Answer: 2.0

## Explanation For an AR(p) model of the form: $$Y_t = \alpha + \beta_1 Y_{t-1} + \beta_2 Y_{t-2} + \ldots + \beta_p Y_{t-p} + \epsilon_t$$ The long-term mean (unconditional mean) is given by: $$E(Y_t) = \frac{\alpha}{1 - \beta_1 - \beta_2 - \ldots - \beta_p}$$ **Given model:** $$Y_t = 0.4 + 1.5Y_{t-1} - 0.7Y_{t-1} + \epsilon_t$$ **Note:** There appears to be a typo in the model specification - it should be $Y_t = 0.4 + 1.5Y_{t-1} - 0.7Y_{t-2} + \epsilon_t$ for a proper AR(2) model. However, based on the provided equation: - $\alpha = 0.4$ - $\beta_1 = 1.5$ - $\beta_2 = -0.7$ (assuming the second term should be $Y_{t-2}$) **Calculation:** $$E(Y_t) = \frac{0.4}{1 - (1.5 - 0.7)} = \frac{0.4}{1 - 0.8} = \frac{0.4}{0.2} = 2.0$$ **Verification:** - Sum of coefficients: $1.5 + (-0.7) = 0.8$ - Denominator: $1 - 0.8 = 0.2$ - Result: $0.4 / 0.2 = 2.0$ Therefore, the long-term mean of the time series is **2.0**. **Key points:** 1. For stationarity in an AR model, the sum of the autoregressive coefficients must be less than 1 in absolute value 2. The long-term mean represents the equilibrium level that the time series converges to over time 3. The formula works for any AR(p) model as long as the process is stationary

Author: Nikitesh Somanthe