In this paper any given risk S (a random variable) is assumed to have a (finite or infinite) mean. We enforce this by imposing E[S−] < ∞.
Let then v(t) be a twice differentiate function with
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0515036100011405/resource/name/S0515036100011405_eqnU1.gif?pub-status=live)
and let z be a constant with o ≤ z ≤ 1.
We define the premium P as follows
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0515036100011405/resource/name/S0515036100011405_eqnU2.gif?pub-status=live)
or equivalently
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0515036100011405/resource/name/S0515036100011405_eqnU3.gif?pub-status=live)
Notation: v−(∞) = ∞.
The definitions (1) and (equivalently) (2) are meaningful because of the
Lemma: a) E[v(S − zQ)] exists for all Q∈(− ∞, + ∞).
b) The set{Q∣ − ∞ < Q < + ∞, E[v(S−zQ)]>v((1−z)Q)} is not empty.
Proof: a) ![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0515036100011405/resource/name/S0515036100011405_eqnU4.gif?pub-status=live)
b) Because of a) E[v(S−zQ)] is always finite or equal to + ∞ If v(− ∞) = − ∞ then E[v(S − zQ)] > v((1 − z)Q) is satisfied for sufficiently small Q. The left hand side of the inequality is a nonincreasing continuous function in P (strictly decreasing if z > 0), while the right hand side is a nondecreasing continuous function in Q (strictly increasing if z > 1).
If v(− ∞) = c finite then E[v(S − zQ)] > c
(otherwise S would need to be equal to − ∞ with probability 1) and again E[v(S − zQ)] > v((1 − z)Q) is satisfied for sufficiently small Q.