Hyun-Woo Jin* and Min-Young Lee**
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1 JOURNAL OF THE CHUNGCHEONG MATHEMATICAL SOCIETY Volue 27, No. 2, May CHARACTERIZATIONS OF THE GAMMA DISTRIBUTION BY INDEPENDENCE PROPERTY OF RANDOM VARIABLES Hyun-Woo Jin* and Min-Young Lee** Abstract. Let {X i, 1 i n} be a sequence of i.i.d. sequence of positive rando variables with coon absolutely continuous cuulative distribution function F (x and probability density function f(x and E(X 2 <. The rando variables X + Y and (X Y 2 (X+Y 2 are independent if and only if X and Y have gaa distributions. In addition, the rando variables S n and (Xi2 i=1 with S n = n i=1 X i are independent for 1 < n if and only if X i has gaa distribution for i = 1,, n. 1. Introduction Let {X i, 1 i n} be independent and identically distributed(i.i.d. non-degenerate and positive rando variables with coon absolutely continuous cuulative distribution function F (x. Let X and Y be two independent non-degenerate positive rando variables. It is known that X/Y and X + Y are independent if and only if X and Y are gaa distributions with the sae scale paraeter as used in Lukacs(1955. By using the oent, Findeisen(1978 characterized the gaa distribution. Also, Hwang and Hu(1999 also proved a characterization of the gaa distribution by the independence of the saple ean and the saple coefficient of variation. Recently, Lee and Li(29 presented characterizations of gaa distribution with the Received January 5, 213; Accepted April 4, Matheatics Subject Classification: Priary 6E5, 62E1. Key words and phrases: independent identically distributed, a statistic scaleinvariant, gaa distribution, characterization, characteristic function. Correspondence should be addressed to Min-Young Lee, leey@dankook.ac.kr.
2 158 Hyun-Woo Jin and Min-Young Lee X property that the rando variables i X j (Σ n k=1 X and Σ n k 2 k=1 X k are independent for 1 i < j n if and only if X i has gaa distribution for i = 1,, n. In this paper, we obtain the characterizations of the gaa distribution by independence property of the quotient of su and difference of rando variables. 2. Results Theore 2.1. Let X and Y be nondegenerate and positive i.i.d. rando variables with coon absolutely continuous cuulative distribution function F (X and E(X 2 <. The rando variables X + Y and are independent if and only if X and Y have gaa (X + Y 2 distributions. Theore 2.2. Let {X i, 1 i n} be nondegenerate and positive i.i.d. rando variables with coon absolutely continuous cuulative distribution function F (X and E(X 2 <. The rando variables S n and Σ i=1 (X i 2 are independent for 1 < n, where S n = Σ n i=1 X i if and only if X i has gaa distribution for i = 1,, n. 3. Proofs Proof of Theore 2.1. Since is a scale-invariant statistic, and X + Y are independent [see Lukacs and Laha(1963]. (X + Y 2 (X + Y 2 We have to prove the reverse. We denote the characteristic functions of X +Y, (X (X + Y 2 and +Y, (X + Y 2 by φ 1 (t, φ 2 (s and φ(t, s, respectively. The independence of X + Y and is equivalent to (X + Y 2 (3.1 φ(t, s = φ 1 (t φ 2 (s. The left hand side of (3.1 becoes φ(t, s = exp (it(x + y + is(x y2 (x + y 2 df (xdf (y.
3 Characterizations of the gaa distribution 159 Also the right hand side of (3.1 becoes φ 1 (t φ 2 (s = exp it(x + y df (xdf (y is(x y 2 exp (x + y 2 df (xdf (y. Then (3.1 gives (3.2 = exp (it(x + y + df (xdf (y is(x y2 (x + y 2 exp it(x + y df (xdf (y is(x y 2 exp (x + y 2 df (xdf (y. The integrals in (3.2 exist not only for reals t and s but also for coplex values t = u + iv, s = u + iv, where u and u are reals, for which v = I(t, v = I(s and they are analytic for all t, s for v = I(t >, v = I(s >, [see, Lukacs(1955]. Differentiating (3.2 two ties with respect to t and then one tie respect to s and setting s =, we get (x y 2 exp it(x + y df (xdf (y (3.3 = θ (x + y 2 exp it(x + y df (xdf (y where [ ] ( X Y 2 θ = E. X + Y The rando variable θ is bounded and the oents exists. Then we know that [ ] [ ( X Y 2 (1 2 ] (3.4 θ = E = E. X + Y 1 + X2 +Y 2 2XY Note that, for x > and y >, < 2xy x 2 + y 2 and the equality on the right hand side occurs only if x = y. By the assued continuity of F (x, we find P(x = y = and 1 < x2 +y 2 2xy <. It follows < θ < 1 by (3.4. Let ϕ(t be the characteristic function of F (x. Then
4 16 Hyun-Woo Jin and Min-Young Lee ϕ (t = i x exp ( itx df (x, ϕ (t = x 2 exp ( itx df (x. Expressing (3.3 as a differential equation for the characteristic function ϕ(t, we get ϕ (tϕ(t 2(ϕ (t 2 + ϕ(tϕ (t = θ[ϕ (tϕ(t + 2(ϕ (t 2 + ϕ(tϕ (t]. That is, ϕ (t ϕ (t = 1 + θ ϕ (t 1 θ ϕ(t, < θ < 1. After integrating with the initial conditions ϕ( = 1, ϕ ( = ie(x, we get (3.5 ϕ (t = ie(x(ϕ(t 1+θ 1 θ, 1 + θ 1 θ > 1. The solution of this differential equation (3.5 with the above initial conditions is ϕ(t = (1 ie(x t λ, λ = 1 θ λ 2θ Consequently, F (x is a gaa distribution. Proof of Theore 2.2. Since Σ i=1 (X i 2 >. is a scale-invariant statistic, Σ i=1 (X i 2 and S n are independent [see Lukacs and Laha(1963]. We have to prove the reverse. We denote the characteristic functions of S n, Σ i=1 (X i 2 and (S n, Σ i=1 (X i 2 by φ 1 (t, φ 2 (s and φ(t, s, respectively. The independence of S n and Σ i=1 (X i 2 is equivalent to (3.6 φ(t, s = φ 1 (t φ 2 (s. The left hand side of (3.6 becoes φ(t, s = exp [it is ( Σ S n + i=1 (X i 2 2 ]df (x 1 df (x n. Sn
5 Characterizations of the gaa distribution 161 Also the right hand side of (3.6 becoes φ 1 (t φ 2 (s = Then (3.6 gives (3.7 = exp [ it(s n ] df (x 1 df (x n [ ( is Σ exp i=1 (X i 2 ] df (x 1 df (x n. exp [it is ( Σ S n + i=1 (X i 2 2 ]df (x 1 df (x n Sn exp [ it(s n ] df (x 1 df (x n [ ( is Σ exp i=1 (X i 2 2 ]df (x 1 df (x n. Sn The integrals in (3.7 exist not only for reals t and s but also for coplex values t = u + iv, s = u + iv, where u and u are reals, for which v = I(t, v = I(s and they are analytic for all t, s for v = I(t >, v = I(s >, [see, Lukacs(1955]. Differentiating (3.7 two ties with respect to t and then one tie respect to s and setting s =, we get (3.8 = θ (Σ i=1(x i 2 exp [ it(s n ] df (x 1 df (x n exp [ it(s n ] df (x 1 df (x n where [ ( Σ i=1 θ = E (X i 2 ]. The rando variable θ is bounded and the oents exist. Then we know that [ ( (X (X +1 2 ] θ = E = = E [ ( (X n (X n 2 [ ( (X (X (X +1 2 = E for i.i.d. rando variables X 1,, X n. Then ( n 1 C 1 Σ (3.9 nc θ = E[ n ] [ i=1 X2 i ( n 1C 1 ] = E 1 + 2Σ 1 i<j nx i X j Σ n i=1 X2 i ] ]
6 162 Hyun-Woo Jin and Min-Young Lee Note that, for x 1, x n >, the relation < 2Σ 1 i<j n x i x j (n 1(x x2 n holds and the equality on the right hand side occurs only if x 1 = = x n. By the assued continuity of F (x we find P(x 1 = = x n = and < 2Σ 1 i<j n x i x j < n 1. It follows that x x2 n < θ < n 2 n by (3.9. Let ϕ(t be the characteristic function of F (x. Then ϕ (t = i x exp[itx]df (x, ϕ (t = x 2 exp[itx]df (x. Expressing (3.8 as a differential equation for the characteristic function ϕ(t, we get ϕ (t(ϕ(t n 1 = θ [ nϕ (t(ϕ(t n nc 2 (ϕ (t 2 (ϕ(t n 2]. That is, ϕ (t ϕ (t = 2 nc 2 θ ϕ (t nθ ϕ(t, n 2 < θ < n. After integrating with the initial conditions ϕ( = 1, ϕ ( = ie(x, we get (3.1 ϕ 2 nc2 θ 2 (t = ie(x(ϕ(t nθ nc 2 θ, nθ > 1. The solution of this differential equation (3.1 with the above initial conditions is ϕ(t = (1 ie(x t λ, λ = nθ λ n 2 θ >. Consequently, F (x is a gaa distribution. References [1] P. Findeisen, A siple proof of a classical theore which characterizes the gaa distribution, Ann. Stat. Math. 6 (1978, no. 5, [2] T. Y. Hwang and C. Y. Hu, On a characterizations of the gaa distribution: The independence of the saple ean and the saple coefficient of variation, Ann. Inst. Stat. Math. 51 (1999, no. 4, [3] M. Y. Lee and E. H. Li, A Characterization of gaa distribution by independent property, J. Chungcheong Math. soc. 22 (29, no. 1, 1-5. [4] E. Lukacs, A Characterization of gaa distribution, Ann. Math. Stat. 17 (1955, [5] E. Lukacs and R. G. Laha, Applications of characteristic function, Charles Griffin. London, 1963.
7 Characterizations of the gaa distribution 163 * Departent of Matheatics Dankook University Cheonan , Republic of Korea E-ail: hwjin@dankook.ac.kr ** Departent of Matheatics Dankook University Cheonan , Republic of Korea E-ail: leey@dankook.ac.kr
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