Expected prediction error












1












$begingroup$


In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
begin{equation}
mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
end{equation}
i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.

Thank you for any help.






statistics







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$endgroup$

















    1












    $begingroup$


    In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
    begin{equation}
    mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
    end{equation}
    i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.

    Thank you for any help.






    statistics







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
      begin{equation}
      mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
      end{equation}
      i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.

      Thank you for any help.






      statistics







      share|cite|improve this question









      $endgroup$




      In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
      begin{equation}
      mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
      end{equation}
      i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.

      Thank you for any help.






      statistics




      statistics






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      asked Dec 24 '18 at 23:05









      yjntyjnt

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          $begingroup$

          $hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.



          Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.






          share|cite|improve this answer









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          • $begingroup$
            oh ok it's clearer now. Thank you
            $endgroup$
            – yjnt
            Dec 24 '18 at 23:24











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          $begingroup$

          $hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.



          Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh ok it's clearer now. Thank you
            $endgroup$
            – yjnt
            Dec 24 '18 at 23:24
















          0












          $begingroup$

          $hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.



          Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh ok it's clearer now. Thank you
            $endgroup$
            – yjnt
            Dec 24 '18 at 23:24














          0












          0








          0





          $begingroup$

          $hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.



          Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.






          share|cite|improve this answer









          $endgroup$



          $hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.



          Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 23:23









          Kenny WongKenny Wong

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          • $begingroup$
            oh ok it's clearer now. Thank you
            $endgroup$
            – yjnt
            Dec 24 '18 at 23:24


















          • $begingroup$
            oh ok it's clearer now. Thank you
            $endgroup$
            – yjnt
            Dec 24 '18 at 23:24
















          $begingroup$
          oh ok it's clearer now. Thank you
          $endgroup$
          – yjnt
          Dec 24 '18 at 23:24




          $begingroup$
          oh ok it's clearer now. Thank you
          $endgroup$
          – yjnt
          Dec 24 '18 at 23:24


















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