Appendix B. Equimeasurable Rearrangements and Supermodularity
The classical theory of monotone equimeasurable rearrangements of Borel-measurable functions on
dates back to the work of Hardy, Littlewood and Pólya [
27], who gave the first integral inequalities involving functions and their rearrangements. Here, the idea of an equimeasurable rearrangement of any element
Y of
with respect to the fixed underlying loss random variable
X is discussed. All of the results in this Appendix are taken from Ghossoub [
26,
28] to which we refer the reader for proofs, additional results and additional references on this topic.
B1. The Nondecreasing Rearrangement Let
be a probability space, and let
be a continuous random variable (i.e.,
is nonatomic) with range
, where
, i.e.,
X is a mapping of
S onto the closed interval
. Denote by Σ the
σ-algebra
, and denote by
φ the law of
X defined by:
for any Borel subset
B of
.
Proposition B1. For any Borel-measurable map there exists a φ-a.s. unique Borel-measurable map , such that:- 1.
is left-continuous and nondecreasing;
- 2.
is φ-equimeasurable with I, in the sense that for any Borel set B, - 3.
If are such that -a.s., then ; and,
- 4.
If denotes the identity function, then .
will be called the nondecreasing φ-rearrangement of I. Now, define and . Since both I and are Borel-measurable mappings of into itself, it follows that . Note also that is nondecreasing in X, in the sense that if are such that , then , and that Y and are P-equimeasurable, that is, for any , . The function will be called a nondecreasing P-rearrangement of Y with respect to X, and it will be denoted by to avoid confusion in case a different measure on is also considered. Note that is P-a.s. unique. Note also that if and are P-equimeasurable and if , then and , for any measurable function ψ, such that the integrals exist.
Similarly to the previous construction, for a given a Borel-measurable
with
, there exists a
φ-a.s. unique (on
B) nondecreasing, Borel-measurable mapping
, which is
φ-equimeasurable with
I on
B, in the sense that for any
,
is called the nondecreasing φ-rearrangement of I on B. Since X is -measurable, there exists , such that , and hence, . Now, define . Since both I and are bounded Borel-measurable mappings, it follows that . Note also that is nondecreasing in X on A, in the sense that if are such that , then , and that Y and are P-equimeasurable on A, that is, for any , . The function will be called a nondecreasing P-rearrangement of Y with respect to X on A, and it will be denoted by to avoid confusion in case a different measure on is also considered. Note that is P-a.s. unique. Note also that if and are P-equimeasurable on A and if , then and , for any measurable function ψ, such that the integrals exist.
Lemma B1. Let , and let be such that and is a Borel set10.
Let be the nondecreasing P-rearrangement of Y with respect to X, and let be the nondecreasing P-rearrangement of Y with respect to X on A. Then -
a.s. B2. Supermodularity and Hardy–Littlewood–Pólya Inequalities A partially-ordered set (poset) is a pair where ≽ is a reflexive, transitive and antisymmetric binary relation on T. For any , denote by (resp. ) the least upper bound (resp. greatest lower bound) of the set . A poset is called a lattice when , for each . For instance, the Euclidean space is a lattice for the partial order ≽ defined as follows: for and , write , when , for each . It is then easy to see that and .
Definition B1. Let
be a lattice. A function
is said to be supermodular if for each
,
In particular, a function
is supermodular if for any
with
and
, one has:
Equation (B2) then implies that a function is supermodular if and only if the function is nondecreasing on , for any and .
Example B1. The following are supermodular functions:
If is concave and , then the function defined by is supermodular. Moreover, if g is strictly concave, then is strictly supermodular.
If are both nonincreasing or both nondecreasing functions, then the function defined by is supermodular.
Lemma B1 (Hardy–Littlewood–Pólya inequalities).
Let , and let be such that and is a Borel set. Let be the nondecreasing P-rearrangement of Y with respect to X, and let be the nondecreasing P-rearrangement of Y with respect to X on A. If L is supermodular, then:- 1.
; and,
- 2.
,
provided the integrals exist. Moreover, if L is strictly supermodular, then equality holds in (1)
if and only if -
a.s. Lemma B2. Let , and let be such that and is a Borel set. Let be the nondecreasing P-rearrangement of Y with respect to X, and let be the nondecreasing P-rearrangement of Y with respect to X on A. Then, the following hold:- 1.
If -a.s., then ; and,
- 2.
If -a.s. on A, then -a.s. on A.
B3. Approximation of the Rearrangement
Lemma B1. If f and are -valued, Σ-measurable functions on S, such that the sequence converges pointwise P-a.s. to f monotonically downwards, then the sequence converges pointwise P-a.s. to monotonically downwards, where is the nondecreasing P-rearrangement of f with respect to X and is the nondecreasing P-rearrangement of with respect to X, for each .
Lemma B2. Let f and be -
valued, Σ-
measurable functions on S. If , for each , and if the sequence converges uniformly to , then:- 1.
The functions and are in , for each , where is the nondecreasing P-rearrangement of f with respect to X and is the nondecreasing P-rearrangement of with respect to X, for each ; and
- 2.
The sequence converges to in the norm.
Appendix C. Proof of Theorem 8
As in
Section 3, there exists a unique pair
of (nonnegative) finite measures on
, such that
,
and
. That is, for all
with
, one has
, and there is some
, such that
. It then also follows that
and
. In the following, the Σ-measurable set
A on which
P is concentrated and
is assumed to be fixed all throughout.
C1. “Splitting” the Initial Problem The idea of splitting the problem into two sub-problems is inspired by the techniques used in Jin and Zhou [
30] (although in a different context and for different purposes), but with some differences that are peculiar to the insurance problem examined here. Now, consider the following three problems:
Problem C1. For a given
,
Note that since the utility function u is continuous (Assumption 1), it is bounded on every closed and bounded subset of . Therefore, since the range of X is closed and bounded, the supremum value of each of the above three problems is finite.
Lemma C1. The feasibility sets of Problems (C1) and (C2) are non-empty.
Proof. Since
P and
Q are not mutually singular, by Assumption 2, and since
, it follows that
. Since
,
and
, it follows from Lemma A1 that there exists some
, such that
,
and
on
B. There are three cases to consider:
- (1)
If , then by Lemma A1, one has -a.s. on A. However, on B. Thus, -a.s. on B. Consequently, there is some , with and , such that on C and . Therefore, . Now, since , for each , it follows that . Thus, by monotonicity of P, . However, , by non-atomicity of (Assumption 2). Therefore, , a contradiction. Hence . Now, for a given , the function is feasible for Problem (C1) with parameter β.
- (2)
If , then is feasible for Problem (C2).
- (3)
If , then , with , is feasible for Problem (C2) with parameter β, for any given .
☐
Lemma C2. If is optimal for Problem (C3), then .
Proof. First note that, as in the proof of Lemma C1, and there exists some such that , and on B. Moreover, since , it follows that , for each , and so, , for each . Consequently, , for each . Therefore, in particular, .
Now, suppose, by way of contradiction, that
is optimal for Problem (C3), and let
be optimal for Problem (C1) with parameter 0, so that
. Since
is optimal for Problem (C3), one has
, for each
. Since
is feasible for Problem (C1) with parameter
, one has
. Now, since
and
, it follows from Lemma A1 that
-a.s. on
A. Moreover, since
on
B and
, it follows that
-a.s. on
B. Define the function
Z by
, and let
. Then,
,
, and
. Therefore, in particular,
is feasible for Problem (C3) and
Z is feasible for Problem (C1) with parameter
. Moreover,
If
, then
and
. Hence, by Lemma A1,
-a.s. on
B. However,
on
B. Thus,
-a.s. on
B. Since
, this yields
-a.s. on
B. Consequently, there is some
, with
and
, such that
on
C and
. Therefore,
. Now, since
, for each
, it follows that
. Thus, by monotonicity of
P,
. However,
, by non-atomicity of
(Assumption 2). Therefore,
, a contradiction. Hence,
. Finally,
This contradicts the optimality of for Problem (C3). Consequently, if is optimal for Problem (C3), then . ☐
The following lemma shows how to combine the solutions of these three problems stated above to obtain a solution to the original problem (Problem (3)).
Lemma C3. If is optimal for Problem (C3), is optimal for Problem (C1) with parameter and is optimal for Problem (C2) with parameter , then is optimal for Problem (3).
Proof. Feasibility of
for Problem (3) is immediate. To show optimality of
for Problem (3), let
be any other feasible solution for Problem (3), and define
. Then,
and
, since
. Moreover,
since
is feasible for Problem (3). Consequently,
α is feasible for Problem (C3). Furthermore,
(resp.
) is feasible for Problem (C1) (resp. Problem (C2)) with parameter
α. Hence,
and
. Now, since
is optimal for Problem (C3), it follows that:
However, and . Therefore, . Hence, is optimal for Problem (3). ☐
The following lemma shows how to obtain monotonicity of an optimal indemnity schedule and how to characterize its distribution.
Lemma C4. Let be an optimal solution for Problem (3)
, and suppose that Q is compatible P. Let be the nondecreasing P-rearrangement of with respect to X. Then:- (1)
is optimal for Problem (3); and
- (2)
-a.s., where is the nondecreasing P-rearrangement of with respect to X on A. In particular, and are identically distributed under P.
Proof. Since the function defined by is supermodular (see Example B1 (1)), it follows from Lemma B1 that Moreover, since , it follows from Lemma B2 that . Finally, since Q is compatible with P, it follows that and so, is optimal for Problem (3). Now, let be the nondecreasing P-rearrangement of with respect to X on A. Since , then by Lemma B1, one has that -a.s. Therefore, and have the same distribution under P. Hence, form the equimeasurability of and , it follows that and have the same distribution under P. ☐
The following lemma shows that a distributional characterization of an optimal indemnity schedule can be reduced to the problem of characterizing the distribution of the solution of Problem (C1).
Lemma C5. Let an optimal solution for Problem (3)
be given by:for some .
Let be the nondecreasing P-rearrangement of with respect to X, and let be the nondecreasing P-rearrangement of with respect to X. Then, -
a.s., and hence, and have the same distribution under P. Proof. Let
be the nondecreasing
P-rearrangement of
with respect to
X on
A. Since
, then by Lemma B1, one has
-a.s. Similarly, let
be the nondecreasing
P-rearrangement of
with respect to
X on
A. Then,
-a.s. Therefore, it suffices to show that
-a.s. Since both
and
are nondecreasing functions of
X on
A, then by the
P-a.s. uniqueness of the nondecreasing rearrangement, it remains to show that they are
P-equimeasurable with
on
A. Now, for each
,
where the first equality follows from the definition of
(equimeasurability), the second equality follows from Equation (B1) and the third equality follows from the definition of
(equimeasurability). Therefore,
-a.s., and hence,
and
have the same distribution under
P. Consequently, by equimeasurability of
and
, it follows that
and
have the same distribution under
P. ☐
Remark C1. By Lemmata C3, C4 and C5, if Q is compatible with P, is optimal for Problem (C3), is optimal for Problem (C1) with parameter and is optimal for Problem (C2) with parameter , then is optimal for Problem (3) and -a.s., where (resp. ) is the P-a.s. unique nondecreasing P-rearrangement of (resp. of ) with respect to X. In particular, and have the same distribution under P.
Henceforth, we focus on solving each problem individually. The solutions can then be combined as per Remark C1.
C2. Solving Problem (C2) Since
, it follows that, for all
, one has
Consequently, any
Y, which is feasible for Problem (C2), with parameter
β is also optimal for Problem (C2) with parameter
β. For instance, define
, where
is chosen such that
. Then
is optimal for Problem (C2).
Remark C1. The choice of
so that
is justified by the following argument. Define the function
by
where
, for each
. Then,
φ is a nonincreasing function of
α. Moreover, by the continuity of the functions
and
, and by Lebesgue’s dominated convergence theorem (e.g., Theorem 2.4.4 of [
23]),
φ is a continuous function of the parameter
α. Now, by the continuity of the functions max and min,
and
. Therefore, by continuity of the function
φ in
α,
and
. Consequently,
φ is a continuous nonincreasing function of
α, such that
and
. Thus, by the intermediate value theorem (e.g., Theorem 4.23 of [
31]), one can always choose
α, such that
, for any
.
C3. Solving Problem (C1). For a fixed parameter , Problem (C1) will be solved “state-wise”, as described below. Moreover, by Lemma C2, one can restrict the analysis to the case where .
Lemma C1. If satisfies the following:- (1)
, for all ;
- (2)
, for some ; and
- (3)
There exists some , such that for all ,
then the function solves Problem (C1)
with parameter β. Proof. Suppose that
satisfies
,
and
above. Then,
is clearly feasible for Problem (C1) with parameter
β. To show optimality of
for Problem (C1), note that for any other
, which is feasible for Problem (C1) with parameter
β, one has, for all
,
Furthermore, since
u is increasing, since
on
A and since
for all
, it follows that for all
,
Consequently,
which completes the proof. ☐
Lemma C2. For any , the function given by:satisfies Conditions and of Lemma C1. Proof. Fix
; fix
; and consider the problem:
Since u is strictly concave (by Assumption 1), so is f, as a function of y. In particular, is a (strictly) decreasing function. Hence the first-order condition on f yields a global maximum for f at . If , then since is decreasing, it is negative on the interval . Therefore, f is decreasing on the interval and hence attains a local maximum of at . If , then since is decreasing, it is positive on the interval . Therefore, f is increasing on the interval and hence attains a local maximum of at . If , then the local maximum of f on the interval is its global maximum . Consequently, the function solves the problem appearing in Equation (B3). Since s and λ were chosen arbitrarily, this completes the proof of Lemma C2. ☐
Lemma C3. For defined in Equation (B2)
, the following holds:Therefore, Proof. Indeed, if
, then
, by Assumption 1. Thus, for each
one has:
The rest then easily follows. ☐
Lemma C4. Define the function as follows: for each ,
Then, φ is a continuous nonincreasing function of the parameter λ. Continuity of
φ is a direct consequence of Lebesgue’s dominated convergence theorem and of the continuity of each of the functions
11 ,
and
. The fact that
φ is nonincreasing in
λ results from the concavity of
u, i.e., from the fact that
is a nonincreasing function. ☐
Lemma C5. Consider the function φ defined above. Then: Proof. By continuity of the functions , and , it follows that for each , . Moreover, as was shown above, Therefore, , for each . Hence, by continuity of the function φ in λ, it follows that . Similarly, by continuity of the functions , and , one has that for each , However, by continuity of the function φ in λ, one has . However, by Assumption 1, , and by Assumption 2, -a.s. Moreover, . Therefore, . ☐
Remark C1. Hence, summing up, the function
φ defined above is a nonincreasing continuous function of the parameter
λ, such that
and
. Therefore,
, and so, by the intermediate value theorem, for each
one can chose
, such that:
Therefore, by Lemmata C1 and C2, the function defined above solves Problem (C1), with parameter β. Finally, let be optimal for Problem (C3); let be chosen for just as was chosen for β in Remark C1; and let be a corresponding optimal solution for Problem (C2) with parameter . The rest then follows from Remark C1. The P-a.s. uniqueness part of Theorem 8 follows from the uniqueness property of the nondecreasing rearrangement. Finally, if the utility function u is strictly concave, then any solution to Problem (3) is such that , P-a.s., by Lemma B1 and Example B1 (1). This concludes the proof of Theorem 8.
Appendix D. Proof of Corollary 11
The idea behind the proof of Corollary 11 is to approximate the solution of Problem (3) characterized above by a sequence of functions that can be characterized. Taking limits then gives us a characterization of the solution of Problem (3).
Fix
, and let
be the corresponding
λ, chosen as in Remark C1. Since
h is nonnegative, Σ-measurable and
P-integrable, there is a sequence
of nonnegative,
P-simple and
P-integrable functions on
that converges monotonically upwards and pointwise to
h (e.g., Proposition 2.1.7 of [
23]). Therefore, since
is continuous, the sequence
, defined by
, for all
, converges pointwise to
, defined by:
Since the sequence
converges monotonically upwards and pointwise to
h and since
is continuous and decreasing, it follows that the sequence
converges monotonically downwards and pointwise to
. Now, for each
, there is some
, a Σ-partition
of
S and some nonnegative real numbers
, for
, such that
. Since
can be written as
; it is then easy to see that:
By the continuity of the functions
and
and since
and
are nondecreasing functions of
t for each
, it follows that the sequence
converges monotonically downwards and pointwise to:
For each
, one can rewrite
as:
where, for
,
and
.
Lemma D1. For each and for each , is either a full insurance indemnity schedule or a deductible indemnity schedule (with a strictly positive deductible) on the set .
Proof. Fix , and fix . If and , then since is decreasing (u is concave), it follows that . Therefore, , and so, , a full insurance indemnity schedule (on ). If , then . However, , by Assumption 1. Therefore, , and so , a full insurance indemnity schedule (on ). If and , then since is strictly decreasing (u is strictly concave), it follows that . Therefore, , and so, , a deductible insurance indemnity schedule (on ) with a strictly positive deductible, where for any , . ☐
The following lemma is a direct consequence of Lemmata B1 and B1, and it is hence stated without a proof.
Lemma D2. If (resp. ) denotes the nondecreasing P-rearrangement of (resp. ) with respect to X, then converges monotonically downwards and pointwise P-a.s. to . Moreover, -a.s., where denotes the nondecreasing P-rearrangement of with respect to X on A.
Let
. Then,
is of the form
12 , for some
. Therefore,
for
.
Lemma D3. Fix . If there exists some such that and , then .
Proof. As in the proof of Lemma D1. ☐
Lemma D4. If P is not absolutely continuous with respect to Q, then for each , there is some , such that .
Proof. Suppose, by way of contradiction, that
P is not absolutely continuous with respect to
Q, but that there is some
, such that
, for each
. Then,
. However, the sequence
converges monotonically upwards and pointwise, to
. Hence, since
, it follows that
, for each
and for each
. Consequently,
. Therefore,
P and
are mutually absolutely continuous (i.e., equivalent – see p. 179 of [
32]). Furthermore, the finite measures
Q,
and
are nonnegative, and hence,
. Thus,
, a contradiction. ☐
Remark D1. Lemmata D3 and D4 imply that if P is not absolutely continuous with respect to Q, then , for each .
Now, let
. Then,
is non-empty (since
and
), and of the form
, where
and
. Indeed, since
and
, it follows that for all
one has
. It is then easily verified that:
Therefore,
. Moreover, one can write
, for some
. Letting
, it follows that
, for each
. Therefore,
One can assume, without loss of generality, that , for all such that . Then, it is easily verified that , because of the concavity of u.
Lemma D5. Let denote the nondecreasing P-rearrangement of with respect to X. Then, there exist , such that , and for P-a.a. ,
where is a nondecreasing and Borel-measurable function, such that for each , and for -
a.a. , one has if and if .
Proof. First note that
, by Lemma B2, since
, by definition of
. Moreover, one has
, for some Borel-measurable function
. Therefore,
, where
is the nondecreasing
-rearrangement of
. Let
. Then,
, for each
, and
is nondecreasing and Borel-measurable. Now, note that:
where the last equality follows form the non-atomicity of
(Assumption 2). Moreover, by equimeasurability, one has that
. However,
where the last equality follows form the non-atomicity of
(Assumption 2). Consequently,
Thus, if , then since is nondecreasing, there exists , such that for P-a.a. , if belongs to or , and if . Therefore, if , for -a.a. . If , then , and so for P-a.a. , if , and if . Thus, with , is P-a.s. of the form Equation (D1), with if , for -a.a. .
Similarly, by equimeasurability, one has that:
However,
where the second-to-last equality follows form the non-atomicity of
(Assumption 2).
Thus, if , then . Therefore, since is nondecreasing, there exists , such that for P-a.a. , if belongs to or , and if . Therefore, if , for -a.a. . If , then , and so, for P-a.a. , such that , one has that . Thus, with , is P-a.s. of the form Equation (D1), with if , for -a.a. .
To show that , suppose, by way of contradiction, that . Since , it follows that . Choose , such that . Then, , since . However, , since , hence contradicting the fact that . Therefore, . ☐
Remark D2. For each , let be the event, such that and is of the form of Equation (D1) on . Let . Then, and, by Lemma A2, . Moreover, for each and for each , is given by Equation (D1).
By Lemma D2, the sequence defined by Equation (D1) converges pointwise P-a.s. to , the nondecreasing P-rearrangement of with respect to X.
Now, let
be an optimal solution to Problem (C2) with parameter
β, as defined previously, and for each
, let:
Finally, let
be optimal for Problem (C3); let
be chosen for
just as
was chosen for
β; and let
be a corresponding optimal solution for Problem (C2) with parameter
. For each
, let:
Then, by Remark C1, the sequence
converges pointwise
P-a.s. to an optimal solution of the initial problem (Problem (3)), which is
P-a.s. nondecreasing in the loss
X. Henceforth,
will denote that optimal solution. Then:
Now, recall that
is the
P-a.s. unique nondecreasing
P-rearrangement of
with respect to
X, where:
Moreover, the sequence
, defined by:
converges pointwise to
. Since the sequence
converges monotonically upwards and pointwise to
h and since
is continuous and decreasing, it follows that the sequence
converges monotonically downwards and pointwise to
. Consequently, one can easily check that the sequence
converges monotonically downwards and pointwise to
, where for each
,
Remark D3. For each , let denote the P-a.s. unique nondecreasing P-rearrangement of with respect to X. Then, by Lemma B1, the sequence converges monotonically downwards and pointwise P-a.s. to . That is, there is some with and , such that for each , the sequence converges monotonically downwards to .
Now, as in Lemma D5, for
P-a.a.
,
for given
, such that
and
, a nondecreasing and Borel-measurable function, such that
for each
,
if
for
-a.a.
, and
if
for
-a.a.
.
Lemma D6. The sequences and are bounded and nondecreasing.
Proof. Since and , the boundedness of the sequences and is clear. We now show that they are nondecreasing. Fix . Since the sequence is nonincreasing pointwise on (as in Remark D3), one has , for each . To show that , first note that if , then . If , let be as in Remark D2, let be as in Remark D3, and choose , such that . Then , and so, . Consequently, , and so . Therefore, . Similarly, if , then . If , choose such that . Then, , and so, . Consequently, , and so, , that is, . ☐
Hence, the sequences
and
are bounded and monotone. Therefore, each has a limit. Let:
Moreover, if there is some , such that , then for each , one has . Furthermore, if there is some , such that , then for each , one has
Lemma D7. With a as defined in Equation (D3) above, one has , and if there is some with .
Proof. Since and , for each , it follows that and . Moreover, since , for each , it follows that . Finally, if there is some , such that , then for each , one has . Therefore, , for each , and so, . ☐
Lemma D8. There exist some and , such that and such that for P-a.a. ,
for some nondecreasing, left-continuous and Borel-measurable function , such that for each .
Proof. Let and , where a and b are as in Equation (D3). Then, . Let be as in Remark D2; let be as in Remark D3; and let . Since , it follows form Lemma A2 that . Suppose that there exists some , such that but . Then, for each one has , since the sequence converges monotonically downwards and pointwise on to and , by definition of . Consequently, , for each . Therefore, , a contradiction. Hence, for each , .
Now, suppose that there exists some
, such that
, but
. Let
. Since the sequence
converges monotonically downwards to
, there is some
, such that for each
one has
. Fix some
, and let
. Then:
Therefore,
, and so
, a contradiction. Consequently
13, for each
,
.
Moreover, , for some bounded, nonnegative, nondecreasing, left-continuous and Borel-measurable function on the range of X (see Section B). Let . One then has, for each , if . Furthermore, since , it follows that , for each . In particular, . This completes the proof of Corollary 11. ☐