ON THE INFINITE DIVISIBILITY OF PROBABILITY DISTRIBUTIONS WITH DENSITY FUNCTION OF NORMED PRODUCT OF CAUCHY DENSITIES

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1 26 HAWAII UNIVERSITY INTERNATIONAL CONFERENCES SCIENCE, TECHNOLOGY, ENGINEERING, ART, MATH & EDUCATION JUNE - 2, 26 HAWAII PRINCE HOTEL WAIKIKI, HONOLULU ON THE INFINITE DIVISIBILITY OF PROBABILITY DISTRIBUTIONS WITH DENSITY FUNCTION OF NORMED PRODUCT OF CAUCHY DENSITIES TAKANO, KATSUO IBARAKI UNIVERSITY JAPAN

2 Dr. Katsuo Takano Ibaraki University Japan. On the Infinite Divisibility of Probability Distributions with Density Function of Normed Product of Cauchy Densities Synopsis: The speaker will talk about the possible infinite divisibility of probability distributions with density function of normd product of the multi-dimensional Cauchy densities by using the soft "Mathematica.

3 ON THE INFINITE DIVISIBILITY OF PROBABILITY DISTRIBUTIONS WITH DENSITY FUNCTION OF NORMED PPRODUCT OF THE CAUCHY DENSITIES Katsuo TAKANO Ibaraki University Mito-city, Ibaraki 3, JAPAN Introduction It is well-known that the Cauchy distribution is infinitely divisible. This fact is coming from the following equality, e itx dx π( + x 2 ) = e t = (e t n ) n = ( e ity n π( n 2 + y 2 ) dy) n for every positive integer n. This shows the Cauchy distribution can be expressed as the (n )-fold convolutions of the Cauchy distribution with itself. It is known that all of the probability distribution with the density function of normed product of the n Cauchy densities c f(, 2,,n; x) = ( + x 2 )(2 2 + x 2 )(3 2 + x 2 ) (n 2 + x 2 ) are infinitely divisible, and it is also known that the density function of normed product of two 3 dimensional Cauchy densities c f(, 2; x) = ( + x 2 ) 2 (2 2 + x 2 ), x =(x,x 2 2,x 3 ) R 3

4 is infinitely divisible. But it seems that it is not known if the probability distriution with the density function of normed product of triple 3 dimensional Cauchy densities is infinitely divisible or not. The goal of this note is to obtain a prospect that the following probability density f(, 2, 3,,n; x) = c ( + x 2 ) 2 (2 2 + x 2 ) 2 (3 2 + x 2 ) 2 (n 2 + x 2 ) 2, x =(x,x 2,x 3 ) R 3 is infinitely divisible. In this note, the author discusses on the infinite divisiblity of the probability distribution with the density function of normed product of the triple 3 dimensional Cauchy densities such as where f(, 2, 3; x) = and c is determined by c ( + x 2 ) 2 (2 2 + x 2 ) 2 (3 2 + x 2 ) 2, () x =(x,x 2,x 3 ) R 3 f(, 2, 3; x)dx =. R 3 We give the definition of an infinitely divisible distribution in the - dimensional case. For the definition of an infinitely divisible distribution in the multi-dimensional case, refer to Sato s book [7]. A probability distribution function F (x) is called an infinitely divisible distribution if for each integer n> there is a probability distribution F n (x) such that F (x) =(F n F n )(x), (* denotes the convolution.). The convolution is defined by (F n F n )(x) = F n (x y)df n (y) In terms of random variable we shall say that the random variable X is infinitely divisible if for every natural number n it can be represented as the sum X = X n + + X nn of n independent identically distributed random variables X n,..., X nn and F n (x) is the probability distribution function P (X nj x). If a probability distribution function F (x) is over the interval (, ) and infinitely divisible, and if we denote the Laplace-Stieltjes taransforms of the probability distributions F (x) and F n (x), L(s) = e sx df (x), L n (s) = 2 e sx df n (x),

5 the equality L(s) =(L n (s)) n hol. It is known that the Laplace-Stieltjes transform of an infinitely divisible probability distribution F (x) over the interval [, ) can be written as follows: L(s) = exp{ (e sx ) x dk(x)} where (c) K(x) is nondecreasing, (c2) K( ) =, (c3) /x dk(x) <. An infinitely divisible probability distribution F (x) on[, ) satisfies an integral equation, x tdf(t) = x F (x t)dk(t), x> and in particular if the probability distribution function F (x) on[, ) has a density function f(x), the density funcion f(x) satisfies the following integral equation: x xf(x) = f(x t)dk(t), x> (cf. F. Steutel [8]). If dk(t) is absolutely continuous, i.e. dk(t) = k(t)dt we obtain the integral equation x xf(x) = f(x t)k(t)t, x >. Making use of the Laplace transform we obtain where L(s) = L (s) =L(s)L(k; s), L(k; s) = exp{ sx}f(x)dx, exp{ st}k(t)dx. And we can find the spectral function k(t) by making use of the inverse Laplace transform on the figure after the references. By computations the author tries to show that the spectral function k(t) for the mixing density g(3; 3; v) is a positive function. 3

6 2 The variance mixture density of the Normal distributions Consider the probability density function with normed product of the triple 3 dimensional Cauchy densities = f(a,a 2,a 3 ; x) c (a 2 + x 2 ) (d+)/2 (a x 2 ) (d+)/2 (a x 2 ) (d+)/2 d =3; a =,a 2 =2,a 3 =3. The number c is normalized such that f(x)dx =. R 3 We have Γ((d +)/2) (a 2 + x 2 ) = exp{ t(a 2 + x 2 )} t (d+)/2 dt (d+)/2 and see that c f(x) = exp{ t {Γ((d +)/2)} 3 (a 2 + x 2 )} t (d+)/2 dt exp{ t 2 (a x 2 )} t (d+)/2 2 dt 2 = exp{ t 3 (a x 2 )} t (d+)/2 c {Γ((d +)/2)} 3 3 dt 3 exp{ (t + t 2 + t 3 ) x 2 a 2 t a 2 2t 2 a 2 3t 3 } (t t 2 t 3 ) (d+)/2 dt dt 2 dt 3 By a change of variables, we have the Jacobian, t = u,t 2 = u 2,t 3 = u 3 u u 2 D(t,t 2,t 3 ) D(u,u 2,u 3 ) = = and we have c f(x) = exp{ u {Γ((d +)/2)} 3 3 x 2 u,u 2,u 3 ;u +u 2 u 3 a 2 u a 2 2 u 2 a 2 3 (u 3 u u 2 )} {u u 2 (u 3 u u 3 )} (d+)/2 du du 2 du 3 4

7 If d = 3 then (d +)/2 = 2 and Γ(2) =. From the above we see that f(x) =c exp{ u 3 x 2 u,u 2,u 3 ;u +u 2 u 3 a 2 u a 2 2u 2 a 2 3(u 3 u u 2 )} {u u 2 (u 3 u u 3 )}du du 2 du 3. Making use of the Fubini theorem since the integrand is non-negative, we can change the order of the three fold integrals such as f(x) =c exp{ u 3 x 2 } [ exp{ a 2 3 u 3 } u,u 2 ;u +u 2 u 3 exp{(a 2 3 a 2 )u +(a 2 3 a 2 2)u 2 } ] {u u 2 (u 3 u u 2 )}du du 2 du3. Make a change of variables, u = u 3 v,u 2 = u 3 v 2, and ignore the boundary set of the volume measure zero in the above three fold integrals {(u,u 2,u 3 ): u =, u 2 u 3 } {(u,u 2,u 3 ):u 2 =, u u 3 } { u,u 2 ; u + u 2 = u 3 }. Then we have f(x) =c exp{ u 3 x 2 } [ exp{ a 2 3 u 3 } <v,v 2 ;v +v 2 < exp{(a 2 3 a2 )u 3v +(a 2 3 a2 2 )u 3v 2 } {u 3 v u 3 v 2 (u 3 u 3 v u 3 v 2 )}u 2 3 dv dv 2 ] du3 = c exp{ u 3 x 2 } [ exp{ a 2 3 u 3 } <v,v 2 ;v +v 2 < exp{((a 2 3 a 2 )v +(a 2 3 a 2 2)v 2 )u 3 } ] {v v 2 ( v v 2 )}dv dv 2 u 5 3 du 3. Make a change of a variable, u 3 =/v and we obtain f(x) = v 2 exp{ x 2 v } [ c v 5 exp{ a2 3 <v /v},v 2 ;v +v 2 < exp{((a 2 3 a2 )v +(a 2 3 a2 2 )v 2)/v} ] {v v 2 ( v v 2 )}dv dv 2 dv 5

8 and we see that f(x) = <v,v 2 ;v +v 2 < cπ 3/2 exp{ x 2 (πv) 3/2 v }dy[ v /2 exp{ a2 3 v } {v v 2 ( v v 2 )}dv dv 2 ]. exp{((a 2 3 a 2 )v +(a 2 3 a 2 2)v 2 )/v} We obtain the variance mixture density function () of the 3 dimensional normal density functions. The mixing density function is g(3; 3; v) = cπ3/2 v /2 exp{ a2 3 v } exp{((a 2 3 a 2 )v +(a 2 3 a 2 2)v 2 )/v} <v,v 2 ;v +v 2 < {v v 2 ( v v 2 )}dv dv 2. If this probability density function is infinitely divisible then the probability density function () is infinitely divisible (cf. Feller [3]). The author is going to show that the mixing density function g(3; 3; v) is possible infinitely divisible. 3 The mixing density function g(3; 3; v) By a change of variables in the above double integrals we have the Jacobian, D(v,v 2 ) D(u,y) = We see that v = u,v 2 =( u )y y u = u. g(3; 3; v) = cπ3/2 v /2 exp{ a2 3 v } exp{((a 2 3 a2 )v +(a 2 3 a2 2 )( v )y) <v <;<y< v } [ v ( v ) 2 y( y) ] ( v )dydv = cπ3/2 v /2 exp{ a2 3 v } exp{((a 2 3 a 2 )v +(a 2 3 a 2 2)( v )y) v } <v <;<y< v ( v ) 3 y( y)dv dy. 6

9 At last, we obtain a simple form of the mixing density function, g(3; 3; v) = cπ3/2 [ v 3/2 (a 2 3 a 2 2) 2 (a 2 3 a 2 ) 2 exp{ a2 3 v } + (a 2 2 a 2 3) 2 (a 2 2 a 2 ) 2 exp{ a2 2 v } + (a 2 a 2 2) 2 (a 2 a 2 3) 2 exp{ a2 v }] + 2cπ3/2 [ ( 2a 2 v /2 (a 2 3 a 2 ) 3 (a 2 3 a 2 2) 3 3 a 2 ) a 2 a2 2 exp{ 3 v } + ( ) 2a 2 (a 2 2 a 2 ) 3 (a 2 2 a 2 3) 3 2 a 2 3 a 2 a 2 exp{ 2 v } + ( ) 2a 2 (a 2 a 2 2) 3 (a 2 a 2 3) 3 a 2 2 a 2 a 2 3 exp{ v }]. (2) 4 The Laplace transform of g(3; 3; v) Denote the Laplace transform of the mixing density function g(3; 3; v) by L(3; 3; s). Then we obtain L(3; 3; s) = exp{ sv}g(3; 3; v)dv = c [ [ + { + } 2 ] s exp{ 2a s} b 2 2b 2 3 a b 2 b 3 + [ + { + } 2 ] s exp{ 2a2 s} b 2 2b 2 23 a 2 b 2 b 23 + [ + { + } 2 ] ] s exp{ 2a3 s} a 3 b 3 b 32 b 2 3b 2 32 where let Re s and a =,a 2 =2,a 3 = 3 and let b 2 = a 2 a2 2, b 3 = a 2 a 2 3,b2 = a 2 2 a 2,b23 = a 2 2 a 2 3,b3 = a 2 3 a 2,b32 = a 2 3 a 2 2. It is L(3; 3; s)b 2 2b 2 3/c = 3 25 exp( 6 s) exp( 4 s) + exp( s) s exp( 6 s) s exp( 4 s) 2 s exp( s) and we take a branch such that L(3; 3; +) =. We will explain the above result. Denote the Laplace-Stieltjes transform of the mixing density g(3; 3; v) 7

10 by L(3; 3; s). We note that under the condition Re s the Laplace transform converges and the right hand side is a holomorphic function of the variable s in the right half complex plane. Then we obtain L(3; 3; s) = exp{ sv}g(3; 3; v)dv = c [ [ + { + } 2 ] s exp{ 2a s} b 2 2b 2 3 a b 2 b 3 + [ + { + } 2 ] s exp{ 2a2 s} b 2 2b 2 23 a 2 b 2 b 23 + [ + { + } 2 ] ] s exp{ 2a3 s} b 2 3b 2 32 a 3 b 3 b 32 By analytic continuation we obtain the Laplace transform L(3; 3; s) in the complex plane except the origin and we take the branch such that L(3; 3; +) =. We obtain L(3; 3; s)b 2 2 b2 3 /c = 3 25 exp( 6 s) exp( 4 s) + exp( 2 s) s exp( 6 s) s exp( 4 s) 2 s exp( 2 s) concretely and write it = 3 25 exp( 6z)+32 exp( 4z) + exp( 2z) exp( 6z)+ exp( 4z) exp( 2z) (3) 5z 375z 2z by replacing s by z. We note that = and ( 6) + ( 4) ( 2) = From these facts it is seen that is bounded and does not vanish in the neighborhood of origin. 5 The inverse Laplace transform and spectral function Next we will talk about the inverse Laplace transform to obtain k(t). We make use of the polar coordinate s = ρe iθ, ( π <θ<π, <ρ). 8

11 Then s = ρe iθ/2 = ρ(cos θ 2 + i sin θ 2 ) and < ρ cos(θ/2). Let us calculate the spectral function k(t) by the formula ξ+ir k(t) = lim e ts ( ) L (3; 3; s) R 2πi ξ ir L(3; 3; s), ( <ξ, ɛ;<t; R = R cos ɛ), on the figure after the References. Since s = z we see that dl(3; 3; s) L(3; 3; s) = d dz 2z and k(t) = lim R ξ+ir 2πi ξ ir e ts ( ) L (3; z) 2 s, (4) From computations we can see that the function does not vanish in the bounded region near at origin in the right half complex plane. And we see that as z 3 25 exp( 6z)+32 exp( 4z) + exp( 2z) 25 and if we put Z := exp( 2z) we obtain at most two roots from the equation 3 25 Z Z += and if z R and R is sufficiently large the function does not have zero points in the right half complex plane. From the Cauchy theorem we obtain e ts ( ) L (3; z) γ 2 s = where γ is a curve of the figure and hence we obtain the following equality = A B B C D G e ts ( ) L (3; z) e ts ( ) L (3; z) e ts ( ) L (3; z) 2 s 2 s C D e ts ( ) L (3; z) 2 s 2 s G H e ts ( ) L (3; z) 2 s 9

12 H E F A e ts ( ) L (3; z) e ts ( ) L (3; z) 2 s E F 2 s. e ts ( ) L (3; z) 2 s It hol that (S) (S2) (S3) (S4) as R and (S5) B C F A C D E F G H e ts ( ) L (3; z) e ts ( ) L (3; z) e ts ( ) L (3; z) e ts ( ) L (3; z) e ts ( ) L (3; z) 2 s, 2 s, 2 s, 2 s 2 s as r. Therefore we obtain ξ+ir k(t) = lim e ts ( ) L (3; z) R 2πi ξ ir = lim { e ts L (3; z) R, r 2πi D G 2 s + 2πi and at last we obtain k(t) = lim { R, r 2πi H E r R e ts L (3; z) 2 s }. 2 s e tρeiπ L (3; ρe iπ/2 ) 2 ρe iπ/2 L(3; ρe iπ/2 ) eiπ dρ + R e tρe iπ L (3; ρe iπ/2 ) 2πi r 2 ρe iπ/2 L(3; ρe iπ/2 ) e iπ dρ} = lim { R e tρ L (3; ρi) R, r 2πi r 2 ρi L(3; ρi) dρ

13 + R e tρ L (3; i ρ) 2πi r 2 ρ( i) L(3; ρi) ( )dρ} = lim { R e tρ L (3; ρi) R, r 2π r 2 ρl(3; ρi) dρ R e tρ L (3; ρi) 2π r 2 ρl(3; ρi) dρ} = lim { R e tρ ( L (3; ρi) R, r 2π r L(3; ρi) + L (3; i ρ) L(3; ρi) ) dρ } 2 ρ From the above we see that k(t) = e tρ 2π ( ) L (3; ρi)l(3; ρi)+l (3; ρi)l(3; ρi) L(3; ρi)l(3; ρi) dρ 2 ρ (5) It is necessary to show that N =( ){L (3; ρi)l(3; ρi)+l (3; ρi)l(3; ρi)} is non-negative and by computations we obtain N =( ){L (3; ρi)l(3; ρi)+l (3; ρi)l(3; ρi)} = 4 ( y ( 2+9y 2 ) cos(2y) 25y 2 + 5( y 2 ) cos(4y) 832 sin(2y) 72 sin(4y) ). (6) Here y equals ρ.it is seen that as y, N/ cos(2y) cos(4y) = (cos(y)) (cos(y))4. (7) By making use of the computer we can see that N is positive in the bounded interval like the interval (,l) and therefore the author concludes that the mixing density function g(3; 3; v) is infinitely divisible and consequently that the variance mixture density of normed product of the triple Cauchy densities () is infinitely divisible.

14 References [] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, New York, Dover, 97. [2] L. Bondesson, Generalized Gamma Convolutions and Related Classes of Distributions and Densities, Lecture Note in Statistics, 76, Springer- Verlag, 992. [3] W. Feller, An Introduction to probability theory and its applications,vol.2, Second edition, Wily, New York, 97. [4] M. J. Goovaerts, L. D Hooge, and N. De Pril, On the infinite divisibility of the product of two Γ-distributed stochastical variables, Applied mathematics and computation, 3 (977), [5] D. H. Kelker, Infinite divisibility and variance mixtures of the normal distribution, Ann. Math. Statist., 42 (97), [6] G. Sansone & J. Gerretsen, Lectures on the theory of functions of a complex variable,. Holomorphic functions, P. Noordhoff-Groningen, 96 [7] K. Sato, Lévy Processes and Infinitely Divisible Distributions, Cambridge Univ. Press, Cambridge, 999. [8] F.W.Steutel, K.van Harn, Infinite divisibility of probability distributions on the real line, Marcel Dekker, 24 [9] F.W.Steutel, Some recent results in infinite divisibility, Stochastic Processes and their Applications, (973), [] K. Takano, Hypergeometric functions and infinite divisibility of probability distributions consisting of Gamma functions, International J. Pure and Applied Math., 2 no.3 (25), [], On Laplace transforms of some probability densities, Applied Mathematics and computations, 87 (27), [2] O. Thorin, On the infinite divisibility of the Pareto distribution, Scand. Acturial. J., (977), 3 4. [3] G. N. Watson, a treatise on the THEORY OF BESSEL FUNCTIONS, Cambridge University Press, Second edition, Reprinted 98. 2

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