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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 = 1 − az 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.

The following figure illustrates the significance of the stationarity condition |a| < 1 of an AR(1)-process. Realizations Yt = aYt−1 + εt , t = 1, . . 5, where ε1 , ε2 , . . , ε10 are independent standard normal in each case and Y0 is assumed to be zero. 5. 2. 5Yt−1 + εt , t = 1, . . , 10, with εt independent standard normal and Y0 = 0.

Show that in a linear regression model yt = β1 xt + β2 , t = 1, . . , n, the squared multiple correlation coefficient R2 based on the least squares estimates βˆ1 , βˆ2 and yˆt := βˆ1 xt + βˆ2 is necessarily between zero and one with R2 = 1 if and only if yˆt = yt , t = 0, . . 12)). 5. (Population2 Data) The following table lists total population numbers of North RhineWestphalia between 1961 and 1979. Suppose a logistic trend for these data and compute the estimators βˆ1 , βˆ3 using PROC REG. Since some observations exceed βˆ3 , use β˜2 from Exercise 3 and do an ex post-analysis.

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