Statistik 3 - Prüfung 6.3.2006 - Beispiel 1

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y_t = \beta_1 x_{t,1} + \beta_2 x_{t,2} + u_t \,

Beispiel 1 a

Restriktion: \beta_1 + \beta_2 = 0 \Leftrightarrow \beta_1 = -\beta_2 \,

y_t = \beta_1 x_{t,1} - \beta_1 x_{t,2} + u_t \,

y_t = \beta_1 (x_{t,1} - x_{t,2}) + u_t \,
X^{\star} = \begin{pmatrix} x_{1,1} - x_{1,2} \\ x_{2,1} - x_{2,2} \\ \vdots \\ x_{n,1} - x_{n,2} \end{pmatrix} \,

\beta_1^{\star} = 
(X^{\star'}X^{\star})^{-1} X^{\star'}y =
\left(\sum_{t=1}^n (x_{t,1} - x_{t,2})^2 \right)^{-1} \left(\sum_{t=1}^n (x_{t,1} - x_{t,2}) y_t\right) =
\frac{\sum_{t=1}^n (x_{t,1} - x_{t,2}) y_t}
{\sum_{t=1}^n (x_{t,1} - x_{t,2})^2}


\beta_2^{\star} = 
- \frac{\sum_{t=1}^n (x_{t,1} - x_{t,2}) y_t}
{\sum_{t=1}^n (x_{t,1} - x_{t,2})^2}
= - \beta_1^{\star}

Beispiel 1 b

Gesucht ist der unrestringierte Schätzer, wenn \sum_{t=1}^n x_{t,1} x_{t,2} = 0 \,

X = \begin{pmatrix} x_{1,1} & x_{1,2} \\ x_{2,1} & x_{2,2} \\ \vdots & \vdots \\ x_{n,1} & x_{n,2} \end{pmatrix} \,

\hat \beta = (X'X)^{-1} X'y = 
\begin{pmatrix}
 \sum_{t=1}^n x_{t,1}^2 &  \sum_{t=1}^n x_{t,1} x_{t,2}  \\
 \sum_{t=1}^n x_{t,2} x_{t,1} & \sum_{t=1}^n x_{t,2}^2 \\
\end{pmatrix}^{-1}

\begin{pmatrix}
 \sum_{t=1}^n x_{t,1} y_t \\
 \sum_{t=1}^n x_{t,2} y_t \\
\end{pmatrix} =

\begin{pmatrix}
 \displaystyle \frac{\sum_{t=1}^n x_{t,1} y_t} {\sum_{t=1}^n x_{t,1}^2} \\
 \displaystyle \frac{\sum_{t=1}^n x_{t,2} y_t} {\sum_{t=1}^n x_{t,2}^2} \\
\end{pmatrix} =

\begin{pmatrix} \hat{\beta}_1 \\ \hat{\beta}_2 \end{pmatrix}