When is an integral stochastic order generated by a poset?
© López-Díaz and López-Díaz; licensee Springer 2012
Received: 17 February 2012
Accepted: 29 October 2012
Published: 14 November 2012
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.
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.
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.
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 .
Let be a σ-algebra on . It will be assumed throughout the paper that for all we have that .
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.
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.
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.
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.
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 .
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.
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.
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, .
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.
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.
In  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 .
So, if ℬ denotes the set of measurable functions with the finite b-norm, ℬ is the set of all measurable and bounded mappings.
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  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. □
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.
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.
being the i th unit vector.
In  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.
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.
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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