How is $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum X_jX_j' - n bar X bar X'$?












1












$begingroup$



Let $X_j$ be a column vectors of the same shape, $(text{_})'$ indicates transposition,
and $bar X = sum X_j / n $



Prove $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum X_jX_j' - n
bar X bar X'$




I see that



$sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum(X_j - bar X) X_j' + (sum(X_j-bar X))(-bar X)' $ but from here I get $sum X_jX_j' - sum bar X X_j' - sum X_j bar X ' + nbar X bar X '$. Is there something I'm missing?






linear-algebra probability







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$endgroup$

















    1












    $begingroup$



    Let $X_j$ be a column vectors of the same shape, $(text{_})'$ indicates transposition,
    and $bar X = sum X_j / n $



    Prove $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum X_jX_j' - n
    bar X bar X'$




    I see that



    $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum(X_j - bar X) X_j' + (sum(X_j-bar X))(-bar X)' $ but from here I get $sum X_jX_j' - sum bar X X_j' - sum X_j bar X ' + nbar X bar X '$. Is there something I'm missing?






    linear-algebra probability







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$



      Let $X_j$ be a column vectors of the same shape, $(text{_})'$ indicates transposition,
      and $bar X = sum X_j / n $



      Prove $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum X_jX_j' - n
      bar X bar X'$




      I see that



      $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum(X_j - bar X) X_j' + (sum(X_j-bar X))(-bar X)' $ but from here I get $sum X_jX_j' - sum bar X X_j' - sum X_j bar X ' + nbar X bar X '$. Is there something I'm missing?






      linear-algebra probability







      share|cite|improve this question









      $endgroup$





      Let $X_j$ be a column vectors of the same shape, $(text{_})'$ indicates transposition,
      and $bar X = sum X_j / n $



      Prove $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum X_jX_j' - n
      bar X bar X'$




      I see that



      $sum_{j=1}^n(X_j - bar X)(X_j - bar X)' = sum(X_j - bar X) X_j' + (sum(X_j-bar X))(-bar X)' $ but from here I get $sum X_jX_j' - sum bar X X_j' - sum X_j bar X ' + nbar X bar X '$. Is there something I'm missing?






      linear-algebra probability




      linear-algebra probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 28 '18 at 13:52









      Oliver GOliver G

      1,4471532




      1,4471532






















          1 Answer
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          2












          $begingroup$

          Notice that for the second term $$sum_j bar{X}X_j'=bar{X} sum_j X_j'=bar{X} (nbar{X})'=nbar{X}bar{X}'.$$



          Similarly for the third term.



          Hopefully you can see how to simplify the equations from here.



          Edit:



          For the third term,



          $$sum_j X_jbar{X}'= left(sum_j X_jright)bar{X}'= (nbar{X})bar{X}'=nbar{X}bar{X}'.$$



          Hence combinining the last three terms, we have



          $$-nbar{X}bar{X}'-nbar{X}bar{X}'+nbar{X}bar{X}'=-nbar{X}bar{X}'$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:11








          • 1




            $begingroup$
            i have included more details in the answer.
            $endgroup$
            – Siong Thye Goh
            Dec 28 '18 at 14:14










          • $begingroup$
            Ah, I see. I factored incorrectly in the third term.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:19











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Notice that for the second term $$sum_j bar{X}X_j'=bar{X} sum_j X_j'=bar{X} (nbar{X})'=nbar{X}bar{X}'.$$



          Similarly for the third term.



          Hopefully you can see how to simplify the equations from here.



          Edit:



          For the third term,



          $$sum_j X_jbar{X}'= left(sum_j X_jright)bar{X}'= (nbar{X})bar{X}'=nbar{X}bar{X}'.$$



          Hence combinining the last three terms, we have



          $$-nbar{X}bar{X}'-nbar{X}bar{X}'+nbar{X}bar{X}'=-nbar{X}bar{X}'$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:11








          • 1




            $begingroup$
            i have included more details in the answer.
            $endgroup$
            – Siong Thye Goh
            Dec 28 '18 at 14:14










          • $begingroup$
            Ah, I see. I factored incorrectly in the third term.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:19
















          2












          $begingroup$

          Notice that for the second term $$sum_j bar{X}X_j'=bar{X} sum_j X_j'=bar{X} (nbar{X})'=nbar{X}bar{X}'.$$



          Similarly for the third term.



          Hopefully you can see how to simplify the equations from here.



          Edit:



          For the third term,



          $$sum_j X_jbar{X}'= left(sum_j X_jright)bar{X}'= (nbar{X})bar{X}'=nbar{X}bar{X}'.$$



          Hence combinining the last three terms, we have



          $$-nbar{X}bar{X}'-nbar{X}bar{X}'+nbar{X}bar{X}'=-nbar{X}bar{X}'$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:11








          • 1




            $begingroup$
            i have included more details in the answer.
            $endgroup$
            – Siong Thye Goh
            Dec 28 '18 at 14:14










          • $begingroup$
            Ah, I see. I factored incorrectly in the third term.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:19














          2












          2








          2





          $begingroup$

          Notice that for the second term $$sum_j bar{X}X_j'=bar{X} sum_j X_j'=bar{X} (nbar{X})'=nbar{X}bar{X}'.$$



          Similarly for the third term.



          Hopefully you can see how to simplify the equations from here.



          Edit:



          For the third term,



          $$sum_j X_jbar{X}'= left(sum_j X_jright)bar{X}'= (nbar{X})bar{X}'=nbar{X}bar{X}'.$$



          Hence combinining the last three terms, we have



          $$-nbar{X}bar{X}'-nbar{X}bar{X}'+nbar{X}bar{X}'=-nbar{X}bar{X}'$$






          share|cite|improve this answer











          $endgroup$



          Notice that for the second term $$sum_j bar{X}X_j'=bar{X} sum_j X_j'=bar{X} (nbar{X})'=nbar{X}bar{X}'.$$



          Similarly for the third term.



          Hopefully you can see how to simplify the equations from here.



          Edit:



          For the third term,



          $$sum_j X_jbar{X}'= left(sum_j X_jright)bar{X}'= (nbar{X})bar{X}'=nbar{X}bar{X}'.$$



          Hence combinining the last three terms, we have



          $$-nbar{X}bar{X}'-nbar{X}bar{X}'+nbar{X}bar{X}'=-nbar{X}bar{X}'$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 28 '18 at 14:14

























          answered Dec 28 '18 at 13:55









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118












          • $begingroup$
            Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:11








          • 1




            $begingroup$
            i have included more details in the answer.
            $endgroup$
            – Siong Thye Goh
            Dec 28 '18 at 14:14










          • $begingroup$
            Ah, I see. I factored incorrectly in the third term.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:19


















          • $begingroup$
            Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:11








          • 1




            $begingroup$
            i have included more details in the answer.
            $endgroup$
            – Siong Thye Goh
            Dec 28 '18 at 14:14










          • $begingroup$
            Ah, I see. I factored incorrectly in the third term.
            $endgroup$
            – Oliver G
            Dec 28 '18 at 14:19
















          $begingroup$
          Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
          $endgroup$
          – Oliver G
          Dec 28 '18 at 14:11






          $begingroup$
          Then for the third term I get $n bar X' bar X$ and the expression is: $sum X_j X_j' -n bar X bar X' - n bar X' bar X + n bar X bar X'$ which reduces to $sum X_j X_j ' - n bar X ' bar X. $ But the last expression isn't equal to $n bar X bar X '$.
          $endgroup$
          – Oliver G
          Dec 28 '18 at 14:11






          1




          1




          $begingroup$
          i have included more details in the answer.
          $endgroup$
          – Siong Thye Goh
          Dec 28 '18 at 14:14




          $begingroup$
          i have included more details in the answer.
          $endgroup$
          – Siong Thye Goh
          Dec 28 '18 at 14:14












          $begingroup$
          Ah, I see. I factored incorrectly in the third term.
          $endgroup$
          – Oliver G
          Dec 28 '18 at 14:19




          $begingroup$
          Ah, I see. I factored incorrectly in the third term.
          $endgroup$
          – Oliver G
          Dec 28 '18 at 14:19


















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