By Michael Falk

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**Additional resources for A First Course on Time Series Analysis : Examples with SAS**

**Example text**

Z − zp ) z z z =c 1− 1− ... 1 − , z1 z2 zp where c := ap (−1)p z1 . . zp . In case of |zi | > 1 we can write for |z| < |zi | 1 1 u u = z , 1 − zzi z i u≥0 where the coefficients (1/zi )u , u ≥ 0, are absolutely summable. In case of |zi | < 1, we have for |z| > |zi | 1 1− 1 z zi =−z zi 1 1− zi z zi =− z ziu z −u u≥0 =− u≤−1 1 zi u zu, where the filter with coefficients −(1/zi )u , u ≤ −1, is not a causal one. In case of |zi | = 1, we have for |z| < 1 1 1− z zi = u≥0 1 zi u zu, where the coefficients (1/zi )u , u ≥ 0, are not absolutely summable.

Difference Filter We have already seen that we can remove a periodic seasonal component from a time series by utilizing an appropriate linear filter. We 28 Elements of Exploratory Time Series Analysis will next show that also a polynomial trend function can be removed by a suitable linear filter. 5. For a polynomial f (t) := c0 + c1 t + · · · + cp tp of degree p, the difference ∆f (t) := f (t) − f (t − 1) is a polynomial of degree at most p − 1. Proof. The assertion is an immediate consequence of the binomial expansion p p (t − 1) = k=0 p k t (−1)p−k = tp − ptp−1 + · · · + (−1)p .

10. Let a ∈ C. The filter (au ) with a0 = 1, a1 = −a and au = 0 elsewhere has an absolutely summable and causal inverse filter (bu )u≥0 if and only if |a| < 1. In this case we have bu = au , u ≥ 0. Proof. The characteristic polynomial of (au ) is A1 (z) = 1−az, z ∈ C. Since the characteristic polynomial A2 (z) of an inverse filter satisfies A1 (z)A2 (z) = 1 on some annulus, we have A2 (z) = 1/(1−az). Observe now that 1 = au z u , if |z| < 1/|a|. , u≥0 |au | < ∞. 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.