# Lecture 28 (April 04, 2022)

### Vector Autoregressive (VAR) model

The VAR($1$) model:

$x_t = \phi_0 + \Phi x_{t-1} + \epsilon_t$

Assume that $x_t$ is weakly stationary

$\mu = E x_t \quad \Rightarrow \mu = \phi_0 + \Phi \mu \quad \Rightarrow \quad \mu = (I-\Phi)^{-1} \phi_0$

Let $\tilde{x}_t = x_t-\mu$.

$\tilde{x}_t = \Phi \tilde{x}_{t-1} + \epsilon_t = \dots = \epsilon_t + \Phi \epsilon_{t-1} + \Phi^2 \epsilon_{t-2} + \dots$

Let $\Sigma = \text{Cov}(\epsilon_t)$. The covariance of $\tilde{x}_t$ is

$\Gamma_0 = \text{Cov}(\epsilon_t + \Phi \epsilon_{t-1} + \Phi^2 \epsilon_{t-2} + \dots) = \sum_{l=0}^\infty \Phi^l \Sigma(\Phi^l)^T$

Also

$\Gamma_l = \text{Cov}[x_t, x_{t-l}] = \text{Cov}[ \Phi x_{t-1} + \epsilon_t, x_{t-l}]= \Phi \Gamma_{l-1}$

which implies that

$\Gamma_l =\Phi^l \Gamma_0$

If $|\Phi | < 1$ then $|\Gamma_l|\leq |\Phi|^l |\Gamma_0| \rightarrow 0$.

 Recall when $k=1$, $\phi <1$ implies weakly stationary of $(x_t)$. when $k>1$, the absolute value of all egienvalues of $\Phi$ less than $1$ will give weakly stationary of $x_t$. The eigen value of $\Phi$ will be
$\det(\lambda I_k - \Phi) = \lambda^k \det (I_k - \frac{1}{\lambda} \Phi)$

This shows the eigenvalue $\lambda$ of $\Phi$, is just the inverse of zeros of the polynomial $\det(I_k-\Phi B)$. Thus, we need the root of the poly nomial to be $>1$.

Remark: $VAR(p)$ model can be recasted to $VAR(1)$ model by augmenting space. For example

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

Let $y_t = (x_{t-p+1}, x_{t-p+2},\dots, x_t)$ and $\eta_t = (0,\dots, 0, \epsilon_t)$ , Then we have

$\begin{pmatrix} x_{t-p+1} \\ x_{t-p+2} \\ \vdots \\ x_t \end{pmatrix} =\begin{pmatrix} 0 & I_k &\dots &0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & I_k \\ \Phi_p & \Phi_{p-1} &\dots & \Phi_1 \end{pmatrix} \begin{pmatrix} x_{t-p} \\ x_{t-p+1} \\ \vdots \\ x_{t-1} \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ \vdots \\ \epsilon_t \end{pmatrix}$

This is just

$y_t = \Phi^* y_{t-1} + \eta_t$

Then the polynomial

$\det(I_{pk} - \Phi^* B) = 0 \Leftrightarrow \det(I_k- \Phi_1 B - \dots -\Phi_p B^p) = 0$

momentum equation:

$\rho_l = \Lambda_1 \rho_{l-1} + \cdots + \Lambda_p \rho_{l-p}$

Where $\Lambda_i = D^{-1/2} \Phi_i D^{1/2}$.

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