Answer: $0.1 + \epsilon_t(0.8L + 0.16L^2 + 1)$

## Explanation Given the MA(2) model: $$Y_t = 0.1 + 0.8\epsilon_{t-1} + 0.16\epsilon_{t-2} + \epsilon_t$$ Using the lag operator $L$ where $LY_t = Y_{t-1}$: - $\epsilon_{t-1} = L\epsilon_t$ - $\epsilon_{t-2} = L^2\epsilon_t$ - $\epsilon_t = L^0\epsilon_t = 1\cdot\epsilon_t$ Substituting into the model: $$Y_t = 0.1 + 0.8(L\epsilon_t) + 0.16(L^2\epsilon_t) + (1\cdot\epsilon_t)$$ Factoring out $\epsilon_t$: $$Y_t = 0.1 + \epsilon_t(0.8L + 0.16L^2 + 1)$$ This matches option C exactly. The constant term 0.1 remains unchanged since the lag operator doesn't affect constants. **Key points:** - The lag operator $L$ shifts the time index backward - Constants are unaffected by the lag operator - The polynomial in $L$ represents the moving average coefficients - The model can be written as $Y_t = \mu + \theta(L)\epsilon_t$ where $\theta(L)$ is the MA polynomial

Author: Tanishq Prabhu