Optimal insurance coverage of low-probability catastrophic risks

Abstract

Catastrophic risks are often characterised by a low probability, a high severity and a large number of affected individuals. Taking these specificities into account, we analyse the capacity of insurance contracts to provide coverage for those risks, independently from the market failures frequently observed in practice. On the demand side, we characterise individual preferences under which the willingness to pay for the coverage of large losses remains significant, although their occurrence probability is very small. On the supply side, the correlation between individual losses affects the insurance pricing through the insurers’ cost of capital. Analysing the interaction between demand and supply yields the key determinants of insurability and of a socially optimal risk sharing strategy.

Appendix: Proofs

1.1 Proof of Proposition 1

We have

$$\begin{aligned} C_{p}^{\prime }(0_+,{\tilde{\ell }})=\frac{u(w)-{\mathbb {E}}u(w-{\tilde{\ell }})}{u^{\prime }(w)}={\mathbb {E}}\left[ \int _{w-{\tilde{\ell }}}^{w}\frac{ u^{\prime }(x)}{u^{\prime }(w)}{\text{d}}x\right] . \end{aligned}$$

Since

$$\begin{aligned} u^{\prime }(x)=u^{\prime }(w)-\int _{x}^{w}u^{\prime \prime }(t){\text{d}}t, \end{aligned}$$

for all $x\in [w-{\bar{L}},w]$, we may write

$$\begin{aligned} C_{p}^{\prime }(0_+,{\tilde{\ell }})\,= & {} {\mathbb {E}}{\tilde{\ell }}-{\mathbb {E}} \left[ \int _{w-{\tilde{\ell }}}^{w}\left[ \int _{x}^{w}\frac{u^{\prime \prime }(t)}{u^{\prime }(w)}{\text{d}}t\right] {\text{d}}x \right] \\\,= & {} {\mathbb {E}}{\tilde{\ell }}+ {\mathbb {E}}\left[ \int _{w- {\tilde{\ell }}}^{w}\left[ \int _{x}^{w}A(t)\frac{u^{\prime }(t)}{u^{\prime }(w)}{\text{d}}t\right] {\text{d}}x \right] , \end{aligned}$$

and thus,

$$\begin{aligned} \theta (0+, {\tilde{\ell }})= & {} \frac{C'_p(0_+,{\tilde{\ell }}) - {\mathbb {E}}{\tilde{\ell }}}{{\mathbb {E}}{\tilde{\ell }}^2} \\= & {} \frac{1}{{\mathbb {E}}{\tilde{\ell }}^{2}}{\mathbb {E}} \left[ \int _{w-{\tilde{\ell }}}^{w} \left[ \int _{x}^{w} A(t) \frac{u'(t)}{u'(w)}{\text{d}}t\right] {\text{d}}x\right] \end{aligned}$$

Integrating by parts three times and making the change of variable $x = w-\ell$ gives

$$\begin{aligned} \theta (0_+,{\tilde{\ell }})= & {} \frac{1}{{\mathbb {E}}{\tilde{\ell }}^{2}}\int _{0}^{{\bar{L}}}\left[ \int _{w-\ell }^{w} (x-w+\ell )A(x)\frac{u^{\prime }(x)}{u^{\prime }(w)}{\text{d}}x\right] f(\ell ) {\text{d}}\ell \end{aligned}$$

(22)

$$\begin{aligned}= & {} \frac{1}{{\mathbb {E}}{\tilde{\ell }}^{2}}\left\{ \int _{w-{\bar{L}}}^{w} (x-w+\ell )A(x)\frac{u^{\prime }(x)}{u^{\prime }(w)}{\text{d}}x \right. \nonumber \\&\qquad \qquad \left. -\int _{0}^{{\bar{L}}} F(\ell )\left[ \int _{w-\ell }^{w} A(x) \frac{u^{\prime }(x)}{u^{\prime }(w)} {\text{d}}x \right] {\text{d}}\ell \right\} \nonumber \\= & {} \frac{1}{{\mathbb {E}}{\tilde{\ell }}^{2}} \left\{ \int _{w-{\bar{L}}}^{w} \left[ x-w+{\bar{L}} - \int _{w-x}^{{\bar{L}}} F(\ell ){\text{d}}\ell \right] A(x) \frac{u'(x)}{u'(w)}{\text{d}}x \right\} . \end{aligned}$$

(23)

Using

$$\begin{aligned} u'(x) = u'(w) \exp \left\{ \int _x^w{A(x) {\text{d}}x} \right\} , \end{aligned}$$

completes the first part of the proof. Integrating by parts twice then gives

$$\begin{aligned} \int _{w-{\bar{L}}}^{w} \left[ x-w+{\bar{L}} - \int _{w-x}^{{\bar{L}}} F(\ell ){\text{d}}\ell \right] {\text{d}}x = \frac{1}{2} {\mathbb {E}}{\tilde{\ell }}^2, \end{aligned}$$

hence

$$\begin{aligned} \int _{w-{\bar{L}}}^{w} k(x) {\text{d}}x = 1. \end{aligned}$$

1.2 Proof of Corollary 1

The corollary follows straightforwardly from

$$\begin{aligned} \exp \left\{ \int _x^w{A(t) {\text{d}}t}\right\} \ge 1. \end{aligned}$$

1.3 Proof of Proposition 2

Let $\nu (x) = x - w + {\bar{L}} - \int _{w-x}^{{\bar{L}}}{F(\ell ) {\text{d}}\ell }> 0$ for all $x \in [w-{\bar{L}},w]$, with $\nu (w) = {\bar{L}} - \int _0^{{\bar{L}}}{F(\ell ){\text{d}}\ell } = {\mathbb {E}}{\tilde{\ell }}$ and $f'({\bar{L}})<0$. Equation (23) is therefore written as

$$\begin{aligned} \theta (0_+,{\tilde{\ell }}) = \frac{1}{{\mathbb {E}}{\tilde{\ell }}^2} \int _{w-{\bar{L}}}^w{\nu (x)A(x) \frac{u'(x)}{u'(w)}{\text{d}}x}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{{\text{d}}[\nu (x) u'(x)]}{{\text{d}}x}= & {} u'(x)\nu '(x) + \nu (x) u''(x) \\= & {} u'(x) \left[ \nu '(x) - \frac{\nu (x)}{x} R(x)\right] \qquad x \in [w-{\bar{L}},w]. \end{aligned}$$

${\text{d}}[\nu (x) u'(x)]/{\text{d}}x >0$ if $\varepsilon (x) \equiv \frac{\nu '(x)x}{\nu (x)}>R(x)$, and

$$\begin{aligned} \varepsilon (x) = \frac{x[1-F(w-x)]}{\int _{w-x}^{{\bar{L}}}{[1-F(\ell )]{\text{d}}\ell }}. \end{aligned}$$

Since ${\mathbb {E}}{\tilde{\ell }} = {\bar{L}} - \int _0^{{\bar{L}}}{F(\ell ){\text{d}}\ell }$, we have $\varepsilon (w) = w/{\mathbb {E}}{\tilde{\ell }}$. Assuming $f(\ell )\rightarrow 0$ when $\ell \rightarrow {\bar{L}}$ and $f'({\bar{L}}) < 0$ and applying L’Hôpital’s rule twice, we obtain $\varepsilon (x) \rightarrow +\infty$ when $x\rightarrow w-{\bar{L}}$. Assuming that $\varepsilon (x)$ is monotonic, and therefore decreasing, implies

$$\begin{aligned} \varepsilon (x)> & {} \varepsilon (w) \qquad \text{ if } \quad x<w. \\= & {} \frac{w}{{\mathbb {E}}{\tilde{\ell }}} \end{aligned}$$

A sufficient condition for ${\text{d}}[\nu (x) u'(x)]/{\text{d}}x >0$ is therefore

$$\begin{aligned} \frac{w}{{\mathbb {E}}{{\tilde{\ell }}}} > {\bar{\gamma }}, \end{aligned}$$

where ${\bar{\gamma }} = \max \{R(x),x\in [w-{\bar{L}},w]\}$. In this case, we have

$$\begin{aligned} \nu (x)u'(x)< & {} \nu (w) u'(w) \qquad \text{ if } \quad x\in [w-{\bar{L}},w] \\\,= & {} {\mathbb {E}}{\tilde{\ell }}u'(w), \end{aligned}$$

and consequently

$$\begin{aligned} \theta (0_+,{\tilde{\ell }})<\frac{{\mathbb {E}}{\tilde{\ell }}}{{\mathbb {E}}{\tilde{\ell }}^2}\int _{w-{\bar{L}}}^w{A(x){\text{d}}x}. \end{aligned}$$

Let ${\bar{A}} = 1/\bar{L}\int _{w-{\bar{L}}}^w{A(x){\text{d}}x}$. We then obtain

$$\begin{aligned} \theta (0_+,{\tilde{\ell }}) < \frac{\bar{L}{\mathbb {E}}{\tilde{\ell }}}{{\mathbb {E}}{\tilde{\ell }}^2} {\bar{A}}. \end{aligned}$$

Using $C_{p}^{\prime \prime }<0$ and $C (0_+,{\tilde{\ell }})=0$ with Eq. (1) allows us to write

$$\begin{aligned} C (p,{\tilde{\ell }})< & {} C^{\prime }(0_+,{\tilde{\ell }})p \\= & {} p{\mathbb {E}}{\tilde{\ell }}[1 + \theta (0_+,{\tilde{\ell }}){\mathbb {E}}{\tilde{\ell }}]\\< & {} p{\mathbb {E}}{\tilde{\ell }}\left[ 1+{\bar{L}}{\bar{A}}\right] . \end{aligned}$$

1.4 Proof of Corollary 3

A simple calculation shows that Proposition 1 and Corollary 1 can be re-written as

Corollary 4

When ${\tilde{\ell }} =L$ with probability 1, we have

$$\begin{aligned} \theta (0_+,{\tilde{\ell }}) =\frac{1}{2} \int _{w-L}^{w}{\left[ k(x) A(x) \exp \left\{ \int _{x}^{w}{A(t){\text{d}}t} \right\} \right] {\text{d}}x} > \frac{1}{2} \int _{w-L}^{w}{k(x) A(x) {\text{d}}x} \end{aligned}$$

where

$$\begin{aligned} k(x) = \frac{2[x- (w-L)]}{L^2} > 0 \end{aligned}$$

and

$$\begin{aligned} \int _{w-L}^{w}k(x){\text{d}}x=1. \end{aligned}$$

Using Lemma 1 then shows that

$$\begin{aligned} d(0_+,L)\le & {} \frac{L}{2} \int _{w-L}^{w}{ \frac{k(x)}{x} R(x){\text{d}}x} \end{aligned}$$

is a sufficient condition for insurance take-up to be positive. If R(x) is non-decreasing, then

$$\begin{aligned} \frac{L}{2}\int _{w-L}^{w}{ \frac{k(x)}{x} R(x){\text{d}}x}\ge & {} \frac{L R(w-L)}{2}\int _{w-L}^{w}{ \frac{k(x)}{x} {\text{d}}x} \\ \,= & {} \frac{ R(w-L)}{L}\int _{w-L}^{w}{ \frac{x-(w-L)}{x} {\text{d}}x} \\ \,= & {} R(w-L) \left[ 1 - \left( \frac{w-L}{L}\right) \ln {\frac{w}{w-L}}\right] \\\equiv & {} \Psi (L) \quad L \in [0,w]. \end{aligned}$$

Noticing that $\lim _{L\rightarrow w}{\psi (L)} = \lim _{x\rightarrow 0}{R(x)}$ provides the result.

1.5 Coefficient of correlation $\rho$

Let ${\tilde{L}}_i$ and ${\tilde{L}}_j$ be two random variables that represent the losses of individuals i and j. Conditionally on a realisation $\kappa$ of the random variable ${\tilde{\kappa }}$, losses are assumed identically and independently distributed, hence

$$\begin{aligned} \tilde{L_i}\tilde{L_j} | \kappa = \left\{ \begin{array}{cl} L^2 &{} \text{ with } \text{ probability } \quad \pi \kappa ^2 \\ 0 &{} \text{ with } \text{ probability } \quad 1 - \pi \kappa ^2 \end{array} \right. . \end{aligned}$$

As a consequence ${\mathbb {E}}(\tilde{L_i} \tilde{L_j}|\kappa ) = L^2 \pi \kappa ^2$ and ${\mathbb {E}}(\tilde{L_i} \tilde{L_j}) = L^2 \pi {\mathbb {E}}{\tilde{\kappa }}^2$. Similarly,

$$\begin{aligned} \tilde{L_i} | \kappa = \left\{ \begin{array}{cl} L &{} \text{ with } \text{ probability } \quad \pi \kappa \\ 0 &{} \text{ with } \text{ probability } \quad 1 - \pi \kappa \end{array} \right. , \end{aligned}$$

for all i, hence ${\mathbb {E}}\tilde{L_i} = L\pi \mu _{\kappa }$ and ${\mathbb {E}}\tilde{L_i} {\mathbb {E}}\tilde{L_j}= (L\pi \mu _{\kappa })^2$. The co-variance between two losses is therefore written as

$$\begin{aligned} \text{ cov }(\tilde{L_i},\tilde{L_j}) = L^2\pi [ {\mathbb {E}}{\tilde{\kappa }}^2 - \pi \mu _{\kappa }^2 ]. \end{aligned}$$

Also, since

$$\begin{aligned} \tilde{L_i}^2 | \kappa = \left\{ \begin{array}{cl} L^2 &{} \text{ with } \text{ probability } \quad \pi \kappa \\ 0 &{} \text{ with } \text{ probability } \quad 1 - \pi \kappa \end{array} \right. , \end{aligned}$$

implies ${\mathbb {E}}(\tilde{L_i}^2) = \pi L^2 \mu _{\kappa }$, we find the variance of $\tilde{L_i}$

$$\begin{aligned} \text{ Var }(\tilde{L_i}) = L^2 \pi \mu _{\kappa } (1- \pi \mu _{\kappa }). \end{aligned}$$

Since $\text{ Var }(\tilde{L_i}) = \text{ Var }(\tilde{L_j})$ for all i, the coefficient of correlation is finally equal to

$$\begin{aligned} \rho= & {} \frac{\text{ cov }(\tilde{L_i},\tilde{L_j})}{\text{ Var }(\tilde{L_i})}\\= & {} \frac{{\mathbb {E}}{\tilde{\kappa }}^2 - \pi \mu _{\kappa }^2}{ \mu _{\kappa } (1- \pi \mu _{\kappa } )} \\= & {} \frac{ \sigma ^2_{\kappa } + \mu _{\kappa }^2 (1-\pi )}{ \mu _{\kappa } (1- \pi \mu _{\kappa } )}, \end{aligned}$$

where the last line is obtained using $\sigma ^2_{\kappa } = {\mathbb {E}}{\tilde{\kappa }}^2 - \mu _{\kappa }^2$.

1.6 Proof of Proposition 4

When $\lambda =0$, inequality (16) is rewritten as

$$\begin{aligned} \frac{{\mathbb {E}}[{\tilde{\kappa }} u'(w - {\tilde{\kappa }}L)]}{\mu _{\kappa }} \le u'(w-L). \end{aligned}$$

(24)

Since $u''<0$, we have

$$\begin{aligned} u'(w- \kappa L) < u'(w-L) \quad \text{ for } \text{ all } \quad \kappa \in [0,1], \end{aligned}$$

and hence

$$\begin{aligned} {\mathbb {E}}[{\tilde{\kappa }} u'(w-{\tilde{\kappa }}L)] < \mu _{\kappa }u'(w-L), \end{aligned}$$

which gives (24). (16) also holds when $\lambda$ is not too large.

1.7 Proof of Proposition 5

Using Proposition 3 with Eqs. (16) and (19) gives $I^*>0$ if

$$\begin{aligned} \varphi (L) \equiv u'(w-L) - x(L,\lambda ,\rho ) > 0, \end{aligned}$$

where $x(L,\lambda ,\rho ) \equiv (1+\lambda )(1 + A(w)L \rho )$. Using $u'(w) = 1$, we obtain $\varphi (0) = u'(w) - x(0,\lambda ,\rho ) = - \lambda <0$, and

$$\begin{aligned} \varphi '(L)= & {} -u''(w-L) - (1+\lambda ) A(w)\rho \\ \varphi ''(L)= & {} u'''(w-L) >0 \end{aligned}$$

The function $\varphi (L)$ is therefore convex with $\varphi (0)<0$ and $\varphi ({\bar{L}})>0$ if $u'(w-{\bar{L}}) > (1+\lambda )[1 + A(w) {\bar{L}} \rho ]$. $\varphi (L) = 0$ hence defines ${\underline{L}}(\lambda ,\rho )$ with ${\bar{L}}(\lambda , \rho )=0$ if $\lambda =0$ and ${\bar{L}}(\lambda , \rho )>0$ if $\lambda >0$. When $L>{\underline{L}}(\lambda ,\rho )$, (i.e. $I^*>0$), Equation (17) gives

$$\begin{aligned} \beta ^* = 1 + \frac{u'^{-1}(x(L)) - w}{L}, \end{aligned}$$

(25)

for given values of $\lambda$ and $\rho$. Let $z(L) = u'^{-1}(x(L))$. Since DARA implies $u'''>0$, $u'^{-1}$ is decreasing and convex and since x(L) is linear, z(L) is also decreasing and convex. From (25), we obtain

$$\begin{aligned} \frac{\partial \beta ^*}{\partial L}= & {} \left[ \frac{Lx'(L)}{u''(u'^{-1}(x(L)))} - u'^{-1}(x(L)) + w \right] /L^2 \\= & {} \frac{Lz'(L) - z(L) + w}{L^2}. \end{aligned}$$

Using $w>=z(0)$ and the convexity of function z(L) yields

$$\begin{aligned} Lz'(L)-z(L)+w>=Lz'(L)-z(L)+z(0)>0, \end{aligned}$$

which gives $\partial \beta ^*/\partial L >0$ when $L\ge {\bar{L}}(\lambda , \rho )$. The comparative statics $\partial \beta ^*/\partial \rho < 0$ and $\partial \beta ^*/\partial \lambda < 0$ are obtained by noticing that an increase in either $\lambda$ or $\rho$ translates into an increase in x(L). $u'^{-1}$ decreasing then provides the results.