Statistik 3 - Prüfung 20.6.2005 - Beispiel 2

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Beispiel 2

Gegeben sie das Modell y_t = a + b x_t + u_t\, mit den Standardannahmen. Sei

a^{\star} = \left[ \frac 1 4 \frac{y_n - y_1}{x_n - x_1} + \frac 3 4 \frac{y_{n-1} - y_1}{x_{n-1} - x_1} \right] (- \bar x) + \bar y \,

Ist dieser Schätzer unverzerrt für a\,? Bestimmen Sie die Varianz von a^{\star}\,.

Erwartungswert

\operatorname{E}(a^{\star}) = 
\operatorname{E}\left(\left[ \frac 1 4 \frac{y_n - y_1}{x_n - x_1} + \frac 3 4 \frac{y_{n-1} - y_1}{x_{n-1} - x_1} \right] (- \bar x) + \bar y \right) =


 = 
\left[ \frac 1 4 \frac{\operatorname{E}(y_n - y_1)}{x_n - x_1} + \frac 3 4 \frac{\operatorname{E}(y_{n-1} - y_1)}{x_{n-1} - x_1} \right] (- \bar x) + \operatorname{E}(\bar y) =

Nebenrechnungen:

\operatorname{E}(y_1) = \operatorname{E}(a + b x_1 + u_1) = a + b x_1 + 0 \,

\operatorname{E}(y_n) = a + b x_n \,
 
\operatorname{E}(y_{n-1}) = a + b x_{n-1} \,

\operatorname{E}(\bar y) = \frac 1 n \sum_{i=1}^n \operatorname{E}(y_i) = \frac 1 n \sum_{i=1}^n \operatorname{E}(a + b x_i + u_i) = a + b \bar x\,
 = 
\left[ \frac 1 4 \frac{a + b x_n - a - b x_1}{x_n - x_1} + \frac 3 4 \frac{a + b x_{n-1} - a - b x_1}{x_{n-1} - x_1} \right] (- \bar x) + (a + b \bar x)=


 = 
\left[ \frac 1 4 \frac{b (x_n - x_1)}{x_n - x_1} + \frac 3 4 \frac{b (x_{n-1} - x_1)}{x_{n-1} - x_1} \right] (- \bar x) + (a + b \bar x) =


 = 
\left[ \frac 1 4 + \frac 3 4  \right] b (- \bar x) + (a + b \bar x) =


 = 
- b \bar x + a + b \bar x = a


D.h. der Schätzer a^{\star}\, ist unverzerrt.

Varianz

\operatorname{Var}(a^{\star}) = \operatorname{E}[ (a^{\star} - \operatorname{E}(a^{\star}))^2 ] = \operatorname{E}[ (a^{\star} - a)^2 ]\,
a^{\star} = \left[ \frac 1 4 \frac{y_n - y_1}{x_n - x_1} + \frac 3 4 \frac{y_{n-1} - y_1}{x_{n-1} - x_1} \right] (- \bar x) + \bar y \,

y_n = a + b x_n + u_n \,

\frac{y_n - y_1}{x_n - x_1} = 
\frac{a + b x_n + u_n - a - b x_1 - u_1}{x_n - x_1} =
\frac{b (x_n - x_1) + (u_n - u_1)}{x_n - x_1} =
b + \frac{u_n - u_1}{x_n - x_1}


a^{\star} = \left[ \frac 1 4 b + \frac 1 4 \frac{u_n - u_1}{x_n - x_1} + \frac 3 4 b + \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1} \right] (- \bar x) + a + b \bar x + \bar u = 

 = - \bar x \left[ \frac 1 4 \frac{u_n - u_1}{x_n - x_1} + \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1} \right] - b \bar x + a + b \bar x + \bar u = \,

 = - \bar x \left[ \frac 1 4 \frac{u_n - u_1}{x_n - x_1} + \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1} \right] + a + \bar u \,
\operatorname{Var}(a^{\star}) =
\operatorname{E}\left[\left(- \bar x \left[ \frac 1 4 \frac{u_n - u_1}{x_n - x_1} + \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1} \right] + a + \bar u - a\right)^2\right] = 


 = \operatorname{E}\left[\left(- \bar x \left[ \frac 1 4 \frac{u_n - u_1}{x_n - x_1} + \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1} \right] + \bar u\right)^2\right] = 

 = \operatorname{E}\left[\left(\bar u - \bar x \frac 1 4 \frac{u_n - u_1}{x_n - x_1} - \bar x \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1}\right)^2\right] = 

 = \operatorname{E}\left[ 
\bar u ^2 - 2 \bar u \bar x \frac 1 4 \frac{u_n - u_1}{x_n - x_1} - 2 \bar u \bar x \frac 3 4 \frac{u_{n-1} - u_1}{x_{n-1} - x_1}
+ \bar{x}^2 \frac{1}{16} \frac{(u_n - u_1)^2}{(x_n - x_1)^2} + 2 \bar{x}^2 \frac{3}{16} \frac{(u_n - u_1) (u_{n-1} - u_1)}{(x_n - x_1)(x_{n-1} - x_1)} 
+ \bar{x}^2 \frac{9}{16} \frac{(u_{n-1} - u_1)^2}{(x_{n-1} - x_1)^2}
\right] = \ldots 
\operatorname{E}(u_i) = 0, \operatorname{E}(u_i^2) = \sigma^2, \operatorname{E}(u_i u_j) = 0, \forall i \ne j \,

\operatorname{E}\left[(u_n-u_1)^2\right] = \operatorname{E}\left[u_n^2-2u_n u_1 + u_1^2\right] = 2 \sigma^2 \,
\operatorname{E}(\bar u^2) = \operatorname{E}\left[ \left(\frac 1 n \sum_{i=1}^n u_i\right)^2 \right] = \,

 = \operatorname{E}\left( \frac{1}{n^2} \sum_{i=1}^n u_i^2 + \frac{1}{n^2} \sum_{i=1, j=1,i \ne j}^n u_i u_j \right) =\,

 = \frac{1}{n^2} \sum_{i=1}^n \operatorname{E}(u_i^2) + \frac{1}{n^2} \sum_{i=1, j=1,i \ne j}^n \operatorname{E}(u_i u_j) = \,

 = \frac{n \sigma^2}{n^2} = \frac{\sigma^2}{n}
\operatorname{E}[\bar u (u_n - u_1)] = \,

= \operatorname{E}\left[\frac 1 n \left( \sum_{i=1}^n u_i u_n - \sum_{i=1}^n u_i u_1 \right) \right] = \,

= \operatorname{E}\left[\frac 1 n \left( \sum_{i=1}^{n-1} u_i u_n - \sum_{i=2}^n u_i u_1 + u_n^2 - u_i^2 \right) \right] = \,

= \frac 1 n \left( \sum_{i=1}^{n-1} \operatorname{E}(u_i u_n) - \sum_{i=2}^n \operatorname{E}(u_i u_1) + \operatorname{E}(u_n^2) - \operatorname{E}(u_i^2) \right) = \,

= \frac{\sigma^2 - \sigma^2}{n} = 0
\operatorname{E}\left[(u_n-u_1)(u_{n-1}-u_1)\right] = 
\operatorname{E}\left[ u_n u_{n-1} - u_{n-1} u_1 - u_n u_1 + u_1^2 \right] = \sigma^2

 \operatorname{Var}(a^{\star}) = \ldots = 
\frac{\sigma^2}{n} + 
\bar{x}^2 \frac{1}{16} \frac{2 \sigma^2}{(x_n-x_1)^2} + 
\bar{x}^2 \frac 3 8 \frac{\sigma^2}{(x_n - x_1)(x_{n-1} - x_1)} + 
\bar{x}^2 \frac{9}{16} \frac{2 \sigma^2}{(x_{n-1} - x_1)^2} =


 =
\sigma^2 \left( 
\frac{1}{n} + 
\frac{1}{8} \frac{\bar{x}^2}{(x_n-x_1)^2} + 
\frac 3 8 \frac{\bar{x}^2}{(x_n - x_1)(x_{n-1} - x_1)} + 
\frac{9}{8} \frac{\bar{x}^2}{(x_{n-1} - x_1)^2}
\right)
\,