Self-insurance and saving under a two-argument utility framework

Abstract

This study analyses self-insurance and saving decisions in a two-period model when the utility function depends on income and health. In this study, we consider an intertemporal and multi-dimensional cost–benefit structure of self-insurance and saving, unlike in the standard one-argument utility model. We show that the impacts of the changes in initial income and health on self-insurance and saving depend on whether an individual is correlation averse or not, the comparison between the absolute risk aversion and the absolute correlation aversion, and the comparison between the income effect and the substitution effect. We also show that self-insurance and saving can be either complements or substitutes.

Appendices

Appendix

Proof of Proposition 1

We prove Proposition 1 (1) in the case of $u_{CA} < 0$. From Eq. (8), we have $\frac{de}{{dy}} > 0$ if and only if:

$$- \frac{{pu_{CA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{\left( {1 - p} \right)u_{CC} \left( {y - f\left( e \right), h} \right) + pu_{CC} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}} < \frac{{f^{\prime}\left( e \right)}}{{l^{\prime}\left( e \right)}}.$$

Using Eq. (7), we have the following inequality:

$$- \frac{{u_{CA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{u_{A} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}} < - \frac{{\left( {1 - p} \right)u_{CC} \left( {y - f\left( e \right), h} \right) + pu_{CC} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{\left( {1 - p} \right)u_{C} \left( {y - f\left( e \right), h} \right) + pu_{C} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}.$$

(A1)

Then, we obtain Eq. (9) by rewriting equation (A1) as follows:

$$- \frac{{u_{CA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{u_{A} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}} < - \frac{{Eu_{CC} \left( {y - f\left( e \right), \tilde{h}} \right)}}{{Eu_{C} \left( {y - f\left( e \right), \tilde{h}} \right)}}.$$

Next, to prove Proposition 1 (2), we totally differentiate Eq. (6) with respect to $h$:

$$\left. {\frac{de}{{dh}}} \right|_{{e = e^{*} }} = - \left\{ { - f^{\prime}\left( e \right)\left[ {\left( {1 - p} \right)u_{CA} \left( {y - f\left( e \right), h} \right) + pu_{CA} \left( {y - f\left( e \right),h - l\left( e \right)} \right)} \right] - pu_{AA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)l^{\prime}\left( e \right)} \right\}/U_{ee} .$$

(A2)

From equation (A2), if $u_{CA} \ge 0$, we have $\frac{de}{{dh}} < 0$. Contrarily, if $u_{CA} < 0$, we have $\frac{de}{{dh}} < 0$ if and only if:

$$- \frac{{\left( {1 - p} \right)u_{CA} \left( {y - f\left( e \right), h} \right) + pu_{CA} \left( {y - f\left( e \right),h - l\left( e \right)} \right)}}{{\left( {1 - p} \right)u_{C} \left( {y - f\left( e \right), h} \right) + pu_{C} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}} < - \frac{{u_{AA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{u_{A} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}.$$

(A3)

We can obtain Eq. (11) by rewriting inequality (A3) as follows:

$$- \frac{{Eu_{CA} \left( {y - f\left( e \right), \tilde{h}} \right)}}{{Eu_{C} \left( {y - f\left( e \right), \tilde{h}} \right)}} < - \frac{{u_{AA} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}{{u_{A} \left( {y - f\left( e \right), h - l\left( e \right)} \right)}}.$$

Proof of Lemma 2

To prove Lemma 2, let us first rearrange Eq. (23) as follows:

$$\frac{{pu_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{R\left[ {\left( {1 - p} \right)u_{C} \left( {y_{1} + Rs, h} \right) + pu_{C} \left( {y_{1} + Rs, h - l\left( e \right)} \right)} \right]}} = - \frac{{f^{\prime}\left( e \right)}}{{l^{\prime}\left( e \right)}}.$$

(A4)

From Eq. (24), the sign of $\frac{de}{{ds}}$ is identical to the sign of $U_{es}$. The condition for $U_{es} < 0$ is as follows:

$$\frac{{\beta R^{2} u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}<- \frac{{u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}}$$

(A5)

Using Eq. (22) and (A4), inequality (A5) can be written as follows:

$$R\frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} < - \frac{{u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}}.$$

(A6)

Proof of Proposition 3

By totally differentiating Eqs. (21) and (22) with respect to $y_{0}$, ${\text{y}}_{1}$ and $h$, and using Cramer’s Rule, we have the following:

$$\left. {\frac{de}{{dy_{0} }}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{{{\text{ss}}}} U_{{ey_{0} }} + U_{es} U_{{sy_{0} }} }}{\left| H \right|},$$

(A7)

$$\left. {\frac{de}{{dy_{1} }}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{{{\text{ss}}}} U_{{ey_{1} }} + U_{es} U_{{sy_{1} }} }}{\left| H \right|},$$

(A8)

$$\left. {\frac{de}{{dh}}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{{{\text{ss}}}} U_{eh} + U_{es} U_{sh} }}{\left| H \right|},$$

(A9)

$$\left. {\frac{ds}{{dy_{0} }}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{ee} U_{{sy_{0} }} + U_{se} U_{{e{\text{y}}_{0} }} }}{\left| H \right|},$$

(A10)

$$\left. {\frac{ds}{{dy_{1} }}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{ee} U_{{sy_{1} }} + U_{se} U_{{ey_{1} }} }}{\left| H \right|},$$

(A11)

$$\left. {\frac{ds}{{dh}}} \right|_{{e = e^{*} , s = s^{*} }} = \frac{{ - U_{ee} U_{sh} + U_{se} U_{eh} }}{\left| H \right|},$$

(A12)

where $e^{*}$ and $s^{*}$ are optimal self-insurance and saving, respectively,

$U_{{ee}} = u_{C} \left( {y_{0} - f\left( e \right) - s,~h} \right)\left( { - f^{\prime\prime}\left( e \right)} \right) + u_{{CC}} \left( {y_{0} - f\left( e \right) - s,~h} \right)\left\{ {f^{\prime}\left( e \right)} \right\}^{2} + \beta pu_{A} \left( {y_{1} + Rs,~h - l\left( e \right)} \right)l^{\prime\prime}\left( e \right) + \beta pu_{{AA}} \left( {y_{1} + Rs,~h - l\left( e \right)} \right)\left\{ {l^{\prime}\left( e \right)} \right\}^{2} < 0$ by the second-order condition,

$$U_{es} = u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)f^{\prime}\left( e \right) - \beta Rpu_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)l^{\prime}\left( e \right),$$

$$U_{{ey_{0} }} = u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)\left( { - f^{\prime}\left( e \right)} \right) > 0,$$

$$U_{{ey_{1} }} = \beta pu_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)\left( { - l^{\prime}\left( e \right)} \right),$$

$$U_{eh} = u_{CA} \left( {y_{0} - f\left( e \right) - s, h} \right)\left( { - f^{\prime}\left( e \right)} \right) - \beta pu_{AA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)l^{\prime}\left( e \right),$$

$$U_{ss} = u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right) + \beta R^{2} \left[ {\left( {1 - p} \right)u_{CC} \left( {y_{1} + Rs, h} \right) + pu_{CC} \left( {y_{1} + Rs, h - l\left( e \right)} \right)} \right] < 0,$$

$$U_{{sy_{0} }} = - u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right) > 0,$$

$$U_{{sy_{1} }} = \beta R\left[ {\left( {1 - p} \right)u_{CC} \left( {y_{1} + Rs, h} \right) + pu_{CC} \left( {y_{1} + Rs, h - l\left( e \right)} \right)} \right] < 0,$$

$$U_{sh} = - u_{CA} \left( {y_{0} - f\left( e \right) - s, h} \right) + \beta R\left[ {\left( {1 - p} \right)u_{CA} \left( {y_{1} + Rs, h} \right) + pu_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)} \right],$$

and $\left| H \right| = U_{ee} U_{ss} - \left( {U_{es} } \right)^{2} > 0$ by the second-order condition. Note that we explicitly assume that the second-order conditions hold, as in the literature (Dionne and Eeckhoudt 1984; Menegatti and Rebessi, 2011; Liu and Menegatti 2019a, b).

4.1 Proof of Proposition 3 (1)

From equation (A7), and by using equation (A4), we can obtain inequality (27), the condition for $\frac{de}{{dy_{0} }} > 0$:

$$- \frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} < - \frac{{Eu_{CC} \left( {y_{1} + Rs, \tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}}.$$

From equation (A10), and since $U_{ee} < 0$, the sign of $\frac{ds}{{dy_{0} }}$ is identical to the sign of the inequality (28):

$$U_{{sy_{0} }} + \left( { - \frac{{U_{{ey_{0} }} }}{{U_{ee} }}} \right)U_{se} .$$

By combining inequality (27) and (28), the following statements prove Proposition 3 (1):

• (i) In the case of $u_{CA} > 0$, we obviously have $\frac{de}{{dy_{0} }} > 0$, and if $\frac{{u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} + R\frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} > 0$, then we have $\frac{ds}{{dy_{0} }} > 0$.

• (ii) In the case of $u_{CA} < 0$, we automatically have $U_{se} < 0$, and if $- \frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} < - \frac{{Eu_{CC} \left( {y_{1} + Rs, \tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}}$, then we have $\frac{de}{{dy_{0} }} > 0$, while the sign of $\frac{ds}{{dy_{0} }}$ is ambiguous.

4.2 Proof of Proposition 3 (2)

The proof is similar to that of Proposition 3 (1), so is omitted here.

4.3 Proof of Proposition 3 (3)

From equation (A9), and since $U_{ss} < 0$, the condition for $\frac{de}{{dh}} < 0$ is as follows:

$$U_{{eh}} + \left( { - \frac{{U_{{sh}} }}{{U_{{ss}} }}} \right)U_{{es}} < 0.$$

(A13)

To examine the sign of (A13), let us first suppose that Lemma 2 holds ($U_{es} < 0$). Then, we have $U_{eh} < 0$, if $u_{CA} \ge 0$, or if $u_{CA} < 0$ and the following inequality holds:

$$- \frac{{u_{{CA}} \left( {y_{0} - f\left( e \right) - s,~h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s,~h} \right)}} < - \frac{{u_{{AA}} \left( {y_{1} + Rs,~h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs,~h - l\left( e \right)} \right)}}.$$

(A14)

Moreover, we have $U_{sh} > 0$ if:

$$- \frac{{u_{{CA}} \left( {y_{0} - f\left( e \right) - s,~h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s,~h} \right)}} > - \frac{{Eu_{{CA}} \left( {y_{1} + Rs,~\tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs,~\tilde{h}} \right)}}. .$$

(A15)

Thus, we obtain $\frac{de}{{dh}} < 0$, if Lemma 2 and inequalities (A14) and (A15) hold.

On the contrary, now we suppose that Lemma 2 does not hold ($U_{es} > 0$). Then, since $U_{eh} < 0$, we have $\frac{de}{{dh}} < 0$ if $U_{sh} < 0$, that is, the following holds:

$$- \frac{{u_{{CA}} \left( {y_{0} - f\left( e \right) - s,~h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s,~h} \right)}} < - \frac{{Eu_{{CA}} \left( {y_{1} + Rs,~\tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs,~\tilde{h}} \right)}}.$$

(A16)

Next, from equation (A12), the sign of $\frac{ds}{{dh}}$ is identical to the sign of the following:

$$U_{{sh}} + \left( { - \frac{{U_{{eh}} }}{{U_{{ee}} }}} \right)U_{{se}} .$$

(A17)

Combining inequality (A13) and (A17), the following statements prove Proposition 3 (3):

• (i) (a) In the case of $u_{CA} > 0$, we automatically have $U_{eh} < 0$. If $\frac{{u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} + R\frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} < 0$ and $- \frac{{u_{CA} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} > - \frac{{Eu_{CA} \left( {y_{1} + Rs, \tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}}$, then we have $U_{sh} > 0$ and $U_{se} < 0$ and, thus, $\frac{de}{{dh}} < 0$ and $\frac{ds}{{dh}} > 0$.

  (b) In the case of $u_{CA} < 0$, we automatically have $U_{es} < 0$, and if $- \frac{{Eu_{CA} \left( {y_{1} + Rs, \tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}} < - \frac{{u_{CA} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} < - \frac{{u_{AA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}$, then we have $U_{eh} < 0$ and $U_{sh} > 0$. Therefore, we obtain $\frac{de}{{dh}} < 0$ and $\frac{ds}{{dh}} > 0$.

• (ii) In the case of $u_{CA} > 0$, if $\frac{{u_{CC} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} + R\frac{{u_{CA} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}}{{u_{A} \left( {y_{1} + Rs, h - l\left( e \right)} \right)}} > 0$ and $- \frac{{u_{CA} \left( {y_{0} - f\left( e \right) - s, h} \right)}}{{u_{C} \left( {y_{0} - f\left( e \right) - s, h} \right)}} < - \frac{{Eu_{CA} \left( {y_{1} + Rs, \tilde{h}} \right)}}{{Eu_{C} \left( {y_{1} + Rs, \tilde{h}} \right)}}$, then we have $U_{eh} < 0$, $U_{sh} < 0$ and $U_{se} > 0$. Therefore, we have $\frac{de}{{dh}} < 0$ and $\frac{ds}{{dh}} < 0$.