We study the problem of conditional expectations in free random variables
and provide closed formulas for the conditional expectation of a resolvent
$\Psi(z) = (1-zP(a_1,\ldots,a_n))^{-1}$  of an arbitrary non–commutative
polynomial $P(a_1,\ldots,a_n)$ in free random variables
$a_1,\ldots,a_n$ onto the subalgebra of an arbitray subset of the
variables $a_1,\ldots,a_n$.
More precisely, given a linearization of $\Psi(z)$, our methods allow to compute a
linearization of its conditional expectation.
The coefficients of the expressions obtained in this process involve certain  Boolean
cumulant functionals which can be computed by solving a system of equations.
On the way towards the main result we introduce a   non–commutative
differential calculus which allows to evaluate
conditional expectations and certain Boolean cumulant functionals $\bbeta{}$
and $\fbeta{}$.
We conclude the paper with several examples which
illustrate the working of the developed machinery.
For completeness two appendices complement the paper.
The first appendix contains a purely algebraic approach to Boolean cumulants
and the second appendix provides a crash course on linearizations of rational

[2020]Primary: 46L54. Secondary:  15B52.

§ INTRODUCTION

§.§ Outline

Free probability was introduced by Voiculescu 40 years ago
[42]
as a means to solve long-standing problems in von Neumann algebras.
Over the years however deep connections to several branches of mathematics came
to light, among others random matrix theory and representation theory of the
symmetric group.
Complementing Voiculescu's analytic approach, Speicher developed a systematic theory of
free cumulants [35], already announced in
In this approach, free independence is characterized by vanishing of mixed free
cumulants, in analogy to classical independence, which can by characterized by
vanishing of mixed classical cumulants.
Indeed, most properties of free cumulants can be obtained from the
corresponding properties of classical cumulants by replacing the lattice of all
set partitions by the lattice of non–crossing partitions,
following the general scheme of multiplicative functions on lattices  [12],
see the standard reference [31] for a detailed treatment of free cumulants.
The discovery of free cumulants triggered a lot of progress in free
and it was the starting point of many deep combinatorial studies of various
structures in free probability
(see [30, 21, 22] and many other).

“Partial cumulants” were introduced by von Waldenfels in order
to simplify certain calculations in mathematical physics
Later they were called Boolean cumulants corresponding to the notion of
Boolean independence which was introduced in [36].
They naturally appear in the guise of first
return probabilities of random walks [48].
Combinatorially Boolean cumulants follow the pattern of classical and free cumulants by
replacing the lattice of set partitions (resp. noncrossing partitions)
with the lattice of interval partitions which is isomorphic to the Boolean lattice.
From a combinatorial point of view Boolean cumulants are the
simplest kind of cumulants.

Recently it was noticed that despite their simplicity
Boolean cumulants are useful for non–commutative probability in general
[18] and free probability in particular
[5, 14, 23, 37].
Boolean cumulants were used for the first combinatorial solution to the problem
of the free anti–commutator in [14] (an analytic solution was found
earlier by Vasilchuk [41]; a solution in terms of free
cumulants was presented recently in [32]),
and for the identification of the coefficients of power series expansion of
subordination functions [23]
(implicitly also in  [48]).

In the present paper we continue these investigations and show that Boolean
cumulants may indeed be used for a systematic study of free random variables.
The first step in this direction is a surprisingly simple characterization of
freeness by the vanishing of some mixed Boolean cumulants.

Let $(\alg{M},\varphi)$ be a tracial non–commutative probability space. Subalgebras $\alg{A}$ and $\alg{B}$ are free if and only if  $\beta_{m}(a_1,a_2,\ldots,a_n)=0$ whenever $n>1$ and  $a_j\in\alg{A}\cup\alg{B}$ for $j=1,2,\ldots,n$ with $a_1$ and $a_n$ coming from different subalgebras.

This property turns out to be the key to an efficient calculation of conditional
expectations in free random variables.
In particular, for free random variables $a_1,a_2,\ldots,a_n$ we determine the
conditional expectation of the resolvent $(1-z P(a_1,a_2,\ldots,a_n))^{-1}$
of an arbitrary non–commutative polynomial $P(a_1,a_2,\dots,a_n)$ onto a subalgebra
generated by some subset of the variables $a_1,a_2,\ldots,a_n$.
To this end we employ the concept of linearizations, a method also used in
[4, 16].
More precisely, given a linearization of the resolvent, we obtain a
linearization of its conditional expectation.
Our discussion is focused on non–commutative rational functions of the form
$(1-z P(a_1,a_2,\ldots,a_n))^{-1}$, however an example at the end of this paper
shows that the methods  are applicable to a wider class of rational functions.

The first step of our approach is based on a recurrence which
can be summarized as a free integration formula and
which in the case of resolvents naturally leads to a system of linear equations
for the conditional expectation
which can be turned into a linearization.
The coefficients  appearing  in said linearization are certain
generating functions of mixed Boolean cumulants of free random variables.

For the sake of simplicity let us consider the case of non–commutative
polynomials in two free random variables $X$ and $Y$.
It is known that for free random variables the conditional expectation of a
polynomial is a polynomial again.
We show how to obtain this result in a recursive way.
The functional $\bbeta{Y}$ is defined on non–commutative polynomials and
depends on the distributions of $X$ and $Y$.
The derivative $\rdelta_{X}$ acts on non–commutative polynomials.
For precise definitions we refer to Section <ref>.

The conditional expectation of a non–commutative polynomial $P\in\IC\langle
satisfies the identity

\begin{align*}
\E_{X}[P]=\bbeta{Y}(P)+(\bbeta{Y}\otimes\E_{X})(\rdelta_{X}(P)).
\end{align*}

This formula has a certain resemblance to classical integration.
$\E_X$ “integrates away” the variable $Y$ and if we denote
$\mathfrak{I}_Y=\bbeta{Y}\otimes \E_X$, then the formula reads

\mathfrak{I}_Y(\rdelta_{X}(P)) = \E_{X}[P] - \bbeta{Y}(P)
= \mathfrak{I}_Y(I\otimes P - P\otimes I).

The formula above allows to calculate conditional expectations in terms of Boolean cumulants. Thus we are faced with the problem to calculate Boolean cumulants of functions
in free random variables.
In order to do this we introduce an algebraic calculus of Boolean cumulants,
based on a number of algebraic rules and devices which allow to establish equations for the generating
functions arising from the previous calculations.
More precisely we introduce two functionals $\bbeta{}$ and
$\fbeta{}$ on the free algebra which evaluate Boolean cumulants in two ways,
blockwise and fully factored.
We then establish mutual recursive equations between these
with the help of the previously developed algebraic devices
which lead to a closed system of algebraic equations.
Although the functionals $\bbeta{}$ and $\fbeta{}$ are defined in
relation to conditional expectations, it appears that they will have much wider
applications, and the calculus we introduce for them is of independent
The calculus for $\bbeta{}$ and $\fbeta{}$, based on generalizations of
observations from [14, 23],
subsumes the combinatorial case-by-case analysis of functions in
free variables into a general algebraic machinery.

First steps towards a similar calculus in terms of free cumulants were done in [10]
and we expect  the unshuffle algebras of [13] to play a role here.
Mixed free cumulants of free random variables vanish and they turn additive free
convolution into a simple addition.
However it turned out that when it comes to multiplicative free convolution,
free cumulants offer no advantage as the formulas are actually identical
The advantage of Boolean cumulants lies in their combinatorial simplicity.
The main steps of our calculations are as follows:

* Turn the simple combinatorics of Boolean cumulants of products of free
variables into an algebraic rule involving certain derivations
which appeared earlier in free probability.

* Apply these derivations to resolvents and use the fact that these
play the role of “eigenfunctions”  analogous to the
exponential function in classical calculus.

* Use the formula for Boolean cumulants with free entries in order to separate
the variables and obtain equations for the generating functions.

We do not prove any new combinatorial results, rather provide algebraic reformulations
of known combinatorial identities and put them
together into an effective machinery to make them available for free analysis.
It allows to establish equations for generating
functions of Boolean cumulants of functions in free variables
which in some cases can be effectively solved
and used to compute conditional expectations and distributions
of arbitrary polynomials in free random variables.

In particular we present an algebraic interpretation
of the formula for Boolean cumulants with products as
entries and a characterization of freeness in terms of Boolean
cumulants from [14] in terms of Voiculescu's free
We refer to Theorem <ref> for the precise statement and
Definitions <ref> and <ref> for definitions of
the functionals $\bbeta{}$ and $\fbeta{}$.

The paper is organized as follows.

The rest of the introduction is devoted to an exposition of the problem of
conditional expectations.

Section <ref> presents results about the basic ingredients: conditional
expectations, Boolean cumulants and derivations.

In Section <ref> we prove the characterization of freeness announced above in Theorem <ref>.

In Section <ref> we provide a method to compute conditional expectations
in terms of Boolean cumulants. In particular we prove Theorem <ref>.

In Section <ref>
we introduce a calculus for the Boolean functionals $\bbeta{}$ and
$\fbeta{}$ which is summarized in Theorem <ref>.

In Section <ref> we show how linearizations allow to solve the
problem for subordinations for general polynomials. In particular we prove Theorem <ref> which we discuss below.

Section <ref>  contains examples.

Appendix <ref> we give self-contained algebraic proofs
of the basic results about Boolean cumulants as well as a reformulation of
these in terms of tensor algebras.

Finally Appendix <ref> contains the essential information
required for the computation of linearizations.

§.§ Subordination for general polynomials

Let $X$ and $Y$ be classically independent random variables with distributions $\mu_X$ and $\mu_Y$,
then their joint distribution is $\mu_{X,Y}=\mu_X\otimes\mu_Y$, i.e.,
the expectation of any integrable function $f(X,Y)$ can in practice
be computed as a double integral

\IE f(X,Y) = \int\int f(x,y)\, d\mu_Y(y)\,d\mu_X(x)

In the non–commutative case there is no such integral representation,
however the inner integral is in fact the conditional expectation

\begin{equation}
\label{eq:intfXy=EfXY}
\int f(X,y) \,d\mu_Y(y) = \IE[f(X,Y)|X]
\end{equation}

and thus

\IE f(X,Y) =
\IE [ \IE[f(X,Y)| X] ]

which does have a non–commutative analogue.
In the present paper we propose a method to compute
this non–commutative conditional expectation

\IE_X[P(X,Y)]

for arbitrary non–commutative polynomials $P$ and more general rational
functions in free random variables.
It is the analogy with (<ref>)
which motivated us to call our endeavour free integral calculus.

This follows ideas of Voiculescu [43] and Biane
[7] who showed that for the sum $a+b$ of two
free random variables $a$, $b$  there exists an analytic self map of
$\omega:\IC^+\to\IC^+$ such that
the conditional  expectation of the resolvent onto the algebra
generated by $a$ is a resolvent again

\begin{equation}
\label{eq:subordination}
\IE_a[(z-a-b)^{-1}]=(\omega(z)-a)^{-1}
\end{equation}

which after application of $\varphi$ yields results in the subordination
relation $G_a(\omega(z))=G_{a+b}(z)$.
Moreover the function $\omega$ is known to satisfy a fixed point equation.

In the present paper we generalize this method to arbitrary polynomials using an algebraic-analytic method based on
the observation from our previous work [14, 23] that Boolean
cumulants turn out to be a convenient tool to “store” the
results of the “partial integral” described above.

Fix a non–commutative probability
space $(\alg{M},\varphi)$.
Given self-adjoint random variables $a_1,a_2,\ldots,a_n\in\alg{M}$ and a
non-commutative polynomial $P\in \IC\langle X_1,X_2,\ldots,X_n\rangle$,
our objective is an explicit formula for the conditional expectation of $
\bigl(1-zP(a_1,a_2,\ldots,a_n)\bigr)^{-1}$
onto the algebra generated by some subset of the variables $a_1,a_2,\ldots,a_n$. Without loss of generality we may assume that the subset consists of $a_1,a_2,\ldots,a_k$, where $k<n$.

In the first step we construct a linearization of the resolvent, i.e.,
matrices $C_1,C_2,\ldots,C_n\in M_N(\IC)$ such that for $L=C_1\otimes a_1+C_2\otimes a_2+\dots +C_n \otimes a_n\in M_N(\IC[z])\otimes \alg{M}$ and $z$ in some neighborhood of zero we have

\begin{equation}
\label{eq:intro:linearization}
(1-z^mP(a_1,a_2,\dots,a_n))^{-1}=u^t (I_N-zL)^{-1}v
\end{equation}

for some vectors $u,v\in\IC^N$, where $m$ is the total degree of $P$.
In general a polynomial may have many linearizations;
in Appendix <ref> we discuss in detail algorithms for finding
linearizations which work for our purposes.
It suffices to say for the moment that in contrast to
[4] some technical issues force us to
work with regular linearizations which are not self-adjoint
and  to  restrict the calculations to the level of formal power series.

In the case when  $\alg{A}$ is the von Neumann subalgebra freely generated by $a_1,a_2,\ldots,a_n$,
we prove the following theorem.
It follows by evaluating  the formal expression  from
Theorem <ref> below in the variables $a_1,a_2,\dots,a_n$.

Given a linearization   (<ref>) for a polynomial
$P\in\IC\langle X_1,X_2,\dots,X_n\rangle$ of degree $m$
we have for $z$ in some neighbourhood of $0$ the following linearization for
the conditional expectation of the resolvent

\begin{multline}
\label{eq:thm:1.1}
\E_{\alg{A}}\left[(1-z^mP(a_1,a_2,\ldots,a_n))^{-1}\right]
\\
=u^t\left(I_N-z\left(C_1 \otimes a_1 +C_2 \otimes a_2 +\dots+C_k\otimes a_k+ (C_{k+1}
F_{k+1}+\dotsm+C_n F_n)\otimes I \right)\right)^{-1}v
\end{multline}

where the matrices $F_1,F_2,\ldots,F_n$ constitute the unique
fixed point of the system of equations

\begin{equation}
\label{eq:thm:1.1:eta}
\biggl(z\Bigl(I_N-z \sum_{j\neq i} C_jF_j\Bigr)^{-1} C_i\biggr)
\qquad\text{ for $i=1,2,\ldots,n$}
\end{equation}

with entries which are analytic at $0$.
Here by  $\widetilde{\eta}_a(z)=\sum_{n=1}^{\infty} \beta_{n}(a)z^{n-1}$
we denote the shifted Boolean cumulant generating function of a random variable $a$.

* From a practical point of view Theorem <ref>
asserts that the evaluation of the conditional expectation of the resolvent

\[
\left(
\right)^{-1}=u^t \left(I_N - z\left(C_1\otimes a_1+C_2\otimes a_2+\dots+ C_n \otimes a_n\right)\right)^{-1}v
\]

onto $\alg{A}$ (i.e., “integrating out” $a_{k+1},\ldots,a_n$)
amounts to  replacing the corresponding summands $I_N\otimes a_i$ with
the matrices $F_i\otimes I$.

* Although only the matrices  $F_{k+1},F_{k+2},\ldots,F_n$ explicitly appear in
the final formula     (<ref>),
the matrices $F_1,F_2,\ldots,F_k$ are required as well in order to extract
the former from the solution of equation (<ref>).
Moreover observe that in general the equations cannot be effectively decoupled
(unless $n=2$, see example <ref> below)
and each matrix from $F_1,F_2,\ldots,F_n$ depends on the distributions of all variables $a_1,a_2,\ldots,a_n$.

The previous theorem generalizes the subordination phenomenon in the
following sense.
For simplicity consider the case of two free variables
$a,b$ and fix a polynomial $P\in \IC \langle X,Y\rangle$ of
degree $m$. We fix an $N\times N$ linearization

\[
=u^t(I_N-z C_1X-z C_2Y)^{-1}
\]

then the above theorem says that the moment
generating function $M_P(z)=\varphi\left((1-z P(a,b))^{-1}\right)$ is
from the linearization via

\[
M_{P(a,b)}(z^m)=u^t\varphi^{(N)} (I_N-z C_1 a-z C_2 b)^{-1} v
\]

where $\varphi^{(N)}$ is the entry–wise application of $\varphi$.
Let $H_1=(I-zC_1 F_1)^{-1}$, then the preceding identity can be written as

\[M_{P(a,b)}(z^m)=u^t\varphi^{(N)} (I_N-z H_1 C_2 b)^{-1} H_1 v\]

Suppose that the matrix $H_1C_2$ is diagonalizable and write $zH_1C_2=Q D Q^{-1}$ with $D=\diag(\lambda_1,\ldots,\lambda_n)$. Then we obtain

\[M_{P(a,b)}(z^m)=(Pu)^t\varphi (I_N-D b)^{-1} P^{-1}H_1
\begin{bmatrix}
M_{b}(\lambda_1) & 0 & \ldots & 0 \\
0&  M_{b}(\lambda_2)   & \ldots &0 \\
& \vdots & \ddots   & \vdots \\
0   & 0 & \ldots &M_{b}(\lambda_n)
\end{bmatrix}
\widehat{v}\]

where $\widehat{u}=Pu$, $\widehat{v}=P^{-1}H_1 v$ and
$M_{b}(z) = \varphi((1-zb)^{-1})$
is the moment generating function of $b$.

Therefore the eigenvalues $\lambda_i=\lambda_i(z)$ of $z H_1 C_2$ can be understood as a generalization of the subordination functions from free additive convolution.
At the time of this writing we do not know whether $H_1 C_2$ is actually diagonalizable in general.
If this turns out  not to be the case, one can use the Jordan canonical form and the
derivatives $M_{b}'$, $M_{b}''$,… will populate the upper triangular
parts fo the  Jordan blocks.

Let us illustrate Theorem <ref> with a quick derivation of the
subordination function for additive free convolution and for anti–commutator.
For further examples see Section <ref>.

The polynomial $P(a,b)=a+b$  is already linear and  we can apply Theorem <ref> with the trivial linearization

\begin{align*}
\E_{a}\bigl[(1-z(a+b))^{-1} \bigr]
=(1-z a-z F_b(z))^{-1}
\end{align*}

\begin{align*}
F_a(z) &= \widetilde{\eta}_{a}\Bigl(\frac{z}{1-zF_b(z)}\Bigr) &
F_b(z) &= \widetilde{\eta}_{b}\Bigl(\frac{z}{1-zF_a(z)}\Bigr)
\end{align*}

Both formulas for the conditional expectation and the two equations can
be easily checked to be equivalent to the well known subordination
results for free additive convolution. In particular in this case it is
straightforward to decouple the equations and obtain separate fixed point equations for $F_1$ and $F_2$
after a simple substitution of one equation into the other.

Let $P(a,b)=a b+b a$ be the anti–commutator, then for $z$
in some neighbourhood of zero
the conditional expectations of the resolvent are

\begin{align*}
\IE_{a}\left[(1-z^2(ab+ba))^{-1}\right]&=\left(1-f_{2,43} z-z^2 a \left(f_{2,33} +f_{2,44} \right)-f_{2,34} a^2 z^3\right)^{-1},\\
\IE_{b}\left[(1-z^2(ab+ba))^{-1}\right]&=\left(1-f_{1,12} z-z^2 b \left(f_{1,11} +fx_{22} \right)-f_{1,21} b^2 z^3\right)^{-1}.
\end{align*}

This result is obtained with the linearization involving the matrices

\[
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0   & 0 \\
1 & 0 & 0   & 0 \\
0   & 1 & 0 & 0
\end{bmatrix}
\qquad
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.\]

Apriori the equations (<ref>) involve a
total of 32 variables $f_{k,ij}$, $k\in\{1,2\},1\leq i,j\leq 4$.
However it is easy to see with the help of the projection
matrices onto $\ker C_1$ and $\ker C_2$
(cf. Remark <ref>)
that many variables vanish and the solution matrices have the form

\[
F_1 =
\begin{bmatrix}
f_{1,11} & f_{1,12} & 0 & 0 \\
f_{1,21} & f_{1,22} & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\,
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & f_{2,33} & f_{2,34} \\
0 & 0 & f_{2,43} & f_{2,44}
\end{bmatrix}
\]

and satisfy the following system of equations

\begin{align*}
\begin{cases}
F_{1}&=\widetilde{\eta}_{a}\left(z Q_1 (1-zC_2 F_2)^{-1}C_1\right),\\
F_{2}&=\widetilde{\eta}_{b}\left(z Q_2 (1-zC_1 F_1)^{-1}C_2\right),
\end{cases}
\end{align*}

where $I-Q_i$ is the projection onto the kernel of the matrix $C_i$,
This system gives the analogue for the anti–commutator of the fixed point equation in the case of additive convolution stated in the previous example.
Note that the explicit formulas for the matrices $H_1=Q_1
(1-zC_2F_2)^{-1}C_1$ and $H_2=Q_2 (1-zC_1 F_1)^{-1}C_2$ are
essentially the same as those for matrices $H_a$ and $H_b$ from
Theorem 6.1 in [14], i.e., linearizations were already
implicitly present in [14].

§ PRELIMINARIES

In this section we introduce the main ingredients of the paper:
conditional expectations, Boolean cumulants and derivations.

§.§ Non–commutative probability spaces

We assume that $\alg{M}$ is a unital $*$-algebra and
$\varphi:\alg{M}\mapsto\IC$ is a positive unital linear functional, commonly
called a state and we usually assume it to be faithful.
We will refer to the pair $(\alg{M},\varphi)$ as a non–commutative probability
For technical reasons we will mostly work with the free associative algebra.

Let $\alg{X}=\{X_1,X_2,\ldots,X_n\}$ be an alphabet.
We denote by $\alg{X}^+=\{X_{i_1}X_{i_2}\cdots X_{i_k}\mid k\in\IN, i_j\in
\{1,2,\dots,n\}\}$
the free semigroup it generates and by
$\alg{X}^*=\alg{X}^+\cup\{1\}$ the free monoid.

We denote by $\IC\langle\alg{X}\rangle= \IC\langle X_1,X_2,\ldots,X_n\rangle$ the free associative
algebra generated by the variables $X_1,X_2,\ldots,X_n$, i.e., the linear
span of $\alg{X}^*$, also known as the algebra of
non–commutative polynomials.

For elements $a_1,a_2,\ldots,a_n\in\alg{M}$ and $P\in\IC\langle
we denote
by $P(a_1,a_2,\ldots,a_n)$ the evaluation of a
polynomial $P\in\IC\langle X_1,X_2,\ldots,X_n\rangle$, i.e., the element
of $\alg{M}$ obtained after substituting every $X_i$ with $a_i$ for

The joint distribution of $a_1,a_2,\ldots,a_n$ is the linear functional
$\mu:\IC\langle X_1,X_2,\ldots,X_n\rangle\to\IC$ given by

\[\mu(P)=\varphi\left(P(a_1,a_2,\ldots,a_n)\right).\]

A family of subalgebras $(\alg{A}_i)_{i\in I}$ of a ncps $(\alg{M},\varphi)$ is called
free or free independent if

\begin{equation*}
\varphi(u_1u_2\dotsm u_n)=  0
\end{equation*}

for any choice of $u_j\in \bigcup_i\alg{A}_i$ such that $\varphi(u_j)=0$ and
$u_j\in\alg{A}_{i_j}$ with $i_j\ne i_{j+1}$ for all $j\in\{1,2,\dots,n-1\}$.

§.§ Conditional expectations

Fix a non–commutative probability space $(\alg{M},\varphi)$
and let $\alg{A}\subseteq \alg{M}$ be a subalgebra.
A conditional expectation
is a state-preserving projection $\E_{\alg{A}}:\alg{M}\to \alg{A}$,
such that $\varphi\circ\E_{\alg{A}}=\varphi$.
In general such a map not necessarily needs to exists,
unless $\alg{M}$ is a finite von Neumann algebra and $\varphi$ is
tracial <cit.>.
If it does exist and  the state $\varphi$ is faithful, then the conditional expectation
is the unique  element $\tilde{u}\in\alg{A}$ such that for any
one has $\varphi(ua)=\varphi(\tilde{u}a)$.
$\E_{\alg{A}}:\alg{M}\to\alg{A}$ is a unital $\alg{A}$-bimodule map, i.e.,
$E_{\alg{A}}[a_1ua_2]=a_1E_{\alg{A}}[u]a_2$, for all $u\in\alg{M}$ and $a_1,a_2\in\alg{A}$.

In the present paper we will always assume that
the algebra $\alg{M}$ is freely generated by two of its subalgebras
or more specifically, $\alg{M}$ is freely generated by
some subalgebras $\alg{A}_i$, $i\in I$
and the subalgebra $\alg{A}$ is generated by some subset of these,
i.e., $\alg{A}=\langle\alg{A}_j\rangle_{j\in J}$ where $J\subseteq I$,
and $\alg{B}=\langle \alg{A}_i\rangle_{i\in I\setminus J}$.
In this case the existence of the conditional expectation $\E_{\alg{A}}$ onto $\alg{A}$
is generally warranted by combinatorial arguments <cit.>.
Alternatively, in the C${}^*$-algebraic context, if the state is faithful,
then $\alg{M}$ is isomorphic to the reduced free product
and the existence of the conditional expectation also follows from  <cit.>.

More precisely, in order to find the conditional expectation of a non–commutative polynomial
in variables $a_1,a_2,\ldots,a_n$ onto the algebra $\alg{A}$ generated by
$a_1,a_2,\ldots,a_k$  one has to find suitable expressions for moments of the form

\begin{equation*}
\varphi\left(a_{i_1}a_{i_2}\dotsm a_{i_r} b\right),
\end{equation*}

where $i_1,i_2,\ldots,i_r\in \{1,\ldots,n\}$ and $b\in\alg{A}$.
It is a fundamental property of freeness
(as one of the  universal notions of independence in the sense of [29])
that all joint moments of
freely independent random variables are uniquely determined by the marginal moments
of the variables in question. Thus for each moment of the form indicated above
there is a universal formula (not depending on the particular choice of the
distributions of $a_1,a_2,\ldots,a_n$) which expresses any joint moment as a sum
of products of marginal moments.

After fixing random variables $a_1,a_2,\ldots,a_n$ all relevant information we need
for finding the conditional expectation is contained in the pair
$(\IC\langle \uXn\rangle,\mu)$ defined in Notation
<ref> and therefore we will mostly work on a formal
level and focus on this non–commutative probability space. Note that such $\mu$ need not to be faithful.

Working with non–commutative polynomials is very useful as it allows us to ignore algebraic relations satisfied by
variables $a_1,a_2,\ldots,a_n$. Another advantage of $\IC\langle \uXn\rangle$
is that it is augmented (see below) and has a natural linear basis, hence we can easily define linear mappings and functional on this algebra by prescribing values on the basis elements. It is also straightforward then, again by linearity, to extend all those mappings to the algebra of formal power series in non–commuting variables $\uXn$, denoted by $\IC\llangle \uXn \rrangle$.

§.§ Main idea – Boolean cumulants and conditional expectations

Boolean cumulants appeared in various contexts and disguises in the literature
[47, 40, 36].
In our previous work [23] we observed
that Boolean cumulants appear naturally in connection with conditional expectations of functions in free
variables. The present paper is an exploration of this connection based on a
recursive reformulation which is suitable for explicit calculations in closed form.
In addition we introduce a non–commutative differential calculus for Boolean
cumulants of polynomials in free variables which reduces the
combinatorial apparatus to a minimum.

Although Speicher's free cumulants are the tool of choice in free probability
[35, 31],
more recently it turned out that for certain questions Boolean cumulants are useful as well.
Indeed some problems like the free anti–commutator [14]
and subordination functions [23, 38] are easier
to describe in terms of Boolean cumulants rather than free cumulants.
The present paper extends and unifies these ideas. Before discussing this,
let us start with a review of some basic facts about Boolean cumulants.
An interval partition is a partition $\pi=\{B_1,B_2,\ldots,B_k\}$ of the set
$\{1,2,\ldots,n\}$ such that all blocks are intervals, i.e.,
for all $1\leq i\leq k$ we have $B_i=\{k,k+1,\ldots,l\}$ for some $k\leq l$ in
The set of all interval partitions of $\{1,2,\ldots,n\}$ is denoted by $\IntPart(n)$.

For any tuple $a_1,a_2,\ldots,a_n\in\alg{M}$ we define the Boolean cumulant
functional $\beta_n:\alg{M}^n\to\IC$ implicitly via the formula

\begin{equation}
\label{eq:phi=sumbeta}
\varphi(a_1a_2\dotsm a_n)=\sum_{\pi\in \IntPart(n)} \beta_\pi(a_1,a_2,\ldots,a_n),
\end{equation}

where $\beta_\pi(a_1,a_2,\ldots,a_n)=\prod_{B\in \pi} \beta_{|B|}\left((a_1,a_2,\ldots,a_n)|B\right)$, and by $\beta_{|B|}\left((a_1,a_2,\ldots,a_n)|B\right)$ we mean that we take the functional $\beta_k$ such that $k=|B|$, and we evaluate it at those $a_i$ for which $i\in B$, and the arguments $a_i$ appear in the natural order (one should note that in general Boolean cumulants are not invariant under permutations of arguments).
One way to  inverting the formula  (<ref>) and to
obtain an explicit formula for Boolean cumulants is Möbius inversion on the
lattice of interval partitions.
We refrain from doing so and rather base our calculations on a well known
recurrence, namely

\begin{equation}
\label{eq:recurrenceboolcum}
\varphi(a_1a_2\dotsm a_n)
= \sum_{k=1}^n \beta_k(a_1,a_2,\dots,a_k)\,\varphi(a_{k+1}a_{k+2}\dotsm a_n)
\end{equation}

or equivalently

\begin{equation}
\label{eq:recurrenceboolcumright}
\varphi(a_1a_2\dotsm a_n)
= \sum_{k=1}^n \varphi(a_1a_2\dotsm a_{k-1})\,\beta_{n-k+1}(a_{k},a_{k+1},\dots,a_n).
\end{equation}

This elementary recurrence is the starting point for our investigation.
It immediately implies a similar recurrence
for conditional expectations of products of free random variables:
Assume that $\{a_1,a_2,\ldots,a_{n}\}\in\alg{A}$ and
$\{b_1,b_2,\ldots,b_{n-1}\}\in\alg{B}$ are two families of variables and
assume that subalgebras $\alg{A}$ and $\alg{B}$ are free.

In order to calculate the conditional expectation of
$a_1b_1a_2\dotsm b_{n-1}a_n\in \alg{M}$ onto the subalgebra $\alg{B}$,
we have to find an element $\IE_{\alg{B}}[a_1b_1a_2\dotsm b_{n-1}a_n]\in\alg{B}$ such that

\[\varphi(a_1b_1a_2\dotsm b_{n-1}a_n b)=\varphi\left(\IE_{\alg{B}}[a_1 b_1a_2\dotsm a_n] b\right)\]

for any $b\in\alg{B}$.
If $\varphi$ is faithful then this element is uniquely determined and can be
found using the following recursive reformulation of  <cit.>.

Assume that $\varphi$ is faithful. Then for $a_i\in\alg{A}$ and
$b_i\in\alg{B}$ the conditional expectation of the alternating product
satisfies the recurrence

\begin{align}\label{eq:introCondExprecurrence}
\IE_{\alg{B}}[a_1b_1a_2\dotsm a_{n-1}b_{n-1} a_n]=\sum_{k=1}^{n}\beta_{2k-1}(a_1,b_1,a_2,\ldots,
a_k) \, b_k \,\IE_{\alg{B}}\left[a_{k+1}b_{k+1}\dotsm a_n\right].
\end{align}

The recurrence (<ref>) yields

\begin{multline*}
\varphi\left(\IE_{\alg{B}}[a_1b_1a_1\dotsm a_n] b\right)=\varphi(a_1b_1a_2\dotsm a_n b)\\
\beta_{2k-1}(a_1,b_1,a_2,\dots, a_k)\,
\varphi(b_{k}a_{k+1}\dotsm b)
\beta_{2k}(a_1,b_1,a_2,\dots, b_k)\,
\varphi(a_{k+1}b_{k+1}\dotsm b)
\end{multline*}

Now it follows from Theorem <ref>
(see also Definition <ref> and
Proposition <ref> below)
that for free random variables the mixed Boolean cumulant
$\beta_{2k}(a_1,b_1,a_2,\ldots, a_k,b_k)$ vanishes for $k=1,2,\ldots,n$
and hence

\begin{align*}
\varphi\left(\IE_{\alg{B}}[a_1b_1a_2\dotsm a_n] b\right)
&=\sum_{k=1}^{n}\beta_{2k-1}(a_1,b_1,a_2,\dots, a_k) \, \varphi(b_{k}a_{k+1}b_{k+1}\dotsm a_nb)
\\
\Bigl(
\sum_{k=1}^{n}\beta_{2k-1}(a_1,b_1,a_2,\dots, a_k)\, b_{k} a_{k+1}b_{k+1}\dotsm       a_nb
\Bigr)
\\
\Bigl(
\sum_{k=1}^{n}\beta_{2k-1}(a_1,b_1,a_2,\dots, a_k)\, b_k \IE_{\alg{B}}\left[a_{k+1}b_{k+1}\dotsm a_n\right]b
\Bigr).
\end{align*}

Observe that vanishing of cumulants of  the form
$\beta_{2k}(a_1,b_1,a_2,\ldots,a_k,b_k)$ is essential in this proof,
because it eliminates $\beta_{2n}(a_1,b_1,a_2,\ldots,a_{n-1},b)$.
Otherwise $b$ would be trapped inside this term
and the recurrence would fail.

We will also make use of the original non-recursive version of the formula for the conditional
expectation found in [23].

Let $(\alg{A},\varphi)$ be a ncps and
$\alg{B}\subseteq \alg{A}$  a subalgebra such that the conditional
expectation $\E_{\alg{B}}:\alg{A}\to\alg{B}$ exists.
Let  $\{b_1,b_2,\ldots,b_{n-1}\}\subseteq\alg{B}$ and
assume that the family $\{a_1,a_2,\ldots,a_n\}\subseteq\alg{A}$ is free
from $\alg{B}$.

\begin{multline}
\label{eq:phiYXY}
\varphi\left(a_1b_1a_2\dotsm b_{n-1}a_nb_n\right)
\\
\sum_{k=0}^{n-1}\sum_{1\leq i_1<i_2<\dots< i_k\leq n-1}
\varphi(b_{i_1}b_{i_2}\dotsm b_{i_k}b_n\prod_{j=0}^{k}\beta_{2(i_{j+1}-i_j)-1}(a_{{i_j}+1},b_{i_{j}+1},a_{{i_j}+2},\ldots,a_{i_{j+1}}),
\end{multline}

* Then the conditional expectation of alternating monomials can be evaluated as follows

\begin{multline}
\label{eq:EYXY}
\E_{\alg{B}}\left[a_1b_1a_2\dotsm b_{n-1}a_n\right]
\\
\sum_{k=0}^{n-1}\sum_{1\leq i_1<i_2<\dots< i_k\leq n-1}
b_{i_1}b_{i_2}\dotsm b_{i_k}\prod_{j=0}^{k}\beta_{2(i_{j+1}-i_j)-1}(a_{{i_j}+1},b_{i_{j}+1},a_{{i_j}+2},\ldots,a_{i_{j+1}}),
\end{multline}

where in the above sum for each sequence $0< i_1<i_2<\dots< i_k<n$ we set $i_0=0$ and $i_{k+1}=n$.

§.§ Boolean cumulants with products as entries

In this subsection we present the basic tools preparing
the non–commutative differential calculus for Boolean cumulants of polynomials in free variables.

The main tool is the formula for Boolean cumulants with products as
entries in terms of cumulants of individual variables.
Such formulas are known for classical cumulants
[24, 34] and free cumulants
[20]
and follow a general pattern <cit.>.
The analogous formula for Boolean cumulants is given below. The proof may go
via a standard argument involving Möbius inversion (see Lecture 14 in
In view of possible generalizations and for the
sake of completeness we present alternative algebraic proofs
based solely on the recurrence  (<ref>) in Appendix <ref>.

Let $a_1,a_2,\dots,a_n\in \alg{A}$ be random variables

\begin{equation}
\label{eq:BoolProd}
\beta_{m+1}(a_1a_2\cdots a_{d_1},a_{d_1+1}a_{d_1+2}\cdots a_{d_2},\ldots,a_{d_{m}+1}a_{d_{m}+2}\cdots a_{n})
=\sum_{\substack{\pi \in \IntPart(n)\\ \pi \vee \rho=1_n}}\beta_\pi(a_1,a_2,\ldots,a_n),
\end{equation}

where $\rho=\{\{1,2,\ldots,d_1\},\{d_1+1,d_1+2,\ldots,d_2\},\ldots,\{d_m+1,\ldots,n\}\}\in \IntPart(n)$, and $\vee$ is the join in the lattice of interval partitions.
The condition $\pi \vee \rho=1_n$ is equivalent to

\begin{equation*}
\pi \geq \{\{1\},\{2\},\dots,\{d_1-1\},\{d_1,d_1+1\},\{d_1+2\},\dots,\{d_m-1\},\{d_m,d_m+1\},\dots,\{d_{n}\}\}.
\end{equation*}

Proposition <ref> can be proved by iteration of the following
lemma and Corollary <ref> below.

\begin{multline}
\label{eq:lem:productformula}
\beta_{n-1}(a_1,a_2,\dots,a_pa_{p+1},a_{p+2},a_{p+3},\dots,a_n)
\\
=  \beta_{p}(a_1,a_2,\dots,a_p)
\,
\beta_{n-p}(a_{p+1},a_{p+2},a_{p+3},\dots,a_n)
\beta_{n}(a_1,a_2,\dots,a_n)
\end{multline}

We will actually mostly make use of the following recursive version of Proposition <ref>.

Let $a_1,a_2,\dots,a_n\in \alg{A}$ be random variables consider the interval partition $\rho=\{\{1,\ldots,d_1\},\{d_1+1,\ldots,d_2\},\ldots,\{d_m+1,\ldots,n\}\}\in \IntPart(n)$. We write $\rho=\{B_1,B_2,\ldots,B_{m+1}\}$, where blocks are ordered in natural order. For $j\in\{1,\ldots,n\}$ denote by $\rho(j)$ the number of block containing $j$, i.e. we have $\rho(j)=k$ if $j\in B_{k}$,

\begin{equation}
\label{eq:RecursiveBoolProd}
\begin{multlined}
\beta_{m+1}(a_1a_2\cdots a_{d_1},a_{d_1+1}a_{d_1+2}\cdots a_{d_2},\ldots,a_{d_{m}+1}a_{d_{m}+2}\cdots a_{n})\\
=\sum_{j\in\{1,\ldots,n\}\setminus\{d_1,d_2,\ldots,d_m\}}\beta_{j}(a_1,a_2,\ldots,a_j)\beta_{m-\rho(j)+1)}(a_{j+1}a_{j+2}\cdots a_{d_{\rho(j)}},\ldots,a_{d_m+1}\cdots a_n).
\end{multlined}
\end{equation}

We record here one immediate consequence of Lemma <ref>
which allows to eliminate units. This also follows from Proposition <ref> below.

For any $n\geq 2$ we have

\begin{align}
\label{eq:boolcumunit:1}
\beta_n(1,a_2,\dots,a_n)
\beta_n(a_1,a_2,\dots,a_{n-1},1)
\label{eq:boolcumunit:3}
\beta_n(a_1,a_2,\dots,a_{k-1},1,a_{k+1},\dots,a_n)
&= \beta_{n-1}(a_1,a_2,\dots,a_{k-1},a_{k+1},\dots,a_n)
\end{align}

§.§ Tensor products and tensor algebras

The tensor product $U\otimes V$ of two vector spaces has the universal property
that every bilinear map $B:U\times V\to W$ has a unique extension to a linear map $B:U\otimes V \to W$.
Consequently, a family of multilinear maps $f_n:V^n\to W$, $n\geq 0$,
corresponds uniquely to a linear map $f:T(V) \to W$ on the tensor algebra
$T(V) = \oplus_{n=0}^\infty V^{\otimes n}$.
Moreover, any linear map on $V$ can be extended to a derivation of the tensor
algebra (Lemma <ref>).

It will be convenient to denote the product operation on the tensor algebra by $\odot$ and extend it to matrices as follows:
Given a matrix $A=[a_{ij}]\in M_n(\alg{A})$ and $a\in\alg{A}$, let $a\odot A = [a\otimes a_{ij}] \in M_n(\alg{A}\otimes \alg{A})$.
Similarly, for two matrices $A,  B\in M_n(\alg{A})$ we denote by $A\odot B\in M_n(\alg{A}\otimes \alg{A})$ the matrix
with entries

(A\odot B)_{ij} = \sum_k a_{ik}\otimes b_{kj}

Note that associativity holds in connection with multiplication with scalar matrices $C\in M_n(\IC)$,
in the sense that  $A\odot CB = AC\odot B$.

§.§ Free products of algebras

A coproduct or  (algebraic) free product of unital algebras $\alg{A}_1$ and $\alg{A}_2$
over a field $\IK$ is a unital algebra $\alg{A}$ with embeddings
and $\iota_2:\alg{A}_2\to\alg{A}$ such that the images generate $\alg{A}$
and every pair of homomorphisms
$h_1:\alg{A}_1\to \alg{B}$ and  $h_2:\alg{A}_2\to \alg{B}$ has a unique
extension to a homomorphism $\alg{A}\to\alg{B}$.

For details about free products and tensor algebras we refer to
and [9].

* The free product is unique and is given by the quotient of the
tensor algebra
$T^+(A_1\oplus A_2)=\sum_{n=1}^\infty (A_1\oplus A_2)^{\otimes n}$ with respect to the ideal generated by all elements of
the form

a_1\otimes b_1 - a_1b_1
\qquad
a_2\otimes b_2 - a_2b_2
\qquad
1_{\alg{A}_1} - 1_{\alg{A}_2}
\qquad

The algebraic free product is denoted by $\alg{A}_1\freeprod\alg{A}_2$.

* <cit.>
Let $\alg{A}_1$ and $\alg{A}_2$ be algebras with 1 over a field $\IK$ with
$\IK$-bases $\{1\}\cup M_1$ and  $\{1\}\cup M_2$.
Then $\{1\}\cup M$ is a $\IK$-basis for $\alg{A}_1\FP \alg{A}_2$ where
$M$ is the set of alternating monomials in letters from $M_1$ and $M_2$.

An algebra $\alg{A}$ is called augmented if it comes with an
augmentation map, i.e., an algebra homomorphism $\epsilon:\alg{A}\to\IC$.
Its kernel $\bar{\alg{A}}=\ker\epsilon$ is called the
augmentation ideal.

Typical examples of augmented algebras are
polynomial algebras (both commutative and non–commutative)
where the augmentation map

\epsilon(P) = P(0) = \text{constant coefficient}

is clearly a homomorphism.

The free product of augmented algebras is clearly augmented.
Moreover, it is isomorphic to a subalgebra of the
tensor algebra and this fact will be helpful for the definition of certain
functionals to be defined in Section <ref> below.

The free product of augmented algebras is isomorphic to

\begin{equation}
\label{eq:freeprodaugmalg}
\alg{A}\freeprod\alg{B} \simeq
\IC1\oplus\bigoplus_{n=1}^\infty T_n(\bar{\alg{A}},\bar{\alg{B}})\oplus T_n(\bar{\alg{B}},\bar{\alg{A}})
\end{equation}

where by

T_n(U,V) = U\otimes V\otimes U\otimes \dotsm \quad\text{($n$ factors)}

we denote the alternating tensor product of two vector spaces.
The multiplication is given by the tensor product modulo the identifications
$a'\otimes a'' = a'a''$
and $b'\otimes b'' = b'b''$ already present in Proposition <ref>.

The free associative algebra $\IC\langle X,Y\rangle$ is the isomorphic to
algebraic free product  $\IC[X]\freeprod\IC[X]$.
The decomposition
(<ref>) corresponds to the decomposition into the
constant term and alternating monomials
$X^{k_1}Y^{l_1}X^{k_2}Y^{l_2}\dotsm$ with exponents $k_j,l_j>0$.

Extending the terminology of the preceding example to free products of general augmented algebras,
alternating products of the form
w = a_1b_1a_2b_2\dotsm
(resp. $w = b_1a_1b_2a_2\dotsm$)
with $a_i\in\bar{\alg{A}}$ and $b_i\in\bar{\alg{B}}$,
i.e., the images of elementary tensors from $T_n(\alg{A},\alg{B})$ (resp.  $T_n(\alg{B},\alg{A})$),
will be called monomials.

The reduced free product of ncps $(\alg{A},\varphi)$ and $(\alg{B},\psi)$
will be denoted by  $(\alg{A},\varphi)*(\alg{B},\psi)$ or
$\alg{A}*\alg{B}$ in short.
is realized as the image of the free product of their GNS representations, see
<cit.>. For our purpose the following subalgebra is sufficient.

Let $(\alg{A}_1,\varphi_1)$ and $(\alg{A}_2,\varphi_2)$ be ncps and denote by
$\bub{A}_i=\ker\varphi_i$ their centered components.
Then the orthogonal direct sum

\IC1\oplus \bigoplus_{n=1}^\infty\oplus \bigoplus_{i_1\ne i_2\ne\cdots\ne i_n}
\bub{\alg{A}}_{i_1}\bub{\alg{A}}_{i_2} \dotsm \bub{\alg{A}}_{i_n}

is a dense subalgebra of the reduced free product.

Unital subalgebras $\alg{A}$, $\alg{B}$ of an algebra $\alg{M}$ are
called algebraically free if they do not satisfy any mutual
algebraic relation,
i.e., if the free extension of the embeddings $\iota_A:\alg{A}\to \alg{M}$,
$\iota_B:\alg{B}\to \alg{M}$ to a homomorphism
$h:\alg{A}\FP\alg{B}\to\alg{M}$ is injective.

Let $(\alg{M},\varphi)$ be a ncps with faitful state $\varphi$
and let $\alg{A}$, $\alg{B}$ be freely independent subalgebras
in the sense of Definition <ref>.
Then  $\alg{A}$ and $\alg{B}$ are algebraically free.

The algebra generated by $\alg{A}$ and $\alg{B}$ is isomorphic
to their reduced free product $\alg{A}*\alg{B}$
and thus by <cit.> contains a faithful copy
of the algebraic free product.

§.§ Derivations

Coincidentally it turns out that similar to classical calculus, integration has
strong ties to derivations.
More precisely, we will be concerned with several  derivations
from an algebra $\alg{A}$ into the bimodule $\mathfrak{M} = \alg{A}\otimes\alg{A}$ with
the natural action

\begin{equation}
\label{eq:AotimAmodule}
a_1\cdot (x\otimes y) \cdot a_2 = (a_1\otimes 1)(x\otimes y)(1\otimes a_2) =
a_1x\otimes ya_2
\end{equation}

Let $\alg{A}$ be an algebra and $\mathfrak{M}$ be an $\alg{A}$-bimodule.
An $\mathfrak{M}$-derivation is a linear map $D:\alg{A}\to\mathfrak{M}$ satisfying
the Leibniz rule

\begin{equation}
\label{eq:leibniz}
D(a_1a_2) = D(a_1)\cdot a_2 + a_1\cdot D(a_2)
\end{equation}

If one modifies the  left and right actions of $\alg{A}$ on
$\alg{A}\otimes\alg{A}$ then
homomorphisms become a rich source of derivations.

Let $\alg{A}$ be a unital algebra and
$\Phi_1,\Phi_2:\alg{A}\to\alg{A}\otimes\alg{A}$ be two homomorphisms,
then $D = \Phi_1-\Phi_2$ is a derivation in the sense that
$D(ab) = D(a)\Phi_2(b) + \Phi_1(a)D(b)$

Let us present some examples of derivations

The “tensor commutators”
mapping $\lnabla a = a\otimes 1-1\otimes a$
is a derivation for any algebra $\alg{A}$.
It appears in <cit.> (see Section <ref> below)
and has the universal property that every derivation factors through it
It will be convenient to denote $\rnabla=-\lnabla$.

The free derivative or free difference quotient on $\IC[x]$ is the map
given by

\begin{equation*}
\partial x^n = \sum_{k=0}^{n-1} x^k\otimes x^{n-k-1}
\end{equation*}

In the natural identification $\IC[x]\otimes\IC[x]\simeq \IC[x,y]$ this
coincides with the difference quotient

\begin{equation*}
\partial p(x) =  \frac{p(x)-p(y)}{x-y}
\end{equation*}

and for this reason is also known as the Newtonian coproduct
It first appeared in free probability in connection with the non–commutative Hilbert transform approach to free entropy

A slight modification of the previous derivation comes from the divided powers coproduct
Let again $\alg{A}=\IC[X]$ be the polynomial algebra,
then the deconcatenation coproduct is a homomorphism

\Delta_d(x^n) = \sum_{k=0}^n x^k\otimes x^{n-k}

and thus both

\begin{align*}
\ldelta p(x) &= \Delta_d p(x) - 1\otimes p(x) = (x\otimes  1)\partial p(x)\\
\rdelta p(x) &= \Delta_d p(x) - p(x)\otimes 1 = (1\otimes  x)\partial p(x)
\end{align*}

are derivations.

For an augmented algebra, the map $\Delta(a) = a-\epsilon(a)$ is a derivation
in the sense of Proposition <ref>
and so are the left and right “block derivatives”

\begin{equation*}
\begin{aligned}
\rDelta:\alg{A}&\to\alg{A}\otimes\alg{A}   &    \lDelta:\alg{A}&\to\alg{A}\otimes\alg{A}   \\
a&\mapsto 1\otimes( a-\epsilon(a))         &   a&\mapsto(a-\epsilon(a))\otimes 1
\end{aligned}
\end{equation*}

Next we discuss free products of derivations.

Let $\alg{A}_1$ and $\alg{A}_2$ be algebras and
$\alg{A}=\alg{A}_1\freeprod\alg{A}_2$ their (unital) free product.
Let $\mathfrak{M}$  an $\alg{A}$-bimodule (equivalently, a bimodule for both
$\alg{A}_1$ and $\alg{A}_2$)
and $D_i:\alg{A}_i\to \mathfrak{M}$, $i=1,2$ be derivations.
Then there exists a unique derivation $D_1\ast D_2:\alg{A}_1\freeprod\alg{A}_2\to\mathfrak{M}$
extending $D_1$ and $D_2$.
Moreover, we have the decomposition

D_1\ast D_2= D_1\ast0 + 0\ast D_2

where $0$ is the trivial derivation.

We first extend $\tilde{D}=D_1\oplus D_2:\alg{A}_1\oplus\alg{A}_2\to\mathfrak{M}$
to the tensor algebra $T^+(\alg{A}_1\oplus\alg{A}_2)$ by the Leibniz rule,
see Lemma <ref>:

\tilde{D}(u_1\otimes u_2\otimes\dotsm \otimes u_n)=\sum_{k=1}^n
u_1\otimes u_2\otimes u_{k-1}\cdot \tilde{D}(u_k)\cdot u_{k+1}\otimes\dotsm \otimes u_n

and then check that it passes to the quotient
(see  Proposition <ref>),
the ideal generated by $1_{\alg{A}}-1_{\alg{B}}$, $a'\otimes a''-a'a''$ for
$a',a''\in\alg{A}_1\cup\alg{A}_2$ is contained in  $\ker\tilde{D}$.
Clearly $\tilde{D}(1)=0$ and on the other hand for $a',a''\in\alg{A}_1$

\begin{equation*}
\begin{multlined}
\tilde{D}(u\otimes(a'\otimes a''-a'a'')\otimes v) =
\tilde{D}(u)\cdot (a'\otimes a''-a'a'')\otimes v)
\\
u\cdot(D_1(a')\cdot a''
+ a'\cdot D_1(a'')   - D_1(a'a''))\cdot v
\\
u\otimes (a'\otimes a''-a'a'')\cdot\tilde{D}(v)
\end{multlined}
\end{equation*}

which is mapped to $0$ in the quotient space
and the Leibniz rule is satisfied by definition.

If we identify the algebra of non–commutative polynomials with the unital free product
$\IC\langle x,y\rangle \simeq \IC[ x]\ast\IC[x]$,
then we can construct the partial derivatives $\ldelta_x$, $\rdelta_x$, $\partial_x$ and
as  free products of the derivations $\lnabla$ (Example <ref>), $\ldelta$, $\rdelta$
(Example <ref>)  and $0$  on $\IC[x]$:

\begin{align*}
\ldelta_x&=\ldelta{}\!\star{} 0 &    \ldelta_y&=0\ast\ldelta \\
\rdelta_x&=\rdelta{}\!\star{} 0 &    \rdelta_y&=0\ast\rdelta\\
\partial_x&=\partial\star 0 &  \partial_y&=0\ast\partial \\
\lnabla_x&=\lnabla{}\!\ast 0 &    \lnabla_y&=0\ast\lnabla \\
\rnabla_x&=\rnabla{}\!\ast 0 &  \rnabla_y&=0\ast\rnabla\\
\end{align*}

moreover, we obtain decompositions $\ldelta = \ldelta_x+\ldelta_y$ and $\lnabla = \lnabla_x+\lnabla_y$.

Let $\alg{A}$ and $\alg{B}$ be augmented algebras and set
Denote by $\lDelta_{\alg{A}}$ and $\lDelta_{\alg{B}}$
(resp. $\rDelta_{\alg{A}}$ and   $\rDelta_{\alg{B}}$) the left (resp. right) block derivatives
from Example <ref> into the corresponding submodules of $\mathfrak{M}$.
The free products   $\rDelta_{\alg{A}}{}\!\star 0,0*\rDelta_{\alg{B}}:\alg{A}\freeprod\alg{B}\to \mathfrak{M}$
are called partial block derivatives.
By abuse of notation we will denote them by $\rDelta_{\alg{A}}$ and
$\rDelta_{\alg{B}}$ as well.
Their action on the augmentation ideal is deconcatenation
on monomials

\begin{align*}
\rDelta_{\alg{A}} (a_1b_1a_1a_2b_2\dotsm a_nb_n)
&= \sum_{k=1}^n  a_1b_2a_2\dotsm b_k \otimes a_k b_{k+1}a_{k+1}\dotsm      a_nb_n
\\
\lDelta_{\alg{A}} (a_1b_1a_1a_2b_2\dotsm a_nb_n)
&= \sum_{k=1}^n  a_1b_2a_2\dotsm a_k \otimes b_ka_{k+1} b_{k+1}\dotsm      a_nb_n
\end{align*}

where $a_i\in\bar{\alg{A}}$ and $b_i\in\bar{\alg{B}}$.
In contrast to the previous derivations, neither the partial block derivatives
nor the deconcatenation operator $\rDelta=\rDelta_{\alg{A}}+\rDelta_{\alg{B}}$
satisfy the Leibniz rule with respect to the natural action
but the “compatibility relation” of unital infinitesimal bialgebras applies <cit.>

\rDelta (uv) = (\rDelta u)(1\otimes v) + (u\otimes 1)\rDelta v - u\otimes v.

Note however that $\rDelta_{\alg{A}}-\lDelta_{\alg{A}}=\rnabla_{\alg{A}}$ is
a derivation.

Let us observe that resolvents work nicely with any derivation

Let $\alg{A}$ be a unital algebra and $D:\alg{A}\to \mathfrak{M}$ be a derivation into an
$\alg{A}$-bimodule $\mathfrak{M}$. Then for any invertible element $a\in\alg{A}$ we have
as a consequence of the Leibniz rule

\begin{equation}
\label{eq:derivationofinverse}
\end{equation}

In particular, the derivation of a resolvent $R=(z-a)^{-1}$ satisfies

\begin{equation*}
D(R) = RD(a)R
\end{equation*}

while for $\Psi=(1-za)^{-1}$ we have

\begin{equation}
\label{eq:derivationofpsi}
D(\Psi) = z\Psi D(a)\Psi
\end{equation}

§ BOOLEAN CUMULANTS OF FREE RANDOM VARIABLES

§.§ Characterization of freeness by Boolean cumulants

In this section we elaborate on on the main property which allowed us to derive
recurrence (<ref>), i.e.,
the fact that mixed Boolean cumulants of free variables which start and end
with mutually free variables vanishes.
The main result is  a surprisingly simple characterization of freeness
which is inspired by an algebraic characterization of freeness found by
Voiculescu in [45].
We are grateful to R. Speicher and J. Mingo for bringing this paper to our attention.

* Subalgebras $\alg{A},\alg{B}$ of a ncps $(\alg{M},\varphi)$ have vanishing
cyclically alternating cumulants  if

\beta_{n}(u_1,u_2,\dots,u_n)=0

whenever $u_i\in\alg{A}\cup\alg{B}$ such that $u_1$ and $u_n$ come from
different algebras.
We will call this Property $(CAC)$.

* Subalgebras $\alg{A},\alg{B}$ of a ncps $(\alg{M},\varphi)$ satisfy
Property $(WCAC)$ (weak $(CAC)$), if they satisfy Property $(CAC)$ for alternating
words,   i.e.

\beta_{n}(a_1,b_1,a_2,b_2,\dots,a_n,b_n)=0

\beta_{n}(b_1,a_1,b_2,a_2,\dots,b_n,a_n)=0

for any choice of $a_i\in\alg{A}$, $b_i\in\alg{B}$.

* Subalgebras $\alg{A},\alg{B}$ of a ncps $(\alg{M},\varphi)$
satisfy property $(\nabla)$ if

\begin{equation}
\label{eq:propertynablaA}
(\varphi\otimes\varphi)\circ\lnabla_{\alg{A}}(a_1b_1a_2b_2\dotsm a_nb_n)=0
\end{equation}

for any choice of $a_i\in\alg{A}$ and $b_i\in\alg{B}$.
Here the expression on the right hand side of (<ref>)
is meant to be evaluated as follows:

* The formal derivative $\lnabla_x x_1y_1x_2y_2\dotsm x_ny_n$ is computed in
the free associative algebra
$\IC\langle x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n\rangle$;

* substitute $x_i=a_i$, $y_i=b_i$ to obtain an element of

* evaluate $\varphi\otimes\varphi$ on the latter.

Since $\lnabla = \lnabla_{\alg{A}}+\lnabla_{\alg{B}}$ and trivially
identity (<ref>) is equivalent to

\begin{equation*}
(\varphi\otimes\varphi)\circ\lnabla_{\alg{B}}(a_1b_1a_2b_2\dotsm a_nb_n)=0
\end{equation*}

Property $(WCAC)$  is equivalent to Property $(CAC)$.

Clearly Property $(CAC)$ implies  Property $(WCAC)$.
The converse can be seen as a special case of Lemma <ref> below,
but here is a proof by induction,  assuming that   Property $(WCAC)$ holds for all orders.

Suppose that Property $(CAC)$ holds  for all orders up to
$n-1$ and
pick a tuple $u_1,u_2,\dots,u_n\subseteq \alg{A}\cup\alg{B}$ such that $u_1$
$u_n$ come from different algebras.

If the tuple is alternating then there is nothing to prove.

Therefore assume that it is not alternating.
Without loss of generality we consider the following configuration (the proof
of the other cases is  analogous):
$u_1=a,u_k=a',u_{k+1}=a''\in\alg{A}$ and $u_n=b\in\alg{B}$.
Thus we have to show that the cumulant
$\beta_n(a,u_2,u_2,\dots,a',a'',u_{k+2},\dots,u_{n-1},b)$ vanishes.
We apply the product formula (<ref>) in the reverse direction and obtain

\begin{multline*}
\beta_n(a,u_2,\dots,u_{k-1},a',a'',u_{k+2},\dots,u_{n-1},b)
\\
\beta_{n-1}(a,u_2,u_2,\dots,u_{k-1},a'a'',u_{k+2},\dots,u_{n-1},b)\\
\end{multline*}

All cumulants on the right hand side are of lower order and all except the
second one satisfy the
assumptions of the induction hypothesis and therefore the right hand side

Property $(WCAC)$  is equivalent to Property $(\nabla)$.

We apply recurrence (<ref>) to the
first factor of the first sum
and (<ref>) to the
second factor of the second sum in the following expression

\begin{multline*}
(\varphi\otimes\varphi)(\lnabla_{\alg{A}} a_1b_1\dotsm a_nb_n)\\
\begin{aligned}[t]
\sum_{k=1}^n\varphi(a_1b_1\dotsm a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
- \varphi(a_1b_1\dotsm a_{k-1}b_{k-1})\,\varphi(a_kb_k\dotsm a_nb_n)
\\
\begin{multlined}[t]
\sum_{k=1}^n \sum_{p=1}^k\varphi(a_1b_1\dotsm b_{p-1})\,\beta_{2(k-p)+1}(a_p,b_p,\dots,a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
\\
+ \sum_{k=1}^n \sum_{p=1}^{k-1}\varphi(a_1b_1\dotsm a_p)\,\beta_{2(k-p)}(b_p,\dots,a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
\\
- \sum_{k=1}^n \sum_{p=k}^n\varphi(a_1b_1\dotsm b_{k-1})\,\beta_{2(p-k)+1}(a_k,b_k,\dots,a_p)\,\varphi(b_pa_{p+1}\dotsm a_nb_n)
\\
- \sum_{k=1}^n \sum_{p=k}^n\varphi(a_1b_1\dotsm b_{k-1})\,\beta_{2(p-k)+2}(a_k,b_k,\dots,b_p)\,\varphi(a_{p+1}\dotsm a_nb_n)
\end{multlined}
\\
\begin{multlined}[t]
\sum_{1\leq p\leq k\leq n}\varphi(a_1b_1\dotsm b_{p-1})\,\beta_{2(k-p)+1}(a_p,b_p,\dots,a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
\\
+ \sum_{k=1}^n \sum_{p=1}^{k-1}\varphi(a_1b_1\dotsm a_p)\,\beta_{2(k-p)}(b_p,\dots,a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
\\
-       \sum_{1\leq k\leq p\leq n}\varphi(a_1b_1\dotsm b_{k-1})\,\beta_{2(p-k)+1}(a_k,b_k,\dots,a_p)\,\varphi(b_pa_{p+1}\dotsm a_nb_n)
\\
- \sum_{k=1}^{n-1} \sum_{p=k}^n\varphi(a_1b_1\dotsm
\\
\end{multlined}
\\
\begin{multlined}[t]
\sum_{k=1}^n \sum_{p=1}^{k-1}\varphi(a_1b_1\dotsm a_p)\,\beta_{2(k-p)}(b_p,\dots,a_k)\,\varphi(b_ka_{k+1}\dotsm a_nb_n)
\\
- \sum_{k=1}^{n-1} \sum_{p=k}^n\varphi(a_1b_1\dotsm
\\
\end{multlined}
\end{aligned}
\end{multline*}

Now all but the last term on the right hand side
involve lower order cumulants which vanish by by induction hypothesis.
Therefore the left hand side vanishes if and only if $\beta_{2n}(a_1,b_1,\dots,a_n,b_n)=0$.

Free subalgebras satisfy Property $(CAC)$.

This is an immediate consequence of the vanishing of mixed free cumulants and
the formula expressing Boolean cumulants as a sum of free cumulants
indexed by irreducible non–crossing partitions
[21, 5],
and is also a special case of Proposition <ref> below.

Here is a direct self-contained proof using only the recurrence
(<ref>) and induction.
By Lemma <ref> it suffices to prove Property $(WCAC)$.

To begin with, in the case $n=2$ the recurrence (<ref>)
immediately resolves into the covariance:

\beta_2(a,b) = \varphi(ab)-\varphi(a)\varphi(b)=0

For the induction step,
we first verify that cumulants of alternating centered words vanish
and then reduce the general case to this.

Assume that the assertion holds for any order up to $k\leq n-1$
and that  the tuple $u_1,u_2,\dots,u_n$ is alternating,
i.e., neigbouring elements come from different algebras,
and that all elements are centered.
Then it follows from recurrence (<ref>) that

\begin{equation*}
\beta_{n}(u_1,u_2,\dots,u_n)
=     \varphi(u_1u_2\dotsm u_n) -
\sum_{k=1}^n\beta_k(u_1,u_2,\dots,u_k)\,\varphi(u_{k+1}u_{k+2}\dotsm u_n)
\end{equation*}

and all terms on the right hand said contain expectations of alternating
words in centered elements. Freeness implies that they vanish and so does
the cumulant on the left hand side.

It remains to show that we can replace the letters of an alternating word
with centered elements.
We will do this step by step using multilinearity and
Corollary <ref>.
The condition that the first and last arguments of the involved cumulants
come from different algebras is not violated throughout the following manipulations.

\begin{align*}
\beta_n(u_1,u_2,\dots,u_n)
&=     \beta_n(\bub{u}_1,u_2,\dots,u_n) +    \varphi(u_1)\,\beta_n(1,u_2,\dots,u_n) \\
&=     \beta_n(\bub{u}_1,u_2,\dots,u_n) &    \text{by    \eqref{eq:boolcumunit:1}}\\
&=     \beta_n(\bub{u}_1,\bub{u}_2,u_3,\dots,u_n) +\varphi(u_2)\,\beta_n(\bub{u}_1,1,u_3,\dots,u_n) \\
&=     \beta_n(\bub{u}_1,\bub{u}_2,u_3,\dots,u_n) +\varphi(u_2)\,\beta_{n-1}(\bub{u}_1,u_3,\dots,u_n)     &\text{by    \eqref{eq:boolcumunit:3}}\\
&=     \beta_n(\bub{u}_1,\bub{u}_2,u_3,\dots,u_n) &  \text{by  induction hypothesis} \\
&\ \ \vdots\\
&=     \beta_n(\bub{u}_1,\bub{u}_2,\dots,\bub{u}_n)
\end{align*}

In conclusion, in the tracial case we have the following extension of Voiculescu's freeness criterion <cit.>.

Let $(\alg{M},\varphi)$ be a tracial non-commutative probability space
and $\alg{A},\alg{B}\subset\alg{M}$  two  subalgebras.
Then the following are equivalent.

*  $\alg{A}$ and $\alg{B}$ are free.

*  $\alg{A}$ and $\alg{B}$ satisfy property $(CAC)$.

*  $\alg{A}$ and $\alg{B}$ satisfy property $(\nabla)$.

Items (<ref>) and (<ref>) are equivalent by
Proposition <ref> even without traciality.

Item (<ref>) implies (<ref>) by
Lemma <ref> and it remains to prove the converse in the tracial case, i.e., Property $(CAC)$
implies freeness.

Recall that in [45] it is shown that in the
tracial case for every $n$ the original freeness condition $(F(n))$

\begin{equation*}
\varphi(a_1b_1\cdots a_n b_n )=0
\end{equation*}

whenever all  elements $a_i$, $b_i$ are centered can be strengthened
to condition $(F(n)')$ allowing one of $a_1$ or $b_n$ to have nonzero expectation.

We will use this equivalence and  proceed by induction
and  show  that for every $n$
condition $(F(k)')$ for $k<n$ together with $(CAC)$ implies condition $(F(n))$.
To this end we pick centered elements $a_1,a_2,\ldots,a_n\in\alg{A}$
and $b_1,b_2,\ldots,b_n\in\alg{B}$ and and show that
$\varphi(a_1b_1\cdots a_n b_n)=0$.

The case $n=1$ is obvious.

For the induction step, observe that we can rewrite recurrence
(<ref>) as follows

\begin{multline*}
\varphi(a_1b_1\dotsm a_n b_n)
\sum_{k=1}^{n} \beta_{2k-1}(a_1,b_1,\ldots, a_k)\,\varphi(b_{k}a_{k+1}b_{k+1}\dotsm b_n)
\\
+\sum_{k=1}^{n}\beta_{2k}(a_1,b_1,\ldots, b_k)\, \varphi(a_{k+1}b_{k+1}\dotsm b_n)
\end{multline*}

All terms in the second sum vanish by property (CAC).
In the first sum  the term corresponding to $k=n$ vanishes because we assumed
that $\varphi(b_n)=0$.
In order to see
that the remaining terms corresponding to $k<n$ vanish as well,
first note that
by traciality   $\varphi(b_{k}a_{k+1}\ldots b_n)=\varphi(a_{k+1}\ldots b_n b_k)$
which is an alterning product with at most $2(n-1)$ factors.
Now the element $b_n b_k$ needs not to be centered but  as discussed above,
the freeness conditions $(F(n-1)')$   and $(F(n-1))$ are equivalent
and thus the expectation vanishes.

Note that traciality of the linear functional is essential for this
characterization. Property $(CAC)$ (and hence Property $(\nabla)$) also holds
for Boolean independent and more generally conditionally free
random variables.

§.§ Mixed Boolean cumulants of free random variables

The following characterization of freeness from [14] generalizes
Property $(CAC)$ and will provide an essential tool for later considerations.

Subalgebras $\alg{A}_1,\alg{A}_2,\ldots,\alg{A}_s\subseteq\alg{M}$ of a
ncps $(\alg{M},\varphi)$ are free if and only if for any colouring
$c:\{1,\ldots,n\}\to\{1,\ldots,s\}$ we have

\begin{equation*}
\beta_n(a_1,a_2,\dots,a_n) = \sum_{\pi\in\NCirr(n) \text{ with VNRP}} \beta_\pi(a_1,a_2,\dots,a_n)
\end{equation*}

whenever  $a_i\in\alg{A}_{c(i)}$.
Here a  partition $\pi\in \NCirr(n)$ is said to have $VNRP$ if $\pi\leq\ker
c$ and every inner block covered nested by a block of different colour, i.e.,
$c$ induces a proper coloring on the nesting tree of $\pi$.

We will only use the preceding theorem in the special case of alternating
which was first explicitly stated in [38].

Let $\{a_1,a_2,\ldots,a_{n}\}$ and $\{b_1,b_2,\ldots,b_{n-1}\}$ be free, $n\ge 1$. Then

\begin{equation}
\label{bocu2}
\begin{multlined}
\beta_{2n-1}(a_1,b_1,\ldots,a_n,b_{n-1},a_{n})\\
\end{multlined}
\end{equation}

Next we record another property which will be important in the following,
namely as we already know a Boolean cumulant which takes as arguments elements
of two free algebras $\alg{A}$ and $\alg{B}$ vanishes
unless it starts and ends with elements from the same algebra.
Suppose this is the case, say the both the first and last argument come from
$\alg{A}$, then the Boolean cumulant does not change under arbitrary splitting
of thee arguments coming from the subalgebra $\alg{B}$ into factors.
This will allow us to write any Boolean cumulant in free variables as an alternating cumulant.

Suppose $\alg{A},\alg{B}\subseteq\alg{M}$ are free and let
$a_1,a_2,\ldots,a_n\in\alg{A}$ and
Assume further that for each $j=1,2,\ldots,n-1$ we have $b_i=c^{(i)}_1\cdots c^{(i)}_{j_i}$ with $c^{(i)}_1,\ldots, c^{(i)}_{j_i}\in\alg{B}$, then we have

\begin{align*}
\beta_{2n-1}(a_1,b_1,a_2,\ldots,b_{n-1},a_n)=\beta_{n+j_1+\ldots+j_{n-1}}(a_1,c^{(1)}_1,\ldots, c^{(1)}_{j_1},a_2,\ldots,c^{(n-1)}_1,\ldots ,c^{(n-1)}_{j_{n-1}},a_n).
\end{align*}

This follows from the formula for Boolean cumulants with products as
entries (<ref>) and  property $(CAC)$.
More precisely,
applying said formula to the Boolean cumulant

\beta_{n+j_1+\dots+j_{n-1}}(a_1,c^{(1)}_1c^{(1)}_2\dotsm  c^{(1)}_{j_1},
a_2,\dots,c^{(n-1)}_1c^{(n-1)}_2\dotsm c^{(n-1)}_{j_{n-1}},a_n)

recall that we have to sum over interval partitions $\pi$ such that $\pi \vee \sigma=1_{n+j_1+\dots+j_{n-1}}$, where $\sigma$ is given by the product structure, hence we sum over $\pi$ greater than $\{\{1,2\},\{3\}\dots,\{j_1\},\{j_1+1,j_1+2\},\dots\{n-1,n\}\}$ or in other words are complementary
to $\tilde{\rho}=(1,j_1,1,j_2,\dots,j_{n-1},1)$.
Observe that the block of $\pi$ containing $a_1$ cannot end in any
$c_{i}^{l}$, as in that case the corresponding  Boolean cumulant  vanishes
by property $(CAC)$. On the other hand the block containing $a_1$ cannot
finish in any of the variables $a_1,a_2,\dots,a_{n-1}$ as this would violate
the complementarity property. Thus the block which contains $a_1$ has to end
in $a_n$ and since we consider only interval partitions, there is only one such partition $\pi=1_{n+j_1+\dots+j_{n-1}}$.

Lemma <ref> allows  to rewrite an arbitrary joint Boolean cumulant
of free random variables as an alternating cumulant, by grouping together terms from the
algebra $\alg{B}$ above.
after regrouping the inner variables into
free blocks, Corollary <ref> allows us to fill the gaps
between elements of $\alg{A}$ with  units  to obtain

\beta_{k}(a_1,a_2,\dots,a_{l},a_{l+1},\dots,a_n)

§ CALCULUS FOR CONDITIONAL EXPECTATIONS

In order to define several objects objects used in this and the next section we require
additional structure on the ncps we work with.

From this point we will assume that $(\alg{M},\varphi)$ is a ncps,
where $\alg{M}$ is an augmented algebra (see
Definition <ref>).
Furthermore we assume
$\alg{A}$, $\alg{B}\subseteq \alg{M}$ are freely independent subalgebras which generate
$\alg{M}$ as an algebra, that is, $\alg{M}=\alg{A}\freeprod\alg{B}$.
The latter is a moderate restriction since it is a dense subalgebra
by   Proposition <ref>.

* This assumption is necessary to make sure that the functionals introduced
below are well   defined.
Indeed in the augmented case
the algebraic free product is canonically isomorphic to a subspace of the tensor
algebra (see Proposition <ref>)
and   thus families of multilinear
functionals can be identified with linear functionals on the latter.

The reader not familiar with these notions should think of the algebras as  polynomial
algebras from Example <ref> in the following. More

\alg{M} = \IC\langle X_1,X_2,\dots,X_m,Y_1,Y_2,\dots,Y_n\rangle
\qquad
\alg{A} = \IC\langle X_1,X_2,\dots,X_m\rangle
\qquad
\alg{B} = \IC\langle Y_1,Y_2,\dots,Y_n\rangle

and $\varphi=\mu$ is some formal distribution, i.e., a unital linear
functional $\mu:\alg{M}\to\IC$.
The augmentation is $\epsilon(P)=P(0)$.

In fact it is sufficient to understand the bivariate case

\alg{M} = \IC\langle X,Y\rangle
\qquad
\alg{A} = \IC\langle X\rangle
\qquad
\alg{B} = \IC\langle Y\rangle

and the general case will be straightforward.

In the case of polynomial algebras $w\in\bar{\alg{M}}$ means that $w$ has zero constant term
and in fact it is sufficient to consider monomials $w=a_1b_2a_2b_2\dotsm$
$a_i$ and $b_i$ come from the monomial basis,
in the case of $\IC\langle X,Y\rangle$ this means that $a_i\in\{X^k|k\geq 1\}$
and  $b_i\in\{Y^k|k\geq 1\}$.

It follows from Proposition <ref> that
in the free product of augmented algebras any  monomial (i.e., simple tensor)
$W\in\bar{\alg{M}}$   has a unique factorization into an
alternating product of elements from $\bar{\alg{A}}$ and $\bar{\alg{B}}$
in one out of four types:

\begin{align*}
\text{type $\alg{A}$--$\alg{A}$}& &\text{if }
w&=a_1b_1a_2\dotsm b_{n-1}a_n,
\\
\text{type $\alg{A}$--$\alg{B}$} && \text{if }
w&=a_1b_1a_2\dotsm a_nb_{n},
\\
\text{type $\alg{B}$--$\alg{A}$} &&\text{if }
w&=b_1a_1b_2\dotsm b_na_{n},
\\
\text{type $\alg{B}$--$\alg{B}$} &&\text{if }
w&=b_1a_1b_2\dotsm a_{n-1}b_n,
\end{align*}

with $a_1,a_2,\dots,a_{n}\in\bar{\alg{A}}$ and
We will call such monomials $\alg{A}$–$\alg{A}$ monomials etc. and
we will refer to the factorization as the $\alg{A}$–$\alg{B}$ block factorization.

At this point augmentation is essential. Otherwise allowing
units in monomials leads to inconsistencies like
(1+a)b = b + ab
$ belonging to the image of $\alg{A}\otimes \alg{B}$ and $(\alg{B}\oplus
\alg{A}\otimes \alg{B})$ simultaneously.
However the direct sum decomposition (<ref>)
is essential to consistently extend the multilinear maps to be defined shortly
(see Definition <ref>)
to linear maps on the free product.

We now take formula (<ref>) as a formal definition of a conditional expectation.

Define a linear mapping $\E_{\alg{B}}:\alg{M}\to \alg{B}$ via the following requirements:

* $\E_{\alg{B}}[1]=1$,

* $\E_{\alg{B}}[bwb']=b\E_{{B}}[w]b'$ for any $w\in\alg{M}$ and any $b,b'\in\alg{B}$.

* We define the conditional expectation of a
monomial of type $\alg{A}$–$\alg{A}$
$w=a_1b_1a_2\dotsm b_{n-1}a_n$ with $a_i\in\bar{\alg{A}}$ and $b_i\in\bar{\alg{B}}$

\begin{multline*}
\E_{\alg{B}}\left[a_1b_1a_2\dotsm b_{n-1}a_n\right]
= \beta_{2n-1}\left(a_1,b_1,a_2,\ldots ,b_{n-1},a_n\right)+\\
\sum_{k=1}^{n-1}\sum_{1\leq i_1<i_2<\dots< i_k\leq n-1}
b_{i_1}b_{i_2}\dotsm b_{i_k}\prod_{j=0}^{k}\beta_{2(i_{j+1}-i_j)-1}(a_{{i_j}+1},b_{i_{j}+1},a_{{i_j}+2},\ldots,a_{i_{j+1}}),
\end{multline*}

* We define the block cumulant functional
$\bbeta{}:\alg{M}\to\IC$ by prescribing the values on monomials

\begin{align*}
\bbeta{}(1)&=1\\
\bbeta{}(u_1u_2\dotsm u_n)
&= \beta_{n}(u_1,u_2,\dots, u_n)
\end{align*}

whenever $u_1u_2\dotsm u_n$ is an alternating word with $u_i\in\bar{\alg{A}}\cup\bar{\alg{B}}$.

* Note that property $(CAC)$ implies $\beta{}(u_1u_2\dotsm u_n)$ vanishes
unless $n$ is odd.

* Augmentation is essential here to make sure that the definition of $\bbeta{}$
does not depend on the choice of basis.
Indeed  if instead we choose orthonormal bases
$\{1\}\cup(a_k)_{k\in\IN}\subseteq \alg{A}$ and
$\{1\}\cup(b_k)_{k\in\IN}\subseteq \alg{B}$ then
it follows from Proposition <ref> that
the alternating words in $a_i$ and $b_j$ together with $1$ form an
orthonormal basis of the reduced free product.
However $\beta(u_1,u_2,\dots,u_n)=0$ for any such word
and thus the corresponding notion of $\bbeta{}$ would coincide with
$\varphi$ in this case.

As an immediate consequence of Proposition <ref> and
we have the following.

Consider the maps defined in Definition <ref> with
Assumption <ref>,
$\varphi$ not necessarily faithful.

* The maps $\E_{\alg{B}}$ and $\bbeta{}$ are well defined.

* The map $\E_{\alg{B}}$ defined above is a conditional expectation.

Note that the above conditional expectation is universal and allows for
calculations on the level of
arbitrary algebras.

Let $(\alg{M},\varphi)$ be an arbitrary ncps
with faithful state $\varphi$
and $\alg{A},\alg{B}\subseteq\alg{M}$ be free subalgebras such that the conditional
expectation $\E_{\alg{B}}$ exists (e.g., when $\alg{B}$ and $\alg{M}$ are von
Neumann algebras or $\alg{M}=\alg{A}*\alg{B}$).
Pick elements   $a_1,a_2,\ldots,a_m\in \alg{A}$
and elements $b_1,b_2,\dots,b_n\in\alg{B}$.

$$\mu:\IC\langle   X_1,X_2,\dots,X_m,Y_1,Y_2,\dots,Y_n\rangle\to\IC$$
the joint distribution of the tuple $(a_1,a_2,\dots,a_m,b_1,b_2,\dots,b_n)$,

\mu(P(X_1,X_2,\dots,Y_n)) = \varphi(P(a_1,a_2,\dots,a_m,b_1,b_2,\dots,b_n))

and let $\E^\mu_Y:\IC\langle X_1,X_2,\dots, X_m,Y_1,Y_2,\dots,Y_n\rangle\to\IC\langle Y_1,Y_2,\dots,Y_n\rangle$ be the conditional expectation
from Definition <ref> onto the subalgebra $\IC\langle Y_1,Y_2,\dots,Y_n\rangle$.
\[
\E_{\alg{B}}[P(a_1,a_2,\dots,a_m,b_1,b_2,\dots,b_n)]
\]

where $(\E_Y^{\mu}[P])(b_1,b_2,\dots,b_n)$
is the
polynomial $\E_Y^{\mu}[P]\in \IC \langle Y_1,Y_2,\dots,Y_n \rangle$ evaluated in $b_1,b_2,\ldots,b_n$.

This is an immediate consequence of the definition of $\mu$ and Corollary <ref>.

It is no surprise that this formal conditional expectation satisfies the
recurrence   (<ref>).

For any $\alg{A}$–$\alg{A}$ monomial $w$ the mapping $\E_{\alg{B}}$ satisfies the following recurrence

\begin{align}
\label{eq:EBrecurrence}
\E_{\alg{B}}[a_1b_1a_2\dotsm b_{n-1}a_n]
\begin{multlined}[t]
\beta_{2n-1}\left(a_1,b_1,a_2,\ldots ,b_{n-1},a_n\right)\\
\end{multlined}
\\
\label{eq:EBrecurrencebbeta}
\begin{multlined}[t]
\bbeta{}\left(a_1b_1a_2\dotsm b_{n-1}a_n\right)\\
+\sum_{k=1}^{n-1}\bbeta{}(a_1b_1a_2\dotsm a_k)\,b_k\E_{\alg{B}}[a_{k+1}b_{k+1}a_{k+2}\dotsm a_n].
\end{multlined}
\end{align}

Observe that in Definition <ref> (iii) on the right hand side
of the equation we sum over all possible choices of subsets of
$\{b_1,b_2,\dots,b_{n-1}\}$ and for variables between the chosen $b_i$'s we apply
Boolean cumulant, where we split $b_i$'s from $a_i$'s. In order to obtain the
recurrence above we reorganize the sum appearing in the definition of
$\E_{\alg{B}}$ according to the value of $i_1$, i.e., we have

\begin{multline*}
\sum_{k=1}^{n-1}\sum_{1\leq i_1<\dots< i_k\leq n-1}
b_{i_1}b_{i_2}\dotsm b_{i_k}\prod_{j=0}^{k}\beta_{2(i_{j+1}-i_j)-1}(a_{{i_j}+1},b_{i_{j}+1},a_{{i_j}+2},\ldots,a_{i_{j+1}})
\\
=\sum_{i_1=1}^{n-1} \beta_{2i_k-1}(a_1,b_1,a_2,\ldots,a_{i_1}) b_{i_1}
\sum_{k=0}^{n-2}\sum_{i_1<i_2<\dots< i_k\leq n-1}
b_{i_2}b_{i_3}\dotsm b_{i_k}
\\
\times
\prod_{j=1}^{k}\beta_{2(i_{j+1}-i_j)-1}(a_{{i_j}+1},b_{i_{j}+1},a_{{i_j}+2},\ldots,a_{i_{j+1}}).
\end{multline*}

To conclude the proof observe that each fixed value of $i_1$ the inner summation over
$i_2,\ldots,i_{n-1}$ reproduces exactly the definition of
$\E_{\alg{B}}[a_{i_1+1}b_{i_1+1}a_{i_1+2}\dotsm a_n]$.

Proposition <ref> offers a nice recursive formula for calculations of
conditional expectations in free variables, however a drawback is that it works
only on monomials starting and ending elements of algebra $\alg{A}$. In
order to make this formula useful for general calculations we need to have a
recurrence which works on any given polynomial.
To this end we split the functional $\bbeta{}$ somewhat analogously to the
partial block derivatives
The latter turns out to be the right tool to capture the structure of the recurrence from Proposition <ref>.

Under Assumption <ref>
we define the block cumulant functional
by prescribing its values on monomials as follows:

\begin{align*}
\bbeta{\alg{A}}(1)&=1\\
\bbeta{\alg{A}}(a_1b_1a_2\dotsm b_{n-1}a_n)
&= \beta_{2n+1}(a_1,b_1,a_2,\dots, b_{n-1},a_n)
\end{align*}

for any $\alg{A}$–$\alg{A}$ monomial $w$ with
$\alg{A}$–$\alg{B}$ factorization $w=a_1b_1a_2\dotsm b_{n-1}a_n$ and
$a_i\in\bar{\alg{A}}$ and $b_i\in\bar{\alg{B}}$.
For monomials $w$ which are not of type $\alg{A}$–$\alg{A}$ we set 	$\bbeta{\alg{A}}(w)=0$.

We define the analogous functional $\bbeta{\alg{B}}$,
which produces the value of Boolean cumulants of “blocked” variables from $\bar{\alg{A}}$ and $\bar{\alg{B}}$,
which vanishes unless a words starts and ends with elements from $\bar{\alg{B}}$.

Note that because of Property $(CAC)$ we have

\bbeta{}-\epsilon=\bbeta{\alg{A}}-\epsilon+\bbeta{\alg{B}}-\epsilon

With the help of these functionals and the block derivatives from Example <ref> we
can now extend recurrence
(<ref>) for $\E_{\alg{B}}$ to arbitrary monomials and
to compress it into an intuitive formula, which is the main result  of this section.

Suppose $\alg{M}=\alg{A}\freeprod\alg{B}$ and Assumption <ref>,
then for any $w\in\alg{M}$

\begin{equation}
\label{eq:condexprec}
\begin{aligned}
\E_{\alg{B}}[w] &= \bbeta{\alg{A}}(w)
+ (\bbeta{\alg{A}}\otimes \E_{\alg{B}}) [\rDelta_{\alg{B}}(w)]\\
&= \bbeta{\alg{A}}(w)
+ (\bbeta{\alg{A}}\otimes\E_{\alg{B}}) [\rnabla_{\alg{B}}(w)]\\
&= \bbeta{\alg{A}}(w) +
\end{aligned}
\end{equation}

Consider the first identity.
By linearity it is enough to consider  monomials. For a monomial of type
the statement is equivalent to Proposition <ref>.

If $w$ is a monomial of type $\alg{B}$–$\alg{B}$ or
$\alg{B}$–$\alg{A}$  then by the definition of
$\bbeta{\alg{A}}$  only one term contributes on the RHS and
we obtain the trivial identity

\E_{\alg{B}}[w]=
\E_{\alg{B}})[1\otimes w]

If $w$ is a monomial of type $\alg{A}$–$\alg{B}$ then it can be written as $w=w'b$, where $w'$ is of type $\alg{A}$–$\alg{A}$ and $b\in\bar{\alg{B}}$ and the right hand side is

\begin{align*}
\bbeta{\alg{A}}(w'b) + (\bbeta{\alg{A}}\otimes E_{\alg{B}}) [\rDelta_{\alg{B}}(w') (1\otimes b) + w'\otimes b]
&= 0 +
\Bigl(
(\bbeta{\alg{A}}\otimes E_{\alg{B}})[ \rDelta_{\alg{B}} w'] + \bbeta{\alg{A}}(w')
\Bigr)b\\
&= \E_{\alg{B}}[w']b,
\end{align*}

where we observed that the term inside the parenthesis
contains  the recurrence (<ref>)
for the conditional expectation of the word $w'$
which is of type $\alg{A}$–$\alg{A}$.

For the second identity
recall that   $\rnabla_{\alg{B}}= \rDelta_{\alg{B}}-\lDelta_{\alg{B}}$.
and observe the additional terms produced by $\lDelta_{\alg{B}}$ are tensors whose left
legs are monomials of type $*$–$\alg{B}$
which are annihilated by $\bbeta{\alg{A}}$.

The last identity is proved using the mirrored version
of the Boolean recurrence.

* It is straightforward to extend the functional $\bbeta{}$ and the recurrence
to  formal power series (provided that the resulting series converge) and in
particular compute conditional expectations of resolvents of
the form $(1-zw)^{-1}$ with $w\in\alg{M}$.

* In the setting of Remark <ref> (<ref>),
$\alg{A}=\IC\langle \alg{X}\rangle$ for some alphabet $\alg{X}=\{X_1,X_2,\dots,X_m\}$
and   $\alg{B}=\IC\langle \alg{Y}\rangle$ for some alphabet
we can also use the partial divided power derivations from Example <ref>

\begin{align*}
\rdelta_{\alg{Y}}(P)&=\sum_{i=1}^n (1\otimes Y_i)\partial_{Y_i}(P),\\
\ldelta_{\alg{Y}}(P)&=\sum_{i=1}^n (Y_i\otimes 1)\partial_{Y_i}(P),
\end{align*}

where $\partial_Y$ is the free different quotient
from Example <ref>
and we can write

\begin{align}
\label{eq:EY=bbetardeltaY}
\E_{\alg{Y}}^\mu[P] &= \bbeta{\alg{X}}(P) + (\bbeta{\alg{X}}\otimes
\E_{\alg{Y}}^\mu)[\rdelta_{\alg{Y}}(P)]
\end{align}

Again the additional terms are annihilated by $\bbeta{\alg{X}}$.

* The map $\rDelta_{\alg{B}}$  produces the least number of terms and
therefore is convenient for explicit calculations.
On the other hand, $\rnabla_{\alg{B}}$
(and $\rdelta_{\alg{Y}}$ in the case of   polynomial algebras)
are derivations, which is a great  advantage in connection with
non–commutative formal power
series,  in particular resolvents  and other rational
Indeed, for simplicity let us consider the resolvent $\Psi =
(1-zP(X,Y))^{-1}$ of a bivariate polynomial $P(X,Y)\in\IC\langle X,Y\rangle$.
From identity   (<ref>) we infer  that

\rnabla_X\Psi = z (\Psi\otimes 1) \rnabla_X P(X,Y) (1\otimes\Psi)

and a similar formula is true for $\rdelta_X\Psi$.
It does not hold for $\rDelta_X$, which does not obey the
Leibniz rule, yet in the case of  recurrence (<ref>)
we can pretend that it does and we have

\begin{align*}
\E_{X}[\Psi]
&= \bbeta{Y}(\Psi)
+ z(\bbeta{Y}\otimes \E_{X}) [(\Psi\otimes
1)(\rdelta_X w)(1\otimes \Psi]\\
&= \bbeta{Y}(\Psi)
+ z(\bbeta{Y}\otimes \E_{X}) [(\Psi\otimes
1)(\rnabla_X w)(1\otimes \Psi]\\
&= \bbeta{Y}(\Psi)
+ z(\bbeta{Y}\otimes \E_{X}) [(\Psi\otimes
1)(\rDelta_X w)(1\otimes \Psi]
\end{align*}

because the extra terms arising in $\rdelta_Xw$ and $\rnabla_Xw$ are annihilated by $\bbeta{Y}$.

For any $w\in\alg{M}$

(\bbeta{\alg{A}}\otimes \E_{\alg{B}})[\rDelta_{\alg{A}}w]
=   (\bbeta{\alg{A}}\otimes \E_{\alg{B}})[\lDelta_{\alg{A}}w]

Observe that from     (<ref>) we obtain the identity

\begin{align*}
(\bbeta{\alg{A}}\otimes \E_{\alg{B}})[\rnabla w]
&= \E_{\alg{B}}[w] - \bbeta{\alg{A}}(w) \\
&= (\bbeta{\alg{A}}\otimes \E_{\alg{B}})[\rnabla_{\alg{B}} w];
\end{align*}

on the other hand, $\rnabla=\rnabla_{\alg{A}} + \rnabla_{\alg{B}}$ and it
follows that

(\bbeta{\alg{A}}\otimes \E_{\alg{B}})[\rnabla_{\alg{A}} w]=0

which is equivalent to the claimed identity.

We illustrate this machinery with the problem of additive free convolution,
done in two ways.

Consider $\Psi=(1-z(X+Y))^{-1}=\sum_{n=0}^{\infty}
\left(z(X+Y)\right)^n$ then $\rdelta_{X}\Psi=z\Psi\otimes X\Psi$ and
the conditional expectation is

\begin{equation*}
\E_{X}[\Psi]
=\bbeta{Y}(\Psi) + z\bbeta{Y}\otimes \E_{X}[\Psi\otimes    X\Psi] = \bbeta{Y}(\Psi)(1+z X \E_{X}[\Psi])
\end{equation*}

\begin{equation}
\label{eq:X+Y:ExPsi=bbetaPsi}
\E_X[\Psi] =\bbeta{Y}(\Psi)
\bigl(
1- z\bbeta{Y}(\Psi)X
\bigr)^{-1}
\bigl(
1/\bbeta{Y}(\Psi)-z X
\bigr)^{-1}.
\end{equation}

The evaluation of $\bbeta{Y}(\Psi)$ will be discussed in Example <ref>.

In order to obtain the usual formulation of free additive subordination one has to consider the
resolvent $R=(z-(X+Y))^{-1}$ instead of $\Psi$.

Let $R=R(z)=(z-X-Y)^{-1}$, then $\rdelta_XR=R\otimes XR$ and the solution of the recurrence for the conditional expectation

\E_X[R] = \bbeta{Y}(R) + \bbeta{Y}(R)X \E_X[R]

is immediately obtained as

\begin{equation*}
\E_X[R] = (\bbeta{Y}(R)^{-1}-X)^{-1}
\end{equation*}

Now $\bbeta{Y}(R)$ is an analytic function of $z$ (via $R=R(z)$)
and its reciprocal $\omega(z)=1/\bbeta{Y}(R)$
is the subordination function from (<ref>). For its further evaluation see Example <ref>.

In fact the recurrence (<ref>)  allows for finding a system of
equations for the conditional expectation of resolvents of arbitrary polynomials.

Let $P\in \IC\langle X,Y\rangle $ and suppose
$\rdelta_X P = \sum_{i=1}^m u_i\otimes  X v_i$,
i.e., $\partial_X P = \sum_{i=1}^m u_i\otimes  v_i$,
and let $\Psi=(1-zP)^{-1}$. Then we have

\begin{align}
\label{eq:EXPsi=sumbPsiuiXEXviPsi}
\IE_X[\Psi] &= \bbeta{Y}(\Psi) + z \sum_{i=1}^m \bbeta{Y}(\Psi {u}_i)
X \IE_X[{v}_i\Psi]
\end{align}

where the conditional expectations on the right hand side satisfy the linear system of equations

\begin{equation}
\label{eq:linearizationForCondExp}
\begin{bmatrix}
\IE_X[v_1\Psi]\\
\IE_X[v_2\Psi]\\
\vdots\\
\IE_X[v_m\Psi]
\end{bmatrix}
\begin{bmatrix}
\bbeta{Y}[v_1\Psi]\\
\bbeta{Y}[v_2\Psi]\\
\vdots\\
\bbeta{Y}[v_m\Psi]
\end{bmatrix}
+H X
\begin{bmatrix}
\IE_X[v_1\Psi]\\
\IE_X[v_2\Psi]\\
\vdots\\
\IE_X[v_m\Psi]
\end{bmatrix}
\end{equation}

$H_{ij} = \bbeta{Y}({v}_i\Psi {u}_j)z+
\bbeta{Y}({u}_{ij})$.

* Let us label each appearance of $X$ in $P(X,Y)$ by consecutive labels $X_1,X_2,\ldots$ and similarly for $Y$ we label them as $Y_1,Y_2,\ldots$. For example for $XY+YX$ we write $X_1Y_1+X_2Y_2$. Then one can see that the entry $H_{ij}$ is exactly the functional $\bbeta{Y}$
evaluated in all
possible subwords of $\Psi$ which occur between $X_i$ and $X_j$. More precisely
if we consider $(1-z P)^{-1}$ then if $X_i$ and $X_j$ are in different
monomials of $P$ then these monomials are of the form
$u_iX_iv_i$ and $u_jX_jv_j$ and all subwords
of $(1-zP)^{-1}$ which are between $X_i$ and $X_j$ are exactly of the
form $v_i(1-zP)^{-1}u_j$ if $X_i$ and $X_j$ are in
the same monomial then this monomial is of the form
$u_iX_i u_{ij} X_j v_j$. Thus in this case we get the
additional term $u_{ij}$ to which we apply $\bbeta{Y}$.

Observe this very matrix $H_Y$ (with slightly
different powers of $z$) appeared in <cit.>
in the computation of the anti–commutator.
However there the goal was to determine the
distribution, not the conditional expectation.

* At this point it is not clear how to determine the matrix $H$ above.
For this it is necessary to evaluate the functional $\bbeta{}$ and this
will be done  in Section <ref>.

* Equation (<ref>) is linear and can
immediately be turned into a  linearization
(in the sense of [16])
of the rational
function $\IE_X[v_i\Psi]$.
We will develop this systematically in Section <ref>.

For the recurrence
we can use either $\rDelta_X$ or $\rdelta_X$.
The latter has the advantage of being a derivation.
In any case

\begin{align}
\IE_X[\Psi]
&= \bbeta{Y}(\Psi)
+ z (\bbeta{Y}\otimes \IE_X[ (\Psi\otimes 1)(\rdelta_X T)(1\otimes\Psi)]
\label{eq:EXpsidelta}
\\
&= \bbeta{Y}(\Psi)
+ z (\bbeta{Y}\otimes \IE_X[ (\Psi\otimes 1)(\rDelta_X T)(1\otimes\Psi)]
\end{align}

because the extra terms  in  (<ref>) are annihilated by $\bbeta{Y}$.

\begin{align*}
\rdelta_X T &= \sum_{i=1}^m u_i\otimes  X v_i\\
\end{align*}

i.e., $\partial_X T = \sum_{i=1}^m u_i\otimes v_i$.
Then recurrence (<ref>) reads

\begin{align*}
\IE_X[\Psi] &= \bbeta{Y}(\Psi) + z \sum_{i=1}^m \bbeta{Y}(\Psi {u}_i)
X \IE_X[{v}_i\Psi]\\
\end{align*}

and we can finally establish a system of equations for the conditional expectations
appearing on the RHS of (<ref>). Again we employ
(<ref>) and get

\begin{align*}
\IE_X[{v}_i\Psi] &= \bbeta{Y}({v}_i\Psi) +\sum_j\bbeta{Y}({u}_{ij})X \IE_X[v_j\Psi] + z \sum_{i=1}^m \bbeta{Y}(v_i\Psi {u}_j)X \IE_X[v_j\Psi].
\end{align*}

Define a matrix $H$ as

\begin{equation*}
H_{ij} = \bbeta{Y}({v}_i\Psi {u}_j)z+
\bbeta{Y}({u}_{ij}).
\end{equation*}

It is straightforward to see that identity
(<ref>) holds with $H_Y$ defined as above.

§ CALCULUS FOR THE BLOCK CUMULANT FUNCTIONAL $\BBETA{}$

The recurrence from Theorem <ref> opens a door to the study of
conditional expectations and in the case of resolvents the resulting linear system
can be solved explicitly to obtain explicit formulas in terms of the functional
Its  evaluation  is the subject of the present section.
However at the time of this writing we  lack sufficient understanding of
Boolean cumulants required to  handle this problem
in the general setting  of Assumption <ref>.
Therefore in the remainder of this paper we will mostly work in the formal setting of polynomial
algebras from Remark <ref>,
which allows the reduction to univariate Boolean cumulants which are well understood.

We will work in a formal ncps,
i.e., the free associative algebra $\IC\langle\alg{X}\rangle$ over some
alphabet $\alg{X}$ and some distribution
As usual the augmentation map $\epsilon:\IC \langle \alg{X} \rangle\to \IC$ is the
constant coefficient map   $\epsilon(P)=P(0)$.

More specifically, we will work in one of the following settings:

$\alg{M}=\IC\langle \alg{X}\cup\alg{Y}\rangle$
where $\alg{X}=\{X_1,X_2,\dots,X_m\}$ and $\alg{Y}=\{Y_1,Y_2,\dots, Y_n\}$
are free from each other with respect to $\mu$ and so are
$\alg{A}=\IC\langle \alg{X}\rangle$ and $\alg{B}=\IC\langle \alg{Y}\rangle$.
In this case we will denote the corresponding conditional expectations,
derivations, and functionals
by $\IE_{\alg{X}}$, $\rDelta_{\alg{X}},\rdelta_{\alg{X}}$, $\bbeta{\alg{X}}$ etc.

* $\alg{M}=\left(\IC\langle\uXn \rangle,\mu\right)$

\begin{align*}
\alg{A}&=\IC\langle X_i \mid i\in I\rangle
\alg{B}&=\IC\langle X_j \mid j \in J\rangle.
\end{align*}

where $I\dot{\cup}J=[n]$ is a partition and we  assume in addition that all variables $X_1,X_2,\ldots,X_n$ are free from
each other with respect to the functional $\mu$.
In this case we will denote the corresponding conditional expectations,
derivations, and functionals by
In this case we will denote conditional expectations
by $\IE_I$, $\rDelta_I$, $\bbeta{I}$ etc.

The typical case is

\begin{align*}
\alg{A}&=\IC\langle \uXk\rangle
\alg{B}&=\IC\langle X_{k+1},X_{k+2},\ldots,X_n\rangle.
\end{align*}

The following functionals provide the key to evaluate the functionals
$\bbeta{}$. They operate by fully splitting monomials into their factors.

On the formal ncps $(\IC\langle\alg{X}\rangle,\mu)$
we define the fully factored Boolean cumulant $\fbeta{}$  by prescribing their values on monomials

\begin{align*}
\fbeta{}(1) &=1,\\
\fbeta{}(X_{i_1}X_{i_2}\dotsm X_{i_k})
&= \beta_{k}(X_{i_1},X_{i_2},\dots,X_{i_k})
\end{align*}

We can decompose the functional $\fbeta{}$ along
free subsets of variables   similar to the functional $\bbeta{}$:

For a subset  $\alg{Y}\subseteq\alg{X}$ of the variables
we define linear functionals $\fbeta{\alg{Y}}$ on monomials as follows:

\begin{align*}
\fbeta{\alg{Y}}(1) &=1,\\
\fbeta{\alg{Y}}(X_{i_1}X_{i_2}\dotsm X_{i_k})
&=  \begin{cases}
\beta_{k}(X_{i_1},X_{i_2},\dots,X_{i_k}) & \text{if $X_{i_1},X_{i_k}\in\alg{Y}$}\\
0 & \text{otherwise}
\end{cases}
\end{align*}

Then $\fbeta{}=\fbeta{\alg{X}}$ and if a set of variables $\alg{Y}$ is the disjoint union of mutually
free (with respect to $\mu$) subsets $\alg{Y}_j$ then thanks to property
we have

\fbeta{\alg{Y}}=\epsilon+\sum (\fbeta{\alg{Y}_j}-\epsilon)

Next we state Lemma <ref> in terms of the functional
which will turn out to be useful in the proof of the next theorem.

Assume the setting from Assumption <ref>
i.e.,   $\alg{M}=\IC\langle \alg{X}\cup\alg{Y}\rangle$ with $\alg{X}$ and
$\alg{Y}$ mutually free.
Then for any letters $X_{i_j}\in\alg{X}$ and any elements $V_j\in\IC\langle\alg{Y}\rangle$
we have

\begin{align*}
\fbeta{\alg{X}}(X_{i_0}V_1X_{i_1}V_2\dotsm V_kX_{i_k})
\end{align*}

We emphasize that it is not required that $V_j$ have vanishing constant term.

For the next theorem we enhance our toolbox with yet another derivation, which
implements a kind of elementary unshuffle coproduct.

For a subset of variables $\alg{Y}=\{Y_1,Y_2,\dots,Y_k\}\subseteq\alg{X}$
define the first order unshuffle operator

\begin{align*}
L_{\alg{Y}}^*\odot \partial_{\alg{Y}} = \sum L_{Y_j}\otimes
\partial_{Y_j}:\IC\langle\alg{X}\rangle &\to
\IC\langle\alg{Y}\rangle\otimes\IC\langle\alg{X}\rangle\otimes\IC\langle\alg{X}\rangle\\
W&\mapsto \sum Y_j\otimes \partial_{Y_j}W
\intertext{Its iterations are defined on the left leg}
(L_{\alg{Y}}^*\odot\partial_{\alg{Y}})^2 W
&= \sum_{i,j} Y_iY_j\otimes (\partial_{Y_i}\otimes\id)\partial_{Y_j} W\\
(L_{\alg{Y}}^*\odot\partial_{\alg{Y}})^p W
&= \sum_{i_1,i_2,\dots,i_p} Y_{i_1}Y_{i_2}\dotsm Y_{i_p}\otimes
(\partial_{Y_{i_1}}\otimes\id^{\otimes p-1})(\partial_{Y_{i_2}}\otimes\id^{\otimes p-2})\dotsm\partial_{Y_{i_p}} W
\end{align*}

In the theorem below we summarize the relations between the functionals
$\bbeta{}$ and $\fbeta{}$
which will allow to evaluate these functionals on some non–commutative power series.

Assume the setting from Assumption <ref>   (<ref>),
i.e.,   $\alg{M}=\IC\langle \alg{X}\cup\alg{Y}\rangle$ with $\alg{X}$ and
$\alg{Y}$ mutually free.

* For any element $P\in \IC\langle \alg{X}\cup\alg{Y} \rangle$ we have

\begin{equation}\label{eq:intro_betas_prod_as_entries}
\begin{aligned}
\bbeta{\alg{X}}(P)
&= \epsilon(P) + (\fbeta{\alg{X}}\otimes\bbeta{\alg{X}})(\ldelta_{\alg{X}} P)
\\
&= \epsilon(P) +
(\bbeta{\alg{X}}\otimes\fbeta{\alg{X}})(\rdelta_{\alg{X}} P)
\end{aligned}
\end{equation}

* For any element $P\in \IC\langle \alg{X}\cup\alg{Y} \rangle$ we have

\begin{equation}
\label{eq:IntroVNRP_for_betas}
\begin{aligned}
\fbeta{\alg{X}}(P)
\fbeta{}\otimes\left[\epsilon\otimes
\left(\bbeta{\alg{Y}}\right)^{\otimes(k-1)}\otimes
\epsilon\right]\left( (L_{\alg{X}}^*\odot\partial_{\alg{X}})^k(P)\right)
\\
&= \epsilon(P)+\sum_{k=1}^\infty
\sum_{i_1,i_2,\dots,i_k} \beta_k(X_{i_1},X_{i_2},\dots, X_{i_k})
\left[\epsilon\otimes
\left(\bbeta{\alg{Y}}\right)^{\otimes(k-1)}\otimes
\epsilon\right]\left(
\partial_{X_{i_1}}\partial_{X_{i_2}}\dotsm\partial_{X_{i_k}} P
\right)
\end{aligned}
\end{equation}

In particular, when $\alg{X}=\{X\}$ consists of a single variable, then

\begin{equation}
\label{eq:IntroVNRP_for_betas:uni}
\fbeta{X}(P)
= \epsilon(P)+\sum_{k=1}^\infty
\beta_k(X)
\left[\epsilon\otimes
\left(\bbeta{\alg{Y}}\right)^{\otimes(k-1)}\otimes
\epsilon\right]\left(
\partial_X^k P
\right)
\end{equation}

* It is straightforward to extend by linearity the functionals
$\bbeta{\alg{X}}$ and $\fbeta{\alg{X}}$
and Theorem <ref> to formal power series.
One the one hand, for elements of the algebra     $\IC\llangle \alg{X}\rrangle$
of formal power series     in
non–commuting variables $\alg{X}=\{X_1,X_2,\ldots,X_n\}$,
provided the resulting series converges;
on the other hand, to the algebra $\IC\langle\alg{X}\rangle((z))$
of formal power series with non–commuting coefficients.

* Formula (<ref>) can also be stated in terms of the
half-shuffle coproduct $\Delta_{\prec}$ of <cit.>,
but this will be dealt with elsewhere.

* Essentially formula (<ref>)
is a reformulation of Corollary <ref> in terms of
the functionals and derivations which we introduced above,
after observing that the extra terms
vanish as a consequence of Property $(CAC)$.

We will only prove the first of the two identities
(<ref>) as the proof of second equation
is essentially the same.
By linearity it suffices to consider monomials  $w\in\IC\langle\alg{X}\cup\alg{Y}\rangle$
and in fact only words $w$ of type $\alg{X}$–$\alg{X}$,
any word of different type being annihilated on both sides of the equation by
definition of $\bbeta{\alg{X}}$.
So suppose $w=u_1v_1u_2\dotsm v_{k} u_{k+1}$ is a free factorization
into words $u_i\in\alg{X}^+$ and $v_i\in\alg{Y}^+$, then
On the other hand

\begin{equation}
\label{eq:deltaAW}
\ldelta_{\alg{X}}
w=\sum_{i=1}^{k+1}u_1v_1u_2\cdots v_{i-1}\cdot\ldelta_{\alg{X}}(u_i)\cdot
v_iu_{i+1}\cdots u_{m+1}.
\end{equation}

This produces the same splittings of the factors from $\alg{X}$ in the
$\alg{X}$–$\alg{Y}$ factorization
as in (<ref>), except for two kinds configurations:

* Formula (<ref>) contains splittings of factors of
$w$ coming from both  $\alg{X}^+$ and $\alg{Y}^+$.
The latter are absent from (<ref>) because whenever
a factor  $v_i=v_i' v_i''$ is split, at least one of the factors of the arising term
$\fbeta{\alg{X}}(u_1v_1u_2\cdots v_i')\bbeta{\alg{X}}(v_i''u_{i+1}\dotsm u_{m+1})$
vanishes by definition of $\fbeta{\alg{X}}$ and $\bbeta{\alg{X}}$.

* In the derivative $\ldelta_{\alg{X}}w$ there are extra terms of
$u_1v_1u_2\dotsm u_k\otimes v_ku_{k+1}v_{k+1}\dotsm u_{m+1}$ which do not
appear in Corollary <ref>,
but the application of
creates the factor $\bbeta{\alg{X}}\left(v_k\cdots u_{m+1}\right)=0$.

* For the proof of (<ref>)
first observe that both sides of this equation are $1$ for the empty word
and  that  both sides are zero unless $w$ is of type $\alg{X}$–$\alg{X}$.
Thus fix a monomial $w\in(\alg{X}\cup\alg{Y})^+$  of the latter type.
We write it as $w=X_{i_1}v_1X_{i_1}v_2\dotsm v_k X_{i_{k+1}}$
with $v_j\in\alg{Y}^*$, i.e., we allow some $v_j=1$.
However the latter will be considered later  and we assume first that $v_j\neq 1$ for
$i=1,2,\ldots,k$. Comparing (<ref>) with formula
(<ref>) observe that according to (<ref>) we have a sum over all
choices of variables from $\alg{X}$, with the restriction that both $X_{i_1}$  and
$X_{i_{k+1}}$ are always selected. In formula (<ref>), all
choices of $p$ $X$'s are given by the $p$-th derivative, the variables
annihilated by the free derivative are put to the outer block,
moreover application of
$\epsilon\otimes\left(\bbeta{\alg{Y}}\right)^{\otimes (k-1)}\otimes
\epsilon$ evaluates to zero unless both $X_{i_1}$ and $X_{i_k}$ are
annihilated by the derivative. Further from (<ref>) we
get a product of cumulants of type
which are extracted between the two $X$'s moved to the outer block and
this is
exactly equal to $\bbeta{\alg{Y}}(v_{p}X_{i_{p+1}}\cdots X_{i_q}v_{q})$.

It remains to consider the case where some $v_q=1$.
Assume that it comes from a pocket determined by $X_{i_p}$ and $X_{i_r}$
which are both chosen to the outer block, i.e., $p\leq q<r$.
The existence of such $p$ and $q$ is guaranteed as we always choose the first
and the last $X$ to the outer block.
We will consider the contribution of this pocket in both formulas
$\eqref{bocu2}$ and (<ref>).
There are three possible cases depending on how $v_i$ is placed between $j$-th and $k$-th $X_1$:

* $p=q$ and $r=p+1$ then the pocket created by these two elements contains only $v_q$ and formula (<ref>) gives $\beta_{1}(v_q)=1$ while from (<ref>) we get $\bbeta{\alg{Y}}(1)=1$.

* $p=q$ and $r>p+1$, then  the pocket contains
$v_pX_{i_{p+1}}\cdots v_{r-1}$ and (<ref>) gives precisely
because the first argument is $1$ and   (<ref>) applies.
On the other hand (<ref>) we get
$\bbeta{\alg{Y}}(v_pX_{i_{p+1}}v_{p+1}X_{i_{p+1}}\dotsm v_{r-1})
=\bbeta{\alg{Y}}(X_{i_{p+1}}v_{p+1}X_{i_{p+1}}\dotsm  v_{r-1})=0$
by definition of the functional $\bbeta{\alg{Y}}$.
The case $p<q=r-1$ is treated similarly.

* Having eliminated all such units, it remains to consider  the case
By (<ref>) the contribution of the pocket containing $v_q$ equals

\begin{multline*}
\beta_{2(k-i)-1}(v_p,X_{i_{p+1}},v_{p+1},X_{i_{p+2}},\dots,X_{i_q},1,X_{i_{q+1}},\dots,X_{i_{r-1}},v_{r-1})
\\
\beta_{2(k-i)-1}(v_p,X_{i_{p+1}},v_{p+1},X_{i_{p+2}},\dots,X_{i_q},X_{i_{q+1}},\dots,X_{i_{r-1}},v_{r-1})
\\
\beta_{2(k-i)-1}(v_p,X_{i_{p+1}},v_{p+1},X_{i_{p+2}},\dots,X_{i_q}X_{i_{q+1}},\dots,X_{i_{r-1}},v_{r-1})
\end{multline*}

because of   (<ref>) and  Lemma <ref>.

Assuming that $v_p,v_{p+1},\dots,v_{r-1}\neq 1$, then $\bbeta{\alg{Y}}(v_p
X_{i_{p+1}}v_{p+1}\dotsm X_{i_q} v_q X_{i_{q+1}}\dotsm
X_{i_{r-1}}v_{r-1})$ evaluates to the same value. If there are $v_l=1$ for
$p<l<r-1$ and $l\neq q$ we repeat the same argument.
Observe that with the previous steps we have already made sure that $v_{p},v_{r-1}\neq 1$.

Let us illustrate this with some examples.

A direct calculation using VNRP
shows that

\begin{align*}
\fbeta{X}(XYX)=\beta_3(X,Y,X)=\beta_2(X)\beta_1(Y)
\end{align*}

Let us calculate the derivatives of $\partial_X(XYX)=1\otimes YX+XY\otimes 1$ of
course both corresponding terms vanish after application of $\bbeta{Y}$ and
thus $n=1$ does not contribute (it never will, effectively the sum starts from
n=2). The second derivative gives $D_2(XYX)=1\otimes Y\otimes 1$, and hence
from the formula in the theorem we also get

\begin{align*}
\fbeta{X}(XYX)=\beta_3(X,Y,X)=\beta_2(X)\beta_1(Y).
\end{align*}

[The additive subordination function]

In Example <ref> we concluded that the additive subordination function for
$R=(z-X-Y)^{-1}$ is $1/\bbeta{Y}(R)$.
Now $\ldelta_YR =RY\otimes R$ and
the recurrence (<ref>) yields

\begin{align*}
\bbeta{Y}(R)
&= \epsilon(R) + (\fbeta{Y}\otimes\bbeta{Y})(\ldelta_YR)\\
&= \frac{1}{z} + \fbeta{Y}(RY)\bbeta{Y}(R)
\end{align*}

and thus
the subordination function is

\omega(z) = 1/\bbeta{Y}(R) = z-z\fbeta{Y}(RY)

cf. <cit.>.

In anticipation of matrix valued formulas arising in Section <ref>
we continue Example <ref>
and consider the simple example where $n=2$
and the power series $\Psi=(1-z(X+Y))^{-1}=\sum_{n=0}^{\infty}\left(z(X+Y)\right)^n\in\IC\langle X,Y\rangle((z))$.
Since we work only with two variables we will write $\bbeta{X},\fbeta{X}$ etc. which should not lead to any confusion.

Clearly we have

\begin{align*}
\rdelta_{X}(\Psi)=1+z\Psi\otimes X \Psi,\qquad\rdelta_{Y}(\Psi)=1+z\Psi\otimes Y \Psi.
\end{align*}

Thus from (<ref>) we obtain

\begin{align*}
\bbeta{X}(\Psi)=1+z\bbeta{X}(\Psi)\fbeta{X}(X\Psi),\qquad\bbeta{Y}(\Psi)=1+z\bbeta{Y}(\Psi)\fbeta{Y}(Y\Psi).
\end{align*}

Hence we obtain

\begin{align*}
\bbeta{X}(\Psi)=\left(1-z\fbeta{X}(X\Psi)\right)^{-1},\qquad\bbeta{Y}(\Psi)=\left(1-z\fbeta{Y}(Y\Psi)\right)^{-1}.
\end{align*}

Comparing with (<ref>) we conclude that

\E_X[\Psi] = (1-z\fbeta{Y}(Y\Psi)-zX)^{-1}

Moreover observe that

\begin{align*}
\partial_{X}^n(X\Psi)=z^{n-1}1\otimes \Psi^{\otimes n}+z^{n}X\Psi\otimes \Psi^{\otimes n},\qquad\partial_{Y}^n(Y\Psi)=z^{n-1}1\otimes \Psi^{\otimes n}+z^{n}Y\Psi\otimes \Psi^{\otimes n}.
\end{align*}

Thus equation (<ref>) gives

\begin{align*}
\fbeta{X}(X\Psi)&=\sum_{n=1}^{\infty} \beta_n(X)\bbeta{Y}(\Psi)^{n-1}z^{n-1}=\widetilde{\eta}_{X}(z\bbeta{Y}(\Psi)),\\
\fbeta{Y}(Y\Psi)&=\sum_{n=1}^{\infty} \beta_n(Y)\bbeta{X}(\Psi)^{n-1}z^{n-1}=\widetilde{\eta}_{Y}(z\bbeta{X}(\Psi)).
\end{align*}

Finally we obtain the following system of equations

\begin{align*}
\fbeta{X}(X\Psi)=\widetilde{\eta}_{X}\left(z\left(1-z\fbeta{Y}(Y\Psi)\right)^{-1}\right),\quad
\fbeta{Y}(Y\Psi)=\widetilde{\eta}_{Y}\left(z\left(1-z\fbeta{X}(X\Psi)\right)^{-1}\right).
\end{align*}

We shall see below that this system of equations has   unique  power
series solutions $\fbeta{X}(X\Psi)$ and $\fbeta{Y}(Y\Psi)$ analytic at 0
and in fact yields the fixed point equation for subordination function for free
additive convolution.
Thus this system determines the function needed in Example <ref>.

§ LINEARIZATION AND CONDITIONAL EXPECTATIONS

The procedure presented  in Proposition <ref>
can be systematized using  linearizations of resolvents from the very beginning.
For the reader not familiar with  rational series and linearizations
the basic facts are collected in  Appendix <ref>.
Here we will adapt and amplify all previously defined operations to the level of matrices
and then lift  Example <ref> to the matrix-valued

§.§ Amplifications of expectations and cumulants

Most concepts considered in the present paper can be generalized to the
operator valued case.
Rather than do this in the general case we
restrict the discussion to amplifications to tensor products.
In fact the matrix valued case
would suffice for  later applications to linearizations,
however the proofs are conceptually simpler in the language of
tensor products.

Let $(\alg{A},\varphi)$ be an ncps and $\alg{C}$ a unital algebra.
We shall consider $\alg{C}\otimes\alg{A}$ as a $\alg{C}$-bimodule with
$c_1\cdot (c\otimes a)\cdot c_2 = c_1cc_2\otimes a$.

In the case where $\alg{C}=M_N(\IC)$ is a matrix algebra
we will denote the elements by $\sum C_iu_i$
where for a matrix $C\in M_N(\IC)$ and $u\in\alg{A}$ we denote by $Cu=uC\in M_N(\alg{A})$
the matrix with entries

(Cu)_{ij} = c_{ij}u

The following lemma is easily verified on elementary tensors.

Let $(\alg{A},\varphi)$ be a ncps and $\alg{C}$ a unital algebra.
and denote by
the amplification of $\varphi$.

* $\varphi_{\alg{C}}$     is a $\alg{C}$-bimodule map:

\begin{equation*}
\varphi_{\alg{C}}(c_1\cdot u\cdot c_2) = c_1 \varphi_{\alg{C}}(u) c_2
\end{equation*}

for all $c_1,c_2\in\alg{C}$ and $u\in\alg{C}\otimes\alg{A}$.

* Let $\alg{B}\subseteq\alg{A}$ be a subalgebra and $\E_{\alg{B}}:\alg{A}\to\alg{B}$ be a
conditional expectation for $\varphi$.
Then its amplification
is a $\alg{C}\otimes\alg{B}$-bimodule map
moreover it is a conditional expectation for the amplification
in the sense that

\begin{equation*}
\varphi_{\alg{C}}((\id_{\alg{C}}\otimes\E_{\alg{B}})[u]v) = \varphi_{\alg{C}}(uv)
\end{equation*}

for any $u\in\alg{C}\otimes\alg{A}$ and any $v\in\alg{C}\otimes\alg{B}$.

* The $\alg{C}$-valued Boolean cumulants defined by the amplification of the

\varphi_\alg{C}(u_1u_2\dotsm u_n)
= \sum_{k=1}^n \beta_k^{\alg{C}}(u_1,u_2,\dots,u_k)\,\varphi_\alg{C}(u_{k+1}u_{k+2}\dotsm u_n)

are given by

\beta_n^{\alg{C}}(c_1\otimes a_1,c_2\otimes a_2,\dots,c_n\otimes a_n)
=  c_1c_2\dotsm c_n \,\beta_n( a_1, a_2,\dots, a_n)

In the case where $\alg{C}=M_N(\IC)$ is a matrix algebra  we can apply the usual identification of
$M_n(\IC)\otimes\alg{A}$ with $M_N(\alg{A})$ and  reformulate the lemma in
terms of matrix operations as follows.

Let $(\alg{A},\varphi)$  be a ncps.

* The amplification     $\varphi^{(N)}:M_N(\alg{A})\to M_N(\IC)$ to the matrix algebra
is the     entry-wise application
$\varphi^{(N)}([a_{ij}]) = [\varphi(a_{ij})]$.
These maps
form a family of matrix bimodule maps:

\varphi^{(k)}(U\cdot A\cdot V) = U \varphi^{(N)}(A) V

holds for any matrix $A\in M_N(\alg{A})$ and any scalar matrices
$U\in M_{k\times N}(\IC)$ and $V\in M_{N\times k}(\IC)$,

* Let $\alg{B}\subseteq\alg{A}$ be a subalgebra and $\E_{\alg{B}}:\alg{A}\to\alg{B}$ be a
conditional expectation for $\varphi$.
Then the entry-wise application maps
\bigl[
\bigr]  =
\bigl[\E_{\alg{B}}[a_{ij}]
\bigr]_{ij}$
form a family of  $M_N(\alg{B})$-bimodule maps:

\begin{equation}
\label{eq:EBnbimodule}
\E_{\alg{B}}^{(k)}[U\cdot A\cdot V] = U \cdot \E_{\alg{B}}^{(N)}[A] \cdot V
\end{equation}

holds for any matrix $A\in M_N(\alg{A})$ and any scalar matrices
$U\in M_{k\times N}(\IC)$ and $V\in M_{N\times k}(\IC)$.
Moreover it is a conditional expectation for the map
in the sense that
$\varphi^{(N)}(\E_{\alg{B}}^{(N)}(A)B) = \varphi^{(N)}(AB)$
for any $A\in M_N(\alg{A})$ and any $B\in M_N(\alg{B})$.

* The entries of the $M_N(\IC)$-valued Boolean cumulants defined by the amplification of the

\varphi^{(N)}(A_1A_2\dotsm A_n)
= \sum_{k=1}^n \beta_k^{(N)}(A_1,A_2,\dots,A_k)\,\varphi^{(N)}(A_{k+1}A_{k+2}\dotsm A_n)

are given by

\beta_n^{(N)}(A_1,A_2,\dots,A_n)_{ij}
=  \sum_{i_1,i_2,\dots,i_{n-1}}\beta_n( a_{ii_1}^{(1)}, a_{i_1i_2}^{(2)},\dots, a^{(n)}_{i_{n-1}j})

The corresponding matrix valued versions of the functionals $\bbeta{}$ and
are defined analogously as

\begin{align*}
\bbetan{\alg{A}}{N}(A)_{ij}
&= \bbeta{\alg{A}}(a_{ij})
\fbetan{\alg{A}}{N}(A)_{ij}
&= \fbeta{\alg{A}}(a_{ij})
\end{align*}

whenever these are defined (Assumption <ref>, resp. Assumption <ref>).

§.§ Amplifications of derivations

Let $\alg{A}$ and $\alg{C}$ be algebras, $D:\alg{A}\to\mathfrak{M}$ be a
derivation into an $\alg{A}$-bimodule $\mathfrak{M}$.
Then $D^{(\alg{C})}=\id_{\alg{C}}\otimes D:\alg{C}\otimes\alg{A}\to \alg{C}\otimes\mathfrak{M}$
is a derivation, where $\alg{C}\otimes\mathfrak{M}$ is an
with action

(c_1\otimes a_1) \cdot (c\otimes\mathfrak{m}) \cdot (c_2\otimes a_2) =  c_1cc_2\otimes(a_1\cdot \mathfrak{m}\cdot a_2)

It suffices to verify the Leibniz rule    (<ref>)
for   $D^{(\alg{C})}$ on elementary tensors:

\begin{align*}
\bigl(
(c_1\otimes a_1)(c_2 \otimes a_2)
\bigr)
&= c_1c_2 \otimes D(a_1a_2)\\
&= c_1c_2 \otimes (D(a_1)\cdot a_2) + c_1c_2 \otimes (a_1\cdot D(a_2))\\
&= D^{(\alg{C})}(c_1 \otimes a_1)\cdot(c_2\otimes a_2) +  (c_1\otimes
a_1)\cdot D^{(\alg{C})}(c_2\otimes a_2)
\end{align*}

This is true in particular for $\alg{C}=M_N(\IC)$
(cf. <cit.>).

Let $\mathfrak{M}$ be an $\alg{A}$-bimodule and  $D:\alg{A}\to\mathfrak{M}$ be  a derivation.
Then $M_N(\mathfrak{M})$ is an $M_N(\alg{A})$-bimodule with action

[a'_{ij}]_{ij}\cdot [\mathfrak{m}_{ij}]_{ij}\cdot [a''_{ij}]_{ij}
\Bigl[
\sum_{k,l} a'_{ik}\cdot \mathfrak{m}_{kl}\cdot a''_{lj}
\Bigr]_{ij}

and the matrix amplification $D^{(N)}:M_N(\alg{A}) \to M_N(\mathfrak{M})$,
i.e., entry-wise application

D^{(N)}([a_{ij}]) = [D(a_{ij})]

satisfies the Leibniz rule (<ref>).

Let $D:\alg{A}\to \mathfrak{M}$ be a derivation where
is the bimodule with action

a'\cdot (a_1\otimes a_2)\cdot a'' = a'a_1\otimes a_2a'' = (a'\otimes
1)(a_1\otimes a_2)(1\otimes a'')

Then using Notation <ref>
the Leibniz rule for the product of two matrices $A_1,A_2\in M_N(\alg{A})$

D^{(N)}(A_1A_2) = D^{(N)}(A_1)(1_{\alg{A}}\odot A_2) + (A_1\odot
\in  M_N(\alg{A}\otimes\alg{A}),

= \sum_k D(a^{(1)}_{ik})(1_{\alg{A}}\otimes  a^{(2)}_{kj})
+ (a^{(1)}_{ik}\otimes 1_{\alg{A}}) D(a^2_{kj})

In particular,
for a resolvent $\bPsi = (I_N-zA)^{-1}$ the derivation results in

D^{(N)}\bPsi = z(\bPsi\odot 1)D^{(N)}(A)(1\odot\bPsi)

and in the case of an elementary tensor $A=C\otimes a$ with $D(a) = \sum u_i\otimes v_i$,

D^{(N)}\bPsi = z\sum \bPsi u_i C\odot v_i\bPsi.

§.§ Computing conditional expectations via linearizations

In the following we will omit the
superscripts from   $\IE_{\alg{B}}^{(N)}$  etc. and write  $\IE_{\alg{B}}$ etc. whenever the context is unambiguous.
Now the bimodule property (<ref>) allows us to rewrite
the conditional expectation of a linearization

(1-z^mP)^{-1} = u^t(I-zL)^{-1}v

\begin{equation}
\label{eq:EB1-zP=utEB1-zLv}
\E_{\alg{B}}
\bigl[
\bigr]
= u^t\E_{\alg{B}}
\bigl[(I-zL)^{-1}\bigr]v
\end{equation}

and the problem is reduced to the computation of conditional expectations of matrix pencils.
We shall see that this accomplished by repeating the computation from
Example <ref> with matrix coefficients.

It is immediate to verify the
amplification of the recurrence
(<ref>), namely

(\id_{\alg{C}}\otimes\E_{\alg{B}})[c\otimes P]
= (\id_{\alg{C}}\otimes\bbeta{\alg{A}})(P)
+ (\id_{\alg{C}}\otimes\bbeta{\alg{A}}\otimes\E_{\alg{B}})[\id_{\alg{C}}\otimes\rDelta_{\alg{B}}(P)]

The matrix analog of
Theorem <ref>
in terms of the amplifications of $\E_{\alg{B}}$, $\bbeta{\alg{A}}$,
$\rDelta_{\alg{B}}$ and $\rnabla_{\alg{B}}$
from Lemmas <ref> and   <ref>
reads as follows.

Let $M\in M_N(\alg{M})$ and $\alg{A},\alg{B}\subseteq\alg{M}$ as in
Assumption <ref>.

\begin{align*}
\E_{\alg{B}}^{(N)}[M] &= \bbetan{\alg{A}}{N}(M)
+ (\bbeta{\alg{A}}\otimes \E_{\alg{B}})^{(N)} [\rDelta_{\alg{B}}^{(N)}(M)]\\
&= \bbetan{\alg{A}}{N}(M)
+ (\bbeta{\alg{A}}\otimes\E_{\alg{B}})^{(N)}
&= \bbetan{\alg{A}}{N}(M)
+ (\E_{\alg{B}}\otimes\bbeta{\alg{A}})^{(N)} [\lDelta_{\alg{B}}^{(N)}(M)]
\end{align*}

In particular, for a linear matrix pencil $L=\sum C_i'a_i+C_j''b_j\in M_N(\alg{M})$
the conditional expectation of the resolvent
$\bPsi =
\bigl(
I_N-z(\sum C_i'a_i+ \sum C''_j b_j)
\bigr)^{-1}$ is

\begin{equation}
\label{eq:EnPsi=1-zHAsumCbHA}
\E_{\alg{B}}^{(N)}[\bPsi] =
\bigl(
I_N- z H_{\alg{A}}\sum C_j'' b_j
\bigr)^{-1} H_{\alg{A}}
\end{equation}

$H_{\alg{A}}= \bbetan{\alg{A}}{N}(\bPsi)\in M_N(\IC)$.

Formula (<ref>) is a matrix valued version of
Example <ref>
and the proof runs along the same lines.

Next we switch to Assumption <ref> and
Theorem <ref>
to the matrix level.

Let $\alg{M}= M_N(\IC\langle\alg{X}\cup\alg{Y}\rangle)$ and $\alg{X},\alg{Y}$
free with respect to $\mu:\alg{M}\to\IC$ as in
Assumption <ref>.

\begin{align*}
\E_{\alg{Y}}^{(N)}[M] &= \bbetan{\alg{X}}{N}(M)
+ (\bbeta{\alg{X}}\otimes \E_{\alg{Y}})^{(N)} [\rdelta_{\!\alg{Y}}^{(N)}(M)]
\end{align*}

Next, identity (<ref>) trivially lifts to
tensor products

\begin{equation*}
\begin{aligned}
(\id_{\alg{C}}\otimes\bbeta{\alg{X}})(c\otimes P)
&= c \epsilon(P) + c\cdot(\fbeta{\alg{X}}\otimes\bbeta{\alg{X}})(\ldelta_{\alg{X}} P)
\\
&= c \epsilon(P) +
c\cdot(\bbeta{\alg{X}}\otimes\fbeta{\alg{X}})(\rdelta_{\alg{X}} P)
\end{aligned}
\end{equation*}

which immediately translates to the case of matrices as follows.

Let $M\in M_N(\alg{M})$, then with Assumption <ref> we have

\begin{equation*}
\begin{aligned}
\bbetan{\alg{X}}{N}(M)
&=  \epsilon^{(N)}(M) + (\fbeta{\alg{X}}\otimes\bbeta{\alg{X}})^{(N)}(\ldelta_{\alg{X}}^{(N)} M)
\\
&=  \epsilon^{(N)}(M) + (\bbeta{\alg{X}}\otimes\fbeta{\alg{X}})^{(N)}(\rdelta_{\alg{X}}^{(N)} M)
\end{aligned}
\end{equation*}

In particular, for a linear matrix pencil $L=\sum C_i'X_i+C_j''Y_j\in M_N(\alg{M})$
the block cumulant functional at
$\bPsi =
\left(
I_N-z(\sum C_i'X_i+ \sum C''_j Y_j)
\right)^{-1}$ is

\begin{equation}
\label{eq:bbetaPsi=1-zsumfbetaPsiaiCi}
\begin{aligned}
\bbetan{\alg{X}}{N}(\bPsi)
\bigl(
I_N- z \sum\fbetan{\alg{X}}{N}(\bPsi X_i)C_i
\bigr)^{-1}
\bigl(
I_N- z \fbetan{\alg{X}}{N}(\bPsi L)
\bigr)^{-1}
\\
\bigl(
I_N- z \sum C_i\fbetan{\alg{X}}{N}(X_i\bPsi)
\bigr)^{-1}
\bigl(
I_N- z \fbetan{\alg{X}}{N}(L\bPsi)
\bigr)^{-1}
\end{aligned}
\end{equation}

Thus in order to conclude the computation
of the conditional expectation   (<ref>)
it remains to evaluate
$\fbetan{\alg{X}}{N}(\bPsi X_i)$ for all $i$.
This can be done if we assume that all variables are free
with the help of   (<ref>) which we now lift to the matrix level.

Let $M\in M_N(\IC\langle X_1,X_2,\dots,X_n\rangle)$, then

\begin{equation}
\label{eq:fbetaX1NM}
\fbetan{X_1}{N}(M)=\epsilon^{(N)}(M)+\sum_{k=1}^\infty\beta_k(X_1)\left[\epsilon\otimes\left(\bbeta{\alg{B}}\right)^{\otimes
(k-1)}\otimes \epsilon\right]^{(N)}\left(
\left.
\partial_{X_1}^k\!\!\!
\right.^{(N)}(M)\right).
\end{equation}

We can now subsume the essence of the previous calculations as follows.
Theorem <ref> follows by evaluating formula (<ref>)
in a specific algebra $\alg{A}$.

Let $\mu:\IC\langle X_1,X_2,\dots,X_n\rangle\to\IC$ be a distribution such
that  $X_1,X_2,\dots,X_n$ are free.
In the following for an index set $I\subseteq \{1,2,\dots,n\}$
the conditional expectation onto  the
subalgebra generated by the subset $\{X_i\}_{ i\in I}$
will be denoted   by $\E_I$.
Suppose that the resolvent of a given polynomial
$P=P(X_1,X_2,\dots,X_n)\in \IC\langle X_1,X_2,\dots,X_n\rangle$
of degree $m$ has linearization

\begin{equation}
\label{eq:psi=utLv}
\Psi=(1-z^mP)^{-1}= u^t\left(I_N-z(C_1X_1+C_2X_2+\dots+ C_n X_n)\right)^{-1}v
\end{equation}

with $C_1,C_2,\ldots,C_n\in M_N(\IC)$.

Then the conditional expectations of $\Psi$ are

\begin{equation}
\label{eq:EIps=utsumCiXiCjFj}
\begin{aligned}
\E_I
[(1-z^m P)^{-1}]
&=u^t \Bigl(I_N-zH_{J}
\Bigl(\sum_{i\in I}
&=u^t \Bigl(I_N-z\Bigl(\sum_{i\in I} C_iX_i - \sum_{j\in J} C_jF_j\Bigr)\Bigr)^{-1}v
\end{aligned}
\end{equation}

where $J=[n]\setminus I$ is the complement and $H_J=
\bigl(
I_N-z\sum_{j\in J} C_j F_j
\bigr)^{-1}$
and the matrices $F_i=\fbetan{X_i}{N}(X_i\bPsi)\in M_N(\IC)$, $i=1,2,\dots,n$
are  the unique solution  of the system of matrix equations

\begin{equation}
\label{eq:Fi=etaXi}
\begin{aligned}
\Bigl(
I_N-z\sum_{j\neq i} C_jF_j
\bigr)^{-1} C_i
\Bigr)
\end{aligned}
\end{equation}

which is analytic at $z=0$.
In particular,

\begin{equation*}
\varphi((1-z^m P)^{-1})
=u^t \Bigl(I_N-z\Bigl(\sum_{j=1}^n C_jF_j\Bigr)\Bigr)^{-1}v
\end{equation*}

\bPsi =\left(I_N-z(C_1X_1+C_2X_2+\dots+ C_n X_n)\right)^{-1}

be the matrix of the linearization  $\Psi=u^t\bPsi v$ from (<ref>).
In the following we abbreviate the functionals

\begin{align*}
\bbeta{I} &= \sum_{i\in I}\bbeta{X_i} &
\fbeta{I} &= \sum_{i\in I}\fbeta{X_i}
\end{align*}

Then from     (<ref>) with
$\alg{B}$ the algebra generated by $\{X_i\}_{i\in I}$
and $\alg{A}$ the algebra   generated by $\{X_j\}_{j\in J}$ we obtain

\E_I[\Psi]
= u^t
\bigl(
I_N-zH_J\sum_{i\in I}C_iX_i
\bigr)^{-1}H_Jv

\begin{equation*}
H_J =\bbetan{J}{N}(\bPsi)
= \bigl(
I_N-z\sum_{j\in J} C_j\fbetan{J}{N}(X_j\bPsi)
\bigr)^{-1}
\end{equation*}

by (<ref>).

Next we apply Proposition <ref>
in order to establish equations for $F_i = \fbetan{X_i}{N}(X_i\bPsi)$,
similar to Example <ref>.
First observe that

\partial_{X_i}\bPsi = z (\bPsi\odot 1)C_i(1\odot\bPsi) = z\bPsi C_i\odot \bPsi

\partial_{X_i}X_i\bPsi = 1\odot\bPsi + z X_i\bPsi C_i\odot \bPsi

and applying iterated derivatives to the left leg

\partial_{X_i}^kX_i\bPsi = z^{k-1}1\odot(\bPsi C_i)^{\odot k-1}\odot\bPsi
+ z^k  X_i(\bPsi C_i)^{\odot k} \odot\bPsi

and (<ref>) becomes

\begin{align*}
\fbetan{X_i}{N}(X_i\bPsi)
\sum_{k=1}^\infty
\beta_k(X_i)
\bbetan{[n]\setminus i}{N}(\bPsi C_i)^{k-1}
\\
&= \tilde{\eta}_{X_i}(zH_{[n]\setminus i}C_i)
\end{align*}

Uniqueness of the  matrices $F_i$ follows from the iteration in
Lemma <ref> below.

We have seen that the matrices
F_i(z) = \fbetan{X_i}{N}(X_i\bPsi(z))
satisfy the fixed point equation     (<ref>)
and it remains to show that the latter has a unique  solution analytic at 0.
To this end we expand $F_i(z)$ into a power series

\begin{equation}
\label{eq:Fiz=series}
F_i(z) = \fbetan{X_i}{N}(X_i\bPsi(z))
= \sum_{k=0}^\infty
\fbetan{X_i}{N}(X_i L^k)z^{km}
\end{equation}

and show that iterating the fixed point equation produces a series
whose coefficients converge to those of the series   (<ref>).

Start with constant matrices

\begin{align*}
F_i^{(0)}(z) &= \beta_1(X_i)I_N
& i = 1,2,\dots,n
\end{align*}

and iterate

\begin{align*}
F_i^{(r+1)}(z) &= \tilde{\eta}_i
\Bigl(
\bigl(
I_N-\sum_{j\ne i}C_jF_j^{(r)}(z)
\bigr)^{-1}C_i
\Bigr)
& i = 1,2,\dots,n
\end{align*}

Then for all $r\in\IN_0$ we have

\begin{align*}
F_i^{(r)}(z) - F_i(z) &= \Ord(z^{r+1})
& i = 1,2,\dots,n
\end{align*}

We proceed by induction.
First observe that if $F_i^{(r)}(z) - F_i(z) = \Ord(z^{r+1})$ for $i=1,2,\dots,n$

\bigl(
I_N-\sum_{j\ne i}C_jF_j^{(r)}(z)
\bigr)^{-1}C_i
\bigl(
I_N-\sum_{j\ne i}C_jF_j(z)
\bigr)^{-1}C_j

as well
and thus

\begin{align*}
F_i^{(r+1)}(z)-  F_i(z)
\begin{multlined}[t]
\sum_{k=0}^{r+1}
\beta_{k+1}(X_i)z^k
\Bigl(
\bigl(
I_N-\sum_{j\ne i}C_jF_j^{(r)}(z)
\bigr)^{-1}C_j
\Bigr)^k
\Ord(z^{r+2})\\
\sum_{k=0}^{r+1}
\beta_{k+1}(X_i)z^k
\Bigl(
\bigl(
I_N-\sum_{j\ne i}C_jF_j(z)
\bigr)^{-1}C_X
\Bigr)^k
+ \Ord(z^{r+2})
\end{multlined}
\\
&= \Ord(z^{r+2})
\end{align*}

When it comes to the practical solution of the system (<ref>)
we are faced with a large number of variables.
It can be reduced by observing that the solutions $F_i$ enter
the final expression (<ref>) for the conditional
expectation only in the products $C_iF_i$ and we can
advantage of the fact that the  coefficient matrices $C_i$
usually are sparse.
Indeed let $P_i$ be the projection onto $\ker C_i$ and
$Q_i=I-P_i$, i.e., $C_iP_i=0$ and $C_iQ_i=C_i$.
Then in the expression (<ref>) we
can replace the matrices  $F_i$ with the matrices
$\tilde{F}_i = Q_iF_i$ and the latter satisfy the
slightly modified system

\begin{equation}
\label{eq:Fit=QietaXi}
\begin{aligned}
\tilde{F}_i
\Bigl(
I_N-z\sum_{j\neq i} C_jF_j
\bigr)^{-1} C_i
\Bigr)
\end{aligned}
\end{equation}

It follows that $\tilde{F}_i=Q_i\tilde{F}_iQ_i$
can be obtained by iterating the fixed point equation
(<ref>) with starting point $\tilde{F}_i^{(0)} = \beta_1(X_i)Q_i$.

§ EXAMPLES

In specific calculations it is always clear which $N$ we need to take in $\E^{(N)}$, thus we will write $\E$ for $\E^{(N)}$ and similarly we will write $\bbeta{},\fbeta{}$ etc. for matricial versions all other mappings.

§.§ The product of free random variables

We start with the example $T=XY$ which can be solved
directly without invoking linearizations.

The derivations are $\rdelta_X\Psi=z\Psi\otimes XY\Psi$ and
$\rdelta_Y\Psi=z\Psi X\otimes Y\Psi$, respectively.
First notice that from the resolvent identities

\begin{equation}
\label{eq:XY:resolventidentityPsi}
\Psi = 1+zXY\Psi  = 1+z\Psi XY
\end{equation}

we immediately infer

\begin{equation}
\label{eq:XY:bbetaPsi=1}
\bbeta{X}(\Psi) =   \bbeta{Y}(\Psi) = 1
\end{equation}

and thus

\begin{equation}
\label{eq:EX(Psi)-XY}
\IE_X[\Psi] = \bbeta{Y}(\Psi) + z \bbeta{Y}\otimes \IE_X[\Psi\otimes XY\Psi] =
1+ zX \IE_X[Y\Psi]
\end{equation}

\IE_X[Y\Psi] = \bbeta{Y}(Y\Psi)(1+zX\IE_X[Y\Psi])

and thus

\begin{equation}
\label{eq:XY:EX(YPsi)}
\IE_X[Y\Psi] = \bbeta{Y}(Y\Psi)
\bigl(
\bigr)^{-1}
\end{equation}

Putting together  (<ref>) and
(<ref>) we find

\IE_X[\Psi] =
\bigl(
\bigr)^{-1}

and the subordination equation

\begin{equation*}
M(z) = M_X(z\bbeta{Y}(Y\Psi))
\end{equation*}

We have thus identified the first subordination function

\omega_1(z) = z\bbeta{Y}(Y\Psi)

The conditional expectation with respect to $Y$ is simpler to obtain,
but the result is the same:

\IE_Y[\Psi] = 1 + z\bbeta{X}(\Psi X)Y\IE_Y[\Psi]

and thus

\begin{align*}
\IE_Y[\Psi] &=
\bigl(
1 - z\bbeta{X}(\Psi X)Y
\bigr)^{-1}\\
M(z) &= M_Y(z\bbeta{X}(\Psi X))
\end{align*}

The second subordination function is

\omega_2(z) = z\bbeta{X}(\Psi X)

To find relations between these functions we will resort
to Theorem <ref>.
It is immediate to see combinatorially that
$\bbeta{X}(\Psi X)=\fbeta{X}(\Psi X)$ but let us verify this
with the recurrence from Proposition <ref>.

\begin{align*}
\bbeta{Y}(Y\Psi)
&= \fbeta{Y}\otimes\bbeta{Y}(Y\otimes \Psi + zY\Psi XY\otimes\Psi) \\
&= \fbeta{Y}(Y(1+z\Psi XY))\bbeta{Y}(\Psi)= \fbeta{Y}(Y\Psi)
\end{align*}

because of the identities   (<ref>) and

\begin{align*}
\bbeta{X}(\Psi X)
&= \fbeta{X}\otimes\bbeta{X}( z\Psi X \otimes Y\Psi X+ \Psi X\otimes 1) = \fbeta{X}(\Psi X).
\end{align*}

Thus we have

\omega_1(z) = z\fbeta{Y}(Y\Psi)
\qquad
\omega_2(z) = z\fbeta{X}(\Psi X)

§.§ Conditional expectations of free commutators and anti–commutators

As an application of Theorem <ref> we show that for symmetric and free
$X,Y$ the conditional expectation of $(1-z^2 P(X,Y))^{-1}$ onto one of the
variables coincide for commutator and anti–commutator.

First we consider the anti–commutator $XY+YX$ in free variables.

The linearization matrix described in Remark <ref> is given by

\[
L =
\begin{bmatrix}
0 & 0 & Y & 0 \\
X & 0 & 0   & Y \\
X & 0 & 0   & Y \\
0   & X & 0 & 0
\end{bmatrix}
= C_X X+C_Y Y
\]

\begin{align*}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0   & 0 \\
1 & 0 & 0   & 0 \\
0   & 1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}

We have

\[
\begin{bmatrix}
0 & 1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\ 1 \\ 1 \\ 0
\end{bmatrix}
\]

It follows  that

\begin{align*}
\E_X[(1-z^2 (XY+YX))^{-1}]
\begin{bmatrix}
0 & 1 & 0 & 0
\end{bmatrix}
\E_X\left[(1-zL)^{-1}\right]
\begin{bmatrix}
0 \\ 1 \\ 1 \\ 0
\end{bmatrix}
\end{align*}

\begin{align*}
\E_X\left[(1-z L)^{-1}\right]
&=(1-zH_X C_X X)^{-1} H_X\\
\E_Y\left[(1-z L)^{-1}\right]
&=(1-zH_Y C_Y Y)^{-1} H_Y
\end{align*}

where the
the matrices $H_X$ and $H_Y$ they satisfy the following system of equations
from Theorem <ref> and Remark <ref>:

\begin{align}\label{eq:main_system}
\begin{cases}
&H_X=(I-zC_Y F_Y)^{-1}\\
&H_Y=(I-zC_X F_X)^{-1}\\
&F_X=Q_1  \widetilde{\eta}_X(z H_X C_X)\\
&F_Y=Q_2  \widetilde{\eta}_Y(z H_Y C_Y)
\end{cases}
\end{align}

\begin{align*}
F_X=\beta^{\delta}_Y \left(\mathrm{diag}(X,X,Y,Y) (1-zL)^{-1}\right),\qquad &F_Y=\beta^{\delta}_X \left(\mathrm{diag}(X,X,Y,Y) (1-zL)^{-1}\right),\\
Q_1=\mathrm{diag}(1,1,0,0),\qquad &Q_2=\mathrm{diag}(0,0,1,1).
\end{align*}

Since $X$ and $Y$ are symmetric, all odd  cumulants vanish and hence

\begin{align*}
\begin{bmatrix}
0 & f_{x,12} & 0 & 0 \\
f_{x,21} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0   & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & f_{y,12} \\
0   & 0 & f_{y,21} & 0
\end{bmatrix}
\end{align*}

Substituting $F_X$ and $F_Y$ into the first half of the system
we obtain

\begin{align}
\begin{bmatrix}\label{eq:HX_ac}
1 & 0 & 0 & f_{y,12} \\
0 & 1 & \frac{f_{y,21}}{1-f_{y,21}} & 0 \\
0 & 0 & \frac{1}{1-f_{y,21}} & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{1-f_{x,12}} & 0 & 0 \\
0 & \frac{f_{x,12}}{1-f_{x,12}} & 1 & 0 \\
f_{x,21} & 0 & 0 & 1 \\
\end{bmatrix}
\end{align}

By assumption  $X$ and $Y$ have symmetric distribution and thus

\begin{align*}
\widetilde{\eta}_X(z)
\widetilde{\eta}_Y(z)&=\sum_{n=1}^{\infty}\beta_{2n}(Y)z^{2n-1}.
\end{align*}

We need to understand how odd powers of $H_XC_X$ and and $H_YC_Y$ behave.

First let us note

\begin{align*}
H_X C_X&=
\begin{bmatrix}
0 & f_{y,12} & 0 & 0 \\
\frac{1}{1-f_{y,21}} & 0 & 0 & 0 \\
\frac{1}{1-f_{y,21}} & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\end{bmatrix}
\left(H_X C_X\right)^2
\begin{bmatrix}
\frac{f_{y,12}}{1-f_{y,21}} & 0 & 0 & 0 \\
0 & \frac{f_{y,12}}{1-f_{y,21}} & 0 & 0 \\
0 & \frac{f_{y,12}}{1-f_{y,21}} & 0 & 0 \\
\frac{1}{1-f_{y,21}} & 0 & 0 & 0 \\
\end{bmatrix}
\end{align*}

This immediately implies that

\begin{align*}
&\left(H_X C_X\right)^{2n-1}=\left(\frac{f_{y,12}}{1-f_{y,21}}\right)^{n-1}H_X C_X
\end{align*}

we obtain

\begin{align*}
&\left(H_Y C_Y\right)^{2n-1}=\left(\frac{f_{x,21}}{1-f_{x,12}}\right)^{n-1}H_Y C_Y.
\end{align*}

This gives

\begin{align*}
\widetilde{\eta}_X(z H_X C_X)&=\sum_{n=1}^{\infty}\beta_{2n}(X)\left(z H_X C_X\right)^{2n-1}=\frac{1-f_{y,21}}{f_{y,12}} H_XC_X \sum_{n=1}^{\infty}\beta_{2n}(X) z^{2n} \left(\frac{f_{y,12}}{1-f_{y,21}}\right)^{n}\\
&=\eta_X\left(z\sqrt{\frac{f_{y,12}}{1-f_{y,21}}}\right) \frac{1-f_{y,21}}{f_{y,12}} H_XC_X,\\
\widetilde{\eta}_Y(z H_Y C_Y)&=\eta_Y\left(z\sqrt{\frac{f_{x,21}}{1-f_{x,12}}}\right) \frac{1-f_{x,12}}{f_{x,21}} H_Y C_Y.
\end{align*}

Substituting this into
the second half of the system
results in two matrix equations
with exactly two non-zero entries which we can rewrite as

\begin{align}\label{eq:final_ac}
\left\{
\begin{aligned}
&		f_{y,12}&=\frac{1}{f_{x,21}}\,\eta_Y\left(z\sqrt{\frac{f_{x,21}}{1-f_{x,12}}}\right)\\
\end{aligned}
\right.
\end{align}

\begin{align*}
\E_X[(1-z^2 (XY+YX))^{-1}]
\begin{bmatrix}
0 & 1 & 0 & 0
\end{bmatrix}
(I-zH_X C_X X)^{-1} H_X
\begin{bmatrix}
0 \\ 1 \\ 1 \\ 0
\end{bmatrix}
\\
\E_Y[(1-z^2 (XY+YX))^{-1}]
\begin{bmatrix}
0 & 1 & 0 & 0
\end{bmatrix}
(I-zH_Y C_Y Y)^{-1} H_Y
\begin{bmatrix}
0 \\ 1 \\ 1 \\ 0
\end{bmatrix}
\end{align*}

A direct calculation with $H_X$ and $H_Y$ as in (<ref>) gives

\begin{equation}
\label{eq:cond_exp_ac}
\begin{aligned}
\E_X[(1-z^2 (XY+YX))^{-1}]
\E_Y[(1-z^2 (XY+YX))^{-1}]
\end{aligned}
\end{equation}

Next we repeat the previous steps with the commutator,
which are very similar.
The linearization matrix has the form

\[
L =
\begin{bmatrix}
0 & 0 & Y & 0 \\
iX & 0 & 0   & -iY \\
iX & 0 & 0   & -iY \\
0   & X & 0 & 0
\end{bmatrix}
= C_X X+C_Y Y.
\]

Again the system of equations reduces to just four equations which after
immediate simplifications boil down to the following:

\begin{align}\label{eq:comutator_final}
\left\{
\begin{aligned}
&		f_{y,12}&=\frac{1}{f_{x,21}}\,\eta_Y\left(z\sqrt{\frac{f_{x,21}}{i+f_{x,12}}}\right)\\
&		f_{y,21}&=(i+f_{x,12})\,\eta_Y\left(z\sqrt{\frac{f_{x,21}}{i+f_{x,12}}}\right)
\end{aligned}
\right.
\end{align}

And the conditional expectation in this case is equal to

\begin{equation}
\label{eq:cond_exp_c}
\begin{aligned}
\E_X[(1-z^2 i(XY-YX))^{-1}]
\E_Y[(1-z^2 i(XY-YX))^{-1}]
\end{aligned}
\end{equation}

We are now ready to prove the following proposition.

Assume that $X,Y$ are free and symmetric, then

\begin{align*}
\E_X[(1-z^2 i(XY-YX))^{-1}]
&=\E_X[(1-z^2 (XY+YX))^{-1}]
\\
\E_Y[(1-z^2 i(XY-YX))^{-1}]
&=\E_Y[(1-z^2 (XY+YX))^{-1}]
\end{align*}

Let us rewrite (<ref>) and (<ref>) as

\begin{align*}
\E_X[(1-z^2 (XY+YX))^{-1}]&=\frac{1}{1-f^{ac}_{y,21}-f_{y,12}^{ac}X^2z^2}\\
\E_X[(1-z^2 i(XY-YX))^{-1}]&=\frac{1}{1+if_{y,21}^{c}-if_{y,12}^{c}X^2z^2}\\
\E_Y[(1-z^2 (XY+YX))^{-1}]&=\frac{1}{1-f^{ac}_{x,12}-f^{ac}_{x,21}Y^2z^2}\\
\E_Y[(1-z^2 i(XY-YX))^{-1}]&=\frac{1}{1-if^{c}_{x,12}+if^c_{x,21}Y^2z^2}.
\end{align*}

Thus if

\begin{equation}\label{eq:substitution}
\left\{
\begin{aligned}
\end{aligned}
\right.
\end{equation}

then the desired equality holds.

It is straightforward to verify that
the  substitution of equations (<ref>)
into the system of equations $\eqref{eq:final_ac}$  for the anti–commutator
results in the system (<ref>) for commutator.
Let us verify this for the first of four equations

\begin{align*}
\end{align*}

Substitution results in

\begin{align*}
\end{align*}

which is equivalent to

\begin{align*}
\end{align*}

and the above is exactly the first equation in (<ref>).

§.§ A Lie polynomial

We consider next the conditional expectation of the Lie polynomial
First we show the general form of the subordination, and next we show that for a special choice of the distributions of $X,Y$ we can make a further step and obtain explicit formulas for the functions present in the conditional expectation. The explicit example is as follows, we take $X$ having semicircular distribution with mean zero and variance one, while $Y$ has distribution $\tfrac{1}{2}\left(\delta_{-1}+\delta_{1}\right)$. In this explicit case we can use the conditional expectation in order to find the distribution of $X+i(XY-YX)$, we obtain the following corollary.

For $a,b$ free, $a$ being semicircle element of variance $1$ and $b$ being symmetric Bernoulli element, the element $a+i(ab-ba)$ has semicircle distribution with variance 3. Moreover elements $a$ and $i(ab+ba)$ are not free, while $i(ab+ba)$ has semicircle distribution with variance 2.

Note that $ab$ and $ba$ are free <cit.>,
but $a$ and $c=i(ab-ba)$ are not: indeed a quick calculation shows that $\kappa_4(a,a,c,c)=1$.
Yet because of some nontrivial cancellations $\kappa_n(a+c)=0$ for all $n\neq 2$ and thus $a+c$ has the same distribution as the sum of free copies of $a$ and $c$.
This is another example of a pair of semicircular elements which are not free
independent, yet any real linear combination has semicircular distribution

Let us describe the linearization for this example we have $u^t=(1,0,0)$, $v^t=(1,0,0)$ and $L=C_X X+C_Y Y$ where

\begin{align*}
\begin{bmatrix}
z & 0 & i  \\
1 & 0 & 0    \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & -i & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix}
\end{align*}

The complementary kernel projections are

\begin{align*}
\begin{bmatrix}
1 & 0 & 0  \\
0 & 0 & 0    \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Now the compressed matrices $\widetilde{F}_X=Q_XF_XQ_X$ and $\widetilde{F}_Y=Q_YF_YQ_Y$ have the form

\begin{align*}
\begin{bmatrix}
fx_{11} & 0 & fx_{13} \\
0 & 0 & 0 \\
fx_{31} & 0 & fx_{33}
\end{bmatrix}
\begin{bmatrix}
fy_{11} & fy_{12} & 0 \\
fy_{21} & fy_{22} & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Hence the conditional expectations are of the form

\begin{align*}
\E_X\left[(1-z^2P)^{-1}\right]
&=-\frac{i}{i - fy_{21} z   -  z^2 (i - fy_{11} + fy_{22} )X+ z^3 fy_{12} X^2},\\
\E_Y\left[(1-z^2P)^{-1}\right]&=\frac{i}{i-i fx_{11} z^2+fx_{31} z-z^2\left(fx_{11}-fx_{33} +i fx_{13} z \right)Y -z^3fx_{13} Y^2}.
\end{align*}

If we evaluate the above for $X=a$ where $a$ has a semicircle distribution and $Y=b$ where b has Bernoulli distribution then we have that $z\widetilde{\eta}_X(z)^2-\widetilde{\eta}_X(z)+z=0$ and $\widetilde{\eta}_Y(z)=z$. Thus we solve the following system of equations

\begin{equation*}
\left\{
\begin{aligned}
\widetilde{F}_Y-z Q_Y (I-z C_X \widetilde{F}_X )^{-1} C_Y&=0,\\
Q_X(I-z C_Y \widetilde{F}_Y )^{-1}C_X\widetilde{F}_X^2+z Q_X(I-z C_Y \widetilde{F}_Y )^{-1}C_X-\widetilde{F}_X&=0.
\end{aligned}
\right.
\end{equation*}

Since all formal power series in $z$ considered here are convergent in some neighbourhood of zero, we obtain explicit solutions

\begin{align*}
\E_b\left[(1-z^2P(a,b))^{-1}\right]
\E_a\left[(1-z^2P(a,b))^{-1}\right]
&=\frac{3 \left(4 z^4+1\right)}{z^4 \left(3 a^2 \left(\alpha-2\right)+\alpha+14\right)-24 a z^6-3 a \alpha z^2+\alpha+2}.
\end{align*}

where $\alpha=\sqrt{1-12 z^4}$.

Since the conditional expectation on $b$ is constant we immediately get the distribution of $a+i(ab-ba)$ as we have

\[\varphi\left((1-zP(a,b))^{-1}\right)=\frac{2}{\sqrt{1-12 z^2}+1}\]

and we can get the moments of $a+i(ab-ba)$ which is A005159 entry at OEIS, we see that $\varphi\left((a+i(ab-ba)^{2n})\right)=3^n C_n$, where $C_n$ is the $n$-th Catalan number and Corollary <ref> follows. We show a histogram of eigenvalues of a matrix approximation together with the graph of the density see Figure <ref>.

Density of $a+i(ab-ba)$ in the case $a$ semicircle and $b$ Bernoulli together with matrix approximation.

§.§ An example with three variables

Next we consider the example of product $XZYZX$. The linearization matrix for $(1-z^5XZYZX)$ is given by

\begin{align*}
\begin{bmatrix}
0 & X & 0 & 0 & 0 \\
0 & 0 & Z & 0 & 0 \\
0 & 0 & 0 & Y & 0 \\
0 & 0 & 0 & 0 & Z \\
X & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}

In order to state an explicit result let us assume that $a,b,c$ are free $a,b$
have distribution $1/2\left(\delta_{-1}+\delta_{1}\right)$ and $c$ has
distribution $1/2\left(\delta_0+\delta_2\right)$. The distribution of $acbca$
is the same as free multiplicative convolution of $c$ and $b$, however the
conditional expectation on $a,b$ is a non-trivial problem.
We omit the lengthy calculation wich results in the expression

\begin{multline*}
\E_{a,b}\left[(1-z^5acbca)^{-1}\right]\\
=1-a\left(\left(2 z^5 \left(8 z^{10}+\gamma\right)-\left(b \left(\gamma-4\right) z^5+4 z^{10}+2\right)^{-1}\left(-8 z^{10}+\gamma\right)\right)^{-1}\right.\\
\left.\left(b \left(\gamma-4\right) z^5+4 z^{10}+2\right)\left(\left(-8 z^{10}+\gamma\right)a+4 a z^{10}\right)+a\right)
\end{multline*}

where $\gamma= \sqrt{1-16 z^{10}}+1$
for the conditional expectation.

§.§ A rational example

Non–commutative rational functions have linearizations just like polynomials,
(see, e.g.,  [16] for excellent
discussion about relations between linearizations and questions in free
probability) and consequently our method is not restricted to
polynomials but also allows to compute conditional expectations and
distributions of non–commutative rational functions in free random variables.
Let us illustrate this with the concrete example of a rational function
$r(a,b)=a(I-a-b)^{-1}a$ which will show that nevertheless
certain technical issues arise.
First we must put conditions on the distributions of $a$ and $b$ in order for
$I-a-b$ to be invertible.
Next we compute the conditional expectation of $(1-z r(a,b))^{-1}$
onto $a$ and $b$ and then integrate the latter in order to determine the
distribution of  $a(I-a-b)^{-1}a$.
Using the method described in Section <ref> we obtain the

\left(1-zX(1-X-Y)^{-1}X\right)^{-1}
= u^t(I_2-L)^{-1}v

where  $L=C_X X+C_Y Y$ with

\begin{align*}
\begin{bmatrix}
0 & z \\
1 & 1
\end{bmatrix}
\quad
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}
\end{align*}

and $u=v=(1,0)^t$.