Answer: 1.0989

## Explanation For an AR(1) process of the form: $$ y_t = c + \phi y_{t-1} + u_t $$ where $u_t$ has variance $\sigma_u^2$, the unconditional variance is given by: $$ \text{Var}(y) = \frac{\sigma_u^2}{1 - \phi^2} $$ In this case: - $\phi = 0.3$ (autoregressive coefficient) - $\sigma_u^2 = 1$ (given disturbances have unit variance) Substituting the values: $$ \text{Var}(y) = \frac{1}{1 - (0.3)^2} = \frac{1}{1 - 0.09} = \frac{1}{0.91} \approx 1.0989 $$ **Key points:** - The constant term (0.2) does not affect the variance calculation - The formula only applies when $|\phi| < 1$ for stationarity - The denominator $1 - \phi^2$ comes from the geometric series expansion of the AR(1) process

Author: Tanishq Prabhu