We prove that the exponential distribution is the only one which satisfies a
regression identity. This identity involves conditional expectation of the
sample mean of record values given two record values outside of the sample

Yanev, George P.

English

arXiv

We prove that the exponential distribution is the only one which satisfies a regression identity. This identity involves conditional expectation of the sample mean of record values given two record values outside of the sample

Yanev, George

Scholarworks@UTRGV Univ. of Texas RioGrande Valley

University of Texas Rio Grande Valley ScholarWorks @ UTRGV Mathematical and Statistical Sciences Faculty Publications and Presentations College of Sciences 2011 Characterizations of Exponential Distribution Via Conditional Expectations of Record Values George Yanev The University of Texas Rio Grande Valley Follow this and additional works at: https://scholarworks.utrgv.edu/mss_fac  Part of the Mathematics Commons Recommended Citation Yanev, George P. 2011. “Characterizations of Exponential Distribution Via Conditional Expectations of Record Values.” Pliska Studia Mathematica Bulgarica 20 (January): 233–42. This Article is brought to you for free and open access by the College of Sciences at ScholarWorks @ UTRGV. It has been accepted for inclusion in Mathematical and Statistical Sciences Faculty Publications and Presentations by an authorized administrator of ScholarWorks @ UTRGV. For more information, please contact justin.white@utrgv.edu, william.flores01@utrgv.edu. Pliska Stud. Math. Bulgar. 20 (2011), 233–242STUDIA MATHEMATICABULGARICACHARACTERIZATIONS OF EXPONENTIALDISTRIBUTION VIA CONDITIONAL EXPECTATIONSOF RECORD VALUESGeorge P. YanevWe prove that the exponential distribution is the only one which satisfiesa regression identity. This identity involves conditional expectation of thesample mean of record values given two record values outside of the sample.1. Discussion of resultsBefore formulating and discussing the obtained results, let us introduce somenotation regarding record values. Let X1,X2, . . . be independent copies of arandom variable X with absolutely continuous cumulative distribution function(c.d.f.) F (x). (Upper) Record value is an observation in a discrete time series,which exceeds all previous observations, i.e., Xj is a (upper) record value if Xj >Xi for all i < j. Define the sequence {Tn, n ≥ 1} of record times by T1 = 1 andTn = min{j : Xj > XTn−1 , j > Tn−1} for n > 1. Then the corresponding recordvalues are Rn = XTn for n = 1, 2, . . . (see Nevzorov (2001)).Let F (x) be the exponential c.d.f. given by(1) F (x) = 1 − e−cx, (x ≥ 0),where c > 0. It is well-known (cf. Arnold et al. (1998), Section 4.2.2), that asingle regression function (conditional expectation) of the form E(Rn+1|Rn) is2000 Mathematics Subject Classification: 62G30, 62E10.Key words: characterizations, exponential distribution, record values.234 George P. Yanevenough to characterize a continuous c.d.f. Several authors have studied charac-terizations of exponential distributions in terms of the regression of one recordvalue with two other record values as covariates, i.e., regression functions givenfor 1 ≤ k ≤ n − 1 and r ≥ 1 by(2) E[g(Rn)|Rn−k = u,Rn+r = v] = ϕ(u, v), (0 ≤ u < v),where the function g satisfies certain regularity conditions. Let ḡ(u, v) denote theaverage value of the function g(x) over the interval [u, v], i.e.,ḡ(u, v) =1v − uv∫ug(t)dt.Bairamov et al. (2005) study the particular case of (2) when both covariatesare adjacent (only one spacing away) to Rn. They prove, under some regularityconditions on F and g and using different notation, that F is exponential if andonly ifE[g(Rn)∣∣∣Rn−1 = u,Rn+1 = v]= ḡ(u, v), (0 ≤ u < v).On the other hand, Yanev et al. (2008) prove the following necessary condition fora continuous c.d.f. to be exponential. If F is exponential then for 1 ≤ k ≤ n − 1and r ≥ 1(3) E[g(k−1+r−1)(Rn)k + r − 1∣∣∣Rn−k = u,Rn+r = v]=(k − 1 + r − 1k − 1)∂k−1+r−1∂ur−1∂vk−1ḡ(u, v),where 0 ≤ u < v. Let us demonstrate (3) using two particular choices of thefunction g(x). If g(x) = xk+r−1/(k + r − 1)!, then (3) becomes(4) E[Rn|Rn−k = u,Rn+r = v] =ru + kvk + r, (0 ≤ u < v).Note that the right-hand side of (4) is the weighted mean of the covariates’ values.The weight of one covariate is proportional to the number of spacings Rn is awayfrom the other covariate. As a result, the closer Rn is to one of the covariates,the more affected its value is from this covariate. Another choice of the functiong(x) could be g(x) = (−1)k+r−2x−2/(k + r − 1)! when (3) is equivalent toE[1Rk+rn∣∣∣Rn−k = u,Rn+r = v]=1urvk, (0 ≤ u < v).Characterizations of exponential distribution . . . 235Whether (3) is not only a necessary, but also a sufficient condition for an ab-solutely continuous F to be exponential for any 1 ≤ k ≤ n − 1 and r ≥ 1 is anopen question. Yanev et al. (2008) obtain an affirmative answer when r = 1, i.e.,one of the two covariates is adjacent to Rn. They show, under some regularityassumptions on F and g, that F (x) is exponential if and only if(5) E[g(k−1)(Rn)k∣∣∣Rn−k = u,Rn+1 = v]=∂k−1∂vk−1ḡ(u, v), (0 ≤ u < v).A different direction for extending the characterization results above is toconsider conditional expectations of sums of record values. Akhundov and Nev-zorov (2007) prove that an absolutely continuous F is exponential if and onlyifE[R2 + R3 + . . . + Rnn − 1|R1 = u,Rn+1 = v]=u + v2.(6)Our first result is one generalization of (6).Theorem 1. Let n ≥ 2 be an integer. Suppose F (x) is absolutely continuous,g(x) is continuous in [0,∞), and g(v) 6= ḡ(0, v) for every v > 0. Then (1) holdsif and only if for 0 ≤ u < v < ∞(7) E[g(R2) + g(R3) + . . . + g(Rn)n − 1|R1 = u,Rn+1 = v]= ḡ(u, v).Next, we extend Theorem 1 to the case when the second covariate is twospacings away from Rn. The general case of non-adjacent covariates remains anopen problem. Denote, for a continuous function g(x) and n = 1, 2, . . .(8) In(u, v) =1∫0g(u + (v − u)z)zn−1dz, (0 ≤ u < v).Note that I1(u, v) = ḡ(u, v). Denote H(x) = − ln(1 − F (x)) for x ≥ 0 andh(x) = H ′(x), i.e., H(x) is the cumulative hazard function of X and h(x) is itshazard (failure) rate.Theorem 2. Suppose F (x) satisfies the following conditions.(i) F (x) is absolutely continuous in [0,∞) and F ′′(x) is continuous in (0,∞);(ii) h(x) is nowhere constant in a small interval (0, ε) for ε > 0.236 George P. YanevAssume that g(x) is continuous in [0,∞), g′(x) is continuous in (0,∞), andg′(0+) 6= 0. Let n ≥ 2 be an integer. Then (1) holds if and only if for 0 ≤ u <v < ∞(9) E[g(R2) + g(R3) + . . . + g(Rn)n − 1|R1 = u,Rn+2 = v]= ḡ(u, v) −nIn(u, v) − ḡ(u, v)n − 1.Setting in Theorem 2, g(x) = x and thus nIn(u, v) = (u + nv)/(n + 1), andtaking into account (9) we obtain the following extension of (6).Corollary 1. Suppose F (x) satisfies the following conditions.(i) F (x) is absolutely continuous in [0,∞) and F ′′(x) is continuous in (0,∞);(ii) h(x) is nowhere constant in a small interval (0, ε) for ε > 0. Let n ≥ 2be an integer. Then (1) holds if and only if for 0 ≤ u < v < ∞E[R2 + R3 + . . . + Rnn − 1|R1 = u,Rn+2 = v]=u + v2−v − u2(n + 1)(10)=(n + 2)u + nvn + 2 + n.Notice that (4) implies that a necessary condition for F to be exponential is(11) E[R2 + R3 + . . . + Rnn − 1|R1 = u,Rn+m = v]=(n + 2m − 2)u + nvn + 2m − 2 + n, m ≥ 1.The results (6) and Corollary 2 prove the sufficiency of (11) for m = 1 and m = 2.2. ProofsLemma 1. Let r and n ≥ 2 be positive integers. If (1) holds, then(12) En∑j=2g(Rj) | R1 = u,Rn+r = v= (N + 1)1∫0g(u + (v − u)z)1 −N∑j=n−1(Nj)zj(1 − z)N−j dz,where N = n + r − 3. As usual we assume∑ij(·) = 0 when i < j.Characterizations of exponential distribution . . . 237P r o o f. It is well-known (e.g., Nevzorov (2001), p.11 and p.69) that if (1)holds, then Rn =∑nj=1 ξj, where {ξj} are i.i.d. exponential with (1). Therefore,for 2 ≤ j ≤ nE [Rj |R1 = u,Rn+r = v]= E [ξ1 + ξ2 + . . . + ξj|ξ1 = u, ξ1 + ξ2 + . . . + ξn+r = v]= E [u + ξ2 + . . . + ξj|ξ2 + . . . + ξn+r = v − u]= E [u + Yj|Zn+r = v − u] , say.We obtain the following density functions related to Yj and Zn+r for 2 ≤ j ≤ nfYj ,Zn+r(y, z) =cn+r−1yj−2(z − y)n+r−j−1(j − 2)!(n + r − j − 1)!e−cz, fZn+r(z) =cn+r−1(N + 1)!zN+1e−cz,fYj |Zn+r(y|z) =(N + 1)!(j − 2)!(n + r − j − 1)!(yz)j−2 (1 −yz)n+r−j−1 1z,where N = n + r − 3. Now we are in a position to prove (7). Indeed,En∑j=2g(Rj)|R1 = u,Rn+r = v = En∑j=2g(u + Yj)|Zn+r = v − u=v−u∫0g(u + t)n∑j=2(N + 1)!(j − 2)!(n + r − j − 1)!(tv − u)j−2(1 −tv − u)n+r−j−1 dtv − u= (N + 1)1∫0g(u + (v − u)z)n∑j=2(Nj − 2)zj−2(1 − z)n+r−j−1dz= (N + 1)1∫0g(u + (v − u)z)n−2∑i=0(Ni + k − 1)zi+k−1(1 − z)N−(i+k−1)dz,which is equivalent to (12). 2.1. Proof of Theorem 1Recall the formula for the conditional density fj|1,n+r(t|u, v), say, of Rj givenR1 = u and Rn+r = v, where r ≥ 1 and 2 ≤ j ≤ n (e.g., Ahsanullah (2004), p.6).238 George P. YanevIt is given for u < t < v and 2 ≤ j ≤ n by(13) fj|1,n+r(t|u, v)=(n+r−2)!(j−2)!(n+r−1−j)![H(t)−H(u)H(v)−H(u)]j−2 [ H(v)−H(t)H(v)−H(u)]n+r−1−j h(t)H(v) − H(u).Sufficiency. Assuming (7), we will prove that F (x) is exponential (1). Using(13) with r = 1 and the binomial formula, we obtain(14)n∑j=2fj|1,n+1(t|u, v) =(n − 1)h(t)H(v) − H(u).Taking into account (14), for the left-hand side of (7) we haveEn∑j=2g(Rj)|R1 = u,Rn+1 = v =v∫un∑j=2g(t)fj|1,n+1(t|u, v) dt=n − 1H(v) − H(u)v∫ug(t)dH(t)and thus (7) is equivalent to1H(v) − H(u)v∫ug(t)dH(t) =1v − uv∫ug(t)dt.Letting u → 0+, and hence H(u) → 0, we obtainv∫0g(t)dH(t) =H(v)vv∫0g(t)dt.Differentiating both sides with respect to v, after rearranging terms, we find[g(v) − ḡ(0, v)]h(v) =H(v)v[g(v) − ḡ(0, v)].Dividing both sides by g(v) − ḡ(0, v) 6= 0, we obtain the equationh(v)H(v)=1v,Characterizations of exponential distribution . . . 239which only solution is the exponential F (x) given by (1). Indeed, integratingboth sides with respect to v, we obtainlnH(x) = ln v + ln c (c > 0)and thus H(x) = − ln(1 − F (x)) = cx and (1) follows.Necessity. If F (x) is exponential, then (7) follows from the lemma settingr = 1.2.2. Proof of Theorem 2Sufficiency. Recall the notation fj|1,n+r(t|u, v) introduced before (13). Now, (13)with r = 2 and the binomial formula imply(15)n∑j=2fj|1,n+2(t|u, v)=nh(t)(H(v) − H(u))nn∑j=2(n − 1j − 2)(H(t) − H(u))j−2(H(v) − H(t))n−j+1=nh(t)(H(v) − H(u))nn−2∑i=0(n − 1i)(H(t) − H(u))i(H(v) − H(t))n−1−i=nh(t)H(v) − H(u)[1 −(H(t) − H(u)H(v) − H(u))n−1].It follows from (15) thatEn∑j=2g(Rj)|R1 = u,Rn+2 = v =v∫ug(t)n∑j=2fj|1,n+2(t|u, v)dt(16)=nH(v) − H(u)v∫ug(t)[1 −(H(t) − H(u)H(v) − H(u))n−1]dH(t).It follows from (9) and (16), letting u → 0 (and hence H(u) → 0) and multiplyingby Hn(v)(n − 1)/n thatv∫0g(t)[Hn−1(v) − Hn−1(t)]dH(t) = Hn(v)[ḡ(v) − In(v)],240 George P. Yanevwhere ḡ(v) = ḡ(0, v) and In(v) = In(0, v). Differentiating with respect to v anddividing by Hn−2(v)h(v) we obtain(n − 1)v∫0g(t)dH(t) = nH(v)[ḡ(v) − In(v)] +H2(v)h(v)[ḡ′(v) − I ′n(v)].Differentiating with respect to v again and dividing by H(v) we have(17) [(n − 1)g(v) − nḡ(v) + nIn(v)]h(v)H(v)=[n + 2 −h′(v)H(v)h2(v)][ḡ′(v) − I ′n(v)] + [ḡ′′(v) − I ′′n(v)]H(v)h(v).Observing that(18) nIn(v) = g(v) − I′n(v)v and ḡ(v) = g(v) − ḡ′(v)v,after some algebra, one can write (17) as[nḡ′(v) − I ′n(v)]h(v)vH(v)=[n + 2 −h′(v)H(v)h2(v)][ḡ′(v) − I ′n(v)] + [(n + 1)I′n(v) − 2ḡ′(v)]H(v)h(v)v(19)Let us make the change of variables(20) w(v) =h(v)H(v)v.Note that w(v) ≡ 1 corresponds to the exponential c.d.f. (1). For simplicity, wewrite w for w(v). After multiplying both sides of (19) by h(v)v/H(v), using (20)andh′(v)vh(v)=w′wv + w − 1,it is not difficult to see that (19) is equivalent to(21) [nḡ′(v) − I ′n(v)](w2 − 1) =[(n + 1)(w − 1) −w′wv][ḡ′(v) − I ′n(v)]Clearly, w(v) ≡ 1 is one solution, which corresponds to the exponential distri-bution. It remains to prove that this is the only solution. Suppose w(v) is aCharacterizations of exponential distribution . . . 241solution and there exists a value v1 > 0 such that w(v1) 6= 1. We want to reacha contradiction. Since F is twice differentiable, we have that w(v) is continuouswith respect to v and hence w(v) 6= 1 for v in an open interval around v1. Letv0 = inf{v|w(v) 6= 1}. Since w(0+) = 1, we have v0 ≥ lF . We shall prove thatv0 = lF . Assume on contrary that v0 > 0. Then w(v) = 1 if 0 < v ≤ v0 andintegration of (20) implies that h(v) is constant-valued in this interval. This con-tradicts the assumption (ii). Therefore v0 = 0 and hence equation (21) holds forall v > 0. Dividing (21) by w − 1 6= 0, we obtain(w + 1)[nḡ′(v) − I ′n(v)] =[n + 1 −w′vw(w − 1)][ḡ′(v) − I ′n(v)].(22)It can be seen thatlimv→0+w′vw(w − 1)= 1.Also, using the L’Hopital’s rule, one can find thatlimv→0+ḡ′(v) =g′(0+)2and limv→0+I ′n(v) =g′(0+)n + 1.Therefore, passing to the limit in (22) as v → 0+, we obtain2(n2−1n + 1)g′(0+) = n(12−1n + 1)g′(0+),which is not possible for n ≥ 2. This contradiction completes the proof.Necessity. This follows from the lemma setting r = 2.REFERE NC ES[1] M. Ahsanullah. Record Values – Theory and Applications. Univ. Press ofAmerica, Lanham, MD, 2004.[2] I. S. Akhundov, V. B. Nevzorov. Conditional distributions of recordvalues and characterizations of distributions. J. Math. Sci., 145 (2007), No2, 4837–4842.[3] B. C. Arnold, N. Balakrishnan, H. N. Nagaraja. Records. Willey &Sons, New York, NW, 1998.242 George P. Yanev[4] I. Bairamov, M. Ahsanullah, A. Pakes. A characterization of contin-uous distributions via regression on pairs of record values. Austr. N. Z. J.Statist., 47 (2005), 543–547.[5] V. B. Nevzorov. Records: Mathematical Theory, AMS, Providence, RI,2001.[6] G. P. Yanev, M. Ahsanullah, M. I. Beg. Characterizations of probabil-ity distributions via bivariate regression of record values. Metrika, 68 (2008),51–64.George P. YanevDepartment of MathematicsUniversity of Texas – Pan AmericanEdinburg, Texas 78539e-mail: yanevgp@utpa.edu

Characterizations of Exponential Distribution Via Conditional Expectations of Record Values

2000 Mathematics Subject Classification: 62G30, 62E10.We prove that the exponential distribution is the only one which satisfies a regression identity. This identity involves conditional expectation of the sample mean of record values given two record values outside of the sample

P. Yanev, George

Bulgarian Digital Mathematics Library at IMI-BAS

Pliska Stud. Math. Bulgar. 20 (2011), 233–242STUDIA MATHEMATICABULGARICACHARACTERIZATIONS OF EXPONENTIALDISTRIBUTION VIA CONDITIONAL EXPECTATIONSOF RECORD VALUESGeorge P. YanevWe prove that the exponential distribution is the only one which satisfiesa regression identity. This identity involves conditional expectation of thesample mean of record values given two record values outside of the sample.1. Discussion of resultsBefore formulating and discussing the obtained results, let us introduce somenotation regarding record values. Let X1,X2, . . . be independent copies of arandom variable X with absolutely continuous cumulative distribution function(c.d.f.) F (x). (Upper) Record value is an observation in a discrete time series,which exceeds all previous observations, i.e., Xj is a (upper) record value if Xj >Xi for all i < j. Define the sequence {Tn, n ≥ 1} of record times by T1 = 1 andTn = min{j : Xj > XTn−1 , j > Tn−1} for n > 1. Then the corresponding recordvalues are Rn = XTn for n = 1, 2, . . . (see Nevzorov (2001)).Let F (x) be the exponential c.d.f. given by(1) F (x) = 1− e−cx, (x ≥ 0),where c > 0. It is well-known (cf. Arnold et al. (1998), Section 4.2.2), that asingle regression function (conditional expectation) of the form E(Rn+1|Rn) is2000 Mathematics Subject Classification: 62G30, 62E10.Key words: characterizations, exponential distribution, record values.234 George P. Yanevenough to characterize a continuous c.d.f. Several authors have studied charac-terizations of exponential distributions in terms of the regression of one recordvalue with two other record values as covariates, i.e., regression functions givenfor 1 ≤ k ≤ n− 1 and r ≥ 1 by(2) E[g(Rn)|Rn−k = u,Rn+r = v] = ϕ(u, v), (0 ≤ u < v),where the function g satisfies certain regularity conditions. Let g¯(u, v) denote theaverage value of the function g(x) over the interval [u, v], i.e.,g¯(u, v) =1v − uv∫ug(t)dt.Bairamov et al. (2005) study the particular case of (2) when both covariatesare adjacent (only one spacing away) to Rn. They prove, under some regularityconditions on F and g and using different notation, that F is exponential if andonly ifE[g(Rn)∣∣∣Rn−1 = u,Rn+1 = v] = g¯(u, v), (0 ≤ u < v).On the other hand, Yanev et al. (2008) prove the following necessary condition fora continuous c.d.f. to be exponential. If F is exponential then for 1 ≤ k ≤ n− 1and r ≥ 1(3) E[g(k−1+r−1)(Rn)k + r − 1∣∣∣Rn−k = u,Rn+r = v]=(k − 1 + r − 1k − 1)∂k−1+r−1∂ur−1∂vk−1g¯(u, v),where 0 ≤ u < v. Let us demonstrate (3) using two particular choices of thefunction g(x). If g(x) = xk+r−1/(k + r − 1)!, then (3) becomes(4) E[Rn|Rn−k = u,Rn+r = v] =ru+ kvk + r, (0 ≤ u < v).Note that the right-hand side of (4) is the weighted mean of the covariates’ values.The weight of one covariate is proportional to the number of spacings Rn is awayfrom the other covariate. As a result, the closer Rn is to one of the covariates,the more affected its value is from this covariate. Another choice of the functiong(x) could be g(x) = (−1)k+r−2x−2/(k + r − 1)! when (3) is equivalent toE[1Rk+rn∣∣∣Rn−k = u,Rn+r = v]=1urvk, (0 ≤ u < v).Characterizations of Exponential Distribution . . . 235Whether (3) is not only a necessary, but also a sufficient condition for an ab-solutely continuous F to be exponential for any 1 ≤ k ≤ n − 1 and r ≥ 1 is anopen question. Yanev et al. (2008) obtain an affirmative answer when r = 1, i.e.,one of the two covariates is adjacent to Rn. They show, under some regularityassumptions on F and g, that F (x) is exponential if and only if(5) E[g(k−1)(Rn)k∣∣∣Rn−k = u,Rn+1 = v]=∂k−1∂vk−1g¯(u, v), (0 ≤ u < v).A different direction for extending the characterization results above is toconsider conditional expectations of sums of record values. Akhundov and Nev-zorov (2007) prove that an absolutely continuous F is exponential if and onlyifE[R2 +R3 + . . .+Rnn− 1|R1 = u,Rn+1 = v]=u+ v2.(6)Our first result is one generalization of (6).Theorem 1. Let n ≥ 2 be an integer. Suppose F (x) is absolutely continuous,g(x) is continuous in [0,∞), and g(v) 6= g¯(0, v) for every v > 0. Then (1) holdsif and only if for 0 ≤ u < v <∞(7) E[g(R2) + g(R3) + . . .+ g(Rn)n− 1|R1 = u,Rn+1 = v]= g¯(u, v).Next, we extend Theorem 1 to the case when the second covariate is twospacings away from Rn. The general case of non-adjacent covariates remains anopen problem. Denote, for a continuous function g(x) and n = 1, 2, . . .(8) In(u, v) =1∫0g(u+ (v − u)z)zn−1dz, (0 ≤ u < v).Note that I1(u, v) = g¯(u, v). Denote H(x) = − ln(1 − F (x)) for x ≥ 0 andh(x) = H ′(x), i.e., H(x) is the cumulative hazard function of X and h(x) is itshazard (failure) rate.Theorem 2. Suppose F (x) satisfies the following conditions.(i) F (x) is absolutely continuous in [0,∞) and F ′′(x) is continuous in (0,∞);(ii) h(x) is nowhere constant in a small interval (0, ε) for ε > 0.236 George P. YanevAssume that g(x) is continuous in [0,∞), g′(x) is continuous in (0,∞), andg′(0+) 6= 0. Let n ≥ 2 be an integer. Then (1) holds if and only if for 0 ≤ u <v <∞(9) E[g(R2) + g(R3) + . . . + g(Rn)n− 1|R1 = u,Rn+2 = v]= g¯(u, v) −nIn(u, v) − g¯(u, v)n− 1.Setting in Theorem 2, g(x) = x and thus nIn(u, v) = (u + nv)/(n + 1), andtaking into account (9) we obtain the following extension of (6).Corollary 1. Suppose F (x) satisfies the following conditions.(i) F (x) is absolutely continuous in [0,∞) and F ′′(x) is continuous in (0,∞);(ii) h(x) is nowhere constant in a small interval (0, ε) for ε > 0. Let n ≥ 2be an integer. Then (1) holds if and only if for 0 ≤ u < v <∞E[R2 +R3 + . . . +Rnn− 1|R1 = u,Rn+2 = v]=u+ v2−v − u2(n+ 1)(10)=(n+ 2)u+ nvn+ 2 + n.Notice that (4) implies that a necessary condition for F to be exponential is(11) E[R2 +R3 + . . .+Rnn− 1|R1 = u,Rn+m = v]=(n+ 2m− 2)u+ nvn+ 2m− 2 + n, m ≥ 1.The results (6) and Corollary 2 prove the sufficiency of (11) for m = 1 andm = 2.2. ProofsLemma 1. Let r and n ≥ 2 be positive integers. If (1) holds, then(12) E n∑j=2g(Rj) | R1 = u,Rn+r = v= (N + 1)1∫0g(u+ (v − u)z)1− N∑j=n−1(Nj)zj(1− z)N−j dz,where N = n+ r − 3. As usual we assume∑ij(·) = 0 when i < j.Characterizations of Exponential Distribution . . . 237P r o o f. It is well-known (e.g., Nevzorov (2001), p.11 and p.69) that if (1)holds, then Rn =∑nj=1 ξj, where {ξj} are i.i.d. exponential with (1). Therefore,for 2 ≤ j ≤ nE [Rj |R1 = u,Rn+r = v]= E [ξ1 + ξ2 + . . .+ ξj|ξ1 = u, ξ1 + ξ2 + . . .+ ξn+r = v]= E [u+ ξ2 + . . .+ ξj|ξ2 + . . .+ ξn+r = v − u]= E [u+ Yj|Zn+r = v − u] , say.We obtain the following density functions related to Yj and Zn+r for 2 ≤ j ≤ nfYj ,Zn+r(y, z) =cn+r−1yj−2(z − y)n+r−j−1(j − 2)!(n + r − j − 1)!e−cz, fZn+r(z) =cn+r−1(N + 1)!zN+1e−cz,fYj |Zn+r(y|z) =(N + 1)!(j − 2)!(n + r − j − 1)!(yz)j−2 (1−yz)n+r−j−1 1z,where N = n+ r − 3. Now we are in a position to prove (7). Indeed,E n∑j=2g(Rj)|R1 = u,Rn+r = v = E n∑j=2g(u + Yj)|Zn+r = v − u=v−u∫0g(u+ t)n∑j=2(N + 1)!(j − 2)!(n + r − j − 1)!(tv − u)j−2(1−tv − u)n+r−j−1 dtv − u= (N + 1)1∫0g(u+ (v − u)z)n∑j=2(Nj − 2)zj−2(1− z)n+r−j−1dz= (N + 1)1∫0g(u+ (v − u)z)n−2∑i=0(Ni+ k − 1)zi+k−1(1− z)N−(i+k−1)dz,which is equivalent to (12). 2.1. Proof of Theorem 1Recall the formula for the conditional density fj|1,n+r(t|u, v), say, of Rj givenR1 = u and Rn+r = v, where r ≥ 1 and 2 ≤ j ≤ n (e.g., Ahsanullah (2004), p.6).238 George P. YanevIt is given for u < t < v and 2 ≤ j ≤ n by(13) fj|1,n+r(t|u, v)=(n+r−2)!(j−2)!(n+r−1−j)![H(t)−H(u)H(v)−H(u)]j−2 [H(v)−H(t)H(v)−H(u)]n+r−1−j h(t)H(v)−H(u).Sufficiency. Assuming (7), we will prove that F (x) is exponential (1). Using(13) with r = 1 and the binomial formula, we obtain(14)n∑j=2fj|1,n+1(t|u, v) =(n− 1)h(t)H(v) −H(u).Taking into account (14), for the left-hand side of (7) we haveE n∑j=2g(Rj)|R1 = u,Rn+1 = v =v∫un∑j=2g(t)fj|1,n+1(t|u, v) dt=n− 1H(v)−H(u)v∫ug(t)dH(t)and thus (7) is equivalent to1H(v)−H(u)v∫ug(t)dH(t) =1v − uv∫ug(t)dt.Letting u→ 0+, and hence H(u)→ 0, we obtainv∫0g(t)dH(t) =H(v)vv∫0g(t)dt.Differentiating both sides with respect to v, after rearranging terms, we find[g(v) − g¯(0, v)]h(v) =H(v)v[g(v) − g¯(0, v)].Dividing both sides by g(v) − g¯(0, v) 6= 0, we obtain the equationh(v)H(v)=1v,Characterizations of Exponential Distribution . . . 239which only solution is the exponential F (x) given by (1). Indeed, integratingboth sides with respect to v, we obtainlnH(x) = ln v + ln c (c > 0)and thus H(x) = − ln(1− F (x)) = cx and (1) follows.Necessity. If F (x) is exponential, then (7) follows from the lemma settingr = 1.2.2. Proof of Theorem 2Sufficiency. Recall the notation fj|1,n+r(t|u, v) introduced before (13). Now, (13)with r = 2 and the binomial formula imply(15)n∑j=2fj|1,n+2(t|u, v)=nh(t)(H(v)−H(u))nn∑j=2(n− 1j − 2)(H(t)−H(u))j−2(H(v)−H(t))n−j+1=nh(t)(H(v)−H(u))nn−2∑i=0(n− 1i)(H(t)−H(u))i(H(v)−H(t))n−1−i=nh(t)H(v)−H(u)[1−(H(t)−H(u)H(v) −H(u))n−1].It follows from (15) thatE n∑j=2g(Rj)|R1 = u,Rn+2 = v =v∫ug(t)n∑j=2fj|1,n+2(t|u, v)dt(16)=nH(v)−H(u)v∫ug(t)[1−(H(t)−H(u)H(v) −H(u))n−1]dH(t).It follows from (9) and (16), letting u→ 0 (and hence H(u)→ 0) and multiplyingby Hn(v)(n − 1)/n thatv∫0g(t)[Hn−1(v)−Hn−1(t)]dH(t) = Hn(v)[g¯(v)− In(v)],240 George P. Yanevwhere g¯(v) = g¯(0, v) and In(v) = In(0, v). Differentiating with respect to v anddividing by Hn−2(v)h(v) we obtain(n− 1)v∫0g(t)dH(t) = nH(v)[g¯(v)− In(v)] +H2(v)h(v)[g¯′(v)− I ′n(v)].Differentiating with respect to v again and dividing by H(v) we have(17) [(n− 1)g(v) − ng¯(v) + nIn(v)]h(v)H(v)=[n+ 2−h′(v)H(v)h2(v)][g¯′(v)− I ′n(v)] + [g¯′′(v)− I ′′n(v)]H(v)h(v).Observing that(18) nIn(v) = g(v) − I′n(v)v and g¯(v) = g(v) − g¯′(v)v,after some algebra, one can write (17) as[ng¯′(v)− I ′n(v)]h(v)vH(v)=[n+ 2−h′(v)H(v)h2(v)][g¯′(v)− I ′n(v)] + [(n + 1)I′n(v) − 2g¯′(v)]H(v)h(v)v.(19)Let us make the change of variables(20) w(v) =h(v)H(v)v.Note that w(v) ≡ 1 corresponds to the exponential c.d.f. (1). For simplicity, wewrite w for w(v). After multiplying both sides of (19) by h(v)v/H(v), using (20)andh′(v)vh(v)=w′wv +w − 1,it is not difficult to see that (19) is equivalent to(21) [ng¯′(v) − I ′n(v)](w2 − 1) =[(n+ 1)(w − 1)−w′wv][g¯′(v) − I ′n(v)].Clearly, w(v) ≡ 1 is one solution, which corresponds to the exponential distri-bution. It remains to prove that this is the only solution. Suppose w(v) is aCharacterizations of Exponential Distribution . . . 241solution and there exists a value v1 > 0 such that w(v1) 6= 1. We want to reacha contradiction. Since F is twice differentiable, we have that w(v) is continuouswith respect to v and hence w(v) 6= 1 for v in an open interval around v1. Letv0 = inf{v|w(v) 6= 1}. Since w(0+) = 1, we have v0 ≥ lF . We shall prove thatv0 = lF . Assume on contrary that v0 > 0. Then w(v) = 1 if 0 < v ≤ v0 andintegration of (20) implies that h(v) is constant-valued in this interval. This con-tradicts the assumption (ii). Therefore v0 = 0 and hence equation (21) holds forall v > 0. Dividing (21) by w − 1 6= 0, we obtain(w + 1)[ng¯′(v)− I ′n(v)] =[n+ 1−w′vw(w − 1)][g¯′(v)− I ′n(v)].(22)It can be seen thatlimv→0+w′vw(w − 1)= 1.Also, using the L’Hopital’s rule, one can find thatlimv→0+g¯′(v) =g′(0+)2and limv→0+I ′n(v) =g′(0+)n+ 1.Therefore, passing to the limit in (22) as v → 0+, we obtain2(n2−1n+ 1)g′(0+) = n(12−1n+ 1)g′(0+),which is not possible for n ≥ 2. This contradiction completes the proof.Necessity. This follows from the lemma setting r = 2.REFERENCES[1] M. Ahsanullah. Record Values – Theory and Applications. Univ. Press ofAmerica, Lanham, MD, 2004.[2] I. S. Akhundov, V. B. Nevzorov. Conditional distributions of recordvalues and characterizations of distributions. J. Math. Sci., 145 (2007), No2, 4837–4842.[3] B. C. Arnold, N. Balakrishnan, H. N. Nagaraja. Records. Willey &Sons, New York, NW, 1998.242 George P. Yanev[4] I. Bairamov, M. Ahsanullah, A. Pakes. A characterization of contin-uous distributions via regression on pairs of record values. Austr. N. Z. J.Statist., 47 (2005), 543–547.[5] V. B. Nevzorov. Records: Mathematical Theory, AMS, Providence, RI,2001.[6] G. P. Yanev, M. Ahsanullah, M. I. Beg. Characterizations of probabil-ity distributions via bivariate regression of record values.Metrika, 68 (2008),51–64.George P. YanevDepartment of MathematicsUniversity of Texas – Pan AmericanEdinburg, Texas 78539e-mail: yanevgp@utpa.edu

Characterizations of Exponential Distribution via Conditional Expectations of Record Values

https://scholarworks.utrgv.edu/cgi/viewcontent.cgi?article=1043&context=mss_fac

Characterizations of exponential distribution via conditional expectations of record values

Abstract

Similar works

Full text

Available Versions

Scholarworks@UTRGV Univ. of Texas RioGrande Valley

Bulgarian Digital Mathematics Library at IMI-BAS