Sum of squared differences question The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Construct a Confidence Interval of $95%$Why is variance squared?From sample mean and variance of $X$ to $sqrtX$Stuck on the cancellation of sums when calculating sample variance estimatesNumerator of Sample Variance ExpectationDivergence of Chi-squared statisticFor what value of $w$ is $(1-w)bar X_1 + wbar X_2$ the minimum variance unbiased estimator of $mu$R squared formula in linear regressionVariance of the SSBCalculating the MSE of a particular estimator

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Sum of squared differences question



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Construct a Confidence Interval of $95%$Why is variance squared?From sample mean and variance of $X$ to $sqrtX$Stuck on the cancellation of sums when calculating sample variance estimatesNumerator of Sample Variance ExpectationDivergence of Chi-squared statisticFor what value of $w$ is $(1-w)bar X_1 + wbar X_2$ the minimum variance unbiased estimator of $mu$R squared formula in linear regressionVariance of the SSBCalculating the MSE of a particular estimator










0












$begingroup$


I'm completing a homework question and I need to prove the following:



$$
sum_i=1^N ||vecy - vecx_i||^2 = sum_i=1^N ||vecx_i - vecbarx||^2 + N||vecy-vecbarx||^2
$$



where $vecbarx = frac1N sum_i=1^N vecx_i$ and $||cdot||$ is the Euclidean norm and these are all vectors.



I think I probably need to do something similar to the proof for the sample variance but not sure how to modify it when I have a $y$ in now.






statistics summation proof-explanation statistical-inference







share|cite|improve this question











$endgroup$











  • $begingroup$
    Is this an Euclidean norm or a general norm?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:10










  • $begingroup$
    This is the Euclidean norm - I should've said, edited the question now
    $endgroup$
    – user-2147482565
    Mar 25 at 10:16






  • 1




    $begingroup$
    If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:24
















0












$begingroup$


I'm completing a homework question and I need to prove the following:



$$
sum_i=1^N ||vecy - vecx_i||^2 = sum_i=1^N ||vecx_i - vecbarx||^2 + N||vecy-vecbarx||^2
$$



where $vecbarx = frac1N sum_i=1^N vecx_i$ and $||cdot||$ is the Euclidean norm and these are all vectors.



I think I probably need to do something similar to the proof for the sample variance but not sure how to modify it when I have a $y$ in now.






statistics summation proof-explanation statistical-inference







share|cite|improve this question











$endgroup$











  • $begingroup$
    Is this an Euclidean norm or a general norm?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:10










  • $begingroup$
    This is the Euclidean norm - I should've said, edited the question now
    $endgroup$
    – user-2147482565
    Mar 25 at 10:16






  • 1




    $begingroup$
    If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:24














0












0








0





$begingroup$


I'm completing a homework question and I need to prove the following:



$$
sum_i=1^N ||vecy - vecx_i||^2 = sum_i=1^N ||vecx_i - vecbarx||^2 + N||vecy-vecbarx||^2
$$



where $vecbarx = frac1N sum_i=1^N vecx_i$ and $||cdot||$ is the Euclidean norm and these are all vectors.



I think I probably need to do something similar to the proof for the sample variance but not sure how to modify it when I have a $y$ in now.






statistics summation proof-explanation statistical-inference







share|cite|improve this question











$endgroup$




I'm completing a homework question and I need to prove the following:



$$
sum_i=1^N ||vecy - vecx_i||^2 = sum_i=1^N ||vecx_i - vecbarx||^2 + N||vecy-vecbarx||^2
$$



where $vecbarx = frac1N sum_i=1^N vecx_i$ and $||cdot||$ is the Euclidean norm and these are all vectors.



I think I probably need to do something similar to the proof for the sample variance but not sure how to modify it when I have a $y$ in now.






statistics summation proof-explanation statistical-inference




statistics summation proof-explanation statistical-inference






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 16:32







user-2147482565

















asked Mar 25 at 10:06









user-2147482565user-2147482565

133




133











  • $begingroup$
    Is this an Euclidean norm or a general norm?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:10










  • $begingroup$
    This is the Euclidean norm - I should've said, edited the question now
    $endgroup$
    – user-2147482565
    Mar 25 at 10:16






  • 1




    $begingroup$
    If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:24

















  • $begingroup$
    Is this an Euclidean norm or a general norm?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:10










  • $begingroup$
    This is the Euclidean norm - I should've said, edited the question now
    $endgroup$
    – user-2147482565
    Mar 25 at 10:16






  • 1




    $begingroup$
    If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:24
















$begingroup$
Is this an Euclidean norm or a general norm?
$endgroup$
– Kavi Rama Murthy
Mar 25 at 10:10




$begingroup$
Is this an Euclidean norm or a general norm?
$endgroup$
– Kavi Rama Murthy
Mar 25 at 10:10












$begingroup$
This is the Euclidean norm - I should've said, edited the question now
$endgroup$
– user-2147482565
Mar 25 at 10:16




$begingroup$
This is the Euclidean norm - I should've said, edited the question now
$endgroup$
– user-2147482565
Mar 25 at 10:16




1




1




$begingroup$
If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 10:24





$begingroup$
If you prove this for $y=0$ you can prove it for any $y$ by changing $x_i$ to $x_i-y$. For $y=0$ use brute force.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 10:24











2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



$$(y-x_i)(y-x_i)=((y-overline x)+(overline x-x_i))((y-overline x)+(overline x-x_i))
\=(y-overline x)(y-overline x)+2(y-overline x)(overline x-x_i)+(overline x-x_i)(overline x-x_i).$$



Now take the average over $i$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
    $endgroup$
    – user-2147482565
    Mar 25 at 12:38










  • $begingroup$
    @RyanChan: there are no cross products here.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:07











  • $begingroup$
    Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
    $endgroup$
    – user-2147482565
    Mar 25 at 13:11











  • $begingroup$
    @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:12











  • $begingroup$
    Sorry I'm just confused - the term is in the hint that you gave?
    $endgroup$
    – user-2147482565
    Mar 25 at 13:13


















0












$begingroup$

beginalign*
sum_i=1^N ||vecy - vecx_i||^2 & = sum_i=1^N ||(vecy - vecbarx_i) + (vecbarx - vecx_i)||^2 \
& = sum_i=1^N ((vecy - vecbarx) + (vecbarx - vecx_i)) ((vecy - vecbarx) + (vecbarx - vecx_i))^T \
& = sum_i=1^N (vecy - vecbarx)(vecy - vecbarx)^T + 2 sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T + sum_i=1^N (vecbarx - vecx_i)(vecbarx - vecx_i)^T \
& = sum_i=1^N || vecbarx - vecx_i ||^2 + N ||vecy - vecbarx||^2
endalign*



since



beginalign*
sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T & = sum_i=1^N sum_j (y_j - barx_j)(barx_j - x_ij) \
& = sum_i=1^N sum_j (y_jbarx_j - y_jx_ij - barx_j^2 + barx_jx_ij) \
& = sum_j Bigg[ Ny_jbarx_j - y_j sum_i=1^Nx_ij - Nbarx_j^2 + barx_jsum_i=1^Nx_ij Bigg] \
& = sum_j Bigg[ Ny_jbarx_j - y_jNbarx_j - N barx_j^2 + barx_jNbarx_j Bigg] \
& = sum_j 0 \
& = 0
endalign*






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:



    $$(y-x_i)(y-x_i)=((y-overline x)+(overline x-x_i))((y-overline x)+(overline x-x_i))
    \=(y-overline x)(y-overline x)+2(y-overline x)(overline x-x_i)+(overline x-x_i)(overline x-x_i).$$



    Now take the average over $i$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
      $endgroup$
      – user-2147482565
      Mar 25 at 12:38










    • $begingroup$
      @RyanChan: there are no cross products here.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:07











    • $begingroup$
      Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
      $endgroup$
      – user-2147482565
      Mar 25 at 13:11











    • $begingroup$
      @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:12











    • $begingroup$
      Sorry I'm just confused - the term is in the hint that you gave?
      $endgroup$
      – user-2147482565
      Mar 25 at 13:13















    1












    $begingroup$

    Hint:



    $$(y-x_i)(y-x_i)=((y-overline x)+(overline x-x_i))((y-overline x)+(overline x-x_i))
    \=(y-overline x)(y-overline x)+2(y-overline x)(overline x-x_i)+(overline x-x_i)(overline x-x_i).$$



    Now take the average over $i$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
      $endgroup$
      – user-2147482565
      Mar 25 at 12:38










    • $begingroup$
      @RyanChan: there are no cross products here.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:07











    • $begingroup$
      Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
      $endgroup$
      – user-2147482565
      Mar 25 at 13:11











    • $begingroup$
      @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:12











    • $begingroup$
      Sorry I'm just confused - the term is in the hint that you gave?
      $endgroup$
      – user-2147482565
      Mar 25 at 13:13













    1












    1








    1





    $begingroup$

    Hint:



    $$(y-x_i)(y-x_i)=((y-overline x)+(overline x-x_i))((y-overline x)+(overline x-x_i))
    \=(y-overline x)(y-overline x)+2(y-overline x)(overline x-x_i)+(overline x-x_i)(overline x-x_i).$$



    Now take the average over $i$.






    share|cite|improve this answer









    $endgroup$



    Hint:



    $$(y-x_i)(y-x_i)=((y-overline x)+(overline x-x_i))((y-overline x)+(overline x-x_i))
    \=(y-overline x)(y-overline x)+2(y-overline x)(overline x-x_i)+(overline x-x_i)(overline x-x_i).$$



    Now take the average over $i$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 25 at 11:06









    Yves DaoustYves Daoust

    133k676231




    133k676231











    • $begingroup$
      But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
      $endgroup$
      – user-2147482565
      Mar 25 at 12:38










    • $begingroup$
      @RyanChan: there are no cross products here.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:07











    • $begingroup$
      Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
      $endgroup$
      – user-2147482565
      Mar 25 at 13:11











    • $begingroup$
      @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:12











    • $begingroup$
      Sorry I'm just confused - the term is in the hint that you gave?
      $endgroup$
      – user-2147482565
      Mar 25 at 13:13
















    • $begingroup$
      But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
      $endgroup$
      – user-2147482565
      Mar 25 at 12:38










    • $begingroup$
      @RyanChan: there are no cross products here.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:07











    • $begingroup$
      Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
      $endgroup$
      – user-2147482565
      Mar 25 at 13:11











    • $begingroup$
      @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
      $endgroup$
      – Yves Daoust
      Mar 25 at 13:12











    • $begingroup$
      Sorry I'm just confused - the term is in the hint that you gave?
      $endgroup$
      – user-2147482565
      Mar 25 at 13:13















    $begingroup$
    But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
    $endgroup$
    – user-2147482565
    Mar 25 at 12:38




    $begingroup$
    But I have norms here, if I take the sum over $i$ and we have norms instead of brackets, how does the cross products cancel out?
    $endgroup$
    – user-2147482565
    Mar 25 at 12:38












    $begingroup$
    @RyanChan: there are no cross products here.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:07





    $begingroup$
    @RyanChan: there are no cross products here.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:07













    $begingroup$
    Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
    $endgroup$
    – user-2147482565
    Mar 25 at 13:11





    $begingroup$
    Sorry, I meant cross product not as in the dot product, but the terms $2||y-barx|| ||barx-x_i||$
    $endgroup$
    – user-2147482565
    Mar 25 at 13:11













    $begingroup$
    @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:12





    $begingroup$
    @RyanChan: there are no such terms in the development. When averaging, the middle dot product cancels out.
    $endgroup$
    – Yves Daoust
    Mar 25 at 13:12













    $begingroup$
    Sorry I'm just confused - the term is in the hint that you gave?
    $endgroup$
    – user-2147482565
    Mar 25 at 13:13




    $begingroup$
    Sorry I'm just confused - the term is in the hint that you gave?
    $endgroup$
    – user-2147482565
    Mar 25 at 13:13











    0












    $begingroup$

    beginalign*
    sum_i=1^N ||vecy - vecx_i||^2 & = sum_i=1^N ||(vecy - vecbarx_i) + (vecbarx - vecx_i)||^2 \
    & = sum_i=1^N ((vecy - vecbarx) + (vecbarx - vecx_i)) ((vecy - vecbarx) + (vecbarx - vecx_i))^T \
    & = sum_i=1^N (vecy - vecbarx)(vecy - vecbarx)^T + 2 sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T + sum_i=1^N (vecbarx - vecx_i)(vecbarx - vecx_i)^T \
    & = sum_i=1^N || vecbarx - vecx_i ||^2 + N ||vecy - vecbarx||^2
    endalign*



    since



    beginalign*
    sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T & = sum_i=1^N sum_j (y_j - barx_j)(barx_j - x_ij) \
    & = sum_i=1^N sum_j (y_jbarx_j - y_jx_ij - barx_j^2 + barx_jx_ij) \
    & = sum_j Bigg[ Ny_jbarx_j - y_j sum_i=1^Nx_ij - Nbarx_j^2 + barx_jsum_i=1^Nx_ij Bigg] \
    & = sum_j Bigg[ Ny_jbarx_j - y_jNbarx_j - N barx_j^2 + barx_jNbarx_j Bigg] \
    & = sum_j 0 \
    & = 0
    endalign*






    share|cite|improve this answer











    $endgroup$

















      0












      $begingroup$

      beginalign*
      sum_i=1^N ||vecy - vecx_i||^2 & = sum_i=1^N ||(vecy - vecbarx_i) + (vecbarx - vecx_i)||^2 \
      & = sum_i=1^N ((vecy - vecbarx) + (vecbarx - vecx_i)) ((vecy - vecbarx) + (vecbarx - vecx_i))^T \
      & = sum_i=1^N (vecy - vecbarx)(vecy - vecbarx)^T + 2 sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T + sum_i=1^N (vecbarx - vecx_i)(vecbarx - vecx_i)^T \
      & = sum_i=1^N || vecbarx - vecx_i ||^2 + N ||vecy - vecbarx||^2
      endalign*



      since



      beginalign*
      sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T & = sum_i=1^N sum_j (y_j - barx_j)(barx_j - x_ij) \
      & = sum_i=1^N sum_j (y_jbarx_j - y_jx_ij - barx_j^2 + barx_jx_ij) \
      & = sum_j Bigg[ Ny_jbarx_j - y_j sum_i=1^Nx_ij - Nbarx_j^2 + barx_jsum_i=1^Nx_ij Bigg] \
      & = sum_j Bigg[ Ny_jbarx_j - y_jNbarx_j - N barx_j^2 + barx_jNbarx_j Bigg] \
      & = sum_j 0 \
      & = 0
      endalign*






      share|cite|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        beginalign*
        sum_i=1^N ||vecy - vecx_i||^2 & = sum_i=1^N ||(vecy - vecbarx_i) + (vecbarx - vecx_i)||^2 \
        & = sum_i=1^N ((vecy - vecbarx) + (vecbarx - vecx_i)) ((vecy - vecbarx) + (vecbarx - vecx_i))^T \
        & = sum_i=1^N (vecy - vecbarx)(vecy - vecbarx)^T + 2 sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T + sum_i=1^N (vecbarx - vecx_i)(vecbarx - vecx_i)^T \
        & = sum_i=1^N || vecbarx - vecx_i ||^2 + N ||vecy - vecbarx||^2
        endalign*



        since



        beginalign*
        sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T & = sum_i=1^N sum_j (y_j - barx_j)(barx_j - x_ij) \
        & = sum_i=1^N sum_j (y_jbarx_j - y_jx_ij - barx_j^2 + barx_jx_ij) \
        & = sum_j Bigg[ Ny_jbarx_j - y_j sum_i=1^Nx_ij - Nbarx_j^2 + barx_jsum_i=1^Nx_ij Bigg] \
        & = sum_j Bigg[ Ny_jbarx_j - y_jNbarx_j - N barx_j^2 + barx_jNbarx_j Bigg] \
        & = sum_j 0 \
        & = 0
        endalign*






        share|cite|improve this answer











        $endgroup$



        beginalign*
        sum_i=1^N ||vecy - vecx_i||^2 & = sum_i=1^N ||(vecy - vecbarx_i) + (vecbarx - vecx_i)||^2 \
        & = sum_i=1^N ((vecy - vecbarx) + (vecbarx - vecx_i)) ((vecy - vecbarx) + (vecbarx - vecx_i))^T \
        & = sum_i=1^N (vecy - vecbarx)(vecy - vecbarx)^T + 2 sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T + sum_i=1^N (vecbarx - vecx_i)(vecbarx - vecx_i)^T \
        & = sum_i=1^N || vecbarx - vecx_i ||^2 + N ||vecy - vecbarx||^2
        endalign*



        since



        beginalign*
        sum_i=1^N (vecy - vecbarx)(vecbarx - vecx_i)^T & = sum_i=1^N sum_j (y_j - barx_j)(barx_j - x_ij) \
        & = sum_i=1^N sum_j (y_jbarx_j - y_jx_ij - barx_j^2 + barx_jx_ij) \
        & = sum_j Bigg[ Ny_jbarx_j - y_j sum_i=1^Nx_ij - Nbarx_j^2 + barx_jsum_i=1^Nx_ij Bigg] \
        & = sum_j Bigg[ Ny_jbarx_j - y_jNbarx_j - N barx_j^2 + barx_jNbarx_j Bigg] \
        & = sum_j 0 \
        & = 0
        endalign*







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 26 at 10:15

























        answered Mar 26 at 9:14









        user-2147482565user-2147482565

        133




        133



























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