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On the Infinite Divisibility of Probability Distributions with Density Function of Normed Product of Cauchy Densities

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On the Infinite Divisibility of Probability Distributions with Density Function of Normed Product of Cauchy Densities 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<v1,v2;v1+v2<1 2 2 2 2 { a − a u3v1 a − a u3v2} exp ( 3 1) +( 3 2) 2 ·{u3v1u3v2 (u3 − u3v1 − u3v2)}u3dv1dv2 du3 ∞ 2 = c exp{−u3|x| } 0 2 · exp{−a3u3} 0<v1,v2;v1+v2<1 { 2 − 2 2 − 2 } exp ((a3 a1)v1 +(a3 a2)v2)u3 5 ·{v1v2(1 − v1 − v2)}dv1dv2 u3du3. 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<v1,v2;v1+v2<1 2 2 2 2 { a − a v1 a − a v2 /v} exp (( 3 1) +( 3 2) ) ·{v1v2(1 − v1 − v2)}dv1dv2 dv 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<v ,v ;v +v <1 1 2 1 2 ·{v1v2 (1 − v1 − v2)}dv1dv2 . We obtain the variance mixture density function (1) of the 3 dimensional normal density functions. The mixing density function is 3/2 2 cπ {−a3 } g(3; 3; v)= 11/2 exp v v 2 2 2 2 · exp{((a3 − a1)v1 +(a3 − a2)v2)/v} 0<v1,v2;v1+v2<1 ·{v1v2 (1 − v1 − v2)}dv1dv2. If this probability density function is infinitely divisible then the probability density function (1) 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 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<v <1;0<y<1 1 v 2 v1(1 − v1) y(1 − y) (1 − v1)dydv1 3/2 2 cπ {−a3 } = 11/2 exp v v 2 2 2 2 1 · exp{((a3 − a1)v1 +(a3 − a2)(1 − v1)y) } 0<v1<1;0<y<1 v 3 v1(1 − v1) y(1 − y)dv1dy.
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