Signaling theory revisited: a very short insurance case

Abstract

In order to reduce informational asymmetry a signaling contract seems to be an efficient solution. Standard theory argues that insurance contracts typically produce separating equilibria. Mostly, this property is based on the assumption of two states of the world only. If however, there is a world with a continuum of states the solution is different. We prove that there is no longer a separating equilibrium. As a result, insurance markets are characterized by pooling equilibria or self-insurance structures.

Appendix

Proposition 1

$$\begin{aligned} \frac{\textit{dP}}{\textit{dZ}}\left| {_{E[u\left| {t_i ]=\textit{const}} \right. } } \right. =-\frac{u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} }{\int \limits _0^Z {u^{\prime }(w_0 -L-P)f(L;t_i )\textit{dL}+u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} } } \end{aligned}$$

Proof

As the insurance customer’s a level indifference curve $P(Z)\left| {_{E[u\left| {t_i ]=\textit{const}=a} \right. }} \right.$ is implicitly given by $F(Z,P;t_i )\!\!=\!U(Z,P;t_i )-a \!\!=\!\!\int \limits _0^\mathrm{Z} {\hbox {u}(\hbox {w}_0 -\hbox {L}-\hbox {P})\hbox {f}(\hbox {L};\hbox {t}_{\mathrm{i}} )\hbox {dL}} +\int \limits _{\mathrm{Z}}^{\mathrm{M}} {\hbox {u}(\hbox {w}_0 -\hbox {Z}-\hbox {P})\hbox {f}(\hbox {L};\hbox {t}_{\mathrm{i}} )\hbox {dL}}$ we get $\frac{\textit{dP}}{\textit{dZ}}\left| {_{E[u\left| {t_i ]=\textit{const}} \right. }} \right. =-\frac{F_1 (Z,P;t_i )}{F_2 (Z,P;t_i )}$ using $F_1 (Z,P;t_i )=-u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}}$ and $F_2 (Z,P;t_i )=\int \limits _0^Z u^{\prime }(w_0 -L-P)(-1)f(L;t_i )\textit{dL}+\int \limits _Z^M{u^{\prime }(w_0 -Z-P)(-1)f(L;t_i )\textit{dL}}$. Then, the type i indifference curves are strictly monotonic decreasing according to $\frac{\textit{dP}}{\textit{dZ}}\left| {_{E[u\left| {t_i ]=\textit{const}} \right. } } \right. =-\frac{F_1 (Z,P;t_i )}{F_2 (Z,P;t_i )}=-\frac{u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} }{\int \limits _0^Z {u^{\prime }(w_0 -L-P)f(L;t_i )\textit{dL}+u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} } }$. $\square$

Proposition 2

$$\begin{aligned} \frac{\textit{dP}_i }{\textit{dZ}}=-p(Z;t_i )=-\int \limits _Z^M {f(L;t_i )\textit{dL}} \end{aligned}$$

Proof

The assertion follows from $\frac{\textit{dP}_i }{\textit{dZ}}=-Z f(Z;t_i )+\int \limits _M^Z {f(L;t_i )\textit{dL}+Z f(Z;t_i )} =\int \limits _M^Z {f(L;t_i )\textit{dL}} \quad =-\int \limits _Z^M {f(L;t_i )\textit{dL}=-p(Z;t_i )}$. $\square$

Proposition 3

The insurance customer’s indifference curves are convex

Proof

Let (i) $Z_i \in [0, M], Z_1 <Z_2$ and $\bar{{Z}}=(Z_1 +Z_2 )2^{-1}$, (ii) $P_1 , P_2$ be chosen such that $(Z_1 ,P_1 ), (Z_2 ,P_2 )$ are parts of the insurance customer’s a level indifference curve, and (iii) (Z, P) be given by $P\le w_0 -Z, P<w_0 -Z$ for $Z<M$ since indifference curves with non-negative expected utility are considered only. Note furthermore (iv) $w_0 -Z$ is the insurer’s highest premium in the case of damage, (v) $(w_0 ,0)$ is the only possible contract in the borderline case$Z=M=w_0$, and (vi) $P_1 >(P_1 +P_2 )2^{-1}>P{ }_2$ since the indifference curves are strictly monotonic decreasing. Given $\bar{{P}}=(P_1 +P_2 )2^{-1}$, we have to show $(\bar{{Z}},\bar{{P}})$ is part of an indifference curve of a lower level than a.

We prove $U(\bar{{Z}},\bar{{P}};t_i )<a$. As $U(\bar{{Z}},\bar{{P}};t_i )<\int \limits _0^{\bar{Z}} {u(w_0 -L-\bar{{P}})f(L;t_i )\textit{dL}} +2^{-1}\int \limits _{\bar{Z}}^M (u( w_0 -Z_1 -P_1 ) +u(w_0 -Z_2 -P_2 ))f(L;t_i )\textit{dL}$ is valid (note: the concavity of $u(\cdot )$ implies the convexity of $\nu (\cdot )$ if $v(X): =u(w_0 -X)) \quad v(2^{-1}(Z_1 +P_1 +Z_2 +P_2 ))<2^{-1}(u(w_0 -Z_1 -P_1 )+u(w_0 -Z_2 -P_2 ))$ is satisfied. Hence,

$$\begin{aligned}&U(Z,P;t_i ) <\int \limits _0^{\bar{Z}} {u(w_0 -L-\bar{{P}})f(L;t_i )\textit{dL}}\\&\quad +\,2^{-1}\int \limits _{\bar{Z}}^M (u( w_0 -Z_1 -P_1 )+u(w_0 -Z_2 -P_2 ))f(L;t_i )\textit{dL}\\&\quad <2^{-1}\left( \int \limits _0^{Z_1 } {u(w_0 -L-P_1 )f(L;t_i )\textit{dL}+\int \limits _{Z_1 }^M {u(w_0 -Z_1 -P{}_1)f(L;t_i )\textit{dL}} }\right. \\&\left. \quad +\,\int \limits _0^{Z_2 } {u(w_0 -L-P_2 )f(L;t_i )\textit{dL}+\int \limits _{Z_2 }^M {u(w_0 -Z_2 -P_2 )f(L;t_i )\textit{dL}} }\right. \\&\left. \quad +\,\int \limits _{\bar{Z}}^{Z_2 } u(w_0 \!-\!Z_2 -P_2 )f(L;t_i )\textit{dL}-\!\int \limits _{\bar{Z}}^{Z_2 } u(w_0 -\!L-\!P_2 )f(L;t_i )\textit{dL}\!\right) \!\!<\!2^{-1}a+2^{-1}a =a. \end{aligned}$$

$\square$

Proposition 4

$\left| {\frac{dP}{dZ}\left| {_{E[u\left| {t_i ]=\textit{const}} \right. } } \right. } \right| >p(Z;t_i )$ for $0<Z<M$

Proof

Using that the marginal utility is descending the assertion follows according to $\left| {\frac{dP}{dZ}|_{E[u|t_i ] = \textit{const}} } \right| \!\!>\frac{u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} }{u^{\prime }(w_0 -Z-P)\int \limits _0^Z {f(L;t_i )\textit{dL}+u^{\prime }(w_0 -Z-P)\int \limits _Z^M {f(L;t_i )\textit{dL}} } } =\!\int \limits _Z^M {f(L;t_i )\textit{dL}\!=p(Z;t_i )}$. $\square$

Proposition 5

$P_1 (Z)\le P_2 (Z)$ for all $Z\in [0;M]$

Proof

Using SSD we get $P_1 (Z)=\bar{{L}}-Z+\int \limits _0^Z {F(L;t_1 )\textit{dL}} \le \bar{{L}}-Z+\int \limits _0^Z {F(L;t_2 )\textit{dL}} =P_2 (Z)$ since $P_i (Z)=\int \limits _Z^M {(L-Z)f(L;t_i )\textit{dL}=} \int \limits _0^Z {Lf(L;t_i )\textit{dL}+\int \limits _Z^M {Lf(L;t_i )\textit{dL}}} -\int \limits _0^Z {Lf(L;t_i )\textit{dL}}-Z\int \limits _Z^M {f(L;t_i )\textit{dL}} =\bar{{L}}-(ZF(Z;t_i )-\int \limits _0^Z F(L;t_i )\textit{dL})-Z(1-\int \limits _0^Z {f(L;t_i )\textit{dL})}=\bar{{L}}-Z+\int \limits _0^Z {F(L;t_i )\textit{dL}}$ holds for.

Since the anti-derivative of $L f(L;t_i )$ is given by $H(L;t_i)\!=\!L F(L;t_i )-\!\int \limits _0^L (\int \limits _0^y {f(u;t_i )du)}dy$ we get $\int \limits _0^Z {L f(L; t_i )\textit{dL}=ZF(Z; t_i )-\int \limits _0^Z {F(L; t_i )\textit{dL}} }$. $\square$