Question: Consider an MA(6) model with 1 = 0.5, 2 = 0.25, 3 = 0.125, 4 = 0.0625, 5 = 0.03125, and 6 = 0.015625. Find

  1. Consider an MA(6) model with 1 = 0.5, 2 = 0.25, 3 = 0.125, 4 = 0.0625, 5 = 0.03125, and 6 =

0.015625. Find a much simpler model that has nearly the same -weights.

2 Let {Yt} be an AR(2) process of the special form Yt = 2Yt 2 + et. Use first principles

to find the range of values of 2 for which the process is stationary.

3 Consider the AR(1) model Yt = Yt 1 + et. Show that if || = 1 the process cannot

be stationary. (Hint: Take variances of both sides.)

4 Consider an MA(7) model with 1 = 1, 2 = 0.5, 3 = 0.25, 4 = 0.125,

5 = 0.0625, 6 = 0.03125, and 7 = 0.015625. Find a much simpler model that

has nearly the same -weights.

5 Consider the model Yt = et 1 et 2 + 0.5et 3.

(a) Find the autocovariance function for this process.

(b) Show that this is a certain ARMA(p,q) process in disguise. That is, identify

values for p and q and for the 's and 's such that the ARMA(p,q) process

has the same statistical properties as {Yt}.

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