Answer: -0.6268

## Explanation For an MA(1) model, the relationship between the first-order autocorrelation coefficient (ρ₁) and the MA parameter (θ₁) is given by: \[ \rho_1 = \frac{\theta_1}{1 + \theta_1^2} \] Given ρ₁ = -0.45, we can set up the equation: \[ -0.45 = \frac{\theta_1}{1 + \theta_1^2} \] Cross-multiplying: \[ -0.45(1 + \theta_1^2) = \theta_1 \] \[ -0.45 - 0.45\theta_1^2 = \theta_1 \] Rearranging terms: \[ 0.45\theta_1^2 + \theta_1 + 0.45 = 0 \] This is a quadratic equation in the form \(a\theta_1^2 + b\theta_1 + c = 0\) where: - \(a = 0.45\) - \(b = 1\) - \(c = 0.45\) Using the quadratic formula: \[ \theta_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ \theta_1 = \frac{-1 \pm \sqrt{1^2 - 4(0.45)(0.45)}}{2(0.45)} \] \[ \theta_1 = \frac{-1 \pm \sqrt{1 - 0.81}}{0.9} \] \[ \theta_1 = \frac{-1 \pm \sqrt{0.19}}{0.9} \] \[ \theta_1 = \frac{-1 \pm 0.43589}{0.9} \] This gives two possible solutions: 1. \(\theta_1 = \frac{-1 + 0.43589}{0.9} = \frac{-0.56411}{0.9} = -0.6268\) 2. \(\theta_1 = \frac{-1 - 0.43589}{0.9} = \frac{-1.43589}{0.9} = -1.5954\) Since the problem states that θ₁ lies between -1 and +1, we select the first solution: \[ \theta_1 = -0.6268 \] Therefore, the Yule-Walker estimate for θ₁ is **-0.6268**.

Author: Nikitesh Somanthe