Test eq var cycle results

so that {Yt} could be called asymptotically stationary. These two results follow from parts (b) and (c). (e) Suppose now that we alter the initial condition and put Y 1 = e 1 / ( 1 ­ c 2 ) . Show that now {Yt} is stationary. This part can be solved using repeated substitution to express Yt explicitly as ct ­ 1 Y t = c ( cY t ­ 2 + e t ­ 1 ) + e t = ... = e t + ce t ­ 1 + c 2 e t ­ 2 + ... + c t ­ 2 e 2 + ----------------- e 1 1 ­ c2 2 e Then show that Var ( Y t ) = ------------and Corr ( Y t , Y t ­ k ) = c k for k > 0 . 1 ­ c2 Exercise Two processes {Zt} and {Yt} are said to be independent if for any time points t1, t2,..., tm and s1, s2,..., sn, the random variables { Z t , Z t , ..., Z t } are independent of the random variables { Y s , Y s , ..., Y s }. 1 2 m 1 2 n Show that if {Zt} and {Yt} are independent stationary processes, then Wt = Zt + Yt is stationary. First, E(W t ) = E(Z t ) + E(Y t ) = Z + Y. Then Cov(W t ,W t - k ) = Cov(Z t + Y t , Z t - k + Y t - k) = Cov(Z t ,Z t - k ) + Cov(Yt,Yt - k) which is free of t since both {Zt} and {Yt} are stationary. Exercise Let {Xt} be a time series in which we are interested. However, because the measurement process itself is not perfect, we actually observe Yt = Xt + et. We assume that {Xt} and {et} are independent processes. We call Xt the signal and et the measurement noise or error process. If {Xt} is stationary with autocorrelation function k, show that {Yt} is also stationary with k for k 1 Corr ( Y t , Y t ­ k ) = -------------------------2 2 1 + e / X

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Test eq var cycle results

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