VAR¶
Linearization¶
In a model with power utility and a one-period asset with gross return $R_t$, the dynamic equilibrium condition looks like $c_t^{-\sigma} = \beta R_{t+1} c_{t+1}^{-\sigma}$, where $c$ is consumption. Linearize that equation.
Answer:
Take natural log of both sides
\begin{align*} c_t^{-\sigma} &= \beta R_{t+1} c_{t+1}^{-\sigma} \\ \rightarrow -\sigma \log(c_t) &= \log \beta + \log(R_{t+1}) - \sigma \log(c_{t+1}) \end{align*}
Define $\log(x_t/\bar{x}) \equiv \hat{x}_t$ and substract off steady-state (i.e., long-run equilibrium)
\begin{align*} -\sigma \log(c_t) &= \log \beta + \log(R_{t+1}) - \sigma \log(c_{t+1}) \\ - [ -\sigma \log(\bar{c}) &= \log \beta + \log(\bar{R}) - \sigma \log(\bar{c})] \\ \rightarrow -\sigma \hat{c}_t &= \hat{R}_{t+1} - \sigma \hat{c}_{t+1} \end{align*}
Structural VAR¶
The following is a linearized New Keynesian monetary policy model with variables for output, $y$, inflation, $\pi$, and the interest rate $i$. Map it to a structural VAR.
\begin{gather*} \hat{y}_{t-1} = \hat{y}_t - \frac{1}{\sigma}(\hat{i}_{t-1} - \hat{\pi}_t) \\ \hat{\pi}_{t-1} = \beta \hat{\pi}_t + \kappa \hat{y}_{t-1} \\ \hat{i}_t = \phi_\pi \hat{\pi}_t + \phi_y \hat{y}_t + \sigma \varepsilon_t \\ \end{gather*}
Answer:
\begin{gather*} \begin{bmatrix} 1 & 1/\sigma & 0 \\ 0 & \beta & 0 \\ \phi_y & \phi_\pi & -1 \end{bmatrix} \begin{bmatrix} \hat{y}_t \\ \hat{\pi}_t \\ \hat{i}_t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 1/\sigma \\ -\kappa & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \hat{y}_{t-1} \\ \hat{\pi}_{t-1} \\ \hat{i}_{t-1} \end{bmatrix}+ \begin{bmatrix} 0 \\ 0 \\ \sigma \end{bmatrix} \varepsilon_t \end{gather*}
Identification¶
Write down a bivariate VAR($1$) where the coefficient matrix on the shocks has been recursively identified (e.g., via a Cholesky decomposition). Why and how does the ordering of the variables matter?
Answer: \begin{gather*} \begin{bmatrix} y_t \\ x_t \end{bmatrix} = \begin{bmatrix} b_{0,1} \\ b_{0,2} \end{bmatrix} + \begin{bmatrix} b_{1,1} & b_{1,3} \\ b_{1,2} & b_{1,4} \\ \end{bmatrix} \begin{bmatrix} y_{t-1} \\ x_{t-1} \end{bmatrix} + \begin{bmatrix} a_{0,1} & 0 \\ a_{0,2} & a_{0,3} \\ \end{bmatrix} \begin{bmatrix} \varepsilon_{y,t} \\ \varepsilon_{x,t}\\ \end{bmatrix} \end{gather*}
The ordering of the variables determines which variable is independent of the other's shock within a time period. Here since $y_t$ appears first, shocks to $x_t$ do not affect $y_t$, i.e., $y_t$ is indepedent from (or orthogonal to) shocks to $x_t$.
Cholesky Decomposition¶
Suppose you decompose the matrix $\mathbf{A}$ into $\mathbf{A} = \mathbf{L} \mathbf{L}'$, where
\begin{gather*} \mathbf{L} = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 3 & 2 \end{bmatrix} \end{gather*}
What was $\mathbf{A}$?
Answer:
\begin{gather*} \mathbf{A} = \mathbf{L} \mathbf{L}' = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 0 & 2 & 3 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 4 \\ 2 & 5 & 8 \\ 4 & 8 & 17 \end{bmatrix} \end{gather*}
VECM¶
Rank¶
Suppose
\begin{gather*} \mathbf{A} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 7 & 9 \end{bmatrix} \end{gather*}
Is $\mathbf{A}$ full rank? Why or why not?
Answer: It is not full rank because Row 3 equals Row 1 plus Row 2, i.e., it is a linear combination of other rows.
Cointegration¶
Suppose you observe the following data
| $t$ | $y_t$ | $x_t$ |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1 | 2 |
| 2 | 3 | 2 |
| 3 | 4 | 4 |
| 4 | 6 | 7 |
| 5 | 7 | 7 |
| 6 | 9 | 9 |
| 7 | 11 | 10 |
| 8 | 13 | 12 |
| 9 | 15 | 15 |
- Do $y_t$ and $x_t$ appear to be stationary? Why or why not?
- What conditions need to be satisified so that $y_t$ and $x_t$ are cointegrated?
- Do you think $y_t$ and $x_t$ are cointegrated? Why or why not?
Answer:
- Both $y_t$ and $x_t$ are increasing over time, so they do not appear stationary, although the sample is very short
- Both $y_t$ and $x_t$ must be non-stationary, I($1$), and there must exist some linear combination, $y_t - \beta x_t$, that is stationary.
- Subtracting $x_t$ from $y_t$ yields a sequence that alternates between $-1$, $0$, and $1$, which might be stationary because there might be a stable mean and variance
Cointegration Test¶
- What are the null and alternative hypotheses of the Johansen cointegration test?
- Suppose you are testing for the number of cointegrating relationships among 3 variables. Describe how to to do that with the Johansen conintegration test.
Answer:
Suppose there are $r_0 = 0$ cointegrating relationships
- Null Hypothesis: $H_0: r \leq r_0$ (i.e., the number of cointegrating relationships $\leq r_0$)
- Alternative Hypothesis: $H_A: r > r_0$
I would put a loop around the above hypothesis test and increment $r_0 = 1,2,3$ until it fails to reject the null, i.e., there is evidence to support that the number of cointegration relationships is $r = r_0$.
VECM Model¶
Suppose in an estimated bivariate VECM with variables $y$ and $x$ (in that order), the cointegrating vector is $[1, -3]$ and the loading on $y$ is $\alpha_1 = -0.5$
- What is the long-run relationship between $y$ and $x$?
- Suppose at some point in time $y_t = 6$ and $x_t = 1.5$. What is the error correcion term at $t$? Is the system in long-run equilibrium?
- Interpret the sign and magnitude of $\alpha_1 = -0.5$.
Answer:
- The long-run relationship is $y_t - 3x_t$, i.e., that linear combination produces a stationary time series.
- At $(y_t,x_t) = (6,1.5)$ the $ECT_t = 6 - 3\times 1.5 = 1.5$ is not zero, so the system is not in long-run equilibrium.
- $\alpha_1 < 0$ means that changes in $y_t$ returnt the system back to the long-run equilibrium and the magnitude determines the speed (e.g., $50\%$ in one period) at which it returns to the long-run equilibrium.