Savings, Marriage, and Work-in-Household

Abstract

In this chapter saving rates are modeled as depending on intra-marriage financial distributions among spouses acting as independent decision-makers. Such distributions could be compensations for Work-In-Household (WiHo, see Chap. 2) supplied by one spouse and of value to the other. In the context of traditional gender roles an inter-temporal overlapping-generations model of individual behavior implies higher saving rates by young (pre-marriage) men than by young women. The opposite is predicted regarding the saving rates of married women relative to those of married men. This model potentially helps explain some cross-country variation in gender differentials in savings behavior. Furthermore, higher sex ratios are expected to be associated with higher savings rates among young unmarried men and lower savings among young women, while the opposite will hold for married men and women. Savings rates are also expected to be associated with combinations (matchings) of traits of husbands and wives, such as age and health. The higher the expected intra-marriage distribution the more WiHo-workers will earn in marriage and the more they will save as a proportion of their earned income. The opposite holds for WiHo-users. Whether a society allows polygamy can also help explain savings.

The original version of the chapter was revised: Co-author names are included in Chaps. 7, 9, 11, and table of contents. The erratum to this chapter is available at http://dx.doi.org/10.1007/978-1-4614-1623-4_12.

Appendix

The determinant of the Jacobian matrix of the first-order conditions with respect to s ₁ and s ₂ is positive, i.e., $Det({{F}_{s}})>0$. This is a direct requirement of the optimization problem in that it relates to the strict concavity of the objective function with respect to the decision variables and satisfies the conditions of the implicit function theorem. To obtain the necessary information for the identification of the effects on savings behavior we totally differentiate F ₁ and F ₂ to obtain:

$$ \frac{\partial {{F}_{1}}}{\partial {{s}_{1}}}={U}''(w-{{s}_{1}})+\delta {{(1+r)}^{2}}\left\{{{p}_{2}}{U}''[w+(1+r){{s}_{1}}-{{s}_{2}}]+(1-{{p}_{2}}){U}''[\gamma w+(1+r){{s}_{1}}-{{s}_{2}}] \right\}<0, $$

(11.10)

$$ \frac{\partial {{F}_{1}}}{\partial {{s}_{2}}}=-\delta (1+r)\left\{{{p}_{2}}{U}''[w+1(1+r){{s}_{1}}-{{s}_{2}}]+(1-{{p}_{2}}){U}''[\gamma w+(1+r){{s}_{1}}-{{s}_{2}}] \right\}>0, $$

(11.11)

or using (11.10)

$$ -\frac{\partial {{F}_{1}}}{\partial {{s}_{1}}}=-{U}''(w-{{s}_{1}})+(1+r)\frac{\partial {{F}_{1}}}{\partial {{s}_{2}}}>\frac{\partial {{F}_{1}}}{\partial {{s}_{2}}}, $$

(11.12)

$$\frac{\partial {{F}_{1}}}{\partial {{p}_{2}}}=\delta (1+r)\left\{ U'\left[ w+(1+r){{s}_{1}}-{{s}_{2}} \right]-U'\left[ \gamma w+(1+r){{s}_{1}}-{{s}_{2}} \right] \right\}>0if\gamma>1\,(<0if\gamma <1),$$

(11.13)

$$ \frac{\partial {{F}_{1}}}{\partial {{p}_{3}}}=0, $$

(11.14)

$$\frac{\partial {{F}_{2}}}{\partial {{S}_{1}}}=\frac{\partial {{F}_{1}}}{\partial {{S}_{2}}}=-\partial (1+r)\left\{ {{p}_{2}}{U}''\left[ w+(1+r){{s}_{1}}-{{s}_{2}} \right]+(1-{{p}_{2}}){U}''\left[ \gamma w+(1+r){{s}_{1}}-{{s}_{2}} \right] \right\}>0,$$

(11.15)

$$ \begin{aligned}& \frac{\partial {{F}_{2}}}{\partial {{S}_{2}}}=\partial \left\{{{p}_{2}}{U}''[w+(1+r){{s}_{1}}-{{s}_{2}}]+(1-{{p}_{2}}){U}''[\gamma w+(1+r){{s}_{1}}-{{s}_{2}}] \right\} \\& ~~~~~+{{\delta }^{2}}{{(1+r)}^{2}}\left\{{{p}_{3}}{U}''[w+(1+r){{s}_{2}}]+(1-{{p}_{3}}){U}''[\gamma w+(1+r){{s}_{2}}] \right\} \\& ~~~~ =-\frac{1}{1+r}\cdot \frac{\partial {{F}_{1}}}{\partial {{S}_{2}}}+{{\delta }^{2}}{{(1+r)}^{2}}\left\{{{p}_{3}}{U}''[w+(1+r){{s}_{2}}]+(1-{{p}_{3}}){U}''[\gamma w+(1+r){{s}_{2}}] \right\}<0, \end{aligned} $$

(11.16)

or using (11.10), (11.11), and (11.16)

$$ \begin{aligned}& \frac{\partial {{F}_{2}}}{\partial {{S}_{2}}}=-\frac{1}{{{(1+r)}^{2}}}\cdot \left[{U}''(w-{{s}_{1}})-\frac{\partial {{F}_{2}}}{\partial {{S}_{2}}}\right] \\& ~~~~+{{\delta }^{2}}{{(1+r)}^{2}}\left\{{{p}_{3}}{U}''[w+(1+r){{s}_{2}}]+(1-{{p}_{3}}){U}''[\gamma w+(1+r){{s}_{2}}] \right\}<0, \end{aligned} $$

(11.17)

where given (11.10) the first term has a negative sign.

$$ \frac{\partial {{F}_{2}}}{\partial {{p}_{2}}}=-\delta \left\{U'[w+(1+r){{s}_{1}}-{{s}_{2}}]-U'[\gamma w+(1+r){{s}_{1}}-{{s}_{2}}] \right\}<0\,if\,\gamma>1\,(>0\,if\,\gamma <1). $$

(11.18)

From (11.13) and (11.18) it follows that $\frac{\partial {{F}_{1}}}{\partial {{p}_{2}}}=-(1+r)\frac{\partial {{F}_{2}}}{\partial {{p}_{2}}}$ and these two derivatives will always have the opposite sign regardless of γ.

$$ \frac{\partial {{F}_{2}}}{\partial {{p}_{3}}}={{\delta }^{2}}\left\{U'[w+(1+r){{s}_{2}}]-U'[\gamma w+(1+r){{s}_{2}}] \right\}>0\,if\,\gamma>1\,(<0\,if\,\gamma <1). $$

(11.19)