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When is an integral stochastic order generated by a poset?
Journal of Inequalities and Applications volume 2012, Article number: 265 (2012)
Abstract
Given a partial order ⪯ on a set , one can consider the class of ⪯-preserving real functions on characterized by implies . Such a class of functions allows us the generation of a binary relation on the set of probabilities associated with by means of when for all ⪯-preserving functions f. In this paper we characterize when for an integral stochastic order on the set of probabilities associated with , there exists a partial order ⪯ on such that the relation generated by the class of ⪯-preserving functions is equal to . The above characterization is related to the maximal generator of , a result which can be applied for the search of maximal generators of stochastic orders generated by posets.
MSC:06A06, 60E15.
1 Introduction
Stochastic orders play a key role in many areas. They have been applied successfully in such fields as reliability theory, economics, decision theory, queueing systems, scheduling problems, medicine, genetics, etc. A stochastic order is defined as a partial order relation on a set of probabilities associated with a certain measurable space, although in some contexts the antisymmetric condition is not considered.
A method for generating binary relations on a class of probabilities associated with a set by means of partial order relations on such a set is proposed in [1]. That method is based on the so-called order-preserving functions. Thus, if is a measurable space and ⪯ is a partial order on , one can take the class of all measurable ⪯-preserving real functions, that is, the set of measurable functions such that if satisfy , then . The partial order ⪯ generates a binary relation on the set of probabilities on defined by
for all measurable ⪯-preserving functions f.
In this paper we characterize when, for a stochastic order , there exists a partial order ⪯ on such that is . The above characterization will be related to the maximal generators of integral stochastic orders. Different examples of stochastic orders generated and not generated by partially ordered sets will be developed.
2 Preliminaries
Let be a set. A binary relation ⪯ on which is reflexive, transitive and antisymmetric is called a partial order. The pair is said to be a poset.
A mapping is said to be ⪯-preserving if given with , then .
The reader is referred, for instance, to [2] and [3] for an introduction to the theory of ordered sets.
Let be a σ-algebra on . It will be assumed throughout the paper that for all we have that .
Let .
Let us denote by the set of probabilities on the measurable space .
A binary relation on is said to be a stochastic order if is a poset.
A stochastic order on is said to be integral if there exists a class ℛ of measurable mappings from to ℝ satisfying that if and only if
for all such that the above integrals exist. The class ℛ is said to be a generator of . It is well known that there could be different generators of the same stochastic order.
The reader is referred to [4, 5] and [6] for a rigorous introduction to stochastic orders and integral stochastic orders.
The class of measurable ⪯-preserving functions generates a binary relation on , denoted by , as follows: given , then when
for all measurable ⪯-preserving functions f for which both integrals exist.
3 Main results
Given a stochastic order on , we will obtain a sufficient and necessary condition for the existence of a partial order ⪯ on such that and are the same order. It is immediately seen that for such a condition, the stochastic order should be integral. Moreover, we connect the above result with maximal generators of integral stochastic orders.
In the first place, we construct a partial order on by means of a generator of the integral stochastic order .
Proposition 1 Let be an integral stochastic order on and ℛ be a generator of the order. We define the binary relation on as follows. Let . Then when for all . It holds that is a poset.
Proof Obviously, is reflexive and transitive. Let with and . The first condition implies that for all , or equivalently,
where denotes the probability distribution degenerated at the point . As a consequence, we obtain that . In a similar way, we have that . Since is antisymmetric, it holds that . Now, note that , which implies that . □
Now, we show that the relation defined in Proposition 1 does not depend on a particular generator of the order, but on the order itself.
Proposition 2 Let be an integral stochastic order on and and be generators of the order. Then and are the same partial order.
Proof Let us suppose that . Thus, for all , or equivalently,
that is, . Since is a generator of , we have that
Therefore, for all , which means that . So, implies . Reasoning in a similar way, we obtain that implies , and so the result is proved. □
Given an integral stochastic order, we will denote by ⪯ the partial order on determined by , that is, when for all mappings of a generator of . Proposition 2 says that such an order is well defined.
Next, we prove that implies .
Proposition 3 Let be an integral stochastic order on . Let ⪯ be the partial order on determined by . Then implies .
Proof Let with , that is,
for all measurable ⪯-preserving functions f. Let ℛ be a generator of . In accordance with the definition of the partial order ⪯, any mapping is ⪯-preserving; as a consequence,
for any , that is, . □
We should note that the converse of the above result is not true in general, that is, and are not equal in general. Let us consider the following example.
Example 1 Consider the measurable space , being the usual Borel σ-algebra on ℝ. Let be the integral stochastic order given by the generator
usually referred to as the convex order. Thus, the partial order ⪯ on ℝ determined by is given by when for all . It is seen that for any . Therefore, any measurable map is ⪯-preserving, and so and are not the same order. In fact, if and only if .
The next proposition shows that if there exists a partial order which generates the stochastic order , then the partial order ⪯ on determined by also generates .
Proposition 4 Let be an integral stochastic order on and ⪯ be the partial order on determined by . Let ⪯′ be a partial order on such that and are equal. Then and are the same stochastic order.
Proof We have that is a generator of . In accordance with Proposition 2, it holds that when for all .
Let us suppose that . It implies that for all and so . Now, note that if and , then and so , that is, . As a consequence, we have that , which derives that implies . Now, Proposition 3 provides the result. □
We introduce the following definition which will be key for our purposes.
Definition 1 Let be a poset. The mappings are said to be ⪯-comonotonic if there are no two elements with , and .
Now, we obtain sufficient and necessary conditions to guarantee that and are the same stochastic order.
Theorem 1 Let be an integral stochastic order on and ⪯ be the partial order on determined by . Then and are the same order if and only if there exists a generator ℛ of satisfying that if is a measurable mapping such that g and f are ⪯-comonotonic for all , then .
Proof In the first place, let us see that and are the same order under the existence of a generator satisfying the above condition.
In Proposition 3 we have shown that implies . Now, let with . Let g be a measurable ⪯-preserving function and . Clearly, f and g are ⪯-comonotonic, thus , and so
for any measurable ⪯-preserving function g; that is, , which shows that implies .
Let us prove the converse. If and are the same order, we have that is a generator of . Let us see that this generator satisfies the condition of the statement.
Let be a measurable mapping satisfying that g and f are ⪯-comonotonic for all . Let us see that given with () there exists a mapping with , which implies that , and so g is ⪯-preserving, that is, .
If the above result is false, there exist with such that and for all . This implies that
The antisymmetric property of implies that , which is a contradiction with since . □
Let us consider examples of integral stochastic orders which are in fact generated by posets.
Example 2 Consider the measurable space . Let be the usual stochastic order, that is, when
for any , where stands for the indicator function of the set A. According to Proposition 2, the partial order ⪯ on ℝ determined by is given by when .
It is known that the class of non-decreasing functions is a generator of . Let ℛ be that class. If g is measurable and ⪯-comonotonic with any map of ℛ, then , and in accordance with Theorem 1, and are the same order.
Example 3 Consider the space . Let be the bidirectional order (see [7] and [8]). It is known that if and only if
for any . The definition of the order provides the partial order ⪯ on ℝ determined by . We obtain that when
In [7] it is shown that the class is a generator of the order. A ⪯-comonotonic function with the maps of ℛ belongs to it. Thus, and are the same order.
We will analyze some examples of multivariate stochastic orders after Theorem 2.
Note that, in general, if ℛ is a generator of a stochastic order and g and f are ⪯-comonotonic functions for all , g is not necessarily an element of ℛ. Consider the generator of Example 1, there are comonotonic functions with the maps of the class of convex functions which are not convex. In fact, any mapping is ⪯-comonotonic with all functions of .
Now we are proving that under an appropriate framework, a generator satisfying the condition of Theorem 1 is the maximal generator of the order .
We briefly describe the concept of a maximal generator of an integral stochastic order (see [4] or [5]). Such a concept is associated with the so-called weight function. That function is a measurable mapping . The weight function determines the space of mappings in which we are looking for the maximal generator. Such a space will be the class of measurable mappings with bounded b-norm, where the b-norm of a mapping is
For our purpose, it is sufficient to consider . In this way, the b-norm of any mapping is equal to
So, if ℬ denotes the set of measurable functions with the finite b-norm, ℬ is the set of all measurable and bounded mappings.
The maximal generator of an integral stochastic order is the set of all functions such that
We will assume that all functions in the following results belong to the class ℬ.
Theorem 2 Let be an integral stochastic order on . Let ℛ be a generator of satisfying that if is a measurable mapping such that g and f are ⪯-comonotonic for all , then . It holds that ℛ is the maximal generator of .
Proof Let us see that ℛ is a convex cone containing the constant functions and closed under pointwise convergence.
Let and . Suppose that there exists such that h and λf are not ⪯-comonotonic. Then there exist with , and . This implies that which is a contradiction with since . Therefore, λf and h are ⪯-comonotonic for all and so . As a consequence, ℛ is a cone.
Now, consider and . Suppose that there exists such that h and are not ⪯-comonotonic. We have that there exist with , and . But and since and . Hence, ℛ is convex.
On the other hand, any constant function and any element of ℛ are ⪯-comonotonic. So, any constant function is in ℛ.
Moreover, let such that tends to pointwise. Let . Suppose that h and f are not ⪯-comonotonic. We have that there exist with , and . Then for n large enough, we have that , which contradicts that . Thus, ℛ is closed under pointwise convergence.
Now, the result is a consequence of Corollary 2.3.9 in [5] which says that a generator which is a convex cone containing the constant functions and is closed under pointwise convergence is the maximal generator of the order. □
Example 4 Consider with the usual Borel σ-algebra on . Let stand for the usual multivariate stochastic order. Thus, two probabilities P and Q on the measurable space are ordered in the usual multivariate stochastic order if
for any bounded increasing mapping , where by increasing we mean that for any with , and denotes the usual componentwise order on .
Let ℛ be the above class of mappings which is a generator of . Such a generator allows to obtain the partial order ⪯ on determined by . It can be seen that this order is the order .
To analyze if the integral stochastic order is generated by ⪯, we apply Theorem 1, but taking into account Theorem 2; that is, if there exists a generator satisfying the condition of Theorem 1, such a generator should be the maximal generator of . It is well known that the class ℛ is the maximal generator of .
It is not hard to prove that any bounded measurable mapping which is ⪯-comonotonic with all the mappings of ℛ belongs to it, and so we conclude that the usual multivariate stochastic order is generated by the componentwise order.
Example 5 Now let stand for the upper orthant order on the class of probabilities associated with the measurable space . Two probabilities P and Q are ordered in the upper orthant order if
for any , where denotes the set with .
Thus, the set is a generator of . This generator leads to the partial order ⪯ on determined by . It can be seen that this order is .
As an immediate consequence, we obtain that there does not exist a partial order generating the upper orthant order, since we have seen in the above example that generates the usual multivariate stochastic order.
However, let us check that the maximal generator of does not satisfy the condition of Theorem 1. That is, there are bounded measurable mappings which are ⪯-comonotonic with all the maps of the maximal generator of and they do not belong to such a generator.
A mapping is said to be Δ-monotone if for every subset and for every , it holds that
for all , where
being the i th unit vector.
In [5] it is proved that the class ℒ of all bounded Δ-monotone functions is the maximal generator of .
Let us see that there are ⪯-comonotonic mappings with all the mappings of ℒ, which do not belong to such a class.
Consider and the mapping with
The mapping g is ⪯-comonotonic with all the elements of ℒ, but g is not Δ-monotone.
To conclude we should point out that Theorem 2 could be applied to obtain maximal generators of integral stochastic orders which have been generated by means of partially ordered sets, since the maximal generator is the unique generator which satisfies the condition required in Theorem 1.
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Acknowledgements
We would like to thank the editor and the referees for their key comments and suggestions. The authors are indebted to the Spanish Ministry of Science and Innovation since this research is financed by Grants MTM2010-18370 and MTM2011-22993.
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MCL-D and ML-D have worked together to obtain the results in this manuscript.
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López-Díaz, M.C., López-Díaz, M. When is an integral stochastic order generated by a poset?. J Inequal Appl 2012, 265 (2012). https://doi.org/10.1186/1029-242X-2012-265
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DOI: https://doi.org/10.1186/1029-242X-2012-265