Zeitreihenanalyse - Aufgabe 5

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Exercise

Use the Euler relation to show that


\begin{align}
\frac{e^{ix}+ e^{-ix}}{2}
& = \frac{\cos(x) + i \sin(x) + \cos{x} - i \sin{x}}{2} \\
& = \cos(x) 
\end{align}
\,

Exercise

\overline{e^{ix}} = \overline{\cos(x) + i \sin{x}} = \cos(x) - i \sin{x} = e^{-ix} \,

Exercise


\begin{align}
\sin(x) \cos(y) + \cos(x) \sin(y)
& = \frac{e^{ix} - e^{-ix}}{2i} \frac{e^{iy} + e^{-iy}}{2} + \frac{e^{ix} + e^{ix}}{2} \frac{e^{iy} - e^{-iy}}{2i} \\
& = 
 \frac{e^{ix + iy} + e^{ix - iy} - e^{iy - ix} - e^{-iy -ix} +
       e^{ix + iy} - e^{ix - iy} + e^{iy - ix} - e^{-iy -ix}}{4i}
\\
& = \frac{2 e^{ix - iy} - 2 e^{-ix - iy}}{4i} \\
& = \frac{e^{ix + iy} - e^{-i (x+y)}}{2i} \\
& = \sin(x+y)
\end{align}
\,

Exercise


\begin{align}
\sin^2(x) + \cos^2(x) 
& = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^2 + \left( \frac{e^{ix} + e^{-ix}}{2} \right)^2\\
& = \left( \frac{e^{2ix} - 2 e^{ix - ix} + e^{-2ix}}{4 i^2} \right) + \left( \frac{e^{2ix} + 2 e^{ix - ix} + e^{-2ix}}{4} \right)\\
& = \frac{ -e^{2ix} + 2 - e^{-2ix} + e^{2ix} + 2 + e^{-2ix}}{4} \\
& = 1
\end{align}

Exercise

Zu zeigen: \sum_{t=0}^n q^t = \frac{1-q^{n+1}}{1-q}\,


\begin{align}
\sum_{t=0}^n q^t & = 1 + q + q^2 + \ldots + q^n \\
(1-q) \sum_{t=0}^n q^t & = (1-q) \cdot 1 + (1-q) \cdot q + (1-q) \cdot q^2 + \ldots + (1-q) \cdot q^n \\
(1-q) \sum_{t=0}^n q^t & = 1 + (- q + q) + ( -q^2 + q^2) + \ldots + (-q^n + q^n) - q^{n+1} \\
(1-q) \sum_{t=0}^n q^t & = 1 - q^{n+1} \\
\sum_{t=0}^n q^t & = \frac{1 - q^{n+1}}{1-q} \\
\end{align}
\,

Exercise

Find another symmetric filter with k=2 that allows a quadratic trend yt=a+bt+ct^2\, to pass without distortion.

Y_t = a + bt +ct^2 +\epsilon_t \Rightarrow \operatorname{E}(Y_t) = a + bt + ct^2 \,


Z_t = \alpha Y_{t-2} + \beta Y_{t-1} + \gamma Y_t + \beta Y_{t+1} + \gamma Y_{t+2} \,



\begin{align}
\operatorname{E}(Z_t) = 
& \alpha [a + b (t-2) + c(t-2)^2] + \\
& \beta[a + b (t-1) + c(t-1)^2] + \\
& \gamma[a + b t + ct^2] + \\
& \beta[a + b (t+1) + c(t+1)^2] + \\
& \alpha [a + b (t+2) + c(t+2)^2] \\
= & a + bt +ct^2
\end{align}


(2 \alpha + 2 \beta + \gamma)a + b [2 \alpha + 2\beta + \gamma] + c[(2 \alpha + 2\beta + \gamma)^2  + 8 \alpha + 2 \beta] = a +  bt + ct^2 \,


I. 2 \alpha + 2 \beta + \gamma  = 1\,

II. 8 \alpha + 2\beta = 0\,
\alpha=\frac{\gamma - 1}{6}, \beta=-\frac 2 3 (\gamma -1 ) \,

Exercise

Sketch g(t)=R \cdot \sin(\omega t+ \varphi)\, for each of the following cases