Extension problem for principles of equivalent utility

We prove that, under the Cumulative Prospect Theory, every principle of equivalent utility, defined on a family of ternary risks, possesses a unique extension to the family of all risks.


Introduction
In insurance mathematics, risks are usually represented by non-negative bounded random variables on a given probability space. A premium principle is a functional assigning to every risk a non-negative real number. There are several methods of defining principles. In this paper we deal with the principle of equivalent utility, belonging to the so-called economic methods of insurance contracts pricing. The principle, introduced by Bühlmann [2], involves the notion of a utility function and postulates a fairness in terms of utility. In order to recall the principle, assume that w ∈ [0, ∞) is an insurance company's initial wealth level and is its preference relation over a family X + of risks. Then the relation in a natural way induces on X + the indifference relation ∼: for X, Y ∈ X + . The principle of equivalent utility for the risk X ∈ X + is a real number H (X) such that This condition has the following interpretation: the company is indifferent between rejecting the contract and entering into it. Thus, H (X) is a minimal price for which the insurance company would be ready to insure the risk X.
In general, the principle of equivalent utility need not exist and, even if it exists, it need not be uniquely determined. Under the Expected Utility theory (1) becomes where u : R → R is a continuous strictly increasing utility function. One can prove that, for every X ∈ X + , there exists a unique real number H u (X) such that (2) is valid. Therefore, Eq. (2) determines a functional on X + , called the principle of equivalent utility. For more details concerning the properties of the principle of equivalent utility under the Expected Utility Theory we refer e.g. to [1,2,8,14].
The principle of equivalent utility under the Rank-Dependent Utility model and under the Cumulative Prospect Theory, has been introduced and considered by Heilpern [7] and Ka luszka and Krzeszowiec [9,10], respectively. Under the first of these models, the premium H (u,g) (X) for a risk X ∈ X + is defined as a solution of the equation where u : R → R is a continuous strictly increasing utility function and E g is the Choquet integral with respect to a probability distortion function g. Let us recall that g : [0, 1] → [0, 1] is called a probability distortion function, provided it is non-decreasing and satisfies the boundary conditions g(0) = 0 and g(1) = 1. For every bounded random variable X, the Choquet integral with respect to the probability distortion function g is defined as follows It has been proved in [4] that, if g is a continuous probability distortion function and u : R → R is a continuous strictly increasing utility function, with u(0) = 0, then for every X ∈ X + the number H (u,g) (X) is uniquely determined by (3). Under the Cumulative Prospect Theory, the premium H (u,g,h) (X) for a risk X ∈ X + is defined as a solution of the equation where, for every bounded random variable X, is the generalized Choquet integral related to the probability distortion functions g (for gains) and h (for losses). According to [4,Theorem 3.1], in the case where w ∈ (0, ∞), for every continuous and strictly increasing function u : R → R satisfying u(0) = 0 and every continuous probability distortion functions g and h, Eq. (5) uniquely determines H (u,g,h) (X) for X ∈ X + . If w = 0, then H (u,g,h) (X) is uniquely determined by (5) for every X ∈ X + if and only if Furthermore (cf. [9]), we have These properties are usually referred to as: non-excessive loading, no unjustified risk loading and translation invariance (or consistency), respectively. Furthermore, as the generalized Choquet integral is monotone (cf. [9, Lemma 1]), from (5) one can easily derive that the principle of equivalent utility under the Cumulative Prospect Theory is monotone, that is It turns out that, under the Expected Utility model, every functional of equivalent utility can be uniquely extended from the family of all binary risks, i.e. risks taking exactly two non-negative values with positive probabilities, to X + (cf. e.g. [5,Theorem 6]). In fact, in [5] only the case w = 0 has been considered but, making a straightforward substitution, from [5,Theorem 6] one can easily derive the analogous result for w > 0. In a recent paper [3], some aspects of the extension problem for functionals of equivalent utility under the Cumulative Prospect Theory have been investigated. In particular, the functionals whose restriction to the family of binary risks reduces either to the net principle or to the exponential principle, have been characterized. It follows from the results in [3] that under the Cumulative Prospect Theory a counterpart of [5, Theorem 6] does not hold. Therefore, the following question naturally arises: does there exist a reasonable, in a sense, family of risks such that, under the Cumulative Prospect Theory, every functional of equivalent utility defined on this family can be uniquely extended to X + . The aim of this paper is to show that the family of ternary risks, taking exactly three non-negative values, one of them being 0, with positive probabilities, possesses such a property.

Auxiliary results
Assume that (Ω, Σ, P ) is a non-atomic probability space and X + is the family of all non-negative bounded random variables on (Ω, Σ, P ). According to (5) the premium for a given risk depends only on a probability distribution of (24)

Furthermore, the left (right) inequality in (21) is strict if and only if so is the right (left) inequality in (22).
Proof. Let X := 0, x, y; p and let d X : R → R be given by Since the generalized Choquet integral is monotone and u is strictly increas- Thus, considering (10) and (11), we obtain respectively. On the other hand, considering (5), from (14) and (15) we derive that Therefore, as d X is strictly increasing, we obtain (i)-(iii) as well as the second part of the assertion.
Proof. Let p ∈ P 3 . In view of (16), it is enough to show that φ p is continuous. Furthermore, it follows from (9) and (14) that φ p is monotone in y for each x. Thus, in order to prove the continuity of φ p , it suffices to show that it is continuous in x and y separately. Since in both cases similar arguments work, we prove only the continuity of φ p in x. Suppose that, for some y ∈ (0, ∞), φ p (·, y) is not continuous at the point x ∈ (0, y). Then there exists a sequence (x n : n ∈ N) of elements of (0, y) such that lim n→∞ Moreover, the following three cases are possible: (20) is valid and so, for sufficiently big k ∈ N, we get Furthermore, applying Corollary 2.2(i), we obtain (25) and (1 − g(p 1 + p 2 ))u(ψ p (x n k , y)) + (g(p 1 + p 2 ) − g(p 1 ))u(φ p (x n k , y)) for sufficiently big k ∈ N. Letting k → ∞ in (31) and subtracting the equality obtained this way from (25), in view of (30), we get On the other hand g is non-decreasing, with g(p) ∈ (0, 1) for p ∈ (0, 1), and u is strictly increasing. Thus, making use of (16), we conclude that the left hand side of the last equality is positive whenever d < φ p (x, y), and it is negative whenever d > φ p (x, y). This yields a contradiction.
Suppose that ψ p (x, y) = 0. Then, in view of (16), φ p (x, y) = y − x > 0 and so, applying Lemma 2.1(i) and Corollary 2.2(i), we obtain and If d > y − x, then making use of (30), we get lim k→∞ ψ p (x n k , y) > 0. Hence, for sufficiently big k ∈ N, we have ψ p (x n k , y) > 0 and so, according to Lemma 2.1(i) and Corollary 2.2(i), (31) holds. Thus, passing in (31) to the limit as k → ∞ and subtracting the obtained equality from (33), in view of (30), we get However, as u is strictly increasing and g is non-decreasing, with g(p) ∈ (0, 1) for p ∈ (0, 1), the left hand side of this equality is negative, which gives a contradiction.
If d < y − x, then in view of (30), we get lim k→∞ ψ p (x n k , y) < 0. Thus ψ p (x n k , y) < 0 for sufficiently big k ∈ N. On the other hand, making use of (32), for sufficiently big k ∈ N, we get whence, by Lemma 2.1(ii), φ p (x n k , y) > 0. Therefore, applying Lemma 2.1(ii) and Corollary 2.2(ii), we get h(p 3 )u(ψ p (x n k , y)) + (g(p 1 + p 2 ) − g(p 1 ))u(φ p (x n k , y)) for sufficiently big k ∈ N. Letting in the last equality k → ∞ and subtracting the obtained equality from (33), in view of (30), we obtain Since d < y − x, arguing as previously, we conclude that the left hand side of this equality is positive, which yields a contradiction. Case 2 Applying Lemma 2.1(ii) and Corollary 2.2(ii), we obtain (22), with both inequalities being strict, and (26). Hence, we have h(p 3 )u(x n k −y)+g(p 1 )u(x n k ) < u(w) < (g(p 1 +p 2 )−g(p 1 ))u(y−x n k )+g(p 1 )u(y) Vol. 93 (2019) Extension problem for principles of equivalent utility 225 for sufficiently big k ∈ N. Therefore, according to Corollary 2.2(ii), for sufficiently big k ∈ N, we get h(p 3 )u(ψ p (x n k , y)) + (g(p 1 + p 2 ) − g(p 1 ))u(φ p (x n k , y)) Thus, making use of (26) and arguing as in the previous case, we obtain which gives a contradiction. Case 3 If φ p (x, y) < 0, then according to Lemma 2.1(iii), we have (24). Thus for sufficiently big k ∈ N. Furthermore, applying Corollary 2.2(iii), we obtain (27) and for sufficiently big k ∈ N. Hence, repeating the arguments from the first case, we get which yields a contradiction.
If φ p (x, y) = 0, then in view of (16), ψ p (x, y) = x − y < 0. Hence, according to Lemma 2.1(ii) and Corollary 2.2(ii), we have and It follows from (35) that, for sufficiently big k ∈ N, whence, in view of Lemma 2.1(iii), ψ p (x n k , y) < 0. Therefore, if d > 0, then for sufficiently big k ∈ N, we have ψ p (x n k , y) < 0 < φ p (x n k , y) and so, according to Lemma 2.1(ii) and Corollary 2.2(ii), (34) holds. Passing to the limit in (34) as k → ∞ and subtracting the equality obtained this way from (36), in view of (30), we get Since d > 0, the left hand side of this equality is negative and so we get a contradiction.
If d < 0, then φ p (x n k , y) < 0 for sufficiently big k ∈ N. Thus, applying Lemma 2.1(iii) and Corollary 2.2(iii), we conclude that for sufficiently big k ∈ N. Hence, arguing as previously, we obtain However, as d < 0, the left hand side of the last equality is positive. So, we have a contradiction. This proves the continuity of φ p (·, y) at x. Proof. Note that, for every y ∈ (0, ∞), we have and lim x→0 + Therefore, if w = 0, then taking an arbitrary y ∈ (0, ∞), from (37) we derive that, for x ∈ (0, y) sufficiently close to y, (24) is valid. Thus, according to Lemma 2.1(iii), (23) holds for some x ∈ (0, y). Furthermore, as u is strictly increasing, making use of (38) and (39), we obtain that, for x ∈ (0, y) sufficiently close to 0, (22) is satisfied, with both inequalities being strict. So, applying Lemma 2.1(ii), we conclude that φ p (x, y) > 0 for some x ∈ (0, y). Hence, as φ p is continuous, 0 belongs to the interior of φ p ((0, y) × {y}).

Lemma 2.6. A family {φ p (T ) : p ∈ P 3 } is a cover of [w, ∞) and a family
and respectively.
Since u is a continuous, strictly increasing function, this implies that If w > 0, then for sufficiently small p 1 , p 2 ∈ (0, 1), (20) holds and so, applying Corollary 2.2(i), we get (1 − g(p 1 + p 2 ))u(ψ (p1,p2,1−p1−p2) (x, y)) Thus, repeating the previous arguments, we obtain (42). In this way we have proved that (42) holds for every (x, y) ∈ T . Therefore, taking x 0 ∈ [w, ∞) and (x, y) ∈ T , with y − x > x 0 − w, in view of (16), we get Hence, there exists p ∈ P 3 such that φ p (x, y) > x 0 . On the other hand, from Lemma 2.4 and Lemma 2.5 we deduce that φ p (T ) is connected and w is its interior point, respectively. Thus x 0 ∈ φ p (T ), which proves (40). Now, we show that (41) holds. To this end, fix x 0 ∈ (−∞, w). Let (x, y) ∈ T be such that x < w and y > max{w, w − x 0 }. Then, for sufficiently small p 2 , p 3 ∈ (0, 1), we have and so, applying Corollary 2.2(ii), we obtain Thus, arguing as previously, we get Hence, considering (16), we obtain So, there exists p ∈ P 3 such that ψ p (x, y) < x 0 . Furthermore, in view of (18), we get lim y→0 + ψ p (y/2, y) = w. Since, in view of Lemma 2.4, ψ p (T ) is connected, this means that x 0 ∈ φ p (T ) and so, (41) is proved. and Therefore, applying Lemma 2.1 and Corollary 2.2, we obtain Thus, as u is strictly increasing and g(p 1 ) > 0, we get Hence, in view of (44), we obtain x 1 = x 2 . Furthermore, considering (16), (44) and (45), we conclude that which completes the proof of the injectivity of F p . Now, as F p is continuous and injective, applying the Invariant Domain Theorem, we obtain that F p (T ) is open.
The following lemma, concerning the solutions of the general linear equation on a region, will play an important role in the proof of our main results. For more details concerning the general linear equation we refer to [11,Chapter 13.10]. a nonconstant continuous function, a, b, A, B ∈ R\{0} and c, C ∈ R. Then f satisfies the equation
From Theorems 3.1 and 3.2 we derive the following result. to X + . It has been already mentioned, in the Introduction, that such a functional need not be uniquely extended from the family of all binary risks. We complete the paper with a suitable example. Publisher's Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.