# Lecture 29 (April 06, 2022)

### Cointegration

THe model VARMA(1,1), $k=2$.

$\begin{pmatrix} x_{1t} \\ x_{2t} \end{pmatrix} - \begin{pmatrix} 0.5 & -1.0 \\ -0.25 & 0.5 \end{pmatrix} \begin{pmatrix} x_{1,t-1} \\ x_{2,t-1} \end{pmatrix} = \begin{pmatrix} \epsilon_{1t} \\ \epsilon_{2t} \end{pmatrix} - \begin{pmatrix} 0.2 & -0.4 \\ -0.1 & 0.2 \end{pmatrix} \begin{pmatrix} \epsilon_{1,t-1} \\ \epsilon_{2,t-1} \end{pmatrix}$

we solve for the characteristic polynomial of $\Phi$. we have

$(\lambda-0.5)^2 - 0.25 =0$

this shows that $\lambda_1 = 1$ and $\lambda_2 = 0$. This shows that $x_t$ is not weakly stationary.

$\begin{pmatrix} 1 - 0.5 B & B \\ 0.25 B & 1-0.5 B \end{pmatrix} \begin{pmatrix} x_{1t} \\ x_{2t} \end{pmatrix} = \begin{pmatrix} 1-0.2 B & 0.4B \\ 0.1B & 1-0.2B \end{pmatrix}\begin{pmatrix}\epsilon_{1t} \\ \epsilon_{2t} \end{pmatrix}$

Multiply the matrix

$\begin{pmatrix} 1-0.5 B & -B \\ - 0.25 B & 1-0.5 B \end{pmatrix}$

This gives that

$\begin{pmatrix} 1-B & 0 \\ 0 & 1- B \end{pmatrix}\begin{pmatrix} x_{1t} \\ x_{2t}\end{pmatrix} = \Delta X_t = \begin{pmatrix}1-0.7B & -0.6B \\ -0.15B & 1-0.7 B \end{pmatrix} \begin{pmatrix} \epsilon_{1t} \\ \epsilon_{2t}\end{pmatrix}$

which gives a VARIMA(0,1,1) model. Marginally, $x_{1t}$ and $x_{2t}$ are unit root non-stationary. This deals with $\lambda=1$ cases. What about the $\lambda=0$ component?

Define

$y_t = \begin{pmatrix}y_{1t} \\ y_{2t} \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0.5 & 1 \end{pmatrix}\begin{pmatrix} x_{1t} \\ x_{2t}\end{pmatrix} = L x_{t}$

The $L$ matrix is invertible and

$L^{-1} = \begin{pmatrix} 0.5 & 1 \\ -0.25 & 0.5 \end{pmatrix}$

we call that

$r_t = \begin{pmatrix} r_{1t} \\ r_{2t}\end{pmatrix} = L\epsilon_t$

Recall VARMA(1,1):

$x_t = \Phi x_{t-1} + \epsilon_t - \Theta \epsilon_{t-1}$

thus, we have

\begin{aligned} y_t &= L x_t = L \Phi x_{t-1} + L\epsilon_t - L\Theta \epsilon_{t-1} \\ &= (L\Phi L^{-1}) y_{t-1} + r_t - (L\Theta L^{-1}) r_{t-1} \end{aligned}

This gives that

$\begin{pmatrix} y_{1t} \\ y_{2t} \end{pmatrix} = \begin{pmatrix} 1& 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix}y_{1,t-1}\\y_{2,t-1} \end{pmatrix} + \begin{pmatrix} r_{1,t} \\r_{2,t} \end{pmatrix} - \begin{pmatrix}0.4 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} r_{1,t-1} \\ r_{2,t-1}\end{pmatrix}$

this shows that

\begin{aligned} &y_{1t} = y_{1,t-1} + r_{1t} - 0.4 r_{1,t-1} \\ &y_{2t} = r_{2t} \end{aligned}

this shows that $y_{1t}$ is a univariate ARIMA(0,1) i.e. $y_{1t}$ is unit root nonstationary. $y_{2t}$ is stationary white noise model and $y_{1t}$ and $y_{2t}$ are decoupled. The model only exhibits a single unit root in the system. This is the vector error correction model.

### Vector Error correction model

$\Delta x_{t} = (\Phi - I) x_{t-1} + \epsilon_t - \Theta \epsilon_{t-1}$

because $\Phi$ has a unit root eigen value, the matrix $\Phi-I$ is a rank $1$ matrix. In the above example we have

$\begin{pmatrix} \Delta x_{1t} \\ \Delta x_{2t} \end{pmatrix} = \begin{pmatrix}-1 \\ 0.5\end{pmatrix}\begin{pmatrix} 0.5 & 1 \end{pmatrix} \begin{pmatrix} x_{1,t-1} \\ x_{2,t-1} \end{pmatrix}+ \begin{pmatrix}\epsilon_{1t}\\ \epsilon_{2t} \end{pmatrix} - \begin{pmatrix} 0.2 & - 0.4\\ -0.1 & 0.2 \end{pmatrix} \begin{pmatrix}\epsilon_{1,t-1} \\ \epsilon_{2,t-1} \end{pmatrix}$

General case: VARMA(p,q) model

$x_t - \sum_{i=1}^p \Phi_i x_{t-i} = \epsilon_t - \sum_{j=1}^q \theta_j \epsilon_{t-j}$

want: VECM for VARMA(p,q) in the form

$\Delta x_t = \alpha \beta^T x_{t-1} + \sum_{i=1}^ {p-1} \Phi_i^* \Delta x_{t-i} + \epsilon_t - \sum_{j=1}^q \theta_j \epsilon_{t-j}$

Updated: