TS Module 5 moving average MA(2) practice problems

(The attached PDF file has better formatting.)

**Exercise 5.1: Variance of moving average process

A moving average process of order 2 is Y_t = e_t – 0.6 e_t-1 – e_t-2 , with ó²_e = 1

A.    What is ã₀, the variance of Y_t?

B.    What is ñ₁, the autocorrelation of lag 1?

C.    What is ñ₂, the autocorrelation of lag 2?

Part A: ã₀ = (1 + è²₁ + è²₂) × ó²_ε = (1 + 0.6² + 1²) × 1 = 2.36

Intuition: Y_t is the sum of three independent random variables, with variances of 1, 0.36, and 1.

See Cryer and Chan, chapter 4, page 62, equation at the bottom of the page.

Part B: See Cryer and Chan, chapter 4, page 63, equation 4.2.3.

 or

(–0.6 + 0.6 × 1) / (1 + 0.6² + 1²) = 0/2.36

ñ₁ is the correlation of e_t – 0.6 e_t-1 – e_t-2 with e_t-1 – 0.6 e_t-2 – e_t-3

        The numerator has two non-zero terms: –0.6 e²_t-1 and +0.6 e²_t-2 

        The error terms have the same variance, so these two terms cancel out.

Jacob: Does ñ₂ = 0 mean that the autocorrelation of lag 2 in this time series is zero?

Rachel: Actual time series are finite, so the autocorrelations have random fluctuations. If this time series has 100 observations, the autocorrelation ñ₂ is distributed as a normal random variable with a mean of zero and a standard error of 1/√100 = 0.10. The observed autocorrelation is not zero. But if the time series is infinite, the autocorrelation of lag 2 approaches zero.

Jacob: How should we think of this? Can one see the correlation intuitively?

Rachel: e_t is correlated with e_t-1 two ways:

        Directly, with a correlation of –è₁.

        Indirectly through e_t-2 and back to e_t-1, with a correlation of –è₂ × –è₁.

Moving from e_t-1 to e_t-2 is like moving from e_t-2 to e_t-1. We deal with these relations in more detail in modules 19 and 20 (seasonal time series), where ñ₁₁ = ñ₁₃. 

Part C: See Cryer and Chan, chapter 4, page 63, equation 4.2.3

(–1) / (1 + 0.6² + 1²) = –1/2.36 = -0.424

Intuition: ñ₂ is the correlation of e_t – 0.6 e_t-1 – e_t-2 with e_t-2 – 0.6 e_t-3 – e_t-4