Exercise 3.1

Full Depreciation

Problem

Consider the model of section 3.1. The law of motion of capital given in equation (3.4) is a special case of the specification

\[ k_{t+1} = (1 - \delta)k_t + i_t, \]

where $\delta$ denotes the depreciation rate. The case considered in section 3.1 obtains when $\delta = 0$, which implies that capital does not depreciate. Show that in response to a permanent positive productivity shock of the type analyzed in section 3.3, the trade balance depreciates on impact. Consider first the polar case of full depreciation, which takes place when $\delta = 1$, and then the general case $\delta \in (0, 1)$.

Answer

An equilibrium are time paths for $y_t$, $k_{t+1}$, $i_t$, $c_t$, $\lambda_t$, and $d_t$ satisfying

\[ y_t = A_t F(k_t) \tag{1} \]

\[ k_{t+1} = (1 - \delta)k_t + i_t \tag{2} \]

\[ U'(c_t) = \lambda_t \tag{3} \]

\[ \lambda_t = \beta (1 + r)\lambda_{t+1} \tag{4} \]

\[ \lambda_t = \beta \lambda_{t+1} [A_{t+1} F'(k_{t+1}) + 1 - \delta] \tag{5} \]

\[ c_t + i_t + (1 + r)d_{t-1} = A_t F(k_t) + d_t \tag{6} \]

\[ \lim_{j \to \infty} \frac{d_{t+j}}{(1 + r)^j} = 0 \tag{7} \]

given $d_{-1}$, $k_0$, and the assumption that $\beta (1 + r) = 1$. The case analyzed thus far is $\delta = 0$. We now solve for a given $\delta \in [0, 1]$. Combining the Euler equations for debt and capital yields for all $t \geq 0$

\[ r + \delta = A_{t+1} F'(k_{t+1}) \]

The equilibrium value of capital depends only on $A_{t+1}$:

\[ k_{t+1} = \kappa \left( \frac{A_{t+1}}{r + \delta} \right), \quad \text{with } \kappa' > 0 \]

With $k_{t+1}$ in hand we can find $i_t$ and $y_t$. To find $c_t$, given $\beta(1 + r) = 1$ we have

\[ c_t = c_0 \quad \text{for all } t \]

To find $c_0$, use the intertemporal budget constraint which now is

\[ c_0 + rd_{-1} = \frac{r}{1 + r} \sum_{t=0}^{\infty} \frac{A_t F(k_t) + (1 - \delta)k_t - k_{t+1}}{(1 + r)^t} \tag{8} \]

With $c_t$ in hand, we have the trade balance: $tb_t = y_t - c_t - i_t$.

Prior to period 0, the economy was in a steady state with $A_t = \bar{A}$ for all $t < 0$. In that steady state:

\[ k_0 = \kappa \left( \frac{\bar{A}}{r + \delta} \right), \quad i_{-1} = \delta k_0, \quad y_{-1} = \bar{A}F(k_0), \quad c_{-1} = -rd_{-1} + y_{-1} - i_{-1}, \quad tb_{-1} = rd_{-1} \]

In period 0, $A_t$ permanently increases from $\bar{A}$ to $A' > \bar{A}$. Find $tb_0 - tb_{-1}$. The capital stock increases permanently, that is, for all $t > 0$

\[ k_t = \kappa \left( \frac{A'}{r + \delta} \right) > \kappa \left( \frac{\bar{A}}{r + \delta} \right) = k_0 \]

Investment increases:

\[ i_t = \delta k_1 > \delta k_0 \quad \forall t > 0 \]

\[ i_0 = k_1 - (1 - \delta)k_0 = (1 - \delta)(k_1 - k_0) + \delta k_1 > \delta k_1 = i_1 \]

With paths for $k_t$ and $i_t$ in hand, find consumption, $c_t = c_0$ for all $t \geq 0$. Use

\[ \begin{aligned} c_0 + rd_{-1} &= \frac{r}{1 + r} \sum_{t=0}^{\infty} \frac{A_t F(k_t) - i_t}{(1 + r)^t} \\ &= \frac{1}{1 + r} \left[ r(A'F(k_0) - i_0) + A'F(k_1) - \delta k_1 \right] \\ &= A'F(k_0) - \delta k_0 + \frac{1}{1 + r} \left[ A'F(k_1) - A'F(k_0) - (\delta + r)(k_1 - k_0) \right] \\ &= y_0 - \delta k_0 + \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] \end{aligned} \]

By concavity of $F$,

\[ \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] > 0 \]

\[ \begin{aligned} tb_0 - tb_{-1} &= y_0 - c_0 - i_0 - rd_{-1} \\ &= - (k_1 - k_0) - \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] < 0 \end{aligned} \]

--- title: "Exercise 3.1" subtitle: "Full Depreciation" --- <a class="return-button" href="index.qmd">⬅ Return</a> ## Problem Consider the model of section 3.1. The law of motion of capital given in equation (3.4) is a special case of the specification $$ k_{t+1} = (1 - \delta)k_t + i_t, $$ where $\delta$ denotes the depreciation rate. The case considered in section 3.1 obtains when $\delta = 0$, which implies that capital does not depreciate. Show that in response to a permanent positive productivity shock of the type analyzed in section 3.3, the trade balance depreciates on impact. Consider first the polar case of full depreciation, which takes place when $\delta = 1$, and then the general case $\delta \in (0, 1)$. ## Answer An equilibrium are time paths for $y_t$, $k_{t+1}$, $i_t$, $c_t$, $\lambda_t$, and $d_t$ satisfying $$ y_t = A_t F(k_t) \tag{1} $$ $$ k_{t+1} = (1 - \delta)k_t + i_t \tag{2} $$ $$ U'(c_t) = \lambda_t \tag{3} $$ $$ \lambda_t = \beta (1 + r)\lambda_{t+1} \tag{4} $$ $$ \lambda_t = \beta \lambda_{t+1} [A_{t+1} F'(k_{t+1}) + 1 - \delta] \tag{5} $$ $$ c_t + i_t + (1 + r)d_{t-1} = A_t F(k_t) + d_t \tag{6} $$ $$ \lim_{j \to \infty} \frac{d_{t+j}}{(1 + r)^j} = 0 \tag{7} $$ given $d_{-1}$, $k_0$, and the assumption that $\beta (1 + r) = 1$. The case analyzed thus far is $\delta = 0$. We now solve for a given $\delta \in [0, 1]$. Combining the Euler equations for debt and capital yields for all $t \geq 0$ $$ r + \delta = A_{t+1} F'(k_{t+1}) $$ The equilibrium value of capital depends only on $A_{t+1}$: $$ k_{t+1} = \kappa \left( \frac{A_{t+1}}{r + \delta} \right), \quad \text{with } \kappa' > 0 $$ With $k_{t+1}$ in hand we can find $i_t$ and $y_t$. To find $c_t$, given $\beta(1 + r) = 1$ we have $$ c_t = c_0 \quad \text{for all } t $$ To find $c_0$, use the intertemporal budget constraint which now is $$ c_0 + rd_{-1} = \frac{r}{1 + r} \sum_{t=0}^{\infty} \frac{A_t F(k_t) + (1 - \delta)k_t - k_{t+1}}{(1 + r)^t} \tag{8} $$ With $c_t$ in hand, we have the trade balance: $tb_t = y_t - c_t - i_t$. Prior to period 0, the economy was in a steady state with $A_t = \bar{A}$ for all $t < 0$. In that steady state: $$ k_0 = \kappa \left( \frac{\bar{A}}{r + \delta} \right), \quad i_{-1} = \delta k_0, \quad y_{-1} = \bar{A}F(k_0), \quad c_{-1} = -rd_{-1} + y_{-1} - i_{-1}, \quad tb_{-1} = rd_{-1} $$ In period 0, $A_t$ permanently increases from $\bar{A}$ to $A' > \bar{A}$. Find $tb_0 - tb_{-1}$. The capital stock increases permanently, that is, for all $t > 0$ $$ k_t = \kappa \left( \frac{A'}{r + \delta} \right) > \kappa \left( \frac{\bar{A}}{r + \delta} \right) = k_0 $$ Investment increases: $$ i_t = \delta k_1 > \delta k_0 \quad \forall t > 0 $$ $$ i_0 = k_1 - (1 - \delta)k_0 = (1 - \delta)(k_1 - k_0) + \delta k_1 > \delta k_1 = i_1 $$ With paths for $k_t$ and $i_t$ in hand, find consumption, $c_t = c_0$ for all $t \geq 0$. Use $$ \begin{aligned} c_0 + rd_{-1} &= \frac{r}{1 + r} \sum_{t=0}^{\infty} \frac{A_t F(k_t) - i_t}{(1 + r)^t} \\ &= \frac{1}{1 + r} \left[ r(A'F(k_0) - i_0) + A'F(k_1) - \delta k_1 \right] \\ &= A'F(k_0) - \delta k_0 + \frac{1}{1 + r} \left[ A'F(k_1) - A'F(k_0) - (\delta + r)(k_1 - k_0) \right] \\ &= y_0 - \delta k_0 + \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] \end{aligned} $$ By concavity of $F$, $$ \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] > 0 $$ So $$ \begin{aligned} tb_0 - tb_{-1} &= y_0 - c_0 - i_0 - rd_{-1} \\ &= - (k_1 - k_0) - \frac{A'}{1 + r} \left[ F(k_1) - F(k_0) - F'(k_1)(k_1 - k_0) \right] < 0 \end{aligned} $$