A Bonus-Malus framework for cyber risk insurance and optimal cybersecurity provisioning

Abstract

The cyber risk insurance market is at a nascent stage of its development, even as the magnitude of cyber losses is significant and the rate of cyber loss events is increasing. Existing cyber risk insurance products as well as academic studies have been focusing on classifying cyber loss events and developing models of these events, but little attention has been paid to proposing insurance risk transfer strategies that incentivise mitigation of cyber loss through adjusting the premium of the risk transfer product. To address this important gap, we develop a Bonus-Malus model for cyber risk insurance. Specifically, we propose a mathematical model of cyber risk insurance and cybersecurity provisioning supported with an efficient numerical algorithm based on dynamic programming. Through a numerical experiment, we demonstrate how a properly designed cyber risk insurance contract with a Bonus-Malus system can resolve the issue of moral hazard and benefit the insurer.

Proofs

Proof of Theorem 4.1

In this proof, we apply the dynamic programming principle and perform backward induction in time to show the optimality of $\pi ^\star$. First, one may check that $d^\star _t,\iota ^\star _t$ are $\mathcal {F}_{t-1}$-measurable, $j^\star _t$ is $\mathcal {F}_t$-measurable, and ${\{\iota ^\star _t=0,j^\star _t=1\}=\emptyset }$ for $t=1,\ldots ,T$. Thus, indeed $\pi ^\star \in \Pi$. For all $\pi =(d_s,\iota _s,j_s)_{s=1:T}\in \Pi$ and $t\in \{0,\ldots ,T\}$, let us define $O_t(\pi )=(\widetilde{d}_s,\widetilde{\iota }_s,\widetilde{j}_s)_{s=1:T}\in \Pi$ as follows:

$$\begin{aligned} \begin{aligned}&\big (b^{O_t(\pi )}_{0},i^{O_t(\pi )}_{0}\big ):=(0,\textrm{no}),\\ \text {for each }s&=1,\ldots ,t, \text { let:}\\&\widetilde{d}_s=d_s,\hspace{2.92cm}\widetilde{\iota }_s=\iota _s, \hspace{2.92cm}\widetilde{j}_s=j_s,\\ \text {for each }s&=t+1,\ldots ,T, \text { let:}\\&\widetilde{d}_s=\widehat{d}_s\big (b_{s-1}^{O_t(\pi )},i_{s-1}^{O_t(\pi )}\big ),\quad \quad \widetilde{\iota }_s=\widehat{\iota }_s\big (b_{s-1}^{O_t(\pi )},i_{s-1}^{O_t(\pi )}\big ),\\&\widetilde{j}_s=\widehat{j}_s\big (b_{s-1}^{O_t(\pi )},i_{s-1}^{O_t(\pi )},W_s\big ). \end{aligned} \end{aligned}$$

(26)

By the definition above, one may check that $O_t(\pi )\in \Pi$ for all $\pi \in \Pi$ and $t=0,\ldots ,T$. In particular, when $t=0$, (26) implies that $O_0(\pi )=\pi ^\star$ for all $\pi \in \Pi$. In addition, notice that $O_{t+s}\big (O_t(\pi )\big )=O_t(\pi )$ for all $\pi \in \Pi$ and $s\ge 0$.

Next, we prove the following statement by induction:

$$\begin{aligned} \begin{aligned} \text {for all }t=0,\ldots ,T,\quad \mathcal {V}_t\big (b_{t}^{O_t(\pi )},i_{t}^{O_t(\pi )}\big )=V^{O_t(\pi )}_t\le V^\pi _t\quad \mathbb {P}\text {-a.s. for all }\pi \in \Pi . \end{aligned} \end{aligned}$$

(27)

To begin, we have by definition that $O_T(\pi )=\pi$ and $\mathcal {V}_T(b^{\pi }_{T},i^{\pi }_{T})=V^{\pi }_T=0$ $\mathbb {P}$-a.s. for all $\pi \in \Pi$. Hence, (27) holds when $t=T$. Now, let us suppose that for some $t\in \{1,\ldots ,T\}$, it holds that $\mathcal {V}_t\big (b_{t}^{O_t(\pi )},i_{t}^{O_t(\pi )}\big )=V^{O_t(\pi )}_t\le V^\pi _t$ $\mathbb {P}$-a.s. for all $\pi \in \Pi$. Let $\pi =(d_s,\iota _s,j_s)_{s=1:T}\in \Pi$ be arbitrary and let $O_{t-1}(\pi ):=(\widetilde{d}_s,\widetilde{\iota }_s,\widetilde{j}_s)_{s=1:T}\in \Pi$. Note that (12) ensures that for every $b\in \mathcal {B},i\in \mathcal {I},w\in \mathcal {W}$,

$$\begin{aligned} \begin{aligned}&g_t(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,w),w) +\mathcal {V}_t\big (f_t(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,w),w)\big )\\&\quad =\min _{j\in \{0,1\}}\Big \{g_t(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),j,w) +\mathcal {V}_t\big (f_t(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),j,w)\big )\Big \}. \end{aligned} \end{aligned}$$

(28)

Combining (28) with the definition of $\widehat{d}_t$ and $\widehat{\iota }_t$ in (11), one can show that for all $b\in \mathcal {B}$, $i\in \mathcal {I}$, $d\in \mathcal {D}$, $\iota \in \{0,1\}$, and $\mathfrak {B}(\mathcal {W})$-measurable $j:\mathcal {W}\rightarrow \{0,1\}$, it holds that

$$\begin{aligned} \begin{aligned}&\mathcal {V}_{t-1}(b,i)\\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,W_t),W_t)\\&\quad +\mathcal {V}_{t}\big (f_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,W_t),W_t)\big )\bigg ]\\&\quad \le e^{-r}\mathbb {E}\bigg [g_{t}(b,i,d,\iota ,j(W_t),W_t)+\mathcal {V}_{t}\big (f_{t}(b,i,d,\iota ,j(W_t),W_t)\big )\bigg ]. \end{aligned} \end{aligned}$$

(29)

By (29), (26), the independence between $\mathcal {F}_{t-1}$ and $\sigma (W_t)$, and the induction hypothesis, it holds $\mathbb {P}$-a.s. that

$$\begin{aligned}&\mathcal {V}_{t-1}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}\big )\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,W_t),W_t)\nonumber \\&\qquad +\mathcal {V}_{t}\big (f_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,W_t),W_t) \big )\bigg ]\bigg |_{\begin{array}{c} b=b^{O_{t-1}(\pi )}_{t-1},\;i=i^{O_{t-1}(\pi )}_{t-1} \end{array}}\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}, \widehat{d}_t(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}), \widehat{\iota }_t(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}),\nonumber \\&\qquad \widehat{j}_t(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},W_t),W_t) +\mathcal {V}_{t}\big (f_{t}(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},\widehat{d}_t(b^{O_{t-1} (\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}),\nonumber \\&\qquad \widehat{\iota }_t(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}), \widehat{j}_t(b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},W_t),W_t)\big )\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}, \widetilde{d}_t,\widetilde{\iota }_t,\widetilde{j}_t,W_t\big )\nonumber \\&\quad +\mathcal {V}_{t}\big (f_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},\widetilde{d}_t, \widetilde{\iota }_t,\widetilde{j}_t,W_t\big )\big )\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}, \widetilde{d}_t,\widetilde{\iota }_t,\widetilde{j}_t,W_t\big ) +\mathcal {V}_{t}\big (b^{O_t(O_{t-1}(\pi ))}_{t},i^{O_t(O_{t-1}(\pi ))}_{t}\big )\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}, \widetilde{d}_t,\widetilde{\iota }_t,\widetilde{j}_t,W_t\big )+V^{O_{t-1}(\pi )}_{t}\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =V^{O_{t-1}(\pi )}_{t-1}. \end{aligned}$$

(30)

Now, let $\pi =(d_s,\iota _s,j_s)_{s=1:T}\in \Pi$ be arbitrary and let $O_{t-1}(\pi ):=(\widetilde{d}_s,\widetilde{\iota }_s,\widetilde{j}_s)_{s=1:T}\in \Pi$. By (26), we have

$$\begin{aligned} \big (b^\pi _{t-1},i^\pi _{t-1}\big )=\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}\big ) =\big (b^{O_t(\pi )}_{t-1},i^{O_t(\pi )}_{t-1}\big )\quad \mathbb {P}\text {-a.s.} \end{aligned}$$

(31)

By (30), (28), (31), the independence between $\mathcal {F}_{t-1}$ and $\sigma (W_t)$, and the induction hypothesis, it holds $\mathbb {P}$-a.s. that

$$\begin{aligned}&V^{O_{t-1}(\pi )}_{t-1}\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [\min _{j\in \{0,1\}}\Big \{g_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),j,W_t)\nonumber \\&\qquad +\mathcal {V}_{t}\big (f_{t}(b,i,\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),j,W_t)\big )\Big \}\bigg ] \bigg |_{\begin{array}{c} b=b^{O_{t-1}(\pi )}_{t-1},\;i=i^{O_{t-1}(\pi )}_{t-1} \end{array}}\nonumber \\&\quad \le e^{-r}\mathbb {E}\bigg [\min _{j\in \{0,1\}}\Big \{g_{t}(b,i,d_t,\iota _t,j,W_t) +\mathcal {V}_{t}\big (f_{t}(b,i,d_t,\iota _t,j,W_t)\big )\Big \}\bigg ]\bigg |_{\begin{array}{c} b=b^{O_{t-1} (\pi )}_{t-1},\;i=i^{O_{t-1}(\pi )}_{t-1} \end{array}}\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [\min _{j\in \{0,1\}}\Big \{g_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1}, d_t,\iota _t,j,W_t\big )\nonumber \\&\qquad +\mathcal {V}_{t}\big (f_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},d_t, \iota _t,j,W_t\big )\big )\Big \}\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad \le e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},{d}_t,{\iota }_t, {j}_t,W_t\big )\nonumber \\&\qquad +\mathcal {V}_{t}\big (f_{t}\big (b^{O_{t-1}(\pi )}_{t-1},i^{O_{t-1}(\pi )}_{t-1},{d}_t,{\iota }_t,{j}_t,W_t\big ) \big )\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{\pi }_{t-1},i^{\pi }_{t-1},{d}_t,{\iota }_t,{j}_t,W_t\big )+\mathcal {V}_{t} \big (b^{O_t(\pi )}_{t},i^{O_t(\pi )}_{t}\big )\bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad \le e^{-r}\mathbb {E}\bigg [g_{t}\big (b^{\pi }_{t-1},i^{\pi }_{t-1},{d}_t,{\iota }_t,{j}_t,W_t\big )+V_t^{\pi } \bigg |\mathcal {F}_{t-1}\bigg ]\nonumber \\&\quad =V_{t-1}^{\pi }. \end{aligned}$$

(32)

Combining (30) and (32), we have shown that $\mathcal {V}_{t-1}\big (b_{t-1}^{O_{t-1}(\pi )},i_{t-1}^{O_{t-1}(\pi )}\big )=V^{O_{t-1}(\pi )}_{t-1}\le V^\pi _{t-1}$ $\mathbb {P}$-a.s. for all $\pi \in \Pi$. By induction, (27) holds for $t=0$. Hence, $\mathcal {V}_0\big (b^{\pi ^\star }_0,i^{\pi ^\star }_0\big )=V^{\pi ^\star }_0\le V^{\pi }_0$ for all $\pi \in \Pi$. Since $(b^{\pi ^\star }_0,i^{\pi ^\star }_0\big )=(0,\textrm{no})$, we have $\mathcal {V}_0\big (b^{\pi ^\star }_0,i^{\pi ^\star }_0\big )=V^{\pi ^\star }_0=\inf _{\pi \in \Pi }V^{\pi }_0=\mathbb {V}_0$. The proof is now complete. $\square$

Proof of Theorem 4.4

To prove statement (i), it suffices to show that Claim 1 and Claim 2 in Remark 4.3 hold true. If these two claims hold, then one can verify that $\widehat{d}_t(b,i),\widehat{\iota }_t(b,i),\widehat{j}_t(b,i,w)$ defined on Line 11 and Line 12 coincide with the definitions (11) and (12), and thus statement (i) holds as a consequence of Theorem 4.1. In Claim 1, by (5) and (6), it holds that

$$\begin{aligned} \begin{aligned}&g_{t}(b,i,d,1,1,w)+\mathcal {V}_{t}\big (f_{t}(b,i,d,1,1,w)\big )\\&\quad<g_{t}(b,i,d,1,0,w)+\mathcal {V}_{t}\big (f_{t}(b,i,d,1,0,w)\big )\\&\quad \Updownarrow \\&\quad \mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,w)),\textrm{on}\big )-\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,w)) <\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big ). \end{aligned} \end{aligned}$$

(33)

Observe that, by the definitions of $\underline{b},\overline{b}$ on Line 4 and the definition of $\alpha _t(b,b')$ on Line 7, we have

$$\begin{aligned} \begin{aligned}&\Big \{c\in \mathbb {R}_+:\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,c),\textrm{on}\big ) -c<\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big )\Big \}\\&\quad =\bigcup _{\underline{b}\le b'\le \overline{b}}\Big \{c\in \mathbb {R}_+:\mathcal {B}\mathcal {M}(b,c) =b',c>\mathcal {V}_{t}(b',\textrm{on})-\mathcal {V}_{t}(\underline{b},\textrm{on})\Big \}\\&\quad =\bigcup _{\underline{b}\le b'\le \overline{b}}\mathcal {L}_t(b,b'). \end{aligned} \end{aligned}$$

(34)

Hence, Claim 1 holds true. In Claim 2, in the case where $\iota =1$, we have, by (5) and (6), that

$$\begin{aligned} \begin{aligned}&\mathbb {E}\bigg [\min _{j\in \{0,1\}}\Big \{g_{t}(b,i,d,\iota ,j,W_t)+\mathcal {V}_{t} \big (f_{t}(b,i,d,\iota ,j,W_t)\big )\Big \}\bigg ]\\&\quad =\mathbb {E}\Big [\beta (d)+\iota p^{\mathcal {B}\mathcal {M}}(b,t)+\delta _{\textrm{in}}(t)\mathbbm {1}_{\{i=\textrm{no},\iota =1\}} +\delta _{\textrm{out}}(t)\mathbbm {1}_{\{i=\textrm{on},\iota =0\}}\\&\qquad +\delta _{\textrm{re}} \mathbbm {1}_{\{i\ne \textrm{on},i\ne \textrm{no},\iota =1\}}+L(d,W_t)\Big ]\\&\qquad +\mathbb {E}\bigg [\min _{j\in \{0,1\}}\Big \{\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,j\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))), \textrm{on}\big )-\iota j\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))\Big \}\bigg ]\\&\quad =\beta (d)+\iota p^{\mathcal {B}\mathcal {M}}(b,t)+\delta _{\textrm{in}}(t)\mathbbm {1}_{\{i=\textrm{no},\iota =1\}} +\delta _{\textrm{out}}(t)\mathbbm {1}_{\{i=\textrm{on},\iota =0\}}\\&\qquad +\delta _{\textrm{re}} \mathbbm {1}_{\{i\ne \textrm{on},i\ne \textrm{no},\iota =1\}}+\mathbb {E}\big [L(d,W_t)\big ]\\&\qquad +\mathbb {E}\Big [\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big ) \wedge \Big (\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))),\textrm{on}\big )\\&\qquad -\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))\Big )\Big ]. \end{aligned} \end{aligned}$$

(35)

Moreover, by the definition of $H_t(b,i,d,\iota )$ on Line 9, the definitions of $\underline{b},\overline{b}$ on Line 4, and the definition of $\alpha _t(b,b')$ on Line 7, we have

$$\begin{aligned}&\mathbb {E}\Big [\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big ) \wedge \Big (\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))), \textrm{on}\big )\nonumber \\&\qquad -\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))\Big )\Big ]\nonumber \\&\quad =\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big )\nonumber \\&\qquad -\mathbb {E}\bigg [\Big (\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))-\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))), \textrm{on}\big )\nonumber \\&\qquad +\mathcal {V}_{t}\big (\mathcal {B}\mathcal {M}(b,0),\textrm{on}\big )\Big )^+\bigg ]\nonumber \\&\quad =\mathcal {V}_{t}(\underline{b},\textrm{on})\nonumber \\&\qquad -\sum _{\underline{b}\le b'\le \overline{b}}\mathbb {E}\bigg [\mathbbm {1}_{\{\mathcal {B}\mathcal {M}(b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t)))=b'\}} \Big (\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d,W_t))\nonumber \\&\qquad -\big [\mathcal {V}_{t}(b',\textrm{on})-\mathcal {V}_{t}(\underline{b},\textrm{on})\big ]\Big )^+\bigg ]\nonumber \\&\quad =H_t(b,i,d,1). \end{aligned}$$

(36)

In the case where $\iota =0$, Claim 2 follows directly from (5) and (6). Thus, Claim 2 holds true.

Now, let us prove statement (ii). Let $(b,i)\in \mathcal {B}\times \mathcal {I}$ be fixed, let $\underline{b},\overline{b}$ be defined by Line 4, and let $\mathcal {L}_t(b,b')$ be defined by Line 7. It follows from (5) that, if $\widehat{\iota }_t(b,i)=0$, then

$$\begin{aligned} \mathbb {P}\big [\big (b^{\pi ^\star }_{t},i^{\pi ^\star }_{t}\big )=\mathcal {B}\mathcal {M}_0(b,i)\big |b^{\pi ^\star }_{t-1} =b,i^{\pi ^\star }_{t-1}=i\big ]=1=P^\star _t\big [(b,i)\rightarrow \mathcal {B}\mathcal {M}_0(b,i)\big ], \end{aligned}$$

thus showing the correctness of Line 20. Now, suppose that $\widehat{\iota }_t(b,i)=1$. Then, by (5), we have that $\mathbb {P}\big [i^{\pi ^\star }_{t}=\textrm{on}\big |b^{\pi ^\star }_{t-1}=b,i^{\pi ^\star }_{t-1}=i\big ]=1$. Let us first examine the case where $b'\ne \underline{b}$. We then have

$$\begin{aligned}&\Big \{\big (b^{\pi ^\star }_{t},i^{\pi ^\star }_{t}\big )=(b',\textrm{on}), \big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i) \Big \}\\&\quad =\Big \{\mathcal {B}\mathcal {M}\Big (b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(d^\star _t,W_t))\Big )=b'\Big \}\\&\qquad \cap \big \{j^\star _t=1\big \} \cap \Big \{\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i)\Big \}\\&\quad =\Big \{\mathcal {B}\mathcal {M}\Big (b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t))\Big )=b'\Big \}\\&\qquad \cap \Big \{\widehat{j}_t(b,i,W_t)=1\Big \}\cap \Big \{\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big ) =(b,i)\Big \}\\&\quad =\Big \{\mathcal {B}\mathcal {M}\Big (b,\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t))\Big )=b'\Big \}\\&\qquad \cap \Big \{\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t))\in \textstyle \bigcup _{\underline{b}\le b''\le \overline{b}}\mathcal {L}_t(b,b'')\Big \} \\&\qquad \cap \Big \{\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i)\Big \}\\&\quad =\Big \{\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t))\in \mathcal {L}_t(b,b')\Big \} \cap \Big \{\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i)\Big \}, \end{aligned}$$

where the first equality is by (5), the second equality is by (13), the third equality is by Claim 1 in Remark 4.3, and the last equality is by Line 7 and the property that $\{\mathcal {L}_t(b,b''):\underline{b}\le b''\le \overline{b}\}$ are disjoint sets. Therefore, since $\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )$ and $W_t$ are independent, we have for any $b'\ne \underline{b}$ that

$$\begin{aligned}&\mathbb {P}\Big [b^{\pi ^\star }_{t}=b',i^{\pi ^\star }_{t}=\textrm{on},b^{\pi ^\star }_{t-1} =b,i^{\pi ^\star }_{t-1}=i\Big ]\\&\quad =\mathbb {P}\Big [\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t)) \in \mathcal {L}_t(b,b')\Big ]\mathbb {P}\Big [b^{\pi ^\star }_{t-1}=b,i^{\pi ^\star }_{t-1}=i\Big ]. \end{aligned}$$

Hence, by Line 17,

$$\begin{aligned}&\mathbb {P}\Big [b^{\pi ^\star }_{t}=b',i^{\pi ^\star }_{t}=\textrm{on}\Big |b^{\pi ^\star }_{t-1} =b,i^{\pi ^\star }_{t-1}=i\Big ]\\&\quad =\mathbb {P}\Big [\lambda ^{\mathcal {B}\mathcal {M}}(b,t,L(\widehat{d}_t(b,i),W_t)) \in \mathcal {L}_t(b,b')\Big ]\\&\quad =P^\star _t\big [(b,i)\rightarrow (b',\textrm{on})\big ]. \end{aligned}$$

The remaining case where $b'=\underline{b}$ follows from

$$\begin{aligned}&\mathbb {P}\Big [\big (b^{\pi ^\star }_{t},i^{\pi ^\star }_{t}\big )=(\underline{b},\textrm{on})\Big | \big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i)\Big ]\\&\quad =1-\sum _{\underline{b}<b'\le \overline{b}}\mathbb {P}\Big [\big (b^{\pi ^\star }_{t},i^{\pi ^\star }_{t}\big ) =(b',\textrm{on})\Big |\big (b^{\pi ^\star }_{t-1},i^{\pi ^\star }_{t-1}\big )=(b,i)\Big ]\\&\quad =1-\sum _{\underline{b}<b'\le \overline{b}}P^\star _t\big [(b,i)\rightarrow (b',\textrm{on})\big ]\\&\quad =P^\star _t\big [(b,i)\rightarrow (\underline{b},\textrm{on})\big ], \end{aligned}$$

thus verifying the correctness of Line 18. The correctness of Line 21 and Line 23 follows from the definition that $\big (b^{\pi ^\star }_{0},i^{\pi ^\star }_{0}\big )=(0,\textrm{no})$ and basic properties of a finite state Markov chain. The proof of statement (ii) is complete.

Finally, statement (iii) also follows from the basic properties of a finite state Markov chain. The proof is complete. $\square$

Proof of Lemma 5.2

Statement (i) follows by checking the following:

$$\begin{aligned} \begin{aligned} F_{X}(x)=\mathbb {P}[\widetilde{X}\le x|\widetilde{X}>0]=\frac{\mathbb {P}[0<\widetilde{X}\le x]}{\mathbb {P}[\widetilde{X}>0]}={\left\{ \begin{array}{ll} \frac{F_{\widetilde{X}}(x)-F_{\widetilde{X}}(0)}{1-F_{\widetilde{X}}(0)} &{} \text {if }x>0,\\ 0 &{} \text {if }x\le 0. \end{array}\right. } \end{aligned} \end{aligned}$$

Statement (ii) can be verified directly by checking that $\mathbb {P}[X_U\le x]=F_X(x)$ for all $x\in \mathbb {R}$.

Finally, statement (iii) can be derived from (23) as follows:

$$\begin{aligned}&\mathbb {E}\big [(X-\gamma )^+\big ]=\int _{\gamma }^{\infty }(x-\gamma ) \,{F_{X}}\!\left( {\textrm{d}x}\right) \!\\&\quad =\frac{1}{1-F_{\widetilde{X}}(0)}\left[ \int _{\gamma }^{\infty }x\,{F_{\widetilde{X}}}\!\left( {\textrm{d}x}\right) \! -\gamma (1-F_{\widetilde{X}}(\gamma ))\right] \\&\quad =\frac{\varsigma }{1-F_{\widetilde{X}}(0)}\int _{Y_{g,h}^{-1}\left( \frac{\gamma -\mu }{\varsigma }\right) }^{\infty }Y_{g,h}(z)\,{\Phi }\!\left( {\textrm{d}z}\right) \!+\frac{(\mu -\gamma )(1-F_{\widetilde{X}} (\gamma ))}{1-F_{\widetilde{X}}(0)}\\&\quad =\frac{\varsigma }{1-F_{\widetilde{X}}(0)}\frac{1}{g}\int _{Y_{g,h}^{-1} \left( \frac{\gamma -\mu }{\varsigma }\right) }^{\infty }(\exp (gz)-1)\exp \left( \frac{hz^2}{2}\right) \frac{1}{\sqrt{2\pi }}\exp \left( -\frac{z^2}{2}\right) \,\textrm{d}{z}\\&\qquad +\frac{(\mu -\gamma )(1-F_{\widetilde{X}} (\gamma ))}{1-F_{\widetilde{X}}(0)}\\&\quad =\frac{\varsigma }{1-F_{\widetilde{X}}(0)}\frac{1}{g\sqrt{2\pi }}\Bigg [\int _{Y_{g,h}^{-1} \left( \frac{\gamma -\mu }{\varsigma }\right) }^{\infty }\exp \left( -\frac{(1-h)z^2}{2}+gz\right) -\exp \left( -\frac{(1-h)z^2}{2}\right) \,\textrm{d}{z}\Bigg ]\\&\qquad +\frac{(\mu -\gamma )(1-F_{\widetilde{X}}(\gamma ))}{1-F_{\widetilde{X}}(0)}\\&\quad =\frac{\varsigma }{1-F_{\widetilde{X}}(0)}\frac{1}{g\sqrt{2\pi }}\Bigg [\int _{Y_{g,h}^{-1} \left( \frac{\gamma -\mu }{\varsigma }\right) }^{\infty }\exp \left( \frac{g^2}{2(1-h)}\right) \exp \left( -\frac{(1-h)}{2}\left( z-\frac{g}{1-h}\right) ^2\right) \,\textrm{d}{z}\\&\qquad -\int _{Y_{g,h}^{-1} \left( \frac{\gamma -\mu }{\varsigma }\right) }^{\infty }\exp \left( -\frac{(1-h)z^2}{2}\right) \textrm{d}{z}\Bigg ] +\frac{(\mu -\gamma )(1-F_{\widetilde{X}}(\gamma ))}{1-F_{\widetilde{X}}(0)}\\&\quad =\frac{\varsigma }{(1-F_{\widetilde{X}}(0))g\sqrt{1-h}}\Bigg [\exp \left( \frac{g^2}{2(1-h)}\right) \Phi \left( \left( \frac{g}{1-h}-Y_{g,h}^{-1}\left( \tfrac{\gamma -\mu }{\varsigma }\right) \right) \sqrt{1-h}\right) \\&\qquad -\Phi \left( -Y_{g,h}^{-1}\left( \tfrac{\gamma -\mu }{\varsigma }\right) \sqrt{1-h}\right) \Bigg ]+\frac{(\mu -\gamma )(1-F_{\widetilde{X}}(\gamma ))}{1-F_{\widetilde{X}}(0)}, \end{aligned}$$

where the last equality is obtained by noticing that both integrals are Gaussian integrals after a change of variable. The proof is now complete. $\square$