ON THE MOMENTS OF ITERATED TAIL

Transcription

1 ON THE MOMENTS OF ITERATED TAIL RADU PĂLTĂNEA and GHEORGHIŢĂ ZBĂGANU The classical distribution in ruins theory has the property that the sequence of the first moment of the iterated tails is convergent We give sufficient conditions under which this property holds and also, we construct a counterexample that shows that this property it is not generally true AMS 2 Subject Classification: 6E5, 4A5 Key words: moments, integrated tail BASIC DEFINITIONS AND STATEMENT OF THE PROBLEM Let Ω, K, P be a probability space and L = p Lp + Ω, K, P Thus, X L iff X as and EX p < for every p < Let M be the set of the distributions of the random variables X L In other words, F M iff F [, = and x p df x < p < The integral x p df x = EX p will be denoted by µ p F It is the pth moment of X We shall denote by F x the distribution function of F and by F x its right tail Precisely, F x will stand for F [, x] and F x for F x, The set of absolutely continuous probability distribution from M will be denoted by M ac The density of F provided that it does exist will be denoted by f F and the hazard rate see eg [2] by λ F The hazard rate of an absolutely continuous distribution F M is defined by λ F x = f F x F x or, alternatively, by 2 F x = e x λ F ydy The characteristic function of F will be denoted by ϕ F and the moment generating function abbreviated as mgf by m F Thus, for t R, 3 ϕ F t = e itx df x, m F t = e tx df x MATH REPORTS 363, 2, 65 74

2 66 Radu Păltănea and Gheorghiţă Zbăganu 2 or, integrating by parts 4 ϕ F t = + it Let t F be defined by e itx F xdx, m F t = + t 5 t F = sup{t R; m F t < } e tx F xdx If t F = we say that F is short tailed; if t F = we call F long tailed and if < t F <, F is medium tailed See eg [4] In renewal and ruin theories the following distribution is of interest: it is called the integrated tail see eg [], [3] or [4] It is defined for F M by x 6 F I x =, for x < and F I x = F ydy, for x F ydy The main result in [3] was the following Theorem 35, Corollary 37 Theorem A Consider the operator T : M ac M ac defined by T F = F I Suppose that the limit λ F does exist Then i λ F = t F, where t F is defined by 5; ii if F is medium tailed, then T n F Expλ F, n ; if F is short tailed, then T n F δ, n ; if F is long tailed, then T n F does not converge at all To be precise, in the case of long tails the result was that the mass of T n F vanishes at infinity: T n F x as n, for x Recall that if λ F λ G, we say that G is HR-dominated by F and write G HR F Obviously, G HR F G st F : the HR-domination implies the stochastic one Further on, if λ F is non-decreasing, then F is called a IFRdistribution and if λ F is non-increasing, then F is a DFR-distribution see eg [2], [8] The main tool used in [3] were the HR-monotonousness properties of T Propositions 25, 26 and 3 We state them in a shortened way: Lemma Let F, G M ac Then i λ F λ G λ T F λ T G ; ii if λ F is non-increasing respectively non-decreasing, then λ T F is non-increasing respectively non-decreasing too; iii if F IFR then T F HR F, and if F DFR then F HR T F In this paper we are concerned with studying the moments of T n F in order to find another property of the moments of a random variable, besides the already known ones see [9] Our result is Theorem B Let F M ac and let µ k = x k df x be the moment of order k Suppose that the limit λ F of the hazard rate does exist

3 3 On the moments of iterated tail 67 Then µ n+ 7 lim = n nµ n λ F with the convention that = Moreover, if λ F does not exist, it is possible that the limit 7 does not exist, too This result can be reformulated in terms of the first moment of T n F 2 MOMENTS OF T n F Proposition Moments of F I Let F M, µ k = µ k F be the moments of F, ϕ be the characteristic function of F I Then 2 x k df I x := µ k F = µ k+, k k + µ Proof We know ϕ I t = ϕt itµ Then µ k F I = ϕ I k i k ϕ I t = lim t i k t k /k! = lim k!ϕt tµ t i k+ t k+ µ If we apply l Hospital s rule k + times, the last limit is µ k+ k+µ Proposition 2 Let F M Let F n = T n F and µ n,k = x k df n x be the moments of F n Let also µ m = x m df x be the moments of F Then 22 µ n,k = µ n+k µn n+k k Proof Let ϕ be the characteristic function of F and ϕ n be the characteristic function of T n F The recurrence between the characteristic functions ϕ n is 23 ϕ n+ t = ϕ nt tϕ n which can be better written as 24 ϕ n t = + tϕ n ϕ n+ t Let z n = ϕ n With our notation, one can easily see that z n = iµ n Let k By differentiating 24 k times and applying Leibniz s formula one gets 25 ϕ n k+ t = z n tϕ n+ k+ t + k + ϕ n+ k t

4 68 Radu Păltănea and Gheorghiţă Zbăganu 4 For t = we find that i k+ µ n,k+ = i k+ µ n, µ n+,k Hence the moments satisfy the recurrence 26 µ n,k+ = k + µ n, µ n+,k µ n+,k = µ n,k+ k + µ n, Then we prove by induction 26a µ n+,k = µ n j+,k+j, j n + µn j+,j k+j j Indeed, for j =, relation 26a coincides with relation 26 If 26a is true for j n, then applying again relation 26 we find µ n+,k = µ n j+,k+j = µn j+,j k+j j µ n j,k+j+ k+j+µ n j, µn j j+ k+j j = µ n j,k+j+ k+j+ j+µ n j, j+ µn j,j+ For j = n + one gets µ n+,k = get 22 µ n+k+ n+k+ n+ µ n+ Replacing n with n we Corollary 3 With the same notation as in Proposition 2, the first moment of F n is 27 µ n, = µ n+ n + µ n Proof Obvious Now we take into account the monotonicity properties stated in Lemma Proposition 4 Let F M ac and let λ be the hazard rate of F i If F IFR then µn+ n+µ n n λ as n ii If F DFR then µn+ n+µ n as n n λ Proof If F IFR then the sequence of random variables F n n, F n = T n F, is HR-decreasing This implies that F n+ HR F n Therefore, µ F n+ µ F n, see eg [8] Thus the sequence µ n, n is non-increasing It must have a limit But we know that F n Expλ By Beppo-Levi s theorem, the first moments of F n should converge to the first moment of Expλ, that is, to λ If F DFR and λ =, then the sequence µn+ n+µ n n cannot have other limit but For, if that limit would be λ R let us say then, by 22, µ n,k = µ n+k would converge to k!λ k, k N Hence F n+k n would k µ n converge to Expλ, contradiction

5 5 On the moments of iterated tail 69 Proof of Theorem B direct part Let λ = λ F Suppose first that < λ < Let 28 λ x = inf λx, and λ x = sup λx,, x Then λ λ λ, λ is non-decreasing, λ is non-increasing Moreover, as λ = λ does exist, λ = λ = λ Let F be the distribution with hazard rate λ and F be the distribution with hazard rate λ Then F IFR, F DFR and F HR F HR F According to the above lemma iii, the sequence T n F n is HR-increasing since F is a DFR distribution and T n F n is HR-decreasing since F is a IFR distribution By the same lemma i we have the inequalities 29 T n F HR T n F HR T n F It follows that µ T n F µ T n F µ T n F According to 27 we obtain the inequalities 2 µ T n F µ n+ n + µ n µ T n F But the sequence µ T n F n is increasing and the sequence µ T n F n is decreasing Moreover, T n F Expλ and T n F Expλ as n, Theorem A and the convergence is monotonous According to Beppo-Levi s theorem, µ T n F µ Expλ and µ T n F µ Expλ The proof is complete since µ Expλ = λ Consider now the case λ = We use the inequality T n F HR T n F from 29 We know Theorem A that T n F is decreasing and converges to µ n+ n+µ n δ Then µ T n F must converge to Hence converges to, too The last case is when λ = We use the inequality T n F HR T n F According to Theorem A, the tails T n F x converge to and the convergence is monotonous Thus µ T n F = T n F xdx is increasing and, again by Beppo-Levi s theorem, its limit is dx = Example The Poisson distribution Let F = Poissonλ It is true that Theorem B cannot be applied as it is stated, since F is not absolutely continuous But if we replace F by T F = F I, we obtain an absolutely continuous distribution with λ = It means that for the Poisson distribution with λ = we have lim n µ n+ nµ n = lim n µ n+ n+µ n = = Remark The Poisson distribution F = Poissonλ has no simple formula for the moments It is known that µ n F = P n λ, where P n λ are the so called Touchard Polynomials in λ see [5] They are obtained by the

6 7 Radu Păltănea and Gheorghiţă Zbăganu 6 recurrence T n+ λ = λ n k= n k Pk λ Our result says that T n+ λ 2 lim n nt n λ = In the particular case λ =, the values T n are famous under the name of Bell numbers They represent the number of possible partitions of a n-point set and are denoted by B n [] A particular case of 2 points out that lim n B n+ nb n = It is funny that we were able to check the last limit using brute force, but not for 2 Indeed, it is known see [7] that B n ne n+ λn lim n [λn] n+5 =, n W n where this time λn = and W is the so called Lambert function: W : [, [, is the inverse of the function x xe x Using these facts we can prove that for the distribution Poisson we have B n+ nb n = C n, where C is some constant The real problem is whether the limit of T n F does always exist if λ is bounded? Is it true that the sequence µn+ n+µ n n has always a limit? If the second question has an answer in the affirmative, the same would hold for the first one But the answer is NO: there exist medium tailed distributions F for which the sequence µn+ is bounded and divergent Of course in this n+µ n n case λ F cannot have any limit to infinity Proof of the second part of Theorem B We construct a distribution F µ M ac for which the limit lim n n n+µ n does not exist For ρ >, c >, µ > and k >, denote 22 Ac, ρ = 3c ρ + 9 ln 3 23 Bc, ρ, ν, k = 3 c k + ρ + 3 2k + ν ln 3 We can choose the numbers < ρ < 2, < c <, < ν < 6 and k N, such that for all k k to have Ac, ρ < and Bc, ρ, µ, k < Indeed, choose two numbers q, q 2 such that 9 ln 3 < q < q 2 < 5 9 We can fix the number < c < such that 3 c < q 2 Then we can find a number η > such that Bc, ρ, ν, k < if ρ, ν, 3 k, η Fix ν, η such that ν < 6 and fix k Nsuch that 3 k, η Finally, we can choose ρ, η such that ρ < 2 and ρ + 9 ln 3 < q It follows Ac, ρ < q q 2 < and Bc, ρ, ν, k <

7 7 On the moments of iterated tail 7 for k k Denote 24 p =: Ac, ρ and r = Bc, ρ, ν, k For k N, define Let M = j= = { 3 k+ρ k even 3 k k odd and b k = 3 j < The function f, given by f = j= M [aj,a j +3 j ] k i= 3 3i+c is a density function since it is positive, integrable and ftdt = The corresponding distribution F given by F x = for x and F x = x ftdt for x > is clear an absolute continuous function Denote S n = µ n+ nµ n, n N, where the moments µ n, n N, are defined as above We show that the sequence S n n has not limit First note that aj +3 j µ 3 k = t 3k dt ak +3 k t 3k dt M Mb k j= and on the other hand µ 3 k Mb k In other words, where Similarly, where ak +3 k µ 3 k = a j t 3k dt Mb k γ k j k µ 3 k + = γ k j k Mb k + ak +3 k 3 k j bk ak +3 k 3 k j bk j k t 3k dt 3 k j bk aj + 3 j t 3k + dt aj + 3 j [ + γ k ], 3 k aj + 3 j 3 k [ + γ k ], 3 k +

8 72 Radu Păltănea and Gheorghiţă Zbăganu 8 We shall prove that 25 lim ρ k =, k For this it is sufficient to show that lim k ρ k = 26 lim k max{γ k, γ k } = Let j, k N Using the inequality < ρ < 2, the inequalities a j + 3 j <, j < k and a j + 3 j >, j > k are immediate, independently if k and j are odd or even It follows that k max{γ k, γk } 3 k j bk Also j= aj + 3 j 3 k := T k + T k 2 + j=k+ 3 k j bk Limit of T k Let j < k Hence k 2 First we have aj + 3 j k b k = 3 k i=j So that we obtain T k 3 3i+c = 3 3 j+ρ + 3 j 3 k k i=j 3 k 3 i+c = 3 2 3c 3 k 3 j aj + 3 j = 3 j k+ρ3k + 3 2j ρ 3 k 3 j k+ρ3k e 3k 2j ρ = 3 j k+ρ3k + ln 3 3k 2j ρ j= 3 3k Gj,k, where 3 k + Gj, k = 2 3c 3 j k + j k + ρ + 3 2j+ρ + k j3 k ln 3 2 3c 3 j k + j k + ρ + 9 ln 3 + k j 9 It is immediate that the function Ψs = 2 3c 3 s 8 9 s+ρ+ 9 ln 3, s, is decreasing By taking into account 26 we obtain Gj, k Ψ = p Then T k k 3 kp and lim T k = k Limit of T2 k Let k < j We have j j b k = 3 3i+c = 3 3 i+c i=k = 3 2 3c 3 k 3 j i=k

9 9 On the moments of iterated tail 73 Also, similarly as above we obtain aj + 3 j 3 k 3 j k+ρ3k ++ ln 3 3k +3 2j ρ Let ν be the number fixed at the beginning It follows T 2 3 3k Hj,k νj k, where Hj, k = 3c j=k+ 2 3j k [ k 3c 2 3j k + ] j k+ρ+ 3 2j+ρ ln k [ j k + ρ + 3 2k ln 3 +k j 3 k ν ] + νj k For a fixed number k, consider the function Θs = 2 3c 3 s k [ s + ρ + 3 2k ln 3] + νs s We have Θ s = 2 3c+s ln k + ν 3 2 ln ν < Taking into account relation 26 we obtain, Hj, k Θ = r for k k It follows, for such integers k, that T2 k 3 3k r νj k 3 p 3k ν 3 3k ν = 3 3k ν j=k+ p= and then lim T 2 k = Relations 25 are proved k +γk Now, we have S 3 k = I k +γ k where But I k = ak +3 k / t 3k + dt 3 k a 3 k k 3 k + 3 k I k + 3 k 3 k ak +3 k t 3k dt ak + 3 k 3k From relations 2 and from lim ak =, we find k +3 k Finally, we deduce lim S 32l = 3ρ while l lim S 3 k k = lim k 3 k lim l S 3 2l+ = Since the sequence S n n has two different limit points it has no limit 3 k

10 74 Radu Păltănea and Gheorghiţă Zbăganu REFERENCES [] S Assmussen, Applied Probabilities and Queues Wiley, New York, 987 [2] RE Barlow and F Prochan, Mathematical Theory of Reliability Wiley, New York, 965 [3] H Cramer, Collective Risk Theory Skandia Jubilee Volume, Stockholm, 955 [4] P Embrechts, C Kluppelberg and T Mikosch, Modelling Extremal Events for Insurance and Finance Springer, Berlin, 997 [5] NL Johnson, S Kotz and AW Kemp, Univariate Discrete Distributions, 2nd Edition Wiley, New York, 993 [6] S Karlin and HM Taylor, A Second Course in Stochastic Processes Academic Press, New York, 98 [7] L Lovász, Combinatorial Problems and Exercises, 2nd Edition North-Holland, Amsterdam, 993 [8] M Shaked, and JG Shantikumar, Stochastic Orders and their Applications Academic Press, New York, 994 [9] JA Shohat and JD Tamarkin, The Problem of Moments American Mathematical Society, New York, 943 [] numbers [] W function [2] polynomials [3] G Zbăganu, On iterated integrated tail To appear in Proceedings of the Romanian Academy Received 7 June 29 Transilvania University Faculty of Mathematics and Informatics 536 Braşov, Romania radupaltanea@yahoocom and University of Bucharest Faculty of Mathematics and Informatics Str Academiei 4 79 Bucharest Romania zbagang@fmiunibucro