gheorghe Munteanu On the other side, t 3 y 2 (y 2 + t 2 ) dµ(t) + y 2 + t dµ(t) 2 y 4 t y 2 + t dµ(t) 2 3 dµ(t) + 2y }{{} y 2 y 2 2+ dµ(t) Iy 3 y 2 (y 2 + t 2 ) dµ(t)+ dµ(t) }{{} µ() Lemma 2.3 (iii) o() + yµ( y 2 2y 2 y ) = o(). y 2 Thus, RG µ (iy) My + o(). The imaginary part y 2 y 2 IG µ (iy) = y y 2 + t dµ(t) = 2 y = dµ(t) + y y }{{} µ(r)= t 2 y 2 ± t 2 y 2 + t 2 dµ(t) t 2 y 2 + t 2 dµ(t) = y + y y 2 + t dµ(t) + 2 y y + t 2 dµ(t) + y 3 Lemma 2.3 (iii) y + y 2 o(). 2y µ() t 2 y 2 + t 2 dµ(t) ). ( But = y 2 = iy iy = iy G µ(iy) iy+m y o() iy+m y o() + My o() iy We say the development in Mac-Laurent series of function x +x x +... Then the reciprocal Cauchy transform is F µ (iy) = ( iy My o() iy ) = iy M y + o(). G µ(iy) =
9 On the law of large numbers Lemma 2.5. Let µ be a probability measure on R such that lim yµ( ) = 0 and z Γ /4. Then as z in Γ /4. d dz F µ(z) = + z o(), Proof. By the Nevanlinna representation according to (963,[2]) of F µ (z), where η(y) = F(z) = a + z tz t z dσ(t), z C+ wich implies IF µ (iy) = y + η(y), y(+t 2 ) y 2 +t 2 dσ(t). By Lemma 2.4, F µ (iy) = iy M y +o() wich implies IF µ (iy) = y. Then, through identicaly results that η(y) = o() as y. Observe that, for y > 0, yt 2 + y, then, η(y) y 2 +t 2 2 yt 2 y 2 +t 2 dσ(t) 2 yσ(). Therefore yσ( ) = o() as y. Again by the Nevanlinna representation we get d dz F µ(z) = + For z = x + iy Γ /4 wich implies x < y 4 and + + t 2 (z t) dσ(t) t 2 z t dσ(t) + t 2 t 2 + y 2 y 2 dσ(t) 2 yt 2 y 2 +t 2 dσ(t) + +t 2 + t 2 (x t) 2 + y 2 dσ(t) since here use the inequality t 2 + y 2 y 2 t2 +y 2 2 for all t. (z t) 2 dσ(t). + t 2 t 2 + y 2 dσ(t),
10 64 Bogdan Gheorghe Munteanu Furthermore, notice that I y +t 2 dσ(t) +y σ( t 2 +y 2 y+y y) = 2 y σ( y) = y y σ( y) = yσ( y y ) = y y o(). Here we used the fact that the function t +t2 is increasing for t > 0 and y >. t 2 +y 2 + t 2 On the other side, dσ(t) σ( t 2 + y 2 y) = y o(). Since y y }{{} z < 7, for z Γ 4 /4, we get the desired estimate. Lemma 2.6. Let µ be a probability measure on R such that lim yµ( ) = 0. Then ψ µ (iy) = M y + o() as y. Proof. By Lemma 2.4, F µ (iy) = iy M y + h(y) with lim h(y) = 0, while by Lemma 2.3(ii), M y = o(lny) if y, follows that F µ (iy) Γ /4 for y large enough. Thus for y large enough, iy = F µ (F µ (iy)) = F µ (iy M y +h(y)). However, let z = iy M y +h(y), then F µ (z) = z+m y h(y) = z+o( z ) as z and z Γ /4. By Lemma 2.5 and Fµ (z) = z + o( z ), we have that d F dz µ (z) = + k(z) with k(z) = o() as z and z Γ /4. Therefore z Fµ (iy M y + h(y)) Fµ (iy) = (iy M y + h(y) iy) d dz F We get that ψ µ (iy) = F µ (iy) iy = [ M y + h(y)] ( ) + k(γ) γ µ (z) z=γ = ( M y + h(y))( + o()) as y. = Fµ (iy M y + h(y)) +M y h(y) + M y o() h(y) o() iy }{{} iy = M y + o() as y.
11 On the law of large numbers Proof of the Theorem 2. Proof. (i) (ii): assume that µ satisfies condition (i) of the theorem. The Nevanlinna representation of F µ (z) implies for y > 0 that, wich is equivalent with F µ (iy) = a + iy + F µ (iy) = a + iy + + iyz t iy dσ(t) t y 2 t + i(yt 2 + y) t 2 + y 2 dσ(t), which means IF µ (iy) = y + η(y), with η(y) = + y(+t 2 ) RF µ (iy) = a + ξ(y), t 2 +y 2 dσ(t) and ξ(y) = + t( y 2 ) t 2 +y 2 dσ(t). By Lemmas 2.2 and 2.4, F µ (iy) = iy M y + o(), IF µ (iy) = y, we have that η(y) = o() if y, and the same argument used in Lemma 2.5, yσ( ) = o() as y. This estimate along with Lemma 2.3(i) allows us to conclude that ξ(y) = o( y) as y. Indeed ξ(y) y y y y y y y y t( y 2 ) t 2 + y 2 dσ(t) = o(), t 2 + y2 dσ(t) = y y y 2 y 2 (t 2 + y 2 ) dσ(t) Here we used the inequality y2 y 2 < for y >. We can now estimate the imaginary part of G µ (iy). Is know the fact that I z = Iz z 2. Then
12 66 Bogdan Gheorghe Munteanu (2.6) IG µ (iy) = I F µ (iy) = IF µ(iy) F µ (iy) 2 = y + o() R 2 F µ (iy) + I 2 F µ (iy) y + o() = (a + o( y + o() = y)) 2 + (y + o()) 2 a 2 + y 2 y + o() < = y 2 y + y o(). 2 By Lemma 2.4, IG µ (iy) = y + y + t 2 y 2 +t 2 dµ(t) wich implies yt 2 dµ(t) = o(). y 2 + t2 By proof of Lemma 2.5 results that yt 2 y 2 + t 2 dµ(t) yt 2 y 2 + t 2 dµ(t) 2 yµ(), which lead at yµ( ) = o() as y that is (ii). (ii) (i): suppose that µ satisfies condition (ii) of the theorem. We have ψ νn (z) = ψ µ (nz) M n where the ν n are defined as in condition (i) of the theorem. Notice that the functions ψ νn are defined on a certain truncated cone Γ α,β for every n. By Lemma 2.6 for every fixed y > β, ψ νn (iy) = ψ µ (iny) M n = M ny M n + o(), if n and y. Assuming without loss of generality that β >, the argument used in
13 On the law of large numbers Lemma 2.3, gives the following estimate M ny M n = t dµ(t) t dµ(t) I ny Therefore, lim ψ νn (iy) = 0. n Moreover I n I ny I n dµ(t) = nyµ( ny) nµ( n ) + nyµ( ny ) nµ( n ) + sup tµ( t ) t [n,ny] ny n ny n t dt µ( t ) dt nyµ( ny ) nµ( n ) + sup tµ( t ) = o(). t [n,ny] lim sup ψ νn (iy) y 2k + k lim sup y 2k lim sup + k lim sup y y = 0. Hence Proposition 2.(c), ν n converges weakly to a measure ν and ψ ν (iy) = 0 for every y > β. The identicaly theorem then implies that ψ ν (z) = 0, for every that z Γ α,β which implies that ν = δ 0. = References [] Bercovici, H. and Voiculescu, D., Free convolution of measures with unbounded support, Indiana U. Math. J. 42, [2] Akhiezer, N.I. and Glazman, I.M., Theory of linear operators in Hilbert space, Frederick Ungar, 963. [3] Maassen, H., Addition of freely independent random variables, J.Funct. Anal., 06(2) (992),
14 68 Bogdan Gheorghe Munteanu [4] Orman, V.G., Măsura si procese aleatoare, Editura Universităţii Transilvania, Braşov, Romania, [5] Voiculescu, D., Symmetrics of some reduced free product C -algebras, Lecture Notes in Mathematics 32, Springer, Buşteni, Romania, 983. [6] Voiculescu, D., Addition of certain non-commuting random variables, J.Funct. Anal., 66 (986), [7] Voiculescu, D.,Free noncommutative random variables, random matrices and the II factors of free groups, University of California, Berkeley, 990. [8] Wigner, E.P., Characteristic vectors of bordered matrices with infinite dimensions, Ann. of Math. 62 (955), [9] Wigner, E.P., On the distributions of the roots of certain symmetric matrices, Ann. of Math. 67 (958), Faculty of Mathematics and Computer Science Transilvania University Str.Iuliu Maniu 50 RO Braşov,România b.munteanu@unitbv.ro
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