MOMENTS EQUALITIES FOR NONNEGATIVE INTEGER-VALUED RANDOM VARIABLES

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1 MOMENTS EQUALITIES FOR NONNEGATIVE INTEGER-VALUED RANDOM VARIABLES MOHAMED I RIFFI ASSOCIATE PROFESSOR OF MATHEMATICS DEPARTMENT OF MATHEMATICS ISLAMIC UNIVERSITY OF GAZA GAZA, PALESTINE Abstract. We preset ad prove two theorems about equaltes for the th momet of oegatve teger-valued radom varables. These equaltes geeralze the well kow equalty for the frst momet of a oegatve teger-valued radom varable, X, terms of ts cumulatve dstrbuto fucto, or terms of P (X > x. 1. Itroducto There s a well-kow equalty for the th momet of a oegatve radom varable, Y, terms of P (Y > y as [3, p. 33], for example. A smlar equalty for the frst momet of a oegatve teger-valued radom varable s also well kow ad used a lot the lterature (See [1, p. 43], for example. I ths paper we are gog to prove ths paper s a geeralzato of these equaltes the dscrete case. I the ext secto we wll prove a geeralzato of the well kow equalty the dscrete case. Our equalty gves a eat formula of the th momet, whe t exsts, for oegatve teger-valued varables. 2. Ma Theorems I ths secto we prove two dettes that wll be used to prove our ma theorems. The frst detty s used to express a product terms of the form (X as a fte sum of products of smlar terms whe the sum rages from 1 to a oegatve teger x. The secod detty s used to express x as a fte sum that rages from 1 to a oegatve teger x. Date: September 3, Mathematcs Subject Classfcato. Prmary 60A99. Secodary 62B99. Key words ad phrases. Expectato, Momets, Equaltes. 1

2 2 MOHAMED I RIFFI, ISLAMIC UNIVERSITY OF GAZA Before we proceed to the ma theorems, we eed the followg lemma. Lemma 2.1. ( x = ( x j. Proof. Remember Pascal s detty, amely ( ( ( x x 1 x 1 = + Apply ths detty to each last term o the rght-had sde of the resultg equato. Cotue ths procedure x 1 tmes to get ( x 1 x ( ( x j = + ( x j =. Lemma 2.2. Let g (x := x =1 ( j ad f (x := j=0 (x j. The (g (x = f (x = (x +1, where (z m = z(z 1... (z m + 1. Proof. Notce that ( x f (x = x(x 1 (x = (x +1 = (!. We also otce that g (x s a fte sum of terms of the form (x j(x j 1(x j 2 of legth. Such a geeral term ca be expressed as (x j for j = 1, 2,...,. The umber of these terms s x. Therefore, g (x = (x j. (2.1 Sce (x j =! ( x j By Lemma 2.1, we have, (2.1 ca be wrtte as. ( x j g (x =! ( x g (x =!.

3 Momets Equaltes for Noegatve Iteger-Valued Radom Varables 3 Ths mples that (g (x = f (x. Remark 2.1. Lemma 2.2 smply says that for x 2, 1 x 1 (x = (y. (2.2 y=1 =1 Theorem 2.1. Let X be a oegatve teger-valued radom varable ad 2. The ( 1 ( 2 E (X = (x P (X > x, (2.3 x= 1 provded that the sum o the rght-had sde of (2.3 exsts. Proof. The proof of ths theorem maly depeds o (2.2 whch s proved Lemma 2.2. ( 1 E (X = 1 (x P (X = x x 0 = x 1 P (X = x (y (by (2.2 x 0 y=1 x=y =1 y=1 =1 1 = (y P (X = x (Fub s Theorem 1 = (y P (X y y=1 =1 1 = (x P (X x = x= =1 x= 1 2 (x P (X > x, (2.4 where we have chaged the dummy varable y to x. Remark 2.2. We may start the sum o the rght-had sde of (2.3 at x = 0, sce the frst 1 terms of the product 2 (x are equal to 0.

4 4 MOHAMED I RIFFI, ISLAMIC UNIVERSITY OF GAZA Remark 2.3. The case = 1 s well kow ad ca be stated separately for otatoal coveece, where the proof ca be foud [1]. E(X = P (X > x. Theorem 2.2 (Rff s Equalty. Let X be a oegatve teger-valued radom varable ad 1. The ( E(X = x P (X > x =1 = [(1 + x x ]P (X > x, (2.5 provded that the sum o the rght-had sdes of (2.5 exsts. Proof. We proceed by ducto o. By Remark 2.3 the equalty holds for = 1. Assume that t holds for all m such that m 1. The we show t holds for m =. By assumpto, E(X m = l m (xp (X > x, m 1, (2.6 where l (x = (1 + x x for 1. By Theorem 2.1 ad Remark 2.2 we have E(X = E(φ (X + ψ (xp (X > x, where ψ (x = 2 (x ad φ (x = x 1 (x. The polyomal ψ (x s of the form 1 ψ (x = b, x, (2.7 =1 where the coeffcets b, s are obtaed for 1 2 ad 2 by 1 ( j b, = d,j, (2.8 j j=0 The coeffcets d,, 1 2 are gve by 1 d, = ( 1 ( j k. (2.9 ad d,0 = 1. j 1,j 2,...,j =1 k=1 j 1 <j 2 < <j

5 Momets Equaltes for Noegatve Iteger-Valued Radom Varables 5 The polyomal φ (x s of the form φ (x = 1 =1 e x, where the e s are tegers. Hece, to fd the expectato of φ (x, we apply the (2.6 to each of ts terms. Therefore, ( 2 E(φ (X = e, l (x P (X > x. (2.10 =1 The sum 2 =1 e,l (x ca be rewrtte as 2 a,x. I other words (2.10 becomes E(φ (X = ( 2 a, x P (X > x. (2.11 Here the costats a, are gve by 1 ( j a, = c,j, (2.12 j where coeffcets c, s are obtaed for 1 1 ad 2 by c, = ( j 1,j 2,...,j =1 k=1 j 1 <j 2 < <j ( j k. (2.13 Notce that we have for 1 2 ad 2 ( a, + b, =. (2.14 To complete the proof, we eed (2.14 to prove the followg lemma. Lemma 2.3. Proof. 2 1 a, x + b, x = =1 =1 ( x. ( a, x + b, x =a,0 + a, x + b, x + b, 1 x 1 =1 2 =1 + sce b, 1 = for all 2. =1 =1 ( =1 x + x 1,

6 6 MOHAMED I RIFFI, ISLAMIC UNIVERSITY OF GAZA Note that ( ( =. So, wrtg the sum 2 ( =1 order, we obta 2 =1 ( 2 x = =1 ad ths completes the proof. ( 1 x = =2 Hece the proof of Theorem 2.2 s complete. Refereces x reverse ( x, (2.16 [1] Ka La Chug, A Course Probablty Theory, Secod Edto, Academc Press, New York, [2] George Casella, Roger L. Berger, Statstcal Iferece, Duxbury Press, Calfora, [3] Rchard Durret, Probablty: Theory ad Examples, Wadsworth, Calfora, Islamc Uversty of Gaza, P.O.Box 108, Gaza, Paleste E-mal address: mrff@hotmal.com