# Lecture 30 (April 08, 2022)

### Vector Error Correction Model (VECM)

General VARMA($p$,$q$) model:

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

THe VECM form:

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

where $\Delta x_t = x_t-x_{t-1}$, $\alpha$, $\beta$ are $k\times m$ matrix with $m\leq k$ and the full rank is $k$. Also, $\Phi_{i}^* = -\sum_{j=i+1}^p \Phi_j$. with $\beta = (\beta_1, \dots, \beta_m)$ which is a $k\times m$ matrix, which is called the cointegration vector.

Computation:

\begin{aligned} \sum_{i=1}^{p-1} \Phi_i^*\Delta x_{t-i} &= \sum_{i=1}^{p-1} \Phi_i^* (x_{t-i} - x_{t-i-1}) \\ &=\sum_{i=1}^{p-1} \Phi_i^* x_{t-i} - \sum_{i=1}^{p-1} \Phi_i^* x_{t-i-1} \\ &=\sum_{i=1}^{p-1} \Phi_i^* x_{t-i} - \sum_{i=2}^p \Phi^*_{i-1} x_{t-i} \\ &= \Phi_1^* x_{t-1} -\Phi_{p-1}^* x_{t-p} + \sum_{i=2}^{p-1} \Delta\Phi_i^* x_{t-i} \\ &= -\sum_{i=2}^p \Phi_i x_{t-1} + \Phi_p x_{t-p} + \sum_{i=2}^{p-1} \Phi_{i} x_{t-i} \\ &= -\sum_{i=2}^p \Phi_i x_{t-1} + \sum_{i=2}^{p} \Phi_i x_{t-i} \end{aligned}

Then the general form of VECM becomes

$x_t - \sum_{i=2}^p\Phi_i x_{t-i} = (I_k + \alpha \beta^T - \sum_{i=2}^p \Phi_i) x_{t-1} + \epsilon_t - \sum_{j=1}^q \Theta_j \epsilon_{t-j}$

We can add $-\Phi_{1} x_{t-1}$ on both sides, we then have

$x_t - \sum_{i=1}^p \Phi_i x_{t-i} = (I_k + \alpha\beta^T - \sum_{i=1}^p\Phi_i) x_{t-1} + \epsilon_t - \sum_{j=1}^q \Theta_j \epsilon_{t-j}$

The left hand side is the the original $AR(p )$ part and thus we need that

$\alpha\beta^T = -I_k + \sum_{i=1}^p \Phi_j = - \Phi(1)$

where $\Phi(B)$ is the AR polynomial $I - \Phi_1B - \Phi_2 B^2-\dots$ It looks like the matrix $\alpha\beta^T$ will be not of full rank if $\sum_{i=1}^p \Phi_j$ has an eigen value of $1$.

Consider a cointegrated $VAR(p)$ model:

$x_t = \mu + \Phi_1 x_{t-1} + \dots + \Phi_p x_{t-p} + \epsilon_t$

Recall: if all zeros of $\det(\Phi(B))$ are outside of the unit circle then $x_t$ is a unit root stationary. If $\det (\Phi(1)) = 0$, then $x_t$ is unit root non-stationary. Unit root stationary is $I(0)$ process and unit root non-stationary is $I(1)$ process.

$\Delta x_t = \mu + \Pi x_{t-1} + \Phi_1^*\Delta x_{t-1} + \dots + \Phi_p^* \Delta x_{t-p} + \epsilon_t$

and $\Pi = \alpha \beta^T = - \Phi(1)$.

When rank $\Pi$ = 0, then $\Pi = 0_{k\times k}$ matrix and then

$\Delta x_t = \mu + \Phi_1^* \Delta x_{t-1} + \dots. + \Phi_p^* \Delta x_{t-p} + \epsilon_t$

when $x_t$ is not cointegrated (i.e. no common stochastic trend), then each $x_{it}$ is independent $I(1)$ process.

When rank $\Pi = k$, Then $\Pi$ is not singular and thus $x_t$ is an $I(0)$ process and no need to use the VECM formulation.

When rank $\Pi <k$ then the model is

$\Delta x_t = \mu + \alpha\beta^T x_{t-1} + \Phi_1^*\Delta x_{t-1} + \dots + \Phi_p^* \Delta x_{t-p} + \epsilon_t$

$x_t$ is cointegrated with m linearly independent cointegration vector $\beta$ and the components of $x_t$ in these cointegration vector is $w_t = \beta^T x_t$ and $k-m$ common stochastic rends in $x_t$.

How to botain the $k-m$ common stochastic trends?

$\alpha: k\times m$. and $\alpha_\perp$ is the $k\times (k-m)$ matrix that is perpendicular to $\alpha$. i.e. the column vecotrs in $\alpha_\perp$ are all perpendicular to $\alpha$ and they are orthonormal in themselves.

Then

$\alpha_\perp^T\Delta x_t = \alpha_\perp^T \mu + \alpha_\perp^T \Pi x_{t-1} + \sum_{i=1}^p \alpha_\perp^T \Phi_i^* \Delta x_{t-i} + \alpha_\perp^T \epsilon_t$

If we let $\Delta y_t = \alpha_\perp^T \Delta x_t$ and $\alpha_\perp^T\alpha \beta^T = 0$ and this shows that

$\Delta y_t = \alpha_\perp^T \mu + \sum_{i=1}^T \alpha_\perp^T \Phi_i^* \alpha_\perp \Delta y_{t-1} + \alpha_\perp^T \epsilon_t$

which is the unit root components. and

$w_t = \beta^T x_t$

is the unit-root stationary part.

If $\Omega \Omega^T = I_m$, then

$\alpha\beta^T = \alpha \Omega \Omega^T \beta^T = (\alpha\Omega) (\beta\Omega)^T$

thus, $\alpha, \beta$ are not unique. We need extra constraints to fix $\alpha$ and $\beta$. Need $\beta$ to be

$\beta^T = [I_m; \beta_c^T] \quad \beta_c^T : m\times (k-m)$

example:

$k=2, m=1$,

$\Delta x_t = \mu + \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} \begin{pmatrix} 1 & \beta_c \end{pmatrix} x_{t-1} + \epsilon_t$

we have $w_t = \beta^T x_t$ and thus,

$\beta^T \Delta x_t = \beta^T \mu + \begin{pmatrix}1 & \beta_c \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} \begin{pmatrix} 1 & \beta_c \end{pmatrix} x_{t-1} + \beta^T \epsilon_t$

This shows that

$\Delta w_t = \beta^T \mu + (\alpha_1 + \beta_c\alpha_2) w_{t-1} + \beta^T \epsilon_t$

which gives

$w_t = \beta^T \mu + (1 + \alpha_1 + \beta_c \alpha_2) w_{t-1} + \beta^T \epsilon_t$

Updated: