The lag operator is applied to the AR times series as follows: $(1 - 0.2L)(1 - 0.6L^4)Y_t = e_t$ What is the resulting time series? | Financial Risk Manager Part 1 Quiz

The lag operator is applied to the AR times series as follows:

$(1 - 0.2L)(1 - 0.6L^4)Y_t = e_t$

What is the resulting time series?

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TTanishq

Explanation:

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:

$Y_t = 0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + e_t$

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Financial Risk Manager Part 1