Efficient computation of incremental standard deviation (removing first value)












2












$begingroup$


I found this link about incremental standard deviation where it computes the standard deviation every time a new element was added to the dataset.



Is there a similar method when adding a new value and removing the first value from the dataset?



For example,



[45, 26, 78, 45, 34, 56] - initial data set 



[26, 78, 45, 34, 56, 74 *(new data)*] - first value 45 was removed





statistics algorithms computational-mathematics standard-deviation







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$endgroup$

















    2












    $begingroup$


    I found this link about incremental standard deviation where it computes the standard deviation every time a new element was added to the dataset.



    Is there a similar method when adding a new value and removing the first value from the dataset?



    For example,



    [45, 26, 78, 45, 34, 56] - initial data set 
    

    
[26, 78, 45, 34, 56, 74 *(new data)*] - first value 45 was removed





    statistics algorithms computational-mathematics standard-deviation







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I found this link about incremental standard deviation where it computes the standard deviation every time a new element was added to the dataset.



      Is there a similar method when adding a new value and removing the first value from the dataset?



      For example,



      [45, 26, 78, 45, 34, 56] - initial data set 
      

      
[26, 78, 45, 34, 56, 74 *(new data)*] - first value 45 was removed





      statistics algorithms computational-mathematics standard-deviation







      share|cite|improve this question











      $endgroup$




      I found this link about incremental standard deviation where it computes the standard deviation every time a new element was added to the dataset.



      Is there a similar method when adding a new value and removing the first value from the dataset?



      For example,



      [45, 26, 78, 45, 34, 56] - initial data set 
      

      
[26, 78, 45, 34, 56, 74 *(new data)*] - first value 45 was removed





      statistics algorithms computational-mathematics standard-deviation




      statistics algorithms computational-mathematics standard-deviation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 1:34









      Alexander Gruber♦

      20.1k25103174




      20.1k25103174










      asked Jan 30 at 1:28









      KrissyKrissy

      112




      112






















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          $begingroup$

          The formula at the link is a special case of the sample variance decomposition formula given in O'Neill (2014) (Result 1). Your problem involves a double-application of this formula.



          Denoting the data-sets used in an obvious way, your goal is to write $s_{2:n+1}^2$ in terms of $s_{1:n}^2$ and any other necessary sample quantities. To do this we can use sample variance decompositions:



          $$s_{2:n+1}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2,$$



          $$s_{1:n}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{1})^2.$$



          Hence, we have:



          $$begin{equation} begin{aligned}
          s_{2:n+1}^2
          &= frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
          &= s_{1:n}^2 - frac{1}{n} (bar{x}_{2:n} - x_{1})^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
          &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n} - x_{1})^2 - (bar{x}_{2:n} - x_{n+1})^2 Big] \[6pt]
          &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{1} + x_{1}^2) - (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{n+1} + x_{n+1}^2) Big] \[6pt]
          &= s_{1:n}^2 - frac{1}{n} Big[ -2 bar{x}_{2:n} x_{1} + x_{1}^2 + 2 bar{x}_{2:n} x_{n+1} - x_{n+1}^2 Big] \[6pt]
          &= s_{1:n}^2 - frac{1}{n} Big[ 2 bar{x}_{2:n} (x_{n+1} - x_{1}) + (x_{1}^2 - x_{n+1}^2) Big]. \[6pt]
          end{aligned} end{equation}$$






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            1 Answer
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            1 Answer
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            active

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            1












            $begingroup$

            The formula at the link is a special case of the sample variance decomposition formula given in O'Neill (2014) (Result 1). Your problem involves a double-application of this formula.



            Denoting the data-sets used in an obvious way, your goal is to write $s_{2:n+1}^2$ in terms of $s_{1:n}^2$ and any other necessary sample quantities. To do this we can use sample variance decompositions:



            $$s_{2:n+1}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2,$$



            $$s_{1:n}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{1})^2.$$



            Hence, we have:



            $$begin{equation} begin{aligned}
            s_{2:n+1}^2
            &= frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
            &= s_{1:n}^2 - frac{1}{n} (bar{x}_{2:n} - x_{1})^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
            &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n} - x_{1})^2 - (bar{x}_{2:n} - x_{n+1})^2 Big] \[6pt]
            &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{1} + x_{1}^2) - (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{n+1} + x_{n+1}^2) Big] \[6pt]
            &= s_{1:n}^2 - frac{1}{n} Big[ -2 bar{x}_{2:n} x_{1} + x_{1}^2 + 2 bar{x}_{2:n} x_{n+1} - x_{n+1}^2 Big] \[6pt]
            &= s_{1:n}^2 - frac{1}{n} Big[ 2 bar{x}_{2:n} (x_{n+1} - x_{1}) + (x_{1}^2 - x_{n+1}^2) Big]. \[6pt]
            end{aligned} end{equation}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The formula at the link is a special case of the sample variance decomposition formula given in O'Neill (2014) (Result 1). Your problem involves a double-application of this formula.



              Denoting the data-sets used in an obvious way, your goal is to write $s_{2:n+1}^2$ in terms of $s_{1:n}^2$ and any other necessary sample quantities. To do this we can use sample variance decompositions:



              $$s_{2:n+1}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2,$$



              $$s_{1:n}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{1})^2.$$



              Hence, we have:



              $$begin{equation} begin{aligned}
              s_{2:n+1}^2
              &= frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
              &= s_{1:n}^2 - frac{1}{n} (bar{x}_{2:n} - x_{1})^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
              &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n} - x_{1})^2 - (bar{x}_{2:n} - x_{n+1})^2 Big] \[6pt]
              &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{1} + x_{1}^2) - (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{n+1} + x_{n+1}^2) Big] \[6pt]
              &= s_{1:n}^2 - frac{1}{n} Big[ -2 bar{x}_{2:n} x_{1} + x_{1}^2 + 2 bar{x}_{2:n} x_{n+1} - x_{n+1}^2 Big] \[6pt]
              &= s_{1:n}^2 - frac{1}{n} Big[ 2 bar{x}_{2:n} (x_{n+1} - x_{1}) + (x_{1}^2 - x_{n+1}^2) Big]. \[6pt]
              end{aligned} end{equation}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The formula at the link is a special case of the sample variance decomposition formula given in O'Neill (2014) (Result 1). Your problem involves a double-application of this formula.



                Denoting the data-sets used in an obvious way, your goal is to write $s_{2:n+1}^2$ in terms of $s_{1:n}^2$ and any other necessary sample quantities. To do this we can use sample variance decompositions:



                $$s_{2:n+1}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2,$$



                $$s_{1:n}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{1})^2.$$



                Hence, we have:



                $$begin{equation} begin{aligned}
                s_{2:n+1}^2
                &= frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
                &= s_{1:n}^2 - frac{1}{n} (bar{x}_{2:n} - x_{1})^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n} - x_{1})^2 - (bar{x}_{2:n} - x_{n+1})^2 Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{1} + x_{1}^2) - (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{n+1} + x_{n+1}^2) Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ -2 bar{x}_{2:n} x_{1} + x_{1}^2 + 2 bar{x}_{2:n} x_{n+1} - x_{n+1}^2 Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ 2 bar{x}_{2:n} (x_{n+1} - x_{1}) + (x_{1}^2 - x_{n+1}^2) Big]. \[6pt]
                end{aligned} end{equation}$$






                share|cite|improve this answer









                $endgroup$



                The formula at the link is a special case of the sample variance decomposition formula given in O'Neill (2014) (Result 1). Your problem involves a double-application of this formula.



                Denoting the data-sets used in an obvious way, your goal is to write $s_{2:n+1}^2$ in terms of $s_{1:n}^2$ and any other necessary sample quantities. To do this we can use sample variance decompositions:



                $$s_{2:n+1}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2,$$



                $$s_{1:n}^2 = frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{1})^2.$$



                Hence, we have:



                $$begin{equation} begin{aligned}
                s_{2:n+1}^2
                &= frac{n-2}{n-1} s_{2:n}^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
                &= s_{1:n}^2 - frac{1}{n} (bar{x}_{2:n} - x_{1})^2 + frac{1}{n} (bar{x}_{2:n} - x_{n+1})^2 \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n} - x_{1})^2 - (bar{x}_{2:n} - x_{n+1})^2 Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{1} + x_{1}^2) - (bar{x}_{2:n}^2 -2 bar{x}_{2:n} x_{n+1} + x_{n+1}^2) Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ -2 bar{x}_{2:n} x_{1} + x_{1}^2 + 2 bar{x}_{2:n} x_{n+1} - x_{n+1}^2 Big] \[6pt]
                &= s_{1:n}^2 - frac{1}{n} Big[ 2 bar{x}_{2:n} (x_{n+1} - x_{1}) + (x_{1}^2 - x_{n+1}^2) Big]. \[6pt]
                end{aligned} end{equation}$$







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                answered Jan 31 at 4:10









                BenBen

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