1.       What is the method of moments estimate for ö?

2.       What is the method of moments estimate for è?

Part A: For an ARMA(1,1) process, r₂ = r₁ × ö ➾ ö = 0.704 / 0.880 = 0.8

Part B: For an ARMA(1,1) process (for k ≥ 1):

We estimated ö as r₂ / r₁. We estimate è from

See Cryer and Chan, equation 7.1.6 on page 151.

In this exercise, 0.880 = (1 – 0.8 è) (0.8 – è) / (1 – 2(0.8 è) + è²).

This is a quadratic equation in θ, with roots of –0.4 and –2.5 (use the formula for roots of a quadratic).

The arithmetic is shown below; most final exam problems use simple numbers.

0.880 = (1 – 0.8 è) (0.8 – è) / (1 – 2(0.8 è) + è²)

0.880 × (1 – 2(0.8 è) + è²) = (1 – 0.8 è) (0.8 – è)

88 × (1 – 2(0.8 è) + è²) = (10 – 8 è) (8 – 10 è)

88 – 140.8 è + 88 è² = 80 – 164 è + 80 è²

8 + 23.2 è + 8 è² = 0

Using the formula for the roots of a quadratic equation gives

(8² ± (23.2² – 4 × 8 × 8) ) / (2 × 8) = -0.4 and -2.5

The formula for the roots of a quadratic equation is incorrect in Part B.  The answers of -0.4 and -2.5 are correct but if you solve the equation using the numbers given, those are not the answers you get.

(8² ± (23.2² – 4 ×8 × 8)) / (2 × 8) does not equal -0.4 and -2.5;  it actually equals 21.64 and -13.64

The correct formula should be (-23.2 ± √(23.2² – 4 × 8 × 8)) / (2 × 8) = -0.4 and -2.5.