A First Course on Time Series Analysis : Examples with SAS by Michael Falk

By Michael Falk

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This implies that k≥1 |Xk − Xk−1 | < ∞ with probability one and hence, the limit limn→∞ k≤n (Xk − Xk−1 ) = limn→∞ Xn = X exists in C almost surely. Finally, we check that X ∈ L2 : E(|X|2 ) = E( lim |Xn |2 ) n→∞ ≤E lim n→∞ k≤n = lim E n→∞ k≤n = lim n→∞ k,j≤n ≤ lim sup n→∞ = k≥1 |Xk − Xk−1 | 2 E(|Xk − Xk−1 | |Xj − Xj−1 |) k,j≤n ||Xk − Xk−1 ||2 ||Xj − Xj−1 ||2 = lim sup n→∞ 2 |Xk − Xk−1 | k≤n ||Xk − Xk−1 ||2 ||Xk − Xk−1 ||2 < ∞. 4. , suppose that Xn ∈ L2 , n ∈ N, has the property that for arbitrary ε > 0 one can find an integer N (ε) ∈ N such that ||Xn −Xm ||2 < ε if n, m ≥ N (ε).

10) The differences (2) Yt (2) := Yt − Sˆt then finally give the seasonally adjusted series. Depending on the length of the Henderson moving average used in (2) step (6), Yt is a moving average of length 165, 169 or 179 of the original data. Observe that this leads to averages at time t of the past and future seven years, roughly, where seven years is a typical length of business cycles observed in economics (Juglar cycle)2 . S. Bureau of Census has recently released an extended version of the X-11 Program called Census X-12-ARIMA.

Yt+k from a time series. A local polynomial estimator of order p < 2k + 1 is the minimizer β0 , . . , βp satisfying k u=−k (yt+u − β0 − β1 u − · · · − βp up )2 = min . 15) 26 Elements of Exploratory Time Series Analysis If we differentiate the left hand side with respect to each βj and set the derivatives equal to zero, we see that the minimizers satisfy the p + 1 linear equations j u u + β1 β0 u=−k u=−k k k k k j+1 + · · · + βp u j+p uj yt+u = u=−k u=−k for j = 0, . . , p. 16) where  1 −k (−k)2 1 −k + 1 (−k + 1)2 X=  ...

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