model-CARA.lyx

#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
\save_transient_properties true
\origin unavailable
\textclass article
\use_default_options true
\maintain_unincluded_children false
\language english
\language_package default
\inputencoding auto
\fontencoding global
\font_roman "times" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format default
\output_sync 0
\bibtex_command default
\index_command default
\paperfontsize default
\spacing single
\use_hyperref false
\papersize default
\use_geometry false
\use_package amsmath 1
\use_package amssymb 1
\use_package cancel 1
\use_package esint 1
\use_package mathdots 1
\use_package mathtools 1
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_minted 0
\index Index
\shortcut idx
\color #008000
\end_index
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\end_header

\begin_body

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
U(N_{1},N_{2})=u(Y_{1})+u(Y_{2})+a(N_{1}+N_{2})\label{eq:U}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u$
\end_inset

 is now CARA 
\begin_inset Formula $u(y)=-e^{-\gamma y}$
\end_inset

, with 
\begin_inset Formula $u'(y)=\gamma e^{-\gamma y}$
\end_inset

.
\end_layout

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

\end_inset

with 
\begin_inset Formula $w(s,h)=sh$
\end_inset

.
\end_layout

\begin_layout Standard
Write the Lagrangian of utility 
\begin_inset Formula $U$
\end_inset

 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:U"
plural "false"
caps "false"
noprefix "false"

\end_inset

) as 
\begin_inset Formula 
\[
\mathcal{L}(N_{1},N_{2},s)=u(Y_{1})+u(Y_{2})+a(N_{1}+N_{2})+\lambda_{1}N_{1}+\lambda_{2}(\frac{1}{b}-s-N_{1})+\lambda_{3}N_{2}+\lambda_{4}(\frac{1}{b}-N_{2})+\mu_{1}s+\mu_{2}(1-s)
\]

\end_inset


\end_layout

\begin_layout Standard
with conditions
\begin_inset Formula 
\begin{align*}
\frac{d\mathcal{L}}{dN_{1}} & =-\gamma be^{-\gamma Y_{1}}+a+\lambda_{1}-\lambda_{2}=0\textrm{ with }\lambda_{1}N_{1}=0,\lambda_{2}(\frac{1}{b}-s-N_{1})=0\\
\frac{d\mathcal{L}}{dN_{2}} & =-\gamma shbe^{-\gamma Y_{2}}+a+\lambda_{3}-\lambda_{4}=0\textrm{ with }\lambda_{3}N_{2}=0,\lambda_{4}(\frac{1}{b}-N_{2})=0\\
\frac{d\mathcal{L}}{ds} & =-\gamma e^{-\gamma Y_{1}}+\gamma h(1-bN_{2})e^{-\gamma Y_{2}}+\mu_{1}-\mu_{2}=0\textrm{ with }\mu_{1}s=0,\mu_{2}(1-s)=0\\
N_{1},N_{2},s,\mathbf{\lambda},\mathbf{\mu} & \ge0\\
 & s\le1,N_{1}\le\frac{1-s}{b},N_{2}\le\frac{1}{b}.
\end{align*}

\end_inset


\end_layout

\begin_layout Subsection

\series bold
Solving for 
\begin_inset Formula $N_{1}$
\end_inset

:
\series default

\begin_inset Formula 
\begin{align}
-\gamma be^{-\gamma Y_{1}}+a & \lesseqqgtr0\textrm{ as }N_{1}=0,N_{1}\in(0,\frac{1-s}{b}),N_{1}=\frac{1-s}{b}\nonumber \\
\frac{a}{\gamma b} & \lesseqqgtr e^{-\gamma Y_{1}}\nonumber \\
\log(\frac{a}{\gamma b}) & \lesseqqgtr-\gamma Y_{1}=-\gamma(1-s-bN_{1})\nonumber \\
\frac{1}{\gamma}\log(\frac{a}{\gamma b}) & \lesseqqgtr bN_{1}+s-1\nonumber \\
\frac{1}{b\gamma}\log(\frac{a}{\gamma b})+\frac{1-s}{b} & \lesseqqgtr N_{1}\textrm{ as }N_{1}=0,N_{1}\in(0,\frac{1-s}{b}),N_{1}=\frac{1-s}{b}.\label{eq:N1cond}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
Note that if 
\begin_inset Formula $a>\gamma b$
\end_inset

 then we must have 
\begin_inset Formula $N_{1}=\frac{1-s}{b}$
\end_inset

.
\end_layout

\begin_layout Standard
Equally if 
\begin_inset Formula $\frac{1}{b\gamma}\log(\frac{a}{\gamma b})+\frac{1}{b}<0\Leftrightarrow\log(\frac{a}{\gamma b})<-\gamma$
\end_inset

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

 for any 
\begin_inset Formula $s$
\end_inset

.
\end_layout

\begin_layout Subsection

\series bold
Solving for 
\begin_inset Formula $N_{2}$
\end_inset

:
\series default

\begin_inset Formula 
\begin{align}
-\gamma shbe^{-\gamma Y_{2}}+a & \lesseqqgtr0\textrm{ as }N_{2}=0,N_{2}\in(0,\frac{1}{b}),N_{2}=\frac{1}{b}\nonumber \\
\frac{a}{\gamma shb} & \lesseqqgtr e^{-\gamma Y_{2}}\nonumber \\
\log(\frac{a}{\gamma shb}) & \lesseqqgtr-\gamma Y_{2}=-\gamma(sh(1-bN_{2}))\nonumber \\
\frac{1}{\gamma sh}\log(\frac{a}{\gamma shb}) & \lesseqqgtr bN_{2}-1\nonumber \\
\frac{1}{b}+\frac{1}{b\gamma sh}\log(\frac{a}{\gamma shb}) & \lesseqqgtr N_{2}\textrm{ as }N_{2}=0,N_{2}\in(0,\frac{1}{b}),N_{2}=\frac{1}{b}.\label{eq:N2cond}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
If 
\begin_inset Formula $a>\gamma shb$
\end_inset

, in particular always if 
\begin_inset Formula $a>\gamma hb$
\end_inset

, then we have 
\begin_inset Formula $N_{2}=\frac{1}{b}$
\end_inset

.
\end_layout

\begin_layout Standard
If 
\begin_inset Formula $\frac{1}{b}+\frac{1}{b\gamma sh}\log(\frac{a}{\gamma shb})<0\Leftrightarrow\log(\frac{a}{\gamma shb})<-\gamma sh$
\end_inset

 and in particular always if 
\begin_inset Formula $\log(\frac{a}{\gamma shb})<-\gamma h$
\end_inset

 
\begin_inset Note Note
status open

\begin_layout Plain Layout
but note 
\begin_inset Formula $s$
\end_inset

 is still in there
\end_layout

\end_inset

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

.
\end_layout

\begin_layout Subsection

\series bold
Solving for 
\begin_inset Formula $s$
\end_inset

:
\series default

\begin_inset Formula 
\begin{align}
-\gamma e^{-\gamma Y_{1}}+\gamma h(1-bN_{2})e^{-\gamma Y_{2}} & \lesseqqgtr0\textrm{ as }s=0,s\in(0,1),s=1\nonumber \\
h(1-bN_{2})e^{-\gamma Y_{2}} & \lesseqqgtr e^{-\gamma Y_{1}}\nonumber \\
\log(h(1-bN_{2}))-\gamma Y_{2} & \lesseqqgtr-\gamma Y_{1}\nonumber \\
\frac{1}{\gamma}\log(h(1-bN_{2})) & \lesseqqgtr Y_{2}-Y_{1}=sh(1-bN_{2})-(1-s-bN_{1})\nonumber \\
\frac{1}{\gamma}\log(h(1-bN_{2})) & \lesseqqgtr s(h(1-bN_{2})+1)-(1-bN_{1})\nonumber \\
\frac{\frac{1}{\gamma}\log(h(1-bN_{2}))+(1-bN_{1})}{1+h(1-bN_{2})} & \lesseqqgtr s\textrm{ as }s=0,s\in(0,1),s=1.\label{eq:scond}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
Note that the above, specifically line 2, implies 
\begin_inset Formula $s=0$
\end_inset

 whenever 
\begin_inset Formula $N_{2}=1/b$
\end_inset

.
\end_layout

\begin_layout Subsection

\series bold
Low 
\begin_inset Formula $h$
\end_inset


\end_layout

\begin_layout Standard
Let's start from low 
\begin_inset Formula $h$
\end_inset

.
 Then 
\begin_inset Formula $a>\gamma hb$
\end_inset

 and so we have 
\begin_inset Formula $N_{2}=\frac{1}{b}$
\end_inset

.
 Note that the second line of the above chain becomes 
\begin_inset Formula $0<e^{0}$
\end_inset

 which is always true, so 
\begin_inset Formula $s=0$
\end_inset

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

 then, like 
\begin_inset Formula $N_{2}$
\end_inset

, solely depends on 
\begin_inset Formula $a,\gamma$
\end_inset

 and 
\begin_inset Formula $b$
\end_inset

.
 So at these low values, 
\begin_inset Formula $N^{*}$
\end_inset

 is constant in 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\begin_layout Standard
What's the maximum for this? (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:scond"
plural "false"
caps "false"
noprefix "false"

\end_inset

) is satisfied for 
\begin_inset Formula $s=0$
\end_inset

 so long as 
\begin_inset Formula $N_{2}=1/b$
\end_inset

.
 The condition from (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:N2cond"

\end_inset

) line 1 for 
\begin_inset Formula $N_{2}=1/b$
\end_inset

 is also satisfied so long as 
\begin_inset Formula $s=0$
\end_inset

.
 (If you're out of the labour market having kids, no point in getting educated,
 and if you didn't get educated, you may as well have kids.
 The condition for 
\begin_inset Formula $N_{1}$
\end_inset

from (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:N1cond"

\end_inset

) does not depend on 
\begin_inset Formula $h$
\end_inset

 except via 
\begin_inset Formula $s$
\end_inset

, so if 
\begin_inset Formula $s=0$
\end_inset

 it's satisfied for any 
\begin_inset Formula $h$
\end_inset

.
 But NB we haven't yet shown global concavity, so this might be only a local
 optimum.
\end_layout

\begin_layout Subsection

\series bold
High 
\begin_inset Formula $h$
\end_inset


\end_layout

\begin_layout Standard
Now let's make 
\begin_inset Formula $h$
\end_inset

 high.
 If it is high enough, then from (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:N2cond"

\end_inset

) we can't have 
\begin_inset Formula $N_{2}=\frac{1}{b}$
\end_inset

.
 That then implies that the term (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:scond"

\end_inset

) is proportionate to 
\begin_inset Formula $\log(h)/h$
\end_inset

 which approaches zero asymptotically.
 I.e.
 
\begin_inset Formula $s^{*}$
\end_inset

 is slowly declining.
 It also implies 
\begin_inset Formula $ds^{*}/dh\rightarrow0$
\end_inset

, I think.
 Since 
\begin_inset Formula $s^{*}$
\end_inset

 is slowly declining 
\begin_inset Formula $N_{1}$
\end_inset

 is slowly increasing, towards 
\begin_inset Formula $\frac{1}{b\gamma}\log(\frac{a}{\gamma b})+\frac{1}{b}$
\end_inset

 if that's a valid number for it.
 (If not it stays at 0.) Similarly 
\begin_inset Formula $N_{2}$
\end_inset

 increases towards 
\begin_inset Formula $1/b$
\end_inset

 since 
\begin_inset Formula $\frac{1}{b\gamma sh}\log(\frac{a}{\gamma shb})<0$
\end_inset

 and approaches 0 as 
\begin_inset Formula $h$
\end_inset

 gets large.
 
\end_layout

\begin_layout Subsection
Interior equilibrium
\end_layout

\begin_layout Standard
Let's seek an interior equilibrium.
 Then
\begin_inset Formula 
\begin{align*}
s & =\frac{\frac{1}{\gamma}\log(h(1-bN_{2}))+(1-bN_{1})}{1+h(1-bN_{2})}\in(0,1);\\
N_{2} & =\frac{1}{b}+\frac{1}{b\gamma sh}\log(\frac{a}{\gamma shb})\in(0,\frac{1}{b});\\
N_{1} & =\frac{1}{b\gamma}\log(\frac{a}{\gamma b})+\frac{1-s}{b}\in(0,\frac{1-s}{b})
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
This clearly requires 
\begin_inset Formula $a<\gamma b$
\end_inset

 and 
\begin_inset Formula $a<\gamma shb$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
indeed in compute-CARA.jl you get a stark switch of regimes if this condition
 is violated, from babies/no education to no babies/education.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
We'll insert 
\begin_inset Formula $N_{1}$
\end_inset

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

 into 
\begin_inset Formula $s$
\end_inset

:
\begin_inset Formula 
\begin{align*}
s & =\frac{\frac{1}{\gamma}\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\frac{1}{\gamma}\log(\frac{a}{\gamma b})+s}{1-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb})}\\
 & =\frac{\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(\frac{a}{\gamma b})+s}{\gamma-\frac{1}{s}\log(\frac{a}{\gamma shb})}\\
s(1-\frac{1}{\gamma-\frac{1}{s}\log(\frac{a}{\gamma shb})}) & =\frac{\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(\frac{a}{\gamma b})}{\gamma-\frac{1}{s}\log(\frac{a}{\gamma shb})}\\
s(\gamma-\frac{1}{s}\log(\frac{a}{\gamma shb})-1) & =\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(\frac{a}{\gamma b})\\
s(\gamma-1)-\log(\frac{a}{\gamma shb}) & =\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(\frac{a}{\gamma b})\\
s(\gamma-1) & =\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(\frac{a}{\gamma b})+\log(\frac{a}{\gamma shb})\\
s(\gamma-1) & =\log(-\frac{1}{\gamma s}\log(\frac{a}{\gamma shb}))-\log(sh)\\
s & =\frac{\log(\frac{1}{\gamma s}\log(\frac{\gamma shb}{a}))-\log(sh)}{\gamma-1}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\frac{d}{ds}\frac{1}{\gamma s}\log(\frac{\gamma shb}{a})=\frac{d}{ds}\log((\frac{\gamma shb}{a})^{1/\gamma s})\textrm{ signed by}\frac{d}{ds}(\frac{\gamma shb}{a})^{1/\gamma s}
\]

\end_inset

which is
\begin_inset Formula 
\[
\log\frac{\gamma shb}{a}(\frac{-\gamma}{(\gamma s)^{2}})\frac{\gamma hb}{a}(\frac{\gamma shb}{a})^{1/\gamma s}<0\textrm{ since }\gamma shb>a.
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
With 
\begin_inset Formula $N_{1}=0$
\end_inset


\end_layout

\begin_layout Standard
x
\end_layout

\begin_layout Standard
x
\end_layout

\begin_layout Standard
In between these values, what if 
\begin_inset Formula $a>\gamma b$
\end_inset

? Then 
\begin_inset Formula $N_{1}=(1-s)/b$
\end_inset

 and (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:scond"

\end_inset

) becomes
\begin_inset Formula 
\begin{align*}
\frac{\frac{1}{\gamma}\log(h(1-bN_{2}))+s}{1+h(1-bN_{2})} & \lesseqqgtr s\textrm{ as }s=0,s\in(0,1),s=1.\\
\frac{1}{\gamma}\log(h(1-bN_{2})) & \lesseqqgtr s(h(1-bN_{2}))\\
\frac{1}{\gamma h(1-bN_{2})}\log(h(1-bN_{2})) & \lesseqqgtr s\textrm{ as }s=0,s\in(0,1),s=1.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Also from (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:N2cond"

\end_inset

):
\begin_inset Formula 
\[
\frac{1}{b}+\frac{1}{b\gamma sh}\log(\frac{a}{\gamma shb})\lesseqqgtr N_{2}\textrm{ as }N_{2}=0,N_{1}\in(0,\frac{1}{b}),N_{2}=\frac{1}{b}.
\]

\end_inset


\end_layout

\end_body
\end_document