By Michael Falk
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Additional info for A First Course on Time Series Analysis : Examples with SAS
13. 1(z − 2)(z − 5), with z1 = 2, z2 = 5 being the roots of A(z) = 0. 11 implies the existence of an absolutely summable inverse causal filter, whose coefficients can be obtained by expanding 1/A(z) as a power series of z: 1 = A(z) 1− 1 z 2 1− 1 2 = v≥0 u+w=v v 1 2 = v≥0 w=0 = v≥0 1 2 v = z 5 u u≥0 v−w 1− 1 5 w 1 5 w u zu w≥0 1 5 w 1 2 v+1 zw zv zv v+1 2 5 1− 1 2 2 5 zv = v≥0 10 3 − 1 5 v+1 zv . The preceding expansion implies that bv := (10/3)(2−(v+1) −5−(v+1) ), v ≥ 0, are the coefficients of the inverse causal filter.
To get a time variable t, the temporarily created SAS variable N is used; it counts the observations. The pas- senger totals are plotted against t with a line joining the data points, which are symbolized by small dots. On the horizontal axis a label is suppressed. The variation of the data yt obviously increases with their height. The logtransformed data xt = log(yt ), displayed in the following figure, however, show no dependence of variability from height. 3a: Logarithm of Airline Data xt = log(yt ).
P 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= ... 1 k k2 ... (−k)p . . (−k + 1)p .. . 17) is the design matrix, β = (β0 , .