%slides_capital.tex
%Slides for Chapter 3 of ``Open Economy Macroeconomics'' by Martín Uribe and Stephanie Schmitt-Grohé, Princeton University Press 2017. 
%© Martín Uribe and Stephanie Schmitt-Grohé.

\input{slides_preamble}

\begin{slide}
\begin{minipage}{5in}
\rightline{\includegraphics[width=4.5in]{USG_cover.eps}}

\centerline{\tiny  Princeton University Press,  2017}
\end{minipage}
\begin{minipage}{4.5in}
\centering 
{\small 
slides 

chapter 3   

an open economy with capital

}

\vspace{1in}

\end{minipage}
\end{slide}

\topleft{\textit{Open Economy Macroeconomics, Chapter 3} }
\topright{M.\ Uribe and S.\ Schmitt-Groh\'e}

\begin{slide}
{\bf \large Motivation}

In this chaper we introduce production and physical capital accumulation. Doing so will allow us to address two important issues.  One  is that for the most commonly used stationary specifications of the shock  process---namely, AR(1) specifications---the endowment economy
model presented in Chapter 2 fails to predict the observed countercyclicality of the trade balance  and the current account (documented in Chapter 1). 

The second is that the assumption that output is an exogenously given stochastic process---maintained throughout Chapter 2---is unsatisfactory if the goal is to understand observed business cycles. For output is perhaps the main variable any theory of the business cycle should aim
to explain. 


To allow for a full characterization of the
equilibrium dynamics using pen and paper we abstract from depreciation and uncertainty, and assume, as in Chapter 2, that $\beta(1+r)=1$. In later chapters we will relax these assumptions. 


\end{slide}

\begin{slide}
{\bf \large Intuition}

The reason  allowing for production and capital accumulation might induce the model to predict a counterycyclical trade balanace, even for AR(1) shock processes, is as follows: 

Suppose persistent AR(1) productivity shocks are the main source of uncertainty. Then the marginal product of capital is expected to be high not just in the period of the shock but also in the next couple of periods. Thus the economy has an incentive to invest more to take advantage of the higher productivity of capital. This increase in domestic demand might be so large that total domestic demand, consumption plus investment, rises by more than output, resulting in a countercyclical impact response of the trade balance. 

\end{slide}

\begin{slide}
We will derive the following 2 principles: 

{\bf Principle I:}  The more persistent productivity shocks are, the more likely an initial deterioration of the trade balance will be.


{\bf Principle II:} The more pronounced are capital adjustment costs, the smaller will be the initial trade balance deterioration in response to a positive and persistent productivity shock.


\end{slide}

\begin{slide}

{\bf \large  3.1 Model\label{sec:capital_basic_model}}

Small open economy, no uncertainty, no depreciation.

Preferences: \begin{equation}
\label{eq:capital_utility_capital_no_adj_cost}
\sum_{t=0}^{\infty} \beta^t U(c_t)
\tag{3.1}
\end{equation}
Sequential  budget  
 constraint of the household: 
\begin{equation}
\label{eq:capital_bc_no_adj_cost}
 c_t + i_t + (1+r)d_{t-1} = y_t + d_t
\tag{3.2}
\end{equation}

{\bf Interpretation:} LHS displays the uses of wealth: purchases of consumption goods ($c_t$); purchases of investment goods ($i_t$); 
payment of principal and interest on 
outstanding debt ($(1+r)d_{t-1}$). RHS  displays the sources of wealth: output ($y_t$) and new debt  issuance ($d_t$).


\end{slide}
 \begin{slide}
Production function: 
\begin{equation}
\label{eq:capital_output_no_adj_cost}
y_t = A_t F(k_t)
\tag{3.3}
\end{equation}
$A_t=$  exogenous and deterministic
 productivity factor\\
$F(\cdot)=$ increasing and concave production function \\
$k_t>0$ physical capital, determined in $t-1$


Law of motion of capital:
\begin{equation}
\label{eq:capital_i_economy_capital}
k_{t+1} = k_t + i_t
\tag{3.4}
\end{equation}



No-Ponzi game constraint:
\begin{equation}
\label{eq:capital_no_Ponzi_no_adj_cost} 
\lim_{j\rightarrow \infty}
\frac{d_{t+j}}{(1+r)^j} \le 0
\tag{3.5}
\end{equation}

\end{slide}

\begin{slide}
Lagrangian of household's problem:
\[
 \mathcal{L} = \sum_{t=0}^{\infty} \beta^t \left\{ U(c_t) + \lambda_t
\left[  A_t F(k_t)+ d_t  - c_t - (k_{t+1} -
k_{t}) -  (1+r)d_{t-1} \right] \right\}.
\]
The first-order conditions corresponding to 
$c_t$, $d_t$, $k_{t+1}$, and $\lambda_t$, respectively,   are
\begin{equation}
\label{eq:capital_la_nocost}
U'(c_t) = \lambda_t,
\tag{3.6}
\end{equation}
\begin{equation}
\tag{3.7}
\label{eq:capital_d_nocost}
\lambda_t = \beta (1+r) \lambda_{t+1},
\end{equation}
\begin{equation}
\label{eq:capital_Euler_k_nocost}
\tag{3.8}
\lambda_t = \beta \lambda_{t+1}
[A_{t+1} F'(k_{t+1}) + 1],
\end{equation}
and
\begin{equation}
\label{eq:capital_sequential_resource_constraint_nocost}
\tag{3.9}
A_t F(k_t) + d_t =    c_t +  k_{t+1} - k_t  + (1+r)d_{t-1} .
\end{equation}
Household optimization implies that the borrowing constraint  holds with equality (transversality condition):
\begin{equation}
\label{eq:capital_transversality_nocost}
\tag{3.10}
\lim_{t\rightarrow \infty} \frac{d_t}{(1+r)^t} = 0.
\end{equation}

\end{slide}
\begin{slide}
Assume that 
\begin{equation}
\label{eq:beta=1+r}
\tag{*}
\beta (1+r)=1
\end{equation}

This assumption together with~\eqref{eq:capital_la_nocost} and 
\eqref{eq:capital_d_nocost} implies that consumption is constant
\begin{equation}
\label{eq:capital_nocost-cc}
c_{t+1} = c_t
\tag{3.11}
\end{equation}


As we will see shortly, consumption is again determined by non-financial permanent income net of interest on initial debt outstanding. 

\end{slide}
\begin{slide}


Assumption~\eqref{eq:beta=1+r} and equilibrium condition~\eqref{eq:capital_Euler_k_nocost} imply that 
\begin{equation}
\label{eq:capital_nocost-Euler}
r=A_{t+1} F'(k_{t+1})
\tag{3.12}
\end{equation}

Households invest in physical capital in period $t$ until the expected marginal product of
capital in period $t+1$ equals the rate of return on foreign debt.

It follows from this equilibrium condition  that next period's level of physical capital, $k_{t+1}$, is an increasing function of the future expected level of productivity, 
$A_{t+1}$, and a decreasing function of the interest rate $r$. 
\begin{equation}
\tag{3.14}
\label{eq:capital_kappa_economy_capital}
k_{t+1} = \kappa\left(\frac{A_{t+1}}{r}\right), 
\end{equation}
with $\kappa'>0$.


\end{slide}

\begin{slide}
{To characterize the equilibrium, it will again be convenient to work with the intertemporal budget constraint. Write the sequential budget constraint for period $t+j$:
\[
A_{t+j} F(k_{t+j}) + d_{t+j} =    c_{t+j} +  k_{t+j+1} - k_{t+j}  + (1+r)d_{t+j-1}
\]
Divide by $(1+r)^j$ and sum for $j=0$ to $j=J$. 
\[
\sum_{j=0}^J \frac{A_{t+j} F(k_{t+j}) }{(1+r)^j} + \frac{d_{t+J}}{(1+r)^J} = \sum_{j=0}^J \frac{ c_{t+j}+  k_{t+j+1} - k_{t+j}}{(1+r)^j} + (1+r) d_{t-1}
\]
Now use the fact that in eqm consumption is constant over time, \eqref{eq:capital_nocost-cc}, and rearrange terms
\[
c_t  \sum_{j=0}^J \frac{1}{(1+r)^j} + (1+r) d_{t-1}
=
\sum_{j=0}^J \frac{ A_{t+j} F(k_{t+j}) -
( k_{t+j+1} - k_{t+j} ) }{(1+r)^j} + \frac{d_{t+J}}{(1+r)^J} 
\]
Take limit for $J\rightarrow\infty$ and use the transversality condition
\eqref{eq:capital_transversality_nocost}
 to obtain}

\end{slide}

\begin{slide}

\begin{equation}
\tag{3.13}
\label{eq:capital_nocost-c}
c_t +r d_{t-1} = y^p_t \equiv 
 \frac{r}{1+r} \sum_{j=0}^{\infty}
\frac{A_{t+j} F(k_{t+j})-(k_{t+j+1} - k_{t+j})}
{(1+r)^j}
\end{equation}


{\bf Interpretation}: The right-hand side of~(\ref{eq:capital_nocost-c})  
is the household's nonfinancial permanent income, $y^p_t$. (It is a natural generalization of a similar expression  obtained in the endowment economy, see equation 2.10).  In the present environment, nonfinancial permanent income is given by a weighted average  of present and future expected output net of investment expenditure.  Thus, equilibrium condition~(\ref{eq:capital_nocost-c})  
 states that each period households allocate their nonfinancial permanent income  to consumption and to servicing their debt. 
\end{slide}
\begin{slide}
A {\bf perfect-foresight equilibrium} is a value $c_0$ and a sequence $\{k_{t+1}\}_{t=0}^{\infty}$ satisfying (\ref{eq:capital_nocost-c}) evaluated at $t=0$, and \eqref{eq:capital_nocost-Euler}
 for all $t\ge0$,  
 given the initial stock of physical capital,  $k_0$, the initial net external debt position, $d_{-1}$, and the sequence of  productivity $\{A_t\}_{t=0}^{\infty}$. 


This is a system we can fully characterize with pen and paper. 


[Obtain  eqm values for $c_t$ from \eqref{eq:capital_nocost-cc}, 
$i_t$ from \eqref{eq:capital_i_economy_capital}, $y_t$ from \eqref{eq:capital_output_no_adj_cost}, and $d_t$ from \eqref{eq:capital_bc_no_adj_cost}]


Note that $k_t$ for $t>0$ is a function of the exogenous variable $A_t$ only. Thus permanent income, $y^p_t$, is a function of productivity only and is increasing in present and future values of productivity.  


\end{slide}
\begin{slide}



{\bf Trade balance}: $tb_t = y_t - c_t - i_t $

To obtain the prediction of a countercyclical trade balance response it is no longer required that consumption increases by more than one-for-one with output. As long as domestic absorption, $c_t + i_t$, increases by more than output, the model will predict a countercyclical trade balance response. 

Next we study adjustment to permanent and temporary productivity shocks and ask whether the model predicts a countercyclical trade balance response. 

\end{slide}

\begin{slide}
{\bf  3.2. Steady State Equilibrium }

Suppose $A_t = \bar{A}$ for all $t\ge 0$, and  $k_0 = \bar{k} \equiv \kappa\left(\frac{\bar{A}}{r}\right)$. 

By \eqref{eq:capital_kappa_economy_capital}, $k_t = \bar{k}$ for all $t>0$

By \eqref{eq:capital_nocost-cc} and \eqref{eq:capital_nocost-c}, $c_t = \bar{c} \equiv  -r d_{-1} +  \bar{A} F(\bar{k})$

and $d_t =  d_{-1}$ for all $t\ge 0$


Output: $y_t = \bar{y} \equiv \bar{A} F(\bar{k})$

Trade balance: $tb_t = \bar{tb} \equiv r d_{-1}$

Current account: $ca_t = d_{t-1}-d_t = 0$ 

\end{slide}
\begin{slide}
{\bf 3.3 Adjustment to a Permanent Unanticipated Increase in Productivity}


\uline{Experiment:} In period 0 it is learned that $A_t$  increases from $\bar{A}$ to $A'>\bar{A}$ for all $t\ge0$. Prior to period 0, $A_t$ was expected to be $\bar{A}$ forever. 


$$
A_t=
\left\{
\begin{array}{ll}
\bar{A}& \mbox{for $t\le - 1$}\\
A'>\bar{A}& \mbox{for $t\ge
0$}
\end{array}
\right. .
$$


\centerline{\includegraphics[width=4in]{capital_permanent_slides.eps}}
%These are just cut and pasted from the page proof. 
%Z:\uribe\book_first_edition\PageProofs\Uribe 2nd Proof.pdf


\end{slide}
\begin{slide}
\begin{minipage}{4in}
{\bf Adjustment of Capital and Investment}

\bigskip

For $t>0$, by \eqref{eq:capital_kappa_economy_capital}
\[
k_{t} = k'\equiv \kappa\left(\frac{A'}{r}\right)>\bar{k}
\] 

Thus positive investment in period $0$ and zero investment thereafter. 
\[
t=0: i_0 = k'-\bar{k}>0
\]
\[ t>0: i_t =0 
\]
\end{minipage}
\begin{minipage}{4in}
\includegraphics[width=5in]{capital_ir_permanent_A_k_i_slides.eps}
%cut and pasted from the page proof. 
%Z:\uribe\book_first_edition\PageProofs\Uribe 2nd Proof.pdf

\end{minipage}
\end{slide}


\begin{slide}
\begin{minipage}{4in}
{\bf Adjustment of Output}

\bigskip
Output increases in period 0 because $A_0$ rises 

and then again in period 1 because $k_1$ is larger: 
\[
t=0: y_0 = A' F(\bar{k}) > \bar{A} F(\bar{k}) = \bar{y}
\]
\[
 t>0: y_t = A'F(k')> A'F(\bar{k}) = y_0
\]

\end{minipage}
\begin{minipage}{6in}
\rightline{\includegraphics[width=5in]{capital_ir_permanent_A_y_slides.eps}}
%cut and pasted from the page proof. 
%Z:\uribe\book_first_edition\PageProofs\Uribe 2nd Proof.pdf
\end{minipage}
\end{slide}


\begin{slide}

What about consumption? Intuitively, it should increase. How to show that it does? By \eqref{eq:capital_nocost-cc} 
\[
c_t = c_0
\]
for all $t\ge0$. Thus, we only need to find $c_0$. 
By~\eqref{eq:capital_nocost-c}, $c_0=y^p_0-rd_{-1}$. If permanent income in period 0 rises, so does consumption. Thus, let's find first the adjustment in $y^p_0$.

\end{slide}
\begin{slide}

From period 1 on, $i_t=0$, thus using the definition of $y^p_0$ we have: 
\begin{eqnarray*}
y^p_0 &=&
 \frac{r}{1+r}\left[ (A'F(\bar{k}) - k'+\bar{k}\right] + 
\frac{1}{1+r} A'F(k')\\
&=& A'F(\bar{k}) +    \frac{1}{1+r}\left[   A'F(k') -  A'F(\bar{k}) - r(k'-\bar{k})\right] \\
&=& A'F(\bar{k}) +    \frac{1}{1+r}\left[   A'F(k') -  A'F(\bar{k}) - A'F'(k')(k'-\bar{k})\right] \\
&>& A'F(\bar{k}) (=y_0) \\
&>& \bar{A}F(\bar{k})(=y^p_{-1}) .
\end{eqnarray*}
(The first inequality follows from the facts that $F(\cdot)$ is increasing and concave and that $k'>\bar{k}$)
\\
Because in period 0 permanent income exceeds current income, we have that $c_0$ increases by more than $y_0$. This by itself---that is, ignoring the increase in $i_0$---leads to a negative trade balance response in period 0.
By contrast, in the endowment economy of Chapter 2 a once-and-for-all increase in the endowment leaves the trade balance unchanged. 
The intuition for this result  is that the path of output is upward sloping in the economy with capital in response to the permanent shock.  
\end{slide}
\begin{slide}
{\bf Adjustment of the Trade Balance}

We just established that the trade balance deteriorates in period 0. But what about period 1. In period 1, output is higher than in period 0, consumption is the same,  and investment is lower. Thus clearly $tb_1>tb_0$.  

For $t>0$, $tb_t = tb'> tb_0$. 

Is  $tb'$  greater or less  than $tb_{-1}$?  By \eqref{eq:capital_bc_no_adj_cost}
for $t>0$
\[
d_t =  (1+r)d_{t-1} - tb'
\]
This will satisfy~\eqref{eq:capital_transversality_nocost}  only if 
\[
tb' =r d_0 
\]
where $d_0 = d_{-1} + y^p_0 - y_0>d_{-1}$.  The new level of debt is permanently higher than it was prior to the productivity shock and therefore the trade balance, which is used to service the interest on the debt, must also be permanently higher. 


\end{slide}
\begin{slide}
\centerline{Summary of Adjustment to Permanent Productivity Shock}


\centerline{\includegraphics[width = 11in]{capital_ir_permanent_A_increase_slides.eps}}
%Produced by running the file z:\uribe\book\capital\ir_permanent_A_increase.m

\end{slide}

\begin{slide}
{\bf 3.4 Adjustment to Temporary Productivity Shocks}

\uline{Experiment:} In period 0 it is learned that $A_0=A'>A_{-1}=\bar{A}$ and that $A_t = \bar{A}$ for all $t>0$.  
\[
A_t=
\left\{
\begin{array}{ll}
\bar{A}&\mbox{for $t\le - 1$}\\
A'>\bar{A}&\mbox{for $t=0$}\\
\bar{A}& \mbox{for $t>0$}\\
\end{array}
\right. 
\]
By \eqref{eq:capital_nocost-Euler}
\[
k_t = \bar{k} ; \mbox{ for all } t>0
\]
By \eqref{eq:capital_i_economy_capital}  
\[
i_t = 0; \mbox{ for all } t\ge 0
\]
By \eqref{eq:capital_output_no_adj_cost} 
\[
y_0 = A'F(\bar{k})> \bar{y}; \quad \mbox{ and } y_t = \bar{y} = \bar{A} F(\bar{k}); \mbox{ for all } t> 0
\]
Note that the adjustment to a purely temporary shock in the economy with capital is thus the same as the adjustment to a purely temporary endowment shock in the economy without capital studied in Chapter 2. 


\end{slide}
\begin{slide}

By \eqref{eq:capital_nocost-cc} 
\[
c_t = c_0 \quad \mbox{for all $t\ge 0$}
\]

By~\eqref{eq:capital_nocost-c}
\[
c_0 = -r d_{-1} + \bar{A} F(\bar{k}) 
+ \frac{r}{1+r} 
\left(
A'F(\bar{k}) - 
\bar{A} 
F(\bar{k}) 
\right)
\]
Recalling that $c_{-1} = -r \bar{d} + \bar{A} F(\bar{k})$ and that $d_{-1} = \bar{d}$ yields 
\[
c_0 - c_1 = \frac{r}{1+r} 
\left(
A'F(\bar{k}) - 
\bar{A} 
F(\bar{k}) 
\right)>0
\]
Thus consumption increases by only a small fraction of the increase in income. 
 
\end{slide}

\begin{slide}
{\small 
From the definition of the trade balance we have
\[
tb_0 - tb_{-1}
 =
(y_0 - y_{-1})-
(c_0 - c_{-1})-
(i_0-i_{-1}) = 
 \frac{1}{1+r} (y_0 - y_{-1}) >0
\]
$\Rightarrow$ procyclical trade balance adjustment in period 0.

For $t>0$:
$c_t$, $y_t$, $i_t$ are all constant. Hence $tb_t$ is also constant. At what level? By same argument as above
\[
tb_t = tb' = r d_0; \mbox{ and } d_t = d_0;  \quad \forall t>0
\]
Because $c_0$ increases by less than $y_0$ and $i_0$ is unchanged (at zero), it must be that $d_0<d_{-1}=\bar{d}$. It follows that 
\[tb'< tb_{-1}< tb_0\]
Finally, the adjustment of the current acount is
\[
ca_0 - ca_{-1} = tb_0 - tb_{-1} >0
\]
and 
\[
ca_t = 0; \quad \forall t>0
\]
}
\end{slide}
\begin{slide}
\centerline{\small \bf Adjustment to  Temporary and Permanent Productivity Increases}
\centerline{\includegraphics[width=9.5in]{capital_ir_temporary_A_increase_slides.eps}}
%Produced by running the file z:\uribe\book\capital\ir_temporary_A_increase.m
\vspace{-1cm}
{\tiny \rule{1.5cm}{1mm}: Temporary Productivity Increase \hfill  {\color{blue}{x\dots x\dots x\dots}}: Permanent Productivity Increase\\
}
{\bf Principle I:  The more persistent productivity shocks are, the more likely an initial deterioration of the trade balance will be.}

\end{slide}


\begin{slide}
{\bf  3.5 Capital Adjustment Costs} 

Motivation: 
Capital adjustment costs are a standard feature of open economy business cycle models. They are used to ensure that the predicted  volatility of investment relative to the volatility of output does not exceed the observed one. 

In the presence of adjustment costs, investment will be spread out over a number of periods. This will have two consequences for the period $0$ adjustment of the trade balance. First, the increase in investment in period 0 will be lower. Second, the increase in permanent income will be lower (because output increases slower to its new permanently higher level) and therefore the consumption response in period 0  will be lower. Both factors contribute to a more muted trade balance response. 

We will show that: 

{\bf Principle II:} The more pronounced are capital adjustment costs, the smaller will be the initial trade balance deterioration in response to a positive and persistent productivity shock.
 


\end{slide}
\begin{slide}

 \[
\mbox{Capital adjustment costs } = 
 \frac12 \,\,\,\frac{i_t^2}{k_t}\]

$\bullet$ If $i_t=0$, then adj costs are nil. 

$\bullet$ adj costs are convex in $i_t$

$\bullet$ these are actual resources lost!

$\bullet$ [insert graph of adjustment costs]

$\bullet$ Slope of adjustment costs: $\frac{\partial \frac{i_t^2}{2 k_t}}{\partial i_t} = \frac{i_t}{k_t}$

$\bullet$ in our model in steady state $i_t=0$, so adjustment costs and marginal adjustment costs are nil in steady state. 

\end{slide}
\begin{slide}
{\small 
With adjustment costs the sequential budget constraint becomes: 
\begin{equation}
\label{eq:capital_bc_cost}
\tag{3.16}
c_t + i_t + {\color{red}\frac12 \frac{i_t^2}{k_t}} + (1+r) d_{t-1} = A_t F(k_t) + d_t 
\end{equation}

Lagrangian: 
\[
\mathcal{L} =
\sum_{t=0}^{\infty} \beta^t \left\{
U(c_t)
+ \lambda_t
\left[
A_t F(k_t) + d_t
-(1+r) d_{t-1}   - c_t - i_t - \frac{1}{2} \frac{i_t^2}{k_t}
+ q_t (k_t+i_t-k_{t+1})
\right]
\right\}
\]

Optimality conditions:
(\ref{eq:capital_i_economy_capital}), 
(\ref{eq:capital_no_Ponzi_no_adj_cost}) holding with equality, 
(\ref{eq:capital_la_nocost}), (\ref{eq:capital_d_nocost}), 
(\ref{eq:capital_bc_cost}),
\begin{equation}
\label{eq:capital_i-q}
\tag{3.17}
1 + \frac{i_t}{k_t} = q_t
\end{equation}
\begin{equation}
\label{eq:capital_k-b-tradeoff}
\tag{3.18}
\lambda_t q_t = \beta \lambda_{t+1} \left[ q_{t+1} +  A_{t+1} F'(k_{t+1}) + \frac{1}{2} \left(\frac{i_{t+1}}{k_{t+1}}\right)^2 \right] 
\end{equation}

$q_t=$ Tobin's
$q$, shadow price of capital in terms of consumption goods 
}
\end{slide}
\begin{slide}

Again assume that $\beta(1+r) =1$, then \eqref{eq:capital_k-b-tradeoff}
be written as 

\begin{equation}
\label{eq:capital_k-b-tradeoff1}
\tag{3.19}
(1+r) q_t  =   A_{t+1} F'(k_{t+1}) +q_{t+1} + \frac{1}{2} \left(\frac{i_{t+1}}{k_{t+1}}\right)^2
\end{equation}

{\bf Interpretation: } 
Suppose you have $q_t$ units of consumption goods. LHS is the return if those are invested in bonds. RHS is the return if those are invested in capital, which is the marginal product of capital, the undepreciated capital, and the reduction in investment adjustment costs. 

\end{slide}
\begin{slide}
As in the case without adjustment costs, we can separate $c_t$ dynamics from $k_t$ or $i_t$ dynamics. 


Solving the sequential budget constraint  (\ref{eq:capital_bc_cost}) forward and using  the no-Ponzi-game constraint  (\ref{eq:capital_no_Ponzi_no_adj_cost}) holding with equality yields
\[
c_t = -rd_{t-1} + \frac{r}{1+r} \sum_{j=0}^{\infty}
\frac{A_{t+j} F(k_{t+j})-i_{t+j} - \frac{1}{2}
(i_{t+j}^2/k_{t+j})
}{(1+r)^j}.
\]
This is by now a familiar expression. Households split 
their nonfinancial permanent income, given by the second term on the right-hand side, to service their outstanding debt and to consume. The definition of nonfinancial permanent income is adapted to include adjustment costs as one additional component of domestic absorption subtracted from the flow of output. The right-hand side of the above expression is known as {\it permanent income} and is given by the sum of net investment income ($-rd_{t-1}$) and nonfinancial permanent income.  
 
\end{slide}
\begin{slide}
{Dynamics of the Capital Stock}

Combine~(\ref{eq:capital_i_economy_capital}), (\ref{eq:capital_i-q}), and (\ref{eq:capital_k-b-tradeoff1}), to obtain two first-order, nonlinear difference equations in $k_t$ and $q_t$:
\begin{equation}
\label{eq:capital_Tobin-q}
k_{t+1} = q_t k_t
\tag{3.20}
\end{equation}
\begin{equation}
\tag{3.21}
\label{eq:capital_eqm-Euler}
q_t = \frac{A_{t+1} F'(q_tk_{t}) +  (q_{t+1}-1)^2/2+q_{t+1}}{1+r}
\end{equation}
\end{slide}
\begin{slide}
Steady state solution: $(q,k)$ \\
Suppose $A_t=\bar{A}$ for all $t$

By \eqref{eq:capital_Tobin-q}, 
\[
  q =1
\]
And using this result in  \eqref{eq:capital_eqm-Euler}
\[
r = \bar{A} F'(k) 
\]

$\rightarrow$ investment adjustment costs play no role for long run values of $k$ and $q$ 

but they do play a role for the short-run dynamics, which we will analyze next
using a phase diagram
\end{slide}

\begin{slide}
Let's plot the locus of pairs $(k_t, q_t)$ such that $k_{t+1}=k_t$. Call it the $\overline{KK'}$ locus.
By \eqref{eq:capital_Tobin-q}
if
\[
\begin{array}{ll}
q_t >1, & k_{t+1}>k_t\\
q_t =1, & k_{t+1}=k_t\\
q_t <1, & k_{t+1}<k_t\\
\end{array}
\]

\centerline{\includegraphics[width=4in]{capital_phase_KK_line.eps}}
%produced with phase_slides.m in the directory 
%z:\uribe\book\capital
\end{slide}

\begin{slide}
Assume that $A_t = \bar{A}$ for all $t$. Plot the locus of pairs
 $(k_t, q_t)$ such that $q_{t+1}=q_t$ in a neighborhood around $q_t=1$. (This is a local analysis.) Call this the $\overline{QQ'}$ locus.
By \eqref{eq:capital_eqm-Euler}, the $\overline{QQ'}$ locus is given by 
\[
r q_t = \bar{A} F'(q_tk_{t}) +  (q_{t}-1)^2/2
\]
\[
\mbox{If  } \quad \left\{
\begin{array}{ll}
(k_t, q_t) \mbox{ above } \overline{QQ'}, & q_{t+1}>q_t\\
(k_t, q_t) \mbox{ on }  \overline{QQ'}, & q_{t+1}=q_t\\
(k_t, q_t) \mbox{ below } \overline{QQ'}, & q_{t+1}<q_t\\
\end{array}\right.
\]

\centerline{\includegraphics[width=4in]{capital_phase_QQ_line.eps}}
%produced with phase_slides.m in the directory 
%z:\uribe\book\capital
\end{slide}

\begin{slide}
This yields the phase diagram: 

\centerline{\includegraphics[width=4in]{capital_phase_QQ_KK_line.eps}}
%produced with phase_slides.m in the directory 
%z:\uribe\book\capital

{\small 

$\bullet$ The intersection of $\overline{KK'}$ and $\overline{QQ'}$ is the steady state pair $(k, q) = (\bar{k}, 1)$ 

$\bullet$ The locus $\overline{SS'}$ is the saddle path. 

$\bullet$ Given the initial capital stock, $k_0$, Tobin's q,  $q_0$, jumps to the saddle path, and  $(k_t, q_t)$ converge monotonically to $(\bar{k},1)$. 

}
\end{slide}

\begin{slide}
\uline{Experiment 1}: Adjustment to a  temporary productivity shock. $\rightarrow$ identical to the economy without capital adjustment costs, as there is no reason to adjust the capital stock. (results as in Section 3.4).

\uline{Experiment 2}: Adjustment to a permanent productivity shock. 
\\In period 0 it is learned that $A_t$  increases from $\bar{A}$ to $A'>\bar{A}$ for all $t\ge0$. Prior to period 0, $A_t$ was expected to be $\bar{A}$ forever. 
$$
A_t=
\left\{
\begin{array}{ll}
\bar{A}& \mbox{for $t\le - 1$}\\
A'>\bar{A}& \mbox{for $t\ge
0$}
\end{array}
\right. .
$$

How can we capture this in the phase diagram? The $\overline{KK'}$ locus does not change. But the $\overline{QQ'}$ locus changes. The new locus is implicitly given by $r q_t = A' F'(q_tk_{t}) +  (q_{t}-1)^2/2$. This means that the $\overline{QQ'}$ locus shifts up and to the right.  The new steady state is $(k_t, q_t)=(k',1)$, where $k'$ solves $r=A'F'(k')$. The initial capital stock is $k_0 = \bar{k}$, hence $k_0<k'$. 
\end{slide}

\begin{slide}
{\small 
The dynamics of the capital stock can be read of the graph below.  

\centerline{\includegraphics[width=4in]{capital_phase.eps}}
%produced with phase.m in the directory 
%z:\uribe\book\capital

In period $0$ the economy jumps to point $a$, where $q_0>1$ and $k_0=\bar{k}$. That is, capital converges monotonically to $k'$ from below and Tobin's q converges monotonically to $1$ from above. Investment is positive during the entire transition, but, importantly,  $i_0<k'-\bar{k}$. It follows that domestic absorption increases by less on impact in the presence of capital adjustment costs. And thus, the deterioration of the trade balance in response to a positive permanent productivity shock is smaller on impact. We summarize these results as follows: 


{\bf Principle II: The more pronounced are capital adjustment costs, the smaller will be the initial trade balance deterioration in response to a positive and persistent productivity shock.}

}

\end{slide}
\begin{slide}
Thus far, to determine the dynamics in the model with capital adjustment costs we used a phase diagram. The phase diagram is a convenient graphical tool to analyze dynamics qualitatively. Specifically, we used the phase diagram to establish that if $k_0$ is below steady state, then  
\\
$\bullet$ the model is saddle path stable
\\
$\bullet$ the price of capital converges to its steady state value from above \\
$\bullet$ capital converges to its steady state value from below. 
\\
$\bullet$ investement is positive along the entire transition. 
\\
$\bullet$ capital adjustment costs dampen the trade balance deterioration in response to a permanent productivity increase. 

We now  consider an alternative method to determine whether the model is saddle path stable and to characterize the adjustment of the economy when $k_0$ is below its steady state value. 
\end{slide}
\begin{slide}
Characterization of Adjustment using a Log-linear approximation

We wish to characterize the dynamics of $q_t$ and $k_t$ described by  
\begin{equation}
\tag{\ref{eq:capital_Tobin-q}R}
k_{t+1} = q_t k_t
\end{equation}
\begin{equation}
\tag{\ref{eq:capital_eqm-Euler}R}
q_t = \frac{A_{t+1} F'(q_tk_{t}) +  (q_{t+1}-1)^2/2+q_{t+1}}{1+r}
\end{equation}

$k_t =$  endogenous predetermined variable\\
$q_t=$ endogenous nonpredetermined variable\\
$A_t=$ exogenous variable 

\end{slide}
\begin{slide}
Consider the dynamics around the steady state associated with  $A_t=A'>\bar{A}$ for all $t\ge0$.  
The steady state solution to \eqref{eq:capital_Tobin-q} 
and \eqref{eq:capital_eqm-Euler} is
\[
q^{ss} =1 
\]
\[
k^{ss} = k'
\]
where $k'$ is the solution to $r = A'F'(k')$. 
\\Let 
\[
\hat{q}_t \equiv \ln \frac{q_t}{q^{ss}}
\]
\[
\hat{k}_t \equiv \ln \frac{k_t}{k^{ss}}
\]
Log-linearize \eqref{eq:capital_Tobin-q} 
and \eqref{eq:capital_eqm-Euler} around the point $(q_t, k_t)= (1, k')$
\end{slide}
\begin{slide}

Since this is the first time we use this technique, we will explain each step. 
Take logs of \eqref{eq:capital_Tobin-q}, then take the total differential: 
\begin{align}
\ln k_{t+1} = &\ln k_t + \ln q_t
\nonumber
\\
(\ln k_{t+1} - \ln k^{ss}) = & (\ln k_t - \ln k^{ss}) + (\ln q_t - \ln q^{ss} )
\nonumber
\\
\label{eq:capital_Tobin-q_linearized}
\tag{\ref{eq:capital_Tobin-q}'} 
\hat{k}_{t+1} = &\hat{q}_t + \hat{k}_t
\end{align}
Applying the same steps to \eqref{eq:capital_eqm-Euler} is a little more complicated. To make the presentation clearer, let $x_{t+1} = A' F'(k_{t+1}) +  (q_{t+1}-1)^2/2+q_{t+1}$. (Note that $x^{ss}= 1+r$).
With this notation in hand, after takings logs of both sides, \eqref{eq:capital_eqm-Euler} becomes 
\begin{align*}
\ln(1+r) + \ln  q_t =& \ln x_{t+1}
\end{align*}
Take total differential with respect to $\ln q_t$ and  $\ln x_{t+1}$
\begin{align*}
\ln  q_t - \ln q^{ss}  =& \ln x_{t+1} - \ln x^{ss}\\
\hat{q}_t   =& \hat{x}_{t+1} 
\end{align*}

\end{slide}
\begin{slide}
To find $\hat{x}_{t+1}$ proceed as follows 
\begin{align*}
x_{t+1} = & A' F'(k_{t+1}) +  (q_{t+1}-1)^2/2+q_{t+1}
\\
\ln x_{t+1} = & \ln [A' F'(k_{t+1}) +  (q_{t+1}-1)^2/2+q_{t+1}]
\end{align*}

{\small Totally differentiate
\begin{align*}
\ln x_{t+1} - \ln x^{ss} = \frac{1}{[A' F'(k^{ss}) +  (q^{ss}-1)^2/2+q^{ss}]}\\
\times \left( A' F''(k^{ss}) k^{ss} (\ln k_{t+1}-\ln k^{ss}) +
(q^{ss}-1) q^{ss} (\ln q_{t+1}-\ln  q^{ss}) + q^{ss} (\ln q_{t+1}- \ln q^{ss})\right)
\\
\hat{x}_{t+1} = \frac{1}{1+r}
\left( A' F''(k^{ss}) k^{ss} \hat{k}_{t+1} + \hat{q}_{t+1} \right)
\end{align*}
Let 
\[
\epsilon_{F'} \equiv - \frac{F''(k^{ss}) k^{ss}}{F'(k^{ss})} >0
\]
\begin{align*}
(1+r) \hat{x}_{t+1} = - r  \epsilon_{F'} \hat{k}_{t+1} + \hat{q}_{t+1} 
\end{align*}
The log-linearized version of \eqref{eq:capital_eqm-Euler} then is 
}
\[
(1+r) \hat{q}_t   = - r \epsilon_{F'} \hat{k}_{t+1} + \hat{q}_{t+1} 
\]
\end{slide}
\begin{slide}
After some rearranging and substituting we have that the 
log-linearization of  \eqref{eq:capital_Tobin-q} 
and \eqref{eq:capital_eqm-Euler} around the steady state  $(q^{ss}, k^{ss})= (1, k')$ is
\begin{equation}
\left[
\begin{array}{c}
\hat{k}_{t+1}
\\
\hat{q}_{t+1} 
\end{array}
\right] 
= 
M 
\left[
\begin{array}{c}
\hat{k}_t
\\
\hat{q}_t 
\end{array}
\right] 
;
\quad
M 
=
\left[
\begin{array}{cc}
1&1
\\
 r \epsilon_{F'} & 1+ r + r \epsilon_{F'}
\end{array}
\right] 
\label{eq:capital_log_linear_terminal}
\tag{***}
\end{equation}
If we knew the initial values $k_0$ and $q_0$ we could trace out the dynamics. We do know $k_0$ as it is an initial condition. But we do not know the initial value of Tobin's q, $q_0$. To obtain it, we impose a terminal condition, we require that the economy converges back to the steady state. Thus our question becomes, does there exist such a solution and if so, is it unique. 
We are interested in solutions such that
\[
\lim_{t\rightarrow \infty}
\left[
\begin{array}{c}
\hat{k}_t
\\
\hat{q}_t 
\end{array}
\right] 
= \left[
\begin{array}{c}
0
\\
0
\end{array}
\right] 
\]
which says that $k_t\rightarrow k^{ss} = k'$ and $q_t \rightarrow q^{ss} = 1$.
\end{slide}
\begin{slide}

By \eqref{eq:capital_log_linear_terminal}
\[
\lim_{t\rightarrow \infty} \left[
\begin{array}{c}
\hat{k}_t
\\
\hat{q}_t 
\end{array}
\right] 
= \lim_{t \rightarrow \infty}
M^t 
\left[
\begin{array}{c}
\hat{k}_0
\\
\hat{q}_0 
\end{array}
\right]
\]
$\bullet$ If both eigenvalues of $\boldsymbol{M}$ lie outside the unit circle, then no equilibrium converging to the steady state exists. \\$\bullet$ If both eigenvalue of $\boldsymbol{M}$ lie inside the unit circle, then for any initial value of $q_0$, an equilibrium converging to the steady state exists, that is, the equilibrium is locally indeterminate.\\ $\bullet$ If one eigenvalue of $\boldsymbol{M}$ lies inside the unit circle and one outside, then a unique value for $q_0$ exists, such that the equilibrium converges to the steady state given some $k_0$ in the neighborhood of the steady state. 

Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $\boldsymbol{M}$. Then the equilibrium is locally unique iff
\[
|\lambda_1| >1 \quad \mbox{ and } \quad |\lambda_2| <1
\]
\end{slide}
\begin{slide}
Is this eigenvalue condition satisfied in our economy?
Yes, it is. To see this note, use that in general for any matrix  
\[
det(M) = \lambda_1 \lambda_2
; \quad \mbox{ and } \quad trace(M) = \lambda_1 + \lambda_2
\]
In our case, 
\[
det(M) = 1 + r >1
; \quad \mbox{ and } \quad trace(M) = 1 + 1 + r + r \epsilon_{F'} > 2+r
\]
(from here it follows that both eigenvalues are positive (or have positive real parts) and that at least one is greater than one in modulus. In turn this implies that if an equilibrium of the type we are looking for exists, then it would be unique.)

{\small 
To find whether it exists, let's first consider the case that the eigenvalues are real. Make a graph with $\lambda_1$ on the x-axis and $\lambda_2$ on the y-axis and plot: 1.) $\lambda_2 = \frac{1+r}{\lambda_1}$ and 2.) $\lambda_2 = 2+r+r\epsilon_{F'} - \lambda_1$. These lines must intersect twice in the positive quadrant because 1.) is positive, decreasing, and becomes arbitrarily large as $\lambda_1 \rightarrow 0$ from above and converges to zero as $\lambda_1\rightarrow \infty$ and at the same time 2.) is positive and finite for $\lambda_1=0$, decreasing, and converges to $-\infty$ as $\lambda_1\rightarrow \infty$.  The question is: is one intersection at $\lambda_1<1$ and the second at $\lambda_1>1$? 
At $\lambda_1=1$ 2.) is: $1+r+r\epsilon_{F'}$, which is greater than 1. It then follows that the conditions for uniqueness are satisfied. [Add the figure], that is, 
we have shown that $\lambda_1 >1$  and $0<\lambda_2 <1$. 

}

\end{slide}
\begin{slide}
{\small
How to find $q_0$?
Let 
\[
y_t = \left[ \begin{array}{cc}\hat{k}_t \\ \hat{q}_t
\end{array}
\right]\]
 Premultiply \eqref{eq:capital_log_linear_terminal}
by the left eigenvector of $M$ associated with the unstable eigenvalue,  $\lambda_1$, denoted $v_1$
\[
v_1 y_{t+1} = v_1 M y_t =  \lambda_1 v_1 y_t
\]
Let $\tilde{y}_t = v_1 y_t$. Then $\tilde{y}_t = \lambda_1^t \tilde{y}_0$. Because $|\lambda_1|>1$, $\lim_{t\rightarrow \infty} \tilde{y}_t= 0$ only if $\tilde{y}_0=0$, that is, if 
\[
0= v_1 \left[
\begin{array}{c}
\hat{k}_0
\\
\hat{q}_0 
\end{array}
\right]
=v_1^1 \hat{k}_0 + v_1^2 \hat{q}_0
\] 
\[
\hat{q}_0 = - \frac{v_1^1}{ v_1^2 } \hat{k}_0  = - (1-\lambda_2) \hat{k}_0
\]
 The last equality follows from 
\begin{align*}
\left[\begin{array}{cc} v _1^1 & v_1^2\end{array} \right] M = &
\lambda_1 \left[\begin{array}{cc} v _1^1 & v_1^2\end{array} \right]
\\
 v _1^1 +  v_1^2 (1+r + r\epsilon_{F'} )
 = &
\lambda_1  v_1^2
\\
 v _1^1 +  v_1^2 (trace(M) -1 )
 = &
\lambda_1  v_1^2
\\
 v _1^1 +  v_1^2 (\lambda_1 + \lambda_2 -1 )
 = &
\lambda_1  v_1^2
\\
\frac{ v _1^1 }{  v_1^2 }
 = &
(1- \lambda_2  )
\end{align*}
}
\end{slide}
\begin{slide}
Now use $\hat{q_t} = -(1-\lambda_2) \hat{k}_t $  in \eqref{eq:capital_Tobin-q_linearized}
 to obtain: 
\begin{align*}
\hat{k}_{t+1} =& \hat{k}_t + \hat{q}_t\\
=& (1-1+\lambda_2) \hat{k}_t\\
=& \lambda_2 \hat{k}_t. 
\end{align*}
Summary of dynamics: 
\begin{align*}
\hat{k}_t =& \lambda_2^t \hat{k}_0\\
\hat{q}_t =& -(1-\lambda_2) \lambda_2^t \hat{k}_0
\end{align*}
$\bullet$ unique saddle path stable eqm exists locally in the neighborhood around $(q^{ss}, k^{ss})$. 
\\
$\bullet$ The adjustment to a permanent increase in productivity induces capital to converge monotonically from below, Tobin's q to converge monotonically from above.
\\
$\bullet$ Because the increase in capital is spread out of many periods, investment is also positive for many periods and because the total increase in capital is independent of the size of the adjustment cost, it follows that adjustment costs dampen the initial increase in investment, and hence the initial deterioration in the trade balance and the current account. 
(Principle II)
\end{slide}

\end{document}





