On coverage limits and deductibles for SAI loss severities

Abstract

This paper studies allocation of coverage limits and deductibles for dependent losses in the frame of utility theory. The optimal allocation of deductibles is derived for SAI losses without frequency impact, and the optimal allocation of coverage limits (deductibles) for SAI loss severities with RWSAI frequencies are proved to be arrayed in ascending (descending) order. Sufficient conditions to exclude the worst allocation of coverage limits are built for comonotonic loss severities with RWSAI frequencies. A real application in house property insurance is presented as well.

Appendix

Before rolling out the proofs of three lemmas, let us present the two inequalities.

• (i) If u is increasing and convex, then, for $\omega _1\ge \nu _1$, $\omega _1\ge \nu _2$, and $\omega _1-\nu _1\ge \nu _2-\omega _2$,

  $$\begin{aligned} u(\omega _1)+u(\omega _2)-u(\nu _1)-u(\nu _2)\ge 0. \end{aligned}$$

  (6.1)

• (ii) If u is increasing and concave, then, for $\omega _1\ge \nu _1$, $\omega _1\le \omega _2$ and $\omega _1-\nu _1\ge \nu _2-\omega _2$,

  $$\begin{aligned} u(\omega _1)+u(\omega _2)-u(\nu _1)-u(\nu _2)\ge 0. \end{aligned}$$

  (6.2)

These two inequalities can be established as follows: Due to the increasing u, we have $u(\omega _1)\ge u(\nu _1)$ for $\omega _1\ge \nu _1$. If $\omega _2\ge \nu _2$, then, we get $u(\omega _2)\ge u(\nu _2)$. By these two facts we invoke (6.1). If $\omega _2<\nu _2$ instead, by $\omega _1-\nu _1\ge \nu _2-\omega _2$, the increasing and convex u implies $u(\omega _1)-u\big (\omega _1-(\nu _2-\omega _2)\big )\ge u(\nu _2)-u(\omega _2)$ and $u\big (\omega _1-(\nu _2-\omega _2)\big )-u(\nu _1)\ge 0$, and then (6.1) follows from adding them up. If $\omega _2\ge \nu _2$, it is easy to get (6.2). If $\omega _2<\nu _2$, then $\omega _1\le \omega _2\le \nu _2$. By $\omega _1-\nu _1\ge \nu _2-\omega _2$, the increasing and concave u implies $u(\omega _1)-u\big (\omega _1-(\nu _2-\omega _2)\big )\ge u(\nu _2)-u(\omega _2)$ and $u\big (\omega _1-(\nu _2-\omega _2)\big )- u(\nu _1)\ge 0$, and thus (6.2) stems from adding up the two inequalities.

1.1 Proof of Lemma 3.1

(i) Denote $\omega _1(z_1,z_2)=z_1(x_1-l_1)_++z_2(x_2-l_2)_+$ and $\nu _1(z_1,z_2)=z_1(x_1-l_2)_++z_2(x_2-l_1)_+$. Note that $\omega _1(z_1,z_3)\ge \omega _1(z_1,z_2)$ for $z_3\ge z_2\ge 0$. Owing to the increasing and convex $x_+$, it holds that, for $l_1\ge l_2$ and $x_2\ge x_1$,

$$\begin{aligned} (x_2-l_2)_+-(x_2-l_1)_+\ge (x_1-l_2)_+-(x_1-l_1)_+\ge 0, \end{aligned}$$

(6.3)

and thus $z_3\big ((x_2-l_2)_+-(x_2-l_1)_+\big )\ge z_1\big ((x_1-l_2)_+-(x_1-l_1)_+\big )$ for $z_3\ge z_1\ge 0$, implying $\omega _1(z_1,z_3)\ge \nu _1(z_1,z_3)$ for $z_3\ge z_2\ge z_1\ge 0$. Since $z_2[(x_2-l_2)_+-(x_2-l_1)_+]$ is increasing in $z_2$, we have, for $z_3\ge z_2\ge z_1\ge 0$,

$$\begin{aligned} \omega _1(z_1,z_3)- \nu _1(z_1,z_3)= & {} z_3[(x_2-l_2)_+-(x_2-l_1)_+] -z_1[(x_1-l_2)_+-(x_1-l_1)_+]\\\ge & {} z_2[(x_2-l_2)_+-(x_2-l_1)_+]-z_1[(x_1-l_2)_+-(x_1-l_1)_+]\\= & {} \omega _1(z_1,z_2)- \nu _1(z_1,z_2). \end{aligned}$$

Thus, by using (6.1) we get, for any increasing and convex u,

$$\begin{aligned}&\alpha _2(z_1,z_3)+\alpha _1(z_1,z_2)-\alpha _2(z_1,z_2)-\alpha _1(z_1,z_3)\\&\quad =u\big (\omega _1(z_1,z_3)\big )+u\big (\nu _1(z_1,z_2)\big )- u\big (\omega _1(z_1,z_2)\big )-u\big (\nu _1(z_1,z_3)\big )\ge 0, \end{aligned}$$

which is equivalent to $\alpha _2(z_1,z_3)-\alpha _1(z_1,z_3)\ge \alpha _2(z_1,z_2)-\alpha _1(z_1,z_2)$, implying that $\alpha _2(z_1,z_2)-\alpha _1(z_1,z_2)$ is increasing in $z_2\in [ z_1,+\infty )$.

By (6.3), it is easy to check that $\omega _1(z_1,z_2)\ge \nu _1(z_1,z_2)$ , $\omega _1(z_1,z_2)\ge \nu _1(z_2,z_1)$ and

$$\begin{aligned} \omega _1(z_1,z_2)-\nu _1(z_1,z_2)= & {} z_2\big ((x_2-l_2)_+-(x_2-l_1)_+\big )-z_1\big ((x_1-l_2)_+-(x_1-l_1)_+\big )\\\ge & {} z_2\big ((x_1-l_2)_+-(x_1-l_1)_+\big )-z_1\big ((x_2-l_2)_+-(x_2-l_1)_+\big )\\= & {} \nu _1(z_2,z_1)-\omega _1(z_2,z_1), \end{aligned}$$

for $z_2\ge z_1\ge 0$. By (6.1), we have, for increasing and convex u,

$$\begin{aligned}&\alpha _2(z_1,z_2)+\alpha _2(z_2,z_1)-\alpha _1(z_1,z_2)-\alpha _1(z_2,z_1)\\&\quad =u\big (\omega _1(z_1,z_2)\big )+u\big (\omega _1(z_2,z_1)\big )-u\big (\nu _1(z_1,z_2)\big )-u\big (\nu _1(z_2,z_1)\big )\ge 0. \end{aligned}$$

(ii) Let us verify the increasing property case by case. For $x_1\le x_2\le l_2\le l_1$, it holds that $\gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)=0$ for $z_2\ge z_1$. For $x_1\le l_2\le x_2\le l_1$, clearly $\gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)=u\big (z_2(x_2-l_2)\big )-u\big (z_1(x_2-l_2)\big )$ is increasing in $z_2\in [ z_1,+\infty )$. For $x_1\le l_2\le l_1\le x_2$, in a similar manner to (i), it can be shown that $u\big (z_2(x_2-l_2)\big )-u\big (z_2(x_2-l_1)\big )$ is increasing in $z_2\in [ z_1,+\infty )$. Then, $\gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)=u\big (z_2(x_2-l_2)\big )-u\big (z_2(x_2-l_1)\big )+u\big (z_1(x_2-l_1)\big )-u\big (z_1(x_2-l_2)\big )$ is increasing in $z_2\in [ z_1,+\infty )$. For $l_2\le x_1\le x_2\le l_1$, it is plain that $\gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)=u\big (z_2(x_2-l_2)\big )+u\big (z_2(x_1-l_2)\big )-u\big (z_1(x_1-l_2)\big )-u\big (z_1(x_2-l_2)\big )$ is increasing in $z_2\in [ z_1,+\infty )$. For $l_2\le x_1\le l_1\le x_2$, similar to (i), $u\big (z_2(x_2-l_2)\big )-u\big (z_1(x_1-l_2)+z_2(x_2-l_1)\big )$ can be shown increasing in $z_2\in [ z_1,+\infty )$. Since $u\big (z_1(x_2-l_1)+z_2(x_1-l_2)\big )$ is increasing in $z_2\in [ z_1,+\infty )$, we conclude that

$$\begin{aligned} \gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)= & {} u\big (z_2(x_2-l_2)\big )+u\big (z_1(x_2-l_1)+z_2(x_1-l_2)\big )\nonumber \\&-u\big (z_1(x_1-l_2)+z_2(x_2-l_1)\big )-u\big (z_1(x_2-l_2)\big ) \end{aligned}$$

is increasing in $z_2\in [ z_1,+\infty )$. For $l_2\le l_1\le x_1\le x_2$, similar to (i), we can show that $u\big (z_1(x_1-l_1)+z_2(x_2-l_2)\big )-u\big (z_1(x_1-l_2)+z_2(x_2-l_1)\big )$ and $u\big (z_1(x_2-l_1)+z_2(x_1-l_2)\big )-u\big (z_1(x_2-l_2)+z_2(x_1-l_1)\big )$ are both increasing in $z_2\in [ z_1,+\infty )$. Consequently, it follows that

$$\begin{aligned} \gamma _2(z_1,z_2)-\gamma _1(z_1,z_2)= & {} u\big (z_1(x_1-l_1)+z_2(x_2-l_2)\big )+u\big (z_1(x_2-l_1)+z_2(x_1-l_2)\big )\\&-u\big (z_1(x_1-l_2)+z_2(x_2-l_1)\big ) -u\big (z_1(x_2-l_2)+z_2(x_1-l_1)\big ) \end{aligned}$$

is increasing in $z_2\in [ z_1,+\infty )$.

(iii) For $b_1\ge b_2$ and $a_1\ge a_2$, the majorization $(b_1,b_2)\preceq _m (a_1,a_2)$ implies $a_1\ge b_1\ge b_2\ge a_2$ and $b_2-a_2=a_1-b_1$, by the increasing and convex $x_+$, it is easy to check that, for $x_2\ge x_1$,

$$\begin{aligned} (x_2-a_2)_+-(x_2-b_2)_+\ge (x_1-b_1)_+-(x_1-a_1)_+\ge 0. \end{aligned}$$

(6.4)

In a similar manner to the proof of (i), one can check that

$$\begin{aligned}&\beta _2(z_1,z_2)-\beta _1(z_1,z_2)\\&\quad =u\big (z_1(x_1-a_1)_++z_2(x_2-a_2)_+\big ) -u\big (z_1(x_1-b_1)_++z_2(x_2-b_2)_+\big ) \end{aligned}$$

is increasing in $z_2\in [ z_1,+\infty )$. For $x_2\ge x_1$, denote

$$\begin{aligned} n(z_1,z_2)=z_1(x_1-a_1)_++z_2(x_2-a_2)_+,\quad m(z_1,z_2)=z_1(x_1-b_1)_++z_2(x_2-b_2)_+.\nonumber \\ \end{aligned}$$

(6.5)

In view of (6.4), it is easy to check that $n(z_1,z_2)\ge m(z_1,z_2)$, $n(z_1,z_2)\ge m(z_2,z_1)$ and $n(z_1,z_2)-m(z_1,z_2)\ge m(z_2,z_1)-n(z_2,z_1)$ for $z_2\ge z_1\ge 0$. By (6.1) and (6.5), we have

$$\begin{aligned}&\beta _2(z_1,z_2)+\beta _2(z_2,z_1)-\beta _1(z_1,z_2)-\beta _1(z_2,z_1)\\&\quad u(n(z_1,z_2))+u(n(z_2,z_1))-u(m(z_1,z_2))-u(m(z_2,z_1))\ge 0, \end{aligned}$$

for any increasing and convex u and $z_2\ge z_1\ge 0$. $\square$

1.2 Proof of Lemma 3.2

It can be proved in a similar way to that of Lemma 3.1. $\square$

1.3 Proof of Lemma 3.3

For $b_1\ge a_1\ge a_2\ge b_2$, $b_1+ b_2=a_1+a_2$ and any $x_1$, the concavity of $x_1\wedge x$ implies

$$\begin{aligned} x_1\wedge a_2-x_1\wedge b_2 \ge x_1 \wedge b_1-x_1 \wedge a_1. \end{aligned}$$

(6.6)

It is easy to check that $x_2\wedge a_2-x_2\wedge b_2$ is increasing in $x_2$ for $a_2\ge b_2$. Then, we have $x_2\wedge a_2-x_2\wedge b_2\ge x_1\wedge a_2-x_1\wedge b_2$. From (6.6) it follows that $x_2\wedge a_2-x_2\wedge b_2 +x_1 \wedge a_1 -x_1 \wedge b_1\ge x_1\wedge a_2-x_1\wedge b_2 +x_1 \wedge a_1 -x_1 \wedge b_1\ge 0$ for $x_2\ge x_1$. Therefore, it holds that

$$\begin{aligned} x_1 \wedge a_1+x_2\wedge a_2\ge x_1 \wedge b_1+x_2\wedge b_2. \end{aligned}$$

(6.7)

Due to (6.6), we have $x_2\wedge a_2-x_2\wedge b_2 \ge x_2 \wedge b_1-x_2 \wedge a_1$ and $x_1\wedge a_1-x_1\wedge b_1 \ge x_1 \wedge b_2-x_1 \wedge a_2$. Adding these two inequalities up, we get

$$\begin{aligned} x_1\wedge a_1+x_2\wedge a_2-[x_1\wedge b_1+x_2\wedge b_2]\ge x_1 \wedge b_2+x_2 \wedge b_1-[x_1 \wedge a_2+x_2 \wedge a_1]. \end{aligned}$$

Since $x\wedge a_1-x\wedge a_2$ is increasing in x for $a_1\ge a_2$, we have, for $x_2\ge x_1$, $x_2\wedge a_1-x_2\wedge a_2\ge x_1\wedge a_1-x_1 \wedge a_2$ and hence $x_1 \wedge a_1+x_2\wedge a_2\le x_1 \wedge a_2+x_2\wedge a_1$. Combining these two inequalities with (6.7) and by (6.2) we get

$$\begin{aligned}&u(x_1 \wedge a_1+x_2\wedge a_2)+u(x_1 \wedge a_2+x_2\wedge a_1)\\&\quad -u(x_1 \wedge b_1+x_2\wedge b_2)-u(x_1 \wedge b_2+x_2\wedge b_1)\ge 0, \end{aligned}$$

for any increasing and concave function u. $\square$