Financial Risk Manager Part 1

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

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

What is the resulting time series?

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NNikitesh

Last updated: February 2, 2026 at 10:22

$0.2Y_{t-1} + 0.6Y_{t-4} - 0.12Y_{t-5} + \epsilon_t$

$Y_{t-2} + 0.6Y_{t-4} - 0.12Y_{t-5} + \epsilon_t$

$Y_{t-2} + 0.6Y_{t-4} - 0.12Y_{t-4} + \epsilon_t$

$0.1Y_{t-1} + 0.6Y_{t-4} - 0.2Y_{t-5} + \epsilon_t$

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{aligned} (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 = \epsilon_t \\ &= Y_t - 0.6Y_{t-4} - 0.2Y_{t-1} + 0.12Y_{t-5} = \epsilon_t \end{aligned}

Rearranging we get:

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

Therefore, option A is correct.

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