TS module 12: Method of moments for ARMA(1,1) process (practice problem)


TS module 12: Method of moments for ARMA(1,1) process (practice...

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TS module 12: Method of moments for ARMA(1,1) process (practice problem)

 

(The attached PDF file has better formatting.)

 

Know how to estimate ö and è for an ARMA(1,1) process by the method of moments. You solve a quadratic equation for è.

 

Exercise 1.2: ARMA(1,1) model and method of moments (Yule-Walker equations)

 

An ARMA(1,1) model is fit to a time series with sample autocorrelations for the first two lags of r1 = 0.880 and r2 = 0.704.

 


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

 


 


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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.

Thanks!
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