Definition 5.5.1. Let ind ∈ {bool, free, mono, mono †}. We define the ind-convolution of two B-valued laws µ and ν as the law of X + Y when X and Y are ind-independent and the law of X is µ and the law of Y is ν. The convolution is denoted by µ indν, or alternatively
µ ] ν (boolean case) µ ν (free case) µ B ν (monotone case) µ C ν (anti-monotone case).
In order to verify this definition makes sense, observe first that using the GNS construc-tion (Theorem 2.6.5) and the product space construcconstruc-tion, there always exist independent operators X and Y in a B-valued probability space (A, E) such that the law of X is µ and the law of Y is ν. And second, the law of X + Y is uniquely determined by µ and ν and the independence of X and Y by Lemma 5.2.8.
More generally, given laws µ1, . . . , µN, we may construct operators X1, . . . , XN which are independent with Xj having the law µj, using the GNS construction and the product space construction. Then it follows from the associativity considerations of the previous section that X1+ · · · + XN has the law µ1ind(µ2ind(. . . (µN) . . . ). Moreover, we also have µ1ind(µ2indµ3) = (µ1indµ2) indµ3, or in other words ind is associative, and thus we may remove the parentheses when expressing an iterated ind-convolution.
Observation 5.5.2.
(1) The operations ], , B, and C are associative.
(2) The operations and ] are commutative.
(3) We have µ B ν = ν C µ.
Proof. The first claim follows from the preceding discussion of associativity. The second claim is true because the conditions defining free and boolean independence do not depend on the order of the subalgebras, while for the third claim, if we were to reverse the order of the indices in monotone independence, then we obtain anti-monotone independence.
Our main task in this section is to develop analytic tools for computing the independent convolution of two laws. In the classical case, this role is played by characteristic function (Fourier transform) of a law given by F µ(ξ) = R eixξdµ(x), since addition of independent random variables or classical convolution of laws corresponds to multiplication of the Fourier transforms. In the non-commutative setting, this role is played by various fully matricial functions related to the Cauchy-Stieltjes transform.
5.5.1 The boolean case
The results of this section can be found in [SW97] [Ber06, Theorem 2.2] for the scalar case and [PV13, §2 and §5.3] in the operator-valued case. The proof we give here is based on analogy with the proof from the free case in the next subsection (where we also explain the history and references).
Definition 5.5.3. For a B-valued law µ, we define the K-transform as Kµ(z) := z − Fµ(z).
Remark 5.5.4. We caution that some authors work instead with Bµ(z) = ˜Kµ(z) or slight variants of this definition. We showed in Theorem 4.5.3 that Kµ(z) = µ(X)(n)+ Gσ(z) for some generalized law σ.
Theorem 5.5.5. Kµ]ν(z) = Kµ(z) + Kν(z) as fully matricial functions.
Proof. Let X and Y be freely independent random variables in (A, E) which realize the laws µ and ν respectively. For z in Mn(B) with kzk < 1/ rad(µ), define
UX(n)(z) = (1 − zX(n))−1− 1 =
∞
X
k=1
(zX(n))k.
This is an A-valued fully matricial function. To simplify the notation, we will suppress all the superscripts (n), so that X will stand for X(n), where n is the size of the matrix z. Note that
1 + E[UX(z)] = E[(1 − zX)−1] = ˜Gµ(z)z−1 or in other words
(1 + E[UX(z)])−1 = z ˜Fµ(z)
Note that UX(z) is in the closed span of BhXi0. Define UY(z) analogously. Then 1 − zX − zY = (1 + UX(z))−1+ (1 + UY(z))−1− 1
Therefore,
(1 − zX − zY )−1 = [(1 + UX(z))−1+ (1 + UY(z))−1− 1]−1
= (1 + UX(z))[1 − UY(z)UX(z)]−1(1 + UY(z))
= (1 + UX(z))
∞
X
k=0
(UY(z)UX(z))k
!
(1 + UY(z)).
Next, we take the expectation. Because UX(z) and UY(z) are in the closures of Mn(BhXi0) and Mn(BhY i0) respectively and because X and Y are Boolean independent, we have
E[(1 − zX − zY )−1] = (1 + E[UX(z)])
∞
X
k=0
(E[UY(z)]E[UX(z)])k
!
(1 + E[UY(z)])
= [(1 + E[UX(z)])−1+ (1 + E[UY(z)])−1− 1]−1 Therefore,
G˜µ]ν(z)z−1 = [(1 + E[UX(z)])−1+ (1 + E[UY(z)])−1− 1]−1 By taking reciprocals,
z ˜Fµ]ν(z) = (1 + E[UX(z)])−1+ (1 + E[UY(z)])−1− 1
= z ˜Fµ(z) + z ˜Fν(z) − 1,
Because z ˜Fµ(z) − 1 = z ˜Kµ(z) and the same holds for Y and X + Y , this means precisely that
z ˜Kµ]ν(z) = z ˜Kµ(z) + z ˜Kν(z)
for z in a neighborhood of 0. By Corollary 3.9.7, we have Kµ]ν = Kµ+ Kν on the upper half plane.
5.5.2 The free case
The following analytic transforms were defined by Voiculescu [Voi86]. In the operator-valued case, the definition was developed by Dykema [Dyk07, §6].
Definition 5.5.6. For a B-valued law µ, we define Fµ(z) = Gµ(z)−1 and Φµ(z) := Fµ−1(z) − z,
where Fµ−1(z) is the functional inverse and z is in the image of Fµ.
Remark 5.5.7. Many authors work with the R-transform Rµ(z) = ˜Φµ(z) = Φµ(z−1). We showed in Lemma 4.5.8 that Φµ is defined for Im z ≥ 2δ whenever δ > kVarµ[1]k1/2 and in Lemma 4.5.9 that Rµ(z) is defined in a fully matricial ball around zero.
The following result on the additivity of the R-transform was discovered in the scalar-valued case by Voiculescu [Voi86]. The original proof by Voiculescu used canonical realiza-tions of a law µ by (non-self-adjoint) random variables on a Fock space, and this was adapted to the operator-valued setting by Dykema [Dyk07, §6]. This theorem can also be proved through the combinatorial apparatus of free cumulants due to Speicher [Spe94, Spe98]. The analytic proof presented here is due (in the scalar-valued setting) to Lehner [Leh01, Theorem 3.1].
Theorem 5.5.8. For Im z ≥ 2δ > 2kVarµ(1) + Varν(1)k1/2, we have Φµν(z) = Φµ(z) + Φν(z).
Also, for z in a fully matricial neighborhood of 0, we have Rµν(z) = Rµ(z) + Rν(z)
Proof. Let X and Y be freely independent random variables in (A, E) which realize the laws µ and ν.
We begin by analyzing Rµ(z) in a neighborhood of the origin. Now z−1+ Rµ(z) is the functional inverse of Gµ(z) in a neighborhood of 0 which means that
E[(z−1+ Rµ(z) − X)−1] = z.
Multiplying by z−1 on the right, we can write rewrite this as E[(1 + zRµ(z) − zX)−1] = 1, or in other words, the A-valued fully matricial function
UX(z) = (1 + zRµ(z) − zX)−1− 1
has expectation zero. (Here, as in the previous case, we suppress the superscripts (n) but UX(z) stands for UX(n)(z) and X denotes X(n) where n is the size of the matrix z). The same holds for the analogously-defined function UY(z). We want to show that z−1− Rµ(z) − Rν(z) is the functional inverse of Gµν, which means that
Gµν(z−1+ Rµ(z) + Rν(z)) = z, which after multiplying by z−1 on the right is equivalent to
E[(1 + zRµ(z) + zRν(z) − zX − zY )−1] = 1.
We will rewrite the left hand side in terms of UX(z) and UY(z) so that we can apply freeness together with the fact that UX(z) and UY(z) have expectation zero. Note that
(1 + zRµ(z) + zRν(z) − zX − zY )−1
= [(1 + UX(z))−1+ (1 + UY(z))−1− 1]−1
= (1 + UX(z))[(1 + UY(z)) + (1 + UX(z)) − (1 + UY(z))(1 + UX(z))]−1(1 + UY(z))
= (1 + UX(z))[1 − UY(z)UX(z)]−1(1 + UY(z)).
Now because UX(0) = 0 = UY(0), we know that if kzk is sufficiently small, then we can expand [1 − UY(z)UX(z)]−1 as a geometric series, and thus for small z,
(1 − zRµ(z) − zRν(z) − zX − zY )−1 = (1 + UX(z))
∞
X
k=0
(UY(z)UX(z))k
!
(1 + UY(z)).
Next, we take the expectation. Because UX(z) and UY(z) have expectation zero and because X and Y are free, all the terms on the right hand side have zero expectation except the term 1 which comes from multiplying together the 1 from 1 + UX(z), the 1 from the geometric series, and the 1 from 1 + UY(z). Therefore, as desired,
E[(1 − zRµ(z) − zRν(z) − zX − zY )−1] = 1.
This shows that
Rµν(z) = Rµ(z) + Rν(z) holds in a neighborhood of zero.
This implies that Φµν = Φµ+ Φν if Im z is sufficiently large, and hence by Corollary 3.9.7, we have Φµν = Φµ + Φν on H+,2δ(B), provided that this lies inside the common domain of Φµν, Φµ, and Φν. Since Varµν(1) = Varµ(1) + Varν(1) and all these elements are positive, we have kVarµν(1)k ≥ max(kVarµ(1)k, kVarν(1)k), and hence it is sufficient that δ > kVarµν(1)k1/2.
5.5.3 The (anti-)monotone case
The following result is due to [Mur00, Theorem 3.1] in the scalar-valued case and [Pop08, Theorems 3.2 and 3.7] in the operator-valued case, whose proof we follow here. Another proof in the scalar case is in [Ber05].
Theorem 5.5.9. We have FµBν(z) = Fµ(Fν(z)) and FµCν(z) = Fν(Fµ(z)) as fully matricial functions.
Proof. Let inv denote the fully matricial function z 7→ z−1 where defined. Since Fµ = inv ◦ ˜Gµ◦ inv and inv is an involution, it suffices to show that ˜GµBν = ˜Gµ◦ ˜Gν.
Let X and Y be monotone independent random variables in (A, E) realizing the laws µ and ν. We know that
E[f0(Y )g1(X)f1(Y ) . . . gn(X)fn(Y )] = E[E[f0(Y )]g1(X)E[f1(Y )] . . . gn(X)E[fn(Y )]]
whenever f (Y ) ∈ BhY i0 and f (X) ∈ BhXi0. However, this also holds trivially if fj(Y ) ∈ B, and thus by linearity it holds when fj(Y ) ∈ BhY i.
Now for kzk sufficiently small, we have G˜µBν(z) = E[(1 − zX − zY )−1z]
= E[(1 − (1 − zY )−1zX)−1(1 − zY )−1z] = E
" ∞ X
k=1
[(1 − zY )−1zX]k(1 − zY )−1z
# .
Note that (1 − zY )−1 is in the closure of Mn(BhY i) and zX ∈ Mn(BhXi0) and hence by monotone independence
G˜µBν(z) = E
" ∞ X
k=1
[E[(1 − zY )−1z]X]kE[(1 − zY )−1z]
#
= E
" ∞ X
k=1
[ ˜Gν(z)X]kG˜ν(z)
#
= ˜Gµ◦ ˜Gν(z).
This equality extends to all z by Corollary 3.9.7. The anti-monotone case follows from the monotone case since µ C ν = ν B µ.