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Extra resources for A First Course on Time Series Analysis Examples with SAS
By repeating the above arguments we obtain bw au z v = A1 (z) A2 (z). A(z) = v u+w=v Suppose now that (au ) and (bu ) are absolutely summable filters with characteristic polynomials A1 (z) and A2 (z), which both exist on some annulus r < z < R, where they satisfy A1 (z)A2 (z) = 1. Since 1 = v cv z v if c0 = 1 and cv = 0 elsewhere, the uniquely determined coefficients of the characteristic polynomial of the product filter of (au ) and (bu ) are given by bw au = u+w=v 1 if v = 0 0 if v = 0. 50 Chapter 2.
2. Annual electricity output, first and second order differences ✞ *** Program 1 _2_2 ***; TITLE1 ’ First and second order differences ’; TITLE2 ’ Electricity Data ’; DATA data1 ( KEEP = year sum delta1 delta2 ); INFILE ’c :\ data \ electric . 5 W =1; GOPTIONS NODISPLAY ; PROC GPLOT DATA = data1 GOUT = fig ; PLOT sum * year / VAXIS = AXIS1 HAXIS = AXIS2 ; PLOT delta1 * year / VAXIS = AXIS1 VREF =0; PLOT delta2 * year / VAXIS = AXIS1 VREF =0; RUN ; GOPTIONS DISPLAY ; PROC GREPLAY NOFS IGOUT = fig TC = SASHELP .
If |a| ≥ 1, then A2 (z) = u≥0 au z u exists for all |z| < 1/|a|, but u≥0 |a|u = ∞, which completes the proof. 11. Let a1 , a2 , . . , ap ∈ C, ap = 0. The filter (au ) with coefficients a0 = 1, a1 , . . , ap and au = 0 elsewhere has an absolutely summable and causal inverse filter if the p roots z1 , . . , zp ∈ C of A(z) = 1 + a1 z + a2 z 2 + . . , |zi | > 1 for 1 ≤ i ≤ p. Proof. We know from the Fundamental Theorem of Algebra that the equation A(z) = 0 has exactly p roots z1 , . . g. 5 in Conway (1975)), which are all different from zero, since A(0) = 1.