Neas-Seminars

TS Module 3: Trends HW


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By NEAS - 12/4/2009 6:01:29 AM

TS Module 3: Trends HW

 

(The attached PDF file has better formatting.)

 

Homework assignment: MA(1) Process: Variance of mean

 

Five MA(1) processes with 50 observations are listed below. The variance of åt is 1.

 


A.     For each process, what is the variance of , the average of the Y observations?

B.     How does the pattern of the first time series differ from that of the last time series?

C.    Explain intuitively why oscillating patterns have lower variances of their means.

 

 


 

1.      Yt = ì + et + et-1

2.      Yt = ì + et + ½ et-1

3.      Yt = ì + et

4.      Yt = ì + et – ½ et-1

5.      Yt = ì + et – et-1


 

 

(See page 50 of the Cryer and Chan text, Exercise 3.2)

 

 
By NEAS - 6/2/2014 8:05:48 AM

Jacob: Formula 3.2.3 applied to the first time series gives y-bar = 0.0792. But mu + et + et-1 from 1 to 50 is (1/2500) × Var(50 mu + å (50) + 2 × e(1) + …+ 2 × e(49) ) =

(1/2500) × (0 + Var(e (50) + 4 × (e(1) + …+ e(49) ) ) ) = (1 + 4 × 49) / 2500 = 0.0788

Rachel:The difference is subtle. The formula in the textbook assumes 50 observations are taken from this time series. Your second method assumes the time series starts at observation #1. The difference is one error term. Your second method doesn’t have an error term for period zero. Including this error term raises the variance of y-bar by 0.004.

Intuition:

 The formula in the textbook assumes the time series process is infinite. We observe fifty values, such as 1.014 through 1.063, and compute the variance of y-bar. Your computation assumes the series starts at observation #1, so the error term e0 is zero.

Use the formula in the textbook, since its assumptions are closer to real life. We may observe only certain observations, but the time series may have existed for much longer.