9.5 DLM with seasonal effect

Let’s add a simple fixed quarter effect to the regression model:

\[\begin{equation} \tag{9.11} y_t = \alpha_t + \beta_t x_t + \gamma_{qtr} + e_t \\ \gamma_{qtr} = \begin{cases} \gamma_{1} & \text{if } qtr = 1 \\ \gamma_{2} & \text{if } qtr = 2 \\ \gamma_{3} & \text{if } qtr = 3 \\ \gamma_{4} & \text{if } qtr = 4 \end{cases} \end{equation}\]

We can write Equation (9.11) in matrix form. In our model for \(\gamma\), we will set the variance to 0 so that the \(\gamma\) does not change with time.

\[\begin{equation} \tag{9.12} y_t = \begin{bmatrix} 1 & x_t & 1 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma_{qtr} \end{bmatrix}_t + e_t \\ \begin{bmatrix} \alpha \\ \beta \\ \gamma_{qtr} \end{bmatrix}_t = \begin{bmatrix} \alpha \\ \beta \\ \gamma_{qtr} \end{bmatrix}_{t-1}+ \begin{bmatrix} w_{\alpha} \\ w_{\beta} \\ 0 \end{bmatrix}_t\\ \Downarrow \\ y_t = \mathbf{Z}_{t}\mathbf{x}_{t}+v_t \\ \mathbf{x}_{t} = \mathbf{x}_{t-1}+\mathbf{w}_{t} \end{equation}\]

How do we select the right quarterly effect? Let’s separate out the quarterly effects and add them to \(\mathbf{x}\). We could then select the right \(\gamma\) using 0s and 1s in the \(\mathbf{Z}_t\) matrix. For example, if \(t\) is in quarter 1, our model would be

\[\begin{equation} \tag{9.13} y_t = \begin{bmatrix} 1 & x_t & 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha_t \\ \beta_t \\ \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \end{bmatrix} \\ \end{equation}\]

While if \(t\) is in quarter 2, the model is

\[\begin{equation} \tag{9.14} y_t = \begin{bmatrix} 1 & x_t & 0 & 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha_t \\ \beta_t \\ \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \end{bmatrix} \\ \end{equation}\]

This would work, but we would have to have a different \(\mathbf{Z}_t\) matrix and it might get cumbersome to keep track of the 0s and 1s. If we wanted the \(\gamma\) to evolve with time, we might need to do this. However, if the \(\gamma\) are fixed, i.e. the quarterly effect does not change over time, a less cumbersome approach is possible.

We could instead keep the \(\mathbf{Z}_t\) matrix the same, but reorder the \(\gamma_i\) within \(\mathbf{x}\). If \(t\) is in quarter 1,

\[\begin{equation} \tag{9.15} y_t = \begin{bmatrix} 1 & x_t & 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha_t \\ \beta_t \\ \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \end{bmatrix} \\ \end{equation}\] While if \(t\) is in quarter 2,

\[\begin{equation} \tag{9.16} y_t = \begin{bmatrix} 1 & x_t & 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha_t \\ \beta_t \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \\ \gamma_1 \end{bmatrix} \\ \end{equation}\]

We can use a non-diagonal \(\mathbf{G}\) to to shift the correct quarter effect within \(\mathbf{x}\).

\[ \mathbf{G} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \]

With this \(\mathbf{G}\), the \(\gamma\) rotate within \(\mathbf{x}\) with each time step. If \(t\) is in quarter 1, then \(t+1\) is in quarter 2, and we want \(\gamma_2\) to be in the 3rd row.

\[\begin{equation} \tag{9.17} \begin{bmatrix} \alpha \\ \beta \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \\ \gamma_1 \end{bmatrix}_{t+1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \end{bmatrix}_{t} + \begin{bmatrix} w_{\alpha} \\ w_{\beta} \\ 0 \\0 \\ 0 \\ 0 \end{bmatrix}_{t} \end{equation}\]

At \(t+2\), we are in quarter 3 and \(\gamma_3\) will be in row 3.

\[\begin{equation} \tag{9.18} \begin{bmatrix} \alpha \\ \beta \\ \gamma_3 \\ \gamma_4 \\ \gamma_1 \\ \gamma_2 \end{bmatrix}_{t+2} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma_2 \\ \gamma_3 \\ \gamma_4 \\ \gamma_1 \end{bmatrix}_{t+1} + \begin{bmatrix} w_{\alpha} \\ w_{\beta} \\ 0 \\0 \\ 0 \\ 0 \end{bmatrix}_{t} \end{equation}\]