What is J in while calculating SST in multiple regression?












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I am little confused what actually is the J in the formula of the SST and SSR for multiple regression



SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$



SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$










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    1












    $begingroup$


    I am little confused what actually is the J in the formula of the SST and SSR for multiple regression



    SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$



    SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am little confused what actually is the J in the formula of the SST and SSR for multiple regression



      SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$



      SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$










      share|cite|improve this question











      $endgroup$




      I am little confused what actually is the J in the formula of the SST and SSR for multiple regression



      SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$



      SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$







      matrices statistics regression linear-regression






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      edited Dec 8 '18 at 16:27









      V. Vancak

      11.3k3926




      11.3k3926










      asked Dec 8 '18 at 4:26









      surbhi groversurbhi grover

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          $begingroup$

          $J$ is the matrix of all $1$s. i.e., let
          $$
          mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
          $$

          then
          $$
          J = mathbf{1}mathbf{1}^T.
          $$

          While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
          $$
          frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
          $$

          which is an essential part of any sum of squares.






          share|cite|improve this answer









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            0












            $begingroup$

            $J$ is the matrix of all $1$s. i.e., let
            $$
            mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
            $$

            then
            $$
            J = mathbf{1}mathbf{1}^T.
            $$

            While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
            $$
            frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
            $$

            which is an essential part of any sum of squares.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $J$ is the matrix of all $1$s. i.e., let
              $$
              mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
              $$

              then
              $$
              J = mathbf{1}mathbf{1}^T.
              $$

              While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
              $$
              frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
              $$

              which is an essential part of any sum of squares.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $J$ is the matrix of all $1$s. i.e., let
                $$
                mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
                $$

                then
                $$
                J = mathbf{1}mathbf{1}^T.
                $$

                While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
                $$
                frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
                $$

                which is an essential part of any sum of squares.






                share|cite|improve this answer









                $endgroup$



                $J$ is the matrix of all $1$s. i.e., let
                $$
                mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
                $$

                then
                $$
                J = mathbf{1}mathbf{1}^T.
                $$

                While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
                $$
                frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
                $$

                which is an essential part of any sum of squares.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 16:24









                V. VancakV. Vancak

                11.3k3926




                11.3k3926






























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