This is a very condensed post, mainly just so I could write down the equations I need for my Time Series Analysis course. 😉
However, it you find it useful – I am happy to hear that!
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Start with an AR(2):
$$ y_t = b + \phi_1 y_{t-1} + \phi_2 y_{t-2} + \varepsilon_t $$
Suppose we create a vector containing both \( y_t \) and \( y_{t -1} \):
$$\begin{bmatrix} y_t \\ y_{t-1} \end{bmatrix}$$
We can write our AR(2) as follows:
$$\begin{bmatrix} y_t \\ y_{t-1} \end{bmatrix} = \begin{bmatrix} b \\ 0 \end{bmatrix} + \begin{bmatrix} \phi_1 & \phi_2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} y_{t-1} \\ y_{t-2} \end{bmatrix} + \begin{bmatrix} \varepsilon_t \\ 0 \end{bmatrix}$$
Exercise: expand the above to see that you get back the original AR(2). Note that the 2nd line just ends up giving you \( y_{t-1} = y_{t-1} \).
The above is just a VAR(1)!
You can see this by letting:
$$ \textbf{z}_t = \begin{bmatrix} y_t \\ y_{t-1} \end{bmatrix}$$
$$ \textbf{b}’ = \begin{bmatrix} b \\ 0 \end{bmatrix} $$
$$ \boldsymbol{\Phi}’_1 = \begin{bmatrix} \phi_1 & \phi_2 \\ 1 & 0 \end{bmatrix} $$
$$ \boldsymbol{\eta}_t = \begin{bmatrix} \varepsilon_t \\ 0 \end{bmatrix}$$.
Then we get:
$$ \textbf{z}_t = \textbf{b}’ + \boldsymbol{\Phi}’_1\textbf{z}_{t-1} + \boldsymbol{\eta}_t$$
Which is a VAR(1).
Now let us try to do the same thing with an AR(3).
$$ y_t = b + \phi_1 y_{t-1} + \phi_2 y_{t-2} + \phi_3 y_{t-3} + \varepsilon_t $$
We can write our AR(3) as follows:
$$\begin{bmatrix} y_t \\ y_{t-1} \\ y_{t-2} \end{bmatrix} = \begin{bmatrix} b \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} \phi_1 & \phi_2 & \phi_3 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} y_{t-1} \\ y_{t-2} \\ y_{t-3} \end{bmatrix} + \begin{bmatrix} \varepsilon_t \\ 0 \\ 0 \end{bmatrix}$$
Note that this is also a VAR(1).
Of course, we can just repeat the same pattern for AR(p).
The cool thing is, we can extend this to VAR(p) as well, to show that any VAR(p) can be expressed as a VAR(1).
Suppose we have a VAR(3).
$$ \textbf{y}_t = \textbf{b} + \boldsymbol{\Phi}_1 \textbf{y}_{t-1} + \boldsymbol{\Phi}_2 \textbf{y}_{t-2} + \boldsymbol{\Phi}_3 \textbf{y}_{t-3} + \boldsymbol{ \varepsilon }_t $$
Now suppose that we create a new vector by concatenating \( \textbf{y}_t \), \( \textbf{y}_{t-1} \), and \( \textbf{y}_{t-2} \). We get:
$$\begin{bmatrix} \textbf{y}_t \\ \textbf{y}_{t-1} \\ \textbf{y}_{t-2} \end{bmatrix} = \begin{bmatrix} \textbf{b} \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} \boldsymbol{\Phi}_1 & \boldsymbol{\Phi}_2 & \boldsymbol{\Phi}_3 \\ I & 0 & 0 \\ 0 & I & 0 \end{bmatrix} \begin{bmatrix} \textbf{y}_{t-1} \\ \textbf{y}_{t-2} \\ \textbf{y}_{t-3} \end{bmatrix} + \begin{bmatrix} \boldsymbol{\varepsilon_t} \\ 0 \\ 0 \end{bmatrix}$$
This is a VAR(1)!