Differential Fertility

Notes for the benchmark model of de la Croix’s endogenous fertility and education model. This post mostly consists of mathematical derivations and important analytical results from De La Croix, D., & Doepke, M. (2003). Inequality and growth: why differential fertility matters. American Economic Review, 93(4), 1091-1113.

Setup

Household’s objective function:

\[U^i_t = \ln(c^i_t) + \gamma \ln(n^i_t \pi(e^i_t))\]

where $U$ is the life-time utility, $c$ is parent’s consumption, $n$ is number of children, $\pi$ is the educational quality of each child. Parents enjoy utility from raising children.

The probability for a child to become skilled (B type):

\[\pi(e^i) = \mu^i (\theta + e^i)^\eta\]

where $\mu$ is parent’s type $(i)$, specifically $\mu^B > \mu^A$ with B being high-skilled and A being low-skilled. High-skilled workers’ children tend to become more educated than the low-skilled counterpart’s. $\theta$ is the natural ability of a kid (without parent’s money for education) and $e$ is education input a child receives.

Since $1-\pi$ would be the prob. of a child becoming unskilled, the evolution of the population of each cohort is:

\[\begin{bmatrix} P^A_{t+1} \\ P^B_{t+1} \end{bmatrix} = \begin{bmatrix} n^A(1-\pi^A(e^A)) & n^B(1-\pi^B(e^B)) \\ n^A \pi^A(e^A) & n^B \pi^B(e^B) \end{bmatrix} \begin{bmatrix} P^A_t \\ P^B_t \end{bmatrix}\]

Parent’s budget constraint:

\[c_t = w(1-\phi n_t) - n_t e_t\]

where $\phi n$ is the fraction of time devoted to raise children and $ne$ is cost of raising children.

Simple production function:

\[Y_t = w^A L^A + w^B L^B\]

where $w$ is the wage-marginal product of labor, $L$ is labor input.

In equilibirum:

\[P^i_t(1-\phi n_t^i) = L^i_t\]

where $P$ is type-specific population. The LHS is labor supply and RHS is labor demand.

Solutions

(temporarily discard the superscript i notation)

Objective function:

\[\max_{c,e,n} \ U:= \ln(c) + \gamma\ln(n\mu(\theta+e)^\eta) \\ \text{s.t. } \ \ c = w(1-\phi n) - ne\]

Lagrangian:

\[\mathcal{L} = \ln(c) + \gamma \ln(n\mu(\theta+e)^\eta) + \lambda (w(1-\phi n) - ne - c)\]

FOC:

\[(\frac{\partial \mathcal{L}}{\partial c}): c = \frac{1}{\lambda} \\ (\frac{\partial \mathcal{L}}{\partial n}): \frac{\gamma}{n} = \lambda(w\theta+e) \iff n = \frac{\gamma}{\lambda(w\phi + e)} \\ (\frac{\partial \mathcal{L}}{\partial e}): \frac{\gamma \eta (\theta+e)^{\eta-1}}{(\theta + e)^\eta} = \lambda n \\ (\frac{\partial \mathcal{L}}{\partial \lambda}): c = w(1-\theta n) - ne\]

Plugging $n$ into $e$:

\[\frac{\gamma n}{\theta + e} = \frac{\gamma}{w\phi + e} \iff e = \frac{\eta w \phi - \theta}{1-\eta}\]

Plug this $e$ and $c$ back to $n$ to get:

\[n = \frac{\gamma c}{w\phi + e} \\ \iff n = \frac{\gamma [w-(w\phi+e)n]}{w\phi +e }\\ \iff n = \frac{\gamma w}{(w\phi +e)(1+\gamma)} \\ \iff n = \frac{\gamma w}{w\phi + \frac{\eta w \phi - \theta}{1-\eta} (1+\gamma)} \\ \iff n = \frac{(1-\eta)\gamma w}{(w\phi - \theta)(1+\gamma)}\]

Now solve for $c$:

\[c=w(1-\phi n) - ne \\ \iff c = w - \left(w\phi + \frac{\eta w \phi - \theta}{1-\eta}\right)\frac{(1-\eta)\gamma w}{(w\phi - \theta)(1+\gamma)} \\ \iff c = w - \frac{\gamma}{1+\gamma}w = \frac{w}{1+\gamma}\]

Comparative statics

We see that:

\[\frac{\partial e}{\partial w} = \frac{\eta \phi}{1-\eta} > 0\]

Parental education spending increases with income.

On the other hand:

\[\frac{\partial n}{\partial w} = - \theta \frac{(1-\eta)(1+\gamma)\gamma}{\left[(w\phi-\theta)(1+\gamma)\right]^2} < 0\]

Fertility decreases with income.

$e > 0$ if

\[w > \frac{\theta}{\eta \phi}\]

otherwise: if

\[( e \leq 0 ) \begin{cases} &e=0 \\ &n = \frac{\gamma w}{w\phi (1+\gamma)} = \frac{\gamma}{\phi(1+\gamma)} \end{cases}\]

From here, we can calculate the range of $n$:

For the lower bound:

\[\lim_{x\to \infty} n = \lim_{x\to \infty} \frac{(1-\eta)\gamma w}{(w\phi - \theta)(1+\gamma)}\\ = \lim_{x\to \infty} \frac{(1-\eta)\gamma}{(\phi - \frac{\theta}{w})(1+\gamma)} = \frac{(1-\eta)\gamma}{\phi(1+\gamma)}\]

For the upper bound:

\[\frac{\lim_{x\to0} n}{\lim_{x\to\infty}n} = \frac{\frac{\gamma}{\phi(1+\gamma)}}{\frac{(1-\eta)\gamma}{\theta(1+\gamma)}} = \frac{1}{1-\eta}\]

$x$ here stands for the relative wage of high-skill to low-skill, i.e. $\equiv \frac{w^H}{w^L}$.

Written on October 11, 2021