Answer: $0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + e_t$

This question requires the application of the properties of the Lag operator. Recall that: $$LY_t = Y_{t-1}$$ And that the lag operator has multiplicative property. So, $$ \begin{align*} (1 - 0.2L)(1 - 0.6L^4)Y_t &= (1 - 0.6L^4 - 0.2L + 0.12L^5)Y_t \\ &= Y_t - 0.6Y_tL^4 - 0.2Y_tL + 0.12Y_tL^5 \\ &= Y_t - 0.6Y_{t-4} - 0.2Y_{t-1} + 0.12Y_{t-5} = e_t \end{align*} $$ Rearranging we get: $$Y_t = 0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + e_t$$

Author: Tanishq Prabhu