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

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.

Author: Nikitesh Somanthe