Growth and inequality: a demographic explanation

Abstract

This paper investigates the relationship between growth and inequality from a demographic point of view. In an extended model of the accidental bequest with endogenous fertility, we analyze the effects of a decrease in old-age mortality rate on the equilibrium growth rate as well as on the income distribution. We show that the relationship between growth and inequality is at first positive and then may be negative in the process of population aging. The results are consistent with the empirical evidence in some developed countries.

Appendix

Proof of Proposition 2

Differentiating Eq. 32 with respect to p, we have

$$\frac{{d\widehat{s}}}{{dp}} = \frac{{\alpha _{1} {\left\{ {{\left( {1 - p} \right)}{\left[ {1 + \alpha _{1} {\left( {1 - p} \right)}} \right]} - {\left( {\beta + p} \right)}} \right\}}}}{{\beta {\left[ {1 + \alpha _{1} {\left( {1 - p} \right)}} \right]}^{2} }}$$

We know

$$\begin{array}{*{20}c} {\frac{{d\widehat{s}}}{{dp}}\left| {_{{p = 1}} } \right. = - \frac{{{\left( {1 + \beta } \right)}\alpha _{1} }}{\beta } < 0} \\ {\frac{{d\widehat{s}}}{{dp}}\left| {_{{p = 0}} } \right. = \frac{{\alpha _{1} {\left( {1 + \alpha _{1} - \beta } \right)}}}{{\beta {\left( {1 + \alpha _{1} } \right)}^{2} }}} \\ \end{array} $$

Since the numerator of $d\widehat{s}/dp$ is a quadratic function of p with a positive coefficient of the quadratic term, the saving rate has an interior maximum if and only if $d\widehat{s}/dp\left| {_{{p = 0}} } \right. > 0$, that is,

$$\beta < 1 + \alpha _{1} $$

Proof of Proposition 3

From Eq. 33, we know

$$1 + g = G{\left( p \right)} = \frac{{\alpha _{1} }}{{\alpha _{2} }}\eta v\phi {\left( p \right)}$$

where

$$\begin{array}{*{20}c} {\phi {\left( p \right)} = {\left( {1 - p} \right)}\varphi {\left( p \right)}} \\ {\varphi {\left( p \right)} = 1 + \frac{{\alpha _{2} {\left( {\beta + p} \right)}}}{{\beta {\left[ {1 + \alpha _{1} {\left( {1 - p} \right)}} \right]}}}} \\ \end{array} $$

Note that G(0) > 0, which contrasts with Fuster (1999) and Cipriani (2000). First, we show that the growth rate is concave with respect to p, that is, G′′ < 0. Then, we show that G′(0) > 0 is equivalent to Eq. 35. Since G(1) = 0 and G(0) > 0, g has an interior maximum if and only if the condition (Eq. 35) is satisfied because of the concavity.

We know

$$\begin{array}{*{20}c} {{\phi \prime {\left( p \right)} = - \varphi {\left( p \right)} + {\left( {1 - p} \right)}\varphi \prime {\left( p \right)}}} \\ {{\phi \prime \prime {\left( p \right)} = - 2\varphi \prime {\left( p \right)} + {\left( {1 - p} \right)}\varphi \prime \prime {\left( p \right)}}} \\ \end{array} $$

and

$$\begin{array}{*{20}c} {{\varphi \prime {\left( p \right)} = \frac{{\alpha _{2} {\left( {1 + \alpha _{1} + \beta \alpha _{1} } \right)}}}{{\beta {\left[ {1 + \alpha _{1} {\left( {1 - p} \right)}} \right]}^{2} }} > 0}} \\ {{\varphi \prime \prime {\left( p \right)} = \frac{{2\alpha _{1} \varphi \prime {\left( p \right)}}}{{1 + \alpha _{1} {\left( {1 - p} \right)}}}.}} \\ \end{array} $$

First, we have

$$\phi \prime \prime {\left( p \right)} = - \frac{{2\varphi \prime {\left( p \right)}}}{{1 + \alpha _{1} {\left( {1 - p} \right)}}} < 0$$

which implies that G is concave.

Second, we have

$$\phi \prime {\left( 0 \right)} = \frac{{\alpha _{2} }}{{\beta {\left( {1 + \alpha _{1} } \right)}}} + \frac{{\alpha _{1} \alpha _{2} }}{{{\left( {1 + \alpha _{1} } \right)}^{2} }} - {\left( {1 + \frac{{\alpha _{2} }}{{1 + \alpha _{1} }}} \right)}$$

Thus, ϕ′(0) > 0 is equivalent to Eq. 35.

Proof of Proposition 4

From Eq. 36, we have γ = (1−p)f ₁(p)f ₂(p), where

$$\begin{array}{*{20}c} {f_{1} {\left( p \right)} = 1 - \frac{\beta }{{\beta + p}}} \\ {f_{2} {\left( p \right)} = 1 - \frac{{\alpha _{2} }}{{1 + \alpha _{1} {\left( {1 - p} \right)} + \alpha _{2} }}} \\ \end{array} $$

First, we have γ(0) = γ(1) = 0.

Second, we have f ₁′ > 0, f ₁″ < 0, f ₂′ < 0, and f ₂″ < 0 for p∈[0, 1], and

$$\begin{array}{*{20}c} {\gamma \prime {\left( p \right)} = - f_{1} {\kern 1pt} f_{2} + {\left( {1 - p} \right)}{\left( {f^{\prime }_{1} {\kern 1pt} f_{2} + f_{1} {\kern 1pt} f^{\prime }_{2} } \right)}} \\ {\gamma \prime \prime {\left( p \right)} = - 2{\left( {f^{\prime }_{1} f_{2} + f_{1} {\kern 1pt} f^{\prime }_{2} } \right)} + {\left( {1 - p} \right)}{\left( {f^{{\prime \prime }}_{1} f_{2} + 2{\kern 1pt} {\kern 1pt} f^{\prime }_{1} {\kern 1pt} f^{\prime }_{2} + f_{1} {\kern 1pt} f^{{\prime \prime }}_{2} } \right)}} \\ \end{array} $$

Since γ(p) is continuous in [0,1] and γ(p) > 0 for ∀p∈(0, 1), there exists at least one solution p*∈(0, 1), such that it maximizes γ . The necessary condition, γ′(p*) = 0, requires

$$f^{\prime }_{1} {\kern 1pt} f_{2} + f_{1} {\kern 1pt} f^{\prime }_{2} = \frac{{f_{1} {\kern 1pt} f_{2} }}{{1 - p*}} > 0$$

which implies γ″(p*) < 0. Therefore, we know that γ does not have any local minimums in (0,1). It proves the uniqueness of p*.