Statistik 3 - Beispiel 31

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Zeigen Sie im Modell y = X \beta + u \, die Beziehung

 \sum_{t=1}^n (y_t - \bar y)^2 = \sum_{t=1}^n (\hat{y}_t - \bar \hat y)^2 + \sum_{t=1}^n \hat{u}_t^2

unter der Annahme, dass X\, den Spaltenvektor (1, \ldots, 1)' \, enthält.

Definitionen:

y_t = \hat{y}_t + \hat{u}_t \,

\hat y = X \hat \beta \,

\hat u = y - X \hat\beta \,

Bedingung 1. Ordnung:  X'y - X'X \hat\beta = 0 \,

X ' \hat{u} = X ' (y - X \hat\beta) = X'y - X'X \hat\beta = 0 \,
Wegen der Annahme, dass X\, den Spaltenvektor (1, \ldots, 1)' \, enthält.

gilt \sum_{t=1}^n \hat{u}_t = 0 \,

Beweis: 


  \begin{matrix}
   X ' \hat{u}       & = 0 & & \\
   x_{.,1} ' \hat{u} & = 0 & \Leftrightarrow & \sum_{t=1}^n x_{t,1} \hat{u}_t = 0 \\
   x_{.,2} ' \hat{u} & = 0 & \Leftrightarrow & \sum_{t=1}^n x_{t,2} \hat{u}_t = 0 \\
   \vdots & & & \\
   x_{.,k} ' \hat{u} & = 0 & \Leftrightarrow & \sum_{t=1}^n x_{t,k} \hat{u}_t = 0 \\
  \end{matrix}
  

x_{.,1} = 1 \Rightarrow \sum_{t=1}^n \hat{u}_t = 0 \,

und natürlich auch \frac 1 n \sum_{t=1}^n \hat{u}_t = \bar \hat u = 0 
Wichtige Schlussfolgerung


 \bar y =
 \frac 1 n \sum_{t=1}^n y_t = \frac 1 n \sum_{t=1}^n (\hat{y}_t + \hat{u}_t) =
 \frac 1 n \left( \sum_{t=1}^n \hat{y}_t + \sum_{t=1}^n \hat{u}_t \right) = 
 \frac 1 n \sum_{t=1}^n \hat{y}_t =
 \bar \hat y
 


\sum_{t=1}^n (y_t - \bar y)^2 = 
\sum_{t=1}^n (\hat{y}_t + \hat{u}_t - \bar y)^2 = 
\sum_{t=1}^n (\hat{y}_t - \bar y + \hat{u}_t )^2 =

 = \sum_{t=1}^n (\hat{y}_t - \bar y)^2 + \sum_{t=1}^n \hat{u}_t^2 + 2 \sum_{t=1}^n (\hat{y}_t - \bar y) \hat{u}_t =

 = \sum_{t=1}^n (\hat{y}_t - \bar \hat y)^2 + \sum_{t=1}^n \hat{u}_t^2 + 2 \left( \sum_{t=1}^n \hat{y}_t \hat{u}_t - \bar y \sum_{t=1}^n \hat{u}_t \right) =

 = \sum_{t=1}^n (\hat{y}_t - \bar \hat y)^2 + \sum_{t=1}^n \hat{u}_t^2

Nebenrechung: 

 
  \begin{matrix}
  \sum_{t=1}^n \hat{y}_t \hat{u}_t = 
  \hat{y}'\hat{u} = 
  (X \beta)'(y - X \beta) =
  (X \beta)'y - (X \beta)'(X \beta) =
  \beta ' X ' y - \beta ' X ' X \beta = 
  \beta ' ( & \underbrace{X'y - X'X \beta} & ) \\
  & 0 & 
  \end{matrix}