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
- 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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