Weights of simple moving average are not adding up to one












3












$begingroup$


This is the definition of linear filter from a book I am reading:




A second procedure for dealing with a trend is to use a linear filter, which converts one time series, ${x_t}$, into another, ${y_t}$, by the linear operation
$$y_t = sum_{r = -q}^{+s} a_r x_{t+r}$$
where ${a_r}$ is a set of weights. In order to smooth out local fluctuations and estimate the local mean, we should clearly choose the weights so that $sum a_r = 1$, and then the operation is often referred to as a moving average. Moving averages are discussed in detail by Kendall et al. (1983, Chapter 46), and we will only provide a brief introduction. Moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. The simplest example of a symmetric smoothing filter is the simple moving average, for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$, and the smoothed value of $x_t$ is given by
$$textrm{Sm}(x_t) = frac{1}{2q + 1}sum_{r=-q}^{+q} x_{t+r}$$


It is said there that ${a_r}$ is a set of weights and in order to call the operation a moving average we should clearly choose the weights so sum of $a_r$ is equal to 1.



Then it is described that the moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. So the simple moving average for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$ is $textrm{Sm}(x_t)$



But when I tried to confirm that the sum of $a_r$ for simple moving average equal 1 I got this:



enter image description here



Is there something I misunderstood?






average







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    This is the definition of linear filter from a book I am reading:




    A second procedure for dealing with a trend is to use a linear filter, which converts one time series, ${x_t}$, into another, ${y_t}$, by the linear operation
    $$y_t = sum_{r = -q}^{+s} a_r x_{t+r}$$
    where ${a_r}$ is a set of weights. In order to smooth out local fluctuations and estimate the local mean, we should clearly choose the weights so that $sum a_r = 1$, and then the operation is often referred to as a moving average. Moving averages are discussed in detail by Kendall et al. (1983, Chapter 46), and we will only provide a brief introduction. Moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. The simplest example of a symmetric smoothing filter is the simple moving average, for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$, and the smoothed value of $x_t$ is given by
    $$textrm{Sm}(x_t) = frac{1}{2q + 1}sum_{r=-q}^{+q} x_{t+r}$$


    It is said there that ${a_r}$ is a set of weights and in order to call the operation a moving average we should clearly choose the weights so sum of $a_r$ is equal to 1.



    Then it is described that the moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. So the simple moving average for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$ is $textrm{Sm}(x_t)$



    But when I tried to confirm that the sum of $a_r$ for simple moving average equal 1 I got this:



    enter image description here



    Is there something I misunderstood?






    average







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      This is the definition of linear filter from a book I am reading:




      A second procedure for dealing with a trend is to use a linear filter, which converts one time series, ${x_t}$, into another, ${y_t}$, by the linear operation
      $$y_t = sum_{r = -q}^{+s} a_r x_{t+r}$$
      where ${a_r}$ is a set of weights. In order to smooth out local fluctuations and estimate the local mean, we should clearly choose the weights so that $sum a_r = 1$, and then the operation is often referred to as a moving average. Moving averages are discussed in detail by Kendall et al. (1983, Chapter 46), and we will only provide a brief introduction. Moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. The simplest example of a symmetric smoothing filter is the simple moving average, for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$, and the smoothed value of $x_t$ is given by
      $$textrm{Sm}(x_t) = frac{1}{2q + 1}sum_{r=-q}^{+q} x_{t+r}$$


      It is said there that ${a_r}$ is a set of weights and in order to call the operation a moving average we should clearly choose the weights so sum of $a_r$ is equal to 1.



      Then it is described that the moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. So the simple moving average for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$ is $textrm{Sm}(x_t)$



      But when I tried to confirm that the sum of $a_r$ for simple moving average equal 1 I got this:



      enter image description here



      Is there something I misunderstood?






      average







      share|cite|improve this question











      $endgroup$




      This is the definition of linear filter from a book I am reading:




      A second procedure for dealing with a trend is to use a linear filter, which converts one time series, ${x_t}$, into another, ${y_t}$, by the linear operation
      $$y_t = sum_{r = -q}^{+s} a_r x_{t+r}$$
      where ${a_r}$ is a set of weights. In order to smooth out local fluctuations and estimate the local mean, we should clearly choose the weights so that $sum a_r = 1$, and then the operation is often referred to as a moving average. Moving averages are discussed in detail by Kendall et al. (1983, Chapter 46), and we will only provide a brief introduction. Moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. The simplest example of a symmetric smoothing filter is the simple moving average, for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$, and the smoothed value of $x_t$ is given by
      $$textrm{Sm}(x_t) = frac{1}{2q + 1}sum_{r=-q}^{+q} x_{t+r}$$


      It is said there that ${a_r}$ is a set of weights and in order to call the operation a moving average we should clearly choose the weights so sum of $a_r$ is equal to 1.



      Then it is described that the moving averages are often symmetric with $s = q$ and $a_j = a_{-j}$. So the simple moving average for which $a_r = 1/(2q + 1)$ for $r = -q, ldots, +q$ is $textrm{Sm}(x_t)$



      But when I tried to confirm that the sum of $a_r$ for simple moving average equal 1 I got this:



      enter image description here



      Is there something I misunderstood?






      average




      average






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 3:45









      Mark H

      1,00368




      1,00368










      asked Dec 15 '18 at 16:41









      Kocur4dKocur4d

      1184




      1184






















          2 Answers
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          7












          $begingroup$

          In your code, you have computed [1/(2 * ele + 1) for ele in r], which (in your example) is
          $$ left[ frac{-1}{9}, frac{-1}{7}, frac{-1}{5}, frac{-1}{3}, -1, 1, frac{1}{3}, frac{1}{5}, frac{1}{7}, frac{1}{9}, frac{1}{11} right] text{.} $$ When you sum those up, you get $frac{1}{11} = 0.overline{09}$, as you observed.



          However, the book specifies that you use $frac{1}{2q+1}$, so [1/(2 * q + 1) for ele in r], which will give
          $$ left[ frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11} right] text{.} $$
          Summing those gives $1$, as expected. Also, the simplest moving average should weight all of its inputs equally, which is what happens here.






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Note that the book defines $a_r = frac{1}{2q + 1}$ where $a_r$ does NOT depend on $r$, but only on $q$. Since $q$ is fixed, all of the $a_r$ should be the same. In your code, you calculate $a_r = frac{1}{2r + 1}$ instead.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              7












              $begingroup$

              In your code, you have computed [1/(2 * ele + 1) for ele in r], which (in your example) is
              $$ left[ frac{-1}{9}, frac{-1}{7}, frac{-1}{5}, frac{-1}{3}, -1, 1, frac{1}{3}, frac{1}{5}, frac{1}{7}, frac{1}{9}, frac{1}{11} right] text{.} $$ When you sum those up, you get $frac{1}{11} = 0.overline{09}$, as you observed.



              However, the book specifies that you use $frac{1}{2q+1}$, so [1/(2 * q + 1) for ele in r], which will give
              $$ left[ frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11} right] text{.} $$
              Summing those gives $1$, as expected. Also, the simplest moving average should weight all of its inputs equally, which is what happens here.






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                In your code, you have computed [1/(2 * ele + 1) for ele in r], which (in your example) is
                $$ left[ frac{-1}{9}, frac{-1}{7}, frac{-1}{5}, frac{-1}{3}, -1, 1, frac{1}{3}, frac{1}{5}, frac{1}{7}, frac{1}{9}, frac{1}{11} right] text{.} $$ When you sum those up, you get $frac{1}{11} = 0.overline{09}$, as you observed.



                However, the book specifies that you use $frac{1}{2q+1}$, so [1/(2 * q + 1) for ele in r], which will give
                $$ left[ frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11} right] text{.} $$
                Summing those gives $1$, as expected. Also, the simplest moving average should weight all of its inputs equally, which is what happens here.






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  In your code, you have computed [1/(2 * ele + 1) for ele in r], which (in your example) is
                  $$ left[ frac{-1}{9}, frac{-1}{7}, frac{-1}{5}, frac{-1}{3}, -1, 1, frac{1}{3}, frac{1}{5}, frac{1}{7}, frac{1}{9}, frac{1}{11} right] text{.} $$ When you sum those up, you get $frac{1}{11} = 0.overline{09}$, as you observed.



                  However, the book specifies that you use $frac{1}{2q+1}$, so [1/(2 * q + 1) for ele in r], which will give
                  $$ left[ frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11} right] text{.} $$
                  Summing those gives $1$, as expected. Also, the simplest moving average should weight all of its inputs equally, which is what happens here.






                  share|cite|improve this answer









                  $endgroup$



                  In your code, you have computed [1/(2 * ele + 1) for ele in r], which (in your example) is
                  $$ left[ frac{-1}{9}, frac{-1}{7}, frac{-1}{5}, frac{-1}{3}, -1, 1, frac{1}{3}, frac{1}{5}, frac{1}{7}, frac{1}{9}, frac{1}{11} right] text{.} $$ When you sum those up, you get $frac{1}{11} = 0.overline{09}$, as you observed.



                  However, the book specifies that you use $frac{1}{2q+1}$, so [1/(2 * q + 1) for ele in r], which will give
                  $$ left[ frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11}, frac{1}{11} right] text{.} $$
                  Summing those gives $1$, as expected. Also, the simplest moving average should weight all of its inputs equally, which is what happens here.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 17:02









                  Eric TowersEric Towers

                  32.4k22268




                  32.4k22268























                      4












                      $begingroup$

                      Note that the book defines $a_r = frac{1}{2q + 1}$ where $a_r$ does NOT depend on $r$, but only on $q$. Since $q$ is fixed, all of the $a_r$ should be the same. In your code, you calculate $a_r = frac{1}{2r + 1}$ instead.






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        Note that the book defines $a_r = frac{1}{2q + 1}$ where $a_r$ does NOT depend on $r$, but only on $q$. Since $q$ is fixed, all of the $a_r$ should be the same. In your code, you calculate $a_r = frac{1}{2r + 1}$ instead.






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Note that the book defines $a_r = frac{1}{2q + 1}$ where $a_r$ does NOT depend on $r$, but only on $q$. Since $q$ is fixed, all of the $a_r$ should be the same. In your code, you calculate $a_r = frac{1}{2r + 1}$ instead.






                          share|cite|improve this answer









                          $endgroup$



                          Note that the book defines $a_r = frac{1}{2q + 1}$ where $a_r$ does NOT depend on $r$, but only on $q$. Since $q$ is fixed, all of the $a_r$ should be the same. In your code, you calculate $a_r = frac{1}{2r + 1}$ instead.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 15 '18 at 16:48









                          AlexAlex

                          1777




                          1777






























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