model-scratchpad.lyx

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\begin_body

\begin_layout Section
Thoughts
\end_layout

\begin_layout Itemize
Model below has 
\series bold
less
\series default
 effect of 
\begin_inset Formula $h$
\end_inset

 on 
\begin_inset Formula $N$
\end_inset

 among low-educated, because low education path does not translate 
\begin_inset Formula $h$
\end_inset

 into wages.
\end_layout

\begin_deeper
\begin_layout Itemize
The reality is that controlling for wages gets rid of the effect of PSEA.
 So, does PSEA have less effect on wages depending on education?
\end_layout

\begin_layout Itemize
Actually, if income_cat is the d.v., then PSEA has about 20% 
\bar under
higher
\bar default
 effect if you didn't go to university.
\end_layout

\begin_layout Itemize
But if first_job_pay (estimated by SOC code) is the d.v., then PSEA effect
 is about 40% 
\bar under
lower
\bar default
 if you didn't go to university; whether or not you use logs.
 For last_job_pay there's no effect; for cur_job_pay the PSEA effect is
 about 50% 
\bar under
higher
\bar default
 if you didn't go to university, but only significant when estimated in
 logs.
\end_layout

\begin_layout Itemize
If you use age_fte_cat instead of university, then more education makes
 the PSEA effect stronger for cur_job_pay; and for first_job_pay; and for
 income_cat; no effect for last_job_pay.
 (Why is age_fte_cat so different from university?)
\end_layout

\begin_layout Itemize
All these are just quick pooled regressions.
\end_layout

\begin_layout Itemize
Overall, evidence broadly favours that more education makes PSEA effect
 on 
\series bold
income
\series default
 steeper.
 
\end_layout

\end_deeper
\begin_layout Itemize
So then 
\series bold
why is PSEA effect on 
\bar under
fertility
\bar default
 steeper with 
\bar under
less
\bar default
 education
\series default
?
\end_layout

\begin_deeper
\begin_layout Itemize
No evidence that gender matters much here.
 Both men and women show very similar effects.
\end_layout

\begin_layout Itemize
One point is that LOTS of the PSEA effect on fertility must go 
\bar under
via
\bar default
 education.
 (Big -ve university dummy.)
\end_layout

\begin_layout Itemize
The effect of first_job_pay on children is 
\series bold
positive
\series default
; and (a) steeper at the bottom; and (b) flat for women, steep for men.
\end_layout

\begin_layout Itemize
The effect of age leaving education is 
\series bold
large
\series default
 and 
\series bold
negative
\series default
; and larger for women; and there's a 
\series bold
bigger gap
\series default
 between <16 and 16-18 than between 16-18 and >18.
 This continues to hold when we control for age, though the gap size is
 more even.
\end_layout

\begin_layout Itemize
Obvious conclusion: education choices, more than wages, drive the PSEA-fertility
 link.
\end_layout

\end_deeper
\begin_layout Section
Model
\end_layout

\begin_layout Standard
First, people observe their own human capital 
\begin_inset Formula $h$
\end_inset

 and choose a low- or high-education career path.
 The low-education career pays a wage 
\begin_inset Formula $w$
\end_inset

 in both periods.
 The high-education career pays 
\begin_inset Formula $e<w$
\end_inset

 in period 1, and 
\begin_inset Formula $h$
\end_inset

 in period 2.
\end_layout

\begin_layout Standard
In each period, people decide how many children to have.
 Each child has a cost 
\begin_inset Formula $b$
\end_inset

 in time and 
\begin_inset Formula $m$
\end_inset

 in money.
 Write 
\begin_inset Formula $N_{1},N_{2}$
\end_inset

 for the number of children in each period.
\end_layout

\begin_layout Standard
Utility is given by 
\begin_inset Formula 
\[
u(W_{1}(1-bN_{1})-mN_{1})+u(W_{2}(1-bN_{2})-mN_{2})+\alpha_{1}N_{1}+\alpha_{2}N_{2}
\]

\end_inset

where 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

 parametrize the payoff from having children in each period and 
\begin_inset Formula $W_{1},W_{2}$
\end_inset

 are the wages in each period.
 
\begin_inset Formula $u$
\end_inset

 is utility of consumption, assumed concave.
 We assume no borrowing is possible.
 As a result the choices of 
\begin_inset Formula $N_{1},N_{2}$
\end_inset

 can be worked out separately.
 In each period 
\begin_inset Formula $t=1,2$
\end_inset

 the FOC gives
\begin_inset Note Note
status open

\begin_layout Plain Layout
easy to add 
\begin_inset Formula $\delta$
\end_inset

 time preference
\end_layout

\end_inset


\begin_inset Formula 
\[
u'(W_{t}(1-bN_{t})-mN_{t})=\frac{\alpha_{t}}{m+bW_{t}}.
\]

\end_inset


\end_layout

\begin_layout Subsection
Log utility
\end_layout

\begin_layout Standard
If 
\begin_inset Formula $u(C)=log(C)$
\end_inset

 then the FOC works out to 
\begin_inset Formula 
\[
W_{t}(1-bN_{t})-mN_{t}=\frac{m+bW_{t}}{\alpha_{t}}
\]

\end_inset

which can be solved for 
\begin_inset Formula $N_{t}$
\end_inset

 explicitly, dropping subscripts:
\begin_inset Formula 
\[
N^{*}=\frac{W(1-b/\alpha)-m/\alpha}{Wb+m}.
\]

\end_inset

Per-period utility can also be written:
\begin_inset Formula 
\begin{align*}
 & \log(W(1-bN)-mN)+\alpha N\\
= & \log(W-(bW+m)N)+\alpha N\\
= & \log(\frac{Wb+m}{\alpha})+\frac{W(\alpha-b)-m}{Wb+m}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Total utility for the high-education path is then
\begin_inset Formula 
\[
\log(\frac{w_{1}b+m}{\alpha})+\frac{w_{1}(\alpha-b)-m}{w_{1}b+m}+\log(\frac{hb+m}{\alpha})+\frac{h(\alpha-b)-m}{hb+m}.
\]

\end_inset


\end_layout

\begin_layout Standard
Total utility for the low-education path is
\begin_inset Formula 
\[
2\left(\log(\frac{wb+m}{\alpha})+\frac{w(\alpha-b)-m}{wb+m}\right).
\]

\end_inset


\end_layout

\begin_layout Standard
There is a value of 
\begin_inset Formula $h$
\end_inset

 where these will be the same.
\begin_inset Note Note
status open

\begin_layout Plain Layout
Explicit solution, probably not.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout

\series bold
Goals
\end_layout

\begin_layout Itemize
if you choose higher education, you have children later.
\end_layout

\begin_layout Itemize
The number of children in period 2 is increasing in 
\begin_inset Formula $h$
\end_inset

.
 (Including both education paths!)
\end_layout

\begin_layout Itemize
On the high education path, the total number of children is increasing in
 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\begin_layout Itemize
Overall [sometimes? May need to let curvature of utility vary, hence isoelastic
 as below] the total number of children is decreasing in 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Isoelastic utility
\end_layout

\begin_layout Standard
If 
\begin_inset Formula $u(C)=\frac{C^{1-\sigma}-1}{1-\sigma}$
\end_inset

 for 
\begin_inset Formula $\sigma\ge0$
\end_inset

 then 
\begin_inset Formula $u'(C)=C^{-\sigma}$
\end_inset

 and the FOC becomes
\begin_inset Formula 
\[
W(1-bN)-mN=\left(\frac{m+bW}{\alpha}\right)^{1/\sigma}.
\]

\end_inset

Rearranging gives
\begin_inset Formula 
\[
N^{*}=\frac{W-\left(\frac{m+bW}{\alpha}\right)^{1/\sigma}}{Wb+m}.
\]

\end_inset


\end_layout

\begin_layout Standard
If 
\begin_inset Formula $m=0$
\end_inset

, then this reduces to 
\begin_inset Formula 
\[
N^{*}=\frac{W-\left(\frac{bW}{\alpha}\right)^{1/\sigma}}{Wb}=\frac{1}{b}-\frac{\left(bW\right)^{(1-\sigma)/\sigma}}{\alpha^{1/\sigma}}.
\]

\end_inset

which is increasing in 
\begin_inset Formula $W$
\end_inset

 for 
\begin_inset Formula $\sigma>1$
\end_inset

, constant for 
\begin_inset Formula $\sigma=1$
\end_inset

, and decreasing in 
\begin_inset Formula $W$
\end_inset

 for 
\begin_inset Formula $\sigma<1$
\end_inset

.
\end_layout

\begin_layout Subsection
Isoelastic subutility functions
\end_layout

\begin_layout Standard
Rewrite the utility function as 
\begin_inset Formula 
\[
u(W_{1}(1-bN_{1})-mN_{1})+\alpha_{1}v(N_{1})+u(W_{2}(1-bN_{2})-mN_{2})+\alpha_{2}v(N_{2})
\]

\end_inset

so the interior FOC in either period becomes:
\begin_inset Formula 
\[
(m+bW)u'(C)=\alpha v'(N)
\]

\end_inset

where 
\begin_inset Formula $C\equiv W(1-bN)-mN$
\end_inset

 is consumption.
 A nice way to write this might be
\begin_inset Formula 
\[
\frac{MV(C)}{MV(N)}=\frac{1}{m+bW}
\]

\end_inset

where 
\begin_inset Formula $MV$
\end_inset

 is the marginal value of consumption or children respectively, and the
 RHS is (1 over) the marginal effect of children on consumption.
 
\end_layout

\begin_layout Standard
Suppose both 
\begin_inset Formula $u$
\end_inset

 and 
\begin_inset Formula $v$
\end_inset

 are isoelastic.
 Note that both of these have Inada conditions, so we are guaranteed an
 interior optimum, with positive 
\begin_inset Formula $W$
\end_inset

 and 
\begin_inset Formula $N$
\end_inset

.
 Then the FOC is
\begin_inset Formula 
\[
(m+bW)C^{-\sigma}=\alpha N^{-\sigma}
\]

\end_inset


\end_layout

\begin_layout Standard
which solves to
\begin_inset Formula 
\[
N=\frac{QW}{1+Q(Wb+m)}
\]

\end_inset

 where 
\begin_inset Formula $Q=\left(\frac{\alpha}{m+bW}\right)^{1/\sigma}$
\end_inset

.
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
(m+bW)C^{-\sigma} & =\alpha N^{-\sigma}\\
(m+bW)N^{\sigma} & =\alpha(W(1-bN)-mN)^{\sigma}\\
\frac{N^{\sigma}}{(W(1-bN)-mN)^{\sigma}} & =\frac{\alpha}{m+bW}\\
\frac{N}{(W(1-bN)-mN)} & =Q
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
N & =Q(W(1-bN)-mN)\\
N & =\frac{QW}{1+Q(Wb+m)}
\end{align*}

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
A single utility for children
\end_layout

\begin_layout Standard
If we have non-linear utility for children it seems natural to have a single
 argument for 
\begin_inset Formula $N\equiv N_{1}+N_{2}$
\end_inset

.
 If so:
\begin_inset Formula 
\[
u(W_{1}(1-bN_{1})-mN_{1})+u(W_{2}(1-bN_{2})-mN_{2})+\alpha v(N).
\]

\end_inset


\end_layout

\begin_layout Standard
The FOC is now unified:
\begin_inset Formula 
\begin{align*}
(W_{1}b+m)u'(C_{1}) & =\alpha v'(N)\textrm{ if }N_{1}>0\\
(W_{2}b+m)u'(C_{2}) & =\alpha v'(N)\textrm{ if }N_{2}>0
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
implying equal marginal utility if you have children in both periods.
 Indeed, if you have 
\begin_inset Formula $W_{1}=W_{2}$
\end_inset

 then you're going to have equal numbers of children in each period.
 If 
\begin_inset Formula $N_{1}=0$
\end_inset

 then the first equality becomes a > sign, similarly for 
\begin_inset Formula $N_{2}$
\end_inset

.
\end_layout

\begin_layout Section
Single utility, no 
\begin_inset Formula $m$
\end_inset


\end_layout

\begin_layout Standard
Let's set 
\begin_inset Formula $m=0$
\end_inset

 and both 
\begin_inset Formula $u$
\end_inset

 and 
\begin_inset Formula $v$
\end_inset

 to be CRRA with risk aversion 
\begin_inset Formula $\sigma$
\end_inset

.
 Utility is
\begin_inset Formula 
\[
u(W_{1}(1-bN_{1}))+u(W_{2}(1-bN_{2}))+\alpha u(N)
\]

\end_inset

with 
\begin_inset Formula $u(x)=\frac{x^{1-\sigma}-1}{1-\sigma}$
\end_inset

.
\end_layout

\begin_layout Standard
The FOC is
\begin_inset Formula 
\begin{align*}
W_{1}bC_{1}^{-\sigma} & \ge\alpha N^{-\sigma}\textrm{, with equality if }N_{1}>0;\\
W_{2}bC_{2}^{-\sigma} & \ge\alpha N^{-\sigma}\textrm{, with equality if }N_{2}>0.
\end{align*}

\end_inset

with 
\begin_inset Formula $C_{t}=W_{t}-bN_{t}$
\end_inset

.
\end_layout

\begin_layout Subsection

\series bold
Low-education career.
 Suppose 
\begin_inset Formula $W_{1}=W_{2}=w$
\end_inset

.
 
\end_layout

\begin_layout Standard
Then suppose 
\begin_inset Formula $N_{1}<N_{2}$
\end_inset

.
 Hence 
\begin_inset Formula $C_{1}>C_{2}$
\end_inset

.
 Hence 
\begin_inset Formula $W_{1}bC_{1}^{-\sigma}<W_{2}bC_{2}^{-\sigma}=\alpha N^{-\sigma}$
\end_inset

, contradicting the above.
 A similar argument works if 
\begin_inset Formula $N_{1}>N_{2}$
\end_inset

.
 
\end_layout

\begin_layout Standard
Thus it must be that 
\begin_inset Formula $N_{1}=N_{2}$
\end_inset

 and hence 
\begin_inset Formula $C_{1}=C_{2}$
\end_inset

.
 Rearranging the FOC gives
\begin_inset Formula 
\begin{align*}
wb(w-bN_{1})^{-\sigma} & =\alpha(2N_{1})^{-\sigma}\\
\left(\frac{wb}{\alpha}\right)^{1/\sigma} & =\frac{w-bN_{1}}{2N_{1}}\\
N_{1}=N_{2} & =\frac{w}{b+2\left(\frac{wb}{\alpha}\right)^{1/\sigma}}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
From this we can work out
\begin_inset Formula 
\[
\frac{dN_{1}}{dw}=\frac{2(\sigma-1)\left(\frac{wb}{\alpha}\right)^{1/\sigma}+b\sigma}{\sigma(b+2\left(\frac{wb}{\alpha}\right)^{1/\sigma})^{2}}.
\]

\end_inset


\end_layout

\begin_layout Standard
This is positive if
\begin_inset Formula 
\begin{align*}
2(\sigma-1)\left(\frac{wb}{\alpha}\right)^{1/\sigma}+b\sigma & >0\\
\Leftrightarrow\sigma & >\frac{\left(\frac{wb}{\alpha}\right)^{1/\sigma}}{\left(\frac{wb}{\alpha}\right)^{1/\sigma}+b/2}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
which always holds for 
\begin_inset Formula $\sigma\ge1$
\end_inset

, and holds for any 
\begin_inset Formula $\sigma$
\end_inset

 when 
\begin_inset Formula $w$
\end_inset

 is small enough .
 
\end_layout

\begin_layout Standard
Note that here fertility doesn't easily decrease with 
\begin_inset Formula $w$
\end_inset

.
 However, on the low-education path, 
\begin_inset Formula $w$
\end_inset

 doesn't vary (across people), so this doesn't matter, and in particular
 
\begin_inset Formula $w$
\end_inset

 doesn't vary with 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\begin_layout Standard

\series bold
Some special cases:
\end_layout

\begin_layout Itemize
For log utility, 
\begin_inset Formula $N=N_{1}+N_{2}=\frac{2w}{b+2\left(\frac{wb}{\alpha}\right)}$
\end_inset

.
\end_layout

\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow\infty$
\end_inset

, 
\begin_inset Formula $N\rightarrow2w/(b+2).$
\end_inset


\end_layout

\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow0$
\end_inset

, if 
\begin_inset Formula $wb>\alpha$
\end_inset

 then 
\begin_inset Formula $N\rightarrow0$
\end_inset

; if 
\begin_inset Formula $wb<\alpha$
\end_inset

 then 
\begin_inset Formula $N\rightarrow2w/b$
\end_inset

.
\end_layout

\begin_layout Subsection

\series bold
High-education career.
 Suppose
\series default
 
\begin_inset Formula $W_{1}=e,W_{2}=h$
\end_inset

.
\end_layout

\begin_layout Standard
The FOC is
\begin_inset Formula 
\begin{align*}
ebC_{1}^{-\sigma} & \ge\alpha N^{-\sigma}\textrm{, with equality if }N_{1}>0;\\
hbC_{2}^{-\sigma} & \ge\alpha N^{-\sigma}\textrm{, with equality if }N_{2}>0.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
What are the conditions for 
\begin_inset Formula $N_{1}=0$
\end_inset

 or 
\begin_inset Formula $N_{2}=0$
\end_inset

? (Can't be both because of Inada-ish utility.)
\end_layout

\begin_layout Standard
If 
\begin_inset Formula $eb/C_{1}^{\sigma}>hb/C_{2}^{\sigma}$
\end_inset

 then we can have 
\begin_inset Formula $N_{1}=0$
\end_inset

 (but not 
\begin_inset Formula $N_{2}=0$
\end_inset

); if its the other way round we can have 
\begin_inset Formula $N_{2}=0$
\end_inset

 but not 
\begin_inset Formula $N_{1}=0$
\end_inset

.
\end_layout

\begin_layout Standard
First 
\series bold
assume both 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

 are positive
\series default
.
 This implies
\begin_inset Formula 
\begin{align*}
eb/(e-bN_{1})^{\sigma} & =hb/(h-bN_{2})^{\sigma}\\
\left(\frac{e}{h}\right)^{1/\sigma} & =\frac{e-bN_{1}}{h-bN_{2}}.
\end{align*}

\end_inset


\end_layout

\begin_layout Itemize
In general as 
\begin_inset Formula $\sigma$
\end_inset

 increases, the LHS goes to 1; as it decreases the LHS goes to zero which
 means (?) that 
\begin_inset Formula $N_{1}$
\end_inset

 approaches 
\begin_inset Formula $e/b$
\end_inset

.
\end_layout

\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow\infty$
\end_inset

 this reduces to 
\begin_inset Formula $e-bN_{1}=h-bN_{2}$
\end_inset

 hence 
\begin_inset Formula $N_{2}=N_{1}+\frac{h-e}{b}.$
\end_inset

 
\end_layout

\begin_layout Itemize
For log utility with 
\begin_inset Formula $\sigma=1$
\end_inset

 we get 
\begin_inset Formula $N_{2}=(h/e)N_{1}$
\end_inset

.
\end_layout

\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow0$
\end_inset

, the left hand side goes to zero; 
\begin_inset Formula $N_{1}\rightarrow e/b;$
\end_inset

 then solving for 
\begin_inset Formula $N_{2}$
\end_inset

 gives 
\begin_inset Formula $N_{2}\rightarrow P+Re/b.$
\end_inset


\end_layout

\begin_layout Standard
The above relates 
\begin_inset Formula $N_{1}$
\end_inset

 to 
\begin_inset Formula $N_{2}$
\end_inset

.
 In fact we can solve explicitly for 
\begin_inset Formula $N_{2}$
\end_inset

 as an affine function of 
\begin_inset Formula $N_{1}$
\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
checked that this reduces to the above for 
\begin_inset Formula $\sigma=\infty,1$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
N_{2}=P+RN_{1}.
\]

\end_inset


\end_layout

\begin_layout Standard
where 
\begin_inset Formula $P=\frac{h-e(h/e)^{1/\sigma}}{b}$
\end_inset

, 
\begin_inset Formula $R=\left(\frac{h}{e}\right)^{1/\sigma}$
\end_inset

.
\end_layout

\begin_layout Standard
Now the conditions for both to be positive are 
\begin_inset Formula 
\begin{align*}
eb(e-bN_{1})^{-\sigma} & =\alpha(N_{1}+N_{2})^{-\sigma}\\
eb(e-bN_{1})^{-\sigma} & =\alpha(N_{1}+N_{2})^{-\sigma}\\
\frac{eb}{\alpha} & =\left(\frac{e-bN_{1}}{N_{1}+N_{2}}\right)^{\sigma}\\
Q & =\frac{e-bN_{1}}{N_{1}+N_{2}}\textrm{ where }Q\equiv\left(\frac{eb}{\alpha}\right)^{1/\sigma}\\
Q(N_{1}+N_{2}) & =e-bN_{1}\\
Q(N_{1}+P+RN_{1}) & =e-bN_{1}\\
N_{1} & =\frac{e-QP}{Q(1+R)+b}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Some special cases:
\end_layout

\begin_layout Itemize
For log utility, 
\begin_inset Formula $Q=\frac{eb}{\alpha},P=0;R=h/e$
\end_inset

.
 
\begin_inset Formula $N_{1}=\frac{e}{b[(e+h)/\alpha+1]}$
\end_inset

 and 
\begin_inset Formula $N_{2}=\frac{h}{b[(e+h)/\alpha+1]}.$
\end_inset

 Total 
\begin_inset Formula $N=\frac{e+h}{b[(e+h)/\alpha+1]}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Differentiating 
\begin_inset Formula $N$
\end_inset

 wrt 
\begin_inset Formula $h$
\end_inset

 gives that it is signed by 
\begin_inset Formula $b$
\end_inset

, hence always positive.
\end_layout

\end_deeper
\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow\infty$
\end_inset

, 
\begin_inset Formula $Q\rightarrow1$
\end_inset

;
\begin_inset Formula $P\rightarrow\frac{h-e}{b}$
\end_inset

; and 
\begin_inset Formula $R\rightarrow1$
\end_inset

.
 Hence 
\begin_inset Formula $N_{1}\rightarrow\frac{e-(h-e)/b}{2+b}$
\end_inset

 and 
\begin_inset Formula $N_{2}\rightarrow N_{1}+\frac{h-e}{b}$
\end_inset

, giving:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $N=2\frac{e-(h-e)/b}{2+b}+\frac{h-e}{b}$
\end_inset

.
 Differentiating wrt 
\begin_inset Formula $h$
\end_inset

 gives 
\begin_inset Formula $dN/dh=\frac{1}{b}(1-2/(2+b))>0.$
\end_inset


\end_layout

\begin_layout Itemize
Intuitively: when you're 
\begin_inset Quotes eld
\end_inset

risk averse
\begin_inset Quotes erd
\end_inset

, having more money makes it much less costly to have children.
\end_layout

\end_deeper
\begin_layout Itemize
As 
\begin_inset Formula $\sigma\rightarrow0$
\end_inset

: 
\begin_inset Note Note
status open

\begin_layout Plain Layout
I think!
\end_layout

\end_inset


\begin_inset Formula $N_{1}\rightarrow e/b$
\end_inset

.
 
\begin_inset Formula $N_{2}$
\end_inset

 goes to zero because 
\begin_inset Formula $P\rightarrow-\infty$
\end_inset

, at which point these equations break down.
\end_layout

\begin_deeper
\begin_layout Itemize
From computing, for low values of 
\begin_inset Formula $\sigma$
\end_inset

, it's possible to get fertility declining in 
\begin_inset Formula $h$
\end_inset

 while both 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

 are positive.
\end_layout

\begin_layout Itemize
e.g.
 for 
\begin_inset Formula $e=0.5,\alpha=.5,b=0.8,\sigma=0.5$
\end_inset

, and 
\begin_inset Formula $h$
\end_inset

 less than about 1.2.
\end_layout

\end_deeper
\begin_layout Standard

\series bold
What will overall profile of fertility look like with respect to 
\begin_inset Formula $h$
\end_inset

?
\end_layout

\begin_layout Itemize
Within the low-education group (which will be low 
\begin_inset Formula $h$
\end_inset

), fertility is flat in 
\begin_inset Formula $h$
\end_inset

 and is given by
\begin_inset Formula 
\[
N_{1}=N_{2}=\frac{w}{b+2\left(\frac{wb}{\alpha}\right)^{1/\sigma}}.
\]

\end_inset


\end_layout

\begin_layout Itemize
Within the high-education group, fertility can be going up or down in 
\begin_inset Formula $h$
\end_inset

.
 
\end_layout

\begin_layout Subsection

\series bold
Special-case to log utility.
\end_layout

\begin_layout Itemize
Within the low-education group, 
\begin_inset Formula $N=\frac{2w}{b+2wb/\alpha}.$
\end_inset


\end_layout

\begin_layout Itemize
Within the high-education group: 
\begin_inset Formula $N_{1}=\frac{e}{b[(e+h)/\alpha+1]}$
\end_inset

 and 
\begin_inset Formula $N_{2}=\frac{h}{b[(e+h)/\alpha+1]}.$
\end_inset

 Total 
\begin_inset Formula $N=\frac{e+h}{b+b(e+h)/\alpha}$
\end_inset

.
 Notice the similarity in form between the two groups.
 In particular, high education have more kids whenever 
\begin_inset Formula $(e+h)/2>w$
\end_inset

.
\end_layout

\begin_layout Itemize
We now have: conditional on low education, 
\begin_inset Formula $N$
\end_inset

 is flat in 
\begin_inset Formula $h$
\end_inset

.
 Conditional on high education, it's increasing.
 But overall, fix 
\begin_inset Formula $h^{*}$
\end_inset

 as the level of human capital at which people choose the high education
 career.
 Then if 
\begin_inset Formula $(e+h^{*})/2<w$
\end_inset

, fertility jumps down at this point.
 It then increases.
 
\end_layout

\begin_layout Standard
When do people switch?
\end_layout

\begin_layout Standard
Utility for the low education track is
\begin_inset Formula 
\begin{align*}
 & 2\log(w(1-bN_{1}))+\alpha\log(2N_{1})\\
= & 2\log w+2\log(1-bN_{1})+\alpha\log(2N_{1})\\
= & 2\log w+2\log(1-\frac{w}{1+2w/\alpha})+\alpha\log(\frac{1}{b}\frac{2w}{1+2w/\alpha})\\
= & 2\log w+2\log(1-\frac{w}{1+2w/\alpha})+\alpha\log(\frac{1}{b})+\alpha\log(\frac{2w}{1+2w/\alpha}).
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Utility for the high education track depends on 
\begin_inset Formula $h$
\end_inset

.
 It is 
\begin_inset Formula 
\begin{align*}
 & \log(e(1-bN_{1}))+\log(h(1-bN_{2}))+\alpha\log(N)\\
= & \log(e(1-\frac{e}{(e+h)/\alpha+1}))+\log(h(1-\frac{h}{[(e+h)/\alpha+1]}))+\alpha\log(\frac{e+h}{b(1+(e+h)/\alpha)})
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Intuition: look at the two utilities and spot a way to make them comparable,
 e.g.
 when 
\begin_inset Formula $e+h=2w$
\end_inset

 the last terms are the same.
 Possibly whenever you change you also have more kids in this setup.
\end_layout

\begin_layout Standard
Other intuition.
 Higher education have more kids whenever 
\begin_inset Formula $(e+h)/2>w$
\end_inset

 equivalently 
\begin_inset Formula $h>2w-e$
\end_inset

.
 So just check when switching point is less than 
\begin_inset Formula $2w-e$
\end_inset

.
 In particular, if you plug 
\begin_inset Formula $h=2w-e$
\end_inset

 into the higher education utility, do you always get more utility than
 for lower education?
\end_layout

\begin_layout Section
One period model
\end_layout

\begin_layout Standard
Here's a very simple model in which people choose number of children overall.
 
\begin_inset Formula 
\[
u(W-bWN-mN)+\alpha u(N)
\]

\end_inset


\end_layout

\begin_layout Standard
Start with 
\begin_inset Formula $u=\log$
\end_inset

.
 Differentiate by 
\begin_inset Formula $N$
\end_inset

 to find the FOC:
\begin_inset Formula 
\[
\frac{bW+m}{W-bWN-mN}=\frac{\alpha}{N}
\]

\end_inset


\end_layout

\begin_layout Standard
Note that if 
\begin_inset Formula $m=0$
\end_inset

 the 
\begin_inset Formula $W$
\end_inset

's cancel out and 
\begin_inset Formula $dN^{*}/dW=0$
\end_inset

.
\end_layout

\begin_layout Standard
If not, then we can solve to:
\begin_inset Formula 
\[
N^{*}=\frac{\alpha}{1+\alpha}\frac{W}{bW+m}
\]

\end_inset


\end_layout

\begin_layout Standard
Differentiating with respect to 
\begin_inset Formula $W$
\end_inset

 gives.
 
\begin_inset Formula 
\[
\frac{dN^{*}}{dW}=\frac{\alpha}{1+\alpha}\frac{m}{(bW+m)^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
This is signed by 
\begin_inset Formula $m$
\end_inset

; and approaches 0 as 
\begin_inset Formula $W$
\end_inset

 increases.
 Thus if 
\begin_inset Formula $m$
\end_inset

 is negative – for example if people receive cash benefits for children
 – then we get the pattern we observe.
 
\end_layout

\begin_layout Standard
More generally for
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
u(x)=\frac{x^{1-\sigma}-1}{1-\sigma}
\]

\end_inset


\end_layout

\begin_layout Standard
the FOC is
\begin_inset Formula 
\begin{align*}
(W-bWN-mN)^{-\sigma}(bW+m) & =\alpha N^{-\sigma}\\
(N/(W-bWN-mN))^{\sigma} & =\alpha/(bW+m)\\
(N/(W-bWN-mN)) & =1/Q\\
Q & =(W-bWN-mN)/N\\
NQ & =(W-bWN-mN)\\
N(Q+bW+m) & =W\\
N^{*} & =W/(Q+bW+m).
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
where 
\begin_inset Formula 
\[
Q\equiv[(bW+m)/\alpha]^{1/\sigma}.
\]

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
Check this for 
\begin_inset Formula $\sigma=1$
\end_inset

: it reduces to the expression for log utility above.
\begin_inset Formula 
\[
N^{*}=\frac{\alpha}{1+\alpha}\frac{W}{bW+m}.
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
The derivative of this is
\begin_inset Formula 
\begin{align*}
\frac{dN^{*}}{dW} & =\frac{1}{(Q+bW+m)^{2}}[Q+bW+m-W(Q'+b)]\\
 & =\frac{1}{(Q+bW+m)^{2}}[Q+m-WQ']
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Where 
\begin_inset Formula 
\begin{align*}
Q'\equiv\frac{dQ}{dW} & =(1/\sigma)(b/\alpha)[(bW+m)/\alpha]^{1/\sigma-1}\\
 & =(1/\sigma)(b/\alpha)Q\frac{\alpha}{bW+m}\\
 & =\frac{1}{\sigma}Q\frac{b}{bW+m}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Hence
\begin_inset Formula 
\[
\frac{dN^{*}}{dW}=\frac{1}{(Q+bW+m)^{2}}[Q+m-\frac{1}{\sigma}Q\frac{bW}{bW+m}]
\]

\end_inset


\end_layout

\begin_layout Standard
We can try a couple of reductions.
\begin_inset Note Note
status open

\begin_layout Plain Layout
First if 
\begin_inset Formula $\sigma=1$
\end_inset

 this becomes
\begin_inset Formula 
\begin{align*}
 & \frac{1}{((1+1/\alpha)(bW+m))^{2}}[(bW+m)/\alpha+m-bW/\alpha]\\
= & \frac{1}{((1+1/\alpha)(bW+m))^{2}}[m(\alpha+1)/\alpha]
\end{align*}

\end_inset

which indeed is signed by 
\begin_inset Formula $m$
\end_inset

 and reduces to the log expression above.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Second, if 
\begin_inset Formula $m=0$
\end_inset

:
\begin_inset Formula 
\begin{align*}
\frac{dN^{*}}{dW} & =\frac{1}{(Q+bW)^{2}}[Q-W(\frac{1}{\sigma}Q\frac{b}{bW})]\\
 & =\frac{1}{(Q+bW)^{2}}\frac{\sigma-1}{\sigma}Q
\end{align*}

\end_inset

This is now signed by 
\begin_inset Formula $\sigma-1$
\end_inset

.
 (Thus, if kids only cost time, then what matters is the degree if risk
 aversion.
 But most estimates of 
\begin_inset Formula $\sigma$
\end_inset

 have it greater than 1.
 A possible argument is that the human capital-fertility relationship is
 more negative for riskloving people, and risk aversion might correlate
 with income.
 That's like the economic interpretation of your 
\begin_inset Quotes eld
\end_inset

alternative argument
\begin_inset Quotes erd
\end_inset

, and you can rule it out – to some degree – with the risk question.) 
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Quotes eld
\end_inset

The magnitude of effects also decrease with 
\begin_inset Formula $W$
\end_inset

, since 
\begin_inset Formula $Q/(Q+bW)^{2}$
\end_inset

 decreases with 
\begin_inset Formula $W$
\end_inset

.
\begin_inset Quotes erd
\end_inset


\end_layout

\begin_layout Plain Layout
Check on the computer.
 Hmm.
 Not for 
\begin_inset Formula $\alpha=0.1,b=0.1,\sigma=0.2,m=0$
\end_inset

 and 
\begin_inset Formula $W<0.5$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Probably could figure out 
\begin_inset Formula $d^{2}N^{*}/dWdm$
\end_inset

.
 Computer shows that it isn't always increasing.
\end_layout

\begin_layout Section
Linear in children
\end_layout

\begin_layout Standard
The logic is that you want to display the income and substitution effects
 very simply.
 For this you only need utility in income to be nonlinear.
 Similarly, get rid of money costs as they aren't needed to display the
 substitution effect.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
u(W-bNW)+aN
\]

\end_inset


\end_layout

\begin_layout Standard
with 
\begin_inset Formula $u$
\end_inset

 CRRA gives FOC
\begin_inset Formula 
\[
\frac{bW}{(W-bNW)^{\sigma}}=a
\]

\end_inset


\end_layout

\begin_layout Standard
Rearrange to
\begin_inset Formula 
\begin{align*}
Q\equiv\left(\frac{b}{a}W\right)^{1/\sigma} & =(W-bNW)\\
N^{*} & =\frac{W-Q}{bW}\\
 & =\frac{1}{b}-\frac{Q}{bW}\\
 & =\frac{1}{b}-\frac{b^{(1-\sigma)/\sigma}}{a^{1/\sigma}}W^{(1-\sigma)/\sigma}
\end{align*}

\end_inset

We can then differentiate 
\begin_inset Formula 
\[
\frac{dN^{*}}{dW}=-\frac{b^{(1-\sigma)/\sigma}}{a^{1/\sigma}}\frac{1-\sigma}{\sigma}W^{(1-2\sigma)/\sigma}
\]

\end_inset

which is signed by 
\begin_inset Formula $\sigma-1$
\end_inset

.
 So, if you're not very risk averse (
\begin_inset Formula $\sigma<1$
\end_inset

) then fertility decreases in wages.
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
NB: if 
\begin_inset Formula $\sigma>1$
\end_inset

 then you may have negative 
\begin_inset Formula $N^{*}$
\end_inset

...
 at 
\begin_inset Formula $\sigma=1$
\end_inset

, 
\begin_inset Formula $N^{*}=1/b-1/a$
\end_inset

.
\end_layout

\end_inset

Also 
\begin_inset Formula 
\[
\frac{d^{2}N^{*}}{dW^{2}}\textrm{ is signed by }(\sigma-1)(1-2\sigma)
\]

\end_inset

where the second term comes from the exponent on 
\begin_inset Formula $W$
\end_inset

.
 For 
\begin_inset Formula $1>\sigma>1/2$
\end_inset

, this is positive, so the effect of fertility on wages shrinks towards
 zero as wages increase.
 For 
\begin_inset Formula $\sigma<1/2$
\end_inset

 the effect gets more negative as wages increase.
\end_layout

\begin_layout Standard
We can also check how 
\begin_inset Formula $N^{*}$
\end_inset

 is affected by other variables.
 Unsurprisingly, it decreases in 
\begin_inset Formula $b$
\end_inset

 and increases in 
\begin_inset Formula $a$
\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
check 
\begin_inset Formula $b$
\end_inset

.
 Presumably even if 
\begin_inset Formula $\sigma>1$
\end_inset

 the 
\begin_inset Formula $1/b$
\end_inset

 term dominates
\end_layout

\end_inset

.
 For 
\begin_inset Formula $\sigma$
\end_inset

 write
\begin_inset Formula 
\[
N^{*}=\frac{1}{b}-\left(\frac{bW}{a}\right)^{1/\sigma}\frac{1}{bW}.
\]

\end_inset

Thus 
\begin_inset Formula $N^{*}$
\end_inset

 increases in 
\begin_inset Formula $\sigma$
\end_inset

 if 
\begin_inset Formula $W>a/b$
\end_inset

 and decreases otherwise.
\begin_inset Note Note
status open

\begin_layout Plain Layout
It's possible for either of these to take place while 
\begin_inset Formula $N^{*}$
\end_inset

 is positive.
\end_layout

\end_inset

 
\end_layout

\begin_layout Section
Education model
\end_layout

\begin_layout Standard
Suppose there are two periods.
 For convenience we ignore time discounting and assume the interest rate
 is 0; we also assume that there are no credit markets.
 Write
\begin_inset Formula 
\[
U(N_{1},N_{2})=u(\bar{W}_{1})+u(\bar{W}_{2})+a(N_{1}+N_{2})
\]

\end_inset

where 
\begin_inset Formula $\bar{W_{t}}=W_{t}(1-bN_{t})$
\end_inset

.
 We continue to assume CRRA utility.
 The choice problems of 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

 are separable, and the per-period optimal fertility is
\begin_inset Formula 
\[
N_{t}^{*}=\max\left\{ \frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}W_{t}^{(1-\sigma)/\sigma}\right),0\right\} 
\]

\end_inset

just as before.
 Write
\begin_inset Formula 
\begin{align*}
W_{t} & =w_{t}(h,e)
\end{align*}

\end_inset

where 
\begin_inset Formula $h\ge0$
\end_inset

 is an individual's human capital and 
\begin_inset Formula $e\in[0,1]$
\end_inset

 is a choice variable representing the level of education.
 We assume that 
\begin_inset Formula $w_{1}$
\end_inset

 is decreasing in 
\begin_inset Formula $e$
\end_inset

 while 
\begin_inset Formula $w_{2}$
\end_inset

 is increasing in 
\begin_inset Formula $e$
\end_inset

; that both 
\begin_inset Formula $w_{1}$
\end_inset

 and 
\begin_inset Formula $w_{2}$
\end_inset

 are weakly increasing in 
\begin_inset Formula $h$
\end_inset

; and that 
\begin_inset Formula $d^{2}w_{t}/dedh\ge0$
\end_inset

.
\end_layout

\begin_layout Standard
Suppose that 
\begin_inset Formula $N_{t}^{*}>0$
\end_inset

.
 Then we can apply the envelope theorem to show that 
\begin_inset Formula 
\[
\frac{dU}{dW_{t}}=\frac{\partial U}{\partial W_{t}}
\]

\end_inset

since 
\begin_inset Formula $N_{t}^{*}$
\end_inset

 is at an interior optimum.
 Writing 
\begin_inset Formula 
\begin{align*}
\frac{\partial u}{\partial W_{t}} & =(1-bN_{t})(W_{t}(1-bN_{t}))^{-\sigma}\\
 & =(1-bN_{t})^{1-\sigma}W_{t}{}^{-\sigma}
\end{align*}

\end_inset

we can calculate
\begin_inset Formula 
\[
\frac{\partial u}{\partial e}=\frac{\partial w_{t}}{\partial e}(e,h)\times(1-bN_{t})^{1-\sigma}w_{t}(e,h){}^{-\sigma}
\]

\end_inset

for 
\begin_inset Formula $t=1,2$
\end_inset

.
 An internal optimum choice of education 
\begin_inset Formula $e$
\end_inset

 then solves
\begin_inset Formula 
\[
\frac{\partial w_{1}}{\partial e}(e,h)\times(1-bN_{1}^{*})^{1-\sigma}w_{1}(e,h){}^{-\sigma}=-\frac{\partial w_{2}}{\partial e}(e,h)\times(1-bN_{2}^{*})^{1-\sigma}w_{2}(e,h){}^{-\sigma}.
\]

\end_inset


\end_layout

\begin_layout Subsection
First try
\end_layout

\begin_layout Standard
Suppose
\begin_inset Formula 
\begin{align*}
w_{1}(e,h) & =1-e\\
w_{2}(e,h) & =1+\gamma e+\eta eh
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Then the above becomes
\begin_inset Formula 
\[
-(1-bN_{1}^{*})^{1-\sigma}(1-e){}^{-\sigma}=-(\gamma+\eta h)(1-bN_{2}^{*})^{1-\sigma}(1+\gamma e+\eta eh){}^{-\sigma}.
\]

\end_inset


\end_layout

\begin_layout Standard
Rewrite this as:
\begin_inset Formula 
\begin{align}
\frac{(1-bN_{1}^{*})^{1/\sigma-1}}{(1-bN_{2}^{*})^{1/\sigma-1}} & =\frac{(1-e)(\gamma+\eta h)^{1/\sigma}}{1+e(\gamma+\eta h)}\label{eq:e-implicit}\\
\nonumber 
\end{align}

\end_inset


\end_layout

\begin_layout Standard
We can also write 
\begin_inset Formula 
\begin{align*}
N_{1}^{*} & =\max\left\{ \frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(1-e)^{(1-\sigma)/\sigma}\right),0\right\} \\
N_{2}^{*} & =\max\left\{ \frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(1+\gamma e+\eta eh)^{(1-\sigma)/\sigma}\right),0\right\} 
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Assume for the moment that both the above are positive, then 
\begin_inset Formula $1-bN_{1}^{*}=\left(\frac{b}{a}\right)^{1/\sigma}(1-e)^{(1-\sigma)/\sigma}$
\end_inset

 and 
\begin_inset Formula $1-bN_{2}^{*}=\left(\frac{b}{a}\right)^{1/\sigma}(1+\gamma e+\eta eh)^{(1-\sigma)/\sigma}$
\end_inset

.
\end_layout

\begin_layout Standard
So plug these in to (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:e-implicit"

\end_inset

) and simplify, to get
\begin_inset Formula 
\[
\frac{(1-e)^{(1-\sigma)^{2}/\sigma^{2}}}{(1+e(\gamma+\eta h))^{(1-\sigma)^{2}/\sigma^{2}}}=\frac{(1-e)(\gamma+\eta h)^{1/\sigma}}{1+e(\gamma+\eta h)}
\]

\end_inset


\end_layout

\begin_layout Standard
and simplify a bit more
\begin_inset Formula 
\[
\frac{(1-e)}{(1+e(\gamma+\eta h))}=(\gamma+\eta h){}^{\sigma/[(1-\sigma)^{2}-\sigma^{2}]}
\]

\end_inset


\end_layout

\begin_layout Standard
Write 
\begin_inset Formula 
\[
R\equiv(\gamma+\eta h){}^{\sigma/(1-2\sigma)}
\]

\end_inset

, which is the RHS of the above.
 Then
\begin_inset Formula 
\begin{align*}
\frac{(1-e)}{(1+e(\gamma+\eta h))} & =R\\
e & =\frac{1-R}{1+R(\gamma+\eta h)}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Note that for 
\begin_inset Formula $\sigma>1/2$
\end_inset

, 
\begin_inset Formula $R$
\end_inset

 is decreasing in 
\begin_inset Formula $\gamma,\eta$
\end_inset

 and 
\begin_inset Formula $h$
\end_inset

.
 Note also that (in this case) 
\begin_inset Formula $e<0$
\end_inset

 if and only if 
\begin_inset Formula $\gamma+\eta h<1$
\end_inset

.
 In fact, we can write
\begin_inset Formula 
\[
e^{*}=\min\left\{ 1,\max\left\{ 0,\frac{1-R}{1+R(\gamma+\eta h)}\right\} .\right\} 
\]

\end_inset


\end_layout

\begin_layout Standard
For 
\begin_inset Formula $\sigma>1/2$
\end_inset

, this is increasing in 
\begin_inset Formula $\gamma,\eta$
\end_inset

 and 
\begin_inset Formula $h$
\end_inset

.
 In particular, if 
\begin_inset Formula $h<\frac{1-\gamma}{\eta}$
\end_inset

 then 
\begin_inset Formula $e^{*}=0$
\end_inset

.
 
\series bold
We don't want to allow that because it ends up with flat wages in 
\begin_inset Formula $h$
\end_inset

, and flat 
\begin_inset Formula $N^{*}$
\end_inset

 in 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\begin_layout Standard
We can assume that 
\begin_inset Formula $\gamma>1$
\end_inset

.
 So even if you have 
\begin_inset Formula $h=0$
\end_inset

, it's still worth getting 
\series bold
some
\series default
 education.
\end_layout

\begin_layout Standard
Plugging 
\begin_inset Formula $e^{*}$
\end_inset

 into wages gives
\begin_inset Formula 
\begin{align*}
W_{1} & =1-e^{*}=\frac{R(1+\gamma+\eta h)}{1+R(\gamma+\eta h)};\\
W_{2} & =1+\gamma e^{*}+\eta e^{*}h=1+(\gamma+\eta h)\frac{1-R}{1+R(\gamma+\eta h)}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
Might also write this in terms of 
\begin_inset Formula $S=\gamma+\eta h$
\end_inset

 as
\begin_inset Formula 
\begin{align*}
R & =S^{\sigma/(1-2\sigma)};\\
e^{*} & =\frac{1-S^{\sigma/(1-2\sigma)}}{1+S^{(1-\sigma)/(1-2\sigma)}};\\
W_{1} & =\frac{1+S}{S+S^{-\sigma/(1-2\sigma)}}=\frac{1+1/S}{1+1/S^{(1-\sigma)/(1-2\sigma)}};\\
W_{2} & =\frac{1+S}{1+S^{(1-\sigma)/(1-2\sigma)}}.
\end{align*}

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Surprisingly, 
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
\frac{de^{*}}{dh} & =\frac{de^{*}}{dR}\frac{dR}{dh}\\
\frac{dR}{dh} & =\eta\frac{\sigma}{1-2\sigma}(\gamma+\eta h){}^{\sigma/(1-2\sigma)-1}=\eta\frac{\sigma}{1-2\sigma}\frac{R}{\gamma+\eta h}\\
\frac{de^{*}}{dR} & =\frac{d}{dR}\frac{1-R}{1+R(\gamma+\eta h)}\\
 & =\frac{1}{[1+R(\gamma+\eta h)]^{2}}[-1(1+R(\gamma+\eta h))-(1-R)(\gamma+\eta h)]\\
 & =\frac{-1-(\gamma+\eta h)}{[1+R(\gamma+\eta h)]^{2}}\\
\frac{de^{*}}{dh} & =\frac{1+(\gamma+\eta h)}{[1+R(\gamma+\eta h)]^{2}}\eta\frac{\sigma}{2\sigma-1}\frac{R}{\gamma+\eta h}\\
 & =\frac{1+(\gamma+\eta h)}{[1+(\gamma+\eta h)^{(1-\sigma)/(1-2\sigma)}]^{2}}\eta\frac{\sigma}{2\sigma-1}(\gamma+\eta h){}^{(3\sigma-1)/(1-2\sigma)}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
Note that for 
\begin_inset Formula $\sigma>0.5$
\end_inset

, this is positive since 
\begin_inset Formula $\frac{\sigma}{2\sigma-1}>0$
\end_inset

; also that 
\begin_inset Formula $dR/dh<0$
\end_inset

 for 
\begin_inset Formula $\sigma>0.5$
\end_inset

.
\end_layout

\begin_layout Plain Layout

\series bold
Oh yeah? So how come when you plot it for e.g.
 
\begin_inset Formula $\gamma=5$
\end_inset

 you get 
\begin_inset Formula $e^{*}$
\end_inset

 downward sloping?
\end_layout

\begin_layout Plain Layout
Note also: rewrite as
\begin_inset Formula 
\begin{align*}
\frac{de^{*}}{dh} & =\frac{R(1+\gamma+\eta h)}{(\gamma+\eta h)[1+R(\gamma+\eta h)]^{2}}\eta\frac{\sigma}{2\sigma-1}\\
 & =\frac{(\gamma+\eta h){}^{\sigma/(1-2\sigma)}(1+\gamma+\eta h)}{(\gamma+\eta h)[1+(\gamma+\eta h){}^{(1-\sigma)/(1-2\sigma)}]^{2}}\eta\frac{\sigma}{2\sigma-1}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
The derivative of the first factor wrt 
\begin_inset Formula $h$
\end_inset

 is signed by
\begin_inset Formula 
\begin{align*}
 & f'g-fg'\\
= & (\gamma+\eta h)[1+(\gamma+\eta h){}^{(1-\sigma)/(1-2\sigma)}]^{2}[f']-(\gamma+\eta h){}^{\sigma/(1-2\sigma)}(1+\gamma+\eta h)[g']\\
= & (\gamma+\eta h)[1+R(\gamma+\eta h)]^{2}[f']-R(1+\gamma+\eta h)[g']
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
where
\begin_inset Formula 
\begin{align*}
f & =(\gamma+\eta h){}^{\sigma/(1-2\sigma)}(1+\gamma+\eta h)\\
 & =(\gamma+\eta h){}^{\sigma/(1-2\sigma)}+(\gamma+\eta h)^{1+\sigma/(1-2\sigma)}\\
f' & =\eta\frac{\sigma}{1-2\sigma}(\gamma+\eta h)^{^{\sigma/(1-2\sigma)-1}}+\eta\frac{1-\sigma}{1-2\sigma}(\gamma+\eta h)^{\sigma/(1-2\sigma)}\\
 & =\eta\frac{\sigma}{1-2\sigma}\frac{R}{\gamma+\eta h}+\eta\frac{1-\sigma}{1-2\sigma}R\\
 & =\eta R\left(\frac{\sigma(1/(\gamma+\eta h)-1)+1}{1-2\sigma}\right)\\
 & <0
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
where the inequality is because 
\begin_inset Formula $\sigma(1/(\gamma+\eta h)-1)>-1$
\end_inset

, while 
\begin_inset Formula $1-2\sigma<0$
\end_inset

, both by 
\begin_inset Formula $0.5<\sigma<1$
\end_inset

.
\end_layout

\begin_layout Plain Layout
Now 
\begin_inset Formula 
\begin{align*}
g & =(\gamma+\eta h)[1+(\gamma+\eta h){}^{(1-\sigma)/(1-2\sigma)}]^{2}\\
g' & =\eta[1+(\gamma+\eta h){}^{(1-\sigma)/(1-2\sigma)}]^{2}+(\gamma+\eta h)2[1+(\gamma+\eta h){}^{(1-\sigma)/(1-2\sigma)}]\eta\frac{1-\sigma}{1-2\sigma}(\gamma+\eta h)^{\sigma/(1-2\sigma)}\\
 & =\eta[1+R(\gamma+\eta h)]\{[1+R(\gamma+\eta h)]+(\gamma+\eta h)2\frac{1-\sigma}{1-2\sigma}R\}\\
 & =\eta[1+R(\gamma+\eta h)]\{1+R(\gamma+\eta h)[1+2\frac{1-\sigma}{1-2\sigma}]\}\\
 & \eta[1+R(\gamma+\eta h)]\{1+R(\gamma+\eta h)[\frac{3-4\sigma}{1-2\sigma}]\}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
Hmph.
\end_layout

\end_inset


\end_layout

\begin_layout Standard

\series bold
Problem with the above: needs limits 
\begin_inset Formula $e\in[0,1]$
\end_inset

 which is a hassle and ends up requiring e.g.
 
\begin_inset Formula $\gamma>1$
\end_inset

.
 Maths unbeautiful though manageable just.
\end_layout

\begin_layout Subsection
Second try
\end_layout

\begin_layout Standard
Write
\begin_inset Formula 
\[
U(N_{1},N_{2})=u(\bar{W}_{1})+u(\bar{W}_{2})+a(N_{1}+N_{2})
\]

\end_inset

We continue to assume CRRA utility.
 But now assume
\begin_inset Formula 
\begin{align*}
\bar{W}_{1} & =1-e-bN_{1}\\
\bar{W}_{2} & =w(e,h)(1-bN_{2})
\end{align*}

\end_inset

where 
\begin_inset Formula $w$
\end_inset

 is increasing in its arguments.
 Thus, some time is spent on education in the first period.
 Again we require 
\begin_inset Formula $e\ge0$
\end_inset

 and 
\begin_inset Formula $1-e-bN_{1}\ge0$
\end_inset

.
\end_layout

\begin_layout Standard
The choice problems of 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

 are separable.
 Rearranging the FOC for 
\begin_inset Formula $N_{1}^{*}$
\end_inset

 gives:
\begin_inset Formula 
\[
N_{1}^{*}=\max\left\{ 0,\frac{1}{b}\left(1-e-\left(\frac{b}{a}\right)^{1/\sigma}\right)\right\} .
\]

\end_inset


\end_layout

\begin_layout Standard
Note that 
\begin_inset Formula $N_{1}^{*}$
\end_inset

 is decreasing in 
\begin_inset Formula $e$
\end_inset

, strictly when positive.
\end_layout

\begin_layout Standard

\end_layout

\begin_layout Standard
For 
\begin_inset Formula $N_{2}^{*}$
\end_inset

 given 
\begin_inset Formula $W_{2}=w(e,h)$
\end_inset

 we have
\begin_inset Formula 
\[
N_{2}^{*}=\max\left\{ 0,\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}w(e,h)^{(1-\sigma)/\sigma}\right)\right\} .
\]

\end_inset


\end_layout

\begin_layout Standard
Assume for now that 
\begin_inset Formula $N_{2}^{*}$
\end_inset

 and 
\begin_inset Formula $N_{1}^{*}$
\end_inset

 are both positive.
 Then by the envelope theorem, 
\begin_inset Formula 
\[
\frac{dU}{de}=\frac{\partial U}{\partial e}
\]

\end_inset

since 
\begin_inset Formula $\partial U/\partial N_{t}^{*}=0$
\end_inset

.
\end_layout

\begin_layout Standard
Write this partial derivative as
\begin_inset Formula 
\[
-(1-e-bN_{1}){}^{-\sigma}+\frac{\partial w}{\partial e}(e,h)(1-bN_{2})[w(e,h)(1-bN_{2})]^{-\sigma}
\]

\end_inset


\end_layout

\begin_layout Standard
and solve for 
\begin_inset Formula $e$
\end_inset

 by setting the FOC to 0:
\begin_inset Formula 
\begin{align*}
\frac{\partial w}{\partial e}(e,h)(1-bN_{2})[w(e,h)(1-bN_{2})]^{-\sigma} & =(1-e-bN_{1}){}^{-\sigma}\\
\frac{\frac{\partial w}{\partial e}(e,h)}{w(e,h)^{\sigma}} & =\left(1-bN_{2}\right)^{\sigma-1}(1-e-bN_{1})^{-\sigma}.
\end{align*}

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
NB: for 
\begin_inset Formula $\sigma=1$
\end_inset

 this is d/dx log 
\begin_inset Formula $w$
\end_inset

 on the left, and reduces to something simple on the right.
\end_layout

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Suppose
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
w(e,h)=he^{\tau}
\]

\end_inset

with 
\begin_inset Formula $0<\tau\le1$
\end_inset

....
 
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Then the FOC becomes 
\begin_inset Formula 
\begin{align*}
(1-e-bN_{1})^{\sigma}\tau h^{1-\sigma}e^{\tau(1-\sigma)-1} & =\left(1-bN_{2}\right)^{\sigma-1}\\
\Leftrightarrow(1-e-bN_{1})^{\sigma}e^{\tau(1-\sigma)-1} & =\left(h(1-bN_{2})\right)^{\sigma-1}\tau^{-1}.
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
Fixing 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

, this implicitly defines a unique solution for 
\begin_inset Formula $e$
\end_inset

, 
\begin_inset Formula $e^{*}$
\end_inset

, since the LHS is decreasing in 
\begin_inset Formula $e$
\end_inset

.
 Furthermore, for 
\begin_inset Formula $\sigma<1$
\end_inset

, since the RHS is decreasing in 
\begin_inset Formula $h$
\end_inset

, we have that 
\begin_inset Formula $e^{*}$
\end_inset

is increasing in 
\begin_inset Formula $h$
\end_inset

.
 Indeed, write
\begin_inset Formula 
\[
F(e,h)\equiv(1-e-bN_{1})^{\sigma}e^{\tau(1-\sigma)-1}-\left(h(1-bN_{2})\right)^{\sigma-1}\tau^{-1}=0
\]

\end_inset

; then
\begin_inset Formula 
\begin{align*}
\frac{de^{*}}{dh} & =-\frac{\partial F/\partial h}{\partial F/\partial e};\\
\partial F/\partial e & =(1-e-bN_{1})^{\sigma}(\tau(1-\sigma)-1)e^{\tau(1-\sigma)-2}-e^{\tau(1-\sigma)-1}\sigma(1-e-bN_{1})^{\sigma-1}\\
 & =e^{\tau(1-\sigma)-2}(1-e-bN_{1})^{\sigma-1}[(\tau(1-\sigma)-1)(1-e-bN_{1})-e\sigma]\\
\partial F/\partial h & =(1-\sigma)\left(1-bN_{2}\right)^{\sigma-1}\tau^{-1}h^{\sigma-2}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
Note that 
\begin_inset Formula $\partial F/\partial h$
\end_inset

 is positive.
 Also that 
\begin_inset Formula $-\partial F/\partial e$
\end_inset

 is positive and constant in 
\begin_inset Formula $h$
\end_inset

.
 Thus if 
\begin_inset Formula $\partial F/\partial h$
\end_inset

 is decreasing (increasing) in 
\begin_inset Formula $h$
\end_inset

, then 
\begin_inset Formula $e^{*}$
\end_inset

 will be concave (convex) in 
\begin_inset Formula $h$
\end_inset

.
 In fact it is easy to see that 
\begin_inset Formula $\partial F/\partial h$
\end_inset

 is decreasing in 
\begin_inset Formula $h$
\end_inset

 when 
\begin_inset Formula $\sigma<1$
\end_inset

.
 Hence 
\begin_inset Formula $e^{*}$
\end_inset

 is concave in 
\begin_inset Formula $h$
\end_inset

 (for fixed 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

).
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
note also that for 
\begin_inset Formula $\sigma=1$
\end_inset

, 
\begin_inset Formula $e^{*}$
\end_inset

 is flat in 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Say that 
\begin_inset Formula 
\[
w(e,h)=ef(h)
\]

\end_inset

 for increasing 
\begin_inset Formula $f$
\end_inset

.
 Then the FOC reduces to
\begin_inset Formula 
\begin{align*}
\frac{f(h)}{e^{\sigma}f(h)^{\sigma}} & =\left(1-bN_{2}\right)^{\sigma-1}(1-e-bN_{1})^{-\sigma}\\
\Leftrightarrow\frac{(1-e-bN_{1})^{\sigma}}{e^{\sigma}} & =\left(1-bN_{2}\right)^{\sigma-1}f(h)^{\sigma-1}\\
\Leftrightarrow\frac{(1-e-bN_{1})}{e} & =\left(\left(1-bN_{2}\right)f(h)\right)^{1-1/\sigma}\\
\Leftrightarrow\frac{1-bN_{1}}{e} & =1+\left(\left(1-bN_{2}\right)f(h)\right)^{1-1/\sigma}\\
\Leftrightarrow e & =\frac{1-bN_{1}}{1+\left((1-bN_{2})f(h)\right)^{1-1/\sigma}}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
We now have three explicit equations for our choice variables, at an interior
 equilibrium:
\begin_inset Formula 
\begin{align*}
N_{1}^{*} & =\frac{1}{b}\left(1-e-\left(\frac{b}{a}\right)^{1/\sigma}\right);\\
N_{2}^{*} & =\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(ef(h))^{(1-\sigma)/\sigma}\right);\\
e^{*} & =\frac{1-bN_{1}}{1+\left((1-bN_{2})f(h)\right)^{1-1/\sigma}}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Note that 
\begin_inset Formula $h$
\end_inset

 only enters in one place in all of this.
 So we may as well identify 
\begin_inset Formula $h$
\end_inset

 with 
\begin_inset Formula $f(h)$
\end_inset

.
 We do that from now on.
 We can also simplify
\begin_inset Formula 
\[
e^{*}=\frac{e^{*}+\left(\frac{b}{a}\right)^{1/\sigma}}{1+\left(\left(\frac{b}{a}\right)^{1/\sigma}e^{*(1-\sigma)/\sigma}h{}^{1/\sigma}\right)^{1-1/\sigma}}
\]

\end_inset


\end_layout

\begin_layout Standard
which simplifies to
\begin_inset Formula 
\[
e^{*}=\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{(1-\sigma)/(2\sigma-1)}
\]

\end_inset


\end_layout

\begin_layout Standard
Again note that for 
\begin_inset Formula $1>\sigma>0.5$
\end_inset

 
\begin_inset Formula $e^{*}$
\end_inset

 increases in 
\begin_inset Formula $h$
\end_inset

.
\begin_inset Note Note
status open

\begin_layout Plain Layout
Why not for 
\begin_inset Formula $\sigma>1$
\end_inset

? Presumably 
\begin_inset Formula $h$
\end_inset

 does you so much good that you don't need to invest in education.
\end_layout

\end_inset

 Furthermore 
\begin_inset Formula $e$
\end_inset

 is convex in 
\begin_inset Formula $h$
\end_inset

 if 
\begin_inset Formula $1/2<\sigma<2/3$
\end_inset

, and concave otherwise.
\end_layout

\begin_layout Standard
Now we can write
\begin_inset Note Note
status open

\begin_layout Plain Layout
Pleasingly CORRECT for positive values of N, checked computationally in
 julia.
\end_layout

\end_inset


\begin_inset Formula 
\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);\\
w(e^{*},h) & =\left(\frac{b}{a}\right)^{1/(2\sigma-1)}h^{\sigma/(2\sigma-1)};\\
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*}

\end_inset

Observe that 
\begin_inset Formula $w(e^{*},h)$
\end_inset

 is increasing in 
\begin_inset Formula $h$
\end_inset

 for 
\begin_inset Formula $\sigma>0.5$
\end_inset

, and convex iff 
\begin_inset Formula $0.5<\sigma<1$
\end_inset

.
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
for 
\begin_inset Formula $\sigma=0.5$
\end_inset

, there's a problem
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
....
 Indeed, if we use the first form, 
\begin_inset Formula $e$
\end_inset

 cancels out
\begin_inset Formula 
\begin{align*}
e^{*} & =\frac{e^{*}+\left(\frac{b}{a}\right)^{2}}{1+\left(\left(\frac{b}{a}\right)^{2}e^{*}h{}^{2}\right)^{-1}};\\
h^{-1} & =\left(\frac{b}{a}\right)^{2}
\end{align*}

\end_inset

which is obv wrong....
\end_layout

\end_inset


\end_layout

\end_inset

 Also observe that 
\begin_inset Formula $N_{1}^{*}<N_{2}^{*}$
\end_inset

 if 
\begin_inset Formula $N_{2}^{*}>0$
\end_inset

.
\end_layout

\begin_layout Standard
While 
\begin_inset Formula $N_{1}^{*}$
\end_inset

 and 
\begin_inset Formula $N_{2}^{*}$
\end_inset

 are positive, they have the same derivative with respect to 
\begin_inset Formula $h$
\end_inset

:
\begin_inset Formula 
\[
\frac{dN_{t}^{*}}{dh}=-\frac{1}{b}\left(\frac{b}{a}\right)^{1/(2\sigma-1)}\frac{1-\sigma}{2\sigma-1}h^{(1-\sigma)/(2\sigma-1)-1}
\]

\end_inset

which is negative for 
\begin_inset Formula $1>\sigma>0.5$
\end_inset

; also 
\begin_inset Formula $N_{t}^{*}$
\end_inset

 is convex for 
\begin_inset Formula $1/2<\sigma<2/3$
\end_inset

, and concave otherwise.
\end_layout

\begin_layout Subsubsection
Case that 
\begin_inset Formula $N_{1}^{*}=0$
\end_inset


\end_layout

\begin_layout Standard
Suppose 
\begin_inset Formula $N_{1}^{*}$
\end_inset

 as derived above is negative.
 Then we consider 
\begin_inset Formula $N_{1}^{*}=0$
\end_inset

.
 We can enter that into the equation for 
\begin_inset Formula $e^{*}$
\end_inset

 and solve it to find
\begin_inset Formula 
\[
e^{*}=\frac{1}{1+\left((1-bN_{2})h\right)^{1-1/\sigma}}.
\]

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
I checked this from the start, it's fine....
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now as before, if 
\begin_inset Formula $N_{2}^{*}$
\end_inset

 is positive, we have 
\begin_inset Formula 
\[
N_{2}^{*}=\frac{1}{b}\left(1-\left(\frac{b}{a}\right)^{1/\sigma}(eh)^{(1-\sigma)/\sigma}\right);
\]

\end_inset

Plugging this into 
\begin_inset Formula $e^{*}$
\end_inset

 gives
\begin_inset Formula 
\begin{align*}
e^{*} & =\frac{1}{1+\left(\frac{bh}{a}\right)^{(\sigma-1)/\sigma^{2}}(e^{*}){}^{-(1-\sigma)^{2}/\sigma^{2}}}
\end{align*}

\end_inset

which can be rearranged to
\begin_inset Note Note
status open

\begin_layout Plain Layout
Proof:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
e & =\frac{e^{(1-\sigma)^{2}/\sigma^{2}}}{e^{(1-\sigma)^{2}/\sigma^{2}}+X}\textrm{ where }X=\left(\frac{bh}{a}\right)^{(\sigma-1)/\sigma^{2}};\\
1 & =\frac{e^{(1-\sigma)^{2}/\sigma^{2}-1}}{e^{(1-\sigma)^{2}/\sigma^{2}}+X};\\
e^{(1-\sigma)^{2}/\sigma^{2}-1} & =e^{(1-\sigma)^{2}/\sigma^{2}}+X\\
e^{(1-2\sigma)/\sigma^{2}}(1-e) & =X
\end{align*}

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\[
(1-e^{*})(e^{*})^{(1-2\sigma)/\sigma^{2}}=\left(\frac{a}{bh}\right)^{(1-\sigma)/\sigma^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
implicitly solving for 
\begin_inset Formula $e^{*}$
\end_inset

.
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
We can then plug the solution back in to the expression for 
\begin_inset Formula $N_{2}^{*}$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
To find how 
\begin_inset Formula $e^{*}$
\end_inset

 changes with 
\begin_inset Formula $h$
\end_inset

, note that the RHS decreases in 
\begin_inset Formula $h$
\end_inset

 for 
\begin_inset Formula $\sigma<1$
\end_inset

.
 So if 
\begin_inset Formula $h$
\end_inset

 goes up the RHS goes down.
 Differentiate the LHS to get
\begin_inset Formula 
\begin{align*}
 & \frac{1-2\sigma}{\sigma^{2}}(1-e)e^{(1-2\sigma)/\sigma^{2}-1}-1(e^{(1-2\sigma)/\sigma^{2}})\\
= & \frac{1-2\sigma}{\sigma^{2}}e^{(1-2\sigma)/\sigma^{2}-1}-\frac{1-2\sigma+\sigma^{2}}{\sigma^{2}}e^{(1-2\sigma)/\sigma^{2}}\\
= & =\frac{1-2\sigma}{\sigma^{2}}e^{(1-2\sigma)/\sigma^{2}-1}-\frac{(1-\sigma)^{2}}{\sigma^{2}}e^{(1-2\sigma)/\sigma^{2}}
\end{align*}

\end_inset

This is negative if
\begin_inset Formula 
\begin{align*}
e & >\frac{1-2\sigma}{(1-\sigma)^{2}}
\end{align*}

\end_inset

which is always true for 
\begin_inset Formula $\sigma>0.5$
\end_inset

.
 Thus the LHS is decreasing in 
\begin_inset Formula $e$
\end_inset

.
 Thus if 
\begin_inset Formula $h$
\end_inset

 increases, then the RHS and LHS decrease, and 
\begin_inset Formula $e$
\end_inset

 must increase.
\end_layout

\begin_layout Subsubsection
Case that 
\begin_inset Formula $N_{1}=N_{2}=0$
\end_inset


\end_layout

\begin_layout Standard
Finding 
\begin_inset Formula $e^{*}$
\end_inset

 as above, and plugging it into 
\begin_inset Formula $N_{2}$
\end_inset

, may give a negative number.
 If so, then it must be that 
\begin_inset Formula $N_{1}=N_{2}=0$
\end_inset

.
\begin_inset Note Note
status open

\begin_layout Plain Layout
Right? Seems so after playing with Julia.
 But the proof is imprecise.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
In this case, 
\begin_inset Formula 
\[
e^{*}=\frac{1}{1+h^{(\sigma-1)/\sigma}}.
\]

\end_inset


\end_layout

\begin_layout Standard
Again 
\begin_inset Formula $e$
\end_inset

 increases in 
\begin_inset Formula $h$
\end_inset

 as we'd expect for 
\begin_inset Formula $\sigma<1$
\end_inset

.
\end_layout

\begin_layout Section
Proving uniqueness
\end_layout

\begin_layout Standard
One approach: 
\begin_inset Quotes eld
\end_inset

we know 
\begin_inset Formula $U$
\end_inset

 is differentiable, and we know 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

 are below (some) upper bounds.
 So it must have a global maximum.
 If we find a unique local maximum (via the necessary conditions) then we
 know that this is the global maximum.
\begin_inset Quotes erd
\end_inset

 This seems to work for 
\begin_inset Formula $\sigma>0.5$
\end_inset

.
\end_layout

\begin_layout Standard
Another approach: recursively show that there's a unique maximum 
\begin_inset Formula $e$
\end_inset

 given 
\begin_inset Formula $N_{1}$
\end_inset

 and 
\begin_inset Formula $N_{2}$
\end_inset

; then a unique maximum 
\begin_inset Formula $(e,N_{2})$
\end_inset

 given 
\begin_inset Formula $N_{1}$
\end_inset

; and then a unique maximum.
 
\end_layout

\begin_layout Standard
Or: prove quasiconcavity via the bordered Hessian.
 D1<0 (ad-bc); D2 > 0; D3 < 0 suffices.
 But has to hold everywhere which seems hard for D1.
\end_layout

\end_body
\end_document