# Time Series: How to convert AR(p) to VAR(1) and VAR(p) to VAR(1)

This is a very condensed post, mainly just so I could write down the equations I need for my Time Series Analysis course. ðŸ˜‰

However, it you find it useful – I am happy to hear that!

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$$y_t = b + \phi_1 y_{t-1} + \phi_2 y_{t-2} + \varepsilon_t$$

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)!