model-appendix.Rmd

\section*{Solution for the one-period model}

Differentiating and setting $\frac{dU}{dN}=0$ gives the first order
condition for an optimal choice of children $N^{*}>0$:

\[
\frac{bW}{(W(1-bN^{*}))^{\sigma}}\ge a\textrm{, with equality if }N^{*}>0.
\]
Rearranging gives
\begin{equation}
N^{*}=\max\left\{ \frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}W^{(1-\sigma)/\sigma}\right),0\right\} .\label{eq:N-one-period}
\end{equation}
Note that when $\sigma<1$, for high enough $W$, $N^{*}=0$. Differentiating
gives the effect of wages on fertility for $N^{*}>0$. This is also
the fertility-human capital relationship:

\begin{equation}
\frac{dN^{*}}{dh}=\frac{dN^{*}}{dW}=-\frac{1}{b}\left(\frac{b}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}W^{(1-2\sigma)/\sigma}.\label{eq:DNdW-one-period}
\end{equation}
This is negative if $\sigma<1$. Also, 
\[
\frac{d^{2}N^{*}}{dW^{2}}=-\frac{1}{b}\left(\frac{b}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}\frac{1-2\sigma}{\sigma}W^{(1-3\sigma)/\sigma}
\]
For $0.5<\sigma<1$, this is positive, so the effect of fertility
on wages shrinks towards zero as wages increase (and becomes 0 when
$N^{*}=0$). Next, we consider the time cost of children $b$:
\[
\frac{d^{2}N^{*}}{dWdb}=-\left(\frac{1}{a}\right)^{1/\sigma}\left(\frac{1-\sigma}{\sigma}\right)^{2}(Wb)^{(1-2\sigma)/\sigma}<0.
\]

Lastly we consider the effect of $a$. From (\ref{eq:N-one-period}),
$N^{*}$ is increasing in $a$. Differentiating (\ref{eq:DNdW-one-period})
by $a$ gives
\[
\frac{d^{2}N^{*}}{dadW}=b^{1/\sigma-1}\frac{1-\sigma}{\sigma^{2}}W^{(1-2\sigma)/\sigma}a^{-1/\sigma-1}
\]
which is positive for $\sigma<1$.

\section*{Solution for the two-period model}

Period 1 and period 2 income are:
\begin{align}
Y_{1} & =1-s-bN_{1}\label{eq:Y1}\\
Y_{2} & =w(s,h)(1-bN_{2})\label{eq:Y2}
\end{align}

Write the Lagrangian of utility $U$ (\ref{eq:U}) as 
\[
\mathcal{L}(N_{1},N_{2},s)=u(Y_{1})+u(Y_{2})+a(N_{1}+N_{2})+\lambda_{1}N_{1}+\lambda_{2}N_{2}+\lambda_{3}(\frac{1}{b}-N_{2})+\mu s
\]

Lemma \ref{lemma-concavity} below shows that if $\sigma>0.5$, this
problem is globally concave, guaranteeing that the first order conditions
identify a unique solution. We assume $\sigma>0.5$ from here on.

Plugging (\ref{eq:Y1}) and (\ref{eq:Y2}) into the above, we can
derive the Karush-Kuhn-Tucker conditions for an optimum $(N_{1}^{*},N_{2}^{*},s^{*})$
as:
\begin{align}
\frac{d\mathcal{L}}{dN_{1}}=-bY_{1}^{-\sigma}+a+\lambda_{1} & =0\textrm{, with }\lambda_{1}=0\textrm{ if }N_{1}^{*}>0;\label{eq:kkt1}\\
\frac{d\mathcal{L}}{dN_{2}}=-bs^{*}hY_{2}^{-\sigma}+a+\lambda_{2} & -\lambda_{3}=0\textrm{, with }\lambda_{2}=0\textrm{ if }N_{2}^{*}>0,\lambda_{3}=0\textrm{ if }N_{2}^{*}<\frac{1}{b};\label{eq:kkt2}\\
\frac{d\mathcal{L}}{ds}=-Y_{1}^{-\sigma}+h(1-bN_{2}^{*})Y_{2}^{-\sigma}+\mu & =0;\label{eq:kkt3}\\
N_{1}^{*},N_{2}^{*},s^{*},\lambda_{1},\lambda_{2},\lambda_{3},\mu & \ge0;N_{2}^{*}\le\frac{1}{b}.
\end{align}
Note that the Inada condition (that marginal utility of income grows without 
bound as income approaches zero, $\lim_{x\rightarrow0}u'(x)=\infty$)
for period 1 rules out $s^{*}=1$ and $N_{1}=1/b$, so we need not
impose these constraints explicitly. Also, so long as $N_{2}^{*}<1/b$,
the same condition rules out $s^{*}=0$. We consider four cases,
of which only three can occur.

\subsection*{Case 1: $N_{1}^{*}>0,N_{2}^{*}>0$}

Rearranging (\ref{eq:kkt1}), (\ref{eq:kkt2}) and (\ref{eq:kkt3})
gives:
\begin{align}
N_{1}^{*} & =\frac{1}{b}\left(1-s^{*}-\left(\frac{b}{a}\right)^{1/\sigma}\right);\label{eq:N11}\\
N_{2}^{*} & =\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(s^{*}h)^{(1-\sigma)/\sigma}\right);\label{eq:N21}\\
s^{*} & =\frac{1-bN_{1}^{*}}{1+\left((1-bN_{2}^{*})h\right)^{1-1/\sigma}}.
\end{align}
Plugging the expressions for $N_{1}^{*}$ and $N_{2}^{*}$ into $s^{*}$ gives
\[
s^{*}=\frac{s^{*}+\left(\frac{b}{a}\right)^{1/\sigma}}{1+\left(\left(\frac{b}{a}\right)^{1/\sigma}s^{*(1-\sigma)/\sigma}h{}^{1/\sigma}\right)^{1-1/\sigma}}
\]
which simplifies to
\begin{equation}
s^{*}=\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{(1-\sigma)/(2\sigma-1)}.\label{eq:s1}
\end{equation}

Plugging the above into (\ref{eq:N11}) and (\ref{eq:N21}) gives:
\begin{align*}
N_{1}^{*} & =\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{(1-\sigma)/(2\sigma-1)}-\left(\frac{b}{a}\right)^{1/\sigma}\right);\\
N_{2}^{*} & =\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{(1-\sigma)/(2\sigma-1)}\right).
\end{align*}
Note that that $N_{1}^{*}<N_{2}^{*}$. For these both to be positive
requires low values of $h$ if $\sigma<1$ and high values of $h$
if $\sigma>1$. Also:
\[
w(s^{*},h)\equiv s^{*}h=\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{\sigma/(2\sigma-1)}.
\]

Observe that $w(s^{*},h)$ is increasing in $h$ for $\sigma>0.5$,
and convex iff $0.5<\sigma<1$.  

While $N_{1}^{*}$ and $N_{2}^{*}$ are positive, they have the same
derivative with respect to $h$:
\begin{equation}
\frac{dN_{t}^{*}}{dh}=-\frac{1}{b}\left(\frac{b}{a}\right)^{1/(2\sigma-1)}\frac{1-\sigma}{2\sigma-1}h^{(1-\sigma)/(2\sigma-1)-1}\label{eq:dNdh1}
\end{equation}
Examining this and expression (\ref{eq:s1}) gives:
\begin{lem}
\label{case-1-lemma}For $\sigma<1$, case 1 holds for $h$ low enough,
and in case 1, $N_{1}^{*}$ and $N_{2}^{*}$ decrease in $h$, while
$s^{*}$ increases in $h$. 

For $\sigma>1$, case 1 holds for $h$ high enough, and in case 1
$N_{1}^{*}$ and $N_{2}^{*}$ increase in $h$, while $s^{*}$ decreases
in $h$. 

$N_{t}^{*}$ is convex in $h$ for $\sigma>2/3$, and concave otherwise.
$s^{*}$ is convex in $h$ if $\sigma<2/3$, and concave otherwise.
\end{lem}


\subsection*{Case 2: $N_{1}^{*}=0,N_{2}^{*}>0$}

Replace $N_{1}^{*}=0$ into the first order condition for $s^{*}$
from (\ref{eq:kkt3}), and rearrange to give:
\[
s^{*}=\frac{1}{1+\left((1-bN_{2})h\right)^{1-1/\sigma}}.
\]

Now since $N_{2}^{*}>0$, we can rearrange (\ref{eq:kkt2}) to give
\begin{equation}
N_{2}^{*}=\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(s^{*}h)^{(1-\sigma)/\sigma}\right).\label{eq:N22}
\end{equation}

Plugging this into $s^{*}$ gives
\begin{align*}
s^{*} & =\frac{1}{1+\left(\frac{bh}{a}\right)^{(\sigma-1)/\sigma^{2}}(s^{*}){}^{-(1-\sigma)^{2}/\sigma^{2}}}
\end{align*}
which can be rearranged to
\begin{equation}
(1-s^{*})(s^{*})^{(1-2\sigma)/\sigma^{2}}=\left(\frac{a}{bh}\right)^{(1-\sigma)/\sigma^{2}}.\label{eq:estar2}
\end{equation}
Differentiate the left hand side of the above to get
\begin{align}
 & \frac{1-2\sigma}{\sigma^{2}}(1-s^{*})(s^{*})^{(1-2\sigma)/\sigma^{2}-1}-(s^{*})^{(1-2\sigma)/\sigma^{2}}\nonumber \\
= & \frac{1-2\sigma}{\sigma^{2}}(s^{*})^{(1-2\sigma)/\sigma^{2}-1}-\frac{\sigma^{2}+1-2\sigma}{\sigma^{2}}(s^{*})^{(1-2\sigma)/\sigma^{2}}\nonumber \\
= & \frac{1-2\sigma}{\sigma^{2}}(s^{*})^{(1-2\sigma)/\sigma^{2}-1}-\frac{(1-\sigma)^{2}}{\sigma^{2}}(s^{*})^{(1-2\sigma)/\sigma^{2}}.\label{eq:destar-lhs}
\end{align}
This is negative if and only if
\begin{align*}
s^{*} & >\frac{1-2\sigma}{(1-\sigma)^{2}}
\end{align*}
which is always true since $\sigma>0.5$. Note also that since $\sigma>0.5$,
then the left hand side of (\ref{eq:estar2}) approaches infinity
as $s^{*}\rightarrow0$ and approaches 0 as $s^{*}\rightarrow1$.
Thus, (\ref{eq:estar2}) implicitly defines the unique solution for
$s^{*}$. 

To find how $s^{*}$ changes with $h$, note that the right hand side
of the above decreases in $h$ for $\sigma<1$, and increases in $h$
for $\sigma>1$. Putting these facts together: for $\sigma<1$, when
$h$ increases the RHS of (\ref{eq:estar2}) decreases, hence the
LHS decreases and $s^{*}$ increases, i.e. $s^{*}$ is increasing
in $h$. For $\sigma>1$, $s^{*}$ is decreasing in $h$.

To find how $N_{2}^{*}$ changes with $h$, we differentiate (\ref{eq:N22}):

\begin{align}
\frac{dN_{2}^{*}}{dh} & =-\frac{1}{b}\left(\frac{b}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}(s^{*}h)^{(1-2\sigma)/\sigma}(s^{*}+h\frac{ds^{*}}{dh})\label{eq:dNdh2}
\end{align}
which is negative for $\sigma<1$, since $\frac{ds^{*}}{dh}>0$ in
this case. 

Differentiating again:

\begin{align*}
\frac{d^{2}N_{2}}{dh^{2}} & =-X[\frac{1-2\sigma}{\sigma}(s^{*}h)^{(1-3\sigma)/\sigma}(s^{*}+h\frac{ds^{*}}{dh})^{2}+(s^{*}h)^{(1-2\sigma)/\sigma}(2\frac{ds^{*}}{dh}+h\frac{d^{2}s^{*}}{dh^{2}})]\\
 & =X(s^{*}h)^{(1-3\sigma)/\sigma}[\frac{2\sigma-1}{\sigma}(s^{*}+h\frac{ds^{*}}{dh})^{2}-(s^{*}h)(2\frac{ds^{*}}{dh}+h\frac{d^{2}s^{*}}{dh^{2}})]
\end{align*}
where $X=\frac{1}{b}\left(\frac{b}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}>0$.
Note that $\frac{d^{2}N_{2}}{dh^{2}}$ is continuous in $\sigma$
around $\sigma=1$. Note also from (\ref{eq:estar2}) that for $\sigma=1$,
$s^{*}$ becomes constant in $\sigma$. The term in square brackets
then reduces to $(s^{*})^{2}>0$. Putting these facts together, for
$\sigma$ sufficiently close to 1, $\frac{d^{2}N_{2}^{*}}{dh^{2}}>0$,
i.e. $N_{2}^{*}$ is convex in $h$. 

This case holds for intermediate values on $h$. Equation (\ref{eq:dNdh2}) 
shows that for $\sigma < 1$, $N_2$ decreases in $h$; the requirement that
$N_2>0$ therefore puts a maximum on $h$. When $\sigma>1$, $N_2$ increases in
$h$ and this puts a minimum on $h$. The requirement $N_1 = 0$ provides the 
other bound. Equation (\ref{eq:kkt1}) 
requires $-bY_{1}^{-\sigma}+a \le 0$ since $\lambda_1$ must be non-negative. 
The LHS is increasing in $Y_1$, and hence decreasing in $s$ as $Y_1 = 1-s$ since 
$N_1=0$. Lastly, optimal choice of education $s^*$ increases in $h$ for 
$\sigma<1$, and decreases for $\sigma > 1$. Hence for $\sigma < 1$, (\ref{eq:kkt1})
puts a minimum on $h$, and for $\sigma > 1$ it puts a maximum on $h$.


Summarizing:
\begin{lem}
\label{case-2-lemma}Case 2 holds for intermediate values of $h$.
In case 2: for $\sigma<1$, $s^{*}$ is increasing in $h$ and $N_{2}^{*}$
is decreasing in $h$. For $\sigma>1$, $s^{*}$is decreasing in $h$.
For $\sigma$ close enough to 1, $N_{2}^{*}$ is convex in $h$.
\end{lem}

\subsection*{Case 3: $N_{1}^{*}=0,N_{2}^{*}=0$}

We can solve for $s^{*}$ by substituting values of $Y_{1}$ and $Y_{2}$
into (\ref{eq:kkt3}):
\[
-(1-s^{*}){}^{-\sigma}+h(s^{*}h)^{-\sigma}=0
\]
which rearranges to 
\begin{equation}
s^{*}=\frac{1}{1+h^{(\sigma-1)/\sigma}}.\label{eq:estar1}
\end{equation}

Conditions (\ref{eq:kkt1}) and (\ref{eq:kkt2}) become:
\begin{align*}
-b(1-s^{*})^{-\sigma}+a & \le0\\
-bs^{*}h(s^{*}h)^{-\sigma}+a & \le0
\end{align*}
equivalently
\begin{align*}
\frac{a}{b} & \le(1-s^{*})^{-\sigma}\\
\frac{a}{b} & \le s^{*}h(s^{*}h)^{-\sigma}
\end{align*}
which can both be satisfied for $a/b$ close enough to zero. Note
from (\ref{eq:estar1}) that as $h\rightarrow\infty$, $s^{*}$ increases
towards 1 for $\sigma<1$, and decreases towards 0 for $\sigma>1$.
Note also that the right hand side of the first inequality above approaches
infinity as $s^{*}\rightarrow1$, therefore also as $h\rightarrow\infty$
for $\sigma<1$. Rewrite the second inequality as
\[
\frac{a}{b}<(s^{*}h)^{1-\sigma}=\left(\frac{h}{1+h^{(\sigma-1)/\sigma}}\right)^{1-\sigma}=\left(h^{-1}+h^{-1/\sigma}\right)^{\sigma-1}
\]
and note that again, as $h\rightarrow\infty$, the RHS increases towards
infinity for $\sigma<1$, and decreases towards zero otherwise. Thus,
for $\sigma<1$, both equations will be satisfied for $h$ high enough.
For $\sigma>1$, they will be satisfied for $h$ low enough. Summarizing
\begin{lem}
\label{case-3-lemma}For $\sigma<1$, case 3 holds for $h$ high enough,
and in case 3, $s^{*}$ increases in $h$. For $\sigma>1$, case 3
holds for $h$ low enough and $s^{*}$ decreases in $h$.
\end{lem}

\subsection*{Case 4: $N_{1}^{*}>0,N_{2}^{*}=0$}

Rearranging the first order conditions (\ref{eq:kkt1}) and (\ref{eq:kkt2})
for $N_{1}^{*}$ and $N_{2}^{*}$ gives
\begin{align*}
\frac{a}{b} & =(1-s^{*}-bN_{1}^{*})^{-\sigma}\\
\frac{a}{b} & \le s^{*}hY_{2}^{-\sigma}
\end{align*}
hence
\begin{align*}
(1-s^{*}-bN_{1}^{*})^{-\sigma} & \le s^{*}hY_{2}^{-\sigma}=(s^{*}h)^{1-\sigma}\\
\Leftrightarrow(1-s^{*}-bN_{1}^{*})^{\sigma} & \ge(s^{*}h)^{\sigma-1}\\
\Leftrightarrow1-s^{*}-bN_{1}^{*} & \ge(s^{*}h)^{1-1/\sigma}
\end{align*}

Now rearrange the first order condition for $s^{*}$ from (\ref{eq:kkt3}),
noting that since $N_{2}^{*}=0$, $s^{*}>0$ by the Inada condition.
\begin{align*}
h^{1/\sigma-1}(1-s^{*}-bN_{1}^{*}) & =s^{*}\\
1-s^{*}-bN_{1}^{*} & =s^{*}h^{1-1/\sigma}
\end{align*}
This, combined with the previous inequality, implies 
\begin{align*}
(s^{*}h)^{1-1/\sigma} & \le s^{*}h^{1-1/\sigma}\\
\Leftrightarrow(s^{*})^{-1/\sigma} & \le1
\end{align*}
which cannot hold since $0<s^{*}<1$. 

\subsection*{Comparative statics}

We can now examine how the fertility-human capital relationship 
\[
\frac{dN^{*}}{dh},\textrm{ where }N^{*}\equiv N_{1}^{*}+N_{2}^{*},
\]

changes with respect to other parameters. We focus on the case $\sigma<1$,
since it gives the closest match to our observations, and since it
also generates "reasonable" predictions in other areas, e.g. that
education levels increase with human capital. Figure \ref{fig:theory}
shows how $N^{*}$ changes with $h$ for $a=0.4,b=0.25,\sigma=0.7$.

\begin{figure}
\begin{centering}
\includegraphics[width=9cm, height=6cm]{N-plot}
\par\end{centering}
\caption{\label{fig:theory}Fertility vs. human capital in the two-period model
with $a=0.4,b=0.25,\sigma=0.7$.}
\end{figure}

\begin{lem}
For $\sigma < 1$ in a neighbourhood of 1, $N^*$ is globally convex in $h$.
\end{lem}
\begin{proof}
From Lemmas \ref{case-1-lemma}, \ref{case-2-lemma} and \ref{case-3-lemma},
as $h$ increases we move from $N_{1}^{*},N_{2}^{*}>0$ to $N_{1}^{*}=0,N_{2}^{*}>0$
to $N_{1}^{*}=N_{2}^{*}=0$. Furthermore, for $\sigma>2/3$, $N_{1}^{*}$
and $N_{2}^{*}$ are convex in $h$ when they are both positive, and
for $\sigma$ close enough to 1, $N_{2}^{*}$ is convex in $h$ when
$N_{1}^{*}=0$. All that remains is to check that the derivative is increasing around
the points where these 3 regions meet. That is trivially satisfied
where $N_{2}^{*}$ becomes 0, since thereafter $\frac{dN^{*}}{dh}$
is zero. The derivative as $N_{1}^{*}$ approaches zero is twice the
expression in (\ref{eq:dNdh1}):
\begin{equation}
-\frac{2}{b}\left(\frac{b}{a}\right)^{1/(2\sigma-1)}\frac{1-\sigma}{2\sigma-1}h^{(1-\sigma)/(2\sigma-1)-1}\label{eq:dNonleft}
\end{equation}
and the derivative to the right of this point is given by (\ref{eq:dNdh2}):
\begin{equation}
-\frac{1}{b}\left(\frac{b}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}(s^{*}h)^{(1-2\sigma)/\sigma}(s^{*}+h\frac{ds^{*}}{dh})\label{eq:dNonright}
\end{equation}

We want to prove that the former is larger in magnitude (i.e. more
negative). Dividing (\ref{eq:dNonleft}) by (\ref{eq:dNonright})
gives
\[
2\frac{\sigma}{2\sigma-1}\left(\frac{b}{a}\right)^{(1-\sigma)/(\sigma(2\sigma-1))}\frac{h^{(1-\sigma)^{2}/(\sigma(2\sigma-1))}}{s^{*}(s^{*}+h\frac{ds^{*}}{dh})}
\]

Examining (\ref{eq:estar2}) shows that as $\sigma\rightarrow1$,
$s^{*}\rightarrow0.5$ and $\frac{ds^{*}}{dh}\rightarrow0$, and therefore
the above approaches
\[
2\frac{1}{(0.5)^{2}}=8.
\]
\end{proof}

We can now gather the theoretical predictions stated in Table \ref{tab:theory}.

\textbf{Prediction 1}: for $\sigma<1$, total fertility $N^{*}\equiv N_{1}^{*}+N_{2}^{*}$
is decreasing in human capital $h$. 

Furthermore, for $\sigma$ close enough to 1, fertility is convex
in human capital, i.e. 

\textbf{Prediction 2 part 1}: the fertility-human capital relationship is
closer to 0 at high levels of $h$.

For $\sigma<1$, education levels $s^{*}$ increase in $h$, and so
therefore do equilibrium wages $w(s^{*},h)$. This, plus fact 1, gives:

\textbf{Prediction 2 part 2}: for $\sigma<1$ and close to 1, the fertility-human
capital relationship is weaker among higher earners.

\textbf{Prediction 4}: for $\sigma<1$ and close to 1, the fertility-human
capital relationship is weaker at high levels of education.

Next, we compare people who start fertility early ($N_{1}^{*}>0$)
versus those who start fertility late ($N_{1}^{*}=0$). Again, for
$\sigma<1$ the former group have lower $h$ than the latter group.
Thus we have:

\textbf{Prediction 5}: for $\sigma<1$ and close to 1, the fertility-human
capital relationship is weaker among those who start fertility late.

Lastly, we prove prediction 3. Differentiating $dN_{t}^{*}/dh$ in (\ref{eq:dNdh1})
with respect to $b$, for when $N_{1}^{*}>0$ gives:
\begin{align*}
\frac{d^{2}N_{t}^{*}}{dhdb} & =\frac{2\sigma-2}{2\sigma-1}b^{(3-4\sigma)/(2\sigma-1)}\left(\frac{1}{a}\right)^{1/(2\sigma-1)}\frac{1-\sigma}{2\sigma-1}h^{(\sigma-1)^{2}/(\sigma(2\sigma-1))}
\end{align*}

which is negative for $0.5<\sigma<1$. When $N_{1}^{*}=0$, differentiating
$dN_{2}^{*}/dh$ in (\ref{eq:dNdh2}) gives:

\[
\frac{d^{2}N_{2}^{*}}{dhdb}=-\frac{1-\sigma}{\sigma}b^{(1-2\sigma)/\sigma}\left(\frac{1}{a}\right)^{1/\sigma}\frac{1-\sigma}{\sigma}(s^{*}h)^{(1-2\sigma)/\sigma}(s^{*}+h\frac{ds^{*}}{dh})
\]
which again is negative for $\sigma<1$. Therefore:

\textbf{Prediction 3}: for $\sigma<1$, the fertility-human capital relationship
is more negative when the burden of childcare $b$ is larger.


\subsection*{Including a money cost}

The model can be extended by adding a money cost $m$ per child. Utility is
then

\[
U = u(1 - s - bN_1 - mN_1) + u(w(s,h)(1 - bN_2) - mN_2) + a(N_1 + N_2)
\]

Figure \ref{fig:N-plot-with-m} shows a computed example with 
$a = 0.4, b = 0.175, \sigma = 0.7, m = 0.075$. Fertility first declines steeply with
human capital, then rises. In addition, for parents with low AFLB ($N_1 > 0$), the
fertility-human capital relationship is negative, while for parents with 
higher AFLB ($N_1 = 0$) it is positive.

\begin{figure}
\begin{centering}
\includegraphics[width=9cm, height=6cm]{N-plot-with-m}
\par\end{centering}
\caption{\label{fig:N-plot-with-m}Fertility vs. human capital in the two-period model
with money costs of children. $a=0.4,b=0.175,\sigma=0.7, m = 0.075$.}
\end{figure}


\subsection*{Concavity}
\begin{lem}
\label{lemma-concavity}For $\sigma>0.5$, $U$ in equation (\ref{eq:U})
is concave in $N_{1},N_{2}$ and $s$.
\end{lem}
\begin{proof}
We examine the Hessian matrix of utility in each period. Note that
period 1 utility is constant in $N_{2}$ and period 2 utility is constant
in $N_{1}$. For period 1 the Hessian with respect to $N_{1}$ and
$s$ is: 
\[
\left[\begin{array}{cc}
d^{2}u/dN_{1}^{2} & d^{2}u/dsdN_{1}\\
d^{2}u/dsdN_{1} & d^{2}u/ds^{2}
\end{array}\right]=\left[\begin{array}{cc}
-\sigma b^{2} & -\sigma b\\
-\sigma b & -\sigma
\end{array}\right]Y_{1}^{-\sigma-1}
\]
with determinant 
\[
(\sigma^{2}b^{2}-\sigma^{2}b^{2})Y_{1}^{-2\sigma-2}=0.
\]
Thus, first period utility is weakly concave. For period 2 with respect
to $N_{2}$ and $s$, the Hessian is:

\[
\left[\begin{array}{cc}
d^{2}u/dN_{2}^{2} & d^{2}u/dsdN_{2}\\
d^{2}u/dsdN_{2} & d^{2}u/ds^{2}dN_{2}
\end{array}\right]=\left[\begin{array}{cc}
-\sigma(bsh)^{2}Y_{2}^{-\sigma-1} & -(1-\sigma)bhY_{2}^{-\sigma}\\
-(1-\sigma)bhY_{2}^{-\sigma} & -\sigma[h(1-bN_{2}^{*})]^{2}Y_{2}^{-\sigma-1}
\end{array}\right]
\]
with determinant
\begin{align*}
 & (-\sigma(bsh)^{2}Y_{2}^{-\sigma-1})(-\sigma[h(1-bN_{2}^{*})]^{2}Y_{2}^{-\sigma-1})-(-(1-\sigma)bhY_{2}^{-\sigma})^{2}\\
= & \sigma^{2}(bsh)^{2}[h(1-bN_{2}^{*})]^{2}Y_{2}^{-2\sigma-2}-(1-\sigma)^{2}(bh)^{2}Y_{2}^{-2\sigma}\\
= & \sigma^{2}(bh)^{2}Y_{2}^{-2\sigma}-(1-\sigma)^{2}(bh)^{2}Y_{2}^{-2\sigma}\textrm{, using that }Y_{2}=(sh)(1-bN)\\
= & (bh)^{2}Y_{2}^{-2\sigma}(\sigma^{2}-(1-\sigma)^{2})
\end{align*}
which is positive if and only if $\sigma>0.5$. Thus, if $\sigma>0.5$
then the Hessian is negative definite and thus utility is concave;
this combined with weak concavity of period 1, and linearity of $a(N_{1}+N_{2})$,
shows that (\ref{eq:U}) is concave.
\end{proof}

\subsection*{Effect of $a$}

\begin{lem}
\label{lemma-a}For $\sigma<1$, $d^2N^*/dadh > 0$, i.e. the effect of $a$
increases at higher levels of human capital.
\end{lem}
\begin{proof}
Differentiating (\ref{eq:dNdh1}) with respect to $a$ gives

\[
\frac{d^2N^*_t}{dadh} = -\frac{1}{b}\frac{1}{1-2\sigma}a^{-1}\left(\frac{a}{b}\right)^{1/(1-2\sigma)}\frac{1-\sigma}{2\sigma-1}h^{(1-\sigma)/(2\sigma-1)-1}
\]

for $t=1,2$ when $N^*_1,N^*_2>0$. For $\sigma > 0.5$ this is positive.

When $N^*_1 = 0,N^*_2>0$, differentiating (\ref{eq:dNdh2}) with respect to $a$ gives

\[
\frac{d^2N_{2}^{*}}{dadh} = \frac{1}{b}\frac{1}{\sigma}a^{-1}\left(\frac{a}{b}\right)^{-1/\sigma}\frac{1-\sigma}{\sigma}(s^{*}h)^{(1-2\sigma)/\sigma}(s^{*}+h\frac{ds^{*}}{dh})
\]

which is $-a^{-1}/\sigma$ times (\ref{eq:dNdh2}) and hence is positive.

\end{proof}