Derivation of the ordinary least squares estimator β1 and the sampling distribution?












0












$begingroup$


I am trying to derive the ordinary least squares and its sampling distribution for the model:



$$y = beta_0 + beta_1 x + epsilon$$



How can I obtain the estimator for $beta_1$






least-squares







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am trying to derive the ordinary least squares and its sampling distribution for the model:



    $$y = beta_0 + beta_1 x + epsilon$$



    How can I obtain the estimator for $beta_1$






    least-squares







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to derive the ordinary least squares and its sampling distribution for the model:



      $$y = beta_0 + beta_1 x + epsilon$$



      How can I obtain the estimator for $beta_1$






      least-squares







      share|cite|improve this question











      $endgroup$




      I am trying to derive the ordinary least squares and its sampling distribution for the model:



      $$y = beta_0 + beta_1 x + epsilon$$



      How can I obtain the estimator for $beta_1$






      least-squares




      least-squares






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 19:07









      Waqas

      15913




      15913










      asked Dec 12 '18 at 19:17









      Alex SimionAlex Simion

      204




      204






















          1 Answer
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          oldest

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          1












          $begingroup$

          You could solve the following optimisation problem:



          $$R=min_{beta_0,beta_1}left(y_i-hat{beta}_0+hat{beta}_1x_iright)^2$$



          Now differentiating the above with respect to $hat{beta}_0$ and $hat{beta}_1$ and setting to zero i.e.



          $$frac{partial R}{partial hat{beta}_0}=sum_{i=1}^N-2(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



          and



          $$frac{partial R}{partial hat{beta}_1}=sum_{i=1}^N-2x_i(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



          By simple algebric manupulation and the fact that $sum_i^N y_i=Nbar{y}$ we get the following:



          $$hat{beta}_0=bar{y}-hat{beta}_1bar{x}$$



          Similarly, by some manipulation you should obtain:



          $$hat{beta}_1=frac{sum_{i=1}^N(x_i-bar{x})(y_i-bar{y})}{sum_{i=1}^N(x_i-bar{x})^2}$$



          You could also use matrix notation instead i.e. $y,epsiloninmathbb{R}^n$, $betainmathbb{R}^{ktimes 1}$ and $Xinmathbb{R}^{ntimes k}$. So, you are solving essentially:



          $$y=Xbeta + epsilon$$



          To obtain the estimator you minimise the squared sum of errors i.e. $epsilon'epsilon=y'y-2hat{beta}'X'y+hat{beta}'X'Xhat{beta}$. By doing so we obtain:



          $$hat{beta}=(X'X)^{-1}X'y$$



          From Gauss-Markov theorem (and assumptions) $hat{beta}$ is normally distributed with mean $beta$ and variance $sigma^2(X'X)^{-1}$. I will stop here. For further details please see this






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            You could solve the following optimisation problem:



            $$R=min_{beta_0,beta_1}left(y_i-hat{beta}_0+hat{beta}_1x_iright)^2$$



            Now differentiating the above with respect to $hat{beta}_0$ and $hat{beta}_1$ and setting to zero i.e.



            $$frac{partial R}{partial hat{beta}_0}=sum_{i=1}^N-2(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



            and



            $$frac{partial R}{partial hat{beta}_1}=sum_{i=1}^N-2x_i(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



            By simple algebric manupulation and the fact that $sum_i^N y_i=Nbar{y}$ we get the following:



            $$hat{beta}_0=bar{y}-hat{beta}_1bar{x}$$



            Similarly, by some manipulation you should obtain:



            $$hat{beta}_1=frac{sum_{i=1}^N(x_i-bar{x})(y_i-bar{y})}{sum_{i=1}^N(x_i-bar{x})^2}$$



            You could also use matrix notation instead i.e. $y,epsiloninmathbb{R}^n$, $betainmathbb{R}^{ktimes 1}$ and $Xinmathbb{R}^{ntimes k}$. So, you are solving essentially:



            $$y=Xbeta + epsilon$$



            To obtain the estimator you minimise the squared sum of errors i.e. $epsilon'epsilon=y'y-2hat{beta}'X'y+hat{beta}'X'Xhat{beta}$. By doing so we obtain:



            $$hat{beta}=(X'X)^{-1}X'y$$



            From Gauss-Markov theorem (and assumptions) $hat{beta}$ is normally distributed with mean $beta$ and variance $sigma^2(X'X)^{-1}$. I will stop here. For further details please see this






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You could solve the following optimisation problem:



              $$R=min_{beta_0,beta_1}left(y_i-hat{beta}_0+hat{beta}_1x_iright)^2$$



              Now differentiating the above with respect to $hat{beta}_0$ and $hat{beta}_1$ and setting to zero i.e.



              $$frac{partial R}{partial hat{beta}_0}=sum_{i=1}^N-2(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



              and



              $$frac{partial R}{partial hat{beta}_1}=sum_{i=1}^N-2x_i(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



              By simple algebric manupulation and the fact that $sum_i^N y_i=Nbar{y}$ we get the following:



              $$hat{beta}_0=bar{y}-hat{beta}_1bar{x}$$



              Similarly, by some manipulation you should obtain:



              $$hat{beta}_1=frac{sum_{i=1}^N(x_i-bar{x})(y_i-bar{y})}{sum_{i=1}^N(x_i-bar{x})^2}$$



              You could also use matrix notation instead i.e. $y,epsiloninmathbb{R}^n$, $betainmathbb{R}^{ktimes 1}$ and $Xinmathbb{R}^{ntimes k}$. So, you are solving essentially:



              $$y=Xbeta + epsilon$$



              To obtain the estimator you minimise the squared sum of errors i.e. $epsilon'epsilon=y'y-2hat{beta}'X'y+hat{beta}'X'Xhat{beta}$. By doing so we obtain:



              $$hat{beta}=(X'X)^{-1}X'y$$



              From Gauss-Markov theorem (and assumptions) $hat{beta}$ is normally distributed with mean $beta$ and variance $sigma^2(X'X)^{-1}$. I will stop here. For further details please see this






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You could solve the following optimisation problem:



                $$R=min_{beta_0,beta_1}left(y_i-hat{beta}_0+hat{beta}_1x_iright)^2$$



                Now differentiating the above with respect to $hat{beta}_0$ and $hat{beta}_1$ and setting to zero i.e.



                $$frac{partial R}{partial hat{beta}_0}=sum_{i=1}^N-2(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



                and



                $$frac{partial R}{partial hat{beta}_1}=sum_{i=1}^N-2x_i(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



                By simple algebric manupulation and the fact that $sum_i^N y_i=Nbar{y}$ we get the following:



                $$hat{beta}_0=bar{y}-hat{beta}_1bar{x}$$



                Similarly, by some manipulation you should obtain:



                $$hat{beta}_1=frac{sum_{i=1}^N(x_i-bar{x})(y_i-bar{y})}{sum_{i=1}^N(x_i-bar{x})^2}$$



                You could also use matrix notation instead i.e. $y,epsiloninmathbb{R}^n$, $betainmathbb{R}^{ktimes 1}$ and $Xinmathbb{R}^{ntimes k}$. So, you are solving essentially:



                $$y=Xbeta + epsilon$$



                To obtain the estimator you minimise the squared sum of errors i.e. $epsilon'epsilon=y'y-2hat{beta}'X'y+hat{beta}'X'Xhat{beta}$. By doing so we obtain:



                $$hat{beta}=(X'X)^{-1}X'y$$



                From Gauss-Markov theorem (and assumptions) $hat{beta}$ is normally distributed with mean $beta$ and variance $sigma^2(X'X)^{-1}$. I will stop here. For further details please see this






                share|cite|improve this answer











                $endgroup$



                You could solve the following optimisation problem:



                $$R=min_{beta_0,beta_1}left(y_i-hat{beta}_0+hat{beta}_1x_iright)^2$$



                Now differentiating the above with respect to $hat{beta}_0$ and $hat{beta}_1$ and setting to zero i.e.



                $$frac{partial R}{partial hat{beta}_0}=sum_{i=1}^N-2(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



                and



                $$frac{partial R}{partial hat{beta}_1}=sum_{i=1}^N-2x_i(y_i-hat{beta}_0-hat{beta}_1x_i)=0$$



                By simple algebric manupulation and the fact that $sum_i^N y_i=Nbar{y}$ we get the following:



                $$hat{beta}_0=bar{y}-hat{beta}_1bar{x}$$



                Similarly, by some manipulation you should obtain:



                $$hat{beta}_1=frac{sum_{i=1}^N(x_i-bar{x})(y_i-bar{y})}{sum_{i=1}^N(x_i-bar{x})^2}$$



                You could also use matrix notation instead i.e. $y,epsiloninmathbb{R}^n$, $betainmathbb{R}^{ktimes 1}$ and $Xinmathbb{R}^{ntimes k}$. So, you are solving essentially:



                $$y=Xbeta + epsilon$$



                To obtain the estimator you minimise the squared sum of errors i.e. $epsilon'epsilon=y'y-2hat{beta}'X'y+hat{beta}'X'Xhat{beta}$. By doing so we obtain:



                $$hat{beta}=(X'X)^{-1}X'y$$



                From Gauss-Markov theorem (and assumptions) $hat{beta}$ is normally distributed with mean $beta$ and variance $sigma^2(X'X)^{-1}$. I will stop here. For further details please see this







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 12 '18 at 20:13

























                answered Dec 12 '18 at 20:07









                WaqasWaqas

                15913




                15913






























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