Covariance proof












1












$begingroup$



Let the covariance random values X,Y be :
$$Cov[X, Y]= E[(X- E[X])(Y - E[Y])] = E[XY]-E[X]E[Y]$$



Prove the following for random values $X_1...X_n$ :
$$Var[sum_{i = 1}^n X_i] = sum_{i = 1}^n Var[X_i] + sum_{i = 1}^n sum_{j = 1, jne i}^nCov[X_i,X_j]$$




How would I go about proving this?






probability probability-theory discrete-mathematics







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$endgroup$








  • 1




    $begingroup$
    This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
    $endgroup$
    – Sean Roberson
    Dec 13 '18 at 4:00










  • $begingroup$
    @SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
    $endgroup$
    – drhab
    Dec 13 '18 at 14:15
















1












$begingroup$



Let the covariance random values X,Y be :
$$Cov[X, Y]= E[(X- E[X])(Y - E[Y])] = E[XY]-E[X]E[Y]$$



Prove the following for random values $X_1...X_n$ :
$$Var[sum_{i = 1}^n X_i] = sum_{i = 1}^n Var[X_i] + sum_{i = 1}^n sum_{j = 1, jne i}^nCov[X_i,X_j]$$




How would I go about proving this?






probability probability-theory discrete-mathematics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
    $endgroup$
    – Sean Roberson
    Dec 13 '18 at 4:00










  • $begingroup$
    @SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
    $endgroup$
    – drhab
    Dec 13 '18 at 14:15














1












1








1





$begingroup$



Let the covariance random values X,Y be :
$$Cov[X, Y]= E[(X- E[X])(Y - E[Y])] = E[XY]-E[X]E[Y]$$



Prove the following for random values $X_1...X_n$ :
$$Var[sum_{i = 1}^n X_i] = sum_{i = 1}^n Var[X_i] + sum_{i = 1}^n sum_{j = 1, jne i}^nCov[X_i,X_j]$$




How would I go about proving this?






probability probability-theory discrete-mathematics







share|cite|improve this question









$endgroup$





Let the covariance random values X,Y be :
$$Cov[X, Y]= E[(X- E[X])(Y - E[Y])] = E[XY]-E[X]E[Y]$$



Prove the following for random values $X_1...X_n$ :
$$Var[sum_{i = 1}^n X_i] = sum_{i = 1}^n Var[X_i] + sum_{i = 1}^n sum_{j = 1, jne i}^nCov[X_i,X_j]$$




How would I go about proving this?






probability probability-theory discrete-mathematics




probability probability-theory discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 3:58









J. LastinJ. Lastin

936




936








  • 1




    $begingroup$
    This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
    $endgroup$
    – Sean Roberson
    Dec 13 '18 at 4:00










  • $begingroup$
    @SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
    $endgroup$
    – drhab
    Dec 13 '18 at 14:15














  • 1




    $begingroup$
    This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
    $endgroup$
    – Sean Roberson
    Dec 13 '18 at 4:00










  • $begingroup$
    @SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
    $endgroup$
    – drhab
    Dec 13 '18 at 14:15








1




1




$begingroup$
This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
$endgroup$
– Sean Roberson
Dec 13 '18 at 4:00




$begingroup$
This is akin to expanding a squared sum. Can you try for two random variables and use inducton? That's my strat.
$endgroup$
– Sean Roberson
Dec 13 '18 at 4:00












$begingroup$
@SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
$endgroup$
– drhab
Dec 13 '18 at 14:15




$begingroup$
@SeanRoberson Induction can indeed be used, but I would not recommend it for this case.
$endgroup$
– drhab
Dec 13 '18 at 14:15










2 Answers
2






active

oldest

votes


















2












$begingroup$

We will use principle of mathematical induction.



Basis



$Var(X_1+X_2)=E((X_1+X_2)^2)-E^2(X_1+X_2)$



$=E(X_1^2+2X_1X_2+X_2^2)-(E(X_1)+E(X_2))^2$



$=E(X_1^2)-E^2(X_1)+E(X_2^2)-E^2(X_2)+2(E(X_1X_2)-E(X_1)E(X_2))$



$=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$



Induction Hypothesis



$Var(sum_{i=1}^k X_i)=sum_{i=1}^k Var(X_i)+sum_{i=1}^k sum_{j=1,ineq j}^k Cov(X_i,X_j)$



Inductive step



$Var(sum_{i=1}^{k+1} X_i)=Var(X_{k+1}+sum_{i=1}^k X_i)$



$=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+ 2Cov(X_{k+1},sum_{i=1}^k X_i)$



$=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+2sum_{i=1}^k Cov(X_{k+1} X_i) $



which gives the required result on using hypothesis.



Hope it is helpful:)






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Let $Z_i=X_i-mathbb EX_i$ for $i=1,dots,n$.



    Then:



    $$mathsf{Var}left(sum_{i=1}^nX_iright)=mathbb Eleft(sum_{i=1}^nX_i-mathbb Esum_{i=1}^nX_iright)^2=mathbb Eleft(sum_{i=1}^nZ_iright)^2=mathbb Esum_{i=1}^nsum_{j=1}^nZ_iZ_j=$$$$sum_{i=1}^nmathbb EZ_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathbb EZ_iZ_j=sum_{i=1}^nmathsf{Var}X_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathsf{Cov}(X_i,X_j)$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      We will use principle of mathematical induction.



      Basis



      $Var(X_1+X_2)=E((X_1+X_2)^2)-E^2(X_1+X_2)$



      $=E(X_1^2+2X_1X_2+X_2^2)-(E(X_1)+E(X_2))^2$



      $=E(X_1^2)-E^2(X_1)+E(X_2^2)-E^2(X_2)+2(E(X_1X_2)-E(X_1)E(X_2))$



      $=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$



      Induction Hypothesis



      $Var(sum_{i=1}^k X_i)=sum_{i=1}^k Var(X_i)+sum_{i=1}^k sum_{j=1,ineq j}^k Cov(X_i,X_j)$



      Inductive step



      $Var(sum_{i=1}^{k+1} X_i)=Var(X_{k+1}+sum_{i=1}^k X_i)$



      $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+ 2Cov(X_{k+1},sum_{i=1}^k X_i)$



      $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+2sum_{i=1}^k Cov(X_{k+1} X_i) $



      which gives the required result on using hypothesis.



      Hope it is helpful:)






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        We will use principle of mathematical induction.



        Basis



        $Var(X_1+X_2)=E((X_1+X_2)^2)-E^2(X_1+X_2)$



        $=E(X_1^2+2X_1X_2+X_2^2)-(E(X_1)+E(X_2))^2$



        $=E(X_1^2)-E^2(X_1)+E(X_2^2)-E^2(X_2)+2(E(X_1X_2)-E(X_1)E(X_2))$



        $=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$



        Induction Hypothesis



        $Var(sum_{i=1}^k X_i)=sum_{i=1}^k Var(X_i)+sum_{i=1}^k sum_{j=1,ineq j}^k Cov(X_i,X_j)$



        Inductive step



        $Var(sum_{i=1}^{k+1} X_i)=Var(X_{k+1}+sum_{i=1}^k X_i)$



        $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+ 2Cov(X_{k+1},sum_{i=1}^k X_i)$



        $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+2sum_{i=1}^k Cov(X_{k+1} X_i) $



        which gives the required result on using hypothesis.



        Hope it is helpful:)






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          We will use principle of mathematical induction.



          Basis



          $Var(X_1+X_2)=E((X_1+X_2)^2)-E^2(X_1+X_2)$



          $=E(X_1^2+2X_1X_2+X_2^2)-(E(X_1)+E(X_2))^2$



          $=E(X_1^2)-E^2(X_1)+E(X_2^2)-E^2(X_2)+2(E(X_1X_2)-E(X_1)E(X_2))$



          $=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$



          Induction Hypothesis



          $Var(sum_{i=1}^k X_i)=sum_{i=1}^k Var(X_i)+sum_{i=1}^k sum_{j=1,ineq j}^k Cov(X_i,X_j)$



          Inductive step



          $Var(sum_{i=1}^{k+1} X_i)=Var(X_{k+1}+sum_{i=1}^k X_i)$



          $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+ 2Cov(X_{k+1},sum_{i=1}^k X_i)$



          $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+2sum_{i=1}^k Cov(X_{k+1} X_i) $



          which gives the required result on using hypothesis.



          Hope it is helpful:)






          share|cite|improve this answer











          $endgroup$



          We will use principle of mathematical induction.



          Basis



          $Var(X_1+X_2)=E((X_1+X_2)^2)-E^2(X_1+X_2)$



          $=E(X_1^2+2X_1X_2+X_2^2)-(E(X_1)+E(X_2))^2$



          $=E(X_1^2)-E^2(X_1)+E(X_2^2)-E^2(X_2)+2(E(X_1X_2)-E(X_1)E(X_2))$



          $=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)$



          Induction Hypothesis



          $Var(sum_{i=1}^k X_i)=sum_{i=1}^k Var(X_i)+sum_{i=1}^k sum_{j=1,ineq j}^k Cov(X_i,X_j)$



          Inductive step



          $Var(sum_{i=1}^{k+1} X_i)=Var(X_{k+1}+sum_{i=1}^k X_i)$



          $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+ 2Cov(X_{k+1},sum_{i=1}^k X_i)$



          $=Var(X_{k+1})+Var(sum_{i=1}^k X_i)+2sum_{i=1}^k Cov(X_{k+1} X_i) $



          which gives the required result on using hypothesis.



          Hope it is helpful:)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 0:16

























          answered Dec 13 '18 at 4:38









          MartundMartund

          1,633213




          1,633213























              2












              $begingroup$

              Let $Z_i=X_i-mathbb EX_i$ for $i=1,dots,n$.



              Then:



              $$mathsf{Var}left(sum_{i=1}^nX_iright)=mathbb Eleft(sum_{i=1}^nX_i-mathbb Esum_{i=1}^nX_iright)^2=mathbb Eleft(sum_{i=1}^nZ_iright)^2=mathbb Esum_{i=1}^nsum_{j=1}^nZ_iZ_j=$$$$sum_{i=1}^nmathbb EZ_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathbb EZ_iZ_j=sum_{i=1}^nmathsf{Var}X_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathsf{Cov}(X_i,X_j)$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let $Z_i=X_i-mathbb EX_i$ for $i=1,dots,n$.



                Then:



                $$mathsf{Var}left(sum_{i=1}^nX_iright)=mathbb Eleft(sum_{i=1}^nX_i-mathbb Esum_{i=1}^nX_iright)^2=mathbb Eleft(sum_{i=1}^nZ_iright)^2=mathbb Esum_{i=1}^nsum_{j=1}^nZ_iZ_j=$$$$sum_{i=1}^nmathbb EZ_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathbb EZ_iZ_j=sum_{i=1}^nmathsf{Var}X_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathsf{Cov}(X_i,X_j)$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $Z_i=X_i-mathbb EX_i$ for $i=1,dots,n$.



                  Then:



                  $$mathsf{Var}left(sum_{i=1}^nX_iright)=mathbb Eleft(sum_{i=1}^nX_i-mathbb Esum_{i=1}^nX_iright)^2=mathbb Eleft(sum_{i=1}^nZ_iright)^2=mathbb Esum_{i=1}^nsum_{j=1}^nZ_iZ_j=$$$$sum_{i=1}^nmathbb EZ_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathbb EZ_iZ_j=sum_{i=1}^nmathsf{Var}X_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathsf{Cov}(X_i,X_j)$$






                  share|cite|improve this answer









                  $endgroup$



                  Let $Z_i=X_i-mathbb EX_i$ for $i=1,dots,n$.



                  Then:



                  $$mathsf{Var}left(sum_{i=1}^nX_iright)=mathbb Eleft(sum_{i=1}^nX_i-mathbb Esum_{i=1}^nX_iright)^2=mathbb Eleft(sum_{i=1}^nZ_iright)^2=mathbb Esum_{i=1}^nsum_{j=1}^nZ_iZ_j=$$$$sum_{i=1}^nmathbb EZ_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathbb EZ_iZ_j=sum_{i=1}^nmathsf{Var}X_i^2+sum_{i=1}^nsum_{j=1,jneq i}^nmathsf{Cov}(X_i,X_j)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 14:09









                  drhabdrhab

                  101k545136




                  101k545136






























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