Review 2A Key

by Professor Throckmorton
for Time Series Econometrics
W&M ECON 408/PUBP 616
Slides

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 \(y_t\) returns 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.