Fuss-Catalan Numbers

in Noncommutative Probability

Wojciech Młotkowski^{1}

Received: January 3, 2010

Communicated by Friedrich Götze

Abstract. We prove that if p, r ∈ ℝ, p ≥ 1 and 0 ≤ r ≤ p then the Fuss-Catalan sequence \binom{mp+r}{m} \frac{r}{mp+r} is positive definite. We study the family of the corresponding probability measures μ(p, r) on ℝ from the point of view of noncommutative probability. For example, we prove that if 0 ≤ 2r ≤ p and r + 1 ≤ p then μ(p, r) is ⊞-infinitely divisible. As a by-product, we show that the sequence \frac{m^{m}}{m!} is positive definite and the corresponding probability measure is ⊠-infinitely divisible.

2010 Mathematics Subject Classification: Primary 46L54; Secondary 44A60, 60C05 Keywords and Phrases: Fuss-Catalan numbers, free, boolean and monotonic convolution

1. Introduction

For natural numbers m, p, r let A_{m}(p, r) denote the number of all sequences (a_{1}, a_{2}, …, a_{mp+r}) such that: (1) a_{i} ∈ {1, 1 − p}, (2) a_{1} + a_{2} + … + a_{s} > 0 for all s such that 1 ≤ s ≤ mp + r and (3) a_{1} + a_{2} + … + a_{mp+r} = r. It turns out that this is given by the two-parameter Fuss-Catalan numbers (2.1) (see [5, 13]). Note that the right hand side of (2.1) allows us to define A_{m}(p, r) for all real parameters p and r. In particular, the Catalan numbers A_{m}(2, 1) are known as moments of the Marchenko–Pastur distribution:

(1.1)  dπ̃(x) = \frac{1}{2π} \sqrt{\frac{4 − x}{x}}dx  on [0, 4],

which in the free probability theory plays the role of the Poisson measure. In this paper we are going to study the question for which parameters p, r ∈ ℝ the sequence {A_{m}(p, r)}_{m=0}^{∞} is positive definite, i.e. is the moment sequence of some probability measure (which we will denote μ(p, r)). Recently T. Banica, S. T. Belinschi, M. Capitaine and B. Collins [1] showed that if p > 1 then {A_{m}(p, 1)}_{m=0}^{∞} is the moment sequence of a probability measure which can be expressed as the multiplicative free power π^{⊠p−1}. We are going to prove that if p, r ∈ ℝ, p ≥ 1 and 0 ≤ r ≤ p then {A_{m}(p, r)}_{m=0}^{∞} is the moment sequence of a unique probability measure μ(p, r) which has compact support contained in [0, ∞). Moreover, if 0 ≤ 2r ≤ p and r + 1 ≤ p then μ(p, r) is infinitely divisible with respect to the free convolution ⊞. In some particular cases we are able to determine the multiplicative free convolution, the boolean power and the monotonic convolution of the measures μ(p, r). We will also prove that if 0 ≤ r ≤ p − 1 then the sequence \{\binom{mp+r}{m}\}_{m=0}^{∞} is positive definite and the corresponding probability measure can be expressed as μ(p−r, 1)^{⊎p} ▷ μ(p, r), where ⊎ and ▷ denote the boolean and the monotonic convolution, respectively. The paper is organized as follows. In Section 2 we prove three combinatorial identities. Then we use them to derive some formulas for the generating functions. In Section 4 we regard the numbers A_{m}(p, r) as moments of a probability quasi-measure μ(p, r) (by this we mean a linear functional μ : ℝ[x] → ℝ satisfying μ(1) = 1). On the class of probability quasi-measures one can introduce the free, boolean and monotonic convolutions in combinatorial way. The class of compactly supported probability measures on ℝ, regarded as a subclass of the former, is closed under these operations. We prove some formulas involving the probability quasi measures μ(p, r), for example we find the free R- and S-transforms (4.8), (4.11), the boolean powers μ(p, 1)^{⊎t} (4.18) and, in special cases, the multiplicative free (4.12), (4.13), (4.14) and the monotonic convolution (4.20) of the measures μ(p, r). In Section 5 we prove that if p ≥ 1 and 0 ≤ r ≤ p then μ(p, r) is a measure (we conjecture that this condition is also necessary for p, r > 0). The proof involves the multiplicative free convolution ⊠. Moreover, we show that if 0 ≤ 2r ≤ p and r + 1 ≤ p then μ(p, r) is ⊞-infinitely divisible. In the final part we extend our results to the dilations of the measures μ(p, r), with parameter h > 0. Taking the limit with h → 0 we prove in particular that the sequence \{\frac{m^{m}}{m!} \}_{m=0}^{∞} is positive definite and the corresponding probability measure ν_{0} is ⊠-infinitely divisible.

1Research supported by MNiSW: N N201 364436, by ToK: MTKD-CT-2004-013389, by 7010 POLONIUM project: “Non-Commutative Harmonic Analysis with Applications to Operator Spaces, Operator Algebras and Probability” and by joint PAN-JSPS project: “Noncommutative harmonic analysis on discrete structures with applications to quantum probability”.

2. Some combinatorial identities

We will work with the two-parameter Fuss-Catalan numbers (see [5, 13]) defined by: A_{0}(p, r) := 1 and

(2.1)  A_{m}(p, r) := \frac{r}{m!} ∏_{i=1}^{m−1}(mp + r − i)

for m ≥ 1, where p, r are real parameters. Note that (2.1) can be written as \binom{mp+r}{m} \frac{r}{mp+r}, unless mp + r = 0. One can check that for m ≥ 0

(2.2)  A_{m}(p, r) = A_{m}(p, r − 1) + A_{m−1}(p, p + r − 1),

under convention that A_{−1}(p, r) := 0, and

(2.3)  A_{m}(p, p) = A_{m+1}(p, 1).

It is also known (see [13]) that

(2.4)  ∑_{k=0}^{m} A_{k}(p, r)A_{m−k}(p, s) = A_{m}(p, r + s).

Now we are going to prove three identities, valid for c, d, p, r, t ∈ ℝ, which will be needed later on.

Proposition 2.1.

(2.5)  ∑_{k=0}^{m} A_{k}(p − r, c)A_{m−k}(p, kr + d) = A_{m}(p, c + d).

Proof. It is easy to check that the formula is true for m = 0 and m = 1. Denoting the left hand side by S_{m}(p, r, c, d) we have from (2.2):

S_{m}(p, r, c, d) = ∑_{k=0}^{m} A_{k}(p − r, c)A_{m−k}(p, kr + d)

= ∑_{k=0}^{m} [A_{k}(p − r, c − 1) + A_{k−1}(p − r, p − r + c − 1)] A_{m−k}(p, kr + d)

= ∑_{k=0}^{m} A_{k}(p − r, c − 1)A_{m−k}(p, kr + d)

+ ∑_{k=1}^{m} A_{k−1}(p − r, p − r + c − 1)A_{m−k}(p, kr + d)

= S_{m}(p, r, c − 1, d) + ∑_{k=0}^{m−1} A_{k}(p − r, p − r + c − 1)A_{m−1−k}(p, kr + r + d)

= S_{m}(p, r, c − 1, d) + S_{m−1}(p, r, p − r + c − 1, r + d),

so that we have

S_{m}(p, r, c, d) = S_{m}(p, r, c − 1, d) + S_{m−1}(p, r, p − r + c − 1, r + d).

Fix m and assume that (2.5) holds for m−1. Now we prove that for m it holds for every natural c. Indeed, it holds for c = 0 and if it does for c − 1 then, by assumption and by (2.2):

S_{m}(p, r, c, d) = S_{m}(p, r, c − 1, d) + S_{m−1}(p, r, p − r + c − 1, r + d)

= A_{m}(p, c + d − 1) + A_{m−1}(p, p + c + d − 1) = A_{m}(p, c + d), which proves that the statement is true for c. Therefore it holds for all natural c. Now we note that both sides of (2.5) are polynomials on c of order m, therefore the formula holds for all c ∈ ℝ, which completes the inductive step. □

Proposition 2.2.

(1 − t) ∑_{l=0}^{m} A_{l}(p, 1) ∑_{j=0}^{m−l} A_{m−l−j}(p, j(p − 1) + r)t^{j} (2.6)

+t ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r)t^{j} = A_{m}(p, r + 1).

Proof. Using first (2.4) and then (2.2) we obtain:

t ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r)t^{j}

+ (1 − t) ∑_{l=0}^{m} A_{l}(p, 1) ∑_{j=0}^{m−l} A_{m−l−j}(p, j(p − 1) + r)t^{j}

= t ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r)t^{j}

+ (1 − t) ∑_{j=0}^{m} ∑_{l=0}^{m−j} A_{l}(p, 1)A_{m−j−l}(p, j(p − 1) + r)t^{j}

= t ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r)t^{j}

+ (1 − t) ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r + 1)t^{j}

= ∑_{j=0}^{m} A_{m−j}(p, j(p − 1) + r + 1)t^{j} − ∑_{j=0}^{m−1} A_{m−j−1}(p, j(p − 1) + r + p)t^{j+1}

= A_{m}(p, r + 1).  □

Proposition 2.3.

(2.7)  ∑_{k=0}^{m} A_{m−k}(p, k(p − 1) + r)p^{k} = \binom{mp + r}{m}.

Proof. Denoting the left hand side by T_{m}(p, r) we use (2.2) and get

T_{m}(p, r) =

= ∑_{k=0}^{m} A_{m−k}(p, k(p − 1) + r)p^{k}

= ∑_{k=0}^{m} [A_{m−k}(p, k(p − 1) + r − 1) + A_{m−1−k}(p, k(p − 1) + p + r − 1)] p^{k}

= T_{m}(p, r − 1) + T_{m−1}(p, p + r − 1).

Now we proceed as in the proof of (2.5), using the binomial identity

\binom{mp + r}{m} = \binom{mp + r − 1}{m} + \binom{mp + r − 1}{m − 1}.  □

3. Generating functions

In this part we are going to study the generating functions

(3.1)  B_{p}(z) := ∑_{m=0}^{∞} A_{m}(p, 1)z^{m},

which are convergent in some neighborhood of 0 (to observe this one can use the inequality

|A_{m}(p, r)| ≤ |r|[m(|p| + 1) + |r|]^{m−1}/m!

and apply the Cauchy’s radical test). From (2.4) and (2.3) we have

(3.2)  B_{p}(z)^{r} = ∑_{m=0}^{∞} A_{m}(p, r)z^{m}

and

(3.3)  B_{p}(z) = 1 + zB_{p}(z)^{p}.

Indeed, denoting the right hand side of (3.2) by A_{p,r}(z) we have A_{p,1}(z) = B_{p}(z) and, by (2.4), A_{p,r}(z) · A_{p,s}(z) = A_{p,r+s}(z), which implies that A_{p,r}(z) = B_{p}(z)^{r}. Taking r = p and applying (2.3) we get (3.3). Now we are going to interpret formulas (2.5), (2.6), (2.7) in terms of these generating functions.

Proposition 3.1. For any real parameters p, r we have

(3.4)  B_{p−r} (zB_{p}(z)^{r}) = B_{p}(z).

Proof. First we note that if A(z) = ∑_{m=0}^{∞} a_{m}z^{m}, B(z) = ∑_{n=1}^{∞} b_{n}z^{n} then

(3.5)  A(B(z)) = a_{0} + ∑_{m=1}^{∞} z^{m} ∑_{k=1}^{m} a_{k} ∑_{\begin{matrix}i_{1},i_{2},...,i_{k}≥1 \\ i_{1}+i_{2}+...+i_{k}=m \end{matrix}} b_{i_{1}}b_{i_{2}} … b_{i_{k}}.

Putting b_{i} := A_{i−1}(p, r) for fixed k, m we have:

∑_{\begin{matrix}i_{1},i_{2},...,i_{k}≥1 \\ i_{1}+i_{2}+...+i_{k}=m \end{matrix}} b_{i_{1}}b_{i_{2}} … b_{i_{k}} = ∑_{\begin{matrix}j_{1},j_{2},...,j_{k}≥0 \\ j_{1}+j_{2}+...+j_{k}=m−k \end{matrix}} A_{j_{1}}(p, r)A_{j_{2}}(p, r) … A_{j_{k}}(p, r)

= A_{m−k}(p, kr),

the coefficient of B_{p}(z)^{kr} at z^{m−k}. Now we put a_{k} := A_{k}(p − r, 1) and applying (2.5), with c = 1, d = 0, we get

(3.6)  ∑_{k=1}^{m} a_{k} ∑_{\begin{matrix}i_{1},i_{2},...,i_{k}≥1 \\ i_{1}+i_{2}+...+i_{k}=m \end{matrix}}b_{i_{1}}b_{i_{2}} … b_{i_{k}}

= ∑_{k=0}^{m} A_{k}(p − r, 1)A_{m−k}(p, kr) = A_{m}(p, 1),

as A_{m}(p, 0) = 0 for m ≥ 1, which completes the proof. □

Note that in the proof we applied (2.5) only with c = 1 and d = 0. For p, r, t ∈ ℝ we denote

(3.7)  D_{p,r,t}(z) := \frac{B_{p}(z)^{1+r}}{(1 − t)B_{p}(z) + t}.

Proposition 3.2. For p, r, t ∈ ℝ we have

(3.8)  D_{p,r,t}(z) = ∑_{m=0}^{∞} z^{m} ∑_{k=0}^{m} A_{m−k}(p, k(p − 1) + r)t^{k},

in particular:

(3.9)  D_{p,r,p}(z) = ∑_{m=0}^{∞} \binom{mp + r}{m}z^{m}.

Moreover

(3.10)  D_{p−r,s,t} (zB_{p}(z)^{r}) B_{p}(z)^{r} = D_{p,r+s,t}(z).

Proof. Using (2.6) we can verify that

[(1 − t)B_{p}(z) + t] · [ ∑_{m=0}^{∞} z^{m} ∑_{k=0}^{m} A_{m−k}(p, k(p − 1) + r)t^{k}] = B_{p}(z)^{1+r}

which proves (3.8). Formulas (3.9) and (3.10) are consequences of (2.7) and (3.4). □

Proposition 3.3. In some neighborhood of 0 we have

(3.11)  B_{p}(z(1 + z)^{−p}) = 1 + z,

and more generally, for r ≠ 0 we have

(3.12)  B_{p} (((1 + z)^{\frac{1}{r}} − 1)(1 + z)^{\frac{−p}{r}} )^{r} = 1 + z.

Proof. Since we have B_{p}(0) = 1 and B′_{p}(0) = 1, there is a function f_{p} defined on a neighborhood of 0 such that f_{p}(0) = 0 and B(f_{p}(z)) = 1+z. Substituting z ↦ f_{p}(z) in (3.3) we obtain f_{p}(z) = z(1 + z)^{−p}. Now we put z ↦ (1 + w)^{1/r} − 1 to (3.11) and taking the r-th power we obtain (3.12). □

Remark. Note that (3.11) leads to an analytic proof of (3.4). Namely, substituting in (3.4) z ↦ z(1 + z)^{−p} we get

B_{p−r} (z(1 + z)^{−p}B_{p} (z(1 + z)^{−p})^{r}) = B_{p−r} (z(1 + z)^{−p}(1 + z)^{r})
= 1 + z = B_{p} (z(1 + z)^{−p}) .

Finally we note a symmetry possessed by our generating functions.

Proposition 3.4. For p, r, t ∈ ℝ we have

(3.13)  B_{p}(−z)^{r} = B_{1−p}(z)^{−r},
(3.14)  D_{p,r,t}(−z) = D_{1−p,−1−r,1−t}(z).

Proof. One can check that (−1)^{m}A_{m}(p, r) = A_{m}(1 − p, −r), which proves (3.13), and by the definition (3.7), (3.13) implies (3.14). □

4. Relations with noncommutative probability

By a probability quasi-measure we will mean a linear functional μ on the set ℝ[x] of polynomials with real coefficients, satisfying μ(1) = 1. In the case when μ is given by μ(P) = ∫ P (t) dμ̃(t) for some probability measure μ̃ on ℝ we will identify μ with μ̃ and say that μ is proper or is just a probability measure. A probability quasi-measure μ is uniquely determined by its moment sequence {μ(x^{m})}_{m=0}^{∞}. It is proper if and only if its moment sequence is positive definite, i.e. if

∑_{i,j=0}^{∞} μ(x^{i+j})α_{i}α_{j} ≥ 0

holds for every sequence {α_{i}}_{i=0}^{∞} of real numbers, with only finitely many nonzero entries. All probability measures encountered in this paper are compactly supported and therefore uniquely determined by their moment sequences. For a probability quasi-measure μ we define its moment generating function, which is the (at least formal) power series

M_{μ}(z) := ∑_{m=0}^{∞} μ(x^{m})z^{m}

and its reflection μ̂ by μ̂(x^{m}) := (−1)^{m}μ(x^{m}) or, equivalently, Mμ̂(z) := M_{μ}(−z). If μ is a probability measure then so is μ̂ and then we have μ̂(X) = μ(−X) for every Borel subset of ℝ.

For p, r, t ∈ ℝ we define probability quasi-measures μ(p, r) and ν(p, r, t) by

(4.1)  μ(p, r)(x^{m}) := A_{m}(p, r),
(4.2)  ν(p, r, t)(x^{m}) := ∑_{k=0}^{m} A_{m−k}(p, k(p − 1) + r)t^{k},

in particular, by (2.7),

(4.3)  ν(p, r, p)(x^{m}) = \binom{mp + r}{m}.

For example, μ(1, 1) = ν(1, 0, 1) = δ_{1} and for every p ∈ ℝ we have μ(p, 0) = ν(0, 0, 0) = δ_{0}. Note that ν(p, r, 0) = μ(p, r) so that the class of probability quasi-measures μ(p, r) is contained in that of ν(p, r, t), we will be interested however mainly in the former. First we note that Proposition 3.4 leads to

Proposition 4.1.

(4.4)  { \atop μ(p, r)} = μ(1 − p, −r),
(4.5)  { \atop ν(p, r, t)} = ν(1 − p, −1 − r, 1 − t).  □

There are several convolutions of probability quasi-measures, apart from the classical one: (μ ∗ ν)(x^{n}) := ∑_{k=0}^{n} \binom{n}{k}μ(x^{k})ν(x^{n−k}), which are related to various notions of independence (namely, the free, boolean and the monotonic independence) in noncommutative probability.

1. Free convolution (see [2, 15, 11]) is defined in the following way. For a probability quasi-measure μ we define its free R-transform (or the additive free transform) R_{μ}(z) by the formula:

(4.6)  M_{μ}(z) = R_{μ}(zM_{μ}(z)) + 1.

The free cumulants r_{m}(μ) are defined as the coefficients of the Taylor expansion R_{μ}(z) = ∑_{k=1}^{∞} r_{k}(μ)z^{k} (combinatorial relation between moments and free cumulants is described in [11] and [4]). Then the free convolution μ ⊞ ν can be defined as the unique probability quasi-measure which satisfies

(4.7)  R_{μ⊞ν}(z) = R_{μ}(z) + R_{ν}(z).

We also define free power μ^{⊞t}, t > 0, by R_{μ^{⊞t}}(z) := tR_{μ}(z). As a consequence of (4.6) and (3.4) we obtain:

Proposition 4.2. For the free additive transform of μ(p, r) we have

(4.8)  R_{μ(p,r)}(z) = B_{p−r}(z)^{r} − 1

so that for the free cumulants we have r_{m}(μ(p, r)) = A_{m}(p − r, r), m ≥ 1. □

The free S-transform (or the free multiplicative transform) of a quasi-measure μ, with μ(x^{1}) ≠ 0, is defined by the relation

(4.9) R_{μ}(zS_{μ}(z)) = z or, equivalently, M_{μ}(z(1 + z)^{−1}S_{μ}(z)) = 1 + z.

Then the multiplicative free convolution μ_{1} ⊠ μ_{2} and the multiplicative free power μ^{⊠t}, t > 0, are defined by

(4.10)  S_{μ_{1}⊠μ_{2}}(z) := S_{μ_{1}}(z)S_{μ_{2}}(z) and S_{μ^{⊠t}}(z) := S_{μ}(z)^{t}.

Proposition 4.3. For r ≠ 0 the S-transform of the measure μ(p, r) is equal to

(4.11)  S_{μ(p,r)}(z) = \frac{(1 + z)^{\frac{1}{r}} − 1}{z}(1 + z)^{\frac{r−p}{r}} .

Consequently

(4.12)  μ(1 + p_{1}, 1) ⊠ μ(1 + p_{2}, 1) = μ(1 + p_{1} + p_{2}, 1),

and more generally

(4.13)  μ(p_{1}, r) ⊠ μ(1 + p_{2}, 1) = μ(p_{1} + rp_{2}, r).

We have also

(4.14)  μ(1 + p, 1)^{⊠t} = μ(1 + tp, 1).

Proof. Formula (4.11) is a consequence of (3.12). In particular

(4.15)  S_{μ(1+p,1)}(z) = (1 + z)^{−p}

which leads to (4.12), (4.13) and (4.14). □

2. The boolean convolution μ_{1} ⊎ μ_{2} and the boolean power μ^{⊎t}, t > 0, (see [14, 3]) can be defined by putting

(4.16)  \frac{1}{M_{μ_{1}⊎μ_{2}}(z)} = \frac{1}{M_{μ_{1}}(z)} + \frac{1}{M_{μ_{2}}(z)} − 1,
(4.17)  M_{μ^{⊎t}}(z) = \frac{M_{μ}(z)}{(1 − t)M_{μ}(z) + t}.

Comparing this with definition (3.7) we get

Proposition 4.4. For p, t ∈ ℝ we have

(4.18)  μ(p, 1)^{⊎t} = ν(p, 0, t). □

3. Monotonic convolution (see [10]) is an associative, noncommuting operation ▷ which is defined by: μ_{1} ▷ μ_{2} = μ iff

(4.19)  M_{μ}(z) = M_{μ_{1}}(zM_{μ_{2}}(z)) · M_{μ_{2}}(z).

Then (3.4) and (3.10) yield

Proposition 4.5. For any parameters a, b, r, t ∈ ℝ we have

(4.20)  μ(a, b) ▷ μ(a + r, r) = μ(a + r, b + r),
(4.21)  ν(a, b, t) ▷ μ(a + r, r) = ν(a + r, b + r, t). □

In the next section we are going to study which of the probability quasimeasures μ(p, r) and ν(p, r, t) are actually probability measures. For this purpose we will use some of the the following facts, which are available in literature (see [15, 11, 14, 10, 6, 7]): The class of all compactly supported probability measures on ℝ is closed under the free, boolean, and monotonic convolution and also under taking the powers μ^{⊞s}, μ^{⊎t}, for s ≥ 1, t > 0. Moreover, the class of probability measures with compact support contained in [0, ∞) is closed under the free, multiplicative free, boolean and monotonic convolution and also under taking the powers μ^{⊞s}, μ^{⊠s} and μ^{⊎t} for s ≥ 1 and t > 0. A probability measure μ on ℝ (resp. on [0, ∞)) is called ⊞-infinitely divisible (resp. ⊠-infinitely divisible) if μ^{⊞t} (resp. μ^{⊠t}) is a probability measure for every t > 0. If μ has compact support and r_{m}(μ) are its free cumulants then μ is ⊞-infinitely divisible if and only if the sequence {r_{m+2}(μ)}_{m=0}^{∞} is positive definite.

5. Positivity

The aim of this section is to study which of the quasi measures μ(p, r) and ν(p, r, t) are actually measures, i.e. for which parameters p, r, t ∈ ℝ the corresponding sequence is positive definite. We start with

Theorem 5.1. If p ≥ 1, 0 ≤ r ≤ p then {A_{m}(p, r)}_{m=0}^{∞} is the moment sequence of a probability measure μ(p, r) with a compact support contained in [0, ∞). If p ≤ 0, p−1 ≤ r ≤ 0 then μ(p, r) is a probability measure which is the reflection of μ(1 − p, −r).

Proof. We know already that π̃ = μ(2, 1) is the free Poisson law (1.1). Then, as was noted in [1], π̃ is ⊠-infinitely divisible and for s > 0 we have π^{⊠s} = μ(1+s, 1). Hence if p ≥ 1 then μ(p, 1) is a probability measure with a compact support contained in [0, ∞). By (2.3) it implies that the sequence A_{m}(p, p) = A_{m+1}(p, 1) is also positive definite, namely we have

∫_{ℝ} f(x) dμ(p, p)(x) = ∫_{ℝ} f(x)x dμ(p, 1)(x)

for any continuous function f on ℝ. Hence μ(p, p), p ≥ 1, is a probability measure with a compact support contained in [0, ∞). For 1 ≤ r ≤ p we apply (4.13) to obtain:

μ(p, r) = μ(r, r) ⊠ μ(p/r, 1),

which proves the first statement for the sector 1 ≤ r ≤ p. For r ∈ (0, 1) the measure μ(1, r) is related to the Euler beta function

(5.1)  B(a, b) := ∫_{0}^{1} x^{a−1}(1 − x)^{b−1} dx.

We will use its well known properties: B(a, 1 − a) = \frac{π}{sin aπ} and B(a, b) = \frac{a−1}{a+b−1}B(a − 1, b). If we define probability measure

(5.2)  μ_{r} := \frac{sin πr}{π} x^{r−1}(1 − x)^{−r} dx

on [0, 1] then we have

∫_{ℝ} x^{m} dμ_{r}(x) = \frac{sin πr}{π} B(m + r, 1 − r) = ∏_{k=1}^{m} \frac{r + i − 1}{i} = A_{m}(1, r).

which means that μ(1, r) = μ_{r}. Now for s ≥ 0 we have

μ(1 + rs, r) = μ(1, r) ⊠ μ(1 + s, 1),

which proves the first statement for (p, r) ∈ [1, +∞)×(0, 1). It remains to note that μ(p, 0) = δ_{0} for every p ∈ ℝ. The second statement is a consequence of (4.4). □

We conjecture that the last theorem fully characterizes the set of parameters p, r ∈ ℝ for which μ(p, r) is a measure (apart from the trivial case μ(p, 0) = δ_{0}). It is easy to check that A_{0}(p, r)A_{2}(p, r) − A_{1}(p, r)^{2} = r(2p − 1 − r)/2, hence a necessary condition for positive definiteness of the sequence A_{m}(p, r) is that r(2p − 1 − r) ≥ 0.

Remark. According to Penson and Solomon [12]:

(5.3)  μ(3, 1) = \frac{\sqrt[6]{108}[2^{1/3} (27 + 3\sqrt{81 − 12x})^{2/3} − 6x^{1/3}]}{12πx^{2/3}(27 + 3\sqrt{81 − 12x})^{1/3}} dx

on [0, 27/4]. More generally, for μ(p, 1) with p ∈ ℕ we refer to [8].

Corollary 5.1. If either 0 ≤ 2r ≤ p, r + 1 ≤ p or p ≤ 2r + 1, p ≤ r ≤ 0 then μ(p, r) is ⊞-infinitely divisible.

Proof. By Theorem 13.16 in [11], a compactly supported probability measure μ, with free cumulants r_{m}(μ), is ⊞-infinitely divisible if and only if the sequence {r_{m+2}(μ)}_{m=0}^{∞} is positive definite. Then it is sufficient to refer to (4.8) and to note that the numbers A_{m+2}(p − r, r) are the moments of the measure x^{2} dμ(p − r, r)(x). □

Corollary 5.2. If 0 ≤ r ≤ p − 1, t > 0 then ν(p, r, t) is a probability measure with a compact support contained in [0, +∞). If p ≤ 1 + r ≤ 0, t < 1 then ν(p, r, t) is a probability measure which is the reflection of ν(1−p, −1−r, 1−t). In particular, if either 0 ≤ r ≤ p − 1 or p ≤ 1 + r ≤ 0 then the sequence \{\binom{mp+r}{m}\}_{m=0}^{∞} is positive definite.

Proof. For 0 ≤ r ≤ p − 1, t > 0 we apply (4.21) and (4.18):

ν(p, r, t) = ν(p − r, 0, t) ▷ μ(p, r) = μ(p − r, 1)^{⊎t} ▷ μ(p, r)

and Theorem 5.1. Then we use (4.5). □

A measure ν on ℝ is called symmetric if ν̂ = ν. For a probability quasi-measure μ define its symmetrization μ^{s} by M_{μ^{s}}(z) := M_{μ} (z^{2}). If μ is a probability measure with support contained in [0, ∞) then μ^{s} is a symmetric measure on ℝ, which satisfies ∫_{ℝ} f(t^{2}) dμ^{s}(t) = ∫_{ℝ} f(t) dμ(t) for every compactly supported continuous function on ℝ. Denote by μ^{s}(p, r) and ν^{s}(p, r, t) the symmetrization of μ(p, r) and ν(p, r, t). Then, by (3.4) and (4.9), for the free additive transform we have

(5.4)  R_{μ^{s}(p,r)}(z) = B_{p−2r}(z^{2})^{r} − 1.

In the same way as Corollary 5.2 one can prove

Corollary 5.3. If p ≥ 1, 0 ≤ r ≤ p then μ^{s}(p, r) is a symmetric probability measure on ℝ. Moreover, if p − 2r ≥ 1 and 0 ≤ 3r ≤ p then μ^{s}(p, r) is ⊞-infinitely divisible. □

Let us record some formulas:

(5.5)  μ^{s}(p, 1)^{⊎t} = ν^{s}(p, 0, t),
(5.6)  μ^{s}(a, b) ▷ μ^{s}(a + 2r, r) = μ^{s}(a + 2r, b + r),
(5.7)  ν^{s}(a, b, t) ▷ μ^{s}(a + 2r, r) = ν^{s}(a + 2r, b + r, t).

5.1. Picture for μ(p, r).

[figure: 1; −1; 2; 1; −1; p; r; Σ_{+}; Σ_{+}^{⊞∞}; Σ_{−}; Σ_{−}^{⊞∞}; 3; Σ_{s}^{⊞∞}]

Here we illustrate the main results concerning the measures μ(p, r).

(1) If μ(p, r) is a probability measure then r(2p − 1 − r) ≥ 0 (the right-top and left-bottom sector between the p-axis and the line r = 2p − 1),

(2) Σ_{+} (including Σ_{+}^{⊞∞} and Σ_{s}^{⊞∞}): μ(p, r) is a probability measure with a compact support contained in [0, ∞),

(3) Σ_{−} (including Σ_{−}^{⊞∞}): μ(p, r) is a probability measure, the reflection of μ(1 − p, −r),

(4) Σ_{+}^{⊞∞} ∪ Σ_{−}^{⊞∞} (including Σ_{s}^{⊞∞}): μ(p, r) is ⊞-infinitely divisible,

(5) Σ_{s}^{⊞∞}: the symmetrization of μ(p, r) is ⊞-infinitely divisible.

6. Dilations

For a probability quasi-measure μ we define its dilation with parameter c > 0 by (D_{c}μ)(x^{m}) := c^{m}μ(x^{m}). Then for the moment generating function we have: M_{D_{c}μ}(z) = M_{μ}(cz) and similarly for the free R-transform: R_{D_{c}μ}(z) = R_{μ}(cz), while for the S-transform we have S_{D_{c}μ}(z) = \frac{1}{c}S_{μ}(z). If μ is proper then we have (D_{c}μ)(X) = μ(\frac{1}{c}X) for every Borel subset X of ℝ. In this part we are going to study dilations of the measures μ(p, r) and ν(p, r, t) and their limits as the parameter goes to 0. For h ≥ 0 and a, p, r ∈ ℝ define sequences

(6.1)  \binom{a}{m}_{h} := \frac{1}{m!} ∏_{i=0}^{m−1}(a − ih),

(6.2)  A_{m}(p, r, h) := \frac{r}{m!} ∏_{i=1}^{m−1}(mp + r − ih),

with A_{0}(p, r, h) := 1. In particular A_{m}(p, r, h) = \frac{r}{mp+r} \binom{mp+r}{m}_{h} whenever mp + r ≠ 0. Then, for h ≥ 0 and p, r, t ∈ ℝ define probability quasi-measures:

(6.3)  μ(p, r, h)(x^{m}) := A_{m}(p, r, h),
(6.4)  ν(p, r, t, h)(x^{m}) := ∑_{k=0}^{m} A_{m−k}(p, k(p − h) + r, h)t^{k}.

and their moment generating functions B_{p,r,h}(z) and D_{p,r,t,h}(z) respectively. Note that if h > 0 then A_{m}(p, r, h) = h^{m}A_{m}(p/h, r/h) and hence these probability quasi measures can be represented as dilations:

(6.5)  μ(p, r, h) = D_{h}μ(p/h, r/h),
(6.6)  ν(p, r, t, h) = D_{h}ν(p/h, r/h, t/h).

Therefore the corresponding moment generating functions are

(6.7)  B_{p,r,h}(z) = B_{p/h}(hz)^{r/h},
(6.8)  D_{p,r,t,h}(z) = D_{p/h,r/h,t/h}(hz) = \frac{hB_{p,h+r,h}(z)}{(h − t)B_{p,h,h}(z) + t}.

These formulas allow us to derive properties of the probability quasi-measures μ(p, r, h) and ν(p, r, t, h) directly from our previous results when h > 0, and, after taking the limit with h → 0, for h = 0.

Proposition 6.1. For h > 0 and p, r, t ∈ ℝ

(6.9)  B_{p,h,h}(z) = 1 + zhB_{p,p,h}(z),
(6.10)  log (B_{p,1,0}(z)) = zB_{p,p,0}(z),
(6.11)  D_{p,r,t,0}(z) = \frac{B_{p,r,0}(z)}{1 − ztB_{p,p,0}(z)}.

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Proof. First formula is a consequence of (3.3) and (6.7). Then we have

\frac{B_{p,1,h}(z)^{h} − 1}{h} = \frac{B_{p,h,h}(z) − 1}{h} = zB_{p,p,h}(z).

Taking the limit with h → 0 we obtain (6.10). For (6.11) we write use (6.8) and (6.9) to get

\frac{1}{h}[(h − t)B_{p,h,h}(z) + t] = 1 − (t − h)\frac{B_{p,h,h}(z) − 1}{h} = 1 − (t − h)zB_{p,p,h}(z)

and then we take limit with h → 0. □

Proposition 6.2. For h ≥ 0 and p, r, s ∈ ℝ we have

(6.12)  B_{p−r,s,h} (zB_{p,r,h}(z)) = B_{p,s,h}(z).  □

Proposition 6.3. For h ≥ 0 and p, r ∈ ℝ we have

(6.13)  ν(p, r, p, h)(x^{m}) = \binom{mp + r}{m}_{h}.

Proof. For h > 0 the formula is a consequence of (6.6). Then we take limit with h → 0. □

Proposition 6.4. For h ≥ 0 and p, r, t ∈ ℝ we have

(6.14)  { \atop μ(p, r, h)} = μ(h − p, −r, h),
(6.15)  { \atop ν(p, r, t, h)} = ν(h − p, −h − r, h − t, h).

Proof. First we note that A_{m}(p, r, h)(−1)^{m} = A_{m}(h − p, −r, h) and then we apply (6.8) and (3.14). □

Proposition 6.5. For the free transforms we have

(6.16)  R_{μ(p,r,h)}(z) = B_{p−r,r,h}(z) − 1
(6.17)  S_{μ(p,r,h)}(z) = \frac{(1 + z)^{h/r} − 1}{hz}(1 + z)^{(r−p)/r} for h > 0,
(6.18)  S_{μ(p,r,0)}(z) = \frac{log(1 + z)}{rz}(1 + z)^{(r−p)/r},
(6.19)  S_{ν(p,0,t,0)}(z) = \frac{1}{t}e^{\frac{−pz}{t(1+z)}} .

In particular ν(p, 0, t, 0) = D_{t} (ν(1, 0, 1, 0)^{⊠p/t}).

Proof. Formulas (6.16), (6.17) are consequences of (6.7), (4.11) and (6.12). Therefore, for h > 0 we have

(6.20)  B_{p,r,h} (\frac{(1 + z)^{h/r} − 1}{h}(1 + z)^{−p/r}) = 1 + z,

which leads to

(6.21)  B_{p,r,0} (\frac{log(1 + z)}{r(1 + z)^{p/r}} ) = 1 + z

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and to (6.18). In particular, substituting (1 + z) ↦ e^{\frac{pz}{t(1+z)}} , we have

(6.22)  B_{p,p,0} (\frac{z}{t(1 + z)}e^{\frac{−pz}{t(1+z)}} ) = e^{\frac{pz}{t(1+z)}}

which, combined with (6.11) gives

{□ \atop (6.23)} D_{p,0,t,0} (\frac{z}{t(1 + z)}e^{\frac{−pz}{t(1+z)}} ) = \frac{1}{1 − \frac{z}{1+z}} = 1 + z.

Proposition 6.6. For h > 0 and p, t ∈ ℝ we have

(6.24)  μ(p, h, h)^{⊎t} = ν(p, 0, th, h),
(6.25)  ν(p, 0, 1, 0)^{⊎t} = ν(p, 0, t, 0).

Proof. Since B_{p,0,0}(z) = 1, formula (6.25) is a consequence of (6.11). □

Proposition 6.7. For h ≥ 0, t > 0, a, b ∈ ℝ we have

(6.26)  μ(a, b, h) ▷ μ(a + r, r, h) = μ(a + r, b + r, h),
(6.27)  ν(a, b, t, h) ▷ μ(a + r, r, h) = ν(a + r, b + r, t, h).  □

Proposition 6.8. Assume that h ≥ 0.

1. If p ≥ h and 0 ≤ r ≤ p then μ(p, r, h) is a probability measure with support contained in [0, ∞). If p ≤ 0, p − h ≤ r ≤ 0 then μ(p, r, h) is a probability measure which is the reflection of μ(h − p, −r, h).

2. If either 0 ≤ 2r ≤ p, r + h ≤ p or p ≤ 2r + h, p ≤ r ≤ 0 then μ(p, r, h) is ⊞-infinitely divisible.

3. If 0 ≤ r ≤ p − h, t > 0 then ν(p, r, t, h) is a probability measure with a compact support contained in [0, +∞). If p ≤ h + r ≤ 0, t < h then ν(p, r, t, h) is a probability measure which is the reflection of ν(h − p, −h − r, h − t, h) In particular, if either 0 ≤ r ≤ p − h or p ≤ h + r ≤ 0 then the sequence \{\binom{mp+r}{m}_{h}\}_{m=0}^{∞} is positive definite. □

We conclude with some remarks on the probability measure ν_{0} := ν(1, 0, 1, 0), for which the moments are ν_{0}(x^{m}) = \binom{m}{m}_{0} = \frac{m^{m}}{m!} . From (4.9), (6.19) we have

(6.28)  S_{ν_{0}}(z) = e^{\frac{−z}{1+z}} ,
(6.29)  R_{ν_{0}}(ze^{\frac{−z}{1+z}} ) = z,
(6.30)  M_{ν_{0}}(\frac{z}{1 + z} e^{\frac{−z}{1+z}} ) = 1 + z.

Theorem 6.1. The sequence \{\frac{m^{m}}{m!} \}_{m=0}^{∞} is positive definite and the corresponding probability measure ν_{0} has compact support contained in [0, e]. Moreover, ν_{0} is ⊠-infinitely divisible.

Proof. First observe that lim_{m→∞} \sqrt[m]{\frac{m^{m}}{m!}} = e, which implies that the support of ν_{0} is contained in [0, e]. Now we recall (see Theorem 3.7.3 in [2]) that a probability measure μ with support contained in [0, ∞) is ⊠-infinite divisible if and only if the function Σ_{μ}(z) := S_{μ}(z(1 − z)^{−1}) can be expressed as Σ_{μ}(z) = e^{v(z)}, where v : ℂ \ [0, ∞) ↦ ℂ is analytic, satisfies v(\overline{z}) = \overline{v(z)} and maps the upper half-plane ℂ^{+} into the lower half-plane ℂ^{−}. In our case Σ_{ν_{0}}(z) = e^{−z} and the function v(z) = −z does satisfy these assumptions. □

Let us briefly reconstruct the way we have obtained the measure ν_{0}. We started with π̃ = μ(2, 1, 1), the free Poisson measure. Then

μ(p, h, h) = D_{h}μ(p/h, 1, 1) = D_{h} (π^{⊠ \frac{p}{h} −1}) ,

so putting h = 1/n, p = 1 and using (6.24) with t = 1/h = n we have

(6.31)  (D_{\frac{1}{n}} (π^{⊠n−1}))^{⊎n} ⟶ ν_{0},  with n → ∞,

where the convergence here means that the m-th moment of (D_{\frac{1}{n}} (π^{⊠n−1}))^{⊎n}

tends to \frac{m^{m}}{m!} . Note also that from (6.29) one can calculate free cumulants of ν_{0}: r_{1} = 1, r_{2} = 1, r_{3} = \frac{1}{2}, r_{4} = \frac{−1}{3}. Since r_{4} < 0, the measure ν_{0} is not ⊞-infinitely divisible.

References

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[3] M. Bożejko, J. Wysoczański, Remarks on t-transformations of measures and convolutions, Annales Institute Henri Poincaré, 37/6 (2001) 737–761.

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[6] T. Hasebe, On monotone convolution and momotone infinite divisibility, arXiv:1002.3430v2.

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[9] N. Muraki, Monotonic convolution and monotonic Levy-Hincin formula, preprint 2000.

[10] N. Muraki, Monotonic independence, monotonic central limit theorem and monotonic law of small numbers, Inf. Dim. Anal. Quantum Probab. Rel. Topics 4 (2001) 39–58.

[11] A. Nica, R. Speicher, Lectures on the Combinatorics of Free Probability, Cambridge University Press, 2006.

[12] K. A. Penson, A. I. Solomon, Coherent states from combinatorial sequences, Quantum theory and symmetries, Kraków 2001, World Sci.Publ., River Edge, NJ, 2002, 527–530.

[13] J. Riordan, Combinatorial identities, Willey 1968.

[14] R. Speicher, R. Woroudi, Boolean convolution, Fields Institute Communications 12 (1997), 267–279.

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Wojciech Młotkowski Mathematical Institute University of Wrocław Pl. Grunwaldzki 2/4 50-384 Wrocław Poland mlotkow@math.uni.wroc.pl 956

Documenta Mathematica 15 (2010)