Effective Premium Discrimination for Designing Cyber Insurance Policies with Rare Losses

Abstract

Cyber insurance like other types of insurance is a method of risk transfer, where the insured pays a premium in exchange for coverage in the event of a loss. As a result of the reduced risk for the insured and the lack of information on the insurer’s side, the insured is generally inclined to lower its effort, leading to a worse state of security, a common phenomenon known as moral hazard. To mitigate moral hazard, a widely employed concept is premium discrimination, i.e., an agent/insured who exerts higher effort pays less premium. This, however, relies on the insurer’s ability to assess the effort exerted by the insured. In this paper, we study two methods of premium discrimination that rely on two different types of assessment: pre-screening and post-screening. Pre-screening occurs before the insured enters into a contract and can be done at the beginning of each contract period; the result of this process gives the insurer an estimated risk on the insured, which then determines the contract terms. The post-screening mechanism involves at least two contract periods whereby the second-period premium is increased if a loss event occurs during the first period.

Prior work shows that both pre-screening and post-screening are generally effective in mitigating moral hazard and increasing the insured’s effort. The analysis in this study shows, however, that the conclusion becomes more nuanced when loss events are rare. Specifically, we show that post-screening is not effective at all with rare losses, while pre-screening can be an effective method when the agent perceives them as rarer than the insurer does; in this case pre-screening improves both the agent’s effort level and the insurer’s profit.

This work is supported by the NSF under grants CNS-1616575, CNS-1739517, and ARO W911NF1810208.

Appendix

Proof

(Lemma 1). Proof by contradiction. Let $(\hat{\pi }_1,\hat{\pi }_2,\hat{\pi }_3)$ be the solution of optimization problem (8), and assume that the (IR) constraint is not binding at the optimal contract $(\hat{\pi }_1,\hat{\pi }_2,\hat{\pi }_3)$. Because the (IR) constraint is not binding, the insurer can increase her utility by increasing $\hat{\pi }_2, \hat{\pi }_3$ while she keeps $\exp \{\gamma \hat{\pi }_2\} - \exp \{\gamma \hat{\pi }_3\}$ fixed. Therefore, based on (9) the agent’s effort inside the contract does not change, but the insurer’s profit increases. As a result, $(\hat{\pi }_1,\hat{\pi }_2,\hat{\pi }_3)$ is not an optimal contract. This is the contradiction implying that the (IR) constraint is binding.    $\blacksquare$

Proof

(Theorem 1). Proof by contradiction: Assume that $\hat{e} = 0$ and $t=1$ and $\left[ \frac{(\alpha - \gamma c)(\exp \{\gamma l\}-1)}{\gamma c }\right] > 1$. First we show that under these assumptions, $\hat{\pi }_1 = \hat{\pi }_2 = \frac{1}{\gamma } \ln (1-u^o) := w^o$. Because $\hat{e} = 0$ and $t = 1$, the optimization problem for finding $(\hat{\pi }_1,\hat{\pi }_2,\hat{\pi }_3)$ is as follows,

$$\begin{aligned} \begin{array}{ll} \max \nolimits _{\{\pi _1,\pi _2,\pi _3\}} \pi _1 + \pi _2 - 2l\\ s{.}t{.},\\ (IR)~ 1-\exp \{\gamma \pi _1\} +1- \exp \{\gamma \pi _2\} = 2u^o \\ (IC)~ 0 = e^{in}(\pi _1 , \pi _2, \pi _3) \end{array} \end{aligned}$$

(24)

By (IR) constraint we have,

$$\begin{aligned} \frac{1}{\gamma }\ln (2-2 u^o-\exp \{\gamma \pi _1\}) = \pi _2 \end{aligned}$$

(25)

Therefore, we re-write the optimization problem (24) as follows,

$$\begin{aligned} \begin{array}{ll} \max \nolimits _{\{\pi _1,\pi _2,\pi _3 \}} \pi _1 + \frac{1}{\gamma }\ln (2-2 u^o-\exp \{\gamma \pi _1\}) - 2l\\ s{.}t{.},\\ (IC) ~ 0 = e^{in}(\pi _1, \pi _2, \pi _3)\\ \qquad \,\,\frac{1}{\gamma }\ln (2-2 u^o-\exp \{\gamma \pi _1\}) = \pi _2 \end{array} \end{aligned}$$

(26)

Because $\pi _3$ does not appear in the objective function, we first find $\pi _1$ and $\pi _2$ such that they maximize the objective function. Then, we pick $\pi _3$ such that (IC) constraint is satisfied. By the first order optimality condition for the objective function, we have,

$$\begin{aligned} \hat{\pi }_1 = \hat{\pi }_2 = \frac{1}{\gamma } \ln (1-u^o) \end{aligned}$$

(27)

Without loss of generality, we set $\hat{\pi }_3 = \frac{1}{\gamma }\ln (\frac{\alpha - \gamma c}{\alpha } (1-u^o) )$. By (9), $\hat{e}= 0$ (Notice that $\frac{\alpha }{\gamma c}\frac{\exp \{\gamma \hat{\pi }_2\} - \exp \{\gamma \hat{\pi }_3\}}{\exp \{\gamma \hat{\pi }_1\}} = 1$ and a slight decrease in $\hat{\pi }_3$, increases the agent’s effort based on (9)).

Now we show that the decrease in $\hat{\pi }_3$ increases the insurer’s payoff. Notice that a slight decrease in $\hat{\pi }_3$, increases the agent’s effort (based on (9)) and improves agents’ utility and the (IR) constraint is not violated. We write the insurer’s objective function as a function of $\pi _3$. Therefore, we have (derivatives in the following equation are left derivatives),

$$\begin{aligned} h(\pi _3)= & {} \hat{\pi }_1 - p(e^{in}(\hat{\pi }_1,\hat{\pi }_2,\pi _3)) (l-\hat{\pi }_2 ) + (1-p(e^{in}(\hat{\pi }_1,\hat{\pi }_2,\pi _3)) ) \pi _3 - l \nonumber \\ \frac{\partial h}{\partial \pi _3} | _ {\pi _3 = \hat{\pi }_3}= & {} \frac{\partial p(e^{in}(\hat{\pi }_1,\hat{\pi }_2,\pi _3)) }{\partial \pi _3} \cdot (\hat{\pi }_2 - l)\nonumber \\ {}- & {} \frac{\partial p(e^{in}(\hat{\pi }_1,\hat{\pi }_2,\pi _3)) }{\partial \pi _3} \cdot \pi _3 + (1-p(e^{in}(\hat{\pi }_1,\hat{\pi }_2,\pi _3)) ) \nonumber \\ {}= & {} \left( \frac{\partial p( e^{in}(\hat{\pi }_1,\hat{\pi }_2,{\pi }_3))}{\partial \pi _3} | _ {\pi _3 = \hat{\pi }_3} \cdot (-l+\hat{\pi }_2 - \hat{\pi }_3) -(1-p(e^{in}_1(\hat{\pi }_1,\hat{\pi }_2,\hat{\pi }_3)) )\right) \nonumber \end{aligned}$$

Because $\left[ \frac{(\alpha - \gamma c)(\exp \{\gamma l\}-1)}{\gamma c }\right] > 1$, (5) implies that $e^o$ is not zero and $\hat{\pi }_2 = \frac{1}{\gamma }\ln (1-u^o) <l$. Moreover, $\frac{\partial p( e^{in}(\hat{\pi }_1,\hat{\pi }_2,{\pi }_3))}{\partial \pi _3} | _ {\pi _3 = \hat{\pi }_3}> 0$ implies that $\frac{\partial h }{\partial \pi _3} |_{\pi _3 = \hat{\pi }_3} < 0$. Therefore, the decrease in $\hat{\pi }_3$ increases the insurer’s payoff. This is a contradiction and the agent exerts non-zero effort in the optimal contract under given assumptions.   $\blacksquare$

Proof

(Theorem 2). By (14), the agent exerts non-zero effort in a contract if $\beta = c$. If the discount factor $\beta = c$, then any positive number satisfies the (IC) constraint. Therefore, if $\beta =c$, then the desired effort maximizes the insurer’s utility. By (14), we have,

$$\begin{aligned} \overline{e} = \arg \max _{e} w^o- ce -{ t l} \exp \{-\alpha \cdot e \}-\gamma c^2 \sigma ^2 ~ \end{aligned}$$

(28)

By the first order condition of optimality, the solution of above optimization problem is $\overline{e} =( \frac{1}{\alpha } \ln ( \frac{\alpha \cdot t \cdot l}{c}))^+$. Moreover, if $\overline{e}>0$, then the maximum insurer’s profit using pre-screening (i.e., $\beta = c$) is given by,

$$\begin{aligned} \begin{array}{ll} \left\{ w^o -\frac{c}{\alpha } \ln ( \frac{\alpha t l }{c}) - \frac{c}{\alpha }- \frac{\gamma c^2 \sigma ^2}{2}\right\} \end{array} \end{aligned}$$

(29)

Without pre-screening (i.e., $\beta = 0$), the agent exerts zero effort and the insurer’s profit is given by,

$$\begin{aligned} w^o - t\cdot l \end{aligned}$$

(30)

Therefore, the insurer uses pre-screening if and only if,

$$\begin{aligned} \begin{array}{ll} \frac{1}{\alpha } \ln ( \frac{\alpha \cdot t \cdot l}{c}) >0\\ w^o -\frac{c}{\alpha } \ln ( \frac{\alpha t l}{c}) - \frac{c}{\alpha }- \frac{\gamma c^2 \sigma ^2}{2} \ge w^o - t l \end{array} \end{aligned}$$

(31)

In other words, the insurer uses pre-screening and the agent exerts non-zero effort if and only if,

$$\begin{aligned}&\frac{\alpha \cdot t \cdot l}{c} >1\nonumber \\&\sigma ^2 \le \frac{2}{\gamma c^2 } (t l - \frac{c}{\alpha }(1+\ln (\frac{\alpha t l }{c})) \end{aligned}$$

(32)

   $\blacksquare$

Proof

(Theorem 3). Assume $\sigma < \sigma '$.

Let $g(\beta ,e,\sigma ) = \left[ w^o - ce - \frac{\gamma \beta ^2\sigma ^2}{2} - p(e) l\right]$. It is easy to see that $g(\beta ,e,\sigma ') \le g(\beta ,e,\sigma )$. Therefore, we have,

$$\begin{aligned} \max _{\beta ,e, IC~constraint} g(\beta ,e,\sigma ') \le \nonumber \max _{\beta ,e, IC ~constraint} g(\beta ,e,\sigma ) \end{aligned}$$

Therefore, $V(\sigma ') \le V(\sigma )$.   $\blacksquare$

Proof

(Theorem 4).

• By (9), the agent exerts zero effort if $t_a \frac{\alpha }{\gamma c} \frac{\exp \{\gamma \pi _2 \} - \exp \{\gamma \pi _3 \} }{\exp \{\gamma \pi _1\}} \le 1$. Because $t_a$ goes to zero, $t_a\frac{\alpha }{\gamma c} \frac{\exp \{\gamma \pi _2 \} - \exp \{\gamma \pi _3 \} }{\exp \{\gamma \pi _1\}}$ also goes to zero. Therefore, the agent exerts zero effort under any insurance contract.

• Because the agent exerts zero effort inside the optimal contract, his utility is given by,

  $$\begin{aligned} \begin{array}{ll} U^{in}(0,\pi _1,\pi _2,\pi _3) = -\exp \{\gamma \pi \} - t_a \exp \{\gamma \pi _2\} -(1-t_a) \exp \{\gamma \pi _3\} \\ \text {(IR) is binding and } t_a \rightarrow 0 \Rightarrow ~1-\exp \{\gamma \pi _1\}+1 - \exp \{\gamma \pi _3\} = 2u^o \end{array} \end{aligned}$$

  (33)

  Therefore, the insurer’s problem (8) can be written as follows,

  $$\begin{aligned} \begin{array}{ll} \max \nolimits _{\pi _1,\pi _2,\pi _3 } \pi _1 + \pi _3 - 2\cdot l_p\\ s{.}t{.}, \exp \{\gamma \pi _1\} + \exp \{\gamma \pi _3\} = 2 -2u^o \end{array} \end{aligned}$$

  (34)

  or

  $$\begin{aligned} \begin{array}{ll} \max \nolimits _{\pi _1} \pi _1 + \frac{1}{\gamma }\ln (2 -2u^o-\exp \{\gamma \pi _1\} ) - 2 l_p\\ \end{array} \end{aligned}$$

  (35)

  The optimal solution for the above optimization problem is $\pi _1 = \pi _3 = \frac{1}{\gamma } \ln (1-u^o)$ and also the value of $\pi _2$ does not affect insurer’s or agent’s utility and can be any positive value.

   $\blacksquare$

Proof

(Theorem 5). The proof is similar to the proof of Theorem 2 except that we should substitute $l_p$ for $t\cdot l$.    $\blacksquare$

Proof

(Theorem 6). As the (IR) constraint is binding in (23), similar to (14) we can re-write optimization problem (23) as follows,

$$\begin{aligned} \begin{array}{ll} R(\sigma ) =\max \nolimits _{\{\beta ,e, \beta ',e'\}} \left[ w^o - c e +b (e -e' )- \gamma \frac{(\beta -\beta ')^2 \sigma ^2+ (\beta ')^2 \sigma ^2}{2} - p(e') l\right] \\ s{.}t{.}, (IC) (e,e') \in \arg \min \nolimits _{(\tilde{e}\ge \tilde{e}')} \gamma (c-b+\beta '-\beta ) \tilde{e} +\gamma (-\beta '+b) \tilde{e}' \end{array} \end{aligned}$$

(36)

First we show that $\hat{e}= \hat{e}'$. Proof by contradiction. Assume $\hat{e} > \hat{e}' \ge 0$. Then, $\beta ' -\beta = b-c$ since otherwise $\hat{e} = \infty$ or $\hat{e} =0$. As $b\le c$, then the objective function of (36) can be improved by decreasing $\hat{e}$ without violating (IC) constraint. This contradiction shows that $\hat{e} = \hat{e}'$.

By IC constraint, it is easy to see that if $\hat{e}=\hat{e}'>0$, then $\beta =c$, and $\beta ' = b$.

Let $\beta = \bar{\beta } ,e= \bar{e}$ be the solution to (14). According to the IC constraint of (14), two cases can happen:

• (i) $\overline{\beta } = 0$ and $\overline{e} = 0$. Then, $(\beta = \beta ' = e= e'=0)$ satisfies the IC constraint in (36) and is a feasible point. We have,

  $$\begin{aligned} w^o - c \overline{e} - \frac{\gamma \overline{\beta } ^2\sigma ^2}{2} - p(\overline{e})l = \nonumber \\ w^o - c \overline{e}+ b (\overline{e} - \overline{e}) - \gamma \frac{(\overline{\beta }-\overline{\beta }')^2+(\overline{\beta }')^2}{2}\sigma ^2- p(\overline{e})l \end{aligned}$$

  (37)

• (ii) $\overline{\beta }= c$. Then $(\beta =c , \beta ' =b, e = e'=\overline{e})$ is a feasible point for (36) and satisfies the IC constraint. We have,

  $$\begin{aligned} w^o - c \overline{e} - \frac{\gamma c^2\sigma ^2}{2} - p(\overline{e})l \le \nonumber \\ w^o - c\cdot \overline{e} + b(\overline{e}- \overline{e} ) - \gamma \frac{(c-b)^2+b^2}{2}\sigma ^2- p(\overline{e})l \end{aligned}$$

  (38)

Note that in this case $(\beta =c , \beta ' =b, e = e'=\overline{e})$ is the solution to (36).

By (37) and (38) we have, $V(\sigma ) \le R(\sigma )$. Notice that if $b=c$, then (36) and (14) are equivalent and $V(\sigma ) = R(\sigma )$ as $\hat{e}=\hat{e}'$.    $\blacksquare$