%slides_endowment.tex
%Slides for Chapter 2 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}{5in}
\centering 
{\small slides 

chapter 2   

an open endowment economy
}

\vspace{1in}
\end{minipage}
\end{slide}


\topleft{\textit{Open Economy Macroeconomics, Chapter 2} }
\topright{M.\ Uribe and S.\ Schmitt-Groh\'e}


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

The purpose of this chapter is to build a canonical dynamic, general
equilibrium model of the  open economy and contrast its predictions with 
some of  the  empirical regularities  documented in
chapter~1.

The model economy developed in this chapter
is simple enough to allow for a full characterization of its
equilibrium dynamics using pen and paper.
The economy is inhabited by households that receive an exogenous but stochastic endowment of perishable goods each period. In addition, households have access to an internationally traded bond, which they use to smooth consumption in response to random income disturbances. 
In turn, consumption smoothing gives rise to equilibrium movements in the trade balance and the current account. 

The framework is called the intertemporal approach to the current account.

\end{slide}

\begin{slide}
{\bf \large The Model}\\
The problem of the representative household  is to choose processes for consumption in period $t$, denoted $c_t$, 
and debt, denoted $d_t$, to maximize
\begin{equation}
\label{eq:endowment_lifetime-utility1}
 E_0 \sum_{t=0}^{\infty} \beta^t  U(c_t)
\tag{2.1}
\end{equation}
subject to the sequential budget constraint
\begin{equation}
\label{eq:endowment_sequential-budget-constraint1}
 c_t + (1+r) d_{t-1} =  y_t + d_t,
\tag{2.2}
\end{equation}
and to the no-Ponzi-game constraint
\begin{equation}
\label{eq:endowment_no-Ponzi1}
\lim_{j\rightarrow \infty} E_t\frac{d_{t+j}}{(1+r)^j} \le 0,
\tag{2.3}
\end{equation}
given $d_{-1}$, where $y_t$ is an exogenous endowment and $r$ is a constant interest rate. At the optimal allocation, the no-Ponzi-game constraint must hold with equality. 

{\bf Comment:} In the closed economy context, this is the  permanent income model.
\end{slide}

\begin{slide}
{\bf \large Optimality Conditions}\\
The Lagrangian associated with the household's problem is
\[
\mathcal{L}_0 = E_0 \sum_{t=0}^{\infty} \beta^t\{U(c_t)+\lambda_t
[d_t    +y_t -  (1+r) d_{t-1}- c_t]\},
\]
where $\lambda_t$ is a Lagrange multiplier. The optimality conditions 
with respect to $c_t$ and $d_t$ are, respectively
\[
U'(c_t) = \lambda_t,
\]
and
\[ 
\lambda_t = \beta (1+r) E_t \lambda_{t+1}.
\]
Combining these two expression, yields the Euler Equation
\begin{equation}
\label{eq:endowment_Euler-c1}
U'(c_t) = \beta (1+r) E_t U'(c_{t+1}).
\tag{2.4}
\end{equation}
{\bf Interpretation:}
at the margin, the household is indifferent between consuming a unit of good today or saving it 
and consuming it the next period along with the interest.

\end{slide}

\begin{slide}
{\small 
{\bf The Intertemporal Resource Constraint}

Writing the sequential budget constraint~\eqref{eq:endowment_sequential-budget-constraint1} for period $t+j$, dividing by $(1+r)^j$, and taking expected values conditional on information available in period $t$ yields: 
\[
\frac{E_t c_{t+j}}
{(1+r)^j}
+ 
\frac{E_t d_{t-1+j}}
{(1+r)^{j-1}}
=
\frac{E_t y_{t+j}} {(1+r)^j}
+ 
\frac{E_t d_{t+j}} {(1+r)^j}.
\]
Then sum for $j=0$ to $j=J$:  
\[
\sum_{j=0}^{J}\frac{E_t c_{t+j}}
{(1+r)^j}
+ 
(1+r) d_{t-1}
=
\sum_{j=0}^{J}\frac{E_t y_{t+j}} {(1+r)^j}
+ 
\frac{E_t d_{t+J}} {(1+r)^J}.
\]
Taking limit for $J\rightarrow \infty$ 
 and using the transversality condition (equation~\eqref{eq:endowment_no-Ponzi1} holding with equality) yields the  intertemporal resource constraint
\begin{equation}
\label{eq:endowment_intertemporal_resource_constraint1}
(1+r) d_{t-1} =  \sum_{j=0}^{\infty} \frac{E_t(y_{t+j}-c_{t+j})}{(1+r)^j}.
\tag{2.5}
\end{equation}
{\bf Interpretation: }  at every point in time, the economy must 
be  expected to put aside a 
stream of resources, $\{E_t y_{t+j}-E_tc_{t+j}\}_{j=0}^{\infty}$,  large enough in present discounted value to cover the outstanding external debt. 

}
\end{slide}
\begin{slide}
{\bf Can An Economy Run A Perpetual Trade Deficit?}\\
Because in this economy there is only one good, the trade balance, denoted $tb_t$,  is 
given by the difference between output and consumption, that is,
\begin{equation}
\label{eq:endowment_tb_definition}
\tag{2.6}
tb_t = y_t-c_t.
\end{equation}
Using this definition, we can rewrite the  intertemporal resource constraint~\eqref{eq:endowment_intertemporal_resource_constraint1}  as 
\[
(1+r) d_{t-1} =  \sum_{j=0}^{\infty} \frac{E_t tb_{t+j}}{(1+r)^j}.
\]
The answer to the above question depends on the country's initial debt position:\\
If  the country is a net debtor ($d_{t-1}>0$),  then it must generate an expected trade surplus in at least one period, $E_t tb_{t+j}>0$ 
for some $j$. That is, if the country starts out as a net debtor, then it cannot run perpetual trade deficits. 
\end{slide}

\begin{slide}
{\bf \large Two Simplifying Assumptions}\\
The following two assumptions are handy because they allow for a closed-form solution of the model.\\
(1) The subjective and market discount rates are equal
\[\beta = \frac1{1+r}.\]
(2) The period utility function is quadratic
\begin{equation}
\label{eq:endowment_quadratic-utility}
\tag{2.7}
U(c) = -\frac12(c-\overline{c})^2,
\end{equation}
where $\overline{c}$ is a large enough number so that $c<\overline{c}$ in equilibrium. 
Under these assumptions, the Euler equation~\eqref{eq:endowment_Euler-c1} becomes
\begin{equation}
\label{eq:endowment_Euler-c-quad1}
\tag{2.8}
c_t = E_tc_{t+1},
\end{equation}
that is, consumption follows a {\bf random walk}. Households expect to maintain their current level of consumption  forever
\[E_t c_{t+j} = c_t\]
for all $j>0$. 
\end{slide}

\begin{slide}
Before presenting the closed form solutions to the model we introduce two  variables that will be convenient in their characterization and interpretation

--- nonfinancial permanent  income, $y^p_t$

--- difference between current income and permanent income, $y_t - y^p_t$

\end{slide}

\begin{slide}

{\bf Nonfinancial permanent income, $\boldsymbol{y^p_t}$}

Define nonfinancial permanent income, denoted $y^p_t$, as the constant level of income that, if received  in all future periods $t+j$ for $j\ge 0$, is equal to the expected present discounted value of the stochastic income process as of time $t$. That is, define $y^p_t$ as the solution to
\[
\sum_{j=0}^{\infty} \frac{y^p_t}{(1+r)^j} = \sum_{j=0}^{\infty} \frac{E_t y_{t+j}}{(1+r)^j}
\]


Solving for $y^p_t$ we obtain
\begin{equation}
\label{eq:endowment_yp}
y^p_t  = \frac{r}{1+r}\sum_{j=0}^{\infty} \frac{E_t y_{t+j}}{(1+r)^j}.
\tag{2.10}
\end{equation}

This equation says that $y^p_t$ is a weighted average of expected future income levels, with weights adding up to one. 

\end{slide}

\begin{slide}
{\bf The difference between current and permanent income,
$\boldsymbol{y_t -y^p_t}$ }
\begin{eqnarray*}
y_t - y^p_t & = & y_t + \frac1r \,  y^p_t - \left(\frac{1+r}{r}\right) y^p_t\\
            & = & y_t + \sum_{j=0}^{\infty} \frac{E_t y_{t+j}}{(1+r)^{j+1}}  - \sum_{j=0}^{\infty} \frac{E_t y_{t+j}}{(1+r)^j}
\\
& = &  \sum_{j=1}^{\infty} \frac{E_t y_{t+j-1}}{(1+r)^{j}}  - \sum_{j=1}^{\infty} \frac{E_t y_{t+j}}{(1+r)^j}
\\
& = &   - \sum_{j=1}^{\infty} \frac{E_t y_{t+j}- E_t y_{t+j-1}}{(1+r)^j}
\end{eqnarray*}

$\Rightarrow$
\begin{equation}
\label{eq:endowment_y-yp_delta}
y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j}
\tag{2.27}
\end{equation}
{\bf Interpretation:} Permanent income exceeds current income when income is expected to grow in the future. 


\end{slide}

\begin{slide}
{\bf \large The Closed-Form Equilibrium Solution}\\

Use the fact that $E_tc_{t+j}=c_t$ to eliminate $E_tc_{t+j}$ from the intertemporal resource constraint~\eqref{eq:endowment_intertemporal_resource_constraint1}: 
\begin{equation}
\label{eq:endowment_c+rd=yp}
\tag{2.11}
c_t = y^p_t - r d_{t-1},
\end{equation}
Because $d_{t-1}$ is predetermined in $t$ and $y^p_t$ is exogenous, equation~\eqref{eq:endowment_c+rd=yp}  represents the closed-form solution for $c_t$.

{\bf Interpretation: } in equilibrium consumption is equal to nonfinancial  permanent income minus  interest on existing initial debt.  
\end{slide}

\begin{slide}
{\bf The Equilibrium Behavior of the Trade Balance}\\
Using the equilibrium level of consumption given in~\eqref{eq:endowment_c+rd=yp}, we obtain the closed-form solution for the 
 trade balance, $tb_t=y_t-c_t$ , as

\begin{equation}
\label{eq:endowment_tb_permanent_income}
\tag{2.16}
tb_t = y_t -  y^p_t
+rd_{t-1}
\end{equation}

$\Rightarrow$ the trade balance responds countercyclically  to changes in current income if permanent income increases by more  than current income in response to increases in current income.  This is an important result, because we saw in chapter 1 that the trade balance is countercyclical. 
To capture this fact in the context of the present  model, the endowment process must be such that permanent income increases by more than one for one  with current income. What type of endowment process satisfies this requirement? We will return to this point soon. 
\end{slide}

\begin{slide}
{\bf The Equilibrium Behavior of  External Debt}

Use~\eqref{eq:endowment_c+rd=yp} again to eliminate $c_t$ from the sequential budget constraint~\eqref{eq:endowment_sequential-budget-constraint1} to get
\begin{equation}
\label{eq:endowment_eqm-d1}
d_t - d_{t-1} =   y^p_t
- y_t. 
\tag{2.12}
\end{equation}
which says that the economy borrows to cover deviations of current income from permanent income. Like consumption, external debt has a unit root. 
\end{slide}

\begin{slide}
{\bf The Equilibrium Behavior of the Current Account}


The current account is defined as  the trade balance minus interest payments on the external debt
\begin{equation}
\label{eq:endowment_ca_definition_tb+rd}
ca_t\equiv tb_t  - r d_{t-1}.
\tag{2.13}
\end{equation}
Using the closed-form solution for $tb_t$ given in~\eqref{eq:endowment_tb_permanent_income}, we get
\begin{equation}
\label{eq:endowment_ca_permanent_income}
ca_t = y_t -  y^p_t.
\tag{2.15}
\end{equation}
In chapter 1, we documented that the current account is countercyclical.  Thus, in order to explain this empirical regularity through this model, 
as in the case of the trade balance, 
permanent income must increase more than one for one with current income. 



Comparing the above expression with~\eqref{eq:endowment_eqm-d1} , we have that 
\begin{equation}
\tag{2.14}
\label{eq:endowment_ca_definition}
ca_t  = -(d_t-d_{t-1}).
\end{equation}
This is known as the {\bf fundamental balance of payments identity}: the current account equals the change in the country's net asset position. 
\end{slide}

\begin{slide}
Let's collect the results obtained thus far: 
When $\beta(1+r)=1$ and $U(c) = -1/2(c-\bar{c})^2$, then the model has the  following closed-form solution: 
\begin{equation}
\tag{\ref{eq:endowment_c+rd=yp}}
c_t = y^p_t- r d_{t-1}
%\tag{2.11}
\end{equation}
\begin{equation}
d_t = d_{t-1} + y^p_t -y_t \tag{\ref{eq:endowment_eqm-d1}}
\end{equation}
\begin{equation}
\tag{\ref{eq:endowment_ca_permanent_income}}
ca_t = y_t -  y^p_t
\end{equation}
\begin{equation}
\tag{\ref{eq:endowment_tb_permanent_income}}
tb_t = y_t - y^p_t + r d_{t-1} 
\end{equation}
where 
\begin{equation}
y^p_t \equiv \frac{r}{1+r} \sum_{j=0}^{\infty} \frac{E_t y_{t+j}}{(1+r)^j} \tag{\ref{eq:endowment_yp}}
\end{equation}
and $d_{-1}$ and the stochastic  process for $y_t$ are exogenously given. 

Memo item: 
\begin{equation}
ca_t = y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j}
\tag{\ref{eq:endowment_y-yp_delta}}
\end{equation}
\end{slide}

\begin{slide}
{\bf \large A General Principle} 

Take another look at the expression $ca_t = y_t-y^p_t$. It says that the current account is used whenever 
the economy experiences temporary deviations of  output from permanent income, that is, whenever $y_t-y^p_t\neq0$. By contrast, permanent output shocks, that is movements in $y_t$ that leave $y_t-y^p_t$ unchanged do not produce movements in the current account.  Thus, the following principle emerges:

\begin{quote}  
{\it
Finance temporary shocks (by running current account surpluses or deficits with little change in consumption) and  adjust to permanent shocks (by changing consumption but not the current account).  }
\end{quote}
\end{slide}


\begin{slide}
{\bf \large The Income Process}

We have established that
in order for the present model to account for the observed  countercyclicality of the current account and the trade balance, it is required that  $y_t-y^p_t$  be countercyclical. In turn, the cyclicality of $y_t-y^p_t$ depends on the particular process assumed for the endowment $y_t.$ For this reason, we now characterize the equilibrium under three representative income processes:

\begin{itemize}\item Two Stationary Processes\\
-- $y_t$ follows an  AR(1) process\\
  --   $y_t$ follows an AR(2) process
   \item  A nonstationary process, $\Delta y_t$ follows an AR(1) process.  \end{itemize}

\end{slide}

\begin{slide}
{\bf \large 2.2 An AR(1) Income Process}\\
Suppose $y_t$ follows the law of motion
\[
y_t -\bar{y}= \rho ( y_{t-1}-\bar{y}) + \epsilon_t,
\]
where $\epsilon_t$ is a white noise with mean zero and variance $\sigma^2_{\epsilon}$ and $\bar{y}$ is a positive constant. The parameter $\rho\in (-1,1)$ measures persistence. Then, the $j$-step-ahead forecast of $y_t$ as of period $t$ is
\[
E_t y_{t+j} = \bar{y} + \rho^j (y_t-\bar{y}).
\]
Thus, if $\rho>0$, which is the case of greatest empirical interest,  the forecast  converges monotonically  to the unconditional  mean of the process, namely $\bar{y}$. The more persistent  the process (the higher $\rho$) is,  the slower the speed of convergence of the forecast to the unconditional mean will be.
\end{slide}

\begin{slide}
{\bf Permanent Income with AR(1) Endowment}

Use $E_ty_{t+j}-\bar{y}=\rho^j(y_t-\bar{y})$ to eliminate $E_ty_{t+j}$ from~\eqref{eq:endowment_yp} to obtain
\begin{equation}
\label{eq:endowment_yp_AR(1)}
\tag{2.17}
y^p_t-\bar{y} = \frac r{1+r-\rho}(y_t-\bar{y}).
\end{equation}

This expression is quite intuitive:

$\bullet$ if the endowment process is temporary ($\rho\rightarrow0$),
 then only a small fraction, $r/(1+r)$, of movements in the endowment are incorporated into  permanent income. 

$\bullet$ if the endowment process is highly persistent ($\rho\rightarrow1$), 
most of movements in the current endowment are reflected in movements in permanent income. 

\end{slide}

\begin{slide}

A key variable is $y_t-y^p_t$, which under the AR(1) process is given by
\begin{equation}
\label{eq:endowment_ar1_y_minus_yp}
\tag{*}
y_t -y^p_t = \frac{1-\rho}{1+r-\rho}(y_t-\bar{y}).
\end{equation}

The coefficient 
\[\frac{1-\rho}{1+r-\rho}\] is positive for all possible values of $\rho$, implying that $y_t-y^p_t$ is procyclical. Therefore the trade balance and the current account are predicted to be procyclical --- which according to the evidence shown in chapter 1 is counterfactual. 
\end{slide}

\begin{slide}
{\bf Consumption Adjustment with AR(1) Income}

By \eqref{eq:endowment_c+rd=yp} and \eqref{eq:endowment_yp_AR(1)} 
\begin{equation}
\label{eq:endowment-c-ar1}
\tag{2.18}
c_t = y^p_t - r d_{t-1} = \bar{y} + \frac r{1+r-\rho}(y_t-\bar{y})
-r d_{t-1}
\end{equation}


$\bullet$ if $\rho=0$ consumption increases less than one for one with current income. Why? Because current income is higher than permanent income.  

$\bullet$ if $\rho\approx 1$, consumption adjusts one-for-one with current output, as current income is equal to permanent income.  


\end{slide}
\begin{slide}
{\bf External Debt Adjustment with AR(1) Income}

By \eqref{eq:endowment_eqm-d1} and  \eqref{eq:endowment_ar1_y_minus_yp}
\begin{equation}
\label{eq:endowment-d-ar1}
 \tag{2.21}
d_t = d_{t-1} +( y^p_t - y_t) = d_{t-1} - \frac{1-\rho}{1+r-\rho}(y_t-\bar{y})
\end{equation}

$\bullet$ if $\rho=0$ debt decreases almost one-for-one with income. Why? The country saves most of the temporary income increase to be able to consume not only more in the current period but also in all future periods. 

$\bullet$ if $\rho\approx 1$, debt is unchanged. Why? Because income will be also higher in the future and hence there is no need to save. 


\end{slide}
\begin{slide}
{\bf  Current Account Adjustment with AR(1) Income}

By \eqref{eq:endowment_ca_permanent_income} and \eqref{eq:endowment_ar1_y_minus_yp}
\begin{equation}
\label{eq:endowment-ca-ar1}
ca_t = y_t - y^p_t  =  \frac{1-\rho}{1+r-\rho}(y_t-\bar{y})
\tag{2.20}
\end{equation}



$\bullet$ if $\rho=0$ the change in the current account is close to the change in current income. Why? The country saves most of the temporary income increase to be able to consume not only more in the current period but also in all future periods. 

$\bullet$ if $\rho\approx 1$, the current account is unchanged (and equal to zero). Why? Because income will be also higher in the future and hence there is no need to save. 


\end{slide}
\begin{slide}
{\bf  Trade Balance Adjustment with AR(1) Income}

By \eqref{eq:endowment_tb_permanent_income} and   \eqref{eq:endowment_ar1_y_minus_yp}
\begin{equation}
tb_t = y_t - y^p_t + r d_{t-1}
\end{equation}

$\bullet$ if $\rho=0$, then most of the change in current income is exported resulting in a trade balance improvement. 

$\bullet$ if $\rho\approx 1$, then none of the change in current income is exported and the trade balance is unchanged. 




\end{slide}

\begin{slide}


The next graph displays the impulse response of $y_t$, $c_t$, $d_t$, $tb_t$, and $ca_t$  to a unit increase in $y_t$ assuming zero initial debt, $d_{t-1}=0$ for a value of $\rho\in (0,1)$. 
\end{slide}

\begin{slide}
\begin{center}
{\bf Response to a positive and persistent endowment shock: AR(1) process}
%\label{fig:endowment_shock} 
\centerline{\includegraphics[width=6in]{endowment_shock.eps}}
\end{center}
%produced with endowment_shock.m
\end{slide}

\begin{slide}
{\bf \large  Counterfactual Predictions with AR(1) Endowment}

The figure on the previous slide illustrates  that a positive endowment shock causes an improvement in the trade balance and the current account.
This is counterfactual. In chapter 1, we documented that both of these variables are countercyclical. The intuition behind the model's prediction is
 provided by the left panel of the following figure\\
\centerline{\includegraphics[width=4in]{endowment_shocks_slide.eps}}\\
%produced with endowment_shocks_slides.m
\end{slide}
\begin{slide}
Under the  assumed AR(1) process, an increase in output in the current period is expected to die out over time. So future output is expected to be lower than current output. As a result, 
consumption smoothing 
households save part of the current increase in output   for future consumption. 
Now suppose we could come up with an output process like the one 
shown on the right panel of the figure. There, an increase in output today creates expectations of even higher output in the future. As a result, consumption today increases by more than output, as  consumption-smoothing households  borrow against future income. 
In this environment, an output shock would tend to worsen the trade balance and the current account, which is more in line with the data. 
What type of endowment processes can give rise to an upward sloping  impulse response for output? 
We turn to this issue next. 
\end{slide}
\begin{slide}
\leftline{\bf \large 2.3 An AR(2) Income Process}
\begin{equation}
\label{eq:endowment_AR(2)}
y_t = \bar{y} + \rho_1 (y_{t-1}-\bar{y})+ \rho_2 (y_{t-2}-\bar{y}) + \epsilon_t
\tag{2.22}
\end{equation}


\begin{minipage}{4.5in}
\rightline{\small \bf Impulse Response of Endowment}
\rightline{\small AR(2) with  $\rho_1=1.5$ and 
$\rho_2=-0.51$.}
\centerline{\includegraphics[width=4in]{endowment_ar2_slides.eps}}
%produced with endowment_ar2.m in the directory: \uribe\book\endowment
\end{minipage}
\begin{minipage}{5.5in}{\small 

\vspace{1cm}

$\bullet$ impulse response is hump-shaped, that is, the peak output response occurs several periods after the shock occurs. 

$\bullet$ current level of output {\bf may} rise by less than  permanent income, that is,  the change in $y_t$ may be less than the change in $y^p_t$. 


$\bullet$ If so, the trade balance and the 
current account will deteriorate in response to an increase in output, bringing the model closer to the data. 


}
\end{minipage}

\end{slide}
\begin{slide}
{ \bf Restrictions on $\boldsymbol{\rho}_1$ and $\boldsymbol{\rho}_2$ to ensure stationarity}

$\bullet$ income process  should be mean reverting (stationary), i.e.,
$E_t y_{t+j}$ exists for all $j\ge0$ and 
\[
\lim_{j\rightarrow \infty} E_t y_{t+j} = \bar{y} \quad \forall t\ge 0
\]
Let 
$
Y_t = \left[\begin{array}{r} y_t - \bar{y}\\y_{t-1} - \bar{y}\end{array}\right]
$
Then write \eqref{eq:endowment_AR(2)} as: 
\[
Y_t = R\, Y_{t-1} + \left[\begin{array}{c}\epsilon_t \\0\end{array}\right]
\quad \mbox{ with } R = \left[\begin{array}{cc}\rho_1 &\rho_2\\ 1 & 0\end{array}\right]
\]
This implies that 
\begin{equation}
\tag{2.23}
\label{eq:endowment_AR(2)_EY}
E_t Y_{t+j} = R^j Y_t
\end{equation}
Thus $\lim_{j\rightarrow\infty} E_t y_{t+j} = \bar{y} $ holds for any initial conditions $y_t$, $y_{t-1}$, if and only if both eigenvalues of $\mathbf{R}$ lie inside the unit circle. 
\end{slide}
\begin{slide}
The eigenvalues of any 2x2 matrix  $\mathbf{M}$ lie inside the unit circle if and only if 
\begin{eqnarray*}
|det(M)|<1 \\
|tr(M)| < 1 + det(M)
\end{eqnarray*}

In our application we have, 
\[
det(R) = -\rho_2 \quad \mbox{ and } \quad tr(M) = \rho_1
\]
It follows that the AR(2) process is mean reverting iff
\begin{eqnarray*}
\rho_2<1-\rho_1\\
\rho_2<1+\rho_1\\
\rho_2>-1
\end{eqnarray*}

Take a look at the graph on the next slide. The set of allowable pairs $(\rho_1, \rho_2)$ are those inside the triangle. 
\end{slide}
\begin{slide}
{ \bf Restrictions on $\boldsymbol{\rho}_1$ and $\boldsymbol{\rho}_2$ to ensure stationarity}

\centerline{\includegraphics[width=8in]{ar2_rho1_rho2_plot.eps}}
%produced by running  ar2_rho1_rho2_plot.m in Z:\uribe\book\endowment
 
\end{slide}
\begin{slide}
{  \bf Restrictions on $\boldsymbol{\rho}_1$ and $\boldsymbol{\rho}_2$ to ensure  countercyclical impact response of trade balance}

Recall that the key variable determining the response of the trade balance and the current account is: $y_t - y^p_t$, the difference between current income and permanent income. 

In particular, we want to know for which pairs $(\rho_1, \rho_2)$ current income increases by less than permanent income in response to a positive income shock in the AR(2) case. 


\end{slide}
\begin{slide}
{\bf Permanent Income with AR(2) Endowment} 

{\small 
Recall the definition of permanent income, $y^p_t$, given in~\eqref{eq:endowment_yp}: 
\[
y^p_t-\bar{y}= \frac{r}{1+r}\sum_{j=0}^{\infty} (1+r)^{-j}E_t (y_{t+j}-\bar{y})
\] 
Use~\eqref{eq:endowment_AR(2)_EY}
\begin{eqnarray*}
E_tY_{t+j} &=& \mathbf{R}^j Y_t\\
(1+r)^{-j}E_tY_{t+j} &=& (\mathbf{R}/(1+r))^j Y_t\\
\sum_{j=0}^{\infty} (1+r)^{-j}E_tY_{t+j} 
&=& \sum_{j=0}^{\infty}
(\mathbf{R}/(1+r))^j Y_t\\
\frac{r}{1+r}\sum_{j=0}^{\infty} (1+r)^{-j}E_tY_{t+j} 
&=& \frac{r}{1+r}[\mathbf{I} - \mathbf{R}/(1+r)]^{-1} Y_t\\
&=& \frac{r}{(1+r)(1+r-\rho_1)-\rho_2} \left[\begin{array}{cc} 1+r & \rho_2\\1&1+r-\rho_1\end{array}\right] Y_t\\
\end{eqnarray*}


From here we obtain
\[
y^p_t - \bar{y} = \frac{r}{(1+r)(1+r-\rho_1)-\rho_2} \left[(1+r)(y_t-\bar{y}) + \rho_2 (y_{t-1}-\bar{y})\right]
\tag{2.24}
\]

$\Rightarrow $ The impact response of permanent income, $y^p_t$, to an increase in $y_t$ is always positive. (Can you show this?) 


}
\end{slide}
\begin{slide}
{ \bf The key variable $\boldsymbol{y_t} - \boldsymbol{y^p_t}$ with AR(2) Endowment } 


{\small 
From (2.24) it follows that 
\[
y_t - y^p_t = \gamma_0  (y_t -\bar{y}) + \gamma_1  (y_{t-1} -\bar{y})
\]
with
\[
\gamma_0 =  \frac{(1-\rho_1-\rho_2)  + r(1-\rho_1)}{(1-\rho_1-\rho_2)  + r(1-\rho_1) + r(1+r)} \quad \mbox{ and } \quad \gamma_1 = \dots
\]

The model therefore predicts a counterycyclical response on impact of $y_t -y^p_t$ iff
\[
\gamma_0 <0.
\]

This requires that the numerator and the denominator of $\gamma_0$ have different signs, which we can ensure by adding the requirement
\[
\rho_2 >(1+r)(1-\rho_1) \tag{**}
\] 
to the requirements of stationarity. 

The graph on the next slide shows there exist combinations of $(\rho_1,\rho_2)$ that satisfy this criterion and the stationarity requirements.
 Thus in principle the model with an AR(2) endowment process can predict a countercyclical response of the trade balance and the current account. 
 It is then an empirical question whether output conforms to those restrictions. (See Exercise 2.8)

}
\end{slide}
\begin{slide}
{  \bf Restrictions on $\boldsymbol{\rho}_1$ and $\boldsymbol{\rho}_2$ to ensure countercyclical   impact response of trade balance with AR(2) income}


\centerline{\includegraphics[width=6in]{ar2_rho1_rho2_plot2.eps}}We need $\rho_1>1$. Why? To have future income to be  higher than current income. And we need $\rho_2<0$ for stationarity, but not too negative to avoid that the initial increase in output is followed by a future decline. 
%produced by running  ar2_rho1_rho2_plot.m in Z:\uribe\book\endowment

\end{slide}
\begin{slide}
{\bf Summary of adjustment with  AR(2) income} 

In response to a positive income shock in period $t$: 
\\
$\bullet$ $y^p_t$ increases 
\\
$\bullet$ By~\eqref{eq:endowment_c+rd=yp}: $c_t = y^p_t - r d_{t-1}$, hence $c_t$ procyclical on impact.

and provided $\rho_2>(1+r)(1-\rho_1)$
\\
$\bullet$  $y^p_t -y_t$ increases, 
\\
$\bullet$
By (2.12): $d_t = d_{t-1} + y^p_t -y_t$, hence $d_t$ increases.  
\\
$\bullet$
By (2.15): $ca_t = y_t - y^p_t$, hence $ca_t$ countercyclical 
\\
$\bullet$
By (2.16): $tb_t = y_t - y^p_t + r d_{t-1}$, hence $tb_t$ countercyclical 

We have shown that it is possible for the model to predict a counter cyclical trade balance adjustment even under a stationary income process. Thus non-stationarity is not a necessary assumption for a countercyclical adjustment. Next we show that a non-stationary income process can give rise to a countercyclical trade balance adjustment as well. 

\end{slide}
\begin{slide}
{\bf \large 2.4 A Nonstationary Income Process}


\begin{equation}
\label{eq:endowment_delta_y}
\Delta y_t \equiv y_t-y_{t-1} \tag{2.25}
\end{equation}


\begin{equation}
\label{eq:endowment_deltay}
\Delta y_t = \rho \Delta y_{t-1} + \epsilon_t 
\tag{2.26}
\end{equation}

\end{slide}
\begin{slide}
{\bf \large 
The key variable $\boldsymbol{y_t-y^p_t}$ with Nonstationary Income}


In general (i.e., not using any particular stochastic for the income process)
\begin{equation}
\label{eq:endowment_y-yp_delta}
y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j}
\tag{2.27}
\end{equation}


When income follows \eqref{eq:endowment_deltay}, then 
\[
E_t \Delta y_{t+j} = \rho^j \Delta y_t
\]

and hence \eqref{eq:endowment_y-yp_delta} can be expressed as
\begin{equation}
\label{eq:endowment_y-yp_nonstationary}
y_t - y^p_t = - \, \frac{ \rho }{1+r-\rho} \, \Delta y_t
\tag{***}
\end{equation}

This equation says that permanent income increases by more than current income if $\boldsymbol{\rho>0}$, that is, then the growth rate shock is persistent. In this case, the level of output is expected to rise in the future. 

\end{slide}
\begin{slide}
{\bf \large Current Account Adjustment with Nonstationary Income}

By \eqref{eq:endowment_tb_permanent_income}  and \eqref{eq:endowment_y-yp_nonstationary}
\[
tb_t = y_t -  y^p_t +rd_{t-1} = - \, \frac{ \rho }{1+r-\rho} \, \Delta y_t + r d_{t-1}
\]

and by \eqref{eq:endowment_ca_permanent_income} and \eqref{eq:endowment_y-yp_nonstationary}
\[
ca_t = y_t - y^p_t =  - \, \frac{ \rho }{1+r-\rho} \, \Delta y_t
\]

Hence, as long as, $\boldsymbol{\rho>0}$, the economy with nonstationary income predicts a countercyclical impact response of the trade balance and of the current account.

Why? Agents increase consumption more than income in anticipation of future income increases. 


\end{slide}
\begin{slide}
{\bf  Excess Consumption Volatility with Nonstationary Income}

Can the model account for the empirical regularity that  consumption changes are more volatile than output changes, as documented in Chapter 1.

Let 
\[
\Delta c_t \equiv c_t-c_{t-1}
\]
and $\sigma_{\Delta c}$ and $\sigma_{\Delta y}$ denote the standard deviation of consumption and output changes, respectively. 

Use 
\[
ca_t = y_t - c_t -rd_{t-1}
\]
Take differences
\[ca_t - ca_{t-1} 
= 
\Delta y_t - 
\Delta c_t -
r(d_{t-1}-d_{t-2}).
\]

\end{slide}
\begin{slide}

Noting that $d_{t-1}-d_{t-2}=-ca_{t-1}$ and solving for $\Delta c_t$, we obtain:
\begin{align}
\nonumber
\Delta c_t
&=   \Delta y_t -ca_t+ (1+r)ca_{t-1}\\
\nonumber
&  = 
\Delta y_t +\frac{\rho}{1+r-\rho}
\Delta y_t-\frac{\rho (1+r)}{1+r-\rho}
\Delta y_{t-1}
\\\nonumber
&  = 
\frac{1+r}{1+r-\rho}
\Delta y_t-\frac{\rho (1+r)}{1+r-\rho}
\Delta y_{t-1}
\\
&  =  
\frac{1+r}{1+r-\rho} \, \, 
\epsilon_t.
\label{eq:endowment_Dc}
\tag{2.29}
\end{align}

The change in consumption is a white noise. Why? By the Euler equation~\eqref{eq:endowment_Euler-c-quad1}. 

\end{slide}
\begin{slide}

By~\eqref{eq:endowment_Dc}  
\[
\sigma_{\Delta c}
=
\frac{1+r}{1+r-\rho}
\sigma_{\epsilon}.
\]
and by equation~\eqref{eq:endowment_deltay} 
\[
\sigma_{\Delta y}\sqrt{1-\rho^2}=\sigma_{\epsilon}
\]
so that 
\begin{equation}
\label{eq:endowment_std_deltac_over_deltay}
\tag{2.30}
\frac{\sigma_{\Delta c}}
{\sigma_{\Delta y}}
=
\left[
\frac{1+r}{1+r-\rho}
\right]
\sqrt{1-\rho^2}
\end{equation}

 When $\rho=0$,   consumption and output changes are equally volatile. 

When $\rho>0$, consumption changes can become more volatile than output changes.  To see this: RHS of~\eqref{eq:endowment_std_deltac_over_deltay}  is increasing in $\rho$ at $\rho=0$. 
Since consumption and output changes are equally volatile at $\rho=0$, 
it follows that there are values of $\rho$ in the interval $(0,1)$ for which the volatility of consumption changes is   higher than that of output changes. 

\end{slide}
\begin{slide}

This property ceases to hold as $\Delta y_t$ becomes highly persistent. This is because as $\rho\rightarrow 1$, the variance of $\Delta y_t$  becomes infinitely large as changes in income become a random walk, whereas, as expression~(\ref{eq:endowment_Dc}) shows,  $\Delta c_t$ follows an i.i.d.\ process with finite variance for all values of $\rho \in [0,1]$. 

\end{slide}
\begin{slide}
{Excess Volatility of Consumption Changes and the Persistence of Output Changes}

\begin{minipage}{3.5in}
{\small Note. This figure plots equation~\eqref{eq:endowment_std_deltac_over_deltay} as a function of $\rho$ for an interest rate of 4 percent ($r=0.04$).


}
%produced by running
%sigmac_over_sigmay.m in
%z:\uribe\book\endowment
\end{minipage}
\begin{minipage}{7in}
\leftline{\includegraphics[width=6.5in]{endowment_sigmac_over_sigmay.eps}}
\end{minipage}

\end{slide}
\begin{slide}
{\bf 2.5  Testing the Intertemporal Approach to the Current Account}

$\bullet$ Hall (1978) initiates a large empirical literature  testing  the random walk hypothesis for consumption implied by the permanent income hypothesis. 

$\bullet$ Campbell (1987) tests the predictions of the permanent income model for savings. 

$\bullet$ Nason and Rogers (2006) test the predictions regarding the current account. 

In the present model savings and the current accounts are equal (because we are abstracting from investment). 

\end{slide}
\begin{slide}
Combining ~\eqref{eq:endowment_ca_permanent_income} and \eqref{eq:endowment_y-yp_delta} yields
\begin{equation}
\label{eq:endowment_CA-PDV}
\tag{2.28}
ca_t = y_t -y^p_t = - \sum_{j=1}^{\infty} (1+r)^{-j}E_t\Delta y_{t+j}
\end{equation}

This equation says that a country runs a current account deficit (i.e., borrows from the rest  of the world), when the present discounted value of future income changes is positive. And the country runs a current account surplus (i.e., lends to the rest of the world), when current income exceeds permanent income. It also says that current account data should forecast income changes. 
\end{slide}
\begin{slide}
{\bf Testable Restrictions of the Intertemporal Approach to the Current Account}
%Literature: Hall (1978); (Campbell (1987); Nason and Rogers (2006)

The testable restriction we focus on is: 
\[
\tag{2.28 R}
 ca_t = - \sum_{j=1}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j} 
\]

We have observations on the LHS of this equality because we have current account data. But we do not have data on the RHS. 
How can we get an estimate of the RHS? The idea is to use a vector autoregression (VAR) model in output changes ($\Delta y_t$) and the current account ($ca_t$) to construct an estimate of the RHS given a calibrated value of $r$. Given the estimate of the RHS, we can test the hypothesis that the LHS is equal to the RHS. 
\end{slide}
\begin{slide}

To construct an estimate of the RHS, proceed as follows. 



Estimate the VAR model:
\begin{equation}
\label{eq:endowment_VAR-PDV} 
x_t = D x_{t-1} + \epsilon_t 
\tag{2.31}
\end{equation}
where 
\[
x_t
=
\left[
\begin{array}{c}
\Delta y_t\\
ca_t \\
\end{array}
\right]
\]
and  $D$ is a 2x2 matrix of coefficients and $\epsilon_t$ is a 2x1  mean-zero i.i.d. process. 

Note: The model may or may not imply that $x_t$ has a VAR representation of this form. 
[An example in which such a representation exists is when  $\Delta y_t$ itself is a  univariate AR(1) process like the one assumed in Section 2.4]. 

\end{slide}
\begin{slide} 
$H_t=$  information contained in the vector $x_t$.

Forecast of $x_{t+j}$ given $H_t$:
\begin{equation*}
\label{eq:endowment_forecast-PDV} 
E_t[x_{t+j}|H_t] = D^j x_t
\end{equation*}

Sum the forecasts: \begin{equation*}
\sum_{j=1}^{\infty}
(1+r)^{-j}E_t [x_{t+j}|H_t]
=
\left[I-\frac{D}{1+r}\right]^{-1}
\frac{D}{1+r}
\left[
\begin{array}{c}
\Delta y_t\\
ca_t
\end{array}
\right].
\end{equation*}
Pre-multiplying this expression by $\left[
\begin{array}{cc}
1&0
\end{array}
\right]
$ we obtain: 
\begin{equation*}
\sum_{j=1}^{\infty}
(1+r)^{-j}E_t [\Delta y_{t+j}|H_t]
=
\left[
\begin{array}{cc}
1&0
\end{array}
\right]
\left[I-\frac{D}{1+r}\right]^{-1}
\frac{D}{1+r}
\left[
\begin{array}{c}
\Delta y_t\\
ca_t
\end{array}
\right].
\end{equation*}
Let 
\[
F\equiv -\left[
\begin{array}{cc}
1&0
\end{array}
\right] 
\left[
I - 
\frac{D}{1+r}
\right]^{-1} \frac{D}{1+r}.
\]
\end{slide}
\begin{slide}
{\small 
Use this expression as the  estimate of the RHS of (2.28)
\[
\widehat{RHS} = F \left[
\begin{array}{c}
\Delta y_t\\
ca_t
\end{array}
\right] 
\]
The left-hand-side of (2.28) can be written as: 
\[
LHS = \left[
\begin{array}{cc}
0&1
\end{array}
\right]
 \left[
\begin{array}{c}
\Delta y_t\\
ca_t
\end{array}
\right]
\]
The Null hypothesis we wish to test is: 
\[
H_0:  F = \left[
\begin{array}{cc}
0&1
\end{array}
\right]
\]
The null hypothesis thus consists of 2 restrictions imposed on (a function of) the coefficients of the matrix D. 

Nason and Rogers (2006) estimate a VAR with 4 lags using  Canadian data on the current account and GDP net of investment and government spending. (The above VAR assumes 1 lag; in the next slide we show how to modify the test to 4 lags.) The sample period is 1963:Q1-1997:Q4. They assume that $r=0.037$ per year. 

Their data strongly {\bf reject the null hypothesis}. They find that the Wald statistic, which reflects the distance  between $F$ and $\left[
\begin{array}{cc}
0&1
\end{array}
\right]$, is 16.1 with an asymptotic $p$-value of 0.04. This $p$-value means that if $H_0$ were true, then the Wald statistic would take a value of 16.1 or higher only 4 out of 100 times. 


}
 
\end{slide}
\begin{slide}
The VAR of Nason and Rogers has 4 lags, but our example above assumed a VAR with one lag. Here we show how to perform the test for a VAR with 4 lags. 

Estimate the following VAR(4): 
\[
\Delta y_t = \sum_{j=1}^4 a_j \Delta y_{t-j} + \sum_{j=1}^4 b_j \Delta ca_{t-j} + \epsilon^1_t
\]
and 
\[ca_t = \sum_{j=1}^4 c_j \Delta y_{t-j} + \sum_{j=1}^4 d_j \Delta ca_{t-j} + \epsilon^2_t
\]

\end{slide}
\begin{slide}
Then let: 
\[
x_t
=
\left[
\begin{array}{cccccccc}
\Delta y_t&
\Delta y_{t-1}&
\Delta y_{t-2}&
\Delta y_{t-3}&
ca_t &
ca_{t-1} &
ca_{t-2} &
ca_{t-3} 
\end{array}
\right]'
\]

The vector $x_t$ evolves over time as 
\[
x_t = D x_{t-1} + u_t
\]
where   $D$ is a 8x8 matrix of coefficients 
\[
D = \left[\begin{array}{cccccccc} 
a_1&a_2&a_3&a_4&b_1&b_2&b_3&b_4\\
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
c_1&c_2&c_3&c_4&d_1&d_2&d_3&d_4\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&1&0\\
\end{array}\right]
\]
and $u_t$ is the 8x1 vector : 
\[
u_t = [\epsilon^1_t \quad 0 \quad 0\quad 0 \quad \epsilon^2_t \quad 0 \quad 0\quad 0]'
\]

\end{slide}
\begin{slide}
Following the same steps as in the AR(1) case, we obtain.
\begin{equation*}
\sum_{j=1}^{\infty}
(1+r)^{-j}E_t [\Delta y_{t+j}|H_t]
=
\left[
\begin{array}{cccccccc}
1&0&0&0&0&0&0&0
\end{array}
\right]
\left[I-\frac{D}{1+r}\right]^{-1}
\frac{D}{1+r}
x_t \end{equation*}

Let: 
\[
F\equiv -\left[
\begin{array}{cccccccc}
1&0&0&0&0&0&0&0
\end{array}
\right] 
\left[
I - 
\frac{D}{1+r}
\right]^{-1} \frac{D}{1+r}.
\]
Then the RHS of 2.28 can be expressed as
\[RHS = F x_t
\]
The LHS of  (2.28) can be written as: 
\[
LHS = \left[
\begin{array}{cccccccc}
0&0&0&0&1&0&0&0
\end{array}
\right]x_t
\]
For the VAR(4) the Null hypothesis becomes: 
\[
H_0:  F = \left[
\begin{array}{cccccccc}
0&0&0&0&1&0&0&0
\end{array}
\right]
\]

\end{slide}

\begin{slide}
{\small A second testable restriction. 

Start again with the following implication of the theoretical model
\[
\tag{2.28 R}
 ca_t = - \sum_{j=1}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j} 
\]
and express it as
\[
 ca_t = \frac{- E_t \Delta y_{t+1}}{1+r} - \sum_{j=2}^{\infty} \frac{E_t\Delta y_{t+j}}{(1+r)^j} = \frac{- E_t \Delta y_{t+1}}{1+r} + \frac{1}{1+r} E_t ca_{t+1}
\]
Rearranging yields: 
\[
 E_t \Delta y_{t+1} -  E_t ca_{t+1}  + (1+r) ca_t  = 0
\]
Let
\[
Z_{t+1} \equiv  \Delta y_{t+1} -   ca_{t+1}  + (1+r) ca_t  
\]
Note that $Z_{t+1}$ is observable. Regress $Z_{t+1}$ on $x_t$ (or past $x_t$ , the information contained in $H_t$). If the theoretical model is true,  all coefficients on $x_t$  should be zero, that is, is should be true that 
\[
\left[
\left[ \begin{array}{cc} 1 & -1 \end{array}\right] D + (1+r)  
\left[ \begin{array}{cc} 0 & 1 \end{array}\right]\right]  = \left[ \begin{array}{cc} 0 & 0 \end{array}\right]
\]
Nason and Rogers find that this hypothesis is rejected with a $p$-value of 0.06. 
}




\end{slide}

\begin{slide}
{\bf Conclusions }

$\bullet$ propagation mechanism invoked by canonical intertemporal  model of the current account does not provide a satisfactory account of the observed current account dynamics.

$\bullet$ for AR(1) output specifications that model predicts the current account to be procyclical, whereas it is countercyclical in the data. 

$\bullet$ AR(2) or nonstationary specifications for the income process can in principle imply a countercyclical adjustment of the current account. Yet, empirical tests rejects the basic mechanism of the intertemporal model of the current account namely that current accounts are equal to future expected income changes. 

$\bullet$ to bring observed and predicted behavior of the current account closer together, in the following chapters we will enrich both the model's sources of fluctuations and its propagation mechanism. 

\end{slide}
\end{document}