Minimizing the Probability of Absolute Ruin Under Ambiguity Aversion

Abstract

In this paper, we consider an optimal robust reinsurance problem in a diffusion model for an ambiguity-averse insurer, who worries about ambiguity and aims to minimize the robust value involving the probability of absolute ruin and a penalization of model ambiguity. It is assumed that the insurer is allowed to purchase per-claim reinsurance to transfer its risk exposure, and that the reinsurance premium is computed according to the mean-variance premium principle which is a combination of the expected-value and variance premium principles. The optimal reinsurance strategy and the associated value function are derived explicitly by applying stochastic dynamic programming and by solving the corresponding boundary-value problem. We prove that there exists a unique point of inflection which relies on the penalty parameter greatly such that the robust value function is strictly concave up to the unique point of inflection and is strictly convex afterwards. It is also interesting to observe that the expression of the optimal robust reinsurance strategy is independent of the penalty parameter and coincides with the one in the benchmark case without ambiguity. Finally, some numerical examples are presented to illustrate the effect of ambiguity aversion on our optimal results.

Appendix

1.1 A. The proof of Theorem 1

Proof

The proof of verification theorem is a modified version of Theorem 3.1 in Luo et al. [11] (also see Theorem 6.1 in Bayraktar and Zhang [8]). Firstly, under the optimal reinsurance policy $R^*$, the surplus process given in (6) becomes

$$\begin{aligned} d\hat{U}^{R^*}_t= & {} \big (r \hat{U}^{R^*}_t - \kappa + \theta \lambda \mathbb {E}\big (R^*_t \big ) + \eta \lambda \mathbb {E}\big (YR^*_t \big ) - \dfrac{\eta }{2} \, \lambda \mathbb {E}\big ((R_t^*)^2 \big ) \big )dt \nonumber \\&+\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2 \big )} \, dB_t. \end{aligned}$$

(67)

Let $\mathbb {Q}\in \mathcal {Q}$ be an arbitrary probability measure with a corresponding probability distortion process $\phi$. By the Girsanov theorem, $B^{\mathbb {Q}}_t=B_t-\int _0^t\phi _sds$ is a $\mathbb {Q}$-Brownian motion. Then we have

$$\begin{aligned}\begin{array}{lll}d\hat{U}^{R^*}_t = \Big (r \hat{U}^{R^*}_t - \kappa + \theta \lambda \mathbb {E}\big (R^*_t \big ) + \eta \lambda \mathbb {E}\big (YR^*_t \big ) - \dfrac{\eta }{2} \, \lambda \mathbb {E}\big ( (R^*_t) ^2\big )\\ ~~~~~~~~~~~+\phi _t\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }\,\Big )dt +\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2 \big )} \, dB^{\mathbb {Q}}_t.\end{array}\end{aligned}$$

Applying Ito’s Lemma to $W(\hat{U}^{R^*}_t)$, we have

$$\begin{aligned} \begin{aligned} dW(\hat{U}^{R^*}_{t})&=\Big [\Big (r \hat{U}^{R^*}_t - \kappa + \theta \lambda \mathbb {E}\big (R^*_t \big ) + \eta \lambda \mathbb {E}\big (YR^*_t \big ) - \dfrac{\eta }{2} \, \lambda \mathbb {E}\big ((R^*_t)^2\big ) +\phi _t\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }\,\Big )\cdot \\&\quad W_u(\hat{U}^{R^*}_{t}) +\displaystyle \frac{1}{2}\lambda \mathbb {E}\big ((R^*_t)^2\big ) W_{uu}(\hat{U}^{R^*}_{t})\Big ]dt+\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }W_u(\hat{U}^{R^*}_{t}) \, dB^{\mathbb {Q}}\\&=\mathcal {A}^{R^*,\phi } W\big (\hat{U}^{R^*}_{t}\big )\,dt+\displaystyle \frac{1}{2\epsilon }\phi _t^2\,dt+\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }W_u(\hat{U}^{R^*}_{t}) \, dB^{\mathbb {Q}}_t. \end{aligned} \end{aligned}$$

By Lemma 1, we have $\tau ^R_{M,N}<\infty$ almost surely for any $-\infty<M<N<u_s$. Integrating the above equation form 0 to $\tau ^{R^*}_{M,N}$ and taking $\mathbb {Q}$-expectation on both sides, we get

$$\begin{aligned} \begin{aligned} \mathbb {E}^\mathbb {Q}\Big (W(\hat{U}^{R^*}_{\tau ^{R^*}_{M,N}})\Big )-W(u)&=\mathbb {E}^\mathbb {Q}\displaystyle \int _0^{\tau ^{R^*}_{M,N}}\mathcal {A}^{R^*,\phi } W\big (\hat{U}^{R^*}_{t}\big )dt+\mathbb {E}^\mathbb {Q}\displaystyle \int _0^{\tau ^{R^*}_{M,N}}\displaystyle \frac{1}{2\epsilon }\phi _t^2\,dt\\&\quad +\mathbb {E}^\mathbb {Q}\displaystyle \int _0^{\tau ^{R^*}_{M,N}}\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }W_u(\hat{U}^{R^*}_{t}) \, dB^{\mathbb {Q}}_t. \end{aligned} \end{aligned}$$

Note that the expectation of the third intergral equals 0 because $W_u$ and $\sqrt{\lambda \mathbb {E}\big ((R^*_t)^2\big ) }$ are bounded. Besides, conditions (ii) and (iv) imply that

$$\begin{aligned} 0=\inf \limits _{R \in \mathcal {D}} \sup \limits _{\phi \in \mathcal {R}}\left\{ \mathcal {A}^{R^*,\phi } W(u)\right\} =\sup _{\phi \in \mathcal {R}}\mathcal {A}^{R^*,\phi } W(\hat{U}^{R^*}_t)\ge \mathcal {A}^{R^*,\phi } W(\hat{U}^{R^*}_t). \end{aligned}$$

(68)

Therefore, we obtain

$$\begin{aligned} W(u)\ge \mathbb {E}^\mathbb {Q}\Big (W(\hat{U}^{R^*}_{\tau ^{R^*}_{M,N}})\Big )-\mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R^*}_{M,N}}\displaystyle \frac{1}{2\epsilon }\phi _t^2\,dt. \end{aligned}$$

(69)

We write

$$\begin{aligned} \mathbb {E}^\mathbb {Q}\big (W(\hat{U}^{R^*}_{\tau ^{R^*}_{M,N}})\big )=W(M)\mathbb {Q}_u(\tau ^{R^*}_{M}<\tau ^{R^*}_N )+W(N)\mathbb {Q}_u(\tau ^{R^*}_{M}\ge \tau ^{R^*}_N).\end{aligned}$$

(70)

Because $W(u_s)=0,$ combining (69) with (70) and letting $N\rightarrow u_s$ yield

$$\begin{aligned}W(u)\ge W(M)\mathbb {Q}_u( \tau ^{R^*}_{M}<\tau ^{R^*}_{u_s} )-\displaystyle \mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R^*}_{M,u_s}}\frac{1}{2\epsilon }\phi _t^2\,dt.\end{aligned}$$

Since the inequality holds for all $\mathbb {Q}\in \mathcal {Q}$, we have

$$\begin{aligned} \begin{aligned} W(u)&\ge \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( \tau ^{R^*}_{M}<\tau ^{R^*}_{u_s} )-\displaystyle \mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R^*}_{M,u_s}}\frac{1}{2\epsilon }\phi _t^2\,dt\Big \}\\&= \inf \limits _{R\in \mathcal {D}}\sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( \tau ^{R}_{M}<\tau ^{R}_{u_s} )-\displaystyle \mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R}_{M,u_s}}\frac{1}{2\epsilon }\phi _t^2\,dt\Big \}\\&=\inf \limits _{R\in \mathcal {D}} \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( {\tau ^{R}_{M}}<\infty )-\displaystyle \frac{1}{\epsilon }h(\mathbb {Q}_{\tau ^{R}_{M,u_s}}|\mathbb {P}_{\tau ^{R}_{M,u_s}}) \Big \}. \end{aligned} \end{aligned}$$

(71)

The equality in (71) follows from (14) and the fact that $\tau ^R_M=\infty$ if $\tau ^{R}_{u_s}\le \tau ^{R}_M$. Note that

$$\begin{aligned} \begin{aligned} V(u)&=\inf \limits _{R \in \mathcal {D}} \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{\mathbb {Q}_u(\liminf \limits _{t\rightarrow \infty }\hat{U}^{R}_t=-\infty )-\displaystyle \frac{1}{\epsilon }h(\mathbb {Q}_{\tau ^R}|\mathbb {P}_{\tau ^R})\Big \}\\&=\inf \limits _{R \in \mathcal {D}} \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{\lim \limits _{M\rightarrow -\infty }\mathbb {Q}_u(\tau ^R_{M}<\infty )-\displaystyle \frac{1}{\epsilon }h(\mathbb {Q}_{\tau ^R}|\mathbb {P}_{\tau ^R}) \big \}. \end{aligned} \end{aligned}$$

(72)

Letting $M\rightarrow -\infty$ in (71), we get $W(u)\ge V(u).$

We next prove $W(u)\le V(u)$. Let R be any admissible reinsurance policy. Under the optimal strategy $\phi ^*=\phi ^*(R,u)$, there exists a measure $\mathbb {Q}^*\in \mathcal {Q}$ satisfying $\frac{d\mathbb {Q}^*_t}{d\mathbb {P}_t}={\varepsilon (\int _0^t\phi ^*_tdB_s)}$, where $\varepsilon$ denotes the stochastic exponential. It follows from the Girsanov theorem that $B^{\mathbb {Q}^*}_t=B_t-\int _0^t\phi ^*_sds$. Then the surplus process becomes

$$\begin{aligned} d\hat{U}^R_t= & {} \Big (r \hat{U}^R_t - \kappa + \theta \lambda \mathbb {E}\big (R_t \big ) + \eta \lambda \mathbb {E}\big (YR_t \big ) - \dfrac{\eta }{2} \, \lambda \mathbb {E}\big (R^2_t\big ) +\phi ^*_t\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) }\,\Big )dt\\&+\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) } \, dB^{\mathbb {Q}^*}_t. \end{aligned}$$

Applying Ito’s lemma to $W(\hat{U}^R_t)$, we have

$$\begin{aligned} \begin{aligned} dW(\hat{U}^R_{t})&=\Big [\Big (r \hat{U}^R_t - \kappa + \theta \lambda \mathbb {E}\big (R_t \big ) + \eta \lambda \mathbb {E}\big (YR_t \big ) - \dfrac{\eta }{2} \, \lambda \mathbb {E}\big (R^2_t\big ) +\phi ^*_t\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) }\,\Big )W_u(\hat{U}^R_{t})\\&\quad +\displaystyle \frac{1}{2}\lambda \mathbb {E}\big (R^2_t\big ) W_{uu}(\hat{U}^R_{t})\Big ]dt+\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) }W_u(\hat{U}^R_{t}) \, dB^{\mathbb {Q}^*}\\&=\mathcal {A}^{R,\phi ^*} W\big (\hat{U}^R_{t}\big )\,dt+\displaystyle \frac{1}{2\epsilon }(\phi _t^*)^2\,dt+\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) }W_u(\hat{U}^R_{t}) \, dB^{\mathbb {Q}^*}_t. \end{aligned} \end{aligned}$$

Integrating the above equation from 0 to $\tau ^R_{M,N}$ and taking $\mathbb {Q}^*$-expectation on both sides yield

$$\begin{aligned} \begin{aligned} \mathbb {E}^{\mathbb {Q}^*}\Big (W(\hat{U}^R_{\tau _{M,N}})\Big )-W(u)&=\mathbb {E}^{\mathbb {Q}^*}\displaystyle \int _0^{\tau ^R_{M,N}}\mathcal {A}^{R,\phi ^*} W\big ((\hat{U}^R_{t})\big )dt+\mathbb {E}^{\mathbb {Q}^*}\displaystyle \int _0^{\tau ^R_{M,N}}\displaystyle \frac{1}{2\epsilon }(\phi ^*_t)^2\,dt\\&\quad +\mathbb {E}^{\mathbb {Q}^*}\displaystyle \int _0^{\tau ^R_{M,N}}\sqrt{\lambda \mathbb {E}\big (R^2_t\big ) }W_u(\hat{U}^R_{t}) \, dB^{\mathbb {Q}^*}_t. \end{aligned} \end{aligned}$$

Similarly, the expectation of the third integral equals 0 due to the fact that $W_u$ and $\sqrt{\lambda \mathbb {E}\big (R_t\big )^2 }$ are bounded. By conditions (ii), (iv) and our definition of $\phi ^*$, we have

$$\begin{aligned} 0=\inf \limits _{R \in \mathcal {D}} \sup \limits _{\phi \in \mathcal {R}}\left\{ \mathcal {A}^{R,\phi } W(u)\right\} \le \sup \limits _{\phi \in \mathcal {R}}\mathcal {A}^{R,\phi } W(\hat{U}^R_t)= \mathcal {A}^{R,\phi ^*} W(\hat{U}^R_t).\end{aligned}$$

(73)

Therefore, we have

$$\begin{aligned} W(u)\le \mathbb {E}^{\mathbb {Q}^*}\big (W(\hat{U}^R_{\tau ^R_{M,N}})\big )-\mathbb {E}^{\mathbb {Q}^*}\int _0^{\tau ^R_{M,N}}\displaystyle \frac{1}{2\epsilon }(\phi ^*_t)^2\,dt.\end{aligned}$$

(74)

Write

$$\begin{aligned} \mathbb {E}^{\mathbb {Q}^*}\big (W(\hat{U}^R_{\tau ^R_{M,N}})\big )=W(M)\mathbb {Q}^*_u(\tau ^R_{M}<\tau ^R_N )+W(N)\mathbb {Q}^*_u(\tau ^R_{M}\ge \tau ^R_N). \end{aligned}$$

(75)

Because $W(u_s)=0,$ combining (74) with (75) and letting $N\rightarrow u_s$ yield

$$\begin{aligned} \begin{aligned} W(u)&\le W(M)\mathbb {Q}^*_u\big ( \tau ^R_{M}<\tau ^R_{u_s} \big )-\displaystyle \mathbb {E}^{\mathbb {Q}^*}\int _0^{\tau ^R_{M,u_s}}\frac{1}{2\epsilon }(\phi ^*_t)^2\,dt.\\&= \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( \tau ^{R}_{M}<\tau ^{R}_{u_s} )-\displaystyle \mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R}_{M,u_s}}\frac{1}{2\epsilon }\phi _t^2\,dt\Big \}\\&= \sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( {\tau ^{R}_{M}}<\infty )-\displaystyle \frac{1}{\epsilon }h(\mathbb {Q}_{\tau ^{R}_{M,u_s}}|\mathbb {P}_{\tau ^{R}_{M,u_s}}) \Big \}. \end{aligned} \end{aligned}$$

(76)

This hold for any $R\in \mathcal {D}$, so we have

$$\begin{aligned} W(u)\le \inf \limits _{R\in \mathcal {D}}\sup \limits _{\mathbb {Q}\in \mathcal {Q}}\Big \{W(M)\mathbb {Q}_u( \tau ^{R}_{M}<\tau ^{R}_{u_s} )-\displaystyle \mathbb {E}^\mathbb {Q}\int _0^{\tau ^{R}_{M,u_s}}\frac{1}{2\epsilon }\phi _t^2\,dt\Big \}\end{aligned}$$

(77)

Letting $M\rightarrow -\infty$ in (77), it follows from (72) that $W(u)\le V(u)$. Therefore, we conclude that $W(u)=V(u)$. Moreover, repeat the analysis above by using $R^*$ and $\phi ^*$ in (73) and (74), (73) through (77) hold with equality, and hence

$$\begin{aligned} W(u)=V(u)=V^{R^*,\mathbb {Q}^*}(u).\end{aligned}$$

$\square$

1.2 B. The proof of Theorem 2

Proof

Suppose that W is a candidate solution to (19). When $u_1 <u\le u_s$, the function $W_1(u)=W(u)$ satisfies the following ordinary differential equation (ODE):

$$\begin{aligned}(\beta _1^*(u)-\eta )W_u(u)+\epsilon W^2_{u}+W_{uu}(u)=0.\end{aligned}$$

Note that (44) is a more explicit version of (34). By letting $y(u)=W_u$ and rearranging terms, the second order ODE is reduced to

$$\begin{aligned} y^{-2} \,\frac{dy}{du}=(\eta -\beta ^*_1(u))y^{-1}-\epsilon ,\end{aligned}$$

(78)

which is in the form of Bernoulli differential equation. Furthermore, setting $z(u)=1/y(u)$ yields

$$\begin{aligned}\frac{dz}{du}=(\beta ^*_1(u)-\eta )z+\epsilon ,\end{aligned}$$

and thus has the following general solution

$$\begin{aligned}z(u)=\displaystyle e^{\int _{u_1}^u(\beta ^*_1(w)-\eta )dw}\Big (C_1+\epsilon \int _{u_1}^u e^{\int _{u_1}^y(\eta -\beta ^*_1(w)dw)}dy\Big ),\end{aligned}$$

from which we obtain

$$\begin{aligned}W_1(u)=\displaystyle \int _{u_1}^u\frac{e^{\int _{u_1}^y(\eta -\beta ^*_1(w))dw}}{C_1+\epsilon \int _{u_1}^y e^{\int _{u_1}^{v}(\eta -\beta ^*_1(w))dw}dv}dy+C_2.\end{aligned}$$

Here, $C_1$ and $C_2$ are constants to be determined. When $-\infty <u\le u_1$, the function $W_2(u)=W(u)$ satisfies the following ODE:

$$\begin{aligned}\big (ru+c-\lambda \mathbb {E}Y\big )W_u(u)+\frac{1}{2}\lambda \mathbb {E}Y^2\Big (\epsilon W^2_{u}(u)+W_{uu}(u)\Big )=0.\end{aligned}$$

Along the same lines, one can obtain

$$\begin{aligned}W_2(u)=\displaystyle \int _{-\infty }^u\frac{e^{\int _{-\infty }^y-\beta ^*_2(w)dw}}{C_3+\epsilon \int _{-\infty }^y e^{\int _{-\infty }^{v}-\beta ^*_2(w)dw}dv}dy+C_4, \end{aligned}$$

where $C_3$ and $C_4$ are constants to be determined. By using the boundary conditions in (39) and the smooth-fit conditions, that is,

$$\begin{aligned}W_1(u_1)=W_2(u_1),~~~~~~~~~~ W'_1(u)|_{u=u_1}= W_2'(u)|_{u=u1},\end{aligned}$$

we get

$$\begin{aligned} \left\{ \begin{array}{ll}C_4=1,~~~~~C_2=-\displaystyle \int _{u_1}^{u_s}\frac{e^{\int _{u_1}^y(\eta -\beta ^*_1(w))dw}}{C_1+\epsilon \int _{u_1}^y e^{\int _{u_1}^{v}(\eta -\beta ^*_1(w))dw}dv}dy,\\ C_2=\displaystyle \int _{-\infty }^{u_1}\frac{e^{\int _{-\infty }^y-\beta ^*_2(w)dw}}{C_3+\epsilon \int _{-\infty }^y e^{\int _{-\infty }^{v}-\beta ^*_2(w)dw}dv}dy+C_4,\\ \displaystyle \frac{1}{C_1}=\frac{e^{\int _{-\infty }^{u_1}-\beta ^*_2(w)dw}}{C_3+\epsilon \int _{-\infty }^{u_1} e^{\int _{-\infty }^{v}-\beta ^*_2(w)dw}dv}. \end{array}\right. \end{aligned}$$

(79)

Using these results and Lemma 5, we can show that W equals the right-hand side of (43). In particular, due to the fact that $C_1$ satisfies conditions (42), it is straightforward to show that W is a non-increasing function with bounded first derivative.

Besides, because of $\eta -\beta _1^*(u_1)=-\beta ^*_2(u_1)=0,$ it is not difficult to verify that $W''_1(u)|_{u=u_1}= W''_2(u)|_{u=u_1}.$ Therefore, the function W satisfies all the conditions of Theorem  1. As a result, the probability of absolute ruin V(u) is given by (43), the optimal robust reinsurance strategy is given by (45), and the optimal probability distortion function is given by (46). $\square$