Review 2 Key - Time Series

by Professor Throckmorton
for Time Series Econometrics
W&M ECON 408

Unit Root Test¶

Write down the null and alternative hypotheses of the augmented Dickey-Fuller test (ADF test). You run an ADF test on a time series and obtain a test statistic of -1.8. If the critical value at the 5% level is -2.86, what is your conclusion?

Answer:

The augmented Dickey-Fuller test (ADF test) has hypotheses
- $h_0$ : The time series has a unit root, indicating it is non-stationary.
- $h_A$ : The time series does not have a unit root, suggesting it is stationary.
To reject the null hypothesis we would need a test statistic that is less than the critical value of -2.86. Since -1.8 is not less than the critical value, then we fail to reject the null hypothesis. The time series probably has a unit root.

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
$\begin{align*} c_t^{-\sigma} &= \beta R_{t+1} c_{t+1}^{-\sigma} \\ \rightarrow -\sigma \log(c_t) &= \log \beta + \log(R_{t+1}) - \sigma \log(c_{t+1}) \end{align*}$
(1)
Define $\log(x_t/\bar{x}) \equiv \hat{x}_t$ and substract off steady-state (i.e., long-run equilibrium)
$\begin{align*} -\sigma \log(c_t) &= \log \beta + \log(R_{t+1}) - \sigma \log(c_{t+1}) \\ - [ -\sigma \log(\bar{c}) &= \log \beta + \log(\bar{R}) - \sigma \log(\bar{c})] \\ \rightarrow -\sigma \hat{c}_t &= \hat{R}_{t+1} - \sigma \hat{c}_{t+1} \end{align*}$
(2)

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*}

(3)

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*}

(4)

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*}

(5)

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*}

(6)

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*}

(7)

VECM¶

Rank¶

Suppose

\begin{gather*} \mathbf{A} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 7 & 9 \end{bmatrix} \end{gather*}

(8)

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.