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Department of Mathematics
On Free Moments and Free Cumulants
Jolanta Pielaszkiewicz, Dietrich von Rosen and Martin Singull
Department of Mathematics Link¨oping University S-581 83 Link¨oping, Sweden.
On Free Moments and Free Cumulants
Jolanta Pielaszkiewicz∗ Dietrich von Rosen,∗
Martin Singull∗ LiTH-MAT-R–2014/05–SE
∗ Department of Mathematics, Link¨oping University, SE–581 83 Link¨oping,
Sweden.
Corresponding author: Jolanta Pielaszkiewicz. Tel.: +46 13 28 1433. E-mail address: jolanta.pielaszkiewicz@liu.se
Department of Energy and Technology, Swedish University of Agricultural
On Free Moments and Free Cumulants
Jolanta Pielaszkiewicz∗, Dietrich von Rosen,∗ and Martin Singull∗
∗Department of Mathematics,
Link¨oping University, SE–581 83 Link¨oping, Sweden. E-mail: jolanta.pielaszkiewicz@liu.se
E-mail: martin.singull@liu.se
Department of Energy and Technology,
Swedish University of Agricultural Sciences, SE–750 07 Uppsala, Sweden.
E-mail: Dietrich.von.Rosen@slu.se
Abstract
The concepts of free cumulants and free moments are indispensably related to the idea of freeness introduced by Voiculescu [Voiculescu, D., Proc. Conf., Bu¸steni/Rom., Lect. Notes Math. 1132(1985), pp. 556– 588] and studied further within Free probability theory. Free probabil-ity theory is of great importance for both the developing mathematical theories as well as for problem solving methods in engineering.
The goal of this paper is to present theoretical framework for free cumulants and moments, and then prove a new free cumulant–moment relation formula. The existing relations between these objects will be given. We consider as drawback that they require the combinatorial understanding of the idea of non–crossing partitions, which has been considered by Speicher [Speicher, R., Math. Ann., 298(1994), pp. 611– 628] and then widely studied and developed by Speicher and Nica [Nica, A. and Speicher, R.: Lectures on the Combinatorics of Free Probability, Cambridge University Press, Cambridge, United Kingdom, 2006]. Furthermore, some formulations are given with additional use of the M¨obius function. The recursive result derived in this paper does not required introducing any of those concepts, instead the calculations of the Stieltjes transform of the underlying measure are essential.
The presented free cumulant–moment relation formula is used to calculate cumulants of degree 1 to 5 as an function of the moments of lower degrees. The simplicity of the calculations can be observed by a comparison with the calculations performed in the classical way
using non–crossing partitions. Then, the particular example of non– commutative space i.e., space of p×p matrices X = (Xij)i,j, where Xij
has finite moments, equipped with functional 1pETrX is investigated. Keywords: R–transform; free cumulants; moments; freeness; asymp-totic freeness; free probability; non–commutative probability space; Stieltjes transform; random matrices.
1
Introduction & Background
Free moments and free cumulants are functionals defined within Free prob-ability theory. The theory was established in the middle of the 80’s by Voiculescu (1985) and together with the result published in 1991 regarding asymptotic freeness of random matrices it has established new branches of theories and tools, among others free cumulants and free moments.
It is of great importance to understand the behavior of free cumulants as they give us essentially the full information about a particular probability measure such as for example the measure connected to the spectral distri-bution of random matrices discussed in section 3. In current and 2th section we will consider a general formulations in order to state precisely the results of the article we fix notation and recall the basic definitions and properties. Let us consider a non–commutative ∗–probability space (A, τ ). Following Nica and Speicher (2006) we recall that by non–commutative ∗–probability space we consider pair of a unital algebra A over the field of complex num-bers C with identity element 1A and functional τ such that: τ : A → C is
linear, τ (1A) = 1 and τ (a∗a) ≥ 0 for all a ∈ A. The algebra is equipped
with a ∗-operation such that ∗ : A → A, (a∗)∗ = a and (ab)∗ = b∗a∗ for all a, b ∈ A. Then, the free moments of a normal element a ∈ A are defined as
τ (ak(a∗)n) := Z
C
zkz¯ndµ(z) and in the case of self–adjoint elements a ∈ A as
mk := τ (ak) :=
Z
R
xkdµ(x) (1)
Free moments characterize a compactly supported ∗–distribution of a. The ∗–distribution is denoted by µ and we consider the space with supp(µ) ⊂ R, where supp(µ) stands for the support of the measure µ. The form of the chosen functional τ determines the ∗–distribution of the element a ∈ A.
To introduce the concept of free cumulants as well as to obtain the free cumulant–moment relation formula we use the Stieltjes transform. It ap-pears among others in formulations of a number of results published within Random matrix theory, see for example, Marˇcenko and Pastur (1967), Girko and von Rosen (1994), Silverstain and Bai (1995), Hachem, Loubaton and Najim (2007).
Definition 1.1. Let µ be a probability measure on R. Then, the Stieltjes (Cauchy-Stieltjes) transform of µ is given by
Gµ(z) =
Z
R
1
z − xdµ(x),
for all z ∈ {z : z ∈ C, =(z) > 0}, where =(z) denotes the imaginary part of the complex z.
Defined in such a way the Stieltjes transform is well defined as we prove in Remark 1.1 and can be inverted on any interval, see Remark 1.2. Fur-thermore it can be given as a series of free moments {mi}∞i=1, what is of the
use in proof of the Theorem 2.1.
Remark 1.1. Note that for z ∈ {z : z ∈ C, =(z) > 0} the Stieltjes transform is well defined and G(z) is analytical for all z ∈ {z : z ∈ C, =(z) > 0}. Proof. The fact that the Stieltjes transform is well defined follows from the fact that for the domain under function z−x1 is bounded.
One can show that G(z) is analytical for all z ∈ {z : z ∈ C, =(z) > 0} using Morera’s theorem (see work by Greene and Krantz, 2006). Then, it is enough to show that the contour integral H
ΓG(z)dz = 0 for all closed
contours Γ in z ∈ {z : z ∈ C, =(z) > 0}. We are allowed to interchange integrals and obtain
Z R I Γ 1 z − xdzdµ(x) = Z R 0dµ(x) = 0,
where the first integral vanishes by the Cauchy’s integral theorem for any closed contour Γ as z−x1 is analytic.
Remark 1.2. For any µ being a probability measure on R and any a < b µ((a, b)) +1 2µ({a}) + 1 2µ({b}) = − 1 π y→0lim Z I =G(x + i y)dx. Proof. We have −1 π y→0lim Z I =G(x + i y)dx = −1 πy→0lim Z I Z R = 1 x + i y − tdµ(t)dx (∗) = 1 πy→0lim Z R Z b a y (t − x)2+ y2dxdµ(t) = 1 πy→0lim Z R arctan b − t y − arctan a − t y dµ(t) () = 1 π Z R lim y→0 arctan b − t y − arctan a − t y dµ(t). The order of integration can be interchanged in (∗) due to continuity of the
the limit in () follows by the Bounded convergence theorem as µ(R) ≤ 1 < ∞ and ∃M arctan b − t y − arctan a − t y < M ∀y ∀t so it is an uniformly bounded real-valued measurable function for all y. Then, using that limy→0arctan
T y = π2sgn(T ) for T ∈ R we get arctan b − t y − arctan a − t y y→0 −−−→ 0 if t < a or t > b π 22 = π if t ∈ (a, b) π 2 if t = a or t = b
which by the Dominated convergence theorem completes the proof. Theorem 1.1. Let the free moments mk =
R
Rx
kdµ(x), k = 1, 2, . . .. Then, a
formal power series representing the Stieltjes transform is given by Gµ(z) = 1 z 1 + ∞ X i=1 z−imi . Proof. Gµ(z) = Z R 1 z − xdµ(x) = 1 z Z R 1 1 −xzdµ(x) = 1 z Z R ∞ X i=0 x z i dµ(x) = 1 z ∞ X i=0 z−i Z R xidµ(x) = 1 z 1 + ∞ X i=1 z−imi ,
which completes the proof.
Although the Stieltjes transform Gµ is a convenient tool, even
bet-ter suited for studying convolution of measure µ on non–commutative ∗– probability spaces is the R–transform. The R–transform linearizes free con-volution and plays the same role as the log of the Fourier transform in classi-cal probability theory. The relation between the R– and Stieltjes transform Gµ, or more precisely G−1µ , which is the inverse with respect to composition,
is often considered as a definition of the R–transform.
Definition 1.2. Let µ be a probability measure and Gµ(z) the related Stieltjes
transform. Then, Rµ(z) = G−1µ (z) − 1 z or equivalently Rµ(Gµ(z)) = z − 1 Gµ(z) , defines the R–transform Rµ(z) for the underlying measure µ.
Then, the free cumulants {ki}∞i=1are given as the coefficients of a power
series expansion of the R–transform.
Definition 1.3. Let µ be a probability measure and Rµ(z) be the related R–
transform. Then for a, which is an element of a non–commutative ∗–algebra A, the free cumulants of a, {ki}∞i=1, are defined by
Rµ(z) = ∞
X
i=0
ki+1(a)zi.
To put our result of free cumulant-moment relation formula in respect to the other free formulas we recall that a combinatorial branch of Free proba-bility theory points out that free cumulants defined by the R–transform, as in Definition 1.3, following Mitchener (2005) and Nica and Speicher (2006), can be defined via non–crossing partitions using the following recursive re-lation
k1(a) = τ (a), τ (a1· . . . · ak) =
X
π∈N C(k)
kπ[a1, . . . , ak], (2)
where τ (a1· . . . · ak) describes mixed free moments of a1, . . . , ak, the sum is
taken over all non–crossing partitions N C(k) of the set {1, 2, . . . , k}, ai∈ A
for all i = 1, 2, . . . , k and kπ[a1, . . . , ak] = r Y i=1 kV (i)[a1, . . . , ak] π = {V (1), . . . , V (r)}, kV[a1, . . . , ak] = ks(av(1), . . . , av(s)) V = (v(1), . . . , v(s)).
Then, for a ∈ A the cumulant of a is defined as kn= kn(a, . . . , a).
Another way to look at free cumulants, see Nica and Speicher (2006), is with use of M¨obius function as well as non–crossing partitions
kπ[a1, . . . , ak] =
X
σ∈N C(k),σ≤π
τσ[a1, . . . , ak]µ(σ, π),
where τk(a1, . . . , ak) := τ (a1, . . . , ak), τπ[a1, . . . , ak] := QV ∈πτV[a1, . . . , ak]
and µ is the M¨obius function on N C(k). For more details about above formulations see Nica and Speicher (2006) and Speicher (1994).
The essential in Free probability theory is an idea of freeness which corresponds to the independence in classical probability theory.
Definition 1.4. Let Chc1, . . . , cmi be a free algebra with generators c1, . . . , cm,
i.e. all polynomials in m non-commutative indeterminants. Then, the vari-ables (a1, a2, . . . , am) and (b1, . . . , bn) are said to be free (freely independent)
if and only if for any (Pi, Qi)1≤i≤p ∈ (Cha1, . . . , ami × Chb1, . . . , bni)p such
that
following equation holds τ Y 1≤i≤p Pi(a1, . . . , am)Qi(b1, . . . , bn) = 0.
The freeness of variables defined as above brings the important conclu-sions with respect to cumulants. As it is stated in the Theorem 1.2 the mixed cumulants vanishes, what results in the additive property of them with respect to the free elements, see Theorem 1.3.
Theorem 1.2. Let a1, a2, . . . , an ∈ A then elements a1, a2, . . . , an are freely
independent if and only if all mixed cumulants vanishes, i.e. for n ≥ 2 and any choice of i1, . . . , ik ∈ {1, . . . , n} if there exist j, k such that j 6= k, but
ij = ik then
kn(ai1, . . . , ain) = 0.
Proof. The proof can be found in Nica and Speicher (2006). Theorem 1.3. Let a, b ∈ A be free, then
kna+b= kan+ kbn, for n ≥ 1.
Proof. The proof of the theorem follows from the fact that for free random variables mixed cumulants are equal zero, see Theorem 1.2.
ka+bn := kn(a + b, a + b, . . . , a + b) = kn(a, a, . . . , a) + kn(b, b, . . . , b).
For a better understanding of the idea with free cumulants we would like to mention that the free and classical cumulants differ by the elements associated with crossing partitions. In the classical case we consider all partitions while in the free cumulant case only non–crossing ones are of interest. Then, obviously, the first three cumulants are the same in free and classical sense, since the sets {1}, {1, 2}, {1, 2, 3} have no crossing partitions. However, for the fourth cumulant and cumulants of the higher order the free and classical cumulants differ.
In the next section we will focus on the proof of free cumulant–moment relation and then compare our recursive formula with the result given by equation (2).
2
Main Result
The main purpose of this paper is to present a recursive formula which is not based on non–crossing partitions.
Theorem 2.1. Let {ki}∞i=1 be the free cumulants and {mi}∞i=1 the free
mo-ments for an element of a non–commutative probability space. Then, the following recursive formula holds
k1 = m1, kt = t−1 X i=0 i+1 X h=1 (−1)h+1i + 1 h m, h t − t−1 X h=2 kh m, h − 1 t − h , t = 2, 3, . . . (3) where m, h t = X i1+i2+...+ih=t mi1mi2 · . . . · mih
and the sum is taken over ij ≥ 0.
Proof. Let us consider a non–commutative ∗–probability space (A, τ ), where A is an unitary ∗–algebra equipped with the functional τ (·). Then, the mi = τ (ai) describes the ith free moment of the element a ∈ A as in (1).
By Theorem 1.1 the Stieltjes transform Gµ(z) is given as Gµ(z) = 1z
1 + P∞ i=1z−imi . Suppose G−1µ (z) = 1 z + ∞ X i=0 ki+1zi,
then it will be shown that ki can be determined by a recursive formula
depending on mj, j = 1, 2, . . . , i. In this case Definition 1.2 and 1.3 imply
that the free cumulants have been found. Now, combining formulas for Gµ(z) and G−1µ (z) the following relation will be utilized
z = G−1µ (Gµ(z)) = 1 Gµ(z) + ∞ X i=0 ki+1Gµ(z)i = z 1 +P∞ i=1z−imi + ∞ X i=0 ki+1 1 z 1 + ∞ X j=1 z−jmj i = z ∞ X j=0 − ∞ X i=1 z−imi j + ∞ X i=0 ki+1 zi 1 + ∞ X j=1 z−jmj i = z + z ∞ X j=1 − ∞ X i=1 z−imi j + ∞ X i=0 ki+1 zi ∞ X j=0 z−jmj i .
By simple arithmetic calculations this relation leads to the equations z ∞ X j=0 (−1)j ∞ X i=1 z−imi j+1 = ∞ X i=0 ki+1 zi ∞ X j=0 z−jmj i ,
z ∞ X j=0 (−1)j ∞ X i=0 z−imi− 1 j+1 = ∞ X i=0 ki+1 zi ∞ X j=0 z−jmj i , z ∞ X j=0 j+1 X l=0 j + 1 l (−1)l+1 ∞ X i=0 z−imi l = ∞ X i=0 ki+1 zi ∞ X j=0 z−jmj i .
The next step will be to apply a formula for the powers of a power series (see, Janjic (2010)) ∞ X i=0 mizi k = ∞ X n=0 m, k n zn, where m,kn = Pi1+i2+...+i
k=nmi1mi2 · . . . · mik and the sum is taken over
it≥ 0. Therefore, z ∞ X j=0 − 1 + j+1 X l=1 j + 1 l (−1)l+1 ∞ X i=0 z−imi l = k1+ ∞ X i=1 ki+1 zi ∞ X j=0 z−jmj i , z ∞ X j=0 − 1 + j+1 X l=1 j + 1 l (−1)l+1 ∞ X t=0 m, l t z−t = k1+ ∞ X i=1 ki+1 zi ∞ X t=0 m, i t z−t, ∞ X j=0 − 1 + j+1 X l=1 j + 1 l (−1)l+1 ∞ X t=0 m, l t z−t = k1 z + ∞ X i=1 ki+1 ∞ X t=0 m, i t z−(t+i+1).
By the identification of coefficients of z−t the cumulants are obtained. Let us denote left hand side and right hand side of the equation by corresponding LHS and RHS. Let t = 0, then
LHS = ∞ X j=0 − 1 + j+1 X l=1 j + 1 l (−1)l+1m, l 0 = ∞ X j=1 j X l=0 j l (−1)l+1 = 0 = RHS.
For t = 1 we get k1 = m1 since
RHS = k1, LHS = ∞ X j=0 j+1 X l=1 j + 1 l (−1)l+1m, l 1 = ∞ X j=1 j X l=1 j l (−1)l+1lm1= m1, k1 = m1.
For t ≥ 2 LHS = ∞ X j=0 j+1 X l=1 j + 1 l (−1)l+1m, l t , RHS = t−1 X i=1 ki+1 m, i t − i − 1 = kt m, t − 1 0 + t−2 X i=1 ki+1 m, i t − i − 1 = kt+ t−2 X i=1 ki+1 m, i t − i − 1 , kt = ∞ X j=0 j+1 X l=1 j + 1 l (−1)l+1m, l t − t−2 X i=1 ki+1 m, i t − i − 1 . Let us now show that
∞ X j=t j+1 X l=1 j + 1 l (−1)l+1m, l t = 0. (4)
Using the fact that m,lt is a polynomial of maximally t-th order of l it is enough to show that for all W = 1, 2, . . . , t
∞ X j=t j+1 X l=1 j + 1 l (−1)l+1lW = 0.
We prove the above equation by showing that each element of the sum is zero, i.e., that for any fixed L such that L ≥ t and for all W = 1, 2, . . . , t we have L+1 X l=1 L + 1 l (−1)l+1lW = 0. Furthermore, the sum can be expressed as
L+1 X l=1 L + 1 l (−1)l+1lW = L X h=0 L + 1 h + 1 (−1)h+2(h + 1)W = L X h=0 L + 1 h + 1 L h (−1)h(h + 1)W = (L + 1) L X h=0 L h (−1)h(h + 1)W −1. We will prove using mathematical induction with respect to L that for all L and all W such that L ≥ t ≥ W , L, W ∈ N \ {0},
L X h=0 L h (−1)h(h + 1)W −1= 0.
Let L = 1, thenP1 h=0 1 h(−1) h(h + 1)W −1 = 1 − 2W −1 = 0 as W ≤ L = 1 and W ∈ N \ {0}. Let L = 2, thenP2 h=0 2 h(−1)h(h + 1)W −1= 1 − 2W + 3W −1 W ={1,2}= 0.
Let assume that the equation holds for L. Then
L+1 X h=0 L + 1 h (−1)h(h + 1)W −1 = 1 + L X h=1 L + 1 h (−1)h(h + 1)W −1+ (−1)L+1(L + 2)W −1 = 1 + L X h=1 L h + L h − 1 (−1)h(h + 1)W −1+ (−1)L+1(L + 2)W −1 = L X h=0 L h (−1)h(h + 1)W −1 | {z } =0 + L−1 X h=0 L h (−1)h+1(h + 2)W −1+ (−1)L+1(L + 2)W −1 = L X h=0 L h (−1)h+1(h + 2)W −1 = 0
and the (4) is proved. Then finally k1 = m1, kt = t−1 X i=0 i+1 X h=1 (−1)h+1i + 1 h m, h t − t−1 X h=2 kh m, h − 1 t − h , t = 2, 3, . . . ,
which completes the proof of the theorem.
The first five free cumulants ki, i = 1, . . . , 5, given as a function of mj,
j = 1, . . . , i are stated in Corollary 2.1.
Corollary 2.1. Let (A, τ ) be a non–commutative ∗–probability space and mi = τ (ai) denotes the ith free moment of element a ∈ A. Then, the first
five free cumulants ki of a are given by
k1 = m1,
k2 = m2− m21,
k3 = m3− 3m2m1+ 2m31,
k4 = m4− 4m3m1− 2m22+ 10m2m21− 5m41,
Proof. By definition k1 = m1. Using the relation (3) we obtain k2 = 1 X i=0 i+1 X h=1 (−1)h+1i + 1 h X j1+...+jh=2 mj1 · . . . · mjh = 1 X i=0 i+1 X h=1 (−1)h+1i + 1 h h 2 m21+ hm2 = m2− m21, k3 = 2 X i=0 i+1 X h=1 (−1)h+1i + 1 h X j1+...+jh=3 mj1 · . . . · mjh − 3−1 X h=2 kh X j1+...+jh−1=3−h mj1 · . . . · mjh−1 = 2 X i=0 i+1 X h=1 (−1)h+1i + 1 h hm3+ 2 h 2 m2m1+ h 3 m31 − k2m1 = m3− 2m2m1+ m31− (m2− m21)m1 = m3− 3m2m1+ 2m31, k4 = 3 X i=0 i+1 X h=1 (−1)h+1i + 1 h X j1+...+jh=4 mj1 · . . . · mjh − 3 X h=2 kh X j1+...+jh−1=4−h mj1 · . . . · mjh−1 = 3 X i=0 i+1 X h=1 (−1)h+1i + 1 h hm4+ 2 h 2 m3m1+ h 2 m22 +3h 3 m2m21+ h 4 m41 − k2m2− 2k3m1 = m4− 2m3m1− m22+ 3m2m21− m41− k2m2− 2k3m1 = m4− 4m3m1− 2m22+ 10m2m21− 5m41, k5 = 4 X i=0 i+1 X h=1 (−1)h+1i + 1 h X j1+...+jh=5 mj1· . . . · mjh − 4 X h=2 kh X j1+...+jh−1=5−h mj1· . . . · mjh−1
= 4 X i=0 i+1 X h=1 (−1)h+1i + 1 h hm5+ 2 h 2 m4m1+ 2 h 2 m3m2+ 3 h 3 m3m21 +3h 3 m22m1+ 4 h 4 m2m31+ h 5 m51 − k2m3− k3(2m2+ m21) − 3k4m1 = m5− 2m4m1− 2m3m2+ 3m3m21+ 3m 2 2m1− 4m2m31+ m 5 1 −(m2− m21)m3− (m3− 3m2m1+ 2m31)(2m2+ m21) −3(m4− 4m3m1− 2m22+ 10m2m21− 5m 4 1)m1 = m5− 5m4m1− 5m3m2+ 15m3m21+ 15m 2 2m1− 35m2m31+ 14m 5 1,
which completes the proof.
The above presented proof of Corollary 2.1 gives examples of direct cal-culations of free cumulants using the result of Theorem 2.1. Now consider the equation (2), which was used to obtain the free cumulants of degree 1 to 5 by the combinatorial approach. The equality k1= m1 is again assumed
to hold. Then, m2: = τ (a, a) = X π∈N C(2) kπ[a, a]. If π ∈ N C(2) then π = {1, 2} or π = {{1}, {2}} hence m2 = k1k1+ k2, k2 = m2− k12= m2− m21.
To obtain the third free cumulant the sum is taken over all non–crossing partitions of the three elements set N C(3). Then,
π ∈ {{1, 2, 3},
` ` ` {{1, 2}, {3}},` ` ` {{1, 3}, {2}},` ` ` {{1}, {2, 3}},` ` ` {{1}, {2}, {3}}}` ` `
Each of the sets is illustrated with simple graph. The elements belonging to the same subset are connected with a line. The crossing partition is indicated by the cross of at least two lines from two distinguish subsets. Hence, m3 : = τ (a, a, a) = X π∈N C(3) kπ[a, a, a] = k3+ k2k1+ k2k1+ k1k2+ k31 = k3+ 3k1k2+ k31, k3 = m3− 3k1k2− k31 = m3− 3m1(m2− m21) − m31 = m3− 3m1m2+ 2m31.
While calculating the fourth free cumulant we notice that there is only one crossing partition indicated by cross of line illustrating subsets {1, 3} and {2, 4}, i.e.,
N C(4) 63{{1, 3}, {2, 4}}. ` ` ` ` m4 = X π∈N C(4) kπ[a, a, a, a] = k4+ 4k3k1+ 2k22+ 6k2k12+ k14, k4 = m4− 4k1k3− 2k22− 6k2k12− k14 = m4− 4m1(m3− 3m1m2+ 2m31) − 2(m2− m21)2− 6m21(m2− m21) −m41= m4− 4m1m3− 2m22+ 10m2m21− 5m41.
The calculations of the fifth free cumulant, by use of (2), demands the sum-ming over N C(5). Consider the crossing partitions of the set {1, 2, 3, 4, 5}. N C(5) 6⊃ {{1, 2, 4}, {3, 5}}, ` ` ` ` ` {{1, 4}, {2, 3, 5}},` ` ` ` ` {{1, 3, 4}, {2, 5}} , ` ` ` ` ` N C(5) 63 {{2, 4}, {1, 3, 5}}, ` ` ` ` ` N C(5) 63 {{1, 3}, {2, 4, 5}},` ` ` ` ` N C(5) 6⊃ {{1}, {2, 4}, {3, 5}}, ` ` ` ` ` {{1, 4}, {2}, {3, 5}},` ` ` ` ` {{1, 3}, {2, 5}, {4}} , ` ` ` ` ` N C(5) 63 {{1, 3}, {2, 4}, {5}}, ` ` ` ` ` N C(5) 63 {{1, 4}, {2, 5}, {3}},` ` ` ` ` Then, m5 = X π∈N C(5) kπ[a, a, a, a, a] = k5+ 5k4k1+ 5 2 − 5 k3k2+ 5 3 k3k21+ 5 1 1 2 4 2 − 5 k22k1 +5 2 k2k13+ k 5 1= k5+ 5k4k1+ 5k3k2+ 10k3k21+ 10k 2 2k1+ 10k2k13+ k 5 1, and k5 = m5− 5k4k1− 5k3k2− 10k3k12− 10k22k1− 10k2k31− k51 = m5− 5(m4− 4m3m1− 2m22+ 10m2m21− 5m41)m1 −5(m3− 3m2m1+ 2m31)(m2− m12) − 10(m3− 3m2m1+ 2m31)m21 −10(m2− m2 1)2m1− 10(m2− m21)m31− m51 = m5− 5m4m1+ 15m3m21+ 15m22m1− 35m2m31− 5m3m2+ 14m51.
The calculations with use of both methods are presented. To some extent we find that summing over the i1, . . . , ih, such that i1+. . .+ih = k is simpler
3
Example of free cumulant–moment calculations
It is important to mention a particular example of a non–commutative ∗– probability space (RMp(R), τ ) as an illustration and due to the extended
engineering applications. Here, A = RMp(R) denotes set of all p × p random
matrices with entries being real random variables on a probability space (Ω, F , P ) with finite moments of any order. Defined in this way RMp(R)
is a ∗–algebra, with the classical matrix product as multiplication and the transpose as ∗–operation. The ∗–algebra is equipped with tracial functional τ defined as expectation of the normalized trace in the following way
τ (X) := E 1 pTrX = 1 pE p X i=1 λi = Z R x1 p p X i=1 δ{λi≤x}dx = Z R xkdµ(x),
where X = (Xij)pi,j=1 ∈ RMp(R), δB denotes Dirac delta function on set
B, λi are ordered eigenvalues of matrix X and µ = 1pPpi=1δ{λi≤x} is ∗–
distribution, usually called the eigenvalue distribution (spectral density) of the matrix X. This set up is of common use, when studying the spectral measure of random matrices. Often related research problems arise within e.g., theoretical physics and wireless communication, see Couillet and Deb-bah (2011) and Tulino and Verd´u (2004).
Let us consider a matrix Mp = 1pXX0, where Xij ∼ N (0, 1), which also
belongs to (RMp(R), τ ). A matrix W = pMp = XX0 ∼ Wp(I, p). For the
Wishart matrix W the following relation holds E(TrWk+1) = kE(TrWk) + X i+j=ki,j≤0 E(TrWiTrWj). Then, τ (Mk+1p ) = 1 pk+2E(TrW k+1) = k pk+2E(TrW k) + 1 pk+2 X i+j=k i,j≤0 E(TrWiTrWj).
and first free moments for the matrix Mp are given by
m1= τ (Mp) = 1 p2E(TrW 1) = 1 p2E(TrW 0TrW0) = p2 p2 = 1, m2= τ (M2p) = 1 p3E(TrW 2) = 1 p3 E(TrW) + X i+j=1 i,j≤0 E(TrWiTrWj) = 1 p3 p2+ 2E(TrW0TrW ) = 1 p3 p2+ 2pE(TrW ) = 1 p3 p2+ 2p3 = 2 + 1 p,
m3= τ (M3p) = 1 p4E(TrW 3) = 1 p4 2E(TrW2) + X i+j=2 i,j≤0 E(TrWiTrWj) = 1 p4 2(p2+ 2p3) + 2E(TrW0TrW2) + E(TrW TrW ) = 1 p4 2p2+ 4p3+ 2pE(TrW2) + E(TrW )2 = 1 p4 2p2+ 4p3+ 2p(p2+ 2p3) + 3p2+ p2(p2− 1) = 4 + 6p + 5p 2 p2 , m4= τ (M4p) = 1 p5E(TrW 4) = 1 p5 3E(TrW3) + X i+j=3 i,j≤0 E(TrWiTrWj) = 1 p5 3(4p2+ 6p3+ 5p4) + 2E(TrW0TrW3) + 2E(TrW TrW2) = 1 p5 3(4p2+ 6p3+ 5p4) + 2pE(TrW3) + 2E(TrW TrW2) = 1 p5 3(4p2+ 6p3+ 5p4) + 2p(4p2+ 6p3+ 5p4) +2(2p5+ p4+ 8p3+ 4p2) = 20 + 42p + 29p 2+ 14p3 p3 .
Then, using a result of Corollary 2.1, the free cumulants for p × p matrix Mp k1 = m1= 1, k2 = m2− m21 = 2 + 1 p − 1 = 1 + 1 p, k3 = m3− 3m2m1+ 2m31 = 4 + 6p + 5p 2 p2 − 3(2 + 1 p) + 2 = 4 + 3p + p2 p2 , k4 = m4− 4m3m1− 2m22+ 10m2m21− 5m41 = = 20 + 42p + 29p 2+ 14p3 p3 − 4 4 + 6p + 5p2 p2 − 2 2 +1 p 2 +10 2 +1 p − 5 = 20 + 24p + 7p 2+ p3 p3 .
Size of matrix Mp m1 m2 m3 m4 m5 . . . 2 × 2 1 2.5 9 41.5 232.5 . . . 3 × 3 1 73 = 2.33 679 = 7.44 78527 = 29.07 132.90 . . . 4 × 4 1 94 = 2.25 274 = 6.75 38716 = 24.19 99.14 . . . . . . . p × p, p → ∞ 1 2 5 14 42 . . .
Table 1: Values of free moments for matrices of different size. Size of matrix Mp k1 k2 k3 k4 k5 . . . 2 × 2 1 1.5 3.5 13 67.75 . . . 3 × 3 1 43 = 1.33 229 = 2.44 18227 = 6.74 26.35 . . . 4 × 4 1 54 = 1.25 2 7316 = 4.56 14.70 . . . . . . . p × p, p → ∞ 1 1 1 1 1 . . .
Table 2: Values of free cumulants for matrices of different size. Reading the free cumulants from the pth row of the Table 2 we obtain R–transform for the desired matrices Mp. While p → ∞ the matrix Mp→∞,
which is an ”infinite matrix” realized by a sequence of matrices of increasing size, has R–transform
RMp→∞(z) = ∞
X
j=0
kj+1zj = 1 + z + z2+ z3+ . . . ,
which by the inverse Stieltjes formula corresponds to the spectral distribu-tion given by Marˇcenko–Pastur law, see Marˇcenko and Pastur (1967), i.e. µ0p→∞(x) = 2πx1 √4x − x2.
4
Conclusions
In this article we prove a new free cumulant–moment recursive relation for-mula using the concepts of Stieltjes and R–transforms. The demonstrated results are not based on the combinatorial idea of non–crossing partitions as in the previous studies. That implies that the relation can be obtained with use of significantly, in our opinion, simpler computations. There is a strong believe that the result can successfully complete already existing knowledge regarding free cumulant–moment relations and in some particular cases replace previously used formulas in order to provide easier calculations or avoid introducing partition related concepts.
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