Exercise 2.9

Expected Output Changes and Permanent Income

Problem

Equation (2.27) expresses the difference between current and permanent income, $y_t - y^p_t$, as the present discounted value of expected future changes in the endowment. Present a step-by-step derivation of equation (2.27) starting from definitions (2.10) and (2.25). Comment on the cyclical properties of $y_t - y^p_t$ depending on whether the level or the change of $y_t$ follows an AR(1) process.

Answer

Multiply equation (2.10) by $(1 + r)/r$.

\[ \frac{1 + r}{r} y^p_t = \sum_{j=0}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} \]

Then split the sum,

\[ \frac{1 + r}{r} y^p_t = y_t + \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} \]

Subtract $y^p_t$ from both sides

\[ \frac{1}{r} y^p_t = y_t - y^p_t + \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} \]

Rearrange

\[ y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} + \frac{1}{r} y^p_t \]

Divide both sides of (2.10) by $r$ and then use the resulting expression to eliminate $\frac{1}{r} y^p_t$ from the above expression. This yields

\[ y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} + \sum_{j=0}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^{j+1}} = - \sum_{j=1}^{\infty} \frac{1}{(1 + r)^j} \left[ \mathbb{E}_t y_{t+j} - \mathbb{E}_t y_{t+j-1} \right] \]

Finally, use (2.25) to replace $\mathbb{E}_t y_{t+j} - \mathbb{E}_t y_{t+j-1}$ with $\mathbb{E}_t \Delta y_{t+j}$ to obtain equation (2.27).

If $y_t$ is AR(1), then as shown in section 2.2

\[ y^p_t = \frac{r}{1 + r - \rho} y_t \]

so that

\[ y_t - y^p_t = \frac{1 - \rho}{1 + r - \rho} y_t \]

Since $\rho \in (-1, 1)$, $\frac{1 - \rho}{1 + r - \rho} > 0$. It follows that $y_t - y^p_t$ is perfectly procyclical, $\text{corr}(y_t - y^p_t, y_t) = 1$. And $y_t - y^p_t$ inherits the serial correlation of $y_t$, $\text{corr}(y_t - y^p_t, y_{t-1} - y^p_{t-1}) = \text{corr}(y_t, y_{t-1}) = \rho$.

If $\Delta y_t$ is AR(1), then $\mathbb{E}_t \Delta y_{t+j} = \rho^j \Delta y_t$ for $j \geq 1$. By (2.27)

\[ y_t - y^p_t = - \frac{\rho}{1 + r - \rho} \Delta y_t \]

Assuming (as in section 2.4) $\rho \in [0, 1)$, $-\frac{\rho}{1 + r - \rho} < 0$. It follows that $y_t - y^p_t$ is perfectly countercyclical, $\text{corr}(y_t - y^p_t, \Delta y_t) = -1$. Now $y_t - y^p_t$ inherits the serial correlation of $\Delta y_t$, $\text{corr}(y_t - y^p_t, y_{t-1} - y^p_{t-1}) = \text{corr}(\Delta y_t, \Delta y_{t-1}) = \rho$.

--- title: "Exercise 2.9" subtitle: "Expected Output Changes and Permanent Income" --- <a class="return-button" href="index.qmd">⬅ Return</a> ## Problem Equation (2.27) expresses the difference between current and permanent income, $y_t - y^p_t$, as the present discounted value of expected future changes in the endowment. Present a step-by-step derivation of equation (2.27) starting from definitions (2.10) and (2.25). Comment on the cyclical properties of $y_t - y^p_t$ depending on whether the level or the change of $y_t$ follows an AR(1) process. ## Answer Multiply equation (2.10) by $(1 + r)/r$. $$ \frac{1 + r}{r} y^p_t = \sum_{j=0}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} $$ Then split the sum, $$ \frac{1 + r}{r} y^p_t = y_t + \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} $$ Subtract $y^p_t$ from both sides $$ \frac{1}{r} y^p_t = y_t - y^p_t + \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} $$ Rearrange $$ y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} + \frac{1}{r} y^p_t $$ Divide both sides of (2.10) by $r$ and then use the resulting expression to eliminate $\frac{1}{r} y^p_t$ from the above expression. This yields $$ y_t - y^p_t = - \sum_{j=1}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^j} + \sum_{j=0}^{\infty} \frac{\mathbb{E}_t y_{t+j}}{(1 + r)^{j+1}} = - \sum_{j=1}^{\infty} \frac{1}{(1 + r)^j} \left[ \mathbb{E}_t y_{t+j} - \mathbb{E}_t y_{t+j-1} \right] $$ Finally, use (2.25) to replace $\mathbb{E}_t y_{t+j} - \mathbb{E}_t y_{t+j-1}$ with $\mathbb{E}_t \Delta y_{t+j}$ to obtain equation (2.27). If $y_t$ is AR(1), then as shown in section 2.2 $$ y^p_t = \frac{r}{1 + r - \rho} y_t $$ so that $$ y_t - y^p_t = \frac{1 - \rho}{1 + r - \rho} y_t $$ Since $\rho \in (-1, 1)$, $\frac{1 - \rho}{1 + r - \rho} > 0$. It follows that $y_t - y^p_t$ is perfectly procyclical, $\text{corr}(y_t - y^p_t, y_t) = 1$. And $y_t - y^p_t$ inherits the serial correlation of $y_t$, $\text{corr}(y_t - y^p_t, y_{t-1} - y^p_{t-1}) = \text{corr}(y_t, y_{t-1}) = \rho$. If $\Delta y_t$ is AR(1), then $\mathbb{E}_t \Delta y_{t+j} = \rho^j \Delta y_t$ for $j \geq 1$. By (2.27) $$ y_t - y^p_t = - \frac{\rho}{1 + r - \rho} \Delta y_t $$ Assuming (as in section 2.4) $\rho \in [0, 1)$, $-\frac{\rho}{1 + r - \rho} < 0$. It follows that $y_t - y^p_t$ is perfectly countercyclical, $\text{corr}(y_t - y^p_t, \Delta y_t) = -1$. Now $y_t - y^p_t$ inherits the serial correlation of $\Delta y_t$, $\text{corr}(y_t - y^p_t, y_{t-1} - y^p_{t-1}) = \text{corr}(\Delta y_t, \Delta y_{t-1}) = \rho$.