k-means cost functions comparison












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$begingroup$


$J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.



I have so far,



$J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
$sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $



$sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$



Then I have,



$2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $



$2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$



I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.



    I have so far,



    $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
    $sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $



    $sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
    $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
    $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
    $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
    $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
    $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$



    Then I have,



    $2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $



    $2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$



    I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.



      I have so far,



      $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
      $sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $



      $sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
      $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$



      Then I have,



      $2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $



      $2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$



      I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.










      share|cite|improve this question











      $endgroup$




      $J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.



      I have so far,



      $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
      $sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $



      $sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
      $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
      $sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$



      Then I have,



      $2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $



      $2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$



      I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.







      summation machine-learning clustering






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 21 at 22:10







      Kayanda

















      asked Jan 21 at 21:08









      KayandaKayanda

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