Does mean centering reduces covariance?












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$begingroup$


It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.






correlation covariance random-vector







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$endgroup$

















    1












    $begingroup$


    It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.






    correlation covariance random-vector







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.






      correlation covariance random-vector







      share|cite|improve this question









      $endgroup$




      It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.






      correlation covariance random-vector




      correlation covariance random-vector






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked 1 hour ago









      lvdplvdp

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          $begingroup$

          If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
          $$
          begin{aligned}
          operatorname{Cov}(X + a, Y + b)
          &= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
          &= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
          &= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
          &= E[(X - E[X])(Y - E[Y])] \
          &= operatorname{Cov}(X, Y).
          end{aligned}
          $$
          Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.




          Also, since correlation is defined as
          $$
          operatorname{Corr}(X, Y)
          = frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
          $$
          we can see that
          $$
          begin{aligned}
          operatorname{Corr}(X + a, Y + b)
          &= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
          &= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
          end{aligned}
          $$
          so in particular, correlation isn't affected by centering either.




          That was the population version of the story. The sample version is the same: If we use
          $$
          widehat{operatorname{Cov}}(X, Y)
          = frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
          $$
          as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
          $$
          begin{aligned}
          widehat{operatorname{Cov}}(X + a, Y + b)
          &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
          &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
          &= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
          &= widehat{operatorname{Cov}}(X, Y)
          end{aligned}
          $$
          for any $a$ and $b$.






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            $begingroup$

            If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
            $$
            begin{aligned}
            operatorname{Cov}(X + a, Y + b)
            &= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
            &= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
            &= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
            &= E[(X - E[X])(Y - E[Y])] \
            &= operatorname{Cov}(X, Y).
            end{aligned}
            $$
            Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.




            Also, since correlation is defined as
            $$
            operatorname{Corr}(X, Y)
            = frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
            $$
            we can see that
            $$
            begin{aligned}
            operatorname{Corr}(X + a, Y + b)
            &= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
            &= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
            end{aligned}
            $$
            so in particular, correlation isn't affected by centering either.




            That was the population version of the story. The sample version is the same: If we use
            $$
            widehat{operatorname{Cov}}(X, Y)
            = frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
            $$
            as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
            $$
            begin{aligned}
            widehat{operatorname{Cov}}(X + a, Y + b)
            &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
            &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
            &= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
            &= widehat{operatorname{Cov}}(X, Y)
            end{aligned}
            $$
            for any $a$ and $b$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
              $$
              begin{aligned}
              operatorname{Cov}(X + a, Y + b)
              &= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
              &= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
              &= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
              &= E[(X - E[X])(Y - E[Y])] \
              &= operatorname{Cov}(X, Y).
              end{aligned}
              $$
              Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.




              Also, since correlation is defined as
              $$
              operatorname{Corr}(X, Y)
              = frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
              $$
              we can see that
              $$
              begin{aligned}
              operatorname{Corr}(X + a, Y + b)
              &= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
              &= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
              end{aligned}
              $$
              so in particular, correlation isn't affected by centering either.




              That was the population version of the story. The sample version is the same: If we use
              $$
              widehat{operatorname{Cov}}(X, Y)
              = frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
              $$
              as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
              $$
              begin{aligned}
              widehat{operatorname{Cov}}(X + a, Y + b)
              &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
              &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
              &= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
              &= widehat{operatorname{Cov}}(X, Y)
              end{aligned}
              $$
              for any $a$ and $b$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
                $$
                begin{aligned}
                operatorname{Cov}(X + a, Y + b)
                &= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
                &= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
                &= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
                &= E[(X - E[X])(Y - E[Y])] \
                &= operatorname{Cov}(X, Y).
                end{aligned}
                $$
                Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.




                Also, since correlation is defined as
                $$
                operatorname{Corr}(X, Y)
                = frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
                $$
                we can see that
                $$
                begin{aligned}
                operatorname{Corr}(X + a, Y + b)
                &= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
                &= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
                end{aligned}
                $$
                so in particular, correlation isn't affected by centering either.




                That was the population version of the story. The sample version is the same: If we use
                $$
                widehat{operatorname{Cov}}(X, Y)
                = frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
                $$
                as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
                $$
                begin{aligned}
                widehat{operatorname{Cov}}(X + a, Y + b)
                &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
                &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
                &= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
                &= widehat{operatorname{Cov}}(X, Y)
                end{aligned}
                $$
                for any $a$ and $b$.






                share|cite|improve this answer











                $endgroup$



                If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
                $$
                begin{aligned}
                operatorname{Cov}(X + a, Y + b)
                &= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
                &= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
                &= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
                &= E[(X - E[X])(Y - E[Y])] \
                &= operatorname{Cov}(X, Y).
                end{aligned}
                $$
                Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.




                Also, since correlation is defined as
                $$
                operatorname{Corr}(X, Y)
                = frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
                $$
                we can see that
                $$
                begin{aligned}
                operatorname{Corr}(X + a, Y + b)
                &= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
                &= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
                end{aligned}
                $$
                so in particular, correlation isn't affected by centering either.




                That was the population version of the story. The sample version is the same: If we use
                $$
                widehat{operatorname{Cov}}(X, Y)
                = frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
                $$
                as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
                $$
                begin{aligned}
                widehat{operatorname{Cov}}(X + a, Y + b)
                &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
                &= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
                &= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
                &= widehat{operatorname{Cov}}(X, Y)
                end{aligned}
                $$
                for any $a$ and $b$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 23 mins ago

























                answered 35 mins ago









                Artem MavrinArtem Mavrin

                33516




                33516






























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