Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
2020, Vol. 56, No. 1, 296–325
https://doi.org/10.1214/19-AIHP962© Association des Publications de l’Institut Henri Poincaré, 2020

www.imstat.org/aihp

Some properties of the free stable distributions
Takahiro Hasebea , Thomas Simonb and Min Wangba Department of Mathematics, Hokkaido University, North 10 West 8, Kita-Ku, Sapporo 060-0810, Japan. E-mail: thasebe@math.sci.hokudai.ac.jp
b Laboratoire Paul Painlevé, Université de Lille, Cité Scientifique, 59655 Villeneuve d’Ascq Cedex, France.

E-mail: thomas.simon@univ-lille.fr; min.wang@math.univ-lille1.fr
Received 3 May 2018; revised 28 November 2018; accepted 29 January 2019

Abstract. We investigate certain analytical properties of the free α-stable densities on the line. We prove that they are all classicallyinfinitely divisible when α ≤ 1 and that they belong to the extended Thorin class when α ≤ 3/4. The Lévy measure is explicitly
computed for α = 1, showing that free 1-stable distributions are not in the Thorin class except in the drifted Cauchy case. In thesymmetric case we show that the free stable densities are not infinitely divisible when α > 1. In the one-sided case we prove, refining
unimodality, that the densities are whale-shaped, that is their successive derivatives vanish exactly once on their support. We also
derive several fine properties of spectrally one-sided free stable densities, including a detailed analysis of the Kanter random variable,
complete asymptotic expansions at zero, and several intrinsic features of whale-shaped functions.
Résumé. Nous étudions certaines propriétés analytiques des densités α-stables libres sur la droite. Nous montrons qu’elles sont classiquement infiniment divisibles pour α ≤ 1 et qu’elles appartiennent à la classe de Thorin étendue pour α ≤ 3/4. La mesure de Lévy est
calculée explicitement pour α = 1 et ce calcul entraîne que les lois 1-stables libres n’appartiennent pas à la classe de Thorin, sauf dansle cas de la loi de Cauchy avec dérive. Dans le cas symétrique, nous montrons que les densités α-stables libres ne sont pas infinimentdivisibles quand α > 1. Dans le cas de signe constant nous montrons que les densités stables libres ont une courbe en baleine, autrement
dit que leurs dérivées successives ne s’annulent qu’une seule fois sur leurs supports, ce qui constitue un raffinement de l’unimodalité.
Nous établissons enfin plusieurs propriétés précises des densités stables libres spectralement de signe constant, parmi lesquelles une
analyse détaillée de la variable aléatoire de Kanter, des expansions asymptotiques complètes en zéro, ainsi que plusieurs propriétés
intrinsèques des courbes en baleine.
MSC: 33E20; 46L54; 60E07; 62E15
Keywords: Free stable distribution; Infinite divisibility; Shape of densities; Wright function

1. Introduction
In this paper, we investigate certain properties of real free stable random variables. We say that a real random variable X
is free stable, if for any a, b > 0 there exists c > 0, d ∈ R such thatdaX1 + bX2 = cX + d,(1)

where X1 , X2 are free independent copies of X. As in the classical framework, when X is not constant it turns out that
there exist solutions to (1) only if c = (a α + bα )1/α for some fixed α ∈ (0, 2] which is called the stability parameter.
We will be mostly concerned with free strictly stable densities, which correspond to the case d = 0. Every free strictly
stable distribution turns out to be equivalent to a distribution whose Voiculescu transform is of the form
φα,ρ (z) = −eiπαρ z−α+1 ,(z) > 0,

where (α, ρ) belongs to the following set A of admissible parameters: A = α ∈ (0, 1], ρ ∈ [0, 1] ∪ α ∈ (1, 2], ρ ∈ [1 − 1/α, 1/α] .

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297

Above, we have used the standard terminology that two measures μ, ν on the line are equivalent if there exist real numbers
a > 0, b ∈ R such that μ(S) = ν(aS + b) for every Borel set S. We refer e.g. to [34] for some background on the free
additive convolution, to [11] for the original solution to the equation (1), and to the introduction of [24] for the above
parametrization (α, ρ), which mimics that of the strict classical framework. Let us also recall that free stable laws appear
as limit distributions of spectra of large random matrices with possibly unbounded variance – see [8,16], and that their
domains of attraction have been fully characterized in [9,10]. In the following, we will denote by Xα,ρ the random variable
whose Voiculescu transform is given by (2), and set fα,ρ for its density. The analogy with the classical case extends tothe fact, observed in Corollary 1.3 of [24], that with our parametrization one has
P[Xα,ρ ≥ 0] = ρ.
For this reason, we will call ρ the positivity parameter of the free strictly random variable Xα,ρ . Clearly one has P[Xα,ρ ≤d0] = 1 − ρ and the Voiculescu transform also shows that Xα,ρ = −Xα,1−ρ . In this paper, some focus will be put on the
one-sided case and we will use the shorter notations Xα,1 = Xα and fα,1 = fα . Throughout, the random variable Xα,ρwill be mostly handled as a classical random variable via its usual Fourier, Laplace and Mellin transforms, except for a
few situations where the free independence is discussed.
Several analytical properties of free stable densities have been derived in the Appendix to [10], where it was shown inparticular that they can be expressed in closed form via the inverse of certain trigonometric functions. It is also indicatedin [10] that every free stable distribution of stability index α = 1 is equivalent to a free strictly stable distribution of thesame index. The density fα,ρ turns out to be a truly explicit function in three specific situations only, which is again
reminiscent of the classical case:√
4−x 2
for x ∈ [−2, 2], (semi-circular density),• f2,1/2 (x) = 2π√

4x−1
• f1/2 (x) = 2πx2 for x ≥ 1/4, (inverse Beta density),• f1,ρ (x) = π(x 2 +2sin(πρ)
for x ∈ R, (standard Cauchy density with drift).
cos(πρ)x+1)

The study of fα,ρ was carried on further in [22,24] where, among other results, several factorizations and series
representations were obtained. Our purpose in this paper is to deduce from these results several new and non-trivial
properties. Our first findings deal with the infinite divisibility of Xα,ρ . Since this random variable is freely infinitely
divisible (FID), it is a natural question whether it is also classically infinitely divisible (ID).
Theorem 1. One has(a) For every α ∈ (0, 1] and ρ ∈ [0, 1], the random variable Xα,ρ is ID.
(b) For every α ∈ (1, 2], the random variable Xα,1/2 is not ID.
Above, the non ID character of X2,1/2 is plain from the compactness of its support. Observe also that by continuityof the law of Xα,ρ in (α, ρ) and closedness in law of the ID property – see e.g. Lemma 7.8 in [38], for every α ∈ (1, 2)
there exists some (α) > 0 such that Xα,ρ is not ID for all ρ ∈ [1/2 − (α), 1/2 + ε(α)]. We believe that one can takeε(α) = 1/α − 1/2, that is our above result is optimal with respect to the ID property. Unfortunately, we found no evidence
for this fact as yet – see Remark 3 for possible approaches.
As it will turn out in the proof, for α ≤ 1 the ID random variables Xα,ρ have no Gaussian component. A natural
question is then the structure of their Lévy measure. We will say that the law of a positive ID random variable is a
generalized Gamma convolution (GGC) if its Lévy measure has a density ϕ such that xϕ(x) is a completely monotonic(CM) function on (0, +∞). There exists an extensive literature on such positive distributions, starting from the seventies
with the works of O. Thorin. The denomination comes from the fact that up to translation, these laws are those of therandom integrals ∞a(t) d t ,0where a(t) is a suitable deterministic function and { t , t ≥ 0} is the standard Gamma subordinator. We refer to [12] for acomprehensive monograph with an accent on the Pick functions representation and to the more recent survey [25] for theabove Wiener–Gamma integral representation, among other topics. See also Chapters 8 and 9 in [39] for their relationshipwith Stieltjes functions. In Chapter 7 of [12], this notion is extended to distributions on the real line. Following (7.1.5)
therein, we will say that the law of a real ID random variable is an extended GGC if its Lévy measure has a density ϕsuch that xϕ(x) and xϕ(−x) are CM as a function of x on (0, +∞). In order to simplify our presentation, we will also
use the notation GGC for extended GGC.

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Theorem 2. For every α ∈ (0, 3/4] and ρ ∈ [0, 1], the law of Xα,ρ is a GGC.
Contrary to the above, we think that this result is not optimal and that the random variable Xα,ρ has a GGC law atleast for every α ∈ (0, 4/5] and ρ ∈ [0, 1] – see Conjecture 1. During our proof, we will see that for every α, ρ < 1 the
GGC character of Xα,ρ is a consequence of that of Xα . Unfortunately this simpler question, which is connected to thehyperbolically completely monotonic (HCM) character of negative powers of the classical positive stable distribution, israther involved. Moreover, we will see in Corollary 1 that the law of Xα is not a GGC for α close enough to 1.
Our next result deals with the case α = 1. According to the Appendix of [10], every free 1-stable distribution isequivalent to a unique distribution whose Voiculescu transform writesφρ (z) = −2ρi −2(1 − 2ρ)log z,π(z) > 0,for some ρ ∈ [0, 1]. By (2), this means that a free 1-stable distribution is equivalent to the law of the free independentsum
dCa,b = aX1,1/2 + bT,
for a ≥ 0, b ∈ R, where T has Voiculescu transform − log z and will be called henceforth the exceptional free 1-stablerandom variable. More precisely, for any a > 0 and b = 0, the random variable Ca,b is equivalent to the 1-free non strictly
stable random variable whose Voiculescu transform is φρ , where ρ = 1/2 is determined bymin(ρ, 1 − ρ)a=·πb
1 − 2ρ
The case b = 0 corresponds to ρ = 1/2 and to the 1-free symmetric strictly stable random variable, which is the standard
Cauchy random variable: one sees from (2) that φ1/2 is the Voiculescu transform of X1,1/2 . Observe also that φ0 is the
Voiculescu transform of π2 (T + log(π/2)) whereas φ1 is that of − π2 (T + log(π/2)). Notice finally that in the above
parametrization of free 1-stable distributions, the parameter ρ is not a positivity parameter. Actually, as in the classicalframework there does not seem to exist a closed formula for P[Ca,b ≥ 0] when b = 0.
The density of Ca,b can be retrieved from Proposition A.1.3 of [10], in an implicit way. In this paper, taking advantage
of a factorization due to Zolotarev for the exceptional classical 1-stable random variable, we obtain the following explicit
result.
Theorem 3. The random variable Ca,b is ID without Gaussian component and with Lévy measure



−1 
1 a
|b−1 x|e−2|b x|
+
|b|
1
−
1
1
dx,
{x
=
0}
{bx<0}−1x2 π
1 − e−|b x|
where the second term is assumed to be zero if b = 0.
This computation implies that the random variable Ca,b is self-decomposable (SD) and that the associated Lévy process has CM jumps, but that its law is not a GGC except for b = 0 – see Remark 7. A key-tool for the proof is an identity
connecting T and the free Gumbel random variable – see Proposition 2, providing an analogue of Zolotarev’s factorizationin the free setting, and which is interesting in its own right.
Our last main result concerns the shape of the densities fα,ρ . It was shown in the Appendix to [10] that the latterare analytic on the interior of their support, and strictly unimodal i.e. they have a unique local maximum. These basic
properties mimic those of the classical stable densities displayed in the monograph [44]. A refinement of strict unimodality
was recently investigated in [29], where it is shown that all classical stable densities are bell-shaped (BS), that is their nthderivative vanishes exactly n times on the interior of their support, as is the case for the standard Gaussian density. The
freely strictly 1-stable density f1,ρ is BS, but it is visually clear that this property is not fulfilled neither by f2,1/2 nor by
f1/2 . Let us introduce the following alternative refinement of strict unimodality.
Definition. A non-negative function f on R is said to be whale-shaped if its support is a closed half-line, if it is smoothin the interior of its support and vanishes at both ends of its support, and if
x ∈ Suppf, f (n) (x) = 0 = 1
for every n ≥ 1.

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299

Fig. 1. The free positive 1/2-stable density (WS).

Fig. 2. The free symmetric 1-stable density (BS).

The denomination comes from the visual aspect of such functions – see Figure 1 and compare with the visual aspect ofa bell-shaped density given in Figure 2. We will denote by WS the whale-shaped property and set WS+ (resp. WS− ) for
those whale-shaped functions whose support is a positive half-line [x0 , +∞) for some x0 ∈ R, resp. a negative half-line(−∞, x0 ]. Observe that if f ∈ WS+ , then x → f (−x) belongs to WS− . It is easy to see that if f ∈ WS+ has support[x0 , +∞), then f is positive on (x0 , +∞), f (n) (+∞) = 0 and (−1)n−1 f (n) (x0 +) > 0 for every n ≥ 1. In particular, theclass WBS0 introduced in the main definition of [40] corresponds to those WS+ functions whose support is (0, +∞).
Observe finally that the sequence of vanishing places of the successive derivatives of a function in WS+ increases, byRolle’s theorem. Other less immediate properties of WS functions will be established in Section 3.3.
Theorem 4. One has(a) For every α ∈ (0, 1), the density fα is WS+ .
(b) For every α ∈ (0, 3/4] and ρ ∈ (0, 1), the density fα,ρ is BS.
(c) The density of T is WS− .
(d) For a = 0 and for b = 0 or ab−1 ∈ πZ, the density of Ca,b is BS.
This result leaves open the question of the exact shape of the density for all α > 1. Observe that the limiting case α = 2
is rather peculiar since it can be elementally shown that its even derivatives never vanish, whereas its odd derivatives
vanish only once and at zero. But since the BS property is not closed under pointwise limits, it might be true that fα,ρis BS whenever its support is R. On the other hand, in spite of Theorem 4(c) we think that for α ∈ (1, 2) the visuallywhale-shaped density fα,1/α , whose support is a negative half-line, is not WS− . Indeed, we will see in Proposition 10 that
otherwise it would be ID, and we know that this is not true at least for α close enough to 2.
Our four theorems are proved in Section 2. In the last section, we derive further results related to the analysis of the
one-sided free stable densities. First, we analyze in more detail the Kanter random variable Kα , which plays an importantrole in the proof of all four theorems. The range α < 1/5 is particularly investigated, and two open questions made in [26]and [14] are answered in the negative. We also derive the full asymptotic expansion of the densities of Xα , Xα,1−1/α and1 − T at the left end of their support, completing the series representation at infinity which were given in (1.16) of [24].
Finally, we display some properties of whale-shaped functions and densities.
ddNotation. Throughout, unless otherwise explicitly stated, in any factorization of the type X = Y + Z or X = Y × Z,
the random variables Y, Z on the right-hand side will be assumed to be classically independent.

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T. Hasebe, T. Simon and M. Wang

2. Proofs of the main results
2.1. Preliminaries
The proofs of all four theorems rely on the following result by Haagerup and Möller [22] who, using a general propertyof the S-transform, have computed the fractional moments of Xα . They obtain 
E Xsα =

(1 − s/α)(2 − (1/α − 1)s) (1 − s)

 

1
(1 − s/α)=×
1 + (1 − 1/α)s)(1 − s) (1 − (1/α − 1)s)

for s < α. Identifying the two factors, we get the following multiplicative identity in lawdXα = U1−1/α × Kα ,(3)where U is uniform on (0, 1) and Kα is the so-called Kanter random variable. The latter appears in the following factorization due to Kanter – see Corollary 4.1 in [27]:dZα = L1−1/α × Kα ,(4)where L has unit exponential distribution and Zα is a classical positive α-stable random variable with Laplace transformαE[e−λZα ] = e−λ and fractional moments 
E Zsα =

(1 − s/α)(1 − s)for s < α. Observe that the random variable Kα has fractional moments
 
E Ksα =

(1 − s/α)(1 − (1/α − 1)s) (1 − s)(5)for s < α, and in particular a support [bα , +∞) which is bounded away from zero, with
1/n

1
,
bα−1 = α −1 (1 − α)1− α = lim E K−nαn→+∞by Stirling’s formula. The density of Kα is explicit for α = 1/2, with
1

d

K1 =
2

4 cos2 (πU/2)d=1
4B 1 , 1

2 2

and where, here and throughout, Ba,b stands for a standard β(a, b) random variable with density(a + b) a−1
x (1 − x)b−1(a) (b)on (0, 1). Plugging this in (3) yields easily
d

X1 =
2

1
4B 1 , 3

2 2

and we retrieve the aforementioned closed expression of f1/2 . Several analytical properties of the density of Kα havebeen obtained in [26]. In particular, Corollary 3.2 in [26] shows that the density of Kα − bα is CM, a fact which we will
use repeatedly in the sequel.
Remark 1.
(a) Specifying Haagerup and Möller’s result to the negative integers yields

 −1


1
nα + 1E X−n, n ≥ 0.
=αnnα −1 + 1

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301

The latter is a so-called Fuss–Catalan sequence, and it falls within the scope of more general positive-definite sequences−1
studied in [30,32]. With the notations of these papers, one has Xα = W1/α,1. This implies that fα can be written explicitly,dalbeit in complicated form, for α = 1/3 and α = 2/3 – see (40) and (41) in [32]. It is also interesting to mention that X−1
12

has Marchenko–Pastur (or free Poisson) distribution, with density
4−x
x

1
2π

on (0, 4). More generally, Proposition A.4.3 in [10] – see also (8) in [32] – shows that X−1
1 is distributed for each n ≥ 2
nas the (n − 1)th free multiplicative convolution power of the Marchenko–Pastur distribution.
(b) The negative integer moments of Kα are given by the simple binomial formula −1 


nα
=
,
E K−n
α
n

n ≥ 0.

This shows that the law of K−1
α is of the type studied in [31], more precisely it is ν(1/α, 0) with the notations therein. By
Gauss’s multiplication formula – see e.g. Theorem 1.5.2 in [1] – and Mellin inversion, this also implies the identity−2
K−1
1 = K 2 = 27B 1 , 2 (1 − B 1 , 2 )
d

d

3

3 3

3

3 3

in terms of a single random variable B 1 , 2 . In particular, the density of Kα can be written in closed form for α = 1/3
3 3
and α = 2/3 as a two-to-one transform of the density of B 1 , 2 – see also Theorems 5.1 and 5.2 in [31]. As seen above,
3 3

K−1
1 = 4B 1 , 1 is arc-sine distributed, with density
d

2 2

2

1
π x(4 − x)√

on (0, 4). It is well-known that this is the distribution of the rescaled free independent sum of two Bernoulli random variables with parameter 1/2. It turns out that in general, K−1
1 is distributed for each n ≥ 2 as the (n − 1)th free multiplicativenconvolution power of a free Bernoulli process at time n/(n − 1) – see (6.9) in [31].
(c) The random variable Kα can be expressed as the following explicit deterministic transformation of a single uniform
variable U on (0, 1):
1−α

d

Kα =

sin(παU) sin α (π(1 − α)U)
1

.

(6)

sin α (πU)

This is Kanter’s original observation – see Section 4 in [27], and it will play an important role in the proof of Theorem 3.
Notice that the deterministic transformation involved in (6) appears in the implicit expression of the densities fα , which
is given in the second part of Proposition A.I.4 in [10] – see also (11) in [32] for the case when α is the reciprocal of an
integer. There does not seem to exist any computational explanation of this fact. We refer to equation (1) in [18], and also
to Proposition 1 and Proposition 2 therein for further results on this transformation.
2.2. Proof of Theorem 1
2.2.1. The case α ≤ 1
We begin with the one-sided situation ρ = 1. We deduce from (3) and the multiplicative convolution formula that, for any
x > 0,fα (x + bα ) =

α
1−α
1
1−α

αbα
=
1−α

 1+ x

bα

1

 1
0

2−α

y − 1−α fKα y −1 (x + bα ) dybα (1 − t)x
f
dt.
bα +
2−α Kαbα + tx(bα + tx) 1−αx302

T. Hasebe, T. Simon and M. Wang

On the one hand, for every t ∈ (0, 1), the functionx→(1 − t)xbα + tx

is a Bernstein function, using the terminology of [39]. On the other hand, by the aforementioned Corollary 3.2 in [26],the function z → fKα (bα + z) is CM. It follows from Theorem 3.7 in [39] that g ◦ f ∈ CM for every g ∈ CM and f ∈Bernstein. This implies that the functionbα (1 − t)x
x → fKα bα +bα + tx
is CM, and so is
− 2−α
1−α

x → (bα + tx)
bα (1 − t)xfKα bα +bα + tx

as the product of two CM functions. Integrating in t shows that x → x −1 fα (x + bα ) is CM on (0, ∞) and it is a standard
and easy fact following from Theorem 1.4 in [39] that this implies the factorizationdXα = bα +  2 × X̃αfor some positive random variable X̃α where, here and throughout,  t stands for a standard (t) random variable withdensityx t−1 e−x(t)on (0, +∞). We can finally deduce from Kristiansen’s well-known  2 theorem – see [28] – that Xα is ID.
To handle the two-sided situation ρ ∈ (0, 1), we appeal to the following identity in law which was observed in [24] –see (2.8) therein:dXα,ρ = X1,ρ × Xα .

(7)
(1,ρ)

Since X1,ρ has a drifted Cauchy law and since the underlying Cauchy process {Xtidentity transforms into
d

(1,ρ)

Xα,ρ = XXα ,, t ≥ 0} is 1-self-similar, the latter(8)

which is a Bochner’s subordination identity. By e.g. Theorem 30.1 in [38], this finally shows that Xα,ρ is ID for everyα ∈ (0, 1] and ρ ∈ [0, 1].
Remark 2.
(a) The above proof shows thats→

E[(Xα − bα )s ](2 + s)is the Mellin transform of some positive random variable. On the other hand, it seems difficult to find a closed formulafor the Mellin transform E[(Xα − bα )s ], except in the case α = 1/2 where 21−2s(3/2 + s) (1/2 − s),
E (X 1 − b 1 )s =22π(b) We believe that Xα − bα is as→s ∈ (−3/2, 1/2).

3/2 -mixture for every α ∈ (0, 1), that isE[(Xα − bα )s ](3/2 + s)

is the Mellin transform of some positive random variable. This more stringent property is actually true for α ≤ 3/4, as aconsequence of the above proof and Theorem 2 – see Remark 10(b).

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303

2.2.2. The case α > 1 and ρ = 1/2
We first derive a closed expression for the Fourier transform of Xα,ρ , which has independent interest. It was alreadyobtained as Theorem 1.8 in [24] in a slightly different manner. Our proof is much simpler and so we include it here.
Introduce the so-called Wright functionzn(b + an)n!

φ(a, b, z) =n≥0with a > −1, b ∈ R, and z ∈ C. This function was thoroughly studied in the [42,43] for various purposes, and is referenced
in Formula 18.1(27) in the encyclopedia [21]. It will play a role in other parts of the present paper.
Lemma 1. For all (α, ρ) ∈ A, one has
E eitXα,ρ = φ α − 1, 2, −(it)α e−iπαρ sgn(t) ,t ∈ R.

Proof. The case α = 1 is an easy and classic computation, since X1,ρ has a drifted Cauchy distribution and φ(0, 2, z) =dez . When α = 1, we first observe that since Xα,ρ = −Xα,1−ρ , it is enough to consider the case t > 0.
Let Zα,ρ be a classical strictly stable random variable having characteristic function
α −iπαρ,E eixZα,ρ = e−(ix) e

x > 0.
d1−1/αUsing the multiplicative factorization Zα,ρ = Xα,ρ ×  2of Corollary 1.5 in [24], we obtain
 ∞
 ∞
α
 1−1/α X  1−1/α X 
2αα −iπαρ1−αα,ρ dt = x 1−αα,ρ dt=te−t E eixt
te−tx
E eit
e−(ix) e00for all x > 0. On the other hand, a straightforward term-by-term integration implies ∞α2αα −iπαρα1−αte−tx
φ α − 1, 2, − it 1−1/α e−iπαρ dt = e−(ix) e, x > 0.
x 1−α
0The result follows then by uniqueness of the Laplace transform.



We can now finish the proof of the case α > 1, ρ = 1/2, where the above lemma reads
E eitXα,1/2 = φ α − 1, 2, −|t|α , t ∈ R.
Applying Theorem 1 in [42] and some trigonometry, we obtain the asymptotic behaviour
φ α − 1, 2, −t α ∼ κα t −3/2 ecos(π/α)α(α−1)

1/α−1 tcos 3π/2α + sin(π/α)α(α − 1)1/α−1 t

as t → +∞, for some κα > 0. This implies that t → E[eitXα,1/2 ] vanishes on R, and hence cannot be the characteristic
function of an ID distribution – see e.g. Lemma 7.5 in [38].
Remark 3.
(a) It was recently shown in Theorem 1 of [5] that for any a, β > 0, the function φ(a, β, −z) has only positive zeroes
on C. Combined with Lemma 1, this entails that the function t → E[eitXα,ρ ] never vanishes on R for α > 1 and ρ = 1/2,
so that the above simple argument cannot be applied. Nevertheless, we conjecture that Xα,ρ is not ID for all α > 1 andρ ∈ [1 − 1/α, 1/α].
(b) When ρ = 1/α, Lemma 1 also gives the moment generating function




λα
E eλXα,1/α = φ α − 1, 2, λα =1+
, λ ≥ 0,λα,nn≥1where 0 < λα,1 < λα,2 . . . are the positive zeroes of φ(α − 1, 2, −z). Above, the product representation is a consequenceof the Hadamard factorization for the entire function φ(α − 1, 2, z) which is of order < 1 – see again Theorem 1 in [42],304

T. Hasebe, T. Simon and M. Wang

whereas the simplicity of the zeroes follows from the Laguerre theorem on the separation of zeroes for φ(α − 1, 2, z),which has genus 0. Consider now the random variable−1/αYα = b1/α − Xα,1/α = α(α − 1)1/α−1 − Xα,1/α ,whose support is (0, ∞) by Proposition A.1.2 in [10], and whose infinite divisibility amounts to that of Xα,1/α . Its logLaplace transform reads




λα
− log E e−λYα = α(α − 1)1/α−1 λ −log 1 +
λα,n
=

 ∞

1−e

−λα x

0



n≥1

(α − 1)1/α−1 x −1/α−
(1 − 1/α)e−λα,n x

n≥1



dx
,
x

(9)

where in the second equality we have used Frullani’s identity repeatedly and the well-known change of variable formula(1) p.viii in [39]. Putting everything together shows that Xα,1/α is ID if and only if the function on the right-hand side isBernstein. Unfortunately, this property seems difficult to check at first sight.
2.3. Proof of Theorem 2
2.3.1. The case ρ = 1 and α ∈ (0, 3/4]
Here, we need to show that the law of Xα is a true GGC. To do so, we first observe that by (3) and the main result of [13]which states that the independent product of two random variables having a GGC distribution has a GGC distribution, it
is enough to show that the law of Kα itself is a GGC. Alternatively, this fact is also a consequence of the formulaα  −λX  αλ 1−α ∞  −xK  1αα x α−1 dx,
Ee=Ee
λ ≥ 0,(10)1−α λwhich follows from (3) and of Theorem 6.1.1 and Properties (iv) and (xi) p.68 in [12]. To analyze the law of Kα , we will
use the identity in law
d

1

−1

α
,
Kα = K1−α(11)an easy consequence of (5) which shows that both random variables have the same fractional moments. Plugging (11)−αagain into (4) implies that the Laplace transform of K1−α is the survival function of the power transformation Zα1−α . In
other words, one has


 − α

Fα (x) := E e−xK1−α = P[L ≥ xK1−α ] = P Zα 1−α ≥ x ,x ≥ 0.

(12)

Setting Fα (x) for the function defined in (12), we next observe that since Kα has a CM density and support [bα , +∞),
this function Fα has by Theorem 9.5 in [39] an analytic extension on C \ (−∞, 0] which is given by ∞z θα (t)dt(13)
Fα (z) = exp − b1−α z +t +z t0
1
for some measurable function θα : (0, ∞) → [0, 1] such that 0 θα (t)t −1 dt < ∞. See also Theorem 51.12 in [38]. Applying now Theorem 9.10 (i) in [39], we see that the GGC property of K1−α is equivalent to the non-decreasing characterof θα on (0, ∞), and the following proposition allows us to conclude the proof of the case ρ = 1.
Proposition 1. The function θα has a continuous version on (0, ∞), which is non-decreasing for every α ∈ [1/4, 1).
Proof. The analysis of θα depends, classically, on the behaviour of Fα near (−∞, 0]. Assume for a moment that θα iscontinuous. For every r > 0 and δ ∈ (0, 1), we have, after some simple rearrangements, ∞Fα (reiπ(1−δ) )sin(π(1 − δ))θα (rt)= exp −2i sin(πδ)b1−α r +dt
1 + 2 cos(π(1 − δ))t + t 2
Fα (reiπ(δ−1) )0= exp − 2i sin(πδ)b1−α r + 2iπE θα r(X1,1−δ )+→ exp −2iπθα (r)Some properties of the free stable distributions305

as δ → 0 since X1,1−δ → 1 in law as δ → 0 and θα is bounded continuous. On the other hand, it follows from the third
expression of Fα in (12) and the first formula of Corollary 1 p.71 in [44], after a change of variable, that ∞1iπα α 1−α−iπα α 1−α dt
Fα (x) = 1 +e−t e−e t x − e−e t x, x > 0.
2iπ 0t
The analytic continuations of Fα near (−∞, 0] are then expressed, changing the variable backwards, as ∞1αα −2iπα duFα reiπ = 1 +e−ru eru − eru e
2iπ 0
u
and
Fα re−iπ = 1 +

1
2iπ

 ∞

e−ru eru eα 2iπα

− eru

α

0

du
= Fα reiπ .
uTherefore, we obtain
Fα (reiπ )= e−2iπηα (r)Fα (re−iπ )for every r > 0, with the notation
ηα (r) =1arg Fα re−iπ .
πSince Fα re−iπ

=

1
2π

 ∞

e−ru  eru − eru eαα 2iπα0du>0ufor every r > 0, the function ηα takes its values in [0, 1] and is clearly continuous. By construction, the functions t −1 ηα (t)and t −1 θα (t) have the same Stieltjes transform, and it follows by uniqueness that θα has a continuous version, which isηα .
It remains to study the monotonous character of ηα on (0, ∞). A first observation is that, expanding the exponentials inside the brackets and using Euler’s reflection formula – see e.g. Theorem 1.2.1 in [1], the following absolutely
convergent series representation holds:
Fα re−iπ =n≥0zn eiπnα= φ −α, 1, zeiπαn! (1 − nα)(14)with z = r 1−α . In particular, the functionr → r α−1  Fα re−iπ

=

1
π

n≥1

(nα) (n−1)(1−α)
rn!

is absolutely monotonous on (0, ∞), and the non-decreasing character of θα will hence be established as soon as r →r α−1 (Fα (re−iπ )) is non-increasing on (0, ∞). We use the representation ∞1
1−1/α uα 2iπα du−iπ Fα re=1+e−r
 eu e2π 0uand divide this last part of the proof into two parts.
• The case α ∈ [1/2, 1). If α = 1/2, we simply have (F1/2 (re−iπ )) ≡ 1. If α > 1/2 we rewrite, using again the firstpart of Corollary 1 p.71 in [44],



1 1
1 ∞ −r 1−1/α u
α −iπρα du Fα re−iπ = +e e−u e
1−
2 2
π 0
u
=



1
1 + P Zα,ρ ≤ r 1−1/α ,2

306

T. Hasebe, T. Simon and M. Wang

where ρ = 2 − 1/α ∈ (0, 1) and Zα,ρ is a real α-stable random variable with positivity parameter ρ as in Lemma 1.
Thus, (Fα (re−iπ )) decreases on (0, ∞) and r α−1 (Fα (re−iπ )) also decreases on (0, ∞), as required.
• The case α ∈ [1/4, 1/2). Contrary to the above, the argument is here entirely analytic. We considerαGα (r) =  Fα r α−1 e−iπ

=1+
=1+
=1+

1
2π
1
2π

 ∞

with gα,r (t) = t −1 ecot(2πα)t−r(sin(2πα))
because α ∈ [1/4, 1/2) and

α 2iπα

0

 ∞

1
2πα

e−ru  eu eduuecos(2πα)u −ru sin sin(2πα)uα
α

0

 ∞

du
u

gα,r (t) sin(t) dt
0−1/α t 1/α. Observe that the positive function t → gα,r (t) decreases on (0, +∞),

 (t)
gα,r
1 r
−1/α 1/α−1t< 0.
= cot(2πα) − − sin(2πα)
gα,r (t)tαFor every k ≥ 0 we have 2(k+1)πgα,r (t) sin(t) dt =

2kπ

 π

g(t + 2kπ) − g t + (2k + 1)πsin(t) dt > 0,0

so that Gα (r) > 1 for every r > 0. We next computer r Gα (r) = αr
Gα (r) + Gα (r)
α


 ∞
r
α
> αr α−1 1 −ecos(2πα)u −ru sin sin(2πα)uα du
2πα 0
 ∞
α
r
α
=
e−ru 2πα − ecos(2πα)u sin sin(2πα)uα du > 0,
2π 0
α





α−1

α

since 2πα > 1 ≥ ecos(2πα)u sin(sin(2πα)uα ) for every u > 0. Changing the variable backwards, this finally shows thatr → r α−1 (Fα (re−iπ )) decreases on (0, ∞).

Remark 4.
− α(a) The above argument shows that the survival function x → P[Zα 1−α ≥ x] is HCM for every α ≥ 1/4, with theterminology of [12]. A consequence of Corollary 2 is that this is not true anymore for α < 1/5, and we believe – see− αConjecture 1 – that the right domain of validity of this property is α ∈ [1/5, 1). The more stringent property that Zα 1−αis a HCM random variable for α ≤ 1/2 was conjectured in [14] and some partial results were obtained in [14,15]. In
Remark 9 below, we will see that this property is not true for α < 1/5.
(b) We do not know if the representation (13) holds for the Laplace transform of Xα . Since the latter is a 2 -mixture
we obtain, similarly as above,
E e−xXα = e−b1−α x ∞0να (dt)(x + t)2

for some positive measure να on [0, +∞). This representation would suffice if we could show that the generalized Stieltjes
functions on the right-hand side is the product of two standard Stieltjes functions, applying Theorem 6.17 in [39] as in the
proof of Theorem 9.5 therein. However, this is not true in general, for example when να is the sum of two Dirac ∞ masses.
Alternatively, because of (10) one would like to prove that if f has representation (13), then so has x → x f (y) dy.
This is true in the GGC case by Property (xi) p.68 in [12], but we were not able to prove this in general.

Some properties of the free stable distributions

307

2.3.2. The case ρ < 1 and α ∈ (0, 3/4]d
The case ρ = 0 follows from Xα,0 = −Xα . For ρ ∈ (0, 1) we appeal to (8), the previous case, and the Huff–Zolotarev
subordination formula which is given e.g. in Theorem 30.1 of [38]. Since the law of Xα is a GGC for α ≤ 3/4, its Laplacetransform reads ∞ −λX dx
−λx
αEe
= exp − bα λ +1−ekα (x)
x0
for some CM function kα . Formula (30.8) in [38] and the closed expression of the density of X1,ρ imply that the Lévy
measure να,ρ of Xα,ρ has densitykα (u)sin(πρ) ∞du2
πx + 2 cos(πρ)xu + u2
0


 ∞
kα (|x|u)sin(πρ) bα
du
+
=
π|x|
|x|
1 + 2 cos(πρ) sgn(x)u + u20ψα,ρ (x) = bα ψ1,ρ (x) +over R∗ , where the closed expression for ψ1,ρ can be deduced e.g. from Theorem 14.10 and Lemma 14.11 in [38]. Bothfunctions xψα,ρ (x) and xψα,ρ (−x) are hence CM on (0, ∞).
Remark 5. For α ∈ (0, 1), since bα > 0 and the ID random variable X1,ρ has no Gaussian component, the Huff–Zolotarev
subordination formula shows that Xα,ρ does not have a Gaussian component either, and that for ρ ∈ (0, 1) its Lévy
measure is such that|x|να,ρ (dx) = +∞.
|x|≤1

With the terminology of [38] – see Definition 11.9 therein, this means that the Lévy process associated with Xα,ρ is oftype C. This contrasts with the classical α-stable Lévy process which is of type B for α < 1. When ρ = 1 and α ≤ 3/4, the
GGC property shows that the Lévy process corresponding to Xα is of type B. We believe that this is true for all α ∈ (0, 1).
2.4. Proof of Theorem 3
It is well-known and easy to see from the Voiculescu transform
φ1,1/2 (z) = −1
that the free independent sum of X1,1/2 with any random variable is also a classical independent sum. Hence, the ID
character of Ca,b follows from that of T, which is a consequence of Theorem 1 and the convergence in law(1 − α)1−α − Xα d
−→ T
1−α

as α ↑ 1,(15)

the latter being easily obtained in comparing the two Voiculescu transforms. More precisely, since for any random variable
X having Voiculescu transform φX (z) and any p, q ∈ R with p = 0 the Voiculescu transform of pX + q is given by
q + pφX (z/p), we haveφ (1−α)1−α −Xα (z) = (1 − α)−α −1−α= (1 − α)1−α1
φX −(1 − α)z
1−α α

1 − z1−α→ log z = φT (z).
1−αThis concludes the first part of the theorem. Moreover, it is clear that neither X1,1/2 nor T, whose support is a half-line by
Proposition A.1.3 in [10], have a Gaussian component, and this property conveys hence to Ca,b . Finally, since the Lévy
measure of X1,1/2 is1
1{x=0}πx 2

308

T. Hasebe, T. Simon and M. Wangas seen in the above proof, we are reduced to show by independence and scaling that the Lévy measure of T has density
1|x|e−2|x|1−
1{x<0} .
1 − e−|x|x2
This last computation will be done in two steps. Consider the random variableW=sin(πU) πU cot(πU)
e
πUand the exceptional 1-stable random variable S characterized by 
E esS = s s , s > 0.
Proposition 2. One has the identitiesd

S = log L + log W andd

T = log U + log W.

Proof. We begin with the first identity. Using (4), we decompose



1 
(1 − α)1−α − Zα d
1 − L1− α(1 − α)1−α − Kα
= Kα ×
+
.
1−α
1−α
1−α

(16)

On the one hand, a comparison of the two moment generating functions yields(1 − α)1−α − Zα d
−→ S
1−α

as α ↑ 1.

On the other hand, the right-hand side of (16) is a deterministic transformation, depending on α, of (L, U) independent.
It is easy to see from (6) that

Kα ×

1

1 − L1− α
1−α



a.s.

−→ log L

as α ↑ 1.

To study the second term, we use the elementary expansions
sin(παU) = sin(πU) + (α − 1)πU cos(πU) + O (1 − α)31−α

sin α π(1 − α)U = 1 + (1 − α) log sin πU(1 − α) + O (1 − α)2 log2 (1 − α)1sin α (πU) = sin(πU) 1 + (1 − α) log sin(πU) + O (1 − α)2(1 − α)1−α = 1 + (1 − α) log(1 − α) + O (1 − α)2 log2 (1 − α) ,which, combined with (6), yield the almost sure asymptotics(1 − α)1−α − Kαsin(πU) πU cot(πU)= loge+ O (1 − α) log2 (1 − α) .
1−απU
Putting everything together completes the proof of the first identity. The second one is derived exactly in the same way,using (3) and (15).
Remark 6.
(a) The first identity in Proposition 2 is actually the consequence of an integral transformation due to Zolotarev – see(2.2.19) with β = 1 in [44]. We have offered a separate proof which is perhaps clearer, and which enhances the similarities
between the free and the classical case. Observe in particular the identityd

S = T + log  2 ,(17)

Some properties of the free stable distributions

309

which is reminiscent of Corollary 1.5 in [24]. The latter identity is also a consequence of Proposition 2 and the standard
identitiesdβLβ = Uβ ×  2(18)valid for every β ∈ R∗ and their limit as β → 0, which isdlog L = log U + log  2 .

(19)(b) It is interesting to look at these standard identities (18) and (19) in the context of extreme value distributions.
Indeed, the three classical extreme value distributions are Fréchet Lβ for β < 0, Weibull −Lβ for β > 0 and Gumbel− log L for β → 0, whereas the free counterparts are Uβ for β < 0, −Uβ for β > 0 and − log U for β → 0 according to
d

1−1/α

the classification of [7]. Besides, recall that a  2 factor also appears in the formula Zα,ρ = Xα,ρ ×  2
which connects
classical and free stable random variables.
(c) Recently Vargas and Voiculescu have introduced Boolean extreme value distributions [41]. The result is the Dagum
distribution, which is indexed by β > 0 and has density function
x 1/β−1β(1 + x 1/β )2on (0, ∞). Hence, the Dagum distribution is the law of
U

−1

−1

 β
L
=
,
L

β d

which is the independent quotient of two Fréchet distributions, and an example of the generalized Beta distribution of thesecond kind (GB2). On the other hand, by Proposition 4.12 (b) in [2], the Boolean α-stable distribution has for α ≤ 1 thelaw of the independent quotient
Zα,ρ
Zαand it is interesting to notice that by Zolotarev’s duality – see (3.3.16) in [44] – and scaling, the positive part of thisrandom variable is distributed as


Zαρ
Zαρ

ρ

 1
L α
−→
L
d

as ρ → 0.

Finding an interpretation about why such quotients appear in those two Boolean cases is left to future work.
(d) The second identity in Proposition 2 can be rewritten asdeT = U × W.
In [3], it is pointed out that the law of eT is the Dykema–Haagerup distribution, which appears as the eigenvalue distribution of A∗N AN as N → ∞, where AN is an N × N upper-triangular random matrix with independent complex Gaussian
entries – see [20].
Our second step is to compute the Mellin transform of W.
Proposition 3. For all s > 0, one has


 
 ∞
 
ss
x
dx
E Ws =
1 − e−sx 1 − x= exp s −.
(1 + s)e−1x20Proof. The first equality follows from     s s = E esS = E Ls E Ws = (1 + s)E Ws ,s > 0,

310

T. Hasebe, T. Simon and M. Wanga consequence of the first identity in Proposition 2. To get the second one, we proceed as in the proof of Lemma 14.11 of[38] and start from the standard Frullani identity
 ∞ −xe − e−sxdx,
log s =
x0
which transforms, dividing the integral at 1 and making an integration by parts, into
 ∞

 ∞
dx
dx
−sx
−x
s log s =ee − 1 + x1{x≤1} 2 .
− 1 + sx1{x≤1} 2 − sxx00On the other hand, it is well-known – see e.g. Formulae 1.7.2(21) and 1.9(1) in [21] – that
 ∞dx
e−sx − 1 + sx,
log (1 + s) = −γ s +x − 1)x(e0where γ = −  (1) is Euler’s constant. Putting everything together yields


 ∞
 
x
dx
e−sx − 1 + sx1{x≤1} 1 − x
log E Ws = cs +e − 1 x2
0


 ∞
x
dx
= c̃s −,
1 − e−sx 1 − x
e − 1 x20
where c, c̃ are two constants to be determined. But it is clear that c̃ is the right end of the support of log W which weknow, by Remark 6(c), to be one. Alternatively, one can use Binet’s formula ∞  −x
e−x
e
γ=
−
dx,
1 − e−xx0which is 1.7.2(22) in [21] for z = 1, and rearrange the different integrals, to retrieve c̃ = 1. This completes the proof. We can now finish the proof of Theorem 3. Putting together Propositions 2 and 3, we get
 
 
 
 
log E esT = log E Us + log E Ws = − log(1 + s) + log E Ws
 ∞xe−2x dx
1 − e−sx 1 −,=s−
1 − e−x x 20where the third equality follows from rearranging Frullani’s identity and the second equality in Proposition 3. All of thisshows that the ID random variable T has support (−∞, 1] – in accordance with Proposition A.1.3 in [10], and that its
Lévy measure has density
1|x|e−2|x|1−
1{x<0}
1 − e−|x|x2
as required.
Remark 7.
(a) The first equality in Proposition 3 shows that W has the distribution ν0 studied in Theorem 6.1 of [30]. This
distribution also appears in Sakuma and Yoshida’s limit theorem – see [37]. Finally, combining this equality and thesecond identity in Proposition 2 implies
 
E esT =

ss
(2 + s)for all s > 0, which was previously obtained in [3] by other methods.
(b) It is easy to see that the function
x→

1
1
−
x ex − 1

Some properties of the free stable distributions

311

decreases from 1/2 to zero on (0, ∞). By Corollary 15.11 in [38], this shows that log W is SD. A further computation
yields
  ∞

1
x
(20)
e−ux u − [u] du, x > 0.
=1−
ex − 1x2
0This implies that the Lévy process associated to log W has CM jumps. By Theorem 3, so does the Lévy process associatedto T, whose Lévy measure has density
  ∞

e−|x|
1
|x|
e−u|x| u − [u − 1]+ du, x < 0.
+ 2 1 − |x|
=
|x|
e −1
x
0
By Theorem 51.12 in [38], the latter computation also implies that the law of the positive random variable 1 − log W has
a CM density.
(c) Making an integration by parts in (20) yields


  ∞

1
x
−ux
e
δn (du) ,
1− x
=
du −
x
e −1
0
n≥1

where δ stands for the Dirac mass. By (7.1.5) in [12], this implies that the law of log W is not a GGC, and the same istrue for T because


  ∞

1
xe−2x
−ux
e
δ
(du)
.
1−
=
du
−
n
x
1 − e−x0n≥2

By (15) and Theorem 7.1.1 in [12], this yields the following negative counterpart to Theorem 2.
Corollary 1. There exists α0 < 1 such that for every α ∈ (α0 , 1), the law of Xα is not a GGC.
This also implies that there is a function δ : (α0 , 1) → [0, 1) such that Xα,ρ is not a GGC for α ∈ (α0 , 1) and ρ ∈ [δ(α), 1].
Observe on the other hand that it does not seem possible to apply our methods to Xα,ρ with a fixed ρ ∈ (0, 1). Indeed, as
in the classical case, the possible limit laws of affine transformations of Xα,ρ with ρ ∈ (0, 1) fixed and α → 1 are given
only in terms of X1,ρ , whose law is a GGC.
2.5. Proof of Theorem 4
2.5.1. The one-sided caseBy (3) and Corollary 3.2 in [26], we have the independent factorisation
Xα = bα U−1/β (1 + X),dwhere 1/β = 1/α − 1 and X has a CM density on (0, ∞). We will now show the WS property for all positive randomvariables of the type
Y = U−1/β (1 + X) − 1
with β > 0 and X having a CM density on (0, ∞). Setting f, g for the respective densities of Y, X, the multiplicative
convolution formula shows that
 x
 1
β
β
β
f (x) =
(y
+
1)
g(y)
dy
=
(y + 1/x)β g(xy) dy(x + 1)β+1 0(1 + 1/x)β+1 0
for every x > 0. In particular, one has f (0+) = f (+∞) = 0. Moreover, the first equality and an induction on n imply
that f is smooth with(x + 1)f (n+1) (x) = βg (n) (x) − (β + n + 1)f (n) (x)(21)

for every n ≥ 0. Hence, we also have f (n) (+∞) = 0 for all n ≥ 0 and a successive application of Rolle’s theorem yields# x ∈ (0, ∞)|f (n) (x) = 0 ≥ 1

312

T. Hasebe, T. Simon and M. Wang
(1)

(2)

for every n ≥ 1. Fix now n ≥ 1 and suppose that there exist 0 < xn < xn < ∞ such thatf (n) xn(1) = f (n) xn(2) = 0.
By (21) and the complete monotonicity of g, we have(−1)n f (n+1) xn(i) > 0(3)

for i = 1, 2. An immediate analysis based on the intermediate value theorem shows then that there must exist xn ∈(1) (2)(xn , xn ) withf (n) xn(3) = 0 and (−1)n f (n+1) xn(3) ≤ 0,
which is impossible again by (21) and the complete monotonicity of g. All in all, we have proved that
# x ∈ (0, ∞)|f (n) (x) = 0 = 1
for all n ≥ 1, which is the WS property.
2.5.2. The two-sided caseWe know by Proposition A.1.4 in [10] that fα,ρ is an analytic integrable function on R, and by Theorem 1.7 in [24] thatit converges to zero at ±∞, decreases near +∞ and increases near −∞. Moreover, we have shown in Theorem 2 that if
α ≤ 3/4, it is the density of an ID distribution on R with Lévy measure ϕα,ρ (x) dx such that xϕα,ρ (x) and xϕα,ρ (−x) are
CM on (0, ∞). We are hence in position to apply Corollary 1.2 in [29], which shows that fα,ρ is BS.
2.5.3. The exceptional 1-stable caseWe use the second identity in Proposition 2, which rewritesd1 − T = (1 − log W) + L.
We have seen in Remark 7(b) that the random variable 1 − log W has a CM density on (0, +∞), in other words that it
belongs to the class ME∗ with the notations of [40]. Applying the Proposition in [40] with n = 1 shows that 1 − T has
a WBS0 density, with the notation of the main definition in [40]. As mentioned in the Introduction, this means that thedensity of T is WS− .
2.5.4. The two-sided 1-stable case with b = 0 or ab−1 ∈ πZ
We may suppose a > 0 by symmetry. If b = 0 the statement is clear since it is elementally shown that the Cauchy density1π(1 + x 2 )is BS – see also Corollary 1.3 in [29]. If b = 0, we may suppose b < 0 by symmetry. By independence, we havelog E e−iξ Ca,b = −a|ξ | + log E ei|b|ξ T , ξ ∈ R.
A further computation using Lemma 14.11 in [38] and Remark 7(b) yields



 
 −iξ C 
1 iξ
1
a,b
= c1 + c2 iξ +−−1R\(−1,1) (s) ϕa,b (s) ds
log E es s2
R iξ + s
for some c1 , c2 ∈ R and
ϕa,b (s) =as + |b|s − |b|s − 1 + 1{s≥0} .
πThis function satisfies (1.1) and (1.2) in [29] and is such that sϕa,b (s) ≥ 0. Moreover, for ab−1 ∈ πZ the function
ϕa,b (s) − k changes its sign only once for every k ∈ Z. Finally, we know from Propositions A.1.3 and A.2.1 in [10]that the density of Ca,b is smooth, converges to zero at ±∞, decreases near +∞ and increases near −∞. We can hence
apply Theorem 1.1 in [29] and conclude the proof.

Some properties of the free stable distributions

313

Remark 8.
(a) If the random variable 1 − log W had a PF∞ density as L does, then the BS character of f1,1/2 and the additive
total positivity arguments used in [29] would show that Ca,b has a BS density on R for a = 0. But 1 − log W cannot havea PF∞ density, since its law is not a GGC – see e.g. Example 3.2.2 in [12].
/ πZ, the function ϕa,b (s) − k changes its sign at least three times for every negative integer k, so that we(b) If ab−1 ∈cannot use Theorem 1.1 in [29]. It is not clear to the authors whether the density of Ca,b is always BS for a = 0, and thecase ab−1 ∈ πZ might be more the exception than the rule.
3. Further results
3.1. Some properties of the function θα
In this paragraph we consider further aspects of the function
θα (r) =1arg Fα re−iπ ,π(22)whose non-decreasing character amounts to the GGC property for the law of K1−α . We first prove the following asymptotic result.
Proposition 4. For every α ∈ [1/5, 1), one haslim  Fα re−iπ

r→+∞

=

1
·
2α

For every α ∈ (0, 1/5), one has
lim inf  Fα re−iπr→+∞= −∞ andlim sup  Fα re−iπ
r→+∞

= +∞.

The second part of this proposition has an immediate corollary, which answers in the negative an open problem statedin [26] – see Conjecture 3.1 therein.
Corollary 2. The function θα is not monotonous on (0, ∞) for α < 1/5. In particular, the law of Kα is not a GGC for
α > 4/5.
Proof of Proposition 4. We have seen during the proof of Proposition 1 that (F1/2 (re−iπ )) ≡ 1 and that Fα re−iπ

→

1
2α

as r → +∞ for all α ∈ (1/2, 1). We next consider the case α ∈ [1/5, 1/2) introducing, as above, the function
 ∞

α
1
α 2iπα duGα (r) =  Fα r α−1 e−iπ = 1 +e−ru+u e.
2πu0Note that the integral converges only in its imaginary part, but we keep this notation. Setting θ = 56 (1 − 2α) ∈ (0, 1/2],
we have 2α + αθ ∈ [1/2, 1) and by Cauchy’s theorem, we can rewrite
 ∞

θ
1
−rueiπθ +uα eiπ(2α+αθ) due.
Gα (r) = 1 + +2 2πu0The latter converges to
 ∞

 ∞

1
θ
1
θ
uα eiπ(2α+αθ) du−ue−iπ(1−2α−αθ) du
e
e

=1+ +

1+ +
2 2π
u
2 2πα
u
0
0
as r → 0. The evaluation of the oscillating integral on the right-hand side is given e.g. in Formula 1.6(36) p.13 in [21],and we finally obtainlim Gα (r) = 1 +

r→0

θ
1
1
+
(1 − 2α − αθ) =
·
2 2α
2α

314

T. Hasebe, T. Simon and M. Wang

We finally consider the case α ∈ (0, 1/5), which is much more technical and requires several steps. Setting θ =2α/(1 − α) ∈ (0, 1/2), we have 2α + αθ = θ and the same argument as above implies
 ∞

1
θ
(−rt+t α )eiπθ dt.
Gα (r) = 1 + +
e
2 2π
t
0
Hence, we are reduced to show that
lim inf Hα (r) = −∞ andr→0

lim sup Hα (r) = +∞x→0with the notations fr (t) = sin(πθ )(−rt + t α ) and
  ∞
 ∞
dt
(−rt+t α )eiπθ dt=·
Hα (r) = eecot(πθ)fr (t) sin fr (t)tt00
Let us begin with the liminf. Setting(1 − α) sin(πθ )rk = α2kπ

1/α−1and mk =2kπ(1 − α) sin(πθ )1/α,it is clear that the function frk (t) increases on (0, mk ) and decreases on (mk , +∞), and that its global maximum equals
frk (mk ) = 2kπ . This yields ∞mkecot(πθ)frk (t) sin frk (t)dt< 0 for every k ≥ 1.
tConsidering now the unique ak ∈ (0, mk ) such that frk (ak ) = π , we have limk→∞ ak = (π/ sin(πθ ))1/α , so that akecot(πθ)frk (t)0dt→sin frk (t)t (π/ sin(πθ))1/ααecos(πθ)t sin t α sin(πθ )

0

dt
<∞
t

as k → +∞. Hence it suffices to show that Ak → −∞ as k → +∞, with
Ak =

 mk
e

cot(πθ)frk (t)

ak

1
dt
=
sin frk (t)tsin(πθ ) 2kππecot(πθ)usin(u) du,−rk ϕk (u) + α(ϕk (u))α

where the second equality comes from a change of variable, having set ϕk (u) for the inverse function of frk on [π, 2kπ]
and writtenϕk (u) =

1
1
=
> 0.
frk (ϕk (u)) sin(πθ )(−rk + α(ϕk (u))α−1 )We next define pk (u) := e− cot(πθ)u (−rk ϕk (u) + α(ϕk (u))α ) and prove its strict unimodality on [π, 2kπ], computingpk (u) = e− cot(πθ)uϕk (u)2−rk t + α 2 t α − cos(πθ ) −rk t + αt α
ϕk (u)

with t = ϕk (u). The strict unimodality of pk (u) on (π, 2kπ) amounts to the fact thatqk (t) = −rk t + α 2 t α − cos(πθ ) −rk t + αt α2

has at most one zero point on [ak , mk ]. It is clear by construction that there exists ck ∈ (0, mk ) such that gk (t) = −rk t +αt αincreases on (0, ck ) and decreases on (ck , mk ), and for all t ∈ (ck , mk ) we have qk (t) = tgk (t) − cos(πθ )(−rk t + αt α )2 <0. On the other hand, the function gk (t) is increasing and concave on [0, ck ), so that its inverse function ψk (v) is increasing
and convex on [0, gk (ck )). Now sinceqk (t) = 0⇔

cos(πθ )v 2 − αv + (1 − α)rk ψk (v) = 0,Some properties of the free stable distributions

315

we see that there are at most two solutions of qk (t) = 0 on [0, ck ), one of them being zero, and hence at most one solution
on [ak , mk ), as required. We now denote by zk the unique mode of pk (u) on [ak , mk ] and, setting lk = inf{l ≥ 1, zk ≤ 2lπ },decomposeAk =1sin(πθ ) 2lk ππ

 2kπ

pk−1 (u) sin(u) du +

2lk π


pk−1 (u) sin(u) du .

Since zk → tan(πθ ) viz. lk → l∞ < +∞ as k → ∞, it is easy to see that the first term in the decomposition is bounded,
and we are finally reduced to show that
Bk =

 2kπ
2lk π

pk−1 (u) sin(u) du → −∞as k → +∞.

Since pk−1 (u) increases on [2lk π, 2kπ], we haveBk =k−1  (2j +1)π
2j π

j =lk

pk−1 (u) − pk−1 (u + π)sin(u) duα
for every k ≥ 1 and since pk (u) → sin(πθ)
ue− cot(πθ)u pointwise as k → +∞, Fatou’s lemma implies

lim sup Bk ≤k→+∞sin(πθ )
α

∞
j =l∞


 (2j +1)π  cot(πθ)uecot(πθ)(u+π)
e−sin(u) du.
u
u+π
2j π

Using the inequality
1 + eπ cot πθeπ cot πθ
≤
,
2u
u+π
+1)
, we deduce that for j∞ large enough, one has
which holds for u ≥ π(eeπ cot πθ −1
π cot πθlim sup Bk ≤k→+∞π sin(πθ )
α

∞
j =j∞

1 − eπ cot πθ2(eπ cot πθ − 1)π sin(πθ )≤−4α (2j +1)π2j π∞ecot(πθ)usin(u) duu 2j π+5π/6

j =j∞ 2j π+π/61
du = −∞.
uAll of this shows that
lim inf Hα (r) = −∞.
r→0

The argument for the limsup follows exactly along the same lines, considering the subsequence(1 − α) sin(πθ )r̃k = α(2k + 1)π

1/α−1.

− αRemark 9. As mentioned in Remark 4(a), the above proof shows that x → P[Zα 1−α ≥ x] is not HCM for every α < 1/5.
− α

By Theorem 6.3.5 in [12], this implies that Zα 1−α is not HCM for α < 1/5 either. This shows that Conjecture 1.2 in [14]is not true in general.
We believe that θa is non-decreasing for α ∈ [1/5, 1), which is equivalent to the following
Conjecture 1. The law of Kα is a GGC if and only if α ≤ 4/5.

316

T. Hasebe, T. Simon and M. Wang

Fig. 3. x −1 G1/5 (x −5 ).

Fig. 4. G̃1/5 (x).

The above Corollary 2 shows the only if part, and in the proof of Theorem 2 we have shown the if part for α ≤ 3/4.
However, it seems that our methods fail to handle the remaining case α ∈ (3/4, 4/5], because some simulations show
α−1that r α−1 Gα (r α ) = r α−1 (Fα (re−iπ )) is not monotonous anymore, at least for α close enough to 1/5 – see Figure 3.
Observe from (14) that the problem can be reformulated in terms of the monotonicity of the ratio of two power series, thenon-decreasing character of θα being equivalent to that ofsin(nπα) n(Fα (xe−iπ ))n≥0 n! (1−nα) x=G̃α : x −→cos(nπα) n(Fα (xe−iπ ))xn≥0n! (1−nα)on (0, ∞). A necessary condition for G̃α to be non-decreasing is that its denominator does not vanish on (0, ∞), which
is false for α < 1/5 by Proposition 4 and true for α ≥ 1/4 by the proof of Theorem 2. But the case α ∈ [1/5, 1/4) still
eludes us. Let us mention that monotonicity properties of ratios of power series are studied in the literature on special
functions – see e.g. Chapter 3.1 in [4]. However, we could not find any clue in this literature for our problem, and it is noteasy to understand why the value α = 1/5 should be critical for the monotonicity of the above ratio. See Figure 4 for asimulation.
We finally turn to the behaviour of Fα (re−iπ ) at infinity, which implies that of θα (r).
Proposition 5. For all α ∈ (0, 1), one has
Fα re−iπ ∼icα eb1−α r√r

as r → +∞,12(α−1)

with cα = √α2π(1−α) . In particular, one has θα (r) → 1/2 as r → +∞.
Proof. From (14), we can write
Fα re−iπ = φ −α, 1, r 1−α eiπα ,r > 0.

We now use the asymptotic expansion for large z ∈ C and a ∈ (−1, 0) of the Wright function φ(a, b, z), which has beenobtained in [43]. Applying therein Theorem 1 for α ≤ 1/3 resp. Theorem 5 for α > 1/3 and taking the first term in (1.3)

Some properties of the free stable distributions

317

implies the required asymptotic for Fα (re−iπ ), since we have here
A0 = √

1
2παand

Y = b1−α re−iπin the notation of [43], the first equality being a consequence of Stirling’s formula. From (22), we then readily deducethat θα (r) → 1/2 as r → +∞.
Remark 10.
(a) Taking the first two terms in the series representation (14) yields at once the asymptotic behaviour of θα (r) at zero,
which isθα (r) ∼r 1−α·(α) (1 − α)2On the other hand, the complete asymptotic expansion (1.3) in [43] has only purely imaginary terms in our framework, sothat we cannot deduce from it the asymptotics of θα (r) − 1/2 at infinity. It follows from Proposition 1 that θα (r) ∈ [0, 1/2)
for α ≥ 1/4, and from Proposition 4 that θα (r) − 1/2 crosses zero an infinite number of times for α < 1/5, as r → +∞.
For α ∈ [1/5, 1/4), we are currently unable to prove that θα (r) ∈ [0, 1/2) for every r > 0, which would be a first step toshow that it increases from 0 to 1/2. Recall that the latter is equivalent to the fact that the denominator of the above G̃αdoes not vanish on (0, ∞).
(b) If α ≤ 3/4, it follows from (13), Theorem 8.2 and Remark 8.3 in [39], and the above proposition, that the Thorin
mass of the GGC random variable Kα equals 1/2. Hence, Kα − bα is a 1/2 -mixture by Theorem√ 4.1.1. in [12], which isa refinement of Corollary 3.2 in [26]. Since this property amounts to the CM character of x → xfKα (bα + x), a perusalof the proof of Theorem 1 shows that Xα − bα is a 3/2 -mixture as soon as α ≤ 3/4. We believe that this is true for everyα ∈ (0, 1).
3.2. Asymptotic expansions for the free extreme stable densities
In this paragraph we derive the full asymptotic expansion at zero of the density fYα of the random variableYα =X α − bα

if α ∈ (0, 1),−1/α d −1/α
Xα,1−1/α + b1/α = b1/α − Xα,1/α

if α ∈ (1, 2],

and Y1 = 1 − T. We will use the standard notation of Definition C.1.1 in [1] for asymptotic expansions. Our expansions
complete the estimates of Proposition A.1.2 in [10] and the series representations of Theorem 1.7 in [24], from which one
can only infer that the random variable Yα is positive. They can also be viewed as free analogues of Linnik’s expansions(14.35) in [38] – see also Theorem 2.5.3 in [44] – for the classical extreme stable distributions. Observe that in the classicalcase, the expansion for α > 1 is deduced from that of the case α ∈ [1/2, 1) by the Zolotarev’s duality which is discussed
in Section 2.3 of [44]. Even though the very same duality relationship holds in the free case – see Proposition A.3.1 in[10] and Corollary 1.4 in [24], for Yα this duality only yields
fY1/α (x) =1 −α−1/α−1
1/α
fYα bα−α − x
− bα
b −x
α α

for every α ∈ [1/2, 1), and does not seem particularly helpful to connect explicitly the two expansions at zero. Whenα = 1, our method hinges on Wright’s original papers [42] for the case α > 1 and [43] for the case α < 1. It is remarkablethat the two expansions turn out to have the same parametrization.
Proposition 6. For every α ∈ (0, 1) ∪ (1, 2], one has∞

fYα (x) ∼an (α)x n+1/2

as x → 0,

n=0

with


 n+1/2
−n−1/2 
d2n
(−1)n
2
α+11
−2
×
an (α) =;v(1 − v) 2 F13απ|α − 1|(n+3/2)/α (2n + 1)! dv 2n

.
v=0

318

T. Hasebe, T. Simon and M. Wang

Proof. We begin with the case α > 1, writing down first fYα with the help of Bromwich’s integral formula1
fYα (x) =2π i

 1+i∞

ezx Lα (z) dz,

1−i∞

where
1




α −1
Lα (z) = E e−zYα = e−α(α−1) z × E ezXα,1/α

is well-defined and analytic on the open right half-plane. Combining next Theorem 1.8 in [24] and Theorem 2 in [42], we
obtainLα (z) = e−α(α−1)

1
α −1 z× φ α − 1, 2, zα = O |z|−3/2

uniformly on the right half-plane. Making a change of variable and applying Cauchy’s theorem, we deduce x+i∞
 1+i∞
1
1
ez Lα zx −1 dz =ez Lα zx −1 dz.
fYα (x) =2π ix x−i∞2πix 1−i∞
Using now the full asymptotic expansion of Theorem 2 in [42], we get∞

fYα (x) ∼an (α)x n+1/2

as x → 0,

n=0

where


 1+i∞(−1)n an(−1)n an1z −3/2−nezdz=an (α) =(α − 1)(n+3/2)/α 2πi 1−i∞
(α − 1)(n+3/2)/α (n + 3/2)
and an is defined at the beginning of p.258 in [42] for ρ = α − 1 and β = 2. Above, the interchanging of the contour integral and the expansion is easily justified – alternatively one can use the generalized Watson’s lemma which is mentioned
at the top of p.615 in [1], whereas the second equality follows from Hankel’s formula – see e.g. Exercise 1.22 in [1]. To
conclude the proof of the case α > 1, it remains to evaluate the coefficients an (α), which is done in observing that thefunction in (1.21) of [42] is here



α+11
;v ,
2 F1
3
and making some simplifications.
We now consider the case α < 1. The argument is analogous but it depends on the expansions of [43] which, the authorsays, cannot be simply deduced from those of [42]. We again write 1+i∞
1
ezx Lα (z) dz,
fYα (x) =2π i 1−i∞where
Lα (z) = eα(1−α)

1
α −1 z1
α −1
× E e−zXα = eα(1−α) z × φ α − 1, 2, −zα = O |z|−3/2

uniformly in the open right half-plane, the second equality following from Theorem 1.8 in [24] and the estimate from theLemma p.39 in [43]. Reasoning as above, we get
1
fYα (x) =2π ix

 1+i∞e Lα zx
z

−1

∞

dz ∼

1−i∞

n=0

where
an (α) =an(α(1 − α)1α −1an (α)x n+1/2)n+3/2 (n + 3/2)

as x → 0,

Some properties of the free stable distributions

319

and an is defined at the bottom of p.38 in [43] for σ = 1 − α and β = 2. After some simplifications, we also obtain therequired expression for an (α).
Remark 11.
(a) It does not seem that a simple closed formula can be obtained for the coefficients an (α) in general. We can compute2
2α 2 − 23α + 47
and a1 (α) = −×.
α
36πα|α − 1|5/(2α)

21×α π|α − 1|3/(2α)

a0 (α) =Observe that a1 (α) is always negative. We believe that in general, one has
2Q2n (α)×α πα n |α − 1|(2n+3)/(2α)

an (α) =for some Q2n ∈ Q2n [X]. This would again mimic the classical situation, save for the fact that here the polynomial Q2ndoes not seem to have symmetric coefficients – see Remark 2 p.101 in [44].
(b) For α = 2, the involved hypergeometric function becomes the standard geometric series and we simply getan (2) =

 
(−1)n
−1
d2n
2n
× 2n (1 − v)n−3/2 v=0 =,nπ(2n + 1)! dv
π(2n − 1)16nwhich is always negative except for n = 0. Of course, this can be retrieved via the binomial theorem for the explicit
density√x
fY2 (x) =
π

x
1− ·
4

(c) For α = 1/2, the involved hypergeometric function simplifies with the help of Exercise 3.39 in [1], and we getan (1/2) =√(−1)n 4n+2(−1)n (n + 1)4n+2
d2n
× 2n (1 + 1 − v)2n+1 (1 − v)−2 v=0 =,2π(2n + 1)! dv
πwhose signs alternate. This again can be retrieved via the binomial theorem for the explicit density
fY1/2 (x) =

√
16 x
·
π(1 + 4x)2(d) As already observed in Remark 1(a), the densities of Y1/3 and Y2/3 can be written in closed form with the help of
Formulae (40) and (41) in [32]. In principle, a full asymptotic expansion can also be derived from these expressions, but
the task seems too painful. Notice that here, the involved hypergeometric functions do not seem to simplify.
(e) The above proof shows that the following functions
λ → e−α(α−1)

1
α −1 λφ α − 1, 2, λαresp. λ → eα(1−α)

1
α −1 λφ α − 1, 2, −λαon (0, ∞), which are obtained in removing Wright’s exponential term at infinity, are CM functions for α ∈ (1, 2] resp. forα ∈ (0, 1).
We now complete the picture and derive the asymptotic expansion of Y1 = 1 − T. To state our result, we need to
introduce the Stirling series {cn , n ≥ 0} appearing in the expansion
 x
e
x

x
(x) ∼
2π

cn x −nas x → +∞,n≥0which is given e.g. in Exercise 23 p.267 of [17]. One has c0 = 1, c1 = 1/12, c2 = 1/288 and c3 = −139/51,840. In
general, cn is a rational number and the corresponding sequences of numerators and denominators are referenced underA00163 and A00164 in the online version of [35].

320

T. Hasebe, T. Simon and M. Wang

Proposition 7. One has∞

fY1 (x) ∼an (1)x n+1/2

as x → 0,

n=0

with
an (1) =(−1)n 22n+1/2 n!(c0 + · · · + cn )·π(2n + 1)!

Proof. Applying Remark 7(a), we first compute the Laplace transform z
 −zY 
1
1
z
1
=
Ee
z(1 + z) e(z)for every z in the open right half-plane. Comparing next (2.15) and (2.21) in [33], we get the expansion
1
E e−zY1 ∼ √(−1)n cn z−n2πz(1 + z) n≥0



1
=√
(−1)n z−n(−1)n cn z−n2πz3 n≥0n≥01(−1)n (c0 + · · · + cn )z−n∼√2πz3 n≥0
as |z| → ∞, uniformly in the open right half-plane | arg(z)| < π/2. Reasoning as in Proposition 6, we finally obtain∞

fY1 (x) ∼an (1)x n+1/2 as x → 0,
n=0

with
an (1) =(−1)n (c0 + · · · + cn ) (−1)n 22n+1/2 n!(c0 + · · · + cn )=·√π(2n + 1)!
2π (n + 3/2)

Remark 12. It is easy to see from (15) thatd(1 − α)−1 Yα −→ Y1

as α ↑ 1,

and it is natural to infer from this and Proposition 6 that
−n−1/2 (−1)n 2n+1/2
d2n
2
1
;v
× 2n (1 − v)−2 2 F1
an (1) =3π(2n + 1)!
dv
v=0


n
2n
2n+1
(−1)
d
v
=
,
×
π(2n + 1)! dv 2n (1 − v)2 (−v − log(1 − v))n+1/2 v=0
except that we cannot interchange a priori the asymptotic expansion at zero and the convergence in law. We have checked
the correspondence for n = 0 and n = 1, with
√
√
2
13 2
and a1 (1) = −a0 (1) =π
18π
to be compared with Remark 11(a). We believe that this formula is true for every n ≥ 1. Observe that this is equivalent to
the following expression of the Stirling series:
cn = bn − bn−1 ,n ≥ 1,

Some properties of the free stable distributions

321

with


d2n
v 2n+1,bn = 2n+1/2 × 2n
2n! dv
(1 − v)2 (−v − log(1 − v))n+1/2 v=01

which is different from the combinatorial expression given in Exercise 23 p.267 of [17], and which we could not locate
in the literature.
3.3. Further properties of whale-shaped functions
In this paragraph we prove five analytical properties of WS functions and densities. Those five easy pieces apply all tothe densities fα , and have an independent interest. We restrict the study to the class WS+ , the corresponding properties
for WS− being deduced at once.
Proposition 8. Let f be a WS+ density with unique mode M. Then f is perfectly skew to the right, that is
f (M + x) > f (M − x) for every x > 0.
Proof. Let x0 be the left-extremity of Suppf and M = x1 < x2 < x3 be the vanishing places of the three first derivativesof f . Suppose first M − x0 > x2 − M. Taylor’s formula with integral remainder implies x(x − t) f  (M + t) − f  (M − t) dt.
f (M + x) − f (M − x) =0

On the one-hand, we have f  (M + t) − f  (M − t) > 0 for all t > x2 − M since f  (M − t) ≤ 0 for all t ≥ 0 and
f  (M + t) > 0 for all t > x2 − M. On the other hand, writing tf (3) (M + s) + f (3) (M − s) ds,
f  (M + t) − f  (M − t) =0which is valid for all t < M − x0 , we also have f  (M + t) − f  (M − t) > 0 for all t ≤ x2 − M since f (3) (u) > 0
for all x0 < u < x3 , by the WS+ property. Putting everything together shows f (M + x) > f (M − x) for all x > 0.
Supposing next M − x0 ≤ x2 − M, the proof is analogous and easier; we just need to delete the corresponding arguments
for t > x2 − M.
Remark 13. If we denote by Mα,ρ the unique mode of fα,ρ , the function
x → fα,ρ (Mα,ρ + x) − fα,ρ (Mα,ρ − x)has constant and possibly zero sign on (0, ∞) for ρ = 0, 1/2, 1 and for α = 1, as seen from the above proposition, theexplicit drifted Cauchy case and the symmetric case. One might wonder if this property of perfect skewness remains truein general. The perfect skewness of classical stable densities is a challenging open problem, which had been stated in the
introduction to [23].
Proposition 9. Let f be a WS+ density and M, m, μ be its respective mode, median and mean. Then f satisfies the strict
mean-median-mode inequality
M < m < μ.
Proof. We use the same notation of the proof of the previous proposition. First, the latter clearly implies M < m. To
obtain the two strict inequalities together, let us now consider the function
g(x) = f (m + x) − f (m − x)
on [0, m − x0 ]. If m < x2 , then the WS+ property implies f (2) (m + x) > f (2) (m) > f (2) (m − x) for every x ∈ (0, m − x0 ],so that g is strictly convex on (0, m − x0 ]. Since g(0) = 0, g  (0) < 0 and g(m − x0 ) > 0, this shows that g vanishes only
once on (0, m − x0 ] and from below, and hence also on the whole (0, ∞). If m ≥ x2 , then g is negative on (0, m − M] andstrictly convex on [m − M, m − x0 ] and we arrive at the same conclusion. We are hence in position to apply Lemma 1.9(a) and (a strict, easily proved version of) Theorem 1.14 in [19], which implies the strict mean-median-mode inequality
for f .


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T. Hasebe, T. Simon and M. Wang

Remark 14.
(a) It is well-known that Xα has infinite mean, which can be seen e.g. from the power series formula for the density ofTheorem 1.7 in [24]. Hence, in this framework the above result only reads M < m < ∞, and it is readily obtained fromthe previous proposition.
(b) In the relevant case α ∈ (1, 2) it is natural to conjecture that the strict mean-median-mode inequality holds, in one
or the other direction, for both free and classical stable densities. Observe that the three parameters clearly coincide forρ = 1/2, whereas for ρ = 1/α, easy computations show that the mean is zero and the mode and median are positive,
so that it is enough to prove m < M. In general, this problem is believed to be challenging and beyond the scope of thepresent paper. We refer to [6] for a series of results on this interesting question, which however do not apply to non-explicit
densities.
Proposition 10. Let f be a WS density on (0, +∞) and X be the corresponding random variable. Then X is a
mixture. In particular, it is ID.

2-

Proof. As in Theorem 1, we need to show that g(x) = x −1 f (x) is a CM function, in other words that (−1)n g (n) (x) > 0
on (0, ∞). By Leibniz’s formula, we first computeng (n) (x) = n!
p=0(−1)p f (n−p) (x)·(n − p)!x p+1

This implies, after some simple rearrangements,hn (x) = (−1)n x n f (n+1) (x),(23)

where hn (x) = (−1)n x n+1 g (n) (x) has the same sign as (−1)n g (n) (x). By the WS property, setting xi for the vanishingplaces of the ith derivative of f for all i ≥ 1, we see thatnhn (x) = n!
p=0(−1)n−p f (n−p) (x) n−px(n − p)!

is positive on [xn , ∞) since (−1)i f (i) (x) > 0 when x ∈ (xi , ∞) for all i ≥ 0. Moreover, it follows from (23) and thewhale-shape that hn (x) > 0 for x ∈ (0, xn+1 ]. It is hence enough to show that hn (0+) = 0 in order to conclude the proof,
because (0, ∞) = (0, xn+1 ] ∪ [xn , ∞). But the whale-shape shows again that0 ≤ (−1)i−1 x i f (i) (x) ≤ 2(−1)i−1 x i−1 f (i−1) (x) − f (i−1) (x/2)
for all x ∈ (0, x1 ] and an induction on i, starting from f (0+) = 0, implies (x i f (i) )(0+) = 0 for all i ≥ 0, so that hn (0+) =0 as well.
Remark 15.
(a) The WS property is not satisfied by all densities ofthat the derivative of the density

2 -mixtures vanishing at zero. A simulation shows for example

f (x) = x ta 2 e−ax + (1 − t)e−x
vanishes three times for a = 20 and t = 4/5. This contrasts with the densities of 1 -mixtures, which are characterized bytheir complete monotonicity – see e.g. Proposition 51.8 in [38].
(b) For a given smooth density f on (0, ∞) and n ≥ 0, let us introduce the following property: one has f ∈ BSn if{x > 0, f (i) (x) = 0} = i for i ≤ n,
{x > 0, f (i) (x) = 0} = n for i > n.
For n ≥ 1, this property was introduced in [40] under the less natural denomination WBSn−1 – see the definition therein.
Clearly, one has BS0 = CM and BS1 = WS for densities on (0, ∞). Since the density of  t has mth derivative m  
t−1 e−x
m
m
−p x
(−1)
(1 − t)p x
(t)
p
p=0

Some properties of the free stable distributions323

on (0, ∞), it is an easy exercise using Rolle’s theorem and Descartes’ rule of signs to show that  t ∈ BSn for t ∈ (n, n+1].
In this respect, the class BSn can be thought of as an extension of the densities of  t for t ∈ (n, n + 1]. Moreover, wehave just seen that the set of densities of n+1 -mixtures contains the class BSn for n = 0, 1. We actually believe that thisis true for all n ≥ 0. In view of the Proposition in [40] and of the general results of [29], it is also natural to conjecturethat the densities in BSn are infinitely divisible for all n ≥ 2, contrary to the densities of n+1 -mixtures. Taking the limitn → ∞, this would imply that all bell-shaped densities on a half-line are ID. On the other hand, the class BSn does notseem to share any interesting property related to perfect skewness, mean-median-mode inequality for n ≥ 2.
(c) The above proposition entails that the case {α ∈ (0, 1], ρ = 1} in Theorem 1(a) is a consequence of Theorem 4(a).
On the other hand, as we saw above, the proof of Theorem 1 also shows that Xα is a 3/2 -mixture for α ≤ 3/4, which is
not a consequence of the whale-shape.
We next study the stability of the WS property under exponential tilting. Within ID densities on R, this transformation
amounts to the multiplication of the Lévy measure by e−c|x| , allowing for models with finite positive moments andanalogous small jumps. This is a particular instance of the general tempering transformation, where the exponential
perturbation is replaced by a CM function, and we refer to [36] for a thorough study on tempered stable densities. If
we restrict to ID densities on a positive half-line, it is seen from the Lévy–Khintchine formula that exponential tilting
amounts to multiplying the density by the same e−cx and renormalizing. In particular, the set of densities of t -mixtureswith t ∈ (0, 2] is also stable under exponential tilting.
Proposition 11. If f ∈ WS+ , then e−x f ∈ WS+ .
Proof. It is enough to consider the case Supp f = (0, ∞). Set g(x) = e−x f (x). Considering hn (x) = (−1)n ex g (n) (x)for each n ≥ 0, we have hn+1 = hn − hn and an easy induction starting from h0 = f implies
(p)

(p)

(−1)p−1 hn+1 (0+) > 0 and hn+1 (+∞) = 0(p)for all n, p ≥ 0. We will now show that hn+1 vanishes once on (0, ∞) for all n, p ≥ 0, and that the sequence {xp,n+1 , p ≥(p)0} defined by hn+1 (xp,n+1 ) = 0 is increasing. This is sufficient for our purpose, in taking p = 0.
(p)

(p)

Consider first the case n = 0, with h1 = f (p) − f (p+1) . It is clear that (−1)p h1 (x) > 0 for x ∈ [xp+1 , ∞) and that
(p+1)
(p)
(p)
(−1)p h1(x) > 0 for x ∈ (0, xp+1 ]. Since (−1)p h1 (0+) < 0, this implies that h1 vanishes once on (0, ∞) for all(p)p ≥ 0, and Rolle’s theorem entails that the sequence {xp,1 , p ≥ 0} defined by h1 (xp,1 ) = 0 is increasing.
(p)
(p)
(p+1)
(p)
The induction step is obtained analogously from hn+2 = hn+1 − hn+1 , since (−1)p hn+2 (0+) < 0 and, by the induc(p)(p+1)

tion hypothesis, (−1)p hn+2 (x) > 0 for x ∈ [xp+1,n+1 , ∞) and (−1)p hn+2 (x) > 0 for x ∈ (0, xp+1,n+1 ].

Remark 16.
(a) The above proposition implies that e−x fα , the “tilted free positive stable density”, is WS+ and ID. It would beinteresting to know if it is also FID.
(b) The class WS+ is not stable under the general tempering transformation introduced in [36]. For example, therandom variable obtained from  2 in multiplying its Lévy measure by te−x is easily seen to be (1/2) 2t , whose density
belongs to WS+ only for t ∈ (1/2, 1].
Proposition 12. Let f ∈ WS+ and {xn , n ≥ 0} be the vanishing places of {f (n) , n ≥ 0}. Then f is analytic on (x0 , ∞)and xn → ∞.
Proof. Again we may suppose x0 = 0. If f is a density, then Proposition 10 implies that f = xg where g is CM andhence analytic on (0, ∞), so that f is analytic on (0, ∞) as well. If f is not a density, then Proposition 11 shows thatg = e−cx f is a WS+ density on (0, ∞) for some normalizing c > 0, and f inherits the analyticity of g on (0, ∞).
The second property is an easy consequence of the first one. Let x∞ be the increasing limit of {xn , n ≥ 0} and supposex∞ < ∞. By the whale-shape, we would then have (−1)n f (n) (x) > 0 for x > x∞ , so that f would be CM on (x∞ , ∞),and hence also on (0, ∞) by Bernstein’s theorem and analytic continuation, a contradiction since f (0+) = 0.
324

T. Hasebe, T. Simon and M. Wang

Overview of random variables used in the paper
Xα,ρ
Xα
Ca,b
T
Kα
U
L
Ba,b
Zα
Zα,ρ
t

free strictly α-stable
free positive α-stable
free 1-stable
exceptional free 1-stable
Kanteruniform on (0, 1)
unit exponential
standard β(a, b)classical positive α-stableclassical strictly α-stable
standard γ (t)Section 1

S
W

exceptional classical 1-stablesin(πU) πU cot(πU)πU e

Section 2.4YαXα − bα if α ∈ (0, 1)
1 − T if α = 1−1/αb1/α − Xα,1/α if α ∈ (1, 2]

Section 3.2(Section 2.2)

Section 2.1

Section 2.2

Acknowledgements
The authors were all supported by a JSPS-MAEDI research program Sakura. TH was financially supported by JSPS
Grant-in-Aid for Young Scientists (B) 15K17549. We are very grateful to a referee for a detailed report, which helped us
improve the presentation of the paper.
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