Minimizing the penalized probability of drawdown for a general insurance company under ambiguity aversion

Abstract

We consider an optimal robust investment and reinsurance problem for a general insurance company which holds shares of an insurance company and a reinsurance company. It is assumed that the decision-maker is ambiguity-averse and does not have perfect information in drift terms of the investment and insurance risks. To capture the ambiguity aversion in the objective function, the criterion of this paper is to minimize a robust value involving the probability of drawdown and a penalization of model uncertainty. By using the technique of stochastic control theory and solving the corresponding boundary-value problems, the closed-form expressions of the optimal strategies are derived explicitly, and a new verification theorem is proved to show that a non-increasing solution to the Hamilton–Jacobi–Bellman equation is indeed our value function. Moreover, we examine theoretically how the level of ambiguity aversion affects the value function and optimal drift distortion. In the end, some numerical examples are exhibited to illustrate the influence of the different investment patterns on our optimal results.

Appendices

Appendix A Auxiliary functions

The functions $g_{1i}(x,m)$ and $f_{1i}(y)~(i=1,2,3)$ are given by

$$\begin{aligned} \left\{ \begin{aligned} g_{11}(x,m)&=\displaystyle \int _{\kappa m}^{x}\exp \bigg \{\int _{\kappa m}^{y}\xi _3(w)dw\bigg \}dy,\\ g_{12}(x,m)&=\displaystyle \int _{\kappa m}^{\kappa m \vee x_1}\exp \bigg \{\int _{\kappa m}^{y}\xi _3(w)dw\bigg \}dy\\&\quad +\displaystyle \int _{\kappa m\vee x_1}^{x}\exp \bigg \{\bigg (\int _{\kappa m}^{\kappa m\vee x_1}\xi _3(w)+ \int _{\kappa m\vee x_1}^{y}\xi _2(w) \bigg )dw\bigg \}dy,\\ g_{13}(x,m)&=\displaystyle \int _{\kappa m}^{\kappa m \vee x_1}\exp \bigg \{\int _{\kappa m}^{y}\xi _3(w)dw\bigg \}dy\\&\quad +\displaystyle \int _{\kappa m\vee x_1}^{\kappa m\vee x_2}\exp \bigg \{\bigg (\int _{\kappa m}^{\kappa m\vee x_1}\xi _3(w)+\int _{\kappa m\vee x_1}^{y}\xi _2(w)\bigg )dw\bigg \}dy\\&\quad +\displaystyle \int _{\kappa m\vee x_2}^{x}\exp \bigg \{\bigg (\int _{\kappa m}^{\kappa m\vee x_1}\xi _3(w)+ \int _{\kappa m\vee x_1}^{\kappa m\vee x_2}\xi _2(w)+\int _{\kappa m\vee x_2}^{y}\xi _1(w)\bigg )dw\bigg \}dy, \end{aligned}\right. \end{aligned}$$

and

$$\begin{aligned} f_{1i}(y)=\left\{ \begin{array}{l} \displaystyle \kappa \bigg [\frac{1}{g_{1i}(y,y)}+\xi _1(\kappa y)\bigg ], ~ \text {if} \quad x_2<\kappa m,\\ \displaystyle \kappa \bigg [\frac{1}{g_{1i}(y,y)}+\xi _2(\kappa y)\bigg ], ~\text {if} \quad x_1\le \kappa m\le x_2,\\ \displaystyle \kappa \bigg [\frac{1}{g_{1i}(y,y)}+\xi _3(\kappa y)\bigg ], ~\text {if} \quad \kappa m< x_1. \end{array}\right. \end{aligned}$$

The functions $g_3(x,m)$ and $f_3(y)$ are given by

$$\begin{aligned} g_3(x,m)=\displaystyle \int _{\kappa m}^{x}\exp \bigg \{\int _{\kappa m}^{y}\xi _1(w)dw\bigg \}dy, ~~~~~f_3(y)=\displaystyle \kappa \bigg [\frac{1}{g_{3}(y,y)}+\xi _1(\kappa y)\bigg ]. \end{aligned}$$

Appendix B Proof of Lemma 3.1

Define the process $H=\{H(t)\}_{t\ge 0}$ by $H(t)=e^{-\delta \hat{X}(t)}$, in which $\hat{X}(t)=\hat{X}^u(t)$. In the expression of H(t), $\delta$ is a positive constant to be chosen later in the proof. By applying Itô’s formula to H(t), we obtain

$$\begin{aligned} dH(t)&=-\delta e^{-\delta \hat{X}(t)}d\hat{X}(t)+\displaystyle \frac{1}{2}\delta ^2 e^{-\delta \hat{X}(t)}d<\hat{X},\hat{X}>_t\\&=\delta e^{-\delta \hat{X}(t)}\bigg ({\mathcal {K}}(t)dt-(\alpha \pi _1(t)\sigma _1 +\beta \pi _2(t)\sigma _2\rho )dB^{\mathbb {Q}}_1(t)\\&\quad -\beta \pi _2(t)\sigma _2 \sqrt{1-\rho ^2}dB_2^{\mathbb {Q}}(t)-[\alpha q(t)+\beta (1-q(t))]\sigma _3dB^{\mathbb {Q}}_3(t)\bigg ), \end{aligned}$$

in which

$$\begin{aligned} {\mathcal {K}}(t)=\alpha (\eta -\theta )\mu _3-r\hat{X}^{u}(t)+{\mathcal {M}}(t)+{\mathcal {N}}(t), \end{aligned}$$

with

$$\begin{aligned} {\mathcal {M}}(t)&=-\alpha \pi _1(t)(\mu _1-r)-\beta \pi _2(t)(\mu _2-r)-(\alpha q(t)\\&\quad +\beta (1-q(t))) \eta \mu _3-(\alpha \pi _1(t)\sigma _1\\&\quad +\beta \pi _2(t)\sigma _2\rho )\phi _1(t)\\&\quad -\beta \pi _2(t)\sigma _2\sqrt{1-\rho ^2}\phi _2(t)-(\alpha q(t)+\beta (1-q(t)))\sigma _3\phi _3(t), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {N}}(t)= & {} \displaystyle \frac{\delta }{2}\bigg [(\alpha \pi _1(t) \sigma _1+\beta \pi _2(t)\sigma _2\rho )^2+\beta ^2\pi _2(t)^2\sigma _2^2 (1-\rho ^2)\\&+[\alpha q(t)+\beta (1-q(t))]^2\sigma _3^2\bigg ]. \end{aligned}$$

We wish to show that, for $\delta$ large enough, ${\mathcal {K}}(t)\ge \frac{\alpha (\eta -\theta )\mu _3-rb}{2}>0$ for all $0\le t\le \tau ^u_{\kappa b}$ with probability 1. First, notice, for $x\le b$, we have $\alpha (\eta -\theta ) \mu _3-rx\ge \alpha (\eta -\theta )\mu _3-rb$. Next, find $\delta$ such that

$$\begin{aligned} {\mathcal {M}}(t)+{\mathcal {N}}(t)\ge -\frac{\alpha (\eta -\theta )\mu _3-rb}{2}, \end{aligned}$$

or equivalently,

$$\begin{aligned} \frac{\delta }{2}\ge \displaystyle \frac{-{\mathcal {M}}(t) -\frac{\alpha (\eta -\theta )\mu _3-rb}{2}}{[\alpha \pi _1(t) \sigma _1+\beta \pi _2(t)\sigma _2\rho ]^2+\beta ^2\pi _2(t)^2 \sigma _2^2(1-\rho ^2)+[\alpha q(t)+\beta (1-q(t))]^2\sigma _3^2}, \end{aligned}$$

(B.1)

for all $u\in {\mathcal {D}}$. Note that, for $-{\mathcal {M}}(t)-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}\le 0$, we can simply set $\delta =0$ and the inequality (B.1) holds. For $-{\mathcal {M}}(t)-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}>0$, by condition (2.8), we deduce that

$$\begin{aligned}&\frac{-{\mathcal {M}}(t)-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}}{(\alpha \pi _1(t)\sigma _1+\beta \pi _2(t)\sigma _2\rho )^2+\beta ^2\pi _2(t)^2\sigma _2^2(1-\rho ^2)+[\alpha q(t)+\beta (1-q(t))]^2\sigma _3^2}\\&\quad \le \frac{-{\mathcal {M}}(t)-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}}{(\alpha \pi _1(t)\sigma _1+\beta \pi _2(t)\sigma _2\rho )^2+\beta ^2\pi _2(t)^2\sigma _2^2(1-\rho ^2)+(\min \{\alpha ,\beta \})^2\sigma _3^2}\\&\quad \le \frac{\alpha \pi _1(t)(\mu _1-r+\sigma _1 K_{{\mathbb {Q}}})+\beta \pi _2(t)[\mu _2-r+\sigma _2(\rho +\sqrt{1-\rho ^2}) K_{{\mathbb {Q}}}]+(\max \{\alpha ,\beta \})(\eta \mu _3+\sigma _3 K_{{\mathbb {Q}}})-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}}{(\alpha \pi _1(t)\sigma _1+\beta \pi _2(t)\sigma _2\rho )^2+\beta ^2\pi _2(t)^2\sigma _2^2(1-\rho ^2)+(\min \{\alpha ,\beta \})^2\sigma _3^2}\\&\quad \le \frac{\mu _1-r+\sigma _1 K_{{\mathbb {Q}}}}{2\min \{\alpha ,\beta \}\sigma _1\sigma _3}+\frac{(\mu _2-r)-\rho [(\mu _1-r)+\sigma _1K_{{\mathbb {Q}}}]+\sigma _2(\rho +\sqrt{1-\rho ^2})K_{{\mathbb {Q}}}}{2\min \{\alpha ,\beta \}\sigma _2\sigma _3\sqrt{1-\rho ^2}}\\&\qquad +\frac{(\max \{\alpha ,\beta \})(\eta \mu _2+\sigma _2 K_{{\mathbb {Q}}})-\frac{\alpha (\eta -\theta )\mu _3-rb}{2}}{(\min \{\alpha ,\beta \})^2\sigma _3^2}=\frac{\delta _0}{2}. \end{aligned}$$

The inequality (B.1) holds for any $\delta \ge \delta _0$; thus, set $\delta =\delta _0>0$, and for this choice of $\delta$, we have ${\mathcal {K}}(t)\ge \frac{\alpha (\eta -\theta )\mu _3-rb}{2}>0$ for all $0\le t\le \tau ^u_{\kappa b}$ with probability 1.

Assume $\hat{X}(0)=x\in (\kappa m,b)$; otherwise, $\tau ^u_{\kappa b}=0$. Apply Itô’s formula to $e^{-\delta \hat{X}(t)}$ to obtain

$$\begin{aligned} e^{-\delta \hat{X}(\tau ^u_{\kappa b}\wedge t)}-e^{-\delta x}&=\displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t}\delta e^{-\delta \hat{X}(s)}{\mathcal {K}}(s)ds-\int _0^{\tau ^u_{\kappa b} \wedge t}\delta e^{-\delta \hat{X}(s)}(\alpha \pi _1(t)\sigma _1+ \beta \pi _2(t)\sigma _2\rho ) dB^{\mathbb {Q}}_1(t)\\&\quad +\int _0^{\tau ^u_{\kappa b}\wedge t}\delta e^{-\delta \hat{X}(s)} \beta \pi _2(t)\sigma _2\sqrt{1-\rho ^2}dB^{\mathbb {Q}}_2(t)\\&\quad -\int _0^{\tau ^u_{\kappa b}\wedge t}\delta e^{-\delta \hat{X}(s)}[\alpha q(t) +\beta (1-q(t))]\sigma _3dB^{\mathbb {Q}}_3(t). \end{aligned}$$

The integrands of the second and third integrals are bounded, so the integrals’ expectations equal zero; thus, if we take the ${\mathbb {Q}}$-expectation of both sides, we get

$$\begin{aligned} E^{{\mathbb {Q}}}_{x,m}(e^{-\delta \hat{X}(\tau ^u_{\kappa b}\wedge t)}) -e^{-\delta x}=E^{{\mathbb {Q}}}_{x,m}\int _0^{\tau ^u_{\kappa b}\wedge t} \delta e^{-\delta \hat{X}(s)}{\mathcal {K}}(s)ds. \end{aligned}$$

Note that $e^{-\delta b}\le e^{-\delta \hat{X}(\tau ^u_{\kappa b}\wedge t)}\le e^{-\delta \kappa m}$ for all $t\ge 0$ with probability 1, so we have the following sequence of (in)equalities

$$\begin{aligned} e^{-\delta \kappa m}-e^{-\delta x}&\ge E^{{\mathbb {Q}}}_{x,m}(e^{-\delta \hat{X} (\tau ^u_{\kappa b}\wedge t)})-e^{-\delta x}=E^{{\mathbb {Q}}}_{x,m} \displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t}\delta e^{-\delta \hat{X}(s)}{\mathcal {K}}(s)ds\\&\ge \delta e^{-\delta b}\displaystyle \frac{\alpha (\eta -\theta )\mu _3-rb}{2} E^{{\mathbb {Q}}}_{x,m}\displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t}1ds\\&=\delta e^{-\delta b}\displaystyle \frac{\alpha (\eta -\theta )\mu _3-rb}{2} \bigg (E^{{\mathbb {Q}}}_{x,m}\displaystyle \int _0^{\tau ^u_{\kappa b}}1ds \cdot \mathbf {1}_{\{\tau ^u_{\kappa b}\le t\}}+E^{{\mathbb {Q}}}_{x,m} \displaystyle \int _0^{t}1ds\cdot \mathbf {1}_{\{\tau ^u_{\kappa b}>t\}}\bigg ) \\&\ge \delta e^{-\delta b}\displaystyle \frac{\alpha (\eta -\theta )\mu _3-rb}{2}t \cdot {\mathbb {Q}}_{x,m}(\tau ^u_{\kappa b}>t). \end{aligned}$$

By letting $t\rightarrow \infty$ in the last expression, we deduce that $\lim \limits _{t\rightarrow \infty }{\mathbb {Q}}_{x,m} (\tau ^u_{\kappa b}>t)=0$, or equivalently, ${\mathbb {Q}}_{x,m} (\tau ^u_{\kappa b}<\infty )=1$. $\Box$

Appendix C Proof of the verification theorem

Under the optimal measure ${\mathbb {Q}}^*$ (corresponding to $\phi ^*$) and an arbitrary strategy $u=(\pi _1,\pi _2,q)$, the wealth process becomes

$$\begin{aligned} d\hat{X}^{u}(t)&=[r\hat{X}^{u}(t)+\alpha (\theta -\eta )\mu _3 +\alpha \pi _1(t)(\mu _1-r)+\beta \pi _2(t)(\mu _2-r)+(\alpha q(t)+\beta (1-q(t)))\eta \mu _3\\&\quad +(\alpha \pi _1(t)\sigma _1+\beta \pi _2(t)\sigma _2\rho )\phi ^*_1(t)\\&\quad +\beta \pi _2^*(t) \sigma _2\sqrt{1-\rho ^2}\phi ^*_2(t)+(\alpha q(t)+\beta (1-q(t)))\sigma _3\phi ^*_3(t)]dt\\&\quad +(\alpha \pi _1(t)\sigma _1+\beta \pi _2(t)\sigma _2\rho )dB^{{\mathbb {Q}}^*}_1(t)\\&\quad +\beta \pi _2(t)\sigma _2\sqrt{1-\rho ^2}dB^{{\mathbb {Q}}^*}_2(t) +[\alpha q(t)+\beta (1-q(t))]\sigma _3dB^{{\mathbb {Q}}^*}_3(t). \end{aligned}$$

Applying Itô’s formula to W(x, m) and integrating from 0 to $\tau ^u_{\kappa b}\wedge t$, we get

$$\begin{aligned}&W(\hat{X}^{u}(\tau _{\kappa b}\wedge t),M^{u}(\tau ^u_{\kappa b}\wedge t))\\&\quad =W(x,m)+\displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t} {\mathcal {L}}^{\phi ^*,u}W(\hat{X}^{u}(s),M^{u}(s))ds+\frac{1}{2\varepsilon } \int _0^{\tau ^u_{\kappa b}\wedge t}||\phi ^{*}(s)||^2ds\\&\qquad +\displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t}W_x(\hat{X}^{u}(s), M^{u}(s))\\&\qquad \bigg ((\alpha \pi _1(s)\sigma _1+\beta \pi _2(s)\sigma _2\rho ) dB^{{\mathbb {Q}}^*}_1(s)+\beta \pi _2(s)\sigma _2\sqrt{1-\rho ^2}dB^{{\mathbb {Q}}^*}_2(s)\\&\qquad +[\alpha q(s)+\beta (1-q(s))]\sigma _3dB^{{\mathbb {Q}}^*}_3(s)\bigg ) +\displaystyle \int _0^{\tau ^u_{\kappa b}\wedge t}W_m(\hat{X}^{u}(s),M^{u}(s))dM^{u}(s). \end{aligned}$$

Taking ${\mathbb {Q}}^*$-expectation on both sides, the equation can be simplified as

$$\begin{aligned}&E_{x,m}^{{\mathbb {Q}}^*}[W(\hat{X}^{u}(\tau ^u_{\kappa b}\wedge t),M^{u} (\tau ^u_{\kappa b}\wedge t))]-W(x,m)\\&\quad =E_{x,m}^{{\mathbb {Q}}^*}\bigg [\displaystyle \int _0^{\tau ^u_{\kappa b} \wedge t}{\mathcal {L}}^{\phi ^*,u}W(\hat{X}^{u}(s),M^{u}(s))ds\bigg ] +\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}^*}\bigg [\int _0^{\tau ^u_{\kappa b} \wedge t}||\phi ^{*}(s)||^2 ds\bigg ]\\&\qquad +E_{x,m}^{{\mathbb {Q}}^*}\bigg [\int _0^{\tau ^u_{\kappa b}\wedge t}W_m (\hat{X}^{u}(s),M^{u}(s))dM^{u}(s)\bigg ]. \end{aligned}$$

From conditions (vi) and (vii), we have

$$\begin{aligned} {\mathcal {L}}^{\phi ^*,u}W(\hat{X}^{u}(s),M^{u}(s)) \ge \inf _{u\in {\mathcal {D}}}{\mathcal {L}}^{\phi ^*,u}W(\hat{X}^{u}(s),M^{u}(s))=0, \end{aligned}$$

and the third integral is zero almost surely because $dM^{u}(t)$ is non-zero only when $M^{u}(t)=X^{u}(t)$. At the same time, $W_m(m,m)=0$ is given by condition (iii). Here, we admit the fact that $M^{u}(t)$ is a non-decreasing process; hence, the first variation process associated with it is finite almost surely and it comes to a conclusion that the cross variation of $M^{u}(t)$ and $\hat{X}^{u}(t)$ is zero almost surely. Then, we obtain

$$\begin{aligned} E_{x,m}^{{\mathbb {Q}}^*}[W(\hat{X}^{u}(\tau ^u_{\kappa b}\wedge t), M^{u}(\tau ^u_{\kappa b}\wedge t))]-W(x,m)\ge \frac{1}{2\varepsilon } E_{x,m}^{{\mathbb {Q}}^*}\bigg [\int _0^{\tau ^u_{\kappa b}\wedge t}||\phi ^{*}(s)||^2ds\bigg ], \end{aligned}$$

or equivalently,

$$\begin{aligned} W(x,m)\le E_{x,m}^{{\mathbb {Q}}^*}[W(\hat{X}^{u}(\tau ^u_{\kappa b}\wedge t), M^{u}(\tau ^u_{\kappa b}\wedge t))]-\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}^*} \bigg [\int _0^{\tau ^u_{\kappa b}\wedge t}||\phi ^{*}(s)||^2ds\bigg ]. \end{aligned}$$

(C.1)

Because $\tau ^u_{\kappa b}<\infty$ with probability 1, it follows from the boundary conditions and inequality (C.1) that

$$\begin{aligned}&W(\hat{X}^{u}({\tau _{\kappa b}}\wedge t),M^{u}({\tau ^u_{\kappa b}}\wedge t))\\&\quad =\mathbf {1}_{\{\tau ^u_{\kappa b}\le t\}}\cdot W(\hat{X}^{u}({\tau _{\kappa b}}), M^{u}({\tau ^u_{\kappa b}}))+\mathbf {1}_{\{\tau ^u_{\kappa b}> t\}}\cdot W(\hat{X}^{u}(t),M^{u}(t))\\&\quad =\mathbf {1}_{\{\tau _{\kappa }^u<\tau _b^u,\tau ^u_{\kappa b}\le t\}} +\mathbf {1}_{\{\tau ^u_{\kappa b}> t\}}\cdot W(\hat{X}^{u}(t),M^{u}(t)). \end{aligned}$$

Then, we have

$$\begin{aligned}&E_{x,m}^{{\mathbb {Q}}^*}[W(\hat{X}^{u}(\tau ^u_{\kappa b}\wedge t), M^{u}(\tau ^u_{\kappa b}\wedge t))]\\&\quad ={\mathbb {Q}}_{x,m}^*{\{\tau _{\kappa }^u <\tau _b^u,\tau ^u_{\kappa b}\le t\}}+E^{{\mathbb {Q}}^*}_{x,m}[\mathbf {1}_{\{\tau ^u_{\kappa b}> t\}} \cdot W(\hat{X}^{u}(t),M^{u}(t))]. \end{aligned}$$

Substituting the above equation into (C.1) and applying the Dominated Convergence Theorem, we obtain

$$\begin{aligned} W(x,m)\le {\mathbb {Q}}_{x,m}^*(\tau ^u_{\kappa }<\tau _b^u) -\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}^*}\bigg [\int _0^{\tau ^u_{\kappa b}}||\phi ^{*} (s)||^2ds\bigg ]. \end{aligned}$$

Hence, the inequality

$$\begin{aligned} W(x,m)\le \sup _{{\mathbb {Q}}\in {\mathcal {Q}}}\bigg \{{\mathbb {Q}}_{x,m} (\tau ^u_{\kappa }<\tau _b^u)-\frac{1}{2\varepsilon }E^{{\mathbb {Q}}}_{x,m} \bigg [\int _0^{\tau ^u_{\kappa b}}||\phi ^{*}(s)||^2ds\bigg ]\bigg \}, \end{aligned}$$

holds for any $u\in {\mathcal {D}}$. Then

$$\begin{aligned} W(x,m)\le \inf _{u\in {\mathcal {D}}} \sup _{{\mathbb {Q}} \in {\mathcal {Q}}}\bigg \{{\mathbb {Q}}_{x,m}(\tau ^u_{\kappa }<\tau _b^u) -\frac{1}{2\varepsilon }E^{{\mathbb {Q}}}_{x,m}\bigg [\int _0^{\tau ^u_{\kappa b}}|| \phi ^{*}(s)||^2ds\bigg ]\bigg \}=V(x,m). \end{aligned}$$

Under the optimal investment and reinsurance policy $u^*$ and an arbitrary probability measure ${\mathbb {Q}}\in {\mathcal {Q}}$, the wealth process follows

$$\begin{aligned} d\hat{X}^{u^*}(t)&=[r\hat{X}^{u^*}(t)+\alpha (\theta -\eta )\mu _3 +\alpha \pi ^*_1(t)(\mu _1-r)+\beta \pi ^*_2(t)(\mu _2-r)+(\alpha q^*(t)\\&\quad +\beta (1-q^*(t)))\eta \mu _3\\&\quad +(\alpha \pi ^*_1(t)\sigma _1+\beta \pi ^*_2(t)\sigma _2\rho )\phi _1(t) +\beta \pi _2^*(t)\sigma _2\sqrt{1-\rho ^2}\phi _2(t)+(\alpha q^*(t)\\&\quad +\beta (1-q^*(t)))\sigma _3\phi _3(t)]dt\\&\quad +(\alpha \pi ^*_1(t)\sigma _1+\beta \pi ^*_2(t)\sigma _2\rho ) dB^{{\mathbb {Q}}}_1(t)+\beta \pi _2^*(t)\sigma _2\sqrt{1-\rho ^2}dB^{{\mathbb {Q}}}_2(t) +[\alpha q^*(t)\\&\quad +\beta (1-q^*(t))]\sigma _3dB^{{\mathbb {Q}}}_3(t).\\ \end{aligned}$$

Applying Itô’s formula to W(x, m) and integrating from 0 to $\tau ^{u^*}_{\kappa b}\wedge t$, we obtain

$$\begin{aligned}&W(\hat{X}^{u^*}(\tau _{\kappa b}\wedge t),M^{u^*}(\tau ^{u^*}_{\kappa b}\wedge t))\\&\quad =W(x,m)+\displaystyle \int _0^{\tau ^{u^*}_{\kappa b}\wedge t}{\mathcal {L}}^{\phi ,u^*} W(\hat{X}^{u^*}(s),M^{u^*}(s))ds+\frac{1}{2\varepsilon }\int _0^{\tau ^{u^*}_{\kappa b} \wedge t}||\phi ^{*}(s)||^2ds\\&\qquad +\displaystyle \int _0^{\tau ^{u^*}_{\kappa b}\wedge t}W_x(\hat{X}^{u^*}(s),M^{u^*}(s))\\&\qquad \bigg ([\alpha \pi ^*_1(s)\sigma _1+\beta \pi ^*_2(s)\sigma _2\rho ] dB^{{\mathbb {Q}}}_1(s) +\beta \pi ^*_2(s)\sigma _2\sqrt{1-\rho ^2}dB^{{\mathbb {Q}}}_2(s)\\&\qquad +[\alpha q^*(s)+\beta (1-q^*(s))]\sigma _3dB^{{\mathbb {Q}}}_3(s)\bigg ) +\displaystyle \int _0^{\tau ^{u^*}_{\kappa b}\wedge t}W_m(\hat{X}^{u^*}(s),M^{u^*}(s))dM^{u^*}(s). \end{aligned}$$

Taking ${\mathbb {Q}}$-expectation on both sides, the equation can be simplified as

$$\begin{aligned}&E_{x,m}^{{\mathbb {Q}}}[W(\hat{X}^{u^*}(\tau ^{u^*}_{\kappa b}),M^{u^*} (\tau ^{u^*}_{\kappa b}))]-W(x,m)\\&\quad =E_{x,m}^{{\mathbb {Q}}}\bigg [\displaystyle \int _0^{\tau ^{u^*}_{\kappa b} \wedge t}{\mathcal {L}}^{\phi ,u^*}W(\hat{X}^{u^*}(s),M^{u^*}(s))ds\bigg ] +\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}}\bigg [\int _0^{\tau ^{u^*}_{\kappa b} \wedge t}||\phi ^{*}(s)||^2ds\bigg ]\\&\qquad +E_{x,m}^{{\mathbb {Q}}}\bigg [\int _0^{\tau ^{u^*}_{\kappa b}\wedge t} W_m(\hat{X}^{u^*}(s),M^{u^*}(s))dM^{u^*}(s)\bigg ]. \end{aligned}$$

From conditions (vi) and (vii), we have

$$\begin{aligned} {\mathcal {L}}^{\phi ,u^*}W(\hat{X}^{u^*}(s),M^{u^*}(s)) \le \sup _{{\mathbb {Q}}\in {\mathcal {Q}}}{\mathcal {L}}^{\phi ,u^*} W(\hat{X}^{u^*}(s),M^{u^*}(s))=0. \end{aligned}$$

Again, it follows that

$$\begin{aligned} E_{x,m}^{{\mathbb {Q}}}[W(\hat{X}^{u^*}(\tau ^{u^*}_{\kappa b}\wedge t), M^{u^*}(\tau ^{u^*}_{\kappa b}\wedge t))]-W(x,m)\le \frac{1}{2\varepsilon } E_{x,m}^{{\mathbb {Q}}}\bigg [\int _0^{\tau ^{u^*}_{\kappa b}\wedge t}||\phi ^{*}(s)||^2ds\bigg ], \end{aligned}$$

or equivalently,

$$\begin{aligned} W(x,m)\ge E_{x,m}^{{\mathbb {Q}}}[W(\hat{X}^{u^*}(\tau ^{u^*}_{\kappa b}\wedge t), M^{u^*}(\tau _{\kappa b}\wedge t))]-\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}} \bigg [\int _0^{\tau ^{u^*}_{\kappa b}\wedge t}||\phi ^{*}(s)||^2ds\bigg ]. \end{aligned}$$

Passing $t\rightarrow \infty$, we have

$$\begin{aligned} W(x,m)\ge {\mathbb {Q}}_{x,m}(\tau ^{u^*}_{\kappa }<\tau _{b}^{u^*}) -\frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}}\bigg [\int _0^{\tau ^{u^*}_{\kappa b}}|| \phi ^{*}(s)||^2ds\bigg ]. \end{aligned}$$

The inequality holds for all ${\mathbb {Q}}\in {\mathcal {Q}}$. So

$$\begin{aligned} \displaystyle W(x,m)&\ge \sup \limits _{{\mathbb {Q}}\in {\mathcal {Q}}} \bigg \{ {\mathbb {Q}}_{x,m}(\tau ^{u^*}_{\kappa }<\tau _{b}^{u^*}) -\displaystyle \frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}} \bigg [\int _0^{\tau ^{u^*}_{\kappa b}}||\phi ^{*}(s)||^2ds\bigg ]\bigg \}\\&=\inf \limits _{u\in {\mathcal {D}}}\sup \limits _{{\mathbb {Q}}\in {\mathcal {Q}}} \bigg \{ {\mathbb {Q}}_{x,m}(\tau ^{u}_{\kappa }<\tau _{b}^{u}) -\displaystyle \frac{1}{2\varepsilon }E_{x,m}^{{\mathbb {Q}}} \bigg [\int _0^{\tau ^{u}_{\kappa b}}||\phi ^{*}(s)||^2ds\bigg ]\bigg \}=V(x,m). \end{aligned}$$

From the above discussion, it yields that $W(x,m)=V(x,m)$, which completes the proof of verification theorem. $\square$