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:

• $\epsilon_{t-1} = L\epsilon_t$
• $\epsilon_{t-2} = L^2\epsilon_t$

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.