Bounds for the stop loss premium for unbounded risks under the variance constraints

Transcription

1 Bounds for the stop loss premium for unbounded risks under the variance constraints V. Bentkus 1,2 Abstract Let f be a function with convex derivative f. We consider the maximization problem max Ef(X) assuming that X satisfies the stochastic boundedness condition X st Z, with some Z with known distribution, say ν = L(Z), as well as that the mean m = E X and the variance σ 2 = var X are known. It turns out that the maximizers are independent of the function f and can be constructively described in terms of m, σ 2, ν. In particular, the description of the maximizers allows to find the maximal stop loss premium E(X h) 2 +. Similar results hold for symmetric X. Using the results, one can consider stop loss premiums for aggregated risks having the form X X n assuming that the sum has a super-martingale structure. The basic result can be applied to derive tight bounds for the tail probabilities P{X X n x}, which generalize and improve the classical inequalities for sums of bounded random variables. Key words and phrases: bounds for stop loss premium, unbounded risks, heavy tails, bounds for tail probabilities, variance, convex ordering, s-convex functions and orderings, Hoeffding inequalities. A.M.S. classification (1991): 60E15. 1 Introduction We are interested in finding of upper bounds for the stop loss premium π α (X, h) = E (X h) α +, α 0, 1 Department of Probability Theory and Statistics, Institute of Mathematics and Informatics, Akademijos str. 4, LT Vilnius, Lithuania. 2 The research partially supported by CRC 701 Spectral Structures and Topological Methods in Mathematics, Bielefeld University. 1

2 where the risk X is interpreted as a random variable, and h R is a non-random retention limit. Here z + = max{0, z}, and z α + = (z +) α. We define π 0 (X, h) = P{X h}. In this paper we consider only the case α 1. There is a great number of papers where this question was addressed, let us mention Pinelis 1998, 1999, 2006, 2007a,b, Bentkus 2002, 2004, 2008a,b, 2009 for results with a probabilistic background, where the most complicated case α = 0 is of the primary interest. In these papers results with α 1 are used to derive bounds in the case α = 0. In finance and insurance mathematics bounds for the stop loss premium (with α 1) were obtained by De Vylder and Goovaerts 1983, Kaas and Goovaerts 1986, Jansen et al 1986, Denuit et al 2005, etc. See Hürlimann 2008 for a rather complete review of the related results. Often the problem is stated as follows. Assume that X is bounded, a X b with some known non random a < b. Assume further that certain moments of X are known, for example its mean m = EX and/or the variance var X, etc. Maximize the stop loss premium π α (X, h). In the case α 1, the maximizers usually have discrete distributions with several non zero atoms. Such formulation of the problem has some not always desirable features. First of all, the maximizers having discrete distributions with several non-zero atoms constitute a class with a quite poor structure, which can make it difficult to construct sufficiently flexible models in order to simulate financial or economical phenomena. Secondly, it is difficult to integrate into such models cases where X is unbounded and has heavy tails. In this article we continue to study a class of models able to include the case of unbounded and heavy tailed X (see B 2008a,b, 2009, where a simpler case of bounds without variance constraints is considered). The models seem to be sufficiently flexible since they are parameterized by moments and certain distribution functions. A nice feature of the models is that as a particular case they include the classical case a X b of bounded X. Finally, the models are solvable in the sense that we can provide a constructive description of the maximizers, and hence of the maximal stop loss premium. Another motivation leading to the models would be to obtain sufficiently sharp inequalities for tail probabilities of sums of unbounded random variables comparable in quality with the classical inequalities for bounded random variables (see, for example Hoeffding 1963) or even with inequalities in Pinelis 1998, 1999, 2006, 2007a,b, B 2002, 2004, 2008a,b. While proving inequalities for tail probabilities of sums of random variables it is of interest to maximize the expectation Ef(S n ), where f runs over a given family of functions, and the summands X k of the sum S n = X X n satisfy some moment, (stochastic) boundedness, and structure (like symmetry, unimodality, etc.) restrictions. The families f h (t) = exp{ht}, h > 0, f h (t) = (t h) α +, h R, with α 0 are of interest (see Hoeffding 1963, Talagrand 1995, Pinelis , B , BKZ 2006 among others). An important aspect is that the maximizers have to be independent of f from the family under the consideration (in the case of the stop loss premium of the retention limit), like in Theorems 1, 5, 8, 10, 11 below. This is the reason why under the variance constraints we have to assume that α 2. 2

3 One can interpret the stochastic boundedness conditions as a natural extension of the standard boundedness conditions. To see this, let us reformulate the boundedness condition a X b as follows. Let Y respectively Z be a random variable such that P{Y = a} = 1 respectively P{Z = b} = 1. Then we can represent the condition a X b as Y st X st Z, (1) where Y st X means that X dominates Y in the stochastic order, that is, P{Y x} P{X x} for all x R. Let us briefly describe results from B 2009 obtained without the variance constraints. Let f be a convex function. Consider the maximization problem max Ef(X) (2) assuming that X satisfies the stochastic boundedness condition (1) with some Y and Z with known distributions, say µ = L(Y ) and ν = L(Z), as well as that the mean m = EX is known. It turns out that the maximizers are independent of the function f and can be constructively described in terms of m, µ, ν. In particular, the description of the maximizers allows to find the maximal stop loss premium. Using the information, one can obtain bounds for the stop loss premiums for aggregated risks having the form X 1 + +X n assuming that the sum has a super-martingale structure. The bounds imply tight bounds for the tail probabilities P{X X n x}, which generalize and improve the classical inequalities (see, for example Hoeffding 1963), and extend to the case of unbounded random variables the inequalities from B 2002, 2004, 2008a,b. The approach in B 2009 is related to that one in B 2008a,b, where the special case 0 X st Z was considered. In this paper we consider counterparts of (2) under additional variance (respectively variance and symmetry) constraints for some classes of s-convex functions f. In turns out that the two sided stochastic boundedness condition Y st X st Z can be relaxed to X st Z. Henceforth numbers p, q 0 satisfy p + q = 1. We say positive, increasing, etc instead of non-negative, non-decreasing, etc. In cases where the strict inequalities hold, we say strictly positive, etc. We write x + = max{0, x}, and x s + = (x + ) s. Furthermore, stands for. By δ a we denote Diracs measure concentrated at point a, and δ = δ 0. We R assume that all functions and sets under consideration are measurable. We write I{A} for the indicator function of event A. Let us pass to a description of the stochastic boundedness condition (1) and of the maximizers in the optimization problem (2). Assume that Y st Z. Then the set D = {X : Y st X st Z} is nonempty since, for example, Z D. Assuming that E Y <, E Z <, 3

4 we write It is clear that M N. M = EY, N = EZ. We have to introduce several definitions related to the distributions µ = L(Y ) and ν = L(Z). With 0 p 1 and ν = L(Z) we associate a positive measure, say ν [p], having the total mass p = ν [p] (R) such that ν and ν [p] coincide in a maximal possible neighborhood of. If G ν (t) = ν([t, )) is the survival function of ν then we define the survival function G [p] ν (t) = ν [p] ([t, )) setting G [p] ν (t) = min{p, G ν (t)}. In a more detailed way, if there exists b such that ν(i) = p, where I = [b, ) or I = (b, ), then we define ν [p] setting ν [p] (A) = ν(i A) for (measurable) A R. If such b does not exist then there exist b such that ν((b, )) < p < ν([b, )), which means that ν has an atom at point b. Then we define ν [p] (A) = εν(a {b}) + ν(a (b, )) choosing ε such that ν [p] (R) = p. Note that in general the point b from the definition of ν [p] is not defined in a unique way. With 0 q 1 and µ = L(Y ) we associate a positive measure, say µ {q}, having the total mass q = µ {q} (R) such that µ and µ {q} coincide in a maximal possible neighborhood of. The definition is similar to that of ν [p], exchanging by. If G µ (t) = µ([t, )) is the survival function of µ then we define the survival function G {q} µ (t) = µ {q} ([t, )) setting G {q} µ (t) = max{0, G µ (t) p}. For example, if there exists a such that µ(i) = q, where I = (, a) or I = (, a], then we define µ {q} setting µ {q} (A) = µ(i A) for A R. If such a does not exist, then we correct the definition of µ {q} splitting the atom of µ at point a. Note that in general a is not defined in a unique way. Recalling that p + q = 1, consider the probability distribution λ [p] = µ {q} + ν [p]. Note that the survival function G λ [p](t) = λ [p] ([t, )) of λ [p] is given by G λ [p] = G {q} µ +G [p] ν. Furthermore, the relations { } { } G λ [p] = max G µ, min{p, G ν } = min G ν, max{p, G µ } are obvious. One can check that the function p m p def = t λ [p] (dt) 4

5 is a continuous increasing function of 0 p 1 such that m 0 = M and m 1 = N. Hence, for any m [M, N] there exists p = p m such that m = m p. By η = η(m, µ, ν) we denote a random variable having the distribution L(η) = λ [pm]. It is clear that η D m, where D m = {X D : EX = m}. Note that in general p = p m, a = a m and b = b m are not defined in a unique way. Since their concrete definition is immaterial for our purposes, one can define them in an arbitrary convenient way. Henceforth we write p, q, a, b instead of p m, q m, a m, b m. The following bound for the stop loss premium under some additional technical type condition was obtained in B Theorem 1. Assume that E f(y ) < and E f(z) <. If f is convex then max X D m Ef(X) = Ef(η), (3) where η = η(m, µ, ν) is the random variable defined above, and D m = {X D : EX = m}. If f is convex and increasing then max t m max X D t Ef(X) = Ef(η). (4) Proofs of all non obvious statements are given in Section 2. Choosing f(t) = (t h) +, Lemma 1 provides sharp upper bound for the stop loss premium E (X h) + for X D t with t m. Example 2. To make the definitions more transparent, we assume for a while that the distributions of Y and Z have densities, say u and v such that: u is a strictly increasing (and therefore strictly positive) continuous function on an interval (, A], and u(t) 0 for t > A; the function v is a strictly decreasing (and therefore strictly positive) continuous function on an interval [B, ), where B A, and v(t) 0 for t < B. Then, for given m, the probability distribution λ [p] of η = η(m, µ, ν) is absolutely continuous and has the density t u(t)i{t a} + v(t)i{t b}. The related numbers p, q, a, b are defined in a unique way via the equations p + q = 1, q = a u(t) dt, p = v(t) dt, (5) b and m = a tu(t) dt + b tv(t) dt. (6) 5

6 Example 3. Another simple example would be the case where Z st X st Z and EX = 0, with some Z 0. Note that symmetric X such that X st Z satisfy the conditions. Then η has the distribution L(η)(A) = ν [p] (A) + ν [p] ( A) with p = 1/2. Let us turn to situations where the variance varx is known. We write V(m, σ 2, ν) for the set of random variables such that EX = m, varx = σ 2, X st Z. If instead of EX = m the inequality EX m holds then we write V( m, σ 2, ν) instead of V(m, σ 2, ν), etc. Define η = η(m, σ 2, ν) as a random variable with the distribution L(η) = qδ a + ν [p] such that To ensure the existence of η we impose the conditions varη = σ 2, Eη = m. (7) Z 0, 0 < EZ 2 <, m 0, σ 2 0. (8) Lemma 4. Assume that (8) holds. Then the random variable η = η(m, σ 2, ν) is well defined and η V(m, σ 2, ν). Recall that f is s-convex if the (s 2)-th derivative f (s 2) is convex, see Section 3 for definitions and properties of s-convex functions. Theorem 5. Assume that (8) is fulfilled. Let X V(m, σ 2, ν). Then E (X h) 2 + E (η h)2 +, for h R, (9) where η = η(m, σ 2, ν). Furthermore, if f is 3-convex (that is, f is convex), E f(z) <, and E f(x) < then Ef(X) Ef(η). (10) Example 6. Let b > 0 be a non random constant, and Z = b. Then (9) and (10) hold for X b such that EX = m 0 and var X = σ 2. The random variable η is a Bernoulli random variable which assumes the values b and a = m σ 2 /(b m) with probabilities p = σ 2 σ 2 + (b m) 2 and q = (b m)2 σ 2 + (b m) 2 6

7 respectively. As special cases, Theorem 5 yields the classical inequalities for bounded random variables. Let us recall related known results. Let m = 0. For f(t) = exp{ht}, h 0 inequality (10) is established by Hoeffding In the case of f(t) = (t h) 2 + with h R, inequality (10) is given in B Example 7. Assume that Z 0 has density, say u, such that u is a strictly decreasing positive continuous function on [B, ) for some B 0, and u(t) = 0 for t < B. Let t 2 u(t) dt <. Then, for given m 0 and σ 2 0, the parameters p, q = 1 p, a, b from the definition of η are defined in a unique way via the system of equations p = b u(t) dt, m = aq + b tu(t) dt, (a m) 2 q + b t 2 u(t) dt = σ 2. For example, in the special case of m = 0 and the Pareto distribution P{Z t} = t 3 for t 1, one can express p = b 3, a = 3b/(2b 3 2), where b is a unique solution of the equation 12b 3 3 = 4σ 2 (b 4 b). One can apply Theorem 5 in cases where the means and variances of X and η coincide. However, often only some upper bounds for EX and varx are available. The next theorem is applicable in such situations. The price to pay is that one has to restrict somewhat the class of the functions f. Theorem 8. Assume that (8) is fulfilled. Let X V( 0, σ 2, ν). Then E (X h) 2 + E (η h) 2 +, for h R, (11) where η = η(0, σ 2, ν). If f is convex increasing and 3-convex (that is, f, f are convex and f 0), E f(z) <, and E f(x) < then Ef(X) Ef(η). (12) Let us pass to the case of symmetric X with known variance. We write S(σ 2, ν) for the set of symmetric random variables such that varx = σ 2, X st Z. If instead of varx = σ 2 the inequality var X σ 2 holds then we write S( σ 2, ν) instead of S(σ 2, ν). Recalling that ν = L(Z), and that the measure ν [p] is defined above, introduce the measure µ [p] such that µ [p] (A) = ν [p] ( A) for A R. Define η = η(σ 2, ν) as a random variable with the distribution L(η) = µ [p] + rδ + ν [p] such that r = 1 2p, var η = σ 2, (13) 7

8 where δ is the Dirac measure concentrated at 0. To ensure the existence of η we impose the conditions Z 0, 0 < EZ 2 <, 0 σ 2 2 t 2 ν [1/2] (dt). (14) Lemma 9. Assume that (14) holds. Then the symmetric random variable η = η(σ 2, ν) is well defined and η S(σ 2, ν). Theorem 10. Assume that (14) is fulfilled. Let X S(σ 2, ν). Then E (X h) 3 + E (η h)3 +, for h R, (15) where η = η(σ 2, ν). Furthermore, if f is 4-convex (that is, f is convex), E f(z) <, then Ef(X) Ef(η). (16) Theorem 11. Assume that (14) is fulfilled. Let X S( σ 2, ν). Then E (X h) 3 + E (η h)3 +, for h R, (17) where η = η(σ 2, ν). If f is convex, 4-convex, and f is increasing (that is, f is convex positive and increasing), E f(z) <, then Ef(X) Ef(η). (18) 2 Proofs Proof of Theorem 1. In B 2009 this theorem is proved under the additional assumption that there exists y R such that Y y Z. However, an inspection of the proof shows that this assumption is superfluous. Then We write m p = t ν [p] (dt), σ 2 p = t 2 ν [p] (dt). (19) (t m) 2 ν [p] (dt) = σ 2 p 2mm p + pm 2. (20) 8

9 Proof of Lemma 4. Since η = η(m, σ 2, ν) has the distribution L(η) = qδ a + ν p, we can rewrite the conditions varη = σ 2 and Eη = m as (a m) 2 q + σ 2 p 2m pm + pm 2 = σ 2, aq + m p = m. (21) It suffices to check that there exist p and a satisfying the equations (21) such that 0 < p 1 and a 0. Using the second equation in (21), we can write a = (m m p )/q. Introduce the function u(p) = (a p m) 2 q + (t m) 2 ν [p] (dt), 0 p 1, (22) R where a p = (m m p )/q. Note that u is a continuous positive function of the argument 0 p < 1 such that u(0) = 0, and u(1) =. Indeed, if p = 0 then q = 1, ν [p] = 0, m p = 0, a p = m. Inserting these values in (22), we get u(0) = 0. If p = 1 then q = 0 and m p = EZ > 0 due to the assumptions Z 0 and EZ 2 > 0. Hence, using a p = (m m p )/q and m 0, we have (a p m) 2 q (pm m p ) 2 /q m 2 p/q =. Let us show that the function p u(p) is strictly increasing. It is clear that the function p (t m) 2 ν [p] (dt) is increasing. Note that (a p m) 2 q (m p pm) 2 /q. The function p (m p pm) 2 is increasing and strictly positive for p > 0 since we assume that EZ 2 > 0 and therefore m p > 0 for p > 0. The function p 1/q is strictly increasing. We conclude that u is strictly increasing. Since the function p u(p) is strictly increasing continuous positive function of the argument 0 p < 1 such that u(0) = 0, and u(1) =, the equation u(p) = σ 2 has a unique solution p = p(m, σ 2 ). The knowledge of p allows to define q = 1 p and m p as in (19). Setting a = (m m p )/q we see that η V(m, σ 2, ν). Proof of Theorem 5. We have to prove (9) and (10). Recall that L(η) = qδ a +ν [p], and that b is a number such that ν((b, )) p ν([b, )). We can assume that b 0 since Z 0. The number a satisfies a 0, see the proof of Lemma 4. Both X and η have equal means m and variances σ 2. If σ 2 = 0 then η = m, X = m, and the assertion of the theorem clearly holds. Therefore henceforth we assume that σ 2 > 0. Let us prove (9), that is, that E (X h) 2 + E (η h)2 +, for h R. (23) 9

10 We consider the cases separately. i) h a, ii) a < h < b, iii) h b i) Using the obvious inequality (t h) 2 + (t h)2, noting that E (η h) 2 = E (η h) 2 + since P{η {a} [b, )} = 1 and h a, we have E (X h) 2 + E (X h)2 = σ 2 + (m h) 2 = E (η h) 2 +. ii) Now a < h < b. The function u(t) = max{c(t a) 2, (t h) 2 + }, c = (b h)2 /(b a) 2, satisfies (t h) 2 + u(t). It is elementary to check that u(t) = c(t a) 2 for t b, and that u(t) = (t h) 2 + for t b. Hence, we can write u(t) = c(t a) 2 + v(t), v(t) = ( (t h) 2 c(t a) 2) I{t b}, whence E (X h) 2 + Eu(X) = c E(X a) 2 + Ev(X). (24) Using m = EX = Eη and σ 2 = varx = varη, we obtain Ec(X a) 2 = c(σ 2 + (m a) 2 ) = Ec(η a) 2. (25) The function v is increasing, v(b) = 0 for t b, and X st Z. In view of a < h both (t h) 2 + and (t a)2 vanish at t = a. Hence E v(x) Ev(Z) = {(t h) 2 c(t a) 2 } ν [p] (dt) = (t h) 2 ν [p] (dt) c(t a) 2 ν [p] (dt) Combining (24) (26), we get E (X h) 2 + E (η h)2 +. = E (η h) 2 + Ec(η a) 2. (26) iii) Now h b. The function t (t h) 2 + is increasing and X st Z. Therefore E (X h) 2 + E (Z h)2 + = (t h) 2 ν [p] (dt) = E (η h) 2 +, which completes the proof of (23) and (9). h Let us prove (10), that is, that Ef(X) Ef(η) for 3-convex f. It suffices to show that E (h X) 2 + E (h η) 2 +, for h R. (27) 10

11 Indeed, any 3-convex function f can be represented as (see Section 3) f(x) = P(x) (h x) 2 + λ(dh) + (x h) 2 + λ(dh), (28) (,x 0 ] (x 0, ) where λ is a positive locally finite Borel measure on R (actually 2λ = f in the generalized sense), the point x 0 R is arbitrary, and P(x) = c 2 x 2 + c 1 x +c 0 is a quadratic polynomial whose coefficients can depend on x 0 and f. Using (23) and (27), the representation (28) clearly implies Ef(X) Ef(η). It remains to prove (27). Using the obvious decomposition (h t) 2 + = (t h)2 (t h) 2 +, inequality (27) is equivalent to E (X h) 2 + E (X h) 2 + E (η h)2 + E (η h) 2 +. (29) Since EX = Eη and varx = var η, it follows that E (X h) 2 = E (η h) 2. Hence, (29) is equivalent to E (X h) 2 + E (η h) 2 + which we have proved earlier as (23). Proof of Theorem 8. We have to prove (11) and (12). Write η = η(m, σ 2, ν), ξ = η(0, σ 2, ν), ζ = η(0, s 2, ν), ϑ = η(m, s 2, ν). In view of Theorem 5 instead of (11) and (12) it suffices to prove that for m 0 E (η h) 2 + E (ξ h)2 +, (30) that for s 2 σ 2 E (ζ h) 2 + E (ξ h)2 +, (31) and that for m 0 and s 2 σ 2 Ef(ϑ) Ef(ξ) (32) if f is convex increasing and 3-convex. Let turn to the proof of (30). During the proof σ 2 is fixed since we are only interested in the dependence of the parameters on m 0, which is reflected in the notation supplying the parameters with the index m. For example, p = p m, a = a m. The function u defined by (22) as a function of m has negative partial derivative with respect to m. Indeed, m u = 2(pm m p )/q, 0 p 1, (33) and m u 0 since m 0 and m p 0. Hence, this function is a decreasing function of m. Recall that we define p = p m as a unique solution of the equation u = σ 2. Therefore it is clear that p m p 0. Hence, from the definition of the measure ν [p], it is clear that ν [pm] st ν [p 0]. 11

12 While proving (30) it is convenient to consider the cases i) h min{a m, a 0 }, ii) a m h a 0, iii) a 0 h a m, iv) h max{a m, a 0 } separately. Note that a m 0 b m and a 0 0 b 0. i) Now h a m. Obviously a m m. Hence m h 0 and (m h) 2 h 2 since h 0. Therefore, using in addition h a 0, we get E (η h) 2 + = E (η h)2 = σ 2 + (m h) 2 σ 2 + h 2 = E (ξ h) 2 = E (ξ h) 2 +. ii) Using a m h and ν [pm] st ν [p 0], we obtain E (η h) 2 + = (t h) 2 + ν[pm] (dt) (t h) 2 + ν[p 0] (dt) a 2 0 q 0 + (t h) 2 + ν[p 0] (dt) = E (ξ h) 2 +. iii) Now a 0 h a m. The function u(t) = max{c(t a 0 ) 2, (t h) 2 + }, c = (b 0 h) 2 /(b 0 a 0 ) 2, satisfies (t h) 2 + u(t). It is easy to verify that u(t) = c(t a 0 ) 2 for t b 0, u(t) = c(t h) 2 + for t b 0. Writing with we have u(t) = c(t a 0 ) 2 + v(t) v(t) = w(t)i{t b 0 }, w(t) = (t h) 2 c(t a 0 ) 2, E (η h) 2 + Eu(η) = c E(η a 0) 2 + Ev(η). Note that a 0 a m m. Hence m a 0 0, and (m a 0 ) 2 a 2 0 since a 0 0 and m 0. Thus we have c E (η a 0 ) 2 = c(σ 2 + (m a 0 ) 2 ) c(σ 2 + a 2 0) = c E (ξ a 0 ) 2. The function v is increasing and ν [pm] st ν [p 0]. Using v(t) = 0 for t b 0, observing that P{ξ {a 0 } [b 0, )} = 1, recalling that v(t) = w(t)i{t b 0 } and w(ξ)i{ξ = a 0 } = (ξ h) 2 I{ξ = a 0 } 0, 12

13 we get Ev(η) = v(t) ν [pm] (dt) v(t) ν [p 0] (dt) = Ev(ξ) = E w(ξ)i{ξ b 0 } Ew(ξ)I{ξ b 0 } + Ew(ξ)I{ξ = a 0 } = Ew(ξ) = E (ξ h) 2 c E(ξ a 0 ) 2. Combining the bounds, we obtain E (η h) 2 + E (ξ h) 2, concluding the proof of (30) in the case iii). iv) Now h max{a m, a 0 }. The function t (t h) 2 + is increasing and ν[pm] st ν [p 0]. Therefore E (η h) 2 + = (t h) 2 + ν[pm] (dt) (t h) 2 + ν [p0] (dt) = E (ξ h) 2 +. The proof of (30) is completed. Let us prove (31). During the proof we are interested in the dependence of the parameters on the variances s 2 and σ 2, which is reflected in the notation supplying the parameters with the index s or σ respectively. For example, p = p σ, a = a σ. Since now m = 0, we can rewrite the equations var η = σ 2, Eη = 0 defining the distribution of η = η(0, σ 2, ν) as a 2 q + σ 2 p = σ 2, aq + m p = 0. (34) It is clear from (34) that p σ is an increasing function of σ. Hence ν [ps] st ν [pσ]. It is clear as well that a σ a s. While proving (31) it is convenient to consider the cases i) h a σ, ii) a σ h a s, iii) h a s separately. i) We have E (ζ h) 2 + = E (ζ h)2 = s 2 + h 2 σ 2 + h 2 = E (ξ h) 2 = E (ξ h) 2 +. ii) Now a σ h a s. The function u(t) = max{c(t a σ ) 2, (t h) 2 + }, c = (b σ h) 2 /(b σ a σ ) 2, 13

14 satisfies (t h) 2 + u(t). Writing u(t) = c(t a σ ) 2 + v(t), with v(t) = w(t)i{t b σ }, w(t) = (t h) 2 c(t a σ ) 2, we have E (ζ h) 2 + Eu(ζ) = c E (ζ a σ) 2 + Ev(ζ). Using s 2 σ 2, we have c E(ζ a σ ) 2 = c(s 2 + a 2 σ) c(σ 2 + a 2 σ) = c E (ξ a σ ) 2. The function v is increasing and ν [ps] st ν [pσ]. Hence Ev(ζ) = v(t) ν [ps] (dt) v(t) ν [pσ] (dt) = Ev(ξ) = Ew(ξ)I{ξ b σ } Ew(ξ)I{ξ b σ } + Ew(ξ)I{ξ = a σ } = Ew(ξ) = E (ξ h) 2 c E (ξ a σ ) 2. Combining the bounds, we get E (ζ h) 2 + E (ξ h) 2. iii) The function t (t h) 2 + is increasing and ν[ps] st ν [pσ]. Hence we have E (ζ h) 2 + = (t h) 2 + ν[ps] (dt) (t h) 2 + ν [pσ] (dt) = E (ξ h) 2 +, completing the proof of (31). Let us prove (32), that is, that Ef(ϑ) Ef(ξ) assuming that f is convex and increasing and has a convex derivative f. The outline of the proof is as follows. Combining (30) and (31), the inequality (32) holds for semi-quadratic f(t) = (t h) 2 +. For general f we prove (32) approximating f by functions which can be represented as mixtures of semiquadratic functions. Let k = 1, 2,.... Introduce the functions f k (t) = f(t) for t k, f k (t) = f( k) + f ( k)(t + k) for t k. The function t f( k) + f ( k)(t + k) is tangent to f at t = k. Since f is convex, it follows that f k f and lim f k = f. A bit later we show that f k can be represented as k f k (x) = P(x) + (x h) 2 + λ k(dh) (35) ( 2k, ) 14

15 with P(x) = f ( k)(x+k)+f( k) and a positive locally finite Borel measure λ k. Using (35) it follows that Ef k (ϑ) Ef k (ξ). To see this, it suffices to use E (ϑ h) 2 + E (ξ h)2 + and to note that EP(ϑ) = (m + k)f ( k) + f( k) kf ( k) + f( k) = EP(ξ) since m 0, and f ( k) 0 by our assumption that f is increasing. Since f k f the inequality Ef k (ϑ) Ef k (ξ) implies that Ef k (ϑ) Ef(ξ). Passing now to the limit as k and using lim k f k = f, we derive Ef(ϑ) Ef(ξ). In order to complete the proof of (32) we have to derive (35). It is clear that the derivative f k is convex since f is convex and f k (t) = f (t) for t k, f k (t) = f ( k) for t k. We conclude that f k is 3-convex. Furthermore, f k (t) = f k (t) = 0 for t < k. Being 3-convex f k has an integral representation of type (52) given in Section 3. Choosing x 0 = 2k, this representation takes the form of (35) with some P(x) = c 2 x 2 +c 1 x+c 0. Note that in our particular case the integral involving (h x) 2 + in (52) vanishes because f k (t) = 0 for t < k. To define the coefficients c 2, c 1, c 0 note that for x < 2k the integral in (35) vanishes since (x h) + = 0 for x < 2k < h. Hence f k (x) = P(x) for x < 2k. Recalling that f k (x) = f( k) + f ( k)(x + k) for x k, it follows that c 2 x 2 + c 1 x + c 0 = f( k) + f ( k)(x + k) for x < 2k, whence c 2 = 0, c 1 = f ( k), c 0 = f( k) + f ( k)k. Let us turn to the proofs related to symmetric random variables. Proof of Lemma 9. Note that X st Z together with the symmetry assumption implies that Z st X. For X such that Z st X st Z and EX = 0, Lemma 1 yields EX 2 σ 2 max, σ2 max def = 2 t 2 ν [1/2] (dt). In order to establish the existence of the symmetric η = η(σ 2, ν) with L(η) = µ [p] +rδ+ν [p] and Eη 2 = σ 2, we rewrite the condition Eη 2 = σ 2 as u(p) def = 2 t 2 ν [p] (dt) = σ 2. The function p u(p) is increasing and continuous, u(0) = 0 and u(1/2) = σ 2 max. Since 0 σ 2 σ 2 max, there exists 0 p 1/2 such u(p) = σ2. Note that in cases where P{Z = 0} > 0 the number p is not necessarily defined in a unique way. Proof of Theorem 10. We have to prove (15) and (16). 15

16 Let us prove prove (15), that is, that E (X h) 3 + E (η h) 3 +. It is convenient to reformulate this inequality as an inequality for positive random variables. Namely, consider the inequality Ef h (Y ) Ef h (ξ), for h R, (36) where f h (t) = [ (t h) ( t h) 3 +] /2, a random variable Y satisfies Y 0, EY 2 = σ 2, P{Y x} 2ν([x, )), (37) and ξ is a positive random variable such that L(Y ) = rδ + 2ν [p]. Actually, one can easily show that (15) is equivalent to (36) using the fact that distributions of the symmetric random variables X and η can be recovered in a unique way from the distributions of X and η. Hence, instead of (15) it suffices to prove (36). In the proof of (36) we use a point b 0 such that ν [p] ((b, )) p ν [p] ([b, )), and the inequality Ev(Y ) Ev(ξ), (38) which holds for increasing differentiable functions v : R R such that v(t) = 0 for t b. Integrating by parts and using the stochastic boundedness condition (37), inequality (38) is equivalent to the obvious P{Y t}v (t) dt 2ν([t, ))v (t) dt. (0, ) While proving (36), we consider the cases separately. i) Now h b, and (0, ) i) h b, ii) b < h < 0, iii) 0 h b, iv) h b f h (t) = u(t) for 0 t h, u(t) def = 3ht 2 h 3 and We can write f h (t) = (t h) 3 /2 for t h. f h (t) = u(t) + v(t), v(t) = 2 1 ( (t h) 3 2u(t) ) I{t h}. It is easy to check that the function v is increasing. Using (38) we have Ev(Y ) Ev(ξ). Since EY 2 = Eξ 2 it follows that Eu(Y ) = Eu(ξ). Therefore Ef h (Y ) = Eu(Y ) + Ev(Y ) Eu(ξ) + Ev(ξ) = Ef h (ξ). 16

17 ii) Now 0 < h < b, and f h (t) = u(t) for 0 t h, u(t) def = 3ht 2 h 3 and Introduce the function f h (t) = (t h) 3 /2 for t h. g(t) = ct 2 h 3, c def = (h 3 + (b h) 3 /2)/b 2. The function g satisfies g(0) = u(0) and g(b) = (b h) 3 /2. Introduce the function w(t) = g(t) f h (t). Then w(0) = w(b) = 0. Let us prove that and w(t) 0 for 0 t b, (39) w(t) 0, w (t) 0 for t b. (40) In the proof of (39) we consider the cases a) 0 t h and b) h t b separately. a) Now w(t) = (c + 3h)t 2, and it suffices to show that c + 3h 0. Elementary transformations show that c + 3h 0 is equivalent to b + h 0 which holds since we assume that h < b. b) In this case w(t) = ct 2 h 3 (t h) 3 /2 and h t b. Elementary transformations show that w( h) = h 2 (b + h) 3 /(2b 2 ), and w( h) > 0. It is clear as well that w(b) = 0. We note that w is a quadratic concave function since w = 3. Using the obvious representation w ( h) = h(b + h) 3 /b 2 > 0 we conclude that w has two roots, say t 1, t 2, such that t 1 < h < t 2. If t 2 b then w (t) 0 for h t b, the function w is increasing on [ h, b], and we conclude that w(t) 0 for h t b since w( h) > 0. If t 2 < b then w is positive on [ h, t 2 ] and negative on [t 2, b]. Hence w is increasing on [ h, t 2 ] and decreasing on [t 2, b]. We conclude that w attains its minimal value at the endpoints of the interval [ h, b]. Since w( h) > 0 and w(b) = 0, we conclude that w is positive on [ h, b]. The proof of (39) is completed. Let us prove (40). Now w(t) = ct 2 h 3 (t h) 3 /2 and t b. As it is proved in b), we have w(b) = 0. Hence it suffices to show that w (t) 0 for t b. We have w (t) = 2ct 3(t h) 2 /2. Since w = 3, the function w is concave. Therefore in order to prove that the inequality w (t) 0 holds, it suffices to check that c) w (b) 0, and d) w (b) 0. c) We have w (b) = 2cb 3(b h) 2 /2. Using the definition of c, we can write w (b) = s(h)/b with s(h) = 2h 3 (h b) 2 (h + b/2). 17

18 The inequality w (b) 0 is equivalent to s(h) 0 for b h 0. We note that s( b) = 0 and s (h) = 3h(h + b) 0 for b h 0. Hence s(h) 0, and w (b) 0. d) We have w (t) = 2c 3(t h) and w (b) = 2c 3(b h) s(h)/b 2, s(h) def = h 3 2b 3 + 3bh 2. Thus, inequality w (b) 0 is equivalent to s(h) 0 for b h 0. Since s( b) = 0 and s(0) = 2b 3 < 0, and s is convex (noting that s (h) = 6(b + h) 0), if follows that s(h) 0 for all b h 0. Introduce the function In view of (39) and (40) we have We can decompose F as follows F(t) = max{g(t), f h (t)}, t 0. F(t) = g(t) for 0 t b, F(t) = f h (t) for t b. F(t) = g(t) + v(t), v(t) = (f h (t) g(t)) I{t b} w(t)i{t b}. (41) Using (40) it is clear that v is increasing positive function such that v(t) = 0 for t b. By (38) we have E v(y ) Ev(ξ). Since EY 2 = Eξ 2 it follows that Eg(Y ) = Eg(ξ). Therefore Ef h (Y ) EF(Y ) = Eg(Y ) + Ev(Y ) Eg(ξ) + Ev(ξ). Noting that f h (0) = g(0) and that L(ξ) = rδ + 2ν [p], we obtain Ev(ξ) = [f h (t) g(t)] 2ν [p] (dt) [b, ) = Ef h (ξ) rf h (0) [b, ) Combining the inequalities we derive Ef h (Y ) Ef h (ξ). g(t) 2ν [p] (dt) = Ef h (ξ) Eg(ξ). (42) iii) The proof is similar to the proof in the case ii), therefore we shall omit technical calculations. Now 0 h b, and f h (t) = 0 for 0 t h, f h (t) = (t h) 3 /2 for t h. Introduce the function g(t) = ct 2, c def = (b h) 3 /(2b 2 ). 18

19 The function g satisfies g(0) = 0 and g(b) = f h (b) = (b h) 3 /2. Let Then F(t) = max{g(t), f h (t)}, t 0. F(t) = g(t) for 0 t b, F(t) = f h (t) for t b. Now we can argue similar to (41) (42). iv) Now h b. Using the stochastic domination condition (37), we obtain Ef h (Y ) f h (t) 2ν(dt) = Ef h (ξ), completing the proof of (15). (h, ) Let us prove (16), that is, that Ef(X) Ef(η) for 4-convex f. It suffices to show that E (h X) 3 + E (h η)3 +, for h R. (43) Indeed, any 4-convex function f can be represented as (see Section 3) f(x) = P(x) + (h x) 3 + λ(dh) + (x h) 3 + λ(dh), (44) (,x 0 ] (x 0, ) where λ is a positive locally finite Borel measure on R (actually 6λ = f (4) in the generalized sense), the point x 0 R is arbitrary, and P(x) = c 3 x 3 + c 2 x 2 + c 1 x + c 0 is a cubic polynomial whose coefficients can depend on x 0 and f. Using (15) and (43), noting that EP(X) = EP(η) due to EX 2 = Eη 2 and symmetry of X and η, the representation (44) clearly implies Ef(X) Ef(η). It remains to prove (43). Using the obvious decomposition (h t) 3 + = (t h) 3 + (t h) 3, inequality (43) is equivalent to E (X h) 3 + E (X h)3 E (η h) 3 + E (η h)3. (45) Since X and η are symmetric and EX 2 = Eη 2, it follows that E (X h) 3 = E (η h) 3. Hence, (45) is equivalent to E (X h) 3 + E (η h)3 + which we have proved earlier as (15). Proof of Theorem 11. We have to prove (17) and (18). During the proof we write η = η(σ 2, ν), ξ = η(s 2, ν), s 2 σ 2. In view of Theorem 10 instead of (17) and (18) it suffices to prove that E (ξ h) 3 + E (η h)3 +, for h R, (46) 19

20 and that assuming that f is convex positive and increasing. Ef(ξ) Ef(η), (47) Let us prove (46). Using the symmetry of ξ and η we can rewrite (46) as Ef h ( ξ ) Ef h ( η ) with 2f h (t) = (t h) + 3 +( t h) 3 +. (48) In order to prove (48) it suffices to note that the function t f h (t) : [0, ) R is increasing and that ξ st η (we omit formal checking of these simple facts). Let us prove (47). Let k = 1, 2,.... Introduce the functions f k (t) = f(t) for t k, f k (t) = P(t) for t k, where P(t) = f( k) + f ( k)(t + k) + f ( k)(t + k) 2 /2. (49) It is clear that lim f k = f. One can check that f k f. A bit later we show that f k can k be represented as f k (x) = P(x) + (x h) 3 + λ k(dh) (50) ( 2k, ) with P defined by (49), and a positive locally finite Borel measure λ k. Using (50) it follows that Ef k (ϑ) Ef k (ξ). To see this, it suffices to use E (ξ h) 3 + E (η h)3 + and to note that EP(ξ) EP(η). The latter inequality can be easily checked using the symmetry of ξ and η, applying E ξ 2 Eη 2, and noting that f ( k) 0 since we assume that f 0. Using f k f, the inequality Ef k (ξ) Ef k (η) implies that Ef k (ξ) Ef(η). Passing now to the limit as k and using lim f k = f, we derive Ef(ξ) Ef(η). k In order to complete the proof of (47) we have to derive (50). It is clear that f k is convex since f is convex increasing and f k (t) = f (t) for t k, f k (t) = f ( k) for t k. We conclude that f k is 4-convex. Furthermore, f k (t) = f(4) k (t) = 0 for t < k. Being 4-convex f k has an integral representation of type (52) given in Section 3. Choosing x 0 = 2k, this representation takes the form of (50) with some P(t) = c 3 t 3 + c 2 t 2 + c 1 t + c 0. Note that in our particular case the integral involving (h x) 3 + in (52) vanishes because f (4) k (t) = 0 for t < k. To define the coefficients c 3, c 2, c 1, c 0 note that for x < 2k the integral in (35) vanishes since (x h) + = 0 for x < 2k < h. Hence f k (x) = P(x) for x < 2k. Therefore we can conclude that P is the polynomial given by (49). 20

21 3 Higher order convex functions Let s 2 be an integer. We say that f : R R is s-convex if f is s 2 times continuously differentiable and f (s 2) is a convex function. According to the definition 2-convex functions are just convex ones. For example, the function e t is s-convex for all s 2, and the function e t is s-convex iff s is an even integer. See Pečarić 1992 for alternative definitions and a discussion related to convexity of the higher order. We are going to give some useful representations of s-convex functions. Let us introduce the related notation. Let M loc (J) stand for the class of signed Borel measures on an interval J R such that λ M loc (J) have finite variation sup λ(b) < on compact subsets A J. The class of positive λ is denoted as M loc + (J). We abbreviate Mloc = M loc (R) B A and M loc + = M loc + (R). Let us recall that a measure λ M loc (J) is generalized (or Sobolev) derivative of µ M loc (J) if for all ϕ C 0 (J) ϕ(t) λ(dt) = ϕ (t) µ(dt). Here as usually C 0 (J) stands for the class of all infinitely many times differentiable functions with compact support. We write λ = µ. We identify absolutely continuous measures with their densities with respect to the Lebesgue measure. In particular, notation λ = f means that λ is the generalized derivative of the measure f(t) dt. Note that the density f as a function is defined almost everywhere. Henceforth, if exists, we always choose a continuous or right continuous version of f. Now we can give another definition of s-convexity. Namely, by the definition a function f is s-convex (s = 1, 2,...) if its generalized derivative f (s) is a positive measure, that is, f (s) M loc + (R). We omit a rather standard proof that the two definitions of s-convex functions are equivalent, for s 2. Note that 1-convex functions are increasing ones, 2-convex functions are convex ones, etc. If f is convex and x 0 R then there exist positive Borel measure λ M loc +, and a linear function P 1 such that f(x) = P 1 (x) + (h x) + λ(dh) + (x h) + λ(dh). (51) (,x 0 ] (x 0, ) Naturally, P 1 in (51) can depend on f and x 0. As λ one can take λ = f (2). Several representations close to (51) are given in B 2008b. The representation (51) has a counterpart for s-convex f. Namely, if f is s-convex and x 0 R then with λ = f (s) /(s 1)! and a polynomial P s 1 = a s 1 x s a 0 of degree at most s 1 we have f(x) = P s 1 (x) + ( 1) s (,x 0 ] (h x) s 1 + λ(dh) + 21 (x 0, ) (x h) s 1 + λ(dh), (52)

22 where P s 1 in (52) can depend on f, s and x 0. By our agreement we consider (if exists) the continuous versions of f. Therefore for s 2 the equality (52) holds for all x R. See (53) for a counterpart of (52) in the case s = 1. To prove (52) one can proceed as follows. Let F stand for the right hand side of (52). Differentiating s times the function F, we see that F (s) = f (s). Therefore the difference F f is a polynomial of degree at most s 1. Hence, changing, if necessary, the coefficients P s 1, we get (52). In the proof (s 1)-th and s-th derivatives one has to understand in the generalized sense. We omit a detailed exposition of technicalities related to the proof of (52). In the special case s = 1 the representation (52) can be written as f(x) = P 0 I{h > x} λ(dh) + I{x h} λ(dh), (53) (,x 0 ] where the constant P 0 in (53) can depend on f and x 0. (x 0, ) References [1] V. Bentkus, A remark on the inequalities of Bernstein, Prokhorov, Bennett, Hoeffding, and Talagrand, Lith. Math. J., 42(3): , [2] V. Bentkus, On Hoeffding s inequalities, Ann. Probab., 32(2): , [3] V. Bentkus, An extension of an inequality of Hoeffding to unbounded random variables, Lith. Math. J., 48(3): , 2008a. [4] Bentkus, V., Addendum to: An extension of an inequality of Hoeffding to unbounded random variables, Lith. Math. J., 48(2): , 2008b. [5] Bentkus, V., On stop loss premium for unbounded risks, manuscript, [6] V. Bentkus, N. Kalosha, and M. van Zuijlen, On domination of tail probabilities of (super)martingales: explicit bounds, Lith. Math. J., 46(1):1 43, [7] M. Denuit, J. Dhaene, M. Goovaerts, and R. Kaas Actuarial theory for dependent risks. Measures, orderrs, and models. J. Viley, New York, [8] De Vylder, F.; Goovaerts, M., Best bounds on the stop-loss premium in case of known range, expectation, variance and mode of the risk, Insurance Math. Econom., 2(4): , [9] W. Hoeffding, Probability inequalities for sums of bounded random variables, J. Am. Statist. Assoc., 58:13 30,

23 [10] Hürlimann, Werner, Extremal moment methods and stochastic orders: application in actuarial science. Chapters I, II and III, Bol. Asoc. Mat. Venez., 15:5 110, [11] Jansen, K.; Haezendonck, J.; Goovaerts, M.J., Upper bounds on stop-loss premiums in case of known moments up to the fourth order J.. Insurance Math. Econom., 5(4): , [12] Kaas, R.; Goovaerts, M. J., Extremal values of stop-loss premiums under moment constraints, Insurance Math. Econom., 5(4): , [13] C. McDiarmid, On the method of bounded differences, Surveys in combinatorics, 1989 (Norwich 1989), London Math. Soc. Lecture Note Ser., 141: , [14] I. Pinelis, Optimal tail comparison based on comparison of moments, High dimensional probability (Oberwolfach, 1996), Progr. Probab., 43: , [15] I. Pinelis, Fractional sums and integrals of r-concave tails and applications to comparison probability inequalities, Advances in stochastic inequalities (Atlanta, GA, 1997), Contemp. Math., 234: , Am. Math. Soc., Providence, RI [16] I. Pinelis, On normal domination of (super)martingales, Electron. J. Prabab, 11(39): , [17] I. Pinelis, Inequalities for sums of asymmetric random variables, with applications, Probab. Theory Related Fields, 139(3 4): , 2007a. [18] I. Pinelis, Toward the best constant factor for the Rademacher-Gaussian tail comparison, ESAIM Probab. Stat., 11: , 2007b. [19] M. Talagrand, The missing factor in Hoeffding s inequalities, Ann. Inst. H. Poincaré Probab. Statist., 31(4): , [20] W. Hoeffding, Probability inequalities for sums of bounded random variables, J. Am. Statist. Assoc., 58:13 30, [21] A. Marshall and I. Olkin, Inequalities: Theory of Majorization and Its Applications, Academic Press, New York San Francisco, [22] I. Pinelis, Optimal tail comparison based on comparison of moments, High dimensional probability (Oberwolfach, 1996), Progr. Probab., 43: , [23] I. Pinelis, Fractional sums and integrals of r-concave tails and applications to comparison probability inequalities, Advances in stochastic inequalities (Atlanta, GA, 1997), Contemp. Math., 234: , Am. Math. Soc., Providence, RI [24] M. Shaked and J. G. Shanthikumar, Stochastic orders, Springer, New York,

24 Vidmantas Bentkus Akademijos str. 4 Institute of Mathematics and Informatics Vilnius, Lithuania address: bentkus@ktl.mii.lt 24