2016 HAWAII UNIVERSITY INTERNATIONAL CONFERENCES SCIENCE, TECHNOLOGY, ENGINEERING, ART, MATH & EDUCATION JUNE 10 - 12, 2016 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 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“. ON THE INFINITE DIVISIBILITY OF PROBABILITY DISTRIBUTIONS WITH DENSITY FUNCTION OF NORMED PPRODUCT OF THE CAUCHY DENSITIES
Katsuo TAKANO Ibaraki University Mito-city, Ibaraki 310, JAPAN
1 Introduction
It is well-known that the Cauchy distribution is infinitely divisible. This fact is coming from the following equality, ∞ 1 eitx dx = e−|t| −∞ π(1 + x2) − |t| n =(e n ) 1 ∞ n ity n = e 1 2 dy −∞ π( n2 + y ) for every positive integer n. This shows the Cauchy distribution can be ex- pressed as the (n−1)-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(1, 2, ···,n; x)= (1 + x2)(22 + x2)(32 + x2) ···(n2 + x2) are infinitely divisible, and it is also known that the density function of normed product of two 3 dimensional Cauchy densities
c 3 f(1, 2; x)= ,x=(x1,x2,x3) ∈ R (1 + |x|2)2(22 + |x|2)2
1 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 dimen- sional Cauchy densities is infinitely divisible or not. The goal of this note is to obtain a prospect that the following probability density c f(1, 2, 3, ···,n; x)= , (1 + |x|2)2(22 + |x|2)2(32 + |x|2)2 ···(n2 + |x|2)2
3 x =(x1,x2,x3) ∈ R is infinitely divisible. In this note, the author discusses on the infinite di- visiblity of the probability distribution with the density function of normed product of the triple 3 dimensional Cauchy densities such as c f(1, 2, 3; x)= , (1) (1 + |x|2)2(22 + |x|2)2(32 + |x|2)2 where 3 x =(x1,x2,x3) ∈ R and c is determined by f(1, 2, 3; x)dx =1. R3 We give the definition of an infinitely divisible distribution in the 1- dimensional case. For the definition of an infinitely divisible distribution in the multi-dimensional case, refer to Sato’s book [7]. A probability distri- bution function F (x) is called an infinitely divisible distribution if for each integer n>1 there is a probability distribution Fn(x) such that
F (x)=(Fn ∗···∗Fn)(x),
(* denotes the convolution.). The convolution is defined by
∞ (Fn ∗ Fn)(x)= Fn(x − y)dFn(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 = Xn1 + ···+ Xnn of n independent identically distributed ran- dom variables X1n,..., Xnnand Fn(x) is the probability distribution function P (Xnj ≤ x). If a probability distribution function F (x) is 0 over the inter- val (−∞, 0) and infinitely divisible, and if we denote the Laplace-Stieltjes taransforms of the probability distributions F (x) and Fn(x),
∞ ∞ −sx −sx L(s)= e dF (x),Ln(s)= e dFn(x), 0 0
2 the equality n L(s)=(Ln(s)) holds. It is known that the Laplace-Stieltjes transform of an infinitely divis- ible probability distribution F (x) over the interval [0, ∞) can be written as follows: ∞ 1 L(s) = exp{ (e−sx − 1) dK(x)} −0 x where
(c1) K(x) is nondecreasing,
(c2) K(−0) = 0,
∞ (c3) 1 1/x dK(x) < ∞. An infinitely divisible probability distribution F (x)on[0, ∞) satisfies an integral equation,
x x tdF(t)= F (x − t)dK(t),x>0 0 0 and in particular if the probability distribution function F (x)on[0, ∞) 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>0 0 (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 > 0. 0 Making use of the Laplace transform we obtain
−L(s)=L(s)L(k; s), where ∞ L(s)= exp{−sx}f(x)dx, 0 ∞ L(k; s)= exp{−st}k(t)dx. 0 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 2 The variance mixture density of the Nor- mal distributions
Consider the probability density function with normed product of the triple 3 dimensional Cauchy densities
f(a1,a2,a3; x) c = 2 2 (d+1)/2 2 2 (d+1)/2 2 2 (d+1)/2 (a1 + |x| ) (a2 + |x| ) (a3 + |x| ) d =3; a1 =1,a2 =2,a3 =3. The number c is normalized such that f(x)dx =1. R3 We have Γ((d +1)/2) ∞ = exp{−t(a2 + |x|2)}·t(d+1)/2−1dt (a2 + |x|2)(d+1)/2 0 and see that ∞ c 2 2 (d+1)/2−1 f(x)= exp{−t1(a1 + |x| )}·t1 dt1 {Γ((d +1)/2)}3 0 ∞ 2 2 (d+1)/2−1 exp{−t2(a2 + |x| )}·t2 dt2 0 ∞ 2 2 (d+1)/2−1 exp{−t3(a3 + |x| )}·t3 dt3 0 ∞ ∞ ∞ c 2 = exp{−(t1 + t2 + t3)|x| {Γ((d +1)/2)}3 0 0 0 2 2 2 (d+1)/2−1 − a1t1 − a2t2 − a3t3}·(t1t2t3) dt1dt2dt3 By a change of variables,
t1 = u1,t2 = u2,t3 = u3 − u1 − u2 we have the Jacobian, 100 D(t1,t2,t3) = 010 =1 D(u1,u2,u3) −1 −11 and we have c {− | |2 f(x)= 3 exp u3 x {Γ((d +1)/2)} 0≤u1,u2,u3;u1+u2≤u3 2 2 2 − a1u1 − a2u2 − a3(u3 − u1 − u2)} (d+1)/2−1 ·{u1u2(u3 − u1 − u3)} du1du2du3
4 If d = 3 then (d +1)/2 = 2 and Γ(2) = 1. From the above we see that 2 f(x)=c exp{−u3|x| 0≤u1,u2,u3;u1+u2≤u3 2 2 2 − a1u1 − a2u2 − a3(u3 − u1 − u2)} ·{u1u2(u3 − u1 − u3)}du1du2du3.
Making use of the Fubini theorem since the integrand is non-negative, we can change the order of the three fold integrals such as ∞ 2 f(x)=c exp{−u3|x| } 0 2 · exp{−a3u3} 0≤u1,u2;u1+u2≤u3 { 2 − 2 2 − 2 } exp (a3 a1)u1 +(a3 a2)u2 ·{u1u2(u3 − u1 − u2)}du1du2 du3.
Make a change of variables, u1 = u3v1,u2 = u3v2, and ignore the boundary set of the volume measure zero in the above three fold integrals {(u1,u2,u3): u1 =0, 0 ≤ u2 ≤ u3}∪{(u1,u2,u3):u2 =0, 0 ≤ u1 ≤ u3}∪{0 ≤ u1,u2; u1 + u2 = u3}. Then we have
∞ 2 f(x)=c exp{−u3|x| } 0 2 · exp{−a3u3} 0 Make a change of a variable, u3 =1/v and we obtain ∞ 1 |x|2 f(x)= exp{− } 0 2 v v · c {− 2 } 5 exp a3/v v 0 5 and we see that ∞ 1 |x|2 cπ3/2 a2 f(x)= exp{− }dy exp{− 3 } 0 3/2 11/2 (πv) v v v 2 2 2 2 · exp{((a3 − a1)v1 +(a3 − a2)v2)/v} 0 3 The mixing density function g(3; 3; v) By a change of variables in the above double integrals v1 = u1,v2 =(1− u1)y we have the Jacobian, D(v1,v2) 10 − = =1 u1. D(u1,y) −y 1 − u1 We see that 3/2 2 cπ {−a3 } g(3; 3; v)= 11/2 exp v v 2 2 2 2 1 · exp{((a3 − a1)v1 +(a3 − a2)(1 − v1)y) } 0 6 At last, we obtain a simple form of the mixing density function, 3/2 2 cπ 1 {−a3 } g(3; 3; v)= 3/2 2 2 2 2 2 2 exp v (a3 − a2) (a3 − a1) v 2 1 {−a2 } + 2 2 2 2 2 2 exp (a2 − a3) (a2 − a1) v 2 1 {−a1 } + 2 2 2 2 2 2 exp (a1 − a2) (a1 − a3) v 3/2 2 2cπ 1 2 − 2 − 2 {−a3 } + 1/2 2 2 3 2 2 3 2a3 a1 a2 exp v (a3 − a1) (a3 − a2) v 2 1 2 − 2 − 2 {−a2 } + 2 2 3 2 2 3 2a2 a3 a1 exp (a2 − a1) (a2 − a3) v 2 1 2 − 2 − 2 {−a1 } + 2 2 3 2 2 3 2a1 a2 a3 exp . (2) (a1 − a2) (a1 − a3) v 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 0 √ 1 1 1 1 √2 {− } = c 2 2 + + exp 2a1 s b12b13 a1 b12 b13 s √ 1 1 1 1 √2 {− } + 2 2 + + exp 2a2 s b21b23 a2 b21 b23 s √ 1 1 1 1 √2 {− } + 2 2 + + exp 2a3 s b31b32 a3 b31 b32 s 2 2 where let Re s ≥ 0 and a1 =1,a2 =2,a3 = 3 and let b12 = a1 − a2,b13 = 2 2 2 2 2 2 2 2 2 2 a1 − a3,b21 = a2 − a1,b23 = a2 − a3,b31 = a3 − a1,b32 = a3 − a2. It is 2 2 L(3; 3; s)b12b13/c 3 √ 32 √ √ = exp(−6 s)+ exp(−4 s) + exp(− s) 25 25 117 √ 256 √ 11 √ + √ exp(−6 s)+ √ exp(−4 s) − √ exp(− s) 500 s 375 s 12 s and we take a branch such that L(3; 3; +0) = 1. We will explain the above result. Denote the Laplace-Stieltjes transform of the mixing density g(3; 3; v) 7 by L(3; 3; s). We note that under the condition Re s ≥ 0 the Laplace trans- form 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 0 √ 1 1 1 1 √2 {− } = c 2 2 + + exp 2a1 s b12b13 a1 b12 b13 s √ 1 1 1 1 √2 {− } + 2 2 + + exp 2a2 s b21b23 a2 b21 b23 s √ 1 1 1 1 √2 {− } + 2 2 + + exp 2a3 s b31b32 a3 b31 b32 s 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; +0) = 1. We obtain 2 2 L(3; 3; s)b12b13/c 3 √ 32 √ √ = exp(−6 s)+ exp(−4 s) + exp(−2 s) 25 25 117 √ 256 √ 11 √ + √ exp(−6 s)+ √ exp(−4 s) − √ exp(−2 s) 500 s 375 s 12 s concretely and write it 3 32 L(3; z)= exp(−6z)+ exp(−4z) + exp(−2z) 25 25 117 256 11 + exp(−6z)+ exp(−4z) − exp(−2z) (3) √ 500z 375z 12z by replacing s by z. We note that 117 256 11 + − =0 500 375 12 and 3 32 117 256 11 37 + +1+ (−6) + (−4) − (−2) = . 25 25 500 375 12 375 From these facts it is seen that L(3; z) is bounded and does not vanish in the neighborhood of origin. 5 The inverse Laplace transform and spec- tral function Next we will talk about the inverse Laplace transform to obtain k(t). We make use of the polar coordinate s = ρeiθ, (−π<θ<π,0 <ρ). 8 Then √ √ √ θ θ s = ρeiθ/2 = ρ(cos + i sin ) 2 2 √ and 0 < ρ cos(θ/2). Let us calculate the spectral function k(t) by the formula 1 ξ+iR1 L (3; 3; s) k(t) = lim ets(−1) ds, R→∞ 2πi ξ−iR1 L(3; 3; s) (0 <ξ,;0 9 L(3; z) ds L(3; z) ds − ets(−1) √ − ets(−1) √ H→E L(3; z) 2 s EF L(3; z) 2 s L(3; z) ds − ets(−1) √ . FA L(3; z) 2 s It holds that (S1) L(3; z) ds ets(−1) √ → 0, BC L(3; z) 2 s (S2) L(3; z) ds ets(−1) √ → 0, FA L(3; z) 2 s (S3) L(3; z) ds ets(−1) √ → 0, CD L(3; z) 2 s (S4) L(3; z) ds ets(−1) √ → 0 EF L(3; z) 2 s as R →∞and (S5) L(3; z) ds ets(−1) √ → 0 GH L(3; z) 2 s as r → 0. Therefore we obtain 1 ξ+iR1 L (3; z) ds k(t) = lim ets(−1) √ R→∞ 2πi ξ−iR1 L(3; z) 2 s 1 L(3; z) ds = lim { ets √ R→∞,r→0 2πi D→G L(3; z) 2 s 1 L(3; z) ds + ets √ }. 2πi H→E L(3; z) 2 s and at last we obtain k(t) √ r iπ/2 1 iπ L (3; ρe ) = lim { etρe √ √ eiπdρ R→∞,r→0 2πi R 2 ρeiπ/2L(3; ρeiπ/2) √ R −iπ/2 1 −iπ L (3; ρe ) + etρe √ √ e−iπdρ} 2πi r 2 ρe−iπ/2L(3; ρe−iπ/2) √ 1 R L(3; ρi) = lim { e−tρ √ √ dρ R→∞,r→0 2πi r 2 ρi L(3; ρi) 10 √ 1 R L(3; −i ρ) + e−tρ √ √ (−1)dρ} 2πi r 2 ρ(−i) L(3; − ρi) √ 1 R L(3; ρi) = lim {− e−tρ √ √ dρ R→∞,r→0 2π r 2 ρL(3; ρi) √ 1 R L(3; − ρi) − e−tρ √ √ dρ} 2π r 2 ρL(3; − ρi) √ √ 1 R L(3; ρi) L(3; −i ρ) dρ = lim {− e−tρ( √ + √ ) √ R→∞,r→0 2π r L(3; ρi) L(3; − ρi) 2 ρ From the above we see that 1 ∞ k(t)= e−tρ 2π √0 √ √ √ L(3; ρi)L(3; − ρi)+L(3; − ρi)L(3; ρi) dρ (−1) √ √ √ (5) L(3; ρi)L(3; − ρi) 2 ρ It is necessary to show that √ √ √ √ N =(−1){L(3; ρi)L(3; − ρi)+L(3; − ρi)L(3; ρi)} is non-negative and by computations we obtain √ √ √ √ N =(−1){L(3; ρi)L(3; − ρi)+L(3; − ρi)L(3; ρi)} 4 2 − 2 = 2 243 + 540y + 64( 2+9y ) cos(2y) 125y +5(−23 + 12y2) cos(4y) − 832 sin(2y) − 172 sin(4y) . (6) √ Here y equals ρ.It is seen that as y →∞, 216 1152 24 N/2 ≈ + cos(2y)+ cos(4y) 25 125 25 48 1344 192 = + (cos(y))2 + (cos(y))4. (7) 125 125 25 By making use of the computer we can see that N is positive in the bounded interval like the interval (0,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 (1) is infinitely divisible. 11 References [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, New York, Dover, 1970. [2] L. Bondesson, Generalized Gamma Convolutions and Related Classes of Distributions and Densities, Lecture Note in Statistics, 76, Springer- Verlag, 1992. [3] W. Feller, An Introduction to probability theory and its applica- tions,Vol.2, Second edition, Wily, New York, 1971. [4] M. J. Goovaerts, L. D’Hooge, and N. 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[11] —–, On Laplace transforms of some probability densities, Applied Math- ematics and computations, 187 (2007), 501–506. [12] O. Thorin, On the infinite divisibility of the Pareto distribution, Scand. Acturial. J., (1977), 31–40. [13] G. N. Watson, a treatise on the THEORY OF BESSEL FUNCTIONS, Cambridge University Press, Second edition, Reprinted 1980. 12 C B R ε r G D E O ξ H −r −R A F 13 PolarMap of L(3; z) 0.03 0.02 0.01 0.08 0.09 0.11 0.12 0.13 0.14 -0.01 -0.02 -0.03 14 0.1 0.05 0.05 0.15 0.2 0.25 -0.05 -0.1 15 Graphics of N(y)/2 1 0.8 0.6 0.4 0.2 0.4 0.6 0.8 1.2 1.4 16 20 15 10 5 2 4 6 8 17 17.5 15 12.5 10 7.5 5 2.5 10 12 14 16 18 18