Financial Risk Manager Part 1

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The MA(2) model is defined as $Y_t = 0.1 + 0.8\epsilon_{t-1} + 0.16\epsilon_{t-2} + \epsilon_t$ . What is the corresponding lag polynomial?

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NNikitesh

Last updated: February 2, 2026 at 10:22

$\epsilon_t(0.4L^2 + 0.16L^2 + 3)$

$\epsilon_t(0.5L + 0.15L^2 + 3)$

$\epsilon_t(0.8L + 0.16L^2 + 1)$

$\epsilon_t(0.2L + 0.14L^2 + 2)$

Explanation:

The correct answer is C.

Explanation:

Recall that the lag operator L is defined such that: $LY_t = Y_{t-1}$

For the MA(2) model: $Y_t = 0.1 + 0.8\epsilon_{t-1} + 0.16\epsilon_{t-2} + \epsilon_t$

We can rewrite this using the lag operator:

Substituting: $Y_t = 0.1 + 0.8L\epsilon_t + 0.16L^2\epsilon_t + \epsilon_t$ $Y_t = 0.1 + \epsilon_t(0.8L + 0.16L^2 + 1)$

Note that the constant term 0.1 is not affected by the lag operator. The lag polynomial is therefore: $\epsilon_t(0.8L + 0.16L^2 + 1)$

This corresponds to option C.

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