2010 Mathematics Subject Classification: 62G30, 62E10.Two characterizations of the exponential distribution based on equalities among order statistics in a random sample of size three are proved. This proves two conjectures stated recently in Arnold and Villaseñor [4]

Chakrabort, Santanu

Yanev, George P.

Bulgarian Digital Mathematics Library at IMI-BAS

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Bulgar. 23 (2013), 237–244STUDIA MATHEMATICABULGARICACHARACTERIZATIONS OF EXPONENTIALDISTRIBUTION BASED ON SAMPLE OF SIZE THREEGeorge P. Yanev, Santanu ChakrabortyAbstract. Two characterizations of the exponential distribution based onequalities among order statistics in a random sample of size three are proved.This proves two conjectures stated recently in Arnold and Villasen˜or [4].1. Introduction. The publications on characterizations of the exponentialdistribution are abundant. Comprehensive surveys can be found in Ahsanullahand Hamedani [1], Arnold and Huang [3], and Johnson, Kotz and Balakrish-nan [5]. The Bulgarian probability school has its contribution with the works ofObretenov [6]-[8]. Recently, Arnold and Villasen˜or [4] obtained a series of char-acterizations based on random sample of size two from a continuous distribution.They also identified a list of conjectures for possible extensions of their resultsto samples of size three and bigger. In this note we confirm that two of theseconjectures are true.Let X1,X2,X3 be a random sample of size three from a parent random vari-able X. Denote X2:2 := max{X1,X2} and X3:3 := max{X1,X2,X3}. We writeX ∼ exp{λ} if the probability density function (pdf) of X equals fX(x) =λe−λxI(x > 0), λ > 0. It is known (e.g., Arnold et al. (2008), p.77) that ifX ∼ exp{λ}, thenX1 +12X2 +13X3d= X3:3 and X2:2 +13X3d= X3:3,(1)2010 Mathematics Subject Classification: 62G30, 62E10.Key words: characterization, exponential distribution, order statistics.238 G. P. Yanev, S. Chakrabortywhered= denotes equality in distribution. Arnold and Villasen˜or [4] conjecturedthat each one of the equalities in (1), under some regularity assumptions onthe cumulative distribution function (cdf) F of X, is a sufficient condition forX ∼ exp(λ) for some λ > 0. The theorem below proves these conjectures.Theorem 1. Let X1,X2,X3 be a random sample from X, which has an ab-solutely continuous cdf F with F (0) = 0. Suppose the pdf f of X is analytic in aneighborhood of 0.(i) IfX2:2 +13X3d= X3:3,(2)then X ∼ exp{λ} for some λ > 0.(ii) IfX1 +12X2 +13X3d= X3:3,(3)then X ∼ exp{λ} for some λ > 0.2. Proofs. We begin with a useful lemma (see also Arnold and Villasen˜or[4]).Lemma 1. If F (0) = 0, the pdf f is analytic in a neighborhood of 0, andf (k)(0) =[f ′(0)f(0)]k−1f ′(0), k = 1, 2, . . . ,(4)then X ∼ exp{λ} for some λ > 0.P r o o f. For the Maclaurin series of f(x), we have for x > 0f(x) =∞∑k=0f (k)(0)k!xk(5)= f(0) +∞∑k=1[f ′(0)f(0)]k−1f ′(0)xkk!= f(0) exp{f ′(0)f(0)x}.Characterizations of exponential distribution 239Since f(x) is a pdf, we have f ′(0)/f(0) < 0. Denoting λ = −f ′(0)/f(0) > 0 andsetting the integral of (5) from 0 to ∞ to be 1, we obtain λ = f(0). Therefore,f(x) = λe−λxI(x > 0), i.e., X ∼ exp{λ}. We continue with the proof of the theorem.P r o o f o f P a r t (i). The pdf of the left-hand side of (2) isfX2:2+X3/3(x) =∫ x0fX3/3(y)fX2:2(x− y) dy(6)=∫ x03f(3y)ddx[F 2(x− y)] dy= 6∫ x0f(3y)F (x− y)f(x− y) dy.For the pdf of the right-hand side of (2), we havefX3:3(x) =ddxF 3(x)(7)= 3F 2(x)f(x)= 6f(x)∫ x0F (y)f(y) dy.Define G(x) := F (x)f(x). Referring to (6) and (7) we rewrite (2) as∫ x0f(3y)G(x− y) dy = f(x)∫ x0G(y) dy.(8)For the nth derivative of the left-hand side of (8), we havedndxn∫ x0f(3y)G(x− y) dy =n−1∑i=0[f(3x)](n−1−i)G(i)(0) +∫ x0f(3y)G(n)(x− y) dy.Applying the Leibnitz rule for the nth derivative of a product of two functionsto the right-hand side of (8), we obtaindndxn[f(x)∫ x0G(y) dy]=n∑i=1(ni)[f(x)](n−i)G(i−1)(x) + [f(x)](n)∫ x0G(y) dy.240 G. P. Yanev, S. ChakrabortyNow, differentiating both sides of (8) n times and evaluating the derivatives atx = 0, we obtainn∑i=13n−if (n−i)(0)G(i−1)(0) =n∑i=1(ni)f (n−i)(0)G(i−1)(0).(9)Since G(0) = 0 and G′(0) = f2(0), the above equation is equivalent to[3n−2 −(n2)]f (n−2)(0)f2(0) =n∑i=3[(ni)− 3n−i]f (n−i)(0)G(i−1)(0),(10)where n ≥ 4. We shall prove that (10) implies (4). Equation (4) is trivially truefor k = 1. To proceed by induction, assume (4) for all 1 ≤ k ≤ n − 3, wheren ≥ 4. We need to prove it for k = n − 2. Using the induction assumption, wehave for j = 1, 2, . . . , n− 2G(j)(0) =j∑i=0(ji)F (i)(0)f (j−i)(0)=j∑i=1(ji)f (i−1)(0)f (j−i)(0)= (j + 1)f (j−1)(0)f(0) +j−1∑i=2(ji)[f ′(0)f(0)]i−2f ′(0)[f ′(0)f(0)]j−i−1f ′(0)=[f ′(0)f(0)]j−1f2(0)(2j − 1).Therefore, using the induction assumption again, we have for i = 3, 4, . . . , n − 1f (n−i)(0)G(i−1)(0) =[f ′(0)f(0)]n−i−1f ′(0)[f ′(0)f(0)]i−2f2(0)(2i−1 − 1)(11)=[f ′(0)f(0)]n−3f ′(0)f2(0)(2i−1 − 1).Substituting (11) in the right-hand side of (10) yields (i = n corresponds to a 0term)[3n−2 −(n2)]f (n−2)(0) =[f ′(0)f(0)]n−3f ′(0)n∑i=3[(ni)− 3n−i](2i−1 − 1).Characterizations of exponential distribution 241Thus, to prove (4) for k = n− 2 it is sufficient to show that3n−2 −(n2)=n∑i=3[(ni)− 3n−i](2i−1 − 1)or, equivalently,n∑i=23n−i(2i−1 − 1) =n∑i=2(ni)(2i−1 − 1),which is easily verified. This completes the proof of (4) by induction. The claimin Part (i) follows from (4) and the Lemma.P r o o f o f P a r t (ii). The pdf of the left-hand side of (3) isfX1+X2/2+X3/3(x) =∫ x0fX1(y)fX2/2+X3/3(x− y) dy(12)=∫ x0fX1(y)∫ x−y0fX2/2(z)fX3/3(x− y − z) dz dy= 6∫ x0f(y)∫ x−y0f(2z)f(3(x− y − z)) dz dy.DenotingH(x− y) :=∫ x−y0f(2z)f(3(x− y − z)) dz(13)and taking into account (7) and (12), we write (3) as∫ x0f(y)H(x− y) dy = f(x)∫ x0G(y) dy.(14)Similarly to the proof of Part (i), differentiating n times with respect to x bothsides of (14) and evaluating the derivatives at x = 0, we haven∑i=1f (n−i)(0)H(i−1)(0) =n∑i=1(ni)f (n−i)(0)G(i−1)(0).Since H(0) = G(0) = 0 and H ′(0) = G′(0) = f2(0), the last equation becomes[1−(n2)]f (n−2)(0)f2(0) =n∑i=3[(ni)G(i−1)(0)−H(i−1)(0)]f (n−i)(0).(15)242 G. P. Yanev, S. ChakrabortyNow we are in a position to prove (4) by induction. Equation (4) is trivially truefor k = 1. Assuming (4) for all 1 ≤ k ≤ n − 3, where n ≥ 4, we will prove itfor k = n − 2. Differentiating (13) n times with respect to x and evaluating thederivative at x = y, we haveH(n)(0) =n∑i=12n−if (n−i)(0)3i−1f (i−1)(0).(16)Under the induction assumption, (16) implies for j = 1, 2, . . . , n− 2H(j)(0) =j∑i=12j−i[f ′(0)f(0)]j−i−1f ′(0)3i−1[f ′(0)f(0)]i−2f ′(0)=[f ′(0)f(0)]j−3(f ′(0))2j∑i=12j−i3i−1=[f ′(0)f(0)]j−1f2(0)(3j − 2j).Therefore, using the induction assumption again, we have for i = 3, 4, . . . , n−1f (n−i)(0)H(i−1)(0) =[f ′(0)f(0)]n−i−1f ′(0)[f ′(0)f(0)]i−2f2(0)(3i−1 − 2i−1)=[f ′(0)f(0)]n−3f ′(0)f2(0)(3i−1 − 2i−1).Recalling (11) from the proof of Part (i) we rewrite (15) as (note that i = ncorresponds to a 0 term)[1−(n2)]f (n−2)(0) =[f ′(0)f(0)]n−3f ′(0)n∑i=3[(ni)(2i−1 − 1)− (3i−1 − 2i−1)]Thus, to prove (4) for k = n− 2 it is sufficient to show that1−(n2)=n∑i=3[(ni)(2i−1 − 1)− (3i−1 − 2i−1)],or equivalentlyn∑i=2[(ni)(2i−1 − 1)− 3i−1 + 2i−1]= 0Characterizations of exponential distribution 243which is easily verified. This proves (4). Now, referring to the Lemma we com-plete the proof of the Theorem. 3. Concluding remarks. The more general cases of samples of size n ≥ 4and relationsXn−1:n−1 +1nXnd= Xn:n and Xn−2:n−2 +1n− 1Xn−1 +1nXnd= Xn:n,where Xj:j := max{X1,X2, ...,Xj} for j = n − 1 and j = n are still underinvestigation. Finally, it is worth noticing that if we assume for i.i.d. randomvariables X1,X2, . . . with E|X1| <∞, that for every n = 1, 2, . . .X1 +12X2 +13X3 + . . .+1nXnd= Xn:n,then the Xi’s have a common exponential distribution (see e.g. Arnold andVillasen˜or [4]).REFERENCES[1] M. Ahsanullah, G. G. Hamedani. Exponential Distribution: Theory andMethods. NOVA Science, New York, 2010.[2] B. C. Arnold, N. Balakrishnan, H. N. Nagaraja. A First Course inOrder Statistics. SIAM Classics in Applied Probab. vol. 54, Philadelphia,2008.[3] B. C. Arnold, J. S. Huang. Chapter 12: Characterizations. In: , TheExponential Distribution: Theory, Methods and Applications (Eds N. Bal-akrishnan, A. P. Basu). Gordon and Breach, Amsterdam, 1995, 185–203.[4] B. C. Arnold, J. A. Villasen˜or. Exponential characterizations moti-vated by the structure of order statistics in sample of size two. Statistics andProbability Letters 83 (2013), 596–601.[5] N. L. Johnson, S. Kotz, N. Balakrishnan. Continuous Univariate Dis-ributions, Vol. 1, 2nd Edn. Wiley, New York, 1994.244 G. P. Yanev, S. Chakraborty[6] A. Obretenov. A property of the exponential distribution. Physics-Mathematics Journal, Bulgarian Acad. Sci. 13 (1970), No 46, 51–53 (inBulgarian).[7] A. Obretenov. Characterizations of exponential and geometric distribu-tions based on order statistics. Pliska Stud. Math. Bulgar. 7 (1984), 97–101.[8] A. Obretenov. A characterization of an exponential distribution based onrenewals. In: Exploring stochastic laws. Festschrift in honour of the 70thbirthday of Academician Vladimir Semenovich Korolyuk (Eds A. V. Sko-rokhod et al.) VSP, Utrecht, 1995, 351–355.George P. Yanev, Santanu ChakrabortyDepartment of MathematicsUniversity of Texas – Pan American1201 W. University DriveEdinburg, Texas 78539e-mail: yanevgp@utpa.eduschakraborty@utpa.edu

Characterizations of Exponential Distribution Based on Sample of Size Three

Two characterizations of the exponential distribution based on equalities among order statistics in a random sample of size three are proved. This proves two conjectures stated recently in Arnold and Villasenor [4]

Yanev, George

Chakraborty, Santanu

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 2013 Characterizations of Exponential Distribution Based On Sample of Size Three George Yanev The University of Texas Rio Grande Valley Santanu Chakraborty Follow this and additional works at: https://scholarworks.utrgv.edu/mss_fac  Part of the Mathematics Commons Recommended Citation Yanev, George and Chakraborty, Santanu, "Characterizations of Exponential Distribution Based On Sample of Size Three" (2013). Mathematical and Statistical Sciences Faculty Publications and Presentations. 41. https://scholarworks.utrgv.edu/mss_fac/41 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. 23 (2013), 237–244STUDIA MATHEMATICABULGARICACHARACTERIZATIONS OF EXPONENTIALDISTRIBUTION BASED ON SAMPLE OF SIZE THREEGeorge P. Yanev, Santanu ChakrabortyAbstract. Two characterizations of the exponential distribution based onequalities among order statistics in a random sample of size three are proved.This proves two conjectures stated recently in Arnold and Villasen˜or [4].1. Introduction. The publications on characterizations of the exponentialdistribution are abundant. Comprehensive surveys can be found in Ahsanullahand Hamedani [1], Arnold and Huang [3], and Johnson, Kotz and Balakrish-nan [5]. The Bulgarian probability school has its contribution with the works ofObretenov [6]-[8]. Recently, Arnold and Villasen˜or [4] obtained a series of char-acterizations based on random sample of size two from a continuous distribution.They also identified a list of conjectures for possible extensions of their resultsto samples of size three and bigger. In this note we confirm that two of theseconjectures are true.Let X1,X2,X3 be a random sample of size three from a parent random vari-able X. Denote X2:2 := max{X1,X2} and X3:3 := max{X1,X2,X3}. We writeX ∼ exp{λ} if the probability density function (pdf) of X equals fX(x) =λe−λxI(x > 0), λ > 0. It is known (e.g., Arnold et al. (2008), p.77) that ifX ∼ exp{λ}, thenX1 +12X2 +13X3d= X3:3 and X2:2 +13X3d= X3:3,(1)2010 Mathematics Subject Classification: 62G30, 62E10.Key words: characterization, exponential distribution, order statistics.238 G. P. Yanev, S. Chakrabortywhered= denotes equality in distribution. Arnold and Villasen˜or [4] conjecturedthat each one of the equalities in (1), under some regularity assumptions onthe cumulative distribution function (cdf) F of X, is a sufficient condition forX ∼ exp(λ) for some λ > 0. The theorem below proves these conjectures.Theorem 1. Let X1,X2,X3 be a random sample from X, which has an ab-solutely continuous cdf F with F (0) = 0. Suppose the pdf f of X is analytic in aneighborhood of 0.(i) IfX2:2 +13X3d= X3:3,(2)then X ∼ exp{λ} for some λ > 0.(ii) IfX1 +12X2 +13X3d= X3:3,(3)then X ∼ exp{λ} for some λ > 0.2. Proofs. We begin with a useful lemma (see also Arnold and Villasen˜or[4]).Lemma 1. If F (0) = 0, the pdf f is analytic in a neighborhood of 0, andf (k)(0) =[f ′(0)f(0)]k−1f ′(0), k = 1, 2, . . . ,(4)then X ∼ exp{λ} for some λ > 0.P r o o f. For the Maclaurin series of f(x), we have for x > 0f(x) =∞∑k=0f (k)(0)k!xk(5)= f(0) +∞∑k=1[f ′(0)f(0)]k−1f ′(0)xkk!= f(0) exp{f ′(0)f(0)x}.Characterizations of exponential distribution 239Since f(x) is a pdf, we have f ′(0)/f(0) < 0. Denoting λ = −f ′(0)/f(0) > 0 andsetting the integral of (5) from 0 to ∞ to be 1, we obtain λ = f(0). Therefore,f(x) = λe−λxI(x > 0), i.e., X ∼ exp{λ}. We continue with the proof of the theorem.P r o o f o f P a r t (i). The pdf of the left-hand side of (2) isfX2:2+X3/3(x) =∫ x0fX3/3(y)fX2:2(x− y) dy(6)=∫ x03f(3y)ddx[F 2(x− y)] dy= 6∫ x0f(3y)F (x− y)f(x− y) dy.For the pdf of the right-hand side of (2), we havefX3:3(x) =ddxF 3(x)(7)= 3F 2(x)f(x)= 6f(x)∫ x0F (y)f(y) dy.Define G(x) := F (x)f(x). Referring to (6) and (7) we rewrite (2) as∫ x0f(3y)G(x− y) dy = f(x)∫ x0G(y) dy.(8)For the nth derivative of the left-hand side of (8), we havedndxn∫ x0f(3y)G(x− y) dy =n−1∑i=0[f(3x)](n−1−i)G(i)(0) +∫ x0f(3y)G(n)(x− y) dy.Applying the Leibnitz rule for the nth derivative of a product of two functionsto the right-hand side of (8), we obtaindndxn[f(x)∫ x0G(y) dy]=n∑i=1(ni)[f(x)](n−i)G(i−1)(x) + [f(x)](n)∫ x0G(y) dy.240 G. P. Yanev, S. ChakrabortyNow, differentiating both sides of (8) n times and evaluating the derivatives atx = 0, we obtainn∑i=13n−if (n−i)(0)G(i−1)(0) =n∑i=1(ni)f (n−i)(0)G(i−1)(0).(9)Since G(0) = 0 and G′(0) = f2(0), the above equation is equivalent to[3n−2 −(n2)]f (n−2)(0)f2(0) =n∑i=3[(ni)− 3n−i]f (n−i)(0)G(i−1)(0),(10)where n ≥ 4. We shall prove that (10) implies (4). Equation (4) is trivially truefor k = 1. To proceed by induction, assume (4) for all 1 ≤ k ≤ n − 3, wheren ≥ 4. We need to prove it for k = n − 2. Using the induction assumption, wehave for j = 1, 2, . . . , n− 2G(j)(0) =j∑i=0(ji)F (i)(0)f (j−i)(0)=j∑i=1(ji)f (i−1)(0)f (j−i)(0)= (j + 1)f (j−1)(0)f(0) +j−1∑i=2(ji)[f ′(0)f(0)]i−2f ′(0)[f ′(0)f(0)]j−i−1f ′(0)=[f ′(0)f(0)]j−1f2(0)(2j − 1).Therefore, using the induction assumption again, we have for i = 3, 4, . . . , n − 1f (n−i)(0)G(i−1)(0) =[f ′(0)f(0)]n−i−1f ′(0)[f ′(0)f(0)]i−2f2(0)(2i−1 − 1)(11)=[f ′(0)f(0)]n−3f ′(0)f2(0)(2i−1 − 1).Substituting (11) in the right-hand side of (10) yields (i = n corresponds to a 0term)[3n−2 −(n2)]f (n−2)(0) =[f ′(0)f(0)]n−3f ′(0)n∑i=3[(ni)− 3n−i](2i−1 − 1).Characterizations of exponential distribution 241Thus, to prove (4) for k = n− 2 it is sufficient to show that3n−2 −(n2)=n∑i=3[(ni)− 3n−i](2i−1 − 1)or, equivalently,n∑i=23n−i(2i−1 − 1) =n∑i=2(ni)(2i−1 − 1),which is easily verified. This completes the proof of (4) by induction. The claimin Part (i) follows from (4) and the Lemma.P r o o f o f P a r t (ii). The pdf of the left-hand side of (3) isfX1+X2/2+X3/3(x) =∫ x0fX1(y)fX2/2+X3/3(x− y) dy(12)=∫ x0fX1(y)∫ x−y0fX2/2(z)fX3/3(x− y − z) dz dy= 6∫ x0f(y)∫ x−y0f(2z)f(3(x− y − z)) dz dy.DenotingH(x− y) :=∫ x−y0f(2z)f(3(x− y − z)) dz(13)and taking into account (7) and (12), we write (3) as∫ x0f(y)H(x− y) dy = f(x)∫ x0G(y) dy.(14)Similarly to the proof of Part (i), differentiating n times with respect to x bothsides of (14) and evaluating the derivatives at x = 0, we haven∑i=1f (n−i)(0)H(i−1)(0) =n∑i=1(ni)f (n−i)(0)G(i−1)(0).Since H(0) = G(0) = 0 and H ′(0) = G′(0) = f2(0), the last equation becomes[1−(n2)]f (n−2)(0)f2(0) =n∑i=3[(ni)G(i−1)(0)−H(i−1)(0)]f (n−i)(0).(15)242 G. P. Yanev, S. ChakrabortyNow we are in a position to prove (4) by induction. Equation (4) is trivially truefor k = 1. Assuming (4) for all 1 ≤ k ≤ n − 3, where n ≥ 4, we will prove itfor k = n − 2. Differentiating (13) n times with respect to x and evaluating thederivative at x = y, we haveH(n)(0) =n∑i=12n−if (n−i)(0)3i−1f (i−1)(0).(16)Under the induction assumption, (16) implies for j = 1, 2, . . . , n− 2H(j)(0) =j∑i=12j−i[f ′(0)f(0)]j−i−1f ′(0)3i−1[f ′(0)f(0)]i−2f ′(0)=[f ′(0)f(0)]j−3(f ′(0))2j∑i=12j−i3i−1=[f ′(0)f(0)]j−1f2(0)(3j − 2j).Therefore, using the induction assumption again, we have for i = 3, 4, . . . , n−1f (n−i)(0)H(i−1)(0) =[f ′(0)f(0)]n−i−1f ′(0)[f ′(0)f(0)]i−2f2(0)(3i−1 − 2i−1)=[f ′(0)f(0)]n−3f ′(0)f2(0)(3i−1 − 2i−1).Recalling (11) from the proof of Part (i) we rewrite (15) as (note that i = ncorresponds to a 0 term)[1−(n2)]f (n−2)(0) =[f ′(0)f(0)]n−3f ′(0)n∑i=3[(ni)(2i−1 − 1)− (3i−1 − 2i−1)]Thus, to prove (4) for k = n− 2 it is sufficient to show that1−(n2)=n∑i=3[(ni)(2i−1 − 1)− (3i−1 − 2i−1)],or equivalentlyn∑i=2[(ni)(2i−1 − 1)− 3i−1 + 2i−1]= 0Characterizations of exponential distribution 243which is easily verified. This proves (4). Now, referring to the Lemma we com-plete the proof of the Theorem. 3. Concluding remarks. The more general cases of samples of size n ≥ 4and relationsXn−1:n−1 +1nXnd= Xn:n and Xn−2:n−2 +1n− 1Xn−1 +1nXnd= Xn:n,where Xj:j := max{X1,X2, ...,Xj} for j = n − 1 and j = n are still underinvestigation. Finally, it is worth noticing that if we assume for i.i.d. randomvariables X1,X2, . . . with E|X1| <∞, that for every n = 1, 2, . . .X1 +12X2 +13X3 + . . .+1nXnd= Xn:n,then the Xi’s have a common exponential distribution (see e.g. Arnold andVillasen˜or [4]).REFERENCES[1] M. Ahsanullah, G. G. Hamedani. Exponential Distribution: Theory andMethods. NOVA Science, New York, 2010.[2] B. C. Arnold, N. Balakrishnan, H. N. Nagaraja. A First Course inOrder Statistics. SIAM Classics in Applied Probab. vol. 54, Philadelphia,2008.[3] B. C. Arnold, J. S. Huang. Chapter 12: Characterizations. In: , TheExponential Distribution: Theory, Methods and Applications (Eds N. Bal-akrishnan, A. P. Basu). Gordon and Breach, Amsterdam, 1995, 185–203.[4] B. C. Arnold, J. A. Villasen˜or. Exponential characterizations moti-vated by the structure of order statistics in sample of size two. Statistics andProbability Letters 83 (2013), 596–601.[5] N. L. Johnson, S. Kotz, N. Balakrishnan. Continuous Univariate Dis-ributions, Vol. 1, 2nd Edn. Wiley, New York, 1994.244 G. P. Yanev, S. Chakraborty[6] A. Obretenov. A property of the exponential distribution. Physics-Mathematics Journal, Bulgarian Acad. Sci. 13 (1970), No 46, 51–53 (inBulgarian).[7] A. Obretenov. Characterizations of exponential and geometric distribu-tions based on order statistics. Pliska Stud. Math. Bulgar. 7 (1984), 97–101.[8] A. Obretenov. A characterization of an exponential distribution based onrenewals. In: Exploring stochastic laws. Festschrift in honour of the 70thbirthday of Academician Vladimir Semenovich Korolyuk (Eds A. V. Sko-rokhod et al.) VSP, Utrecht, 1995, 351–355.George P. Yanev, Santanu ChakrabortyDepartment of MathematicsUniversity of Texas – Pan American1201 W. University DriveEdinburg, Texas 78539e-mail: yanevgp@utpa.eduschakraborty@utpa.edu

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Characterizations of Exponential Distribution Based On Sample of Size Three

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Characterizations of Exponential Distribution Based on Sample of Size Three

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