Optimal insurance for a prudent decision maker under heterogeneous beliefs

Abstract

In this paper we extend some of the results in the literature on optimal insurance under heterogeneous beliefs in the presence of the no-sabotage condition, by allowing the likelihood ratio function to be non-monotone. Under the assumption of prudence and a mild smoothness condition on the likelihood ratio function, we first partition the whole domain of loss into disjoint regions and then obtain an explicit parametric form for the optimal indemnity function over each piece, by resorting to the marginal indemnity function formulation. The case where there exists belief singularity between the decision maker and the insurer is also studied. As an illustration, we consider a special case of our setting in which the premium principle is a distortion premium principle. We then obtain a closed-form characterization of the optimal indemnity for the cases where premia are determined by Value-at-Risk and Tail Value-at-Risk. Our study complements the literature and provides new insights into several similar problems.

Appendix

1.1 Proof of Theorem 2.1

Let $J(I):={\mathbb {E}}[u(w-X+I(X)-\Pi )]-\lambda {\mathbb {E}}^{{\mathbb {Q}}}[I(X)]$. Suppose that $I^*$ is the optimal indemnity function to Problem 1b, then for any given $I\in {\mathcal {I}}$, we have $\epsilon I^*+(1-\epsilon )I\in {\mathcal {I}}$ for any $\epsilon \in [0,1]$. Since

$$\begin{aligned} \begin{aligned} \frac{d^2J(\epsilon I^*+(1-\epsilon )I)}{d\epsilon ^2} \, = \, {\mathbb {E}}[u''(w-X+\epsilon I^*(X)+(1-\epsilon )I(X)-\Pi )(I^*(X)-I(X))^2] \, < \, 0, \end{aligned} \end{aligned}$$

the function $J(\epsilon I^*+(1-\epsilon )I)$ is concave with respect to $\epsilon$. As such, $I^*$ is the solution to Problem 1b if and only if

$$\begin{aligned} \begin{aligned}&\frac{dJ(\epsilon I^*+(1-\epsilon )I)}{d\epsilon }\Big |_{\epsilon =1}\ge 0 \\&\quad \Longrightarrow {\mathbb {E}}[u'(w-X+I^*(X)-\Pi )(I^*(X)-I(X))]-\lambda {\mathbb {E}}^{{\mathbb {Q}}}[I^*(X)-I(X)]\ge 0\\&\quad \Longrightarrow {\mathbb {E}}[u'(w-X+I^*(X)-\Pi )I^*(X)]-\lambda {\mathbb {E}}^{{\mathbb {Q}}}[I^*(X)]\\&\quad \ge {\mathbb {E}}[u'(w-X+I^*(X)-\Pi )I(X)]-\lambda {\mathbb {E}}^{{\mathbb {Q}}}[I(X)]. \end{aligned} \end{aligned}$$

This implies that $I^*$ is also the solution to the following problem:

$$\begin{aligned} \max _{I\in {\mathcal {I}}}\ {\mathbb {E}}[u'(w-X+I^*(X)-\Pi )I(X)]-\lambda {\mathbb {E}}^{{\mathbb {Q}}}[I(X)]. \end{aligned}$$

The objective function of the above problem could be further simplified to be

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}[u'(w-X+I^*(X)-\Pi )I(X)]-\lambda \pi (I(X)) \\&\quad =\int _{0}^{M}u'\left( w-x+I^*(x)-\Pi \right) I(x)dF_X^{{\mathbb {P}}}(x)-\lambda \int _{0}^{M}I(x)dF_X^{{\mathbb {Q}}}(x) \\&\quad = \int _{0}^{M}u'\left( w-x+I^*(x)-\Pi \right) \left\{ \int _{0}^{x}\eta (t)dt\right\} dF_X^{{\mathbb {P}}}(x)-\lambda \int _{0}^{M}\left\{ \int _{0}^{x}\eta (t)dt\right\} dF_X^{{\mathbb {Q}}}(x) \\&\quad = \int _{0}^{M}\left\{ \int _{t}^{x}u'(w-x+I^*(x)-\Pi )dF_X^{{\mathbb {P}}}(x)\right\} \eta (t)dt-\lambda \int _{0}^{M}S_X^{{\mathbb {Q}}}(t)\eta (t)dt \\&\quad = \int _{0}^{M}\left\{ \int _{t}^{x}u'(w-x+I^*(x)-\Pi )dF_X^{{\mathbb {P}}}(x)-\lambda S_X^{{\mathbb {Q}}}(t)\right\} \eta (t)dt, \end{aligned} \end{aligned}$$

where the third equation is due to the Fubini’s theorem. Therefore, to maximize the above integral we only need to maximize its integrand function at each $t\in [0,M]$, i.e.

$$\begin{aligned} \eta ^*(t)=\left\{ \begin{aligned}&1,&\quad&L(t;\; I^*)>0, \\&\xi (t),&\quad&L(t;\; I^*)=0, \\&0,&\quad&L(t;\; I^*)<0, \end{aligned} \right. \end{aligned}$$

where $L(t;\; I^*):=\int _{t}^{x}u'(w-x+I^*(x)-\Pi ) \, dF_X^{{\mathbb {P}}}(x)-\lambda S_X^{{\mathbb {Q}}}(t)$ and $\xi (t)\in [0,1]$ is such that $\eta ^*\in {\mathcal {I}}_0$. $\square$

1.2 Proof of Theorem 3.1

For ease of presentation, we simply write $w-x+I^*(x)+\Pi$ as W(x). The first order derivative of $L(t;\; I^*)$ is given by

$$\begin{aligned} L'(t;\; I^*)=-u'\left( W(t)\right) f_X(t)+\lambda f_X^{{\mathbb {Q}}}(t)=f_X(t)\, K(t), \end{aligned}$$

(A.1)

where $K(t):=\lambda LR(t)-u'(W(t))$. The proofs of statements $(1)\sim (6)$ are given in (I)–(VI) below.

(I) Proving (1) is equivalent to proving that ${I^*}'(x)={\mathbbm {1}}_{[t_1,M]}(x)$ for some $t_1\in [x_{m-1},M]$. We first note that $\psi _{\lambda }'(x)> 0$ is equivalent to $LR'(x)< 0$ over $S_{m,1}$. Since $\psi _{\lambda }'(x)>0$, it is impossible for $L(t;\; I^*)=0$ over any sub-intervals within $S_{m,1}$. We next prove through contradiction that $L(t;\; I^*)$ cannot down-cross the t-axis anywhere in $S_{m,1}$. In other words, there does not exist a point $t^*$ such that

$$\begin{aligned} L(t^*;\; I^*)<0 \quad \text {and}\quad L'(t^*;\; I^*)\le 0. \end{aligned}$$

(A.2)

If such $t^*$ exists, then from (A.1) we know that $K(t^*)\le 0$. Moreover,

$$\begin{aligned} K'(t)=\lambda LR'(t)-u''(W(t))({I^*}'(t)-1)<0, \end{aligned}$$

then $K(t)\le 0$ for $t\in [t^*,M]$. This implies that $L'(t;\; I^*)\le 0$ for $t\in [t^*,M]$, which further implies that

$$\begin{aligned} L(t^*;\; I^*)=\int _{t^*}^{M}-L'(t;\; I^*)dt\ge 0. \end{aligned}$$

This contradicts with $L(t^*;\; I^*)<0$. Therefore, such $t^*$ does not exist. Let

$$\begin{aligned} {\mathcal {S}}_1:=\left\{ t\in [x_{m-1},M]: L(t;\; I^*)\ge 0\right\} ,\quad t_1:=\inf {\mathcal {S}}_1, \end{aligned}$$

with the convention $t_1=M$ if ${\mathcal {S}}_1=\varnothing$. Since $L(t;\; I^*)$ cannot down-cross the t-axis, $L(t;\; I^*)<0$ on $[x_{m-1},t_1)$ and $L(t;\; I^*)\ge 0$ on $[t_1,M]$. Applying Theorem 2.1 gives ${I^*}'(x)={\mathbbm {1}}_{[t_1,M]}(x)$.

(II) Proving (2) is equivalent to proving that ${I^*}'(x)={\mathbbm {1}}_{[x_{m-1},t_2)}(x)+I_{\lambda }'(x){\mathbbm {1}}_{[t_2,M]}(x)$ or ${I^*}'(x)=I_{\lambda }'(x){\mathbbm {1}}_{[t_2,M]}(x)$ for some $t_2\in [x_{m-1},M]$. Note that

$$\begin{aligned} \begin{aligned} \psi _{\lambda }'(x)\in [-1,0]&\ \Longrightarrow \ \frac{\lambda LR'(x)}{u''\left( (u')^{-1}(\lambda LR(x))\right) }\in [-1,0] \\&\ \Longrightarrow \ LR'(x)\ge 0,\quad \lambda LR'(x)+u''\left( (u')^{-1}(\lambda LR(x))\right) \le 0. \end{aligned} \end{aligned}$$

We first prove that there does not exist a point $t^*$ as described in (A.2). If such $t^*$ exists, then

$$\begin{aligned} K(t^*)\le 0\ \Longrightarrow \ \lambda LR(t^*)\le u'(W(t^*))\ \Longrightarrow \ (u')^{-1}(\lambda LR(t^*))\ge W(t^*). \end{aligned}$$

As per Theorem 2.1, ${I^*}'(t^*)=0$. As such,

$$\begin{aligned} \begin{aligned} K'(t^*)&=\lambda LR'(t^*)-u''(W(t^*))({I^*}'(t^*)-1)\\&=\lambda LR'(t^*)+u''\left( (u')^{-1}(\lambda LR(t^*))\right) +u''(W(t^*))-u''\left( (u')^{-1}(\lambda LR(t^*))\right) \le 0, \end{aligned} \end{aligned}$$

where the last inequality holds due to $u'''(x)\ge 0$. Thus, $K(t)\le 0$ for $t\in [t^*,M]$. This implies that $L'(t;\; I^*)\le 0$ for $[t^*,M]$ and therefore $L(t^*;\; I^*)=\int _{t^*}^{M}-L'(t;\; I^*)dt\ge 0$, which contradicts the fact that $L(t^*;\; I^*)<0$. Therefore, such $t^*$ does not exist.

Next, we prove that there does not exist a point $t^{**}$ such that

$$\begin{aligned} L(t^{**};\; I^*)>0\quad \text {and}\quad L'(t^{**};\; I^*)\ge 0. \end{aligned}$$

(A.3)

If such $t^{**}$ exists, then as per Theorem 2.1, ${I^*}'(t^{**})=1$. As such

$$\begin{aligned} K'(t^{**})=\lambda LR'(t^{**})-u''(W(t^{**}))({I^*}'(t^{**})-1)=\lambda LR'(t^{**})\ge 0. \end{aligned}$$

This implies that $K(t)\ge 0$ for $t\in [t^{**},M]$ and therefore $L'(t;\; I^*)\ge 0$ over $[t^{**},M]$. This further implies that $L(t^{**};\; I^*)=\int _{t^{**}}^{M}-L'(x;\; I^*)dx\le 0$, which contradicts with $L(t^{**};\; I^*)>0$. Therefore, such $t^{**}$ does not exist.

Based on the above findings, we conclude that: first, the function $L(t;\; I^*)$ can never cross the t-axis as otherwise the point $t^*$ or $t^{**}$ exists; second, if there exists a point ${\tilde{t}}$ such that $L({\tilde{t}};\; I^*)=0$, then $L(t;\; I^*)=0$ over $[{\tilde{t}},M]$ as otherwise the point $t^*$ or $t^{**}$ exists. Let ${\mathcal {S}}_2:=\left\{ t\in [x_{m-1},M]: L(t;\; I^*)=0\right\}$ and

$$\begin{aligned} t_2:=\left\{ \begin{aligned}&\inf {\mathcal {S}}_2,&\quad&\text {if}\ {\mathcal {S}}_2\ne \varnothing ,\\&M,&\quad&\text {if}\ {\mathcal {S}}_2=\varnothing , \end{aligned} \right. \end{aligned}$$

then $L(t;\; I^*)\lessgtr 0$ for $t\in [x_{m-1},t_2)$ and $L(t;\; I^*)=0$ for $t\in [t_2,M]$, which leads to the result of this case under Theorem 2.1.

(III) Proving (3) is equivalent to proving that ${I^*}'(x)={\mathbbm {1}}_{[x_{m-1},t_3]}(x)$ for some $t_3\in [x_{m-1},M]$. First note that

$$\begin{aligned} \begin{aligned} \psi _{\lambda }'(x)<-1&\ \Longrightarrow \ \frac{\lambda LR'(x)}{u''\left( (u')^{-1}(\lambda LR(x))\right) }< -1 \\&\ \Longrightarrow \ LR'(x)>0,\quad \lambda LR'(x)+u''\left( (u')^{-1}(\lambda LR(x))\right) >0, \end{aligned} \end{aligned}$$

and that $L(t;\; I^*)=0$ cannot hold on any sub-intervals of $[x_{m-1},M]$. We next prove that there does not exist a point $t^{**}$ as described in (A.3). If such $t^{**}$ exists, then $K(t^{**})\ge 0$ and

$$\begin{aligned}\begin{aligned} K'(t^{**})=&\lambda LR'(t^{**})-u''(W(t^{**}))({I^*}'(t^{**})-1) \\ =&\lambda LR'(t^{**})>0. \end{aligned} \end{aligned}$$

This implies that $K(t)\ge 0$ for $t\in [t^{**},M]$ and that $L'(t;\; I^*)\ge 0$ for any $t\in [t^{**},M]$. Therefore $L(t;\; I^*)=\int _{t^{**}}^{M}-L'(x;\; I^*)dx\le 0$, which contradicts the fact that $L(t^{**};\; I^*)>0$. This shows that $L(t;\; I^*)$ cannot up-cross the t-axis.

Now let $t_3:=\inf \left\{ t\in [x_{m-1},M]: L(t;\; I^*)<0\right\}$, we have $L(t;\; I^*)\ge 0$ over $[x_{m-1},t_3]$ and $L(t;\; I^*)<0$ over $(t_3,M]$, which leads to the result of this case by applying Theorem 2.1.

To prove statements (4)–(6), first note that

$$\begin{aligned} L(t;\; I^*)=\int _{t}^{M}-L'(x;\; I^*)dx=\int _{t}^{x_i}-L'(x;\; I^*)dx+L(x_i;\; I^*), \end{aligned}$$

where $L'(x;\; I^*)=f_X(x)\, K(t)$ with $K(t):=\lambda LR(x)-u'(W(x))$.

(IV) Proving (4) is equivalent to proving that ${I^*}'(x)={\mathbbm {1}}_{[t_4,t_5]}(x)$ where $x_{i-1}\le t_4\le t_5\le x_i$. We first note that on $[x_{i-1},x_i]$

$$\begin{aligned} K'(t)=\lambda LR'(t)-u''(W(t))({I^*}'(t)-1)<0, \end{aligned}$$

due to $\psi _{\lambda }'(x)<0$. Denote by $t_r$ the root of $K(t)=0$, if $t_r\in (x_{i-1},x_{i})$, then $K(t)\ge 0$ over $[x_{i-1},t_r]$ and $K(t)< 0$ over $(t_r,x_i]$. This implies that $L'(t;\; I^*)\ge 0$ over $[x_{i-1},t_r]$ and $L'(t;\; I^*)<0$ over $(t_r,x_i]$. Since $L(t;\; I^*)=0$ cannot hold on any sub-intervals of $[x_{i-1},x_i]$, $L(t;\; I^*)$ can cross the t-axis at most twice over $[x_{i-1},x_i]$. Let

$$\begin{aligned} t_4:=\inf \left\{ t\in [x_{i-1},x_i]: L(t;\; I^*)\ge 0\right\} \quad \text {and}\quad t_5:=\inf \left\{ t_5\in [t_4,x_i]: L(t;\; I^*)\le 0\right\} , \end{aligned}$$

then $L(t;\; I^*)<0$ for $t\in [x_{i-1},t_4)$, $L(t;\; I^*)\ge 0$ for $t\in [t_4,t_5]$ and $L(t;\; I^*)<0$ for $t\in (t_5,x_i]$. Applying Theorem 2.1 leads to the result of this case.

(V) Proving (5) is equivalent to proving that one of the following cases is true:

• (i) ${I^*}'(x)={\mathbbm {1}}_{[x_{i-1},t_6)\cup (t_7,x_i]}(x)+I'_{\lambda }(x){\mathbbm {1}}_{[t_6,t_7]}(x)$.

• (ii) ${I^*}'(x)={\mathbbm {1}}_{[x_{i-1},t_6)}(x)+I'_{\lambda }(x){\mathbbm {1}}_{[t_6,t_7]}(x)$.

• (iii) ${I^*}'(x)={\mathbbm {1}}_{(t_7,x_i]}(x)+I'_{\lambda }(x){\mathbbm {1}}_{[t_6,t_7]}(x)$.

• (iv) ${I^*}'(x)=I'_{\lambda }(x){\mathbbm {1}}_{[t_6,t_7]}(x)$.

Similar to the proof of (2), on $[x_{i-1},x_i]$ we first show that if there exists a point $t'\in [x_{i-1},x_i)$ such that $L(t';\; I^*)>0$ and $L'(t';\; I^*)\ge 0$, then $K'(t')\ge 0$. This implies that $L'(t;\; I^*)\ge 0$ over $[t',x_{i}]$. This leads to $L(t;\; I^*)>0$ over $[t',x_i]$.

If there exists a point $t''\in [x_{i-1},x_i)$ such that $L(t'';\; I^*)<0$ and $L'(t'';\; I^*)\le 0$, then $K'(t'')\le 0$. This implies that $L'(t;\; I^*)\le 0$ over $[t'',x_{i}]$. This leads to $L(t;\; I^*)<0$ over $[t'',x_i]$.

Next we show that $L(t;\; I^*)=0$ can only hold on at most one sub-interval of $[x_{i-1},x_i]$. Suppose there are two disjoint sub-intervals, e.g. [a, b] and [c, d] where $b<c$, of $[x_{i-1},x_i]$ such that $L(t;\; I^*)=0$ for $t\in [a,b]\cup [c,d]$. Then there must exist a point $t'\in (b,c)$ such that $L(t';\; I^*)>0$ and $L'(t';\; I^*)\ge 0$ or a point $t''\in (b,c)$ such that $L(t'';\; I^*)<0$ and $L'(t'';\; I^*)\le 0$. However, according to the earlier derivation, $L(t;\; I^*)>0$ on $[t',x_i]$ or $L(t;\; I^*)<0$ over $[t'',x_i]$, which contradicts with $L(t;\; I^*)=0$ over $[c,d]\subseteq [t',x_i]$ or $[t'',x_i]$. Suppose $L(t;\; I^*)=0$ over $[t_6,t_7]$ where $x_{i-1}\le t_6\le t_7\le x_i$, then applying Theorem 2.1 leads to the result of this case.

(VI) Proving (6) is equivalent to proving that ${I^*}'(x)={\mathbbm {1}}_{[x_{i-1},t_8]}(x)+{\mathbbm {1}}_{[t_9,x_i]}(x)$ where $x_{i-1}\le t_8\le t_9\le x_i$. We again first note that $L(t;\; I^*)=0$ cannot hold on any sub-intervals of $[x_{i-1},x_i]$. Similar to the proof of (3), if there exists a point $t'''\in [x_{i-1},x_i)$ such that $L'(t''';\; I^*)\ge 0$, then we have $K'(t''')>0$. This implies that $L'(t;\; I^*)\ge 0$ over $[t''',x_i]$. Now let ${\mathcal {S}}_2:=\left\{ t\in [x_{i-1},x_i): L'(t;\; I^*)\ge 0\right\}$ and

$$\begin{aligned} {\hat{t}}:=\left\{ \begin{aligned}&\inf {\mathcal {S}}_2,&\quad&\text {if}\ {\mathcal {S}}_2\ne \varnothing ,\\&x_i,&\quad&\text {if}\ {\mathcal {S}}_2=\varnothing , \end{aligned} \right. \end{aligned}$$

then $L'(t;\; I^*)< 0$ over $[x_{i-1},{\hat{t}})$ and $L'(t;\; I^*)\ge 0$ over $[{\hat{t}},x_i]$. In other words, $L(t;\; I^*)$ can cross the t-axis at most twice. Let

$$\begin{aligned} t_8:=\inf \left\{ t\in [x_{i-1},x_i]: L(t;\; I^*)\le 0\right\} \quad \text {and}\quad t_9:=\inf \left\{ t\in [t_8,x_i]: L(t;\; I^*)\ge 0\right\} , \end{aligned}$$

then $L(t;\; I^*)\ge 0$ for $t\in [x_{i-1},t_8]$, $L(t;\; I^*)<0$ for $t\in (t_8,t_9)$ and $L(t;\; I^*)\ge 0$ for $t\in [t_9,x_i]$. Applying Theorem 2.1 leads to the result of this case, which concludes the proof. $\square$

Remark 1

The assumption $u'''(x)\ge 0$ is a sufficient condition for the monotonicity of K(t) in Cases (2) and (5), with which we are able to understand the general shape of $L(t;\; I^*)$ on that sub-interval. Without this assumption, it would be much more complicated to discuss the monotonicity of K(t) and also the shape of $L(t;\; I^*)$.

1.3 Proof of Proposition 4.1

First of all, since $u'(\cdot )>0$, for any $t\in (\text {VaR}_{\alpha }(X),M]$,

$$\begin{aligned} L(t;\; I^*)=\int _{t}^{M}u'(w-x+I^*(x)-\Pi )dF_X^{{\mathbb {P}}}(x)>0. \end{aligned}$$

Therefore, as per Theorem 2.1, ${I^*}'(x)=1$ over $(\text {VaR}_{\alpha }(X),M]$.

Second, for $t\in [0,\text {VaR}_{\alpha }(X)]$

$$\begin{aligned} L'(t;\; I^*)=-u'(w-t+I^*(t)-\Pi )f_X^{{\mathbb {P}}}(t)<0. \end{aligned}$$

Let $t_0$ be the root of

$$\begin{aligned}\displaystyle \int _{t}^{M}u'(w-x+I^*(x)-\Pi )dF_X^{{\mathbb {P}}}(x)-\lambda \end{aligned}$$

over $[-\infty ,M]$⁷

and define

$$\begin{aligned} \gamma :=\min \left\{ \text {VaR}_{\alpha }(X),\max \left\{ 0,t_0\right\} \right\} , \end{aligned}$$

then $L(t;\; I^*)>0$ over $[0,\gamma )$⁸ and $L(t;\; I^*)<0$ over $(\gamma ,\text {VaR}_{\alpha }(X)]$. As per Theorem 2.1, ${I^*}'(x)=1$ for $x\in [0,\gamma )$ and ${I^*}'(x)=0$ for $x\in (\gamma ,\text {VaR}_{\alpha }(X)]$. Utilizing the basic formula $I^*(x)=\int _{0}^{x}{I^*}'(t)dt$ leads to the result in Proposition 4.1. $\square$

1.4 Proof of Proposition 4.2

First, note that over $[0,\text {VaR}_{\alpha }(X)]$,

$$\begin{aligned} L'(t;\; I^*)=-u'(w-t+I^*(t)-\Pi )f_X^{{\mathbb {P}}}(t)<0. \end{aligned}$$

Therefore, $L(t;\; I^*)$ is strictly decreasing over $[0,\text {VaR}_{\alpha }(X)]$. Let $t_0$ be the root of $\int _{t}^{M}u'(w-x+I^*(x)-\Pi )dF_X^{{\mathbb {P}}}-\lambda$ over $[-\infty ,M]$ and define

$$\begin{aligned} \gamma :=\min \left\{ \text {VaR}_{\alpha }(X),\max \left\{ 0,t_0\right\} \right\} , \end{aligned}$$

then $L(t;\; I^*)>0$ for $t\in [0,\gamma )$ and $L(t;\; I^*)<0$ for $t\in (\gamma ,\text {VaR}_{\alpha }(X)]$. As per Theorem 2.1, we have ${I^*}'(x)=1$ for $x\in [0,\gamma )$ and ${I^*}'(x)=0$ for $x\in (\gamma ,\text {VaR}_{\alpha }(X)]$. This leads to $I^*(x)=\int _{0}^{x}{I^*}'(t)dt=\min \left\{ x,\gamma \right\}$ for $x\in [0,\text {VaR}_{\alpha }(X)]$.

Over $(\text {VaR}_{\alpha }(X),M]$, $L'(t;\; I^*)=-k(t)\, f_X^{{\mathbb {P}}}(t)$ where $k(t):=u'(w-t+I^*(t)-\Pi )-\frac{\lambda }{1-\alpha }$. We next show by contradiction that there does not exist $t_1$ and $t_2$ such that

$$\begin{aligned}&L(t_1;\; I^*)>0,\quad \text {and}\quad L'(t_1;\; I^*)\ge 0,\\&L(t_2;\; I^*)<0,\quad \text {and}\quad L'(t_2;\; I^*)\le 0. \end{aligned}$$

If such $t_1$ exists, then as per Theorem 2.1

$$\begin{aligned} L(t_1;\; I^*)>0\ \Longrightarrow \ {I^*}'(t_1)=1. \end{aligned}$$

Therefore,

$$\begin{aligned} k'(t_1)=u''(w-t_1+I^*(t_1)-\Pi )({I^*}'(t_1)-1)=0. \end{aligned}$$

This implies that $L'(t;\; I^*)\ge 0$ for any $t\in [t_1,M]$. However, note that $L(t_1;\; I^*)=\int _{t_1}^{M}-L'(t;\; I^*)dt\le 0$, which contradicts with $L(t_1;\; I^*)>0$. Thus, $t_1$ does not exist. Similarly, we can prove that $t_2$ does not exist either.

Based on the above findings, $L(t;\; I^*)$ cannot cross the x-axis. Furthermore, if $L(t;\; I^*)=0$ over any sub-intervals of $(\text {VaR}_{\alpha }(X),M]$, then over these sub-intervals

$$\begin{aligned} L'(t;\; I^*)=0\ \Longrightarrow \ I^*(t)=I_{\lambda }(t), \end{aligned}$$

where $I_{\lambda }(t)=t-w+\Pi +(u')^{-1}\left( \frac{\lambda }{1-\alpha }\right)$. This implies that ${I^*}'(t)=I'_{\lambda }(t)$ when $L(t;\; I^*)=0$.

Let

$$\begin{aligned}t_3:=\inf \left\{ t\in (\text {VaR}_{\alpha }(X),M]: L(t;\; I^*)=0\right\} ,\end{aligned}$$

then we have the following two situations:

• (1) $L(t;\; I^*)>0$ for $t\in (\text {VaR}_{\alpha }(X),t_3)$ and $L(t;\; I^*)=0$ for $t\in (t_3,M]$. This leads to ${I^*}'(x)={\mathbbm {1}}_{(\text {VaR}_{\alpha }(X),t_3)}(x)+{I_{\lambda }}'(x)\,{\mathbbm {1}}_{(t_3,M]}(x)={\mathbbm {1}}_{(\text {VaR}_{\alpha }(X),t_3)}(x)+{\mathbbm {1}}_{(t_3,M]}(x)$.

• (2) $L(t;\; I^*)<0$ for $t\in (\text {VaR}_{\alpha }(X),t_3)$ and $L(t;\; I^*)=0$ for $t\in (t_3,M]$. This leads to ${I^*}'(x)={I_{\lambda }}'(x)\,{\mathbbm {1}}_{(t_3,M]}(x)={\mathbbm {1}}_{(t_3,M]}(x)$.

Applying the basic formula $I^*(x)=I^*(\text {VaR}_{\alpha }(X))+\int _{\text {VaR}_{\alpha }(X)}^{x}{I^*}'(t)dt$ leads to

$$\begin{aligned} I^*(x)=I^*(\text {VaR}_{\alpha }(X))+\min \left\{ (x-\text {VaR}_{\alpha })_+,\max \left\{ 0,I_{\lambda }(x){\mathbbm {1}}_{\{x>\text {VaR}_{\alpha }(X)\}}(x)\right\} \right\} \end{aligned}$$

for $x\in (\text {VaR}_{\alpha }(X),M]$. This concludes the proof. $\square$