Calculations Regarding Time Series Analysis

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2.11 Suppose C0v(X,,X,_ k) = yk is free oft but that E(X,) = 3:.
(a) Is {X,} stationary?
(b) Let Y, = 7 — 3! +X,. Is {Y,} stationary?
Solution
a) In order to show that {XI} is stationary, we have to show that its mean and autocovariance
functions are free of t.

It is given that the autocovariance function C01/(XI, X ,_k) = 7/k is free of t.
However, E (X I) = 3t , which varies with t.

Since E (X I) is not free of t, {Xt} is not stationary.
b) In order to show that {I/I} is stationary, we have to show that its mean and autocovariance
functions are free of t.

We have,
E(Y,) = E(7—3t+ XI)
: 7 — 3t + E (X I)
= 7—3t+3t (Since E(X,)=3t)
= 7, which is free of t.
Also,

C0v(Y,,Y,_k) = C0v(7 —3t+X,,7 —3(t—k)+X,_k)
= C0v(X[,X,_k)
= 7/,6 , which is free of t.
Therefore, the mean and autocovariance functions of are free of t.
Thus, is stationary.
[Note that: If a and b are constants, then C01/(a + X ,b + Y) = Cot/(X ,Y) . Applying this result, we
get C01/(7 —3t+ X,,7 —3(t—k)+ X,_k)= Cot/(XI, X,_k) since 7 —3t and 7 —3(t—k) are
constants.]

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