Optimal life-cycle fertility in a Barro-Becker economy

Abstract

Parenthood postponement is a key demographic trend of the last three decades. In order to rationalize that stylized fact, we extend the canonical model by Barro and Becker (Econometrica 57:481–501, 1989) to include two—instead of one—reproduction periods. We examine how the cost structure of early and late children in terms of time and goods affects the optimal fertility timing. Then, we identify conditions that guarantee the existence and uniqueness of a stationary equilibrium with a stationary cohort size. Finally, we examine how the model can rationalize the observed postponement of births, and we highlight two plausible causes: (1) a general rise in the cost of children in terms of goods and (2) a decline in the degree of family altruism.

Appendix

Proof of Lemma 1

In order to compare the derivatives $\frac {\partial {N}_{t+1+j}}{\partial {n}_{t} }$ and $\frac {\partial {N}_{t+1+j}}{\partial m_{t}}$, remember that ${N}_{t+1}={N}_{t}g_{t+1}={N}_{t}\left ({n}_{t}+\frac {m_{t}{N}_{t-1}}{{N}_{t}}\right )$. Hence, we have

$$\begin{array}{@{}rcl@{}} \frac{\partial {N}_{t+1}}{\partial {n}_{t}} &=&{N}_{t} \\ \frac{\partial {N}_{t+1}}{\partial m_{t}} &=&{N}_{t-1} \end{array}$$

$$\begin{array}{@{}rcl@{}} \frac{\partial {N}_{t+2}}{\partial {n}_{t}} {\kern-1.5pt}&=&{\kern-1.5pt}{N}_{t}g_{t+2}+{N}_{t+1}\left(\frac{ -m_{t+1}{N}_{t}{N}_{t}}{{N}_{t+1}^{2}}\right) ={N}_{t}\left(g_{t+2}-\frac{m_{t+1}}{ g_{t+1}}\right) ={N}_{t}{n}_{t+1} \\ \frac{\partial {N}_{t+2}}{\partial m_{t}}{\kern-1.5pt} &=&{\kern-1.5pt}{N}_{t-1}g_{t+2}+{N}_{t+1}{\kern-1.5pt}\left(\frac{-m_{t+1}{N}_{t}{N}_{t-1}}{{N}_{t+1}^{2}}\right){} ={\kern-1.5pt}{N}_{t-1}{\kern-1.5pt}\left(g_{t+2}-\frac{ m_{t+1}}{g_{t+1}}\right){\kern-1.5pt} ={\kern-1.5pt}{N}_{t-1}{n}_{t+1} \\ \frac{\partial {N}_{t+3}}{\partial {n}_{t}} &=&{N}_{t}{n}_{t+1}g_{t+3}+{N}_{t+2}\left(\frac{m_{t+2}{N}_{t}{N}_{t+2}-m_{t+2}{N}_{t+1}{N}_{t}{n}_{t+1}}{{N}_{t+2}^{2}}\right) \\ &=&{N}_{t}\left[ {n}_{t+1}{n}_{t+2}+m_{t+2}\right] \\ \frac{\partial {N}_{t+3}}{\partial m_{t}} {\kern-1.5pt}&=&{\kern-1.5pt}{N}_{t-1}{n}_{t+1}g_{t+3}{\kern-1.5pt}+{\kern-1.5pt}m_{t+2}{N}_{t-1}{\kern-1.5pt}\left(\frac{{N}_{t+2}-{N}_{t+1}{n}_{t+1}}{ {N}_{t+2}}\right) \\ &=&{N}_{t-1}\left[ {n}_{t+1}{n}_{t+2}+m_{t+2}\right] \\ &&... \end{array}$$

or, in general terms,

$$\frac{\partial {N}_{t+1+j}}{\partial {n}_{t}}=\frac{{N}_{t}}{{N}_{t-1}}\frac{ \partial {N}_{t+1+j}}{\partial m_{t}}$$

Thus, whether the marginal impact of raising early fertility is larger than the marginal impact of raising late fertility depends on whether the cohort size is growing over time or not. If N _t > N _{t − 1} or g _t > 1, we necessarily have, everything else being unchanged, that $\frac {\partial {N}_{t+1+j}}{\partial {n}_{t}}>\frac {\partial {N}_{t+1+j}}{\partial m_{t}}$.

Proof of Proposition 1

From the condition

$$1=\left[ \alpha \frac{bA(1-\eta )\kappa^{\eta }+e}{\left[ \frac{BA(1-\eta )\kappa^{\eta }+E}{A\eta \kappa^{\eta -1}}\right] }\right]^{\frac{1}{ \varepsilon }},$$

the existence of a stationary equilibrium with a stationary cohort size requires κ to satisfy the condition:

$$A\kappa^{\eta }(1-\eta )B+E-\alpha \eta A\kappa^{\eta -1}e=\alpha \eta (1-\eta )bA^{2}\kappa^{2\eta -1}$$

Let us denote the LHS by φ(κ)≡A κ ^η(1 − η)B + E − α η A κ ^{η − 1} e and the RHS by ψ(κ)≡α η(1 − η)b A ² κ ^{2η − 1}.

Three cases can arise, depending on $\eta \gtreqqless 1/2$.

When η < 1/2, we have φ(0) = −∞ and φ(+∞) = +∞. We also have φ′(κ) = A η κ ^{η − 1}(1 − η)B − (η − 1)α η A κ ^{η − 2} e > 0.

Regarding ψ(κ), we have ψ(0) = + ∞ and ψ(+∞) = 0. We also have ψ′(κ) = α η(1 − η)(2η − 1)b A ² κ ^{2η − 2} < 0.

Given that φ(0) < ψ(0) and that φ(+∞) > ψ(+∞), it follows that there exists, by continuity, a level of κ such that φ(κ) = ψ(κ). Moreover, given the strict monotonicity of φ(κ) and ψ(κ), the intersection of φ(κ) and ψ(κ) is unique.

When η = 1/2, we have φ(0) = −∞ and φ(+∞) = +∞. We also have φ′(κ) = A η κ ^{η − 1}(1 − η)B − (η − 1)α η A κ ^{η − 2} e > 0.

Regarding ψ(κ), we have ψ(0) = α η(1 − η)b A ² < 0 and ψ(+∞) = α η(1 − η)b A ² < 0. We also have ψ′(κ) = 0.

Given that φ(0) < ψ(0) and that φ(+∞) > ψ(+∞), it follows that there exists, by continuity, a level of κ such that φ(κ) = ψ(κ). Moreover, given the strict monotonicity of φ(κ), the intersection of φ(κ) and ψ(κ) is unique.

When η > 1/2, we have φ(0) = −∞ and φ(+∞) = +∞. We also have φ′(κ) = A η κ ^{η − 1}(1 − η)B − (η − 1)α η A κ ^{η − 2} e > 0 and φ″(κ) = A η(η − 1)κ ^{η − 2}(1 − η)B − (η − 1)(η − 2)α η A κ ^{η − 3} e < 0.

Regarding ψ(κ), we have ψ(0) = 0 and ψ(+∞) = +∞. We also have ψ′(κ) = α η(1 − η)(2η − 1)b A ² κ ^{2η − 2} < 0 and ψ″(κ) = α η(1 − η)(2η − 1)(2η − 2)b A ² κ ^{2η − 3} < 0.

Note that, when κ → 0, we have φ′(κ) = A η κ ^{η − 1}(1 − η)B − (η − 1)α η A κ ^{η − 2} e = +∞ > ψ′(κ) = α η(1 − η)(2η − 1)b A ² κ ^{2η − 2} = 0.

Note also that, when κ → +∞, we have φ′(κ) = A η κ ^{η − 1}(1 − η)B − (η − 1)α η A κ ^{η − 2} e = 0 < ψ′(κ) = α η(1 − η)(2η − 1)b A ² κ ^{2η − 2} < 0.

We thus have φ(0) < ψ(0), with φ′(κ) > ψ′(κ) when κ → 0 and φ′(κ) < ψ′(κ) when κ → +∞.

Given that φ(0) < ψ(0), as well as φ′(κ) > 0, φ″(κ) < 0, ψ′(κ) > 0 and ψ″(κ) < 0, three distinct cases can arise: either no intersection of φ(κ) and ψ(κ), or a single intersection of φ(κ) and ψ(κ), or a double intersection of φ(κ) and ψ(κ). In the first case, we have that, for all κ > 0, it is the case that φ(κ) < ψ(κ). The two other cases can be described as follows. For low levels of κ, we have φ(κ) < ψ(κ), but the higher slope of φ(κ) for low levels of κ makes the φ(κ) curve intersect the ψ(κ) from below. That intersection occurs for low levels of κ, for which φ′(κ) > ψ′(κ). Let us denote that intersection by κ ⁻. At that intersection, φ(κ) crosses ψ(κ) from below, since we initially have φ(0) < ψ(0). Then, two cases can arise. First, it is possible that the two curves φ(κ) and ψ(κ) do not intersect any more, the curve ψ(κ) remaining is always below φ(κ). Alternatively, φ(κ) and ψ(κ) may intersect a second time. Those two cases can arise, since we know that, although φ′(κ) > ψ′(κ) at κ = κ ⁻, we have φ′(κ) = 0 < ψ′(κ) when κ → +∞. Hence, it can be the case that there exists a second intersection, which can be denoted by κ ⁺.

Let us now concentrate on condition (36). The existence of a stationary equilibrium with a stationary cohort size requires condition (36) to be satisfied. Since the stationarity of the cohort size requires κ = κ ^∗, it follows that there exists a stationary equilibrium with stationary population size if and only if there exists some interior level of early fertility n ^∗ ∈ ]0, 1[ such that the following condition holds:

$$\kappa^{\ast }=\frac{\left[ \begin{array}{l} \left[ A(1-\eta )\kappa^{\ast \eta }(1-bn^{\ast })-n^{\ast }e\right] \left[ (A\eta \kappa^{\ast \eta -1})\beta \right]^{\frac{1}{1-\sigma }} \\ -A(1-\eta )\kappa^{\ast \eta }(1-B(1-n^{\ast }))+(1-n^{\ast })E \end{array} \right] }{(2-bn^{\ast }-B(1-n^{\ast }))\left[ A\eta \kappa^{\ast \eta -1}+ \left[ (A\eta \kappa^{\ast \eta -1})\beta \right]^{\frac{1}{1-\sigma }} \right] }$$

In general terms, that condition can be rewritten as

$$\begin{array}{@{}rcl@{}} &&{}\kappa^{\ast }(2-bn-B(1-n))\left[ A\eta \kappa^{\ast \eta -1}+\left[ (A\eta \kappa^{\ast \eta -1})\beta \right]^{\frac{1}{1-\sigma }}\right] \\ &=&\left[ A(1-\eta )\kappa^{\ast \eta }(1-bn)-ne\right] \left[ (A\eta \kappa^{\ast \eta -1})\beta \right]^{\frac{1}{1-\sigma }}\\&&-A(1-\eta )\kappa^{\ast \eta }(1-B(1-n))+(1-n)E \end{array}$$

Further simplifications yield

$$\begin{array}{@{}rcl@{}} &&{}A\eta \kappa^{\ast \eta }(2-bn-B(1-n))+\left[ (A\eta )\beta \right]^{ \frac{1}{1-\sigma }}\kappa^{\ast \frac{\eta -\sigma }{1-\sigma } }(2-bn-B(1-n)) \\ & {\kern10pt} =&A^{\frac{2-\sigma }{1-\sigma }}(1-\eta )\left(\eta \beta \right)^{\frac{ 1}{1-\sigma }}(1-bn)\kappa^{\ast \frac{2\eta -1-\eta \sigma }{1-\sigma }}-ne \left[ (A\eta )\beta \right]^{\frac{1}{1-\sigma }}\kappa^{\ast \frac{\eta -1}{1-\sigma }} \\ &&-A(1-\eta )\kappa^{\ast \eta }(1-B(1-n))+(1-n)E \end{array}$$

Denote $A^{\frac {2-\sigma }{1-\sigma }}(1-\eta )\left (\eta \beta \right )^{ \frac {1}{1-\sigma }}$by Γ and $\left [ \eta A\beta \right ]^{\frac {1}{ 1-\sigma }}$ by Θ. That condition can be rewritten as follows. There exists some interior level of early fertility n ∈ ]0, 1[ such that

$$\begin{array}{@{}rcl@{}} && {} A\eta \kappa^{\ast \eta }(2-bn-B(1-n))+\Theta \kappa^{\ast \frac{\eta -\sigma }{1-\sigma }}(2-bn-B(1-n)) \\ &&+ne\Theta \kappa^{\ast \frac{\eta -1}{1-\sigma }}+A(1-\eta )\kappa^{\ast \eta }(1-B(1-n))-(1-n)E =\kappa^{\ast \frac{2\eta -\eta \sigma -1}{1-\sigma }}(1-bn)\Gamma \end{array}$$

Denote the LHS by

$$\begin{array}{@{}rcl@{}} \chi \left(n\right) &\equiv &A\eta \kappa^{\ast \eta }(2-bn-B(1-n))+\Theta \kappa^{\ast \frac{\eta -\sigma }{1-\sigma }}(2-bn-B(1-n)) \\ &&+ne\Theta \kappa^{\ast \frac{\eta -1}{1-\sigma }}+A(1-\eta )\kappa^{\ast \eta }(1-B(1-n))-(1-n)E \end{array}$$

and the RHS by

$$\phi \left(n\right) \equiv \kappa^{\ast \frac{2\eta -\eta \sigma -1}{ 1-\sigma }}(1-bn)\Gamma$$

We have $\chi \left (0\right ) =A\eta \kappa ^{\ast \eta }(2-B)+\Theta \kappa ^{\ast \frac {\eta -\sigma }{1-\sigma }}(2-B)+A(1-\eta )\kappa ^{\ast \eta }(1-B)-E\gtrless 0$ and $\chi \left (1\right ) =A\eta \kappa ^{\ast \eta }(2-b)+\Theta \kappa ^{\ast \frac {\eta -\sigma }{1-\sigma }}(2-b)+e\Theta \kappa ^{\ast \frac {\eta -1}{1-\sigma }}+A(1-\eta )\kappa ^{\ast \eta }>0$.

Note also that, under B ≥ b, we have

$$\chi^{\prime }\left(n\right) =A\eta \kappa^{\ast \eta }(B-b)+\Theta \kappa^{\ast \frac{\eta -\sigma }{1-\sigma }}(B-b)+e\Theta \kappa^{\ast \frac{\eta -1}{1-\sigma }}+A(1-\eta )\kappa^{\ast \eta }B+E\geq 0$$

We have also $\phi \left (0\right ) =\kappa ^{\ast \frac {2\eta -\eta \sigma -1 }{1-\sigma }}\Gamma >0$ and $\phi (1)=\kappa ^{\ast \frac {2\eta -\eta \sigma -1}{1-\sigma }}(1-b)\Gamma >0$, as well as

$$\phi^{\prime }\left(n\right) =\kappa^{\ast \frac{2\eta -\eta \sigma -1}{ 1-\sigma }}(-b)\Gamma \leq 0$$

Hence, two sufficient conditions for the existence and uniqueness of n ^∗ are χ(0) < ϕ(0) and χ(1)>ϕ(1). In formulas

$$\left(A\eta \kappa^{\ast \eta }+\Theta \kappa^{\ast \frac{\eta -\sigma }{ 1-\sigma }}\right) (2-B)+A(1-\eta )\kappa^{\ast \eta }(1-B)-E<\kappa^{\ast \frac{2\eta -\eta \sigma -1}{1-\sigma }}\Gamma$$

and

$$\left(A\eta \kappa^{\ast \eta }+\Theta \kappa^{\ast \frac{\eta -\sigma }{ 1-\sigma }}\right) (2-b)+e\Theta \kappa^{\ast \frac{\eta -1}{1-\sigma } }+A(1-\eta )\kappa^{\ast \eta }>\kappa^{\ast \frac{2\eta -\eta \sigma -1}{ 1-\sigma }}(1-b)\Gamma.$$

When those two conditions are satisfied, we have that there exists a stationary equilibrium with a stationary cohort size.

Proof of Corollary 1

Let us start from expression (36):

$$\kappa =\frac{\left[ \begin{array}{l} \left[ A(1-\eta )\kappa^{\eta }(1-bn)-ne\right] \left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{1-\sigma }} \\ -A(1-\eta )\kappa^{\eta }(1-B(1-n))+(1-n)E \end{array} \right] }{(2-bn-B(1-n))\left[ A\eta \kappa^{\eta -1}+\left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{1-\sigma }}\right] }$$

Isolating n yields

$$n=\frac{{}\left[A(1{}-{}\eta )\kappa^{\eta }\right] \left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{1-\sigma }}{}-{}A(1{}-{}\eta )\kappa^{\eta }(1{}-{}B){}+{}E{}-{}\kappa (2{}-{}B){}\left[ A\eta \kappa^{\eta -1}{}+{}\left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{1-\sigma }}{}\right] }{\left[{}\kappa \left(B{}-{}b\right) \left[{}A\eta \kappa^{\eta -1}{}+{}\left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{1-\sigma }}\right] {}+{}\left[ A(1-\eta )\kappa^{\eta }b{}+{}e\right] \left[ (A\eta \kappa^{\eta -1})\beta \right]^{\frac{1}{ 1-\sigma }}{}+{}A(1{}-{}\eta )\kappa^{\eta }B{}+{}E\right] }$$

Given that κ = κ ^∗ at the equilibrium, we obtain n ^∗ by replacing κ by κ ^∗ in that expression. Similar substitutions can be made for the special cases of no time cost and no resource cost of children.

Proof of Lemma 2

Regarding $\frac {\partial \frac {{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}}$, we have, for s = 0:

$$\frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}}=\frac{\partial \frac{{N}_{t-1}}{{N}_{t}}}{\partial {n}_{t}}=0$$

For s = 1, s = 2, and s = 3:

$$\begin{array}{@{}rcl@{}} \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}} &=&\frac{\partial \frac{{N}_{t}}{{N}_{t+1}}}{\partial {n}_{t}}=\frac{-{N}_{t}{N}_{t}}{{N}_{t+1}^{2}} \\ \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}} {} &=& {}\frac{\partial \frac{{N}_{t+1}}{{N}_{t+2}}}{\partial {n}_{t}}{}={}\frac{{N}_{t}{N}_{t+2}-{N}_{t+1}\left({n}_{t+1}{N}_{t}\right) }{{N}_{t+2}^{2}}{}={}\frac{{N}_{t}\left({N}_{t+2}-{N}_{t+1}{n}_{t+1}\right) }{{N}_{t+2}^{2}}{}={}\frac{{{N}_{t}^{2}}m_{t+1}}{ {N}_{t+2}^{2}} \\ \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}} {}&=&{}\frac{\partial \frac{{N}_{t+2}}{{N}_{t+3}}}{\partial {n}_{t}}{}={}\frac{{N}_{t+3}\left({n}_{t+1}{N}_{t}\right) {}-{}{N}_{t+2}\left({n}_{t+2}{n}_{t+1}{N}_{t}+m_{t+2}{N}_{t}\right) }{ {N}_{t+3}^{2}}\\ &=&{}\frac{-{{N}_{t}^{2}}m_{t+1}m_{t+2}}{{N}_{t+3}^{2}} \end{array}$$

Hence, for s > 1, we have

$$\frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial {n}_{t}}=\frac{\left(-1\right)^{s}{{N}_{t}^{2}}m_{t+1}m_{t+2}...m_{t+s-1}}{{N}_{t+s}^{2}}$$

Regarding $\frac {\partial \frac {{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}}$, we have, for s = 0:

$$\frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}}=\frac{\partial \frac{{N}_{t+1}}{{N}_{t}}}{\partial {n}_{t}}=\frac{{{N}_{t}^{2}}}{{{N}_{t}^{2}}}=1$$

For s = 1, s = 2, and s = 3:

$$\begin{array}{@{}rcl@{}} \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}} &=&\frac{\partial \frac{{N}_{t+2}}{{N}_{t+1}}}{\partial {n}_{t}}=\frac{ {n}_{t+1}{N}_{t}{N}_{t+1}-{N}_{t+2}{N}_{t}}{{N}_{t+1}^{2}}=\frac{-{{N}_{t}^{2}}m_{t+1}}{ {N}_{t+1}^{2}} \\ \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}} {} &=& {}\frac{\partial \frac{{N}_{t+3}}{{N}_{t+2}}}{\partial {n}_{t}}{}={}\frac{{N}_{t+2}\left({n}_{t+2}{n}_{t+1}{N}_{t}{}+{}m_{t+2}{N}_{t}\right) {}-{}{N}_{t+3}\left({n}_{t+1}{N}_{t}\right) }{ {N}_{t+2}^{2}}{}={}\frac{{{N}_{t}^{2}}m_{t+1}m_{t+2}}{{N}_{t+2}^{2}} \\ \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}} &=&\frac{\partial \frac{{N}_{t+4}}{{N}_{t+3}}}{\partial {n}_{t}}=m_{t+3}\frac{\left({n}_{t+1}{N}_{t}\right) {N}_{t+3}-{N}_{t+2}\left({n}_{t+2}\left({n}_{t+1}{N}_{t}\right) +m_{t+2}{N}_{t}\right) }{{N}_{t+3}^{2}} \\ &=&\frac{-{{N}_{t}^{2}}m_{t+1}m_{t+2}m_{t+3}}{{N}_{t+3}^{2}} \end{array}$$

Hence, for s > 0, we have

$$\frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial {n}_{t}}=\frac{\left(-1\right)^{s}{{N}_{t}^{2}}m_{t+1}...m_{t+s}}{{N}_{t+s}^{2}}$$

Regarding $\frac {\partial \frac {{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}}$, we have, for s = 0:

$$\frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}}=\frac{{\partial} {\frac{{N}_{t-1}}{{N}_{t}}}}{\partial m_{t}}=0$$

For s = 1, s = 2, and s = 3:

$$\begin{array}{@{}rcl@{}} \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}} &=&\frac{\partial \frac{{N}_{t}}{{N}_{t+1}}}{\partial m_{t}}=\frac{-{N}_{t}{N}_{t-1}}{{N}_{t+1}^{2}} \\ \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}} &=&\frac{\partial \frac{{N}_{t+1}}{{N}_{t+2}}}{\partial m_{t}}=\frac{{N}_{t-1}{N}_{t+2}-{N}_{t+1}\left({n}_{t+1}{N}_{t-1}\right) }{{N}_{t+2}^{2}}=\frac{{N}_{t}{N}_{t-1}m_{t+1}}{{N}_{t+2}^{2}} \\ \frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}} &=&\frac{\partial \frac{{N}_{t+2}}{{N}_{t+3}}}{\partial m_{t}}=\frac{\left({n}_{t+1}{N}_{t-1}\right) {N}_{t+3}-{N}_{t+2}\left({n}_{t+2}\left({n}_{t+1}{N}_{t-1}\right) +m_{t+2}{N}_{t-1}\right) }{{N}_{t+3}^{2}} \\ &=&\frac{-{N}_{t}{N}_{t-1}m_{t+1}m_{t+2}}{{N}_{t+3}^{2}} \end{array}$$

Hence, for s > 1, we have

$$\frac{\partial \frac{{N}_{t+s-1}}{{N}_{t+s}}}{\partial m_{t}}=\frac{\left(-1\right)^{s}{N}_{t}{N}_{t-1}m_{t+1}...m_{t+s}}{{N}_{t+s}^{2}}$$

Finally, regarding $\frac {\partial \frac {{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}}$, we have, for s = 0:

$$\frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}}=\frac{\partial \frac{{N}_{t+1}}{{N}_{t}}}{\partial m_{t}}=\frac{{N}_{t-1}}{{N}_{t}}$$

For s = 1, s = 2, and s = 3:

$$\begin{array}{@{}rcl@{}} \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}} &=&\frac{\partial \frac{{N}_{t+2}}{{N}_{t+1}}}{\partial m_{t}}=\frac{-m_{t+1}{N}_{t}{N}_{t-1}}{{N}_{t+1}{~}^{2}} \\ \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}} {} &=& {}\frac{\partial \frac{{N}_{t+3}}{{N}_{t+2}}}{\partial m_{t}}{}={}m_{t+2}\frac{({N}_{t-1})\left({N}_{t+2}\right) {}-{}{N}_{t+1}\left({n}_{t+1}m_{t}{N}_{t-1}\right) }{\left({N}_{t+2}\right)^{2}}{}={}\frac{{N}_{t-1}{N}_{t}m_{t+1}m_{t+2}}{{N}_{t+2}^{2}} \\ \frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}} {}&=&{}\frac{\partial \frac{{N}_{t+4}}{{N}_{t+3}}}{\partial m_{t}}{}={}\frac{{}\left[ \begin{array}{l} {N}_{t+3}\left({n}_{t+3}\left({n}_{t+2}{n}_{t+1}{N}_{t-1}{}+{}m_{t+2}{N}_{t-1}\right) {}+{}m_{t{}+{}3}\left({n}_{t{}+{}1}{N}_{t-1}\right) \right) \\ -{N}_{t+4}\left({n}_{t+2}\left({n}_{t+1}{N}_{t-1}\right) {}+{}m_{t+2}{N}_{t-1}\right) \end{array} \right] }{{N}_{t+3}^{2}} \\ &=&\frac{-{N}_{t-1}{N}_{t}m_{t+1}m_{t+2}m_{t+3}}{{N}_{t+3}^{2}} \end{array}$$

Hence, for s > 0, we have

$$\frac{\partial \frac{{N}_{t+1+s}}{{N}_{t+s}}}{\partial m_{t}}=\frac{\left(-1\right)^{s}{N}_{t-1}{N}_{t}m_{t+1}m_{t+2}..m_{t+s}}{\left({N}_{t+s}\right)^{2} }$$

Proof of Proposition 2

Take the expression for κ _t+1 at g = 1, x _{t − 1} = x _t = x _t+1 and κ = κ ^∗. Isolating n yields

$$n=\frac{\left[ \begin{array}{l} \frac{\left[ A(1-\eta )\kappa^{\ast \eta }+xA\eta \kappa^{\ast \eta -1} \right] \left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{ 1-\sigma }}}{A\eta \kappa^{\ast \eta -1}}-\frac{A(1-\eta )\kappa^{\ast \eta }(1-B)}{A\eta \kappa^{\ast \eta -1}} \\ +\frac{E+x}{A\eta \kappa^{\ast \eta -1}}+\left(x-\kappa^{\ast }\left[ 1+(1-B)\right] \right) \left(1+\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}} \right) \end{array} \right] }{\left[ \begin{array}{l} \kappa \left[ -b+B\right] \left(1+\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}} \right) +A(1-\eta )\kappa^{\ast \eta }b\left(\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}}\right) \\ +\frac{A(1-\eta )\kappa^{\ast \eta }B}{A\eta \kappa^{\ast \eta -1}}+\frac{E }{A\eta \kappa^{\ast \eta -1}}+e\left(\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}} \right) \end{array} \right] }$$

When B − b ≥ 0, that expression is unambiguously increasing in x:

$$\frac{\partial n}{\partial x}=\frac{1+\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}{}+{}\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}+1}{A\eta \kappa^{\ast \eta -1} }}{\left[ \begin{array}{l} \kappa^{\ast }\left[ B{}-{}b\right] \left(1{}+{}\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}} \right) {}+{}A(1{}-{}\eta )\kappa^{\ast \eta }b\left(\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}}\right) \\ +\frac{A(1-\eta )\kappa^{\ast \eta }B}{A\eta \kappa^{\ast \eta -1}}+\frac{E }{A\eta \kappa^{\ast \eta -1}}+e\left(\frac{\left((A\eta \kappa^{\ast \eta -1})\beta \right)^{\frac{1}{1-\sigma }}}{A\eta \kappa^{\ast \eta -1}} \right) \end{array} \right] }>0$$