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\[E(\hat{\beta}_{j})=\beta_{j},\quad j=0,1\] (Eget bevis (Sätt in \(E(y…
\[E(\hat{\beta}_{j})=\beta_{j},\quad j=0,1\]
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Eget bevis
Sätt in \(E(y_{i})=\beta_{0}+\beta_{1}x_{i}\) i uttrycken för \(\beta_0,\,\beta_1\) #
\(\color{red}{E(y_{i}-\overline{y})=\left(\beta_{0}+\beta_{1}x_{i}\right)-\left(\beta_{0}-\beta_{1}\overline{x}\right)=\beta_{1}\left(x_{i}-\overline{x}\right)}\)
Detta ger ju\[E(\hat{\beta}_{1})=\beta_{1}\frac{\sum(x_{i}-\overline{x}){\color{red}{(x_{i}-\overline{x})}}}{\sum(x_{i}-\overline{x})^{2}}=\beta_{1}\]
\[E(\hat{\beta}_{0})=E(\overline{y})-{\color{red}{E(\hat{\beta}_{1})}\overline{x}}=\beta_{0}+\beta_{1}\overline{x}-{\color{red}{\beta_{1}}}\overline{x}=\beta_{0}\]
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