aa r X i v : . [ m a t h . S T ] J a n ON POSITIVE CLASSICAL AND FREE STABLE LAWS
N. Demni Abstract.
We derive the representative Bernstein measure of the density of( X α ) − α/ (1 − α ) , < α <
1, where X α is a positive stable random variable, asa Fox-H function. When 1 − α = 1 /j for some integer j ≥
2, the Fox H-function reduces to a Meijer G-function so that the Kanter’s random variable(see below) is closely related to a product of ( j −
1) independent Beta randomvariables. When α tends to 0, the Bernstein measure becomes degeneratethereby agrees with Cressie’s result for the asymptotic behaviour of stabledistributions for small values of α . Coming to free probability, our resultmakes more explicit that of Biane on the density of its free analog. The paperis closed with analytic arguments explaining the occurence of the Kanter’srandom variable in both the classical and the free settings. Motivation: Kanter’s random variable
The study of stable random variables has known a considerable growth duringapproximately the last five decades, since they are shown to be an efficient modelfor various phenomena occuring in biology, quantum physics, market finance ([16]).Unfortunately, their densities has no closed formulas except in few cases. Considerfor instance, the stable random variable X α of index 0 < α < , ∞ ) ([9], p.50). Its Laplace transform is given by e − t α , t > − πx ∞ X k =1 ( − k k ! sin( kπα )Γ(1 + kα ) 1 x kα , x > . The positivity of this series is not trivial and follows for instance from the integralrepresentation displayed p.74 in [16]. Amazing and not trivial as well is the com-plete monotonicity of the density of ( X α ) − r , r ≥ α/ (1 − α ) (in particular its infinitedivisibility, [12]) which follows from the following representation ([11]): there ex-ists a standard exponential random variable L and an infinitely divisible randomvariable V α,r , both variables being independent and( X α ) r d = e V α,r L .
Recall that the proof of this fact relies on a suitable integral representation of theGamma function and that the random variable V α,r is characterized by its Laplacetransform. For r = α/ (1 − α ), the above representation is more precise and tracesback implicitely to Chernin and Ibragimov’s paper [4] (see also Ch.II in [9]). In IRMAR, Rennes 1 University, France. Email: [email protected]
Keywords : Stable laws, free probability, Fox H-function.
AMS Classification : 60E07; 33E12; 60B20. fact ([10] p.703) ( X α ) α/ (1 − α ) d = a α ( U ) L , where U is a uniform random variable on (0 , π ) and a α ( u ) = [ b α ( u )] / (1 − α ) , b ( u ) = [ c α ( u )] α [ c − α ( u )] (1 − α ) , c α ( u ) := sin( αu )sin( u ) . We shall refer to a α ( U ) as the Kanter’s random variable. From the injectivity ofthe Laplace transform, one gets V α, − α d = log[ a α ( U )] . Mysteriously, the function a α is intimately related to the density of the free stablelaw of index 0 < α < ρ = 1 derived p.1053 in [1], thatwe shall call shortly the positive free stable law. More precisely, if U is defined onsome probability space (Ω , F , P ), then P ( a α ( U ) ≤ u ) = 1 π a − α ( u ) , u ≥ a α (0) = (1 − α ) α α/ (1 − α ) since a α is strictly increasing ([10] p.704), while the density of the positive freestable distribution takes the form ([1]):(1) 1 πx sin( a − − α ( x )) sin( αa − − α ( x ))sin((1 − α ) a − − α ( x )) , for x ≥ a − α (0). That is why we thought it is reasonable to start with seekingan expression for the density of a α ( U ) as a special function and more generally forthat of e V α,r . The easiest way to proceed is to invert the Mellin transform of thedensity of e − V α,r given by ([11] p.292): E ( e − sV α,r ) = Γ( rs/α + 1)Γ( s + 1)Γ( sr + 1) , ℜ ( s ) > − α/r ≥ − (1 − α ) . However, the Mellin’s inversion formula(2) 12 iπy Z + i ∞− i ∞ Γ( rs/α + 1)Γ( s + 1)Γ( sr + 1) y − s ds, applies for any r > α/ (1 − α ) (we take ℜ ( s ) = 0 as a path of integration) and doesnot when r = α/ (1 − α ). The latter fact is seen from the following estimate of | Γ( a + ib ) | , a, b ∈ R , | b | → ∞ displayed in formula 1.24. p.4 from [8]:(3) | Γ( a + ib ) | ∼ √ π | b | a − / e − a − π | b | / , | b | → ∞ so that the absolute value of the integrand in (2) is equivalent to | s | − / (up to aconstant). Note however that when r > α/ (1 − α ), the integral displayed in (2) isnothing else but a so-called Fox H-function ([8] p.3, see below):12 iπy Z + i ∞− i ∞ Γ( rs/α + 1)Γ( s + 1)Γ( sr + 1) y − s ds = 1 y H , , h y | (1 , , (1 ,r )(1 ,r/α ) i so that the density of e V α,r reads1 y H , , (cid:20) y | (1 , , (1 ,r )(1 ,r/α ) (cid:21) , y > , r > α − α . N POSITIVE CLASSICAL AND FREE STABLE LAWS 3
The main result of this paper states then that in the pathological case r = α/ (1 − α ),the density of a α ( U ) is a H-function too represented through a path integral re-sembling to (2) but with a different contour (Hankel path). When α approacheszero, this density tends to zero pointwisely, a fact that agrees with Cressie’s re-sult implying that the Kanter’s random variable becomes degenerate in the limit α → + . When 1 − α = 1 /j for some integer j ≥
2, the derived H-function reducesto the so-called Meijer G-function so that a α ( U ) is closely related to a product ofsuitably chosen j − a α ( U ) is computed yielding an ex-plicit expression for the density of a positive free stable law. Using the free analogof Zolotarev’s duality ([1]), a similar result holds for a free stable law with index1 < α < ρ = 0. The paper is closed with supplyinganalytic arguments that explain the occurence of a α in both probabilistic settings.It turns out that it is resumed in the inverse formula for the Fourier transform ofthe density of X α together with deformations of paths of integration. Nevertheless,we think that a group theoretical argument exists but seems to be very hidden.For sake of completeness, some basic needed facts on the H-function are collectedin the next section. A good reference to them is the monograph [8].2. On the Fox H-function
The Fox H-function is defined as a Mellin-Barnes integral: H m,np,q h z | ( a i ,A i ) ≤ i ≤ p ( b i ,B i ) ≤ i ≤ q i = Z L Θ( s ) z − s ds where 1 ≤ m ≤ q, ≤ n ≤ p, a i , b i ∈ R , A i , B i > s ) = Q mi =1 Γ( b i + B i s ) Q ni =1 Γ(1 − a i − A i s ) Q qi = m +1 Γ(1 − b i − B i s ) Q pi = n +1 Γ( a i + A i s ) , the principal determination of the power function is taken (though it is not necessaryhere), an empty product being equal one and L is a suitable contour separating thepoles of both products of the meromorphic (Gamma) functions in the numeratorof Θ ([8] p.3,4). The choice of L and the domain of convergence of the Mellin-Barnes integral defining the H-function for each contour depend on the parameters a i , b i , A i , B i . Eight cases are discussed in [8] p.4. and we will only make use of fiveof them in the sequel that we shall refer to whenever needed:i) q ≥ µ := q X i =1 B i − p X i =1 A i > , then the H-function exists in the punctured complex plane.ii) p ≥ , µ = 0, the H-function exists in the domain | z | > β where β := p Y i =1 A − A i i Y i =1 B B i i and L = L + ∞ described in [8] p.3. ON POSITIVE CLASSICAL AND FREE STABLE LAWS iii) Let Ω be the real number defined byΩ := n X i =1 A i + m X i =1 B i − p X i = n +1 A i − q X i = m +1 B i . Then the Fox H-function exists for all z = 0 such that | arg( z ) | < π Ω / , Ω > L is the infinite semi-circle γ − i ∞ , γ + i ∞ for a suitable real γ .iv) When q ≥ , µ = 0, the H-function exists in the open disc | z | < β and L = L −∞ is described in [8] p.3.v) Let δ := q X i =1 b i − p X i =1 a i + p − q . If µ = 0 , δ < −
1, then the H-function is well-defined for complex numberslying on the circle | z | = β .The most crucial fact when looking at these definitions is that whenever morethan one definition make sense, they lead to the same value of the integral. Infact, the integral is evaluated by means of Cauchy’s residue Theorem applied totruncated contours together with asymptotic estimates at infinity, and the contourof integration can be deformed into another one provided that the deformationis admissible ([6], [14], see also CH.V. in [7] for a similar discussion on Meijer’sG-function which fits the Fox H-function when A i = B j = 1 , ≤ i ≤ p, ≤ j ≤ q ). Remark 1.
In the sequel, the complex variable z will take only strictly positivevalues thereby s z − s is defined through the real-valued logarithm function. The density of a α ( U )According to Kanter’s representation, there exists a density h α supported by a α (0 , π ) = ( a α (0) , ∞ ) such that − − ααπx ∞ X k =1 ( − k k ! sin( kπα )Γ(1 + kα ) x k (1 − α ) = Z ∞ e − xy yh α ( y ) dy. The main result of this paper is stated as:
Proposition 1.
The density of the Kanter’s variable is expressed as: h α ( y ) = 1 y H , , (cid:20) y | (1 ,α/ (1 − α )) , (1 , , / (1 − α )) (cid:21) = 2(1 − α ) y H , , (cid:20) y − α ) | (1 , α ) , (1 , − α ))(1 , (cid:21) for y ∈ [ a α (0) , ∞ [ .Proof : We shall start from the density of ( X α ) − α which has the expansion − απx ∞ X k =1 ( − k k ! sin( kπα )Γ(1 + kα ) x k . Using the mirror formula satisfied by the Euler’s Gamma function ([7] p.3),Γ(1 + kα )Γ( − kα ) = − π sin( kαπ ) , the density of X − αα is transformed to1 αx ∞ X k =1 ( − k k ! x k Γ( − kα ) N POSITIVE CLASSICAL AND FREE STABLE LAWS 5 and the infinite summation may start from k = 0 since z = 0 is a pole of the Gammafunction. According to A.6 p.218 in [8], this is expressed through the H-functionas follows : − απx ∞ X k =1 ( − k k ! sin( kπα )Γ(1 + kα ) x k = 2 αx H , , h x | (0 , α )(0 , i for all x > z = 0 in the complexplane by i) since q ≥ , µ = 2 − α > X α ) − α/ (1 − α ) reads 2(1 − α ) αx H , , h x − α ) | (0 , α )(0 , i . According to iii), one has H , , h x − α ) | (0 , α )(0 , i = 12 iπ Z γ + i ∞ γ − i ∞ Γ(2 s )Γ(2 αs ) dsx − α ) s where ℜ ( s ) = γ >
0. With the help of the Gamma integral2(1 − α ) αx H , , h x − α ) | (1 , α )(1 , i = 2(1 − α )2 αiπ Z γ + i ∞ γ − i ∞ Γ(2 s )Γ(2 αs )Γ(2 + 2(1 − α ) s ) Z ∞ e − xy y − α ) s +1 dyds = 2(1 − α )2 iπ Z γ + i ∞ γ − i ∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(2 + 2(1 − α ) s ) Z ∞ e − xy y − α ) s +1 dyds. In order to change the order of integration, one needs to check that Z + ∞−∞ | Γ(2 is + 2 γ + 1) || Γ(2 iαs + 2 αγ + 1)Γ(2(1 − α ) s + 2(1 − α ) γ + 2) | Z ∞ e − xy y − α ) γ dyds < ∞ . Performing an integration with respect to y , it amounts to check Z + ∞−∞ | Γ(2 is + 2 γ + 1) || Γ(2 iαs + 2 αγ + 1)Γ(2 i (1 − α ) s + 2(1 − α ) γ + 2) | ds < ∞ which follows again from the estimate of | Γ( a + ib ) | , | b | → ∞ (see (3)). Thus,Fubini’s Theorem yields Z ∞ e − xy yh α ( y ) dy = 2(1 − α ) Z ∞ xe − xy Z γ + i ∞ γ − i ∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(2 + 2(1 − α ) s ) y − α ) s +1 dsdy = 2(1 − α ) Z ∞ xe − xy yH , , (cid:20) y − α ) | (1 , α ) , (2 , − α ))(1 , (cid:21) dy. (4)Note that the H-function displayed in (4) may be represented for y > (1 − α ) α α/ (1 − α ) as a path integral through the loop L −∞ according to iv), it vanishes for 0 < y < (1 − α ) α α/ (1 − α ) since according to ii) the contour is a loop L + ∞ that contains nopole of the meromorphic function s Γ(1 + 2 s ), and it takes a finite value at z = (1 − α ) α α/ (1 − α ) regarding v) since µ = 0 , δ = − / < − It is a special kind of what V. M. Zolotarev called incomplete hypergeometric function in [17].
ON POSITIVE CLASSICAL AND FREE STABLE LAWS of the Kanter’s random variable is h α ( y ) = 2(1 − α ) y ddy (cid:26) yH , , (cid:20) y − α ) | (1 , α ) , (2 , − α ))(1 , (cid:21)(cid:27) = 2(1 − α ) y ddy Z L −∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(2 + 2(1 − α ) s ) y − α ) s +1 ds = 2(1 − α ) y Z L −∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(2(1 − α ) s + 1) y − α ) s ds = 2(1 − α ) y H , , (cid:20) y − α ) | (1 , α ) , (1 , − α ))(1 , (cid:21) = 1 y Z L ′−∞ Γ( s/ (1 − α ) + 1)Γ( αs/ (1 − α ) + 1)Γ( s + 1) y s ds = 1 y H , , (cid:20) y | (1 ,α/ (1 − α )) , (1 , , / (1 − α )) (cid:21) , where L ′−∞ is the image of L −∞ under the map s − α ) s . The density h α iswell defined for y > (1 − α ) α α/ (1 − α ) by iv) and vanishes for 0 < y < (1 − α ) α α/ (1 − α ) according to ii) (see also 1.33 p.6 in [8]). (cid:4) Special indices
The / -stable case. When α = 1 /
2, the density of the positive stable ran-dom variable reads ([16] p.66) 12 √ π x − / e − / (4 x ) { x> } . According to Kanter’s representation, the image of the above density under themap x /x , that is 12 √ πx e − x/ { x> } = 12 π r πx { x> } e − x/ , should be the Laplace transform of y yh / ( y ) which takes the form (after theuse of the mirror formula): yh / ( y ) = 12 π ∞ X k =0 sin [( k + 1) π/ [( k + 1) /
2] ( − k k ! 1 y ( k +1) / , y > / . Note that x x − / is completely monotone since x x / is a Bernstein function,and that the representing measure is easily computed using the Gamma integral as dy √ πy { y> } . Now, since sin( nπ ) = 0 for any integer n , then yh / ( y ) = 12 π ∞ X k =0 sin [(2 k + 1) π/ [(2 k + 1) / k !) 1 y (2 k +1) / = 12 π y / ∞ X k =0 Γ ( k + 1 / k + 1) 1 y k for y > /
4. Using Legendre’s duplication formula ([7] p.5) √ π Γ(2 k + 1) = 2 k Γ( k + 1 / k + 1) , N POSITIVE CLASSICAL AND FREE STABLE LAWS 7 one gets yh / ( y ) = 12 πy / ∞ X k =0 (1 / k k ! 1(4 y ) k = 12 πy / p − / (4 y ))= 12 π p y − / , y > / , which is nothing but 12 π √ y { y> } ⋆ δ / ( dy )where ⋆ is the classical convolution of positive measures.4.2. Relation to product of independent Beta variables.
Let1 − α = 1 j , α = jj − j ≥
2, then E (cid:20) a − /j ( U )] s (cid:21) = Γ( js + 1)Γ( s + 1)Γ(( j − s + 1) ℜ ( s ) > − /j. The multiplication Theorem (generalization of Legendre’s duplication formula, [7]p.4) Γ( js + 1) = 1( √ π ) j − j js +1 / j Y k =1 Γ (cid:18) s + kj (cid:19) yields E (cid:20) a − /j ( U )] s (cid:21) = 1 √ π s jj − j js ( j − ( j − s j − Y k =1 Γ ( s + k/j )Γ( s + k/ ( j − . Now, let β , β , · · · , β j − be independent Beta random variables of the first kindsuch that the density of β k reads:1 B ( k/j + 1 , k/ ( j ( j − t k/j (1 − t ) k/ ( j − − k/j − (0 , ( t )where B ( · , · ) stands for the Beta function. According to formula (4.11) p.122 in [8], E [( β · · · β j − ) s − ] = j − Y k =1 Γ( k/ ( j −
1) + 1)Γ( k/j + 1) j − Y k =1 Γ( s + k/j )Γ( s + k/ ( j − √ π s j − j j j ( j − ( j − Γ( j − j ) j − Y k =1 Γ( s + k/j )Γ( s + k/ ( j − √ π s j ( j − j j ( j − ( j − j − Y k =1 Γ( s + k/j )Γ( s + k/ ( j − E (cid:20) a − /j ( U )] s (cid:21) = j (cid:20) j j ( j − ( j − (cid:21) s − E [( β · · · β j − ) s − ] . ON POSITIVE CLASSICAL AND FREE STABLE LAWS
When s = 0, the last equality may be written as Z ∞ u s − P ( a − /j ( U ) < /u ) du = j E "(cid:18) j j ( j − ( j − V β · · · β j − (cid:19) s − where V is uniformly distributed in (0 ,
1) and is independent from the β k s. Remark 2.
The relation to product of independent Beta variables is not surprisingsince the H-function involved in the density of h − /j reduces, when − α = 1 /j toa Meijer’s G-function ( [8] p.123). It somewhat matches with Williams result ( [15] )if one keeps in mind the Gamma-Beta algebra. In fact, /X α is distributed, when α = 1 /j, j ≥ as a product of j − Gamma random variables γ k , ≤ k ≤ j − with densities k/j ) t k/j − e − t { t> } , ≤ k ≤ j − . Moreover, since [ a α ( θ )] (1 − α ) /α = a − α ( θ ) , then Williams result implies j j γ · · · γ k d = 1 X /j d = L j − a − /j ( U ) . Note also that, for α = 1 / ( jd ) , j ≥ d, d ≥ , a more general representation ofpositive stable laws is due to T. Simon ( [13] ) and involves independent Gamma andBeta variables. Free positive stable distribution
Free stable distributions may be defined in a similar way as in the classicalsetting, when substituting the classical convolution of probability measures byVoiculescu’s free convolution (see [1]). As mentioned in the introduction, the in-verse function (in the composition’s sense) of a − α provides an explicit expressionof the density of a stable law of index 0 < α < ρ = 1in the free probability setting. More precisely, if V is a uniform random variableon (0 ,
1) defined on some probability space (Ω , F , P ), then1 π a − − α ( x ) = P ( a − α ( πV ) ≤ x )for x > α (1 − α ) (1 − α ) /α . When α = 1 /
2, it simplifies to a − / ( x ) = 12 Z x / dyy p y − / Z x − / dy ( y + 1 / √ y = 2 Z x − / dy y = 2 arctan " r x − . Using the trigonometric identitysin( θ ) = 2 tan( θ/ ( θ/ πx sin[ a − / ( x )] = 2 πx tan( a − / ( x ) / ( a − / ( x ) /
2) = 2 πx √ x −
11 + 4( x − /
4) = √ x − πx . N POSITIVE CLASSICAL AND FREE STABLE LAWS 9
For general α ∈ (0 , H , , (cid:20) y α | (1 , α ) , (1 , − α ))(1 , (cid:21) = 12 iπ Z L −∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) y αs ds. Now, assume for a while that Fubini’s Theorem applies so that1 π a − − α ( x ) = 2 α iπ Z L −∞ Γ(2 s + 1)Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) Z xα (1 − α ) (1 − α ) /α y αs − dyds = 22 iπ Z L −∞ Γ(2 s )Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) h x αs − ( α α (1 − α ) − α ) ) s i ds. Since the poles of s Γ(2 s ) are s = − k/ , k ∈ N and simple, and since the pole s = 0 lies outside L −∞ , then Cauchy’s Residue Theorem yields12 iπ Z L −∞ Γ(2 s )Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) x αs ds = H , , (cid:20) x α | (1 , α ) , (1 , − α ))(0 , (cid:21) − Res( x, x,
0) is the residue of the meromorphic function s Γ(2 s )Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) x αs at s = 0 for fixed parameter x > α (1 − α ) (1 − α ) /α . This residue is easily computedas lim s → s Γ(2 s )Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) x αs = 12 , x > α α (1 − α ) − α ) . Since the last H-function is defined on the whole real line, in particular at α (1 − α ) (1 − α ) /α by the virtue of v) since µ = 0 and δ < − / < −
1, then1 π a − − α ( x ) = 2 H , , (cid:20) x α | (1 , α ) , (1 , − α ))(0 , (cid:21) − H , , (cid:20) α α (1 − α ) − α ) | (1 , α ) , (1 , − α ))(0 , (cid:21) Regarding the continuity of the H-function at α (1 − α ) (1 − α ) /α , it vanishes there(since it vanishes for x < α (1 − α ) (1 − α ) /α ) and one gets1 π a − − α ( x ) = 2 H , , (cid:20) x α | (1 , α ) , (1 , − α ))(0 , (cid:21) , x > α (1 − α ) (1 − α ) /α and zero otherwise. Finally, in order to fit in the density of the free positive stablelaw displayed in (1), we can rewrite the last H-function as1 α H , , (cid:20) x | (1 , , (1 , (1 − α ) /α )(0 , /α ) (cid:21) = 11 − α H , , (cid:20) x α/ (1 − α ) | (1 ,α/ (1 − α ) , (1 , , / (1 − α )) (cid:21) . Note that the duality relation given in [1] allows to derive the density of a freestable distribution of index 1 < α < ρ = 0 (the case ofan asymmetry parameter ρ = 1 is obtained by the simple variable change x
7→ − x ).Coming back to the validity of Fubini’s Theorem, one has to prove, after integratingwith respect to the variable y , the convergence of Z L −∞ | Γ(2 s + 1) || Γ(2 αs + 1)Γ(1 + 2(1 − α ) s ) | h x α ℜ ( s ) − ( α α (1 − α ) − α ) ) ℜ ( s ) i | ds ||ℜ ( s ) | . This is easily cheked from the estimate of Γ( a + ib ) for large a (formula 1.23 p.4.in [8]): | Γ( a + ib ) | ∼ √ π | a | a − / e − a − a (1 − sgn( b )) / , | a | → ∞ which shows that the integrand is equivalent to C |ℜ ( s ) | / e − ( α log( α )+(1 − α ) log(1 − α )) ℜ ( s ) , |ℜ ( s ) | → ∞ for some constant C = C ( ℑ ( s )) depending only on ℑ ( s ).5.1. Behaviour for small indices: Cressie’s result and its free analog.
Inthis section, we investigate the limiting behaviour of h α as α →
0. In the clas-sical setting, the density tends to zero pointwisely therefore agrees somehow withCressie’s result we recall below. The latter implies that a α ( U ) becomes degener-ate when α approaches zero. In the free setting, the limiting density is an infinitepositive measure whose image under the map x /x is the Haar measure on thecompact interval [0 , h α is given by (A.6 p.218 in [8]) h α ( y ) = (1 − α ) ∞ X k =0 − α − kα )Γ( α − k (1 − α )) ( − k k ! 1 y ( k +1)(1 − α )+1 for y > (1 − α ) α α/ (1 − α ) . The mirror formulaΓ( α − k (1 − α ))Γ((1 − α )( k + 1)) = π sin((1 − α )( k + 1) π )transforms h α to h α ( y ) = 1 − απ ∞ X k =0 sin[(1 − α )( k + 1) π ]Γ[(1 − α )( k + 1)]Γ(1 − α − kα ) ( − k k ! 1 y ( k +1)(1 − α )+1 . The defining term of the last series converges as α → − k sin(( k + 1) π ) y k +2 = 0for any integer k ≥ − α ) α α/ (1 − α ) , ∞ ) of h α decreases to[1 , ∞ ) as α →
0. Since Γ[(1 − α )( k + 1)]Γ(1 − α − kα ) k ! → α → α , then Lebesgue’sconvergence Theorem ensures the convergence of h α ( y ), for fixed y >
1, to zero asthe positive index α does. According to [5], (see also [3]), a stable variable withshape parameter c = 1 and zero shift γ = 0 satisfies( X α ) α d → L , α → a α ( U ) converges in distribution to the Dirac measure δ . For a positive freerandom variable, we start from the image of its density function under the map x x α : a − − α ( x /α ) πx sin( αa − − α ( x /α )) αa − − α ( x /α ) sin( a − − α ( x /α ))sin((1 − α ) a − − α ( x /α )) , N POSITIVE CLASSICAL AND FREE STABLE LAWS 11 for x > [ a − α (0)] α = α α (1 − α ) − α ≥ /
2. Then, we expand a − − α ( x /α ) π = 2 H , , (cid:20) x α | (1 , α ) , (1 , − α ))(0 , (cid:21) = ∞ X k =0 − kα )Γ(1 − k (1 − α )) ( − k k ! 1 x k . With the help of the mirror formula,Γ(1 − k (1 − α ))Γ((1 − α ) k ) = π sin( k (1 − α ) π ) , k ≥ , one easily deriveslim α → π a − − α ( x /α ) = 1 + 1 π lim α → ∞ X k =1 sin( k (1 − α ) π )Γ( k (1 − α ))Γ(1 − kα ) ( − k k ! 1 x k = 1for fixed x >
1. Since α α (1 − α ) − α → α →
0, thenlim α → a − − α ( x /α ) πx sin( αa − − α ( x /α )) αa − − α ( x /α ) sin( a − − α ( x /α ))sin((1 − α ) a − − α ( x /α )) = 1 x for any x >
1. Note that Scheff´e’s Lemma does not apply as in the classical setting,nevertheless the image of dxx { x> } under the map x /x is the Haar measure on [0 , dxx {
N POSITIVE CLASSICAL AND FREE STABLE LAWS 13
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