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Commun. Korean Math. Soc. 25 (2010), No. 1, pp. 119–128 DOI 10.4134/CKMS.2010.25.1.119 SOME RESULTS ON ASYMPTOTIC BEHAVIORS OF RANDOM SUMS OF INDEPENDENT IDENTICALLY DISTRIBUTED RANDOM VARIABLES Tran Loc Hung and Tran Thien Thanh Abstract. Let {Xn , n ≥ 1} be a sequence of independent identically distributed (i.i.d.) random variables (r.vs.), defined on a probability space (Ω, A, P ), and let {Nn , n ≥ 1} be a sequence of positive integer-valued r.vs., defined on the same probability space (Ω, A, P ). Furthermore, we assume that the r.vs. Nn , n ≥ 1 are independent of all r.vs. Xn , n ≥ 1. In present paper we are interested in asymptotic behaviors of the random sum SNn = X1 + X2 + · · · + XNn , S0 = 0, where the r.vs. Nn , n ≥ 1 obey some defined probability laws. Since the appearance of the Robbins’s results in 1948 ([8]), the random sums SNn have been investigated in the theory probability and stochastic processes for quite some time (see [1], [4], [2], [3], [5]). Recently, the random sum approach is used in some applied problems of stochastic processes, stochastic modeling, random walk, queue theory, theory of network or theory of estimation (see [10], [12]). The main aim of this paper is to establish some results related to the asymptotic behaviors of the random sum SNn , in cases when the Nn , n ≥ 1 are assumed to follow concrete probability laws as Poisson, Bernoulli, binomial or geometry. 1. Introduction Let {Xn , n ≥ 1} be a sequence of independent identically distributed (i.i.d.) random variables (r.vs.), defined on a probability space (Ω, A, P ) and let {Nn , n ≥ 1} be a sequence of positive integer-valued r.vs., defined on the same probability space (Ω, A, P ). Furthermore, we assume that the r.vs. Nn , n ≥ 1 are independent of all i.i.d.r.vs. Xn , n ≥ 1. From now on, the random sum is defined by (1) SNn = X1 + X2 + · · · + XNn , S0 = 0. Received October 16, 2008; Revised August 13, 2009. 2000 Mathematics Subject Classification. 60F05, 60G50. Key words and phrases. random sum, independent identically distributed random variables, asymptotic behavior, Poisson law, Bernoulli law, binomial law, geometric law. c °2010 The Korean Mathematical Society 119 120 TRAN LOC HUNG AND TRAN THIEN THANH Since the appearance of the Robbins’s results in 1948 (see [8] for more details), the random sums SNn have been investigated in the theory probability and stochastic processes for quite some time (see [5], [1], [2] and [12] for the complete bibliography). In the classical theory of limit theorems we can consider the non-random index n in the sum Sn = X1 + X2 + · · · + Xn as a random variable degenerated at a point n. Therefore, the replacement of the number n of the sum Sn by the positive integer-valued r.vs. is natural. In simple terms, a random sum is a sum of a random number of r.vs.. The number of the terms Nn , n ≥ 1 in the sum, as well as the individual terms, can obey various probability laws. In stochastic theory, Nn , n ≥ 1 are often assumed to follow Poisson law or geometric law. In general, the r.vs. Nn , n ≥ 1 should satisfy any conditions. To illustrate this, we can recall three types of classical conditions for the r.vs. Nn as follows (2) E(Nn ) → +∞ (3) Nn P − →1 n or (4) P Nn − →∞ as n → ∞, as n → ∞, as n → ∞. The condition (2) was used in well-known results of H. Robbins’s in 1948 (see details in [8]). The condition (3) was applied in Feller’s theorem for random sums (cf. [1]), while the last condition in (4) was presented in various papers like [4], [5], [9], [6], [7],. . . . It is to be noticed that the conditions (2), (3) and (4) are in following relationship (3) ⇒ (4) ⇒ (2). This paper is organized as follows. In Section 2 we present the main results related to the asymptotic behaviors of random sums in (1), when the r.vs. Nn , n ≥ 1 belong to some discrete probability laws. The proofs of these main results are presented in Section 3. 2. Main results From now on, the random variable of standard normal law N (0, 1) will be d denoted by X ∗ , the notation − → will mean the convergence in distribution and P − → will denote the convergence in probability. Theorem 2.1. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs., Xj ∼ Bernoulli(p), p ∈ (0, 1), j = 1, 2, . . . , n. Moreover, suppose that {Nn , n ≥ 1} is a sequence of positive integer-valued r.vs. independent on all Xj , j = 1, 2, . . . , n. Then, (i) SNn ∼ P oisson(λp), ¡ ifq Nn¢ ∼ P oisson(λ), λ > 0, n ≥ 1. , if Nn ∼ Geometry(q), q ∈ (0, 1), p + q = (ii) SNn ∼ Geometry p+q−pq 1, n ≥ 1 with P(Nn = k) = q(1 − q)k , k = 0, 1, . . . SOME RESULTS ON ASYMPTOTIC BEHAVIORS OF RANDOM SUMS 121 (iii) SNn ∼ Binomial(n, pq), if Nn ∼ Binomial(n, q), q ∈ (0, 1), p + q = 1, n ≥ 1. It is well known that the sum of independent r.vs. from Bernoulli law will belong to the same law. But in cases (i) and (ii) the random sums of independent r.vs. of Bernoulli law will not obey the Bernoulli. The final distributions of the random sums SNn depend on the distribution of r.vs. Nn , n ≥ 1. The same concludes would be found in B. Gnedenko’s or V. Kruglov’s and V. Korolev’s papers, in cases when the r.vs. Xj , j = 1, 2, . . . , n were independent standard normal distributed while the r.vs. Nn , n ≥ 1 were uniformly distributed (see for more details in [2], [3] and [5]). The conclusion in (iii) is a very interesting result when the random sum SNn and the r.vs. Nn , n ≥ 1 are identically distributed. Furthermore, we can receive the Poisson’s Approximation Theorem by using the Theorem 2.1(i) as follows Theorem 2.2. Assume that for each n = 1, 2, . . ., {Xnk , k = 1, 2, . . . , n} be a sequence of independent and identically Bernoulli distributed r.vs. with parameter pn ∈ (0, 1) and pn → 0, npn → λ (λ > 0) as n → ∞. Then, as n → ∞, n X d Sn = Xnk − → P oisson(λ). k=1 Theorem 2.3. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs.. Suppose that {Nn , n ≥ 1} is a sequence of positive integer-valued r.vs. independent on all Xj , j = 1, 2, . . . , n. Furthermore, assume that Nn ∼ Geometry(p), n ≥ 1. Then, we have (i) SNn ∼ Exp(λp), when Xj ∼ Exp(λ), j = 1, 2, . . . , n. (ii) SNn ∼ Geometry(pq), when Xj ∼ Geometry(q), j = 1, 2, . . . , n. Theorem 2.4. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E(X1 ) = 0, V ar(X1 ) = 1. Moreover, suppose that {Nn , n ≥ 1} is a sequence of r.vs. belonging to Poisson law P oisson(λn ) and independent of all Xn , n ≥ 1. If λn → 1 as n → ∞, n then SNn d √ − → X ∗ as n → ∞. n Theorem 2.5. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E(X1 ) = µ1 , E(X12 ) = µ2 . Suppose that {Nn , n ≥ 1} is a sequence of r.vs. from Poisson law P oisson(λn ) and independent of all Xn , n ≥ 1. Assume that λn → 1 as n → ∞ n and µ ¶ √ λn n − 1 → 0 as n → ∞. n 122 Then TRAN LOC HUNG AND TRAN THIEN THANH SNn − nµ1 d − → X∗ √ nµ2 as n → ∞. Theorem 2.6. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E(X1 ) < ∞, 0 < E(X12 ) < ∞. Furthermore, assume that the {Nn , n ≥ 1} is a sequence of r.vs. from Poisson law P oisson(λn ) and independent of all Xn , n ≥ 1. If λn → +∞ then as SNn − λn E(X1 ) d p − → X∗ λn E(X12 ) n → ∞, as n → ∞. Theorem 2.7. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E(X1 ) = 0, V ar(X1 ) = 1. Moreover, suppose that the {Nn , n ≥ 1} is a sequence of r.vs. from Binomial law B(n, p) and they are independent of all Xn , n ≥ 1. Then S d p Nn − → X ∗ as n → ∞. E(Nn ) Theorem 2.8. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. with finite moments. Suppose that {Nn , n ≥ 1} is a sequence of r.vs. from Binomial law B(n, p) and they are independent of all Xn , n ≥ 1. Then SNn − E(SNn ) d p − → X∗ V ar(SNn ) as n → ∞. Theorem 2.9. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E(X) = 0 and V ar(X) = 1. We assume that {Nn , n ≥ 1} is a sequence of random variables of Geometry law Geometry(pn ) and they are independent of all Xn , n ≥ 1. If pn → 0 as n → ∞, then S d p Nn − → Z as n → ∞, E(Nn ) where Z does not belong to N (0, 1). Remark 1. It is worth pointing out that from Theorem 2.9, we have E(Nn ) = 1/pn → ∞, as n → ∞ but the random sum SNn does not obey central limit theorem. Therefore, the condition E(Nn ) → ∞, as n → ∞ is not sufficient for satisfying the central limit theorem for random sum SN . Theorem 2.10. Let {Xn , n ≥ 1} be a sequence of independent standard normal distributed r.vs.. Suppose that {Nn , n ≥ 1} is a sequence of positive integervalued r.vs. such that the condition (2) is true, and (5) E|Nn − E(Nn )| →0 E(Nn ) as n → ∞. SOME RESULTS ON ASYMPTOTIC BEHAVIORS OF RANDOM SUMS Then p SNn d E(Nn ) − → X∗ as 123 n → ∞. p Remark 2. Based on the fact that E|Nn − E(Nn )| ≤ V ar(Nn ) we can conar(Nn ) clude that the condition VE(N → 0, as n → ∞ in Robbins’s results [8] will n) be stronger than the condition (5). Theorem 2.11. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs. such that E|X1 | < ∞, E(X1 ) = µ and suppose that {Nn , n ≥ 1} is a sequence of positive integervalued r.vs. independent of all Xn , n ≥ 1. Assume that the condition (3) true, then SNn P − → µ as n → ∞. n Theorem 2.12. Let {Xn , n ≥ 1} be a sequence of i.i.d.r.vs., and assume that Nn ∼ Binomial(n, pn ), n ≥ 1, satisfying npn → λ, as n → ∞. Then, d SNn − → SN as n → ∞, where N ∼ P oisson(λ). 3. Proofs Proof of Theorem 2.1. (i) It is clear that the generating function of r.vs. Nn , n ≥ 1 is g(t) = eλ(t−1) and characteristic function of the r.vs. Xn , n ≥ 1 is ϕ(t) = 1 + p(eit − 1). Then, the characteristic function of random sum SNn is given by ψ(t) = g(ϕ(t)) = eλp(e it −1) . Thus SNn ∼ P oisson(λp). q (ii) Let us denote g(t) = 1−(1−q)t the generating function of the r.vs. Nn and let the characteristic function of the r.vs. Xn , n ≥ 1 be ϕ(t) = 1 + p(eit − 1). Then, the characteristic function of random sum SNn can be calculated by ψ(t) = g(ϕ(t)) = 1− q p+q−pq ¢ . ¡ q 1 − p+q−pq eit q Geometry( p+q−pq ). n Therefore SNn ∼ (iii) Let g(t) = [1 + q(t − 1)] and ϕ(t) = 1 + p(eit − 1) be the generating function and characteristic function of r.vs. Nn , n ≥ 1 and Xj , j = 1, 2, . . . , respectively. Then, characteristic function of random sum SNn will be defined by ψ(t) = g(ϕ(t)) = [1 + pq(eit − 1)]n . By this way, SNn ∼ Binomial(n, pq). ¤ 124 TRAN LOC HUNG AND TRAN THIEN THANH Proof of Theorem 2.2. Let consider Nn ∼ P oisson(n). According to Theorem 2.1(i) we have SNn ∼ P oisson(npn ). Then, P(SNn = k) = e−λ λk e−npn (npn )k → k! k! as n → ∞, d or SNn − → P oisson(λ). Therefore, we must only to show that ∆n (t) = |ϕSNn (t) − ϕSn (t)| → 0. We contend that ∆n (t) = |enpn (e it −1) − [1 − pn (1 − eit )]n | ≤ n|epn (e it −1) − 1 − pn (eit − 1)|. Clearly, Re(pn (eit − 1)) ≤ 0, applying the inequality |eα − 1 − α| ≤ Re(α) ≤ 0, we obtain n|pn (eit − 1)|2 pn |eit − 1|2 = npn →0 2 2 This completes the proof. ∆n (t) ≤ as |α|2 2 with n → ∞. ¤ pt Proof of Theorem 2.3. (i) Let us denote the generating function g(t) = 1−(1−p)t λ and the characteristic function ϕ(t) = λ−it of Nn , n ≥ 1 and Xj , j = 1, 2, . . . respectively. Then, the characteristic function of random sum SNn is given by ψ(t) = g(ϕ(t)) = λp . λp − it We conclude that SNn ∼ Exp(λp). pt (ii) Let g(t) = 1−(1−p)t be the generating function of Nn , n ≥ 1 and denote it qe ϕ(t) = 1−(1−q)e it the characteristic function of Xj , j = 1, 2, . . . . Then, the characteristic function of random sum SNn , n ≥ 1 will be given by ψ(t) = g(ϕ(t)) = pqeit . 1 − (1 − pq)eit Hence, SNn ∼ Geometry(pq). ¤ Proof of Theorem 2.4. Denote gn (t) = eλn (t−1) the generating function of Nn , S n ≥ 1 and ϕ, ψn are characteristic functions of X1 , √Nnn , respectively. Then, £ ¡ t ¢ ¤ µ ¶ µ µ ¶¶ t t −1 λn ϕ √ n ψn (t) = ϕSNn √ , = gn ϕ √ =e n n ¡ ¢ ¡ ¢ t2 where ϕ √tn = 1 − 2n + o n1 . Hence ³ 1 ´i ³λ ´ h t2 λn t 2 n +o =− · +o . ln ψn (t) = λn − 2n n n 2 n 2 According to assumptions, as n → ∞, we have then ln ψn (t) → − t2 . This finishes the proof. ¤ SOME RESULTS ON ASYMPTOTIC BEHAVIORS OF RANDOM SUMS 125 Proof of Theorem 2.5. Let gn (t) = eλn (t−1) be generating function of Nn , n ≥ S −nµ 1 and denote ϕ, ψn are characteristic functions of X1 , N√nnµ2 1 , respectively. Then, £ ¡ t ¢ ¤ µ ¶ nµ nµ −it √nµ1 −it √nµ1 λn ϕ √nµ −1 t 2 · ϕ 2 · e 2 ψn (t) = e = e , √ SN n nµ2 t here ϕ( √nµ )=1+ 2 From this ln ψn (t) itµ √ 1 nµ2 − t2 2n + o( n1 ). ³λ ´ λn µ1 λn t2 nµ1 n + it √ − · +o = −it √ nµ2 nµ2 n 2 n ´ λ ³λ ´ 2 µ 1 √ ³ λn n t n = it √ n −1 − · +o . µ2 n n 2 n 2 On account of assumptions, when n → ∞, we get ln ψn (t) → − t2 . The proof is straightforward. ¤ Proof of Theorem 2.6. Putting µ1 = E(X1 ), µ2 = E(X12 ) and let gn (t) = eλn (t−1) be a generating function of Nn , n ≥ 1 and ϕ, ψn be are characteristic S −λ µ functions of X1 , N√nλ µn 1 , respectively. Then, n 2 £ ¡ t ¢ ¤ µ ¶ λ µ λ µ t −it √ n 1 −it √ n 1 λn ϕ √ −1 λn µ2 · ϕ λn µ2 · e λn µ2 √ ψn (t) = e =e , SN n λn µ 2 ³ ´ ³ ´ t2 1 1 where ϕ √λt µ = 1 + √itµ − + o 2λn λn . λn µ2 n 2 Hence h ³ 1 ´i t2 t2 ln ψn (t) = λn − +o = − + o(1). 2λn λn 2 2 If n → ∞, then ln ψn (t) → − t2 . We obtain the proof. ¤ Proof of Theorem 2.7. It is easy to see that the Nn , n ≥ 1 have the generating function gn (t) = [1+p(t−1)]n and E(Nn ) = np. Let ϕ, ψn be are characteristic S functions of X1 and √ Nn , respectively. Then, E(Nn ) µ µ ¶¶ µ · µ ¶ ¸¶n t t = 1+p ϕ √ −1 , ψn (t) = gn ϕ √ np np ³ ´ ¡ ¢ t2 where ϕ √tnp = 1 − 2np + o n1 . Therefore, ³ 1 ´in h t2 +o . ψn (t) = 1 − 2n n t2 If n → ∞, then ψn (t) → e− 2 . We have the complete proof. ¤ 126 TRAN LOC HUNG AND TRAN THIEN THANH Proof of Theorem 2.8. Let us denote µ1 = E(X1 ); µ2 = E(X12 ); δ = µ2 − µ21 p. Then, E(SNn ) = npµ1 , V ar(SNn ) = np(µ2 − µ21 p) = npδ. Clearly, gn (t) = [1 + p(t − 1)]n is the generating function of Nn , n ≥ 1. We S n −npµ1 denote ϕ and ψn the characteristic functions of X1 and N√ , respectively. npδ Then, ¶ µ ¶ ¸¶n · µ µ √ √ t t ψn (t) = e−itµ1 np/δ · ϕSNn √ = e−itµ1 np/δ 1 + p ϕ √ −1 , npδ npδ ³ ´ ¡ ¢ t2 µ 2 t t where ϕ √npδ − 2npδ + o n1 . = 1 + iµ1 √npδ Therefore r h ³ 1 ´in √ t2 (δ + µ21 p) p −itµ1 np/δ ψn (t) = e 1 + itµ1 − +o nδ 2nδ n r h ³ 1 ´in √ 2 2 2 p t µ p t 1 − − +o = e−itµ1 np/δ 1 + itµ1 nδ 2nδ 2n n h ³ 1 ´in √ √ 2 t = e−itµ1 np/δ eitµ1 p/(nδ) − +o 2n n h ³ ´i t2 −itµ1 √p/(nδ) 1 n = 1− e +o . 2n n t2 By letting n → ∞, ψn (t) → e− 2 . We have the proof. ¤ pn t Proof of Theorem 2.9. Let gn (t) = 1−(1−p be a generating function of the n )t random sum Nn , n ≥ 1 and suppose that E(Nn ) = p1n . We denote ϕ and ψn √ the characteristic functions of X1 and pn SNn , respectively. Then, √ pn ϕ(t pn ) √ ψn (t) = gn (ϕ(t pn )) = , √ 1 − (1 − pn )ϕ(t pn ) √ where ϕ(t pn ) = 1 − pn t2 /2 + o(pn ). Therefore 1 − pn t2 /2 + o(pn ) ψn (t) = . 1 + t2 /2 − o(1 − pn ) From the assumptions, by letting n → ∞, then ψn (t) → the complete proof. 2 t2 +2 t2 6= e− 2 . We get ¤ Proof of Theorem 2.10. Suppose that Xj , j = 1, 2, . . . having the characteristic t2 function ϕ(t) = e− 2 . Let us put gn (t) = E(tNn ), pk = P(Nn = k), an = S E(Nn ). Moreover, suppose that ψn (t) is a characteristic function of √Nann . Then, µ µ ¶¶ X ∞ t2 k t − 2a n . pk e ψn (t) = gn ϕ √ = an k=0 SOME RESULTS ON ASYMPTOTIC BEHAVIORS OF RANDOM SUMS 127 Therefore ∞ ∞ ¯ ¯ t2 k ¯X ³ t2 k X t2 ¯ t2 ´¯¯ t2 ¯¯ − ¯ ¯ ¯ − − ¯ pk e 2an − e− 2 ¯ ≤ pk ¯e 2an − e− 2 ¯. ¯ψn (t) − e 2 ¯ = ¯ k=0 k=0 In another way, using the Lagrange’s Theorem for the function h(x) = ¡ − t2 ¢x e 2 continuing on [ akn , 1] or [1, akn ], we have ¯ t2 k ¯ µk ¶ ¯ ¯k ¯ t2 ¯ ¯ − 2an ¯ ¯ ¯ ¯ −2 ¯ −e − h(1)¯ = ¯ − 1¯|h0 (c)| (for any c) ¯e ¯ = ¯h an an ¯ ¯ t2 ¯¯ k t2 ¯¯ k ¯ ¯ = ¯ − 1¯ · h(c) ≤ ·¯ − 1¯. 2 an 2 an (because of c ≥ 0, h(x) is decreasing) Then ∞ ¯ ¯k ¯ t2 X t2 ¯ t2 E|Nn − an | ¯ ¯ ¯ −2 ¯ (t) − e ≤ pk ¯ − 1¯ = . ¯ψn ¯ an 2 2 an k=0 t2 According to assumptions, by letting n → ∞, then ψn (t) → e− 2 . We have the proof. ¤ Proof of Theorem 2.11. According to the Weak Law of Large Numbers, we P have Snn − → µ. Because of assumptions, using the considerations on relaP tionships among the conditions (3) and (4), it follows Nn − → ∞. Since, for ² > 0, ∃n0 , ∀n > n0 : P(| Snn − µ| > ²) < ², and ¯ ¯ µ¯ ¶ µ¯ ¶ X n0 ∞ ∞ X X ¯ SN ¯ ¯ ¯ Sk P ¯¯ n − µ¯¯ > ² − µ¯¯ > ² = = P(Nn = k)P ¯¯ + Nn k k=1 k=1 k=n0 +1 ≤ P(Nn ≤ n0 ) + ², it is easy to derive SNn Nn P − → µ. Then, SNn SNn Nn P − → µ. = · n Nn n We obtain the proof. ¤ Proof of Theorem 2.12. Let gn (t) = [1 + pn (t − 1)]n be a generating function of Nn and suppose that ϕ and ψn are characteristic functions of X1 and SNn , respectively. Then ´n ¡ ¢n ³ npn ψn = gn (ϕ(t)) = 1 + pn [ϕ(t) − 1] = 1 + [ϕ(t) − 1] . n λ[ϕ(t)−1] By putting n → ∞, then ψn (t) → e . The proof is finished. ¤ Acknowledgements. The authors would like to express their gratitude to referees for several constructive suggestions. 128 TRAN LOC HUNG AND TRAN THIEN THANH References [1] W. Feller, An Introduction to Probability Theory and Its Applications, volume II, 2nd edition, John Wiley & Sons, New York, 1971. [2] B. 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Borkowski, Random sum estimators and their efficiency, Technical Report, Department of Mathematical Science, Montana State University, (2004), 1–20. [13] D. Szasz, Limit theorems for the distributions of the sums of a random number of random variables, Ann. Math. Statist. 43 (1972), 1902–1913. [14] P. Vellaisamy and B. Chaudhuri, Poisson and compound Poisson approximations for random sums of random variables, J. Appl. Probab. 33 (1996), 127–137. Tran Loc Hung Department of Mathematics Hue University 77 Nguyen Hue, Hue city, +84-54, Vietnam E-mail address: [email protected] Tran Thien Thanh Department of Mathematics Hue University 77 Nguyen Hue, Hue city, +84-54, Vietnam E-mail address: [email protected]