How to show that the simple least square estimators minimizes the SSE using the second partial derivative...












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I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:



$frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:



$2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?






partial-derivative least-squares







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    $begingroup$


    I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:



    $frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:



    $2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?






    partial-derivative least-squares







    share|cite|improve this question









    $endgroup$















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      1





      $begingroup$


      I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:



      $frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:



      $2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?






      partial-derivative least-squares







      share|cite|improve this question









      $endgroup$




      I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:



      $frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:



      $2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?






      partial-derivative least-squares




      partial-derivative least-squares






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      asked Dec 29 '18 at 20:08









      YandleYandle

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          $begingroup$

          It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.



          $$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$



          This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$



          $$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$



          Adding at subtracting $2overline x^2$



          $$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$



          $$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$



          $$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$



          $$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$



          $$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$



          Using the first binomial fomumla we get another common version of the variance



          $$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$



          A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.






          share|cite|improve this answer









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            $begingroup$

            It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.



            $$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$



            This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$



            $$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$



            Adding at subtracting $2overline x^2$



            $$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$



            $$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$



            $$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$



            $$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$



            $$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$



            Using the first binomial fomumla we get another common version of the variance



            $$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$



            A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.



              $$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$



              This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$



              $$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$



              Adding at subtracting $2overline x^2$



              $$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$



              $$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$



              $$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$



              $$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$



              $$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$



              Using the first binomial fomumla we get another common version of the variance



              $$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$



              A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.



                $$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$



                This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$



                $$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$



                Adding at subtracting $2overline x^2$



                $$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$



                $$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$



                Using the first binomial fomumla we get another common version of the variance



                $$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$



                A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.






                share|cite|improve this answer









                $endgroup$



                It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.



                $$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$



                This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$



                $$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$



                Adding at subtracting $2overline x^2$



                $$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$



                $$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$



                $$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$



                Using the first binomial fomumla we get another common version of the variance



                $$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$



                A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 21:20









                callculuscallculus

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