Cost-efficient contingent claims with market frictions

Abstract

In complete frictionless securities markets under uncertainty, it is well-known that in the absence of arbitrage opportunities, there exists a unique linear positive pricing rule, which induces a state-price density (e.g., Harrison and Kreps in J Econ Theory 20(3):381–408, 1979). Dybvig (J Bus 61(3):369–393, 1988; Rev Financ Stud 1(1):67–88, 1988) showed that the cheapest way to acquire a certain distribution of a consumption bundle (or security) is when this bundle is anti-comonotonic with the state-price density, i.e., arranged in reverse order of the state-price density. In this paper, we look at extending Dybvig’s ideas to complete markets with imperfections represented by a nonlinear pricing rule (e.g., due to bid-ask spreads). We consider an investor in a securities market where the pricing rule is “law-invariant” with respect to a capacity (e.g., Choquet pricing as in Araujo et al. in Econ Theory 49(1):1–35, 2011; Chateauneuf et al. in Math Financ 6(3):323–330, 1996; Chateauneuf and Cornet in Submodular financial markets with frictions, 2015; Cerreia-Vioglio et al. in J Econ Theory 157:730–762, 2015). The investor holds a security with a random payoff X and his problem is that of buying the cheapest contingent claim Y on X, subject to some constraints on the performance of the contingent claim and on its level of risk exposure. The cheapest such claim is called cost-efficient. If the capacity satisfies standard continuity and a property called strong diffuseness introduced in Ghossoub (Math Op Res 40(2):429–445, 2015), we show the existence and monotonicity of cost-efficient claims, in the sense that a cost-efficient claim is anti-comonotonic with the underlying security’s payoff X. Strong diffuseness is satisfied by a large collection of capacities, including all distortions of diffuse probability measures. As an illustration, we consider the case of a Choquet pricing functional with respect to a capacity and the case of a Choquet pricing functional with respect to a distorted probability measure. Finally, we consider a simple example in which we derive an explicit analytical form for a cost-efficient claim.

Appendices

Appendix 1: The Choquet integral

Definition 7.1

For a given capacity $\nu$ on a measurable space $\left( S, \mathcal {G}\right)$ and a given $\psi \in B^{+}\left( \mathcal {G}\right)$, the Choquet integral $\int \psi \ d\nu$ of $\psi$ with respect to $\nu$ is defined by

$$\begin{aligned} \int \psi \ d\nu := \int _{0}^{+\infty } \nu \big (\{ s \in S: \psi \left( s\right) > t \}\big ) \ dt = \int _{0}^{+\infty } G_{\nu ,\psi }\left( t\right) \ dt. \end{aligned}$$

If $\phi \in B\left( \mathcal {G}\right)$, then the Choquet integral $\int \phi \ d\nu$ of $\phi$ with respect to $\nu$ is defined by

$$\begin{aligned} \int \phi \ d\nu := \int _{0}^{+\infty } \nu \big (\{ s \in S: \phi \left( s\right) > t \}\big ) \ dt + \int _{-\infty }^{0} \left[ \nu \big (\{ s \in S: \phi \left( s\right) > t \}\big ) - 1\right] \ dt. \end{aligned}$$

As a result, if $\phi _{1}, \phi _{2} \in B\left( \mathcal {G}\right)$ have the same upper-distribution with respect to $\nu$ then $\int \phi _{1} \ d\nu = \int \phi _{2} \ d\nu$. That is, the Choquet integral with respect to a given capacity $\nu$ is a $\nu$-upper-law-invariant function on $B\left( \mathcal {G}\right)$.

The Choquet integral with respect to a measure is simply the usual Lebesgue integral with respect to that measure [29, p. 59]. Moreover, for any capacity $\nu$ on $\left( S, \mathcal {G}\right)$ and for any $\phi \in B\left( \mathcal {G}\right)$, the following holds [29, p. 60]:

$$\begin{aligned} \int \psi \ d\nu = \int _{0}^{+\infty } \nu \Big (\{ s \in S: \phi \left( s\right) \geqslant t \}\Big ) \ dt + \int _{-\infty }^{0} \Bigg [\nu \big (\{ s \in S: \phi \left( s\right) \geqslant t \}\big ) - 1\Bigg ] \ dt. \end{aligned}$$

Finally, as a functional on $B\left( \mathcal {G}\right)$, the Choquet integral (with respect to some given capacity) is supnorm-continuous, being Lipschitz continuous [29, Prop. 4.11]. The following proposition gathers some other properties of the Choquet integral.

Proposition 7.1

Let $\nu$ be a capacity on $\left( S, \mathcal {G}\right)$.

• (1) If $\phi _{1}, \phi _{2} \in B\left( \mathcal {G}\right)$ are comonotonic, then $\int \left( \phi _{1} + \phi _{2}\right) \ d\nu = \int \phi _{1} \ d\nu + \int \phi _{2} \ d\nu$.

• (2) If $\phi \in B\left( \mathcal {G}\right)$ and $c \in \mathbb {R}$, then $\int \left( \phi + c\right) \ d\nu = \int \phi \ d\nu + c$.

• (3) If $A \in \mathcal {G}$ then $\int \mathbf {1}_{A} \ d\nu = \nu \left( A\right)$.

• (4) If $\phi \in B\left( \mathcal {G}\right)$ and $a \geqslant 0$, then $\int a \ \phi \ d\nu = a \ \int \phi \ d\nu$.

• (5) If $\phi _{1}, \phi _{2} \in B\left( \mathcal {G}\right)$ are such that $\phi _{1} \leqslant \phi _{2}$, then $\int \phi _{1} \ d\nu \leqslant \int \phi _{2} \ d\nu$.

• (6) If $\nu$ is submodular, then for any $\phi _{1}, \phi _{2} \in B\left( \mathcal {G}\right)$, $\int \left( \phi _{1} + \phi _{2}\right) \ d\nu \leqslant \int \phi _{1} \ d\nu + \int \phi _{2} \ d\nu$.

Proof

[29, Th. 4.3,Th. 4.6,Prop. 4.11] or [13, Prop. 5.1, Prop. 6.3]. $\square$

For more about capacities and the Choquet integral we refer to Marinacci and Montrucchio [29].

Appendix 2: Monotone equimeasurable rearrangements with respect to a strongly nonatomic capacity

All results of this section are taken from Ghossoub [19] to which we refer the reader for proofs and additional analysis. Consider the setting of Sect. 2. In particular, S is a set of states of the world and $X: S \rightarrow \left[ 0,M\right]$ is an underlying security’s random payoff. Let $\nu$ be a capacity on the measurable space $\left( S, \Sigma \right)$, where $\Sigma = \sigma \{X\}$. Contingent claims are the elements of $B^{+}\left( \Sigma \right)$, that is, the functions $Y: S \rightarrow \mathbb {R}$ of the form $Y = I \circ X$, for some bounded, nonnegative and Borel-measurable map $I : X\left( S\right) \rightarrow \mathbb {R}^{+}$. For each $I: X\left( S\right) \rightarrow \mathbb {R}^{+}$, let $\Vert I\Vert _{sup} := \sup \ \{I\left( x\right) : x \in X\left( S\right) \}$. Similarly, for each $Y \in B^{+}\left( \Sigma \right)$, let $\Vert Y\Vert _{sup} := \sup \ \{Y\left( s\right) : s \in S\}$. Then for any $Y = I \circ X \in B^{+}\left( \Sigma \right)$, $\Vert Y\Vert _{sup} = \Vert I\Vert _{sup}$. If $I, I_{n} : \left[ 0,M\right] \rightarrow \left[ 0,M\right]$, for each $n \geqslant 1$, we will write $I_{n} \uparrow I$ to signify that the sequence $\{I_{n}\}_{n}$ is a nondecreasing sequence of functions and that $\underset{n \rightarrow +\infty }{\lim }I_{n}\left( t\right) = I\left( t\right)$, for all $t \in \left[ 0,M\right]$.

Assumption 8.1

$\nu$ is continuous and strongly diffuse with respect to X.

Assumption 8.1 implies that $\nu \circ X^{-1}$ is a continuous and diffuse capacity. Now, for a given $Y = I \circ X \in B^{+}\left( \Sigma \right)$, define the map

$$\begin{aligned} G_{\nu ,X,I}:&\mathbb {R} \rightarrow \left[ 0,1\right] \nonumber \\&t \mapsto G_{\nu ,X,I}\left( t\right) := \nu \circ X^{-1} \big ( \{z \in \left[ 0,M\right] : I\left( z\right) > t\} \big ) \end{aligned}$$

(8.1)

to be the upper-distribution of I with respect to $\nu \circ X^{-1}$. Then $G_{\nu ,X,I}$ is nonincreasing and right-continuous, due to Assumption 8.1. Let $Y = I \circ X \in B^{+}\left( \Sigma \right)$, with $I : \left[ 0,M\right] \rightarrow \mathbb {R}^{+}$ bounded and Borel-measurable.

Definition 8.2

Define the function ${\widetilde{I}}: \mathbb {R}^{+} \rightarrow \mathbb {R}^{+}$ by

$$\begin{aligned} {\widetilde{I}} \left( t\right) := \inf \Big \{z \in \mathbb {R}^{+}: G_{\nu ,X,I}\left( z\right) \leqslant \nu \circ X^{-1} \big ( \left[ 0,t\right] \big ) \Big \}. \end{aligned}$$

(8.2)

The following proposition gives some properties of the map ${\widetilde{I}}$.

Proposition 8.3

Under Assumption 8.1, the following holds:

• (1) ${\widetilde{I}}$ is nonincreasing and Borel-measurable.

• (2) ${\widetilde{I}}$ is right-continuous.

• (3) For all $t \in \mathbb {R}^{+}$, $G_{\nu ,X,I}\left( {\widetilde{I}}\left( t\right) \right) \leqslant \nu \circ X^{-1} \big ( \left[ 0,t\right] \big )$.

• (4) If $I_{1}, I_{2} : \left[ 0,M\right] \rightarrow \mathbb {R}^{+}$ are such that $I_{1} \leqslant I_{2}$, then ${\widetilde{I}}_{1} \leqslant {\widetilde{I}}_{2}$.

• (5) I and ${\widetilde{I}}$ have the same upper-distribution with respect to $\nu \circ X^{-1}$.

• (6) If $\Vert I\Vert _{sup} = N \ (< +\infty )$, then $\Vert {\widetilde{I}}\Vert _{sup} \leqslant N$.

• (7) If {$I_{n}\}_{n}$ is a sequence of bounded Borel-measurable functions from $\left[ 0,M\right]$ into $\mathbb {R}^{+}$ such that $I_{n} \uparrow I$, for some bounded Borel-measurable function $I : \left[ 0,M\right] \rightarrow \mathbb {R}^{+}$, then ${\widetilde{I}}_{n} \uparrow {\widetilde{I}}$.

Definition 8.4

For each $Y = I \circ X \in B^{+}\left( \Sigma \right)$, define the function the function ${\widetilde{Y}}$ by ${\widetilde{Y}} := {\widetilde{I}} \circ X$.

When Assumption 8.1 holds, Proposition 8.3 implies that the function ${\widetilde{Y}}$ is bounded, $\Sigma$-measurable, anti-comonotonic with X and has the same upper-distribution as Y with respect to $\nu$. In particular, $\int {\widetilde{Y}} \ d\nu = \int Y \ d\nu$. Moreover, $\Vert {\widetilde{Y}}\Vert _{sup} \leqslant \Vert Y\Vert _{sup}$. The function ${\widetilde{Y}}$ will be called a nonincreasing $\nu$-upper-equimeasurable rearrangement of Y with respect to X.

Appendix 3: Proofs

1.1 Proof of Proposition 3.9

Let $Y \in \mathcal {F}$ be a feasible contingent claim for the investor and let ${\widetilde{Y}}$ be a nonincreasing $\nu$-upper-equimeasurable rearrangement of Y with respect to X. By Assumption 2.5, $\mathcal {C}$ is $\nu$-upper-law-invariant. Hence, $\mathcal {C}$, $\mathcal {P}$ and $\mathcal {R}$ are all $\nu$-upper-law-invariant. Therefore, it follows from Definition 2.4 and Proposition 8.3 that ${\widetilde{Y}}$ is a feasible contingent claim for the investor, and that $\mathcal {C}({\widetilde{Y}}) = \mathcal {C}(Y)$. $\square$

1.2 Proof of Proposition 3.10

Suppose that Problem (3.1) admits a solution $Y^{*}$. Take ${\widetilde{Y}}^{*}$ to be a nonincreasing $\nu$-upper-equimeasurable rearrangement of $Y^{*}$ with respect to X. The rest follows form Assumption 2.5, Definition 2.4 and Proposition 8.3 (5) and (6), as in the proof of Proposition 3.9 above. Now, to show that any strictly cost-efficient claim is anti-comonotonic with X, suppose, per contra, that there exists a strictly cost-efficient claim $Y_{1}$ that is not anti-comonotonic with X. Let ${\widetilde{Y}}_{1}$ to be a nonincreasing $\nu$-upper-equimeasurable rearrangement of $Y_{1}$ with respect to X. Then by Proposition 3.9, $C\left( Y_{1}\right) = C\left( {\widetilde{Y}}_{1}\right)$, which contradicts the fact that $Y_{1}$ is strictly cost-efficient.

$\square$

1.3 Proof of Theorem 3.12

This proof is similar to that of Theorem 6.3 of Ghossoub [19]. Let $\mathcal {F} := \Big \{ Y \in B^{+}\left( \Sigma \right) : Y \leqslant N, \ \mathcal {P}\left( Y\right) \geqslant P_{0}, \ \mathcal {R}\left( Y\right) \leqslant R_{0} \Big \}$ be the feasibility set for Problem (3.1) and assume that $\mathcal {F} \ne \varnothing$. Denote by $\mathcal {F}^{\downarrow }$ the collection of all elements of $\mathcal {F}$ that are anti-comonotonic with X. Then $\mathcal {F}^{\downarrow } \ne \varnothing$, by a proof identical to that of Proposition 3.10. Moreover, by a proof identical to that of Proposition 3.10, for any $Y \in B^{+}\left( \Sigma \right)$ which is feasible for Problem (3.1), there is a ${\widetilde{Y}} \in B^{+}\left( \Sigma \right)$ which is not only feasible for Problem (3.1) and anti-comonotonic with X, but is such that $\mathcal {C}({\widetilde{Y}}) = \mathcal {C}\left( Y\right)$. Hence, one can choose a minimizing sequence $\{Y_{n}\}_{n}$ in $\mathcal {F}^{\downarrow }$ for Problem (3.1). That is,

$$\begin{aligned}\underset{n \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{n}\right) = H := \inf _{Y \in \mathcal {F}} \ \mathcal {C}\left( Y\right) . \end{aligned}$$

Since $0 \leqslant Y_{n} \leqslant N$, for each $n \geqslant 1$, the sequence $\{Y_{n}\}_{n}$ is uniformly bounded. Moreover, for each $n \geqslant 1$ one has $Y_{n} = I_{n} \circ X$. Consequently, the sequence $\{I_{n}\}_{n}$ is a uniformly bounded sequence of nonincreasing Borel-measurable functions.

Lemma 9.1

If $\left( f_{n}\right) _{n}$ is a uniformly bounded sequence of nonincerasing real-valued functions on some closed interval $\mathcal {I}$ in $\mathbb {R}$, with bound N (i.e. $|f_{n}\left( x\right) | \leqslant N, \ \forall x \in \mathcal {I}, \ \forall n \geqslant 1$), then there exists a nonincerasing real-valued bounded function $f^{*}$ on $\mathcal {I}$, also with bound N, and a subsequence of $\left( f_{n}\right) _{n}$ that converges pointwise to $f^{*}$ on $\mathcal {I}$.

The proof of this lemma can be found in [14, pp. 165-166]. Now, by Lemma 9.1, there is a nonincreasing function $I^{*}: \left[ 0,M\right] \rightarrow \left[ 0,N\right]$ and a subsequence $\{I_{m}\}_{m}$ of $\{I_{n}\}_{n}$ such that $\{I_{m}\}_{m}$ converges pointwise on $\left[ 0,M\right]$ to $I^{*}$. Hence, $I^{*}$ is also Borel-measurable, and so $Y^{*} := I^{*} \circ X \in B^{+}\left( \Sigma \right)$ is such that $0 \leqslant Y^{*} \leqslant N$ and $Y^{*}$ is anti-comonotonic with X. Moreover, the sequence $\{Y_{m}\}_{m}$, defined by $Y_{m} := I_{m} \circ X$, converges pointwise to $Y^{*}$. Thus, by the assumption that the mappings $\mathcal {P}$ and $\mathcal {R}$ preserve uniformly bounded pointwise convergence, it follows that $Y^{*} \in \mathcal {F}^{\downarrow }$. Now, by the assumption that the mapping $\mathcal {C}$ preserves uniformly bounded pointwise convergence, one has

$$\begin{aligned} \mathcal {C}\left( Y^{*}\right) = \underset{m \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{m}\right) = \underset{n \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{n}\right) = H. \end{aligned}$$

Hence $Y^{*}$ solves Problem (3.1), that is, $Y^{*}$ is cost-efficient. $Y^{*}$ is also anti-comonotonic with X.

It remains to show that any strictly cost-efficient claim is anti-comonotonic with X. Suppose, per contra, that there exists a strictly cost-efficient claim $Y_{0}$ such that $Y_{0}$ is not anti-comonotonic with X. Then for all $Z \in \mathcal {F}$, we have that $\mathcal {C}\left( Y_{0}\right) < \mathcal {C}\left( Z\right)$. However, by Proposition 3.9, there exists ${\widetilde{Y}}_{0} \in \mathcal {F}$ such that $\mathcal {C}({\widetilde{Y}}_{0}) = \mathcal {C}\left( Y_{0}\right)$. Hence, since ${\widetilde{Y}}_{0} \in \mathcal {F}$ and $Y_{0}$ is strictly cost-efficient, we have that $\mathcal {C}({\widetilde{Y}}_{0}) = \mathcal {C}\left( Y_{0}\right) < \mathcal {C}({\widetilde{Y}}_{0})$, a contradiction. Therefore, any strictly cost-efficient claim is anti-comonotonic with X. $\square$

1.4 Proof of Theorem 5.3

Let $\mathcal {F}$ denote the feasibility set of Problem (5.4), assumed to be nonempty. Let $\mathcal {F}^{\downarrow }$ denote the collection of all elements of $\mathcal {F}$ that are anti-comonotonic with X.

Lemma 9.2

For each $Y \in \mathcal {F}$, there exists ${\widetilde{Y}} \in \mathcal {F}^{\downarrow }$ such that $\mathcal {C}(Y) = \mathcal {C}({\widetilde{Y}})$, $\mathcal {P}(Y) = \mathcal {P}({\widetilde{Y}})$ and $\mathcal {R}(Y) = \mathcal {R}({\widetilde{Y}})$.

Proof

Fix $Y \in \mathcal {F}$. Since P is a probability measure and $P \circ X^{-1}$ is diffuse, it follows that P is continuous and strongly diffuse with respect to X. We can then define ${\widetilde{Y}}$ to be a nonincreasing P-upper-equimeasurable rearrangement of Y with respect to X. Then ${\widetilde{Y}} \leqslant N$, by Proposition 8.3. Moreover,

$$\begin{aligned} \mathcal {C}(Y)= & {} \Xi \left( \int Y d T_{1} \circ P \right) = \Xi \left( \int _{0}^{+\infty } T_{1}\left( P\big (\{s \in S: Y\left( s\right) > t\}\right) \big ) dt \right) \\= & {} \Xi \left( \int _{0}^{+\infty } T_{1}\left( P\left( \{s \in S: {\widetilde{Y}}\left( s\right) > t\}\right) \right) dt \right) = \Xi \left( \int {\widetilde{Y}} d T_{1} \circ P \right) \\= & {} \mathcal {C}({\widetilde{Y}}), \end{aligned}$$

where the third equality follows from the P-upper-equimeasurability of ${\widetilde{Y}}$ and Y. Similarly, one can show that $\mathcal {P}(Y) = \mathcal {P}({\widetilde{Y}})$ and $\mathcal {R}(Y) = \mathcal {R}({\widetilde{Y}})$. $\square$

In particular, Lemma 9.2 says that if $\mathcal {F} \ne \varnothing$, then $\mathcal {F}^{\downarrow } \ne \varnothing$. Moreover, in light of Lemma 9.2, one can choose a minimizing sequence $\{Y_{n}\}_{n}$ in $\mathcal {F}^{\downarrow }$ for Problem (5.4). That is,

$$\begin{aligned} \underset{n \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{n}\right) = H := \inf _{Y \in \mathcal {F}} \ \mathcal {C}\left( Y\right) . \end{aligned}$$

Since $0 \leqslant Y_{n} \leqslant N$, for each $n \geqslant 1$, the sequence $\{Y_{n}\}_{n}$ is uniformly bounded. Moreover, for each $n \geqslant 1$ one has $Y_{n} = I_{n} \circ X$. Consequently, the sequence $\{I_{n}\}_{n}$ is a uniformly bounded sequence of nonincreasing Borel-measurable functions. Thus, by Lemma 9.1, there is a nonincreasing function $I^{*}: \left[ 0,M\right] \rightarrow \left[ 0,N\right]$ and a subsequence $\{I_{m}\}_{m}$ of $\{I_{n}\}_{n}$ such that $\{I_{m}\}_{m}$ converges pointwise on $\left[ 0,M\right]$ to $I^{*}$. Hence, $I^{*}$ is also Borel-measurable, and so $Y^{*} := I^{*} \circ X \in B^{+}\left( \Sigma \right)$ is such that $0 \leqslant Y^{*} \leqslant N$ and $Y^{*}$ is anti-comonotonic with X. Moreover, the sequence $\{Y_{m}\}_{m}$, defined by $Y_{m} := I_{m} \circ X$, converges pointwise to $Y^{*}$.

Now, since the distortion functions $T_{1}$, $T_{2}$ and $T_{3}$ are differentiable (Assumption 5.1), they are continuous. Moreover, since P is a probability measure, it is a continuous capacity. Therefore, it follows that $T_{1} \circ P$, $T_{2} \circ P$ and $T_{3} \circ P$ are continuous capacities on $\left( S, \Sigma \right)$. Since the functions p and r are continuous and nondecreasing (Definition 5.2), they are also Borel-measurable and bounded on any closed and bounded subset of $\mathbb {R}$. Since the Choquet integral with respect to some continuous capacity is an operator on $B^{+}\left( \Sigma \right)$ that preserves uniformly bounded pointwise convergence (Definition 3.11) by [30, Th. 7.16], since the functions $\Xi$, $\Psi$, $\Phi$, p and r are all continuous (by assumption), and since the functions p and r are bounded on any closed and bounded subset of $\mathbb {R}$, it follows that $\mathcal {C}$, $\mathcal {P}$ and $\mathcal {R}$ all preserve uniformly bounded pointwise convergence.

Since the mappings $\mathcal {P}$ and $\mathcal {R}$ preserve uniformly bounded pointwise convergence, it follows that $Y^{*} \in \mathcal {F}^{\downarrow }$. Now, since the mapping $\mathcal {C}$ preserves uniformly bounded pointwise convergence, one has

$$\begin{aligned} \mathcal {C}\left( Y^{*}\right) = \underset{m \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{m}\right) = \underset{n \rightarrow +\infty }{\lim }\mathcal {C}\left( Y_{n}\right) = H. \end{aligned}$$

Hence $Y^{*}$ solves Problem (5.4), that is, $Y^{*}$ is cost-efficient. $Y^{*}$ is also anti-comonotonic with X.

It remains to show that any strictly cost-efficient claim is anti-comonotonic with X. Suppose, per contra, that there exists a strictly cost-efficient claim $Y_{0}$ such that $Y_{0}$ is not anti-comonotonic with X. Then for all $Z \in \mathcal {F}$, we have that $\mathcal {C}\left( Y_{0}\right) < \mathcal {C}\left( Z\right)$. However, by Lemma 9.2, there exists ${\widetilde{Y}}_{0} \in \mathcal {F}$ such that $\mathcal {C}({\widetilde{Y}}_{0}) = \mathcal {C}\left( Y_{0}\right)$. Hence, since ${\widetilde{Y}}_{0} \in \mathcal {F}$ and $Y_{0}$ is strictly cost-efficient, we have that $\mathcal {C}({\widetilde{Y}}_{0}) = \mathcal {C}\left( Y_{0}\right) < \mathcal {C}({\widetilde{Y}}_{0})$, a contradiction. Therefore, any strictly cost-efficient claim is anti-comonotonic with X. $\square$

1.5 Proof of Proposition 5.4

Let $\mathcal {F}$ denote the feasibility set of Problem (5.4), assumed to be nonempty. Let $\mathcal {F}^{\downarrow }$ denote the collection of all elements of $\mathcal {F}$ that are anti-comonotonic with X.

Lemma 9.3

• (1) $U := F_{X}\left( X\right)$ is a random variable on the probability space $\left( S, \Sigma , P\right)$ with a uniform distribution on $\left( 0,1\right)$ (and hence so is $1-U$),

• (2) $X = F_{X}^{-1}\left( U\right) , \ P$-a.s.

• (3) For each $Y \in \mathcal {F}$, the function $Y^{*}$ defined by $Y^{*} := F_{Y}^{-1}\left( 1-F_{X}\left( X\right) \right) = F_{Y}^{-1}\left( 1-U\right)$ is such that:

  • (a) $Y^{*} \in \mathcal {F}^{\downarrow }$,

  • (b) $Y \underset{P}{\sim }Y^{*}$ and $Y \overset{P}{\sim }Y^{*}$,

  • (c) $\mathcal {C}(Y) = \mathcal {C}(Y^{*})$, $\mathcal {P}(Y) = \mathcal {P}(Y^{*})$ and $\mathcal {R}(Y) = \mathcal {R}(Y^{*})$.

Proof

(1) and (2) follow from the diffuseness of $P \circ X^{-1}$ ([18, Lemma A.21] or [20, Lemma 4.1]). Now fix $Y \in B^{+}\left( \Sigma \right)$, let ${\widetilde{Y}}$ denote a nonincreasing P-upper-equimeasurable rearrangement of Y with respect to X and let $Y^{*} := F_{Y}^{-1}\left( 1-F_{X}\left( X\right) \right)$. Since P is a probability measure, it is easy to verify that $Y \underset{P}{\sim }Y^{*}$ and $Y \overset{P}{\sim }Y^{*}$. By Lemma 9.2 and its proof, in order to show the rest of (3) it suffices to show that $Y^{*} = {\widetilde{Y}}$.

Since $Y \in B^{+}\left( \Sigma \right)$, it is of the form $Y = I \circ X$, for some bounded, nonnegative and Borel-measurable map $I : X\left( S\right) \rightarrow \mathbb {R}^{+}$. Moreover, as in Definition 8.4, ${\widetilde{Y}}$ is given by ${\widetilde{Y}} = {\widetilde{I}} \circ X$, where ${\widetilde{I}}$ is the function defined as in equation (8.2) by

$$\begin{aligned} {\widetilde{I}} \left( t\right) = \inf \Big \{z \in \mathbb {R}^{+}: P \circ X^{-1} \big ( \{v \in \left[ 0,M\right] : I\left( v\right) > z\} \big )\leqslant P \circ X^{-1} \big ( \left[ 0,t\right] \big ) \Big \}. \end{aligned}$$

Therefore, for each $s_{0} \in S$,

$$\begin{aligned} {\widetilde{Y}}\left( s_{0}\right)= & {} {\widetilde{I}} \left( X\left( s_{0}\right) \right) \\= & {} \inf \Big \{z \in \mathbb {R}^{+}: P \circ X^{-1} \big ( \{t \in \left[ 0,M\right] : I\left( t\right) > z\} \big ) \leqslant P \circ X^{-1} \big ( \left[ 0,X\left( s_{0}\right) \right] \big ) \Big \}. \end{aligned}$$

Now, $P \circ X^{-1} \big ( \left[ 0,X\left( s_{0}\right) \right] \big ) = F_{X}\left( X\left( s_{0}\right) \right) = F_{X}\left( X\right) \left( s_{0}\right)$. Moreover,

$$\begin{aligned} P \circ X^{-1} \big ( \{t \in \left[ 0,M\right] : I\left( t\right) > z\} \big )= & {} 1- P \circ X^{-1} \big ( \{t \in \left[ 0,M\right] : I\left( t\right) \leqslant z\} \big ) \\= & {} 1 - P \big ( \{t \in \left[ 0,M\right] : I\left( X\left( t\right) \right) \leqslant z\} \big ) \\= & {} 1 - P \big ( \{t \in \left[ 0,M\right] : Y\left( t\right) \leqslant z\} \big ) \\= & {} 1 - F_{Y}\left( z\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} {\widetilde{Y}}\left( s_{0}\right)= & {} \inf \Big \{z \in \mathbb {R}^{+}: 1 - F_{Y}\left( z\right) \leqslant F_{X}\left( X\right) \left( s_{0}\right) \Big \} \\= & {} \inf \Big \{z \in \mathbb {R}^{+}: F_{Y}\left( z\right) \geqslant 1 - F_{X}\left( X\right) \left( s_{0}\right) \Big \} \\= & {} F_{Y}^{-1}\left( 1-F_{X}\left( X\right) \right) \left( s_{0}\right) . \end{aligned}$$

Since $s_{0} \in S$ was chosen arbitrarily, it follows that ${\widetilde{Y}} = F_{Y}^{-1}\left( 1-F_{X}\left( X\right) \right) = Y^{*}$, which concludes the proof of Lemma 9.3. $\square$

Now, since the distortion functions are differentiable (Assumption 5.1), we have⁴ that for each $Y \in B^{+}\left( \Sigma \right)$,

$$\begin{aligned} \mathcal {C}\left( Y\right)= & {} \Xi \left( \int Y d T_{1} \circ P \right) \\= & {} \Xi \left( \int T_{1}^{\prime }\left( 1-U\right) F_{Y}^{-1}\left( U\right) d P \right) = \Xi \left( \int T_{1}^{\prime }\left( U\right) F_{Y}^{-1}\left( 1-U\right) d P \right) \\= & {} \Xi \left( \int _{0}^{1} T_{1}^{\prime }\left( 1-t\right) F_{Y}^{-1}\left( t\right) d t \right) = \Xi \left( \int _{0}^{1} T_{1}^{\prime }\left( t\right) F_{Y}^{-1}\left( 1-t\right) d t \right) , \\ \end{aligned}$$

and

$$\begin{aligned} \mathcal {P}\left( Y\right)= & {} \Psi \left( \int p\left( Y\right) d T_{2} \circ P \right) = \Psi \left( \int T_{2}^{\prime }\left( 1-U\right) F_{p\left( Y\right) }^{-1}\left( U\right) d P \right) \\= & {} \Psi \left( \int T_{2}^{\prime }\left( 1-U\right) p\left( F_{Y}^{-1}\right) \left( U\right) d P \right) = \Psi \left( \int T_{2}^{\prime }\left( U\right) p\left( F_{Y}^{-1}\right) \left( 1-U\right) d P \right) \\= & {} \Psi \left( \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( F_{Y}^{-1}\right) \left( t\right) d t \right) = \Psi \left( \int _{0}^{1} T_{2}^{\prime }\left( t\right) p\left( F_{Y}^{-1}\right) \left( 1-t\right) d t \right) , \\ \end{aligned}$$

where the third equality follows from the monotonicity of the function⁵ p. Similarly, we have

$$\begin{aligned} \mathcal {R}\left( Y\right)= & {} \Phi \left( \int r\left( Y\right) d T_{3} \circ P \right) = \Phi \left( \int T_{3}^{\prime }\left( 1-U\right) F_{r\left( Y\right) }^{-1}\left( U\right) d P \right) \\= & {} \Phi \left( \int T_{3}^{\prime }\left( 1-U\right) r\left( F_{Y}^{-1}\right) \left( U\right) d P \right) = \Phi \left( \int T_{3}^{\prime }\left( U\right) r\left( F_{Y}^{-1}\right) \left( 1-U\right) d P \right) \\= & {} \Phi \left( \int _{0}^{1} T_{3}^{\prime }\left( 1-t\right) r\left( F_{Y}^{-1}\right) \left( t\right) d t \right) = \Phi \left( \int _{0}^{1} T_{3}^{\prime }\left( t\right) r\left( F_{Y}^{-1}\right) \left( 1-t\right) d t \right) . \end{aligned}$$

We now come to the formal proof of Proposition 5.4. Recall that $\mathcal {Q}$ denotes the collection of all quantile functions:

$$\begin{aligned} \mathcal {Q} := \Big \{f: \left( 0,1\right) \rightarrow \mathbb {R} \ \Big | \ f \hbox { is nondecreasing and left-continuous}\Big \}. \end{aligned}$$

Recall also that $\mathcal {Q}^{*}$ denotes the collection of all quantile functions $f \in \mathcal {Q}$ of functions $Y \in B^{+}\left( \Sigma \right)$ such that $0 \leqslant Y \leqslant N$. That is,

$$\begin{aligned} \mathcal {Q}^{*} = \Big \{ f \in \mathcal {Q} \ \Big | \ 0 \leqslant f\left( z\right) \leqslant N, \hbox { for each } 0 < z < 1\Big \}. \end{aligned}$$

Now, let $f^{*}$ be optimal for Problem (5.7). We show that $f^{*}\left( 1-F_{X}\left( X\right) \right) = f^{*}\left( 1-U\right)$ is cost-efficient and anti-comonotonic with X, i.e., that $f^{*}\left( 1-U\right)$ is optimal for Problem (5.4) and anti-comonotonic with X. Let $Z^{*} \in B^{+}\left( \Sigma \right)$ be a function such that $f^{*}$ is the quantile of $Z^{*}$. Hence, $0 \leqslant Z^{*} \leqslant N$, and the feasibility of $f^{*}$ for Problem (5.7) yields:

$$\begin{aligned} \mathcal {P}\left( Z^{*}\right)= & {} \Psi \left( \int p\left( Z^{*}\right) d T_{2} \circ P \right) \\= & {} \Psi \left( \int _{0}^{1} T_{2}^{\prime }\left( t\right) p\left( F_{Z^{*}}^{-1}\right) \left( 1-t\right) d t \right) = \Psi \left( \int _{0}^{1} T_{2}^{\prime }\left( t\right) p\left( f^{*}\left( 1-t\right) \right) dt \right) \geqslant P_{0},\end{aligned}$$

and

$$\begin{aligned} \mathcal {R}\left( Z^{*}\right)= & {} \Phi \left( \int r\left( Z^{*}\right) d T_{3} \circ P \right) \\= & {} \Phi \left( \int _{0}^{1} T_{3}^{\prime }\left( t\right) r\left( F_{Z^{*}}^{-1}\right) \left( 1-t\right) d t \right) = \Phi \left( \int _{0}^{1} T_{3}^{\prime }\left( t\right) r\left( f^{*}\left( 1-t\right) \right) dt \right) \leqslant R_{0}. \end{aligned}$$

Therefore, $Z^{*}$ is feasible for Problem (5.4). Hence, by Lemma 9.3, the function ${\widetilde{Z}}^{*} \in B^{+}\left( \Sigma \right)$ defined by ${\widetilde{Z}}^{*} := f^{*}\left( 1-U\right) = F^{-1}_{Z^{*}}\left( 1-F_{X}\left( X\right) \right)$ is feasible for Problem (5.4), anti-comonotonic with X and is such that $\mathcal {C}\left( Z^{*}\right) = \mathcal {C}({\widetilde{Z}}^{*})$, $\mathcal {P}\left( Z^{*}\right) = \mathcal {P}({\widetilde{Z}}^{*})$ and $\mathcal {R}\left( Z^{*}\right) = \mathcal {R}({\widetilde{Z}}^{*})$.

To show optimality of ${\widetilde{Z}}^{*}$ for Problem (5.4), let Z be any other feasible function for Problem (5.4). Then, by Lemma 9.3, the function ${\widetilde{Z}} \in B^{+}\left( \Sigma \right)$ defined by ${\widetilde{Z}} := F_{Z}^{-1}\left( 1-F_{X}\left( X\right) \right)$ is also feasible for Problem (5.4) and it is anti-comonotonic with X. Moreover, $F_{Z} = F_{{\widetilde{Z}}}$. It then remains to show that $\Xi \left( \int {\widetilde{Z}}^{*} d T_{1} \circ P \right) \leqslant \Xi \left( \int {\widetilde{Z}} d T_{1} \circ P \right)$. Let $f := F_{Z}^{-1}$ so that ${\widetilde{Z}} = f\left( 1-U\right)$. Since ${\widetilde{Z}}$ is feasible for Problem (5.4), we have

$$\begin{aligned} P_{0} \leqslant \mathcal {P}\left( {\widetilde{Z}}\right) = \Psi \left( \int p\left( {\widetilde{Z}}\right) d T_{2} \circ P \right) = \Psi \left( \int _{0}^{1} T_{2}^{\prime }\left( t\right) p\left( f\left( 1-t\right) \right) dt \right) , \end{aligned}$$

and

$$\begin{aligned} R_{0} \geqslant \mathcal {R}\left( {\widetilde{Z}}\right) = \Phi \left( \int r\left( {\widetilde{Z}}\right) d T_{3} \circ P \right) = \Phi \left( \int _{0}^{1} T_{3}^{\prime }\left( t\right) r\left( f\left( 1-t\right) \right) dt \right) . \end{aligned}$$

Consequently, the function f is feasible for Problem (5.7). Since $f^{*}$ is optimal for Problem (5.7), it follows that

$$\begin{aligned} \Xi \left( \int {\widetilde{Z}}^{*} d T_{1} \circ P \right)= & {} \Xi \left( \int _{0}^{1} T_{1}^{\prime }\left( t\right) f^{*}\left( 1-t\right) d t \right) \leqslant \Xi \left( \int _{0}^{1} T_{1}^{\prime }\left( t\right) f\left( 1-t\right) d t \right) \\= & {} \Xi \left( \int {\widetilde{Z}} d T_{1} \circ P \right) . \end{aligned}$$

This concludes the proof of Proposition 5.4. $\square$

1.6 Proof of Theorem 5.9

The existence of a cost-efficient claim that is anti-comonotonic with X is given by Theorem 5.3, since $P \circ X^{-1}$ is diffuse by hypothesis. Now, recall that $\mathcal {Q}$ denotes the collection of all quantile functions. That is,

$$\begin{aligned} \mathcal {Q} := \Big \{f: \left( 0,1\right) \rightarrow \mathbb {R} \ \Big | \ f \hbox { is nondecreasing and left-continuous}\Big \}. \end{aligned}$$

Recall also that $\mathcal {Q}^{*}$ denotes the collection of all quantile functions $f \in \mathcal {Q}$ of functions $Y \in B^{+}\left( \Sigma \right)$ such that $0 \leqslant Y \leqslant N$. That is,

$$\begin{aligned} \mathcal {Q}^{*} = \Big \{ f \in \mathcal {Q} \ \Big | \ 0 \leqslant f\left( z\right) \leqslant N, \hbox { for each } 0 < z < 1\Big \}. \end{aligned}$$

Consider the following problem.

$$\begin{aligned} \inf _{f \in \mathcal {Q}^{*}} \Bigg \{ \int _{0}^{1} T_{1}^{\prime }\left( t\right) f\left( 1-t\right) dt : \int _{0}^{1} T_{2}^{\prime }\left( t\right)&p\left( f\left( 1-t\right) \right) dt \geqslant P_{0} \Bigg \}. \end{aligned}$$

(9.3)

By a similar proof to that of Proposition 5.4, if $f^{*}$ is optimal for Problem (9.3), then the function $Y^{*} := f^{*}\left( 1-F_{X}\left( X\right) \right)$ is cost-efficient (i.e., it solves for Problem (5.8)) and anti-comonotonic with X. It then suffices to solve Problem (9.3).

Lemma 9.4

If $f^{*} \in \mathcal {Q}^{*}$ is such that:

• (i) $\int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p \left( f^{*}\left( t\right) \right) dt = P_{0}$; and,

• (ii) there exists $\lambda > 0$ such that for each $t \in \left( 0,1\right) ,$

  $$\begin{aligned}f^{*}\left( t\right) = \underset{ 0 \leqslant y \leqslant N}{\arg \min }\Bigg [ T_{1}^{\prime }\left( 1-t\right) y - \lambda T_{2}^{\prime }\left( 1-t\right) p\left( y\right) \Bigg ], \end{aligned}$$

then $f^{*}$ solves Problem (9.3).

Proof

Suppose that $f^{*} \in \mathcal {Q}^{*}$ satisfies condition (i). Then $f^{*}$ is feasible for Problem (9.3). Now, suppose that $f^{*}$ also satisfies condition (ii). To show that $f^{*}$ is optimal for Problem (9.3), let f be any other feasible function for Problem (9.3). Then:

$$\begin{aligned} \int _{0}^{1} T_{2}^{\prime }\left( t\right) p\left( f\left( 1-t\right) \right) dt = \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( f\left( t\right) \right) dt \geqslant P_{0} = \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( f^{*}\left( t\right) \right) dt. \end{aligned}$$

Moreover, since $f^{*}$ satisfies condition (ii), it follows that there is some $\lambda > 0$ such that for any $t \in \left( 0,1\right)$,

$$\begin{aligned} T_{1}^{\prime }\left( 1-t\right) f^{*}\left( t\right) - \lambda T_{2}^{\prime }\left( 1-t\right) p\left( f^{*}\left( t\right) \right) \leqslant T_{1}^{\prime }\left( 1-t\right) f\left( t\right) - \lambda T_{2}^{\prime }\left( 1-t\right) p\left( f\left( t\right) \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} T_{1}^{\prime }\left( 1-t\right) f^{*}\left( t\right) - T_{1}^{\prime }\left( 1-t\right) f\left( t\right) \leqslant \lambda \Bigg [T_{2}^{\prime }\left( 1-t\right) p\left( f^{*}\left( t\right) \right) - T_{2}^{\prime }\left( 1-t\right) p\left( f\left( t\right) \right) \Bigg ]. \end{aligned}$$

Integration yields

$$\begin{aligned} \int _{0}^{1}T_{1}^{\prime }&\left( 1-t\right) f^{*}\left( t\right) dt - \displaystyle \int _{0}^{1} T_{1}^{\prime }\left( 1-t\right) f\left( t\right) dt \leqslant \lambda \left[ P_{0} - \int _{0}^{1}T_{2}^{\prime }\left( 1-t\right) p\left( f\left( t\right) \right) dt \right] . \end{aligned}$$

Since $\int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( f\left( t\right) \right) dt \geqslant P_{0}$, it follows that

$$\begin{aligned} \int _{0}^{1}T_{1}^{\prime }\left( 1-t\right) f^{*}\left( t\right) dt \leqslant \int _{0}^{1} T_{1}^{\prime }\left( 1-t\right) f\left( t\right) dt. \end{aligned}$$

$\square$

In light of Lemma 9.4, we can start by solving the problem:

$$\begin{aligned} \underset{0 \leqslant f_{\lambda }\left( t\right) \leqslant N}{\min }\left[ T_{1}^{\prime }\left( 1-t\right) f_{\lambda }\left( t\right) - \lambda T_{2}^{\prime }\left( 1-t\right) p\left( f_{\lambda }\left( t\right) \right) \right] , \end{aligned}$$

(9.4)

for a fixed $t \in \left( 0,1\right)$ and a given $\lambda > 0$. By concavity of the function p, the first-order conditions are sufficient for a (global) minimum, and they imply that for a given $\lambda > 0$ and a fixed $t \in \left( 0,1\right)$, a solution to Problem (9.4) is given by

$$\begin{aligned} f_{\lambda }^{*}\left( t\right) = \max \Bigg \{0,\min \left[ N, \left( p^{\prime }\right) ^{-1}\left( \frac{T^{\prime }_{1}\left( 1-t\right) }{\lambda T^{\prime }_{2}\left( 1-t\right) }\right) \right] \Bigg \}. \end{aligned}$$

Lemma 9.5

For each $\lambda > 0$, the function $t \mapsto \frac{T^{\prime }_{1}\left( 1-t\right) }{\lambda T^{\prime }_{2}\left( 1-t\right) }$ is nonincreasing on its domain.

Proof

By hypothesis, $T_{2}$ is less ambiguity averse than $T_{1}$. That is, for each $t \in \left[ 0,1\right] \setminus \{t \in \left[ 0,1\right] : T^{\prime }_{1}\left( 1-t\right) = T^{\prime }_{2}\left( 1-t\right) = 0\}$,

$$\begin{aligned} \frac{T^{\prime \prime }_{2}\left( 1-t\right) }{T^{\prime }_{2}\left( 1-t\right) } \leqslant \frac{T^{\prime \prime }_{1}\left( 1-t\right) }{T^{\prime }_{1}\left( 1-t\right) }. \end{aligned}$$

It then follows that

$$\begin{aligned} \frac{T^{\prime }_{1}\left( 1-t\right) T^{\prime \prime }_{2}\left( 1-t\right) - T^{\prime }_{2}\left( 1-t\right) T^{\prime \prime }_{1}\left( 1-t\right) }{\left[ T^{\prime }_{2}\left( 1-t\right) \right] ^{2}} \leqslant 0. \end{aligned}$$

That is,

$$\begin{aligned} \frac{d}{dt} \left( \frac{T^{\prime }_{1}\left( 1-t\right) }{T^{\prime }_{2}\left( 1-t\right) }\right) \leqslant 0. \end{aligned}$$

Therefore, for each $\lambda > 0$, the function $t \mapsto \frac{T^{\prime }_{1}\left( 1-t\right) }{\lambda T^{\prime }_{2}\left( 1-t\right) }$ is nonincreasing on its domain. $\square$

By Assumption 5.5, the function p is continuously differentiable, strictly increasing and strictly concave. This then implies by the Inverse Function Theorem [32, pp. 221-223] that the function $\left( p^{\prime }\right) ^{-1}$ is continuous and strictly decreasing. Also, since Borel-continuity implies Borel-measurability, the function p is measurable. Moreover, by Lemma 9.5, the function $t \mapsto \frac{T^{\prime }_{1}\left( 1-t\right) }{\lambda T^{\prime }_{2}\left( 1-t\right) }$ is nonincreasing, for any given $\lambda > 0$. Therefore, for any $\lambda > 0$, $f_{\lambda }^{*}$ is a nondecreasing function of t. Now, Assumption 5.8 implies that both $T^{\prime }_{1}$ and $T^{\prime }_{2}$ are continuous. Finally, the continuity of the function $\left( p^{\prime }\right) ^{-1}$ and of the functions $T_{1}^{\prime }$ and $T_{2}^{\prime }$ yield the left-continuity of the function $f_{\lambda }^{*}$. Hence, $f_{\lambda }^{*} \in \mathcal {Q}^{*}$.

It remains to show that there exists a $\lambda ^{*} > 0$ such that $\int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( f^{*}_{\lambda ^{*}}\left( t\right) \right) dt = P_{0}$. To this end, define the function $\Psi : \mathbb {R}^{+} \setminus \{0\} \longrightarrow \mathbb {R}$ by

$$\begin{aligned} \Psi \left( \lambda \right):= & {} \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( f^{*}_{\lambda }\left( t\right) \right) dt \\= & {} \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( \max \Bigg \{0,\min \left[ N, \left( p^{\prime }\right) ^{-1}\left( \frac{T^{\prime }_{1}\left( 1-t\right) }{\lambda T^{\prime }_{2}\left( 1-t\right) }\right) \right] \Bigg \}\right) dt. \end{aligned}$$

Since the function $\left( p^{\prime }\right) ^{-1}$ is decreasing, it follows from the monotonicity of the Lebesgue integral that $\Psi$ is a nondecreasing function of $\lambda$. Moreover, since the function p is continuous by Assumption 5.5, it is bounded on every closed and bounded subset of $\mathbb {R}$. Similarly, Assumption 5.8 implies that $T^{\prime }_{2}$ is continuous, and therefore bounded on every closed and bounded subset of $\mathbb {R}$. Consequently, since the functions $\left( p^{\prime }\right) ^{-1}$, $\min \left( N,.\right)$ and $\max \left( 0,.\right)$ are continuous, it follows from Lebesgue’s Dominated Convergence Theorem [12, Th. 2.4.4] that $\Psi$ is a continuous function of $\lambda$. Furthermore, Assumption 5.5 implies that

$$\begin{aligned} \underset{\lambda \rightarrow 0}{\lim }\Psi \left( \lambda \right) = \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( 0\right) dt = 0. \end{aligned}$$

and

$$\begin{aligned} \underset{\lambda \rightarrow +\infty }{\lim }\Psi \left( \lambda \right) = \int _{0}^{1} T_{2}^{\prime }\left( 1-t\right) p\left( N\right) dt = p\left( N\right) \left[ T_{2}\left( 1\right) - T_{2}\left( 0\right) \right] = p\left( N\right) . \end{aligned}$$

Therefore, by Assumption 5.6 and by the Intermediate Value Theorem [32, Theorem 4.23], there exists some $\lambda ^{*} > 0$ such that $\Psi \left( \lambda ^{*}\right) = P_{0}$.

Finally, the fact that the cost-efficient claim $Y^{*} = f^{*}_{\lambda ^{*}}\left( 1-F_{X}\left( X\right) \right)$ given in Theorem 5.9 is anti-comonotonic with X is an immediate consequence of Lemma 9.5 and the fact that the function $\left( p^{\prime }\right) ^{-1}$ is decreasing. $\square$