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
          2






          active

          oldest

          votes


















          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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041694%2fweights-of-simple-moving-average-are-not-adding-up-to-one%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041694%2fweights-of-simple-moving-average-are-not-adding-up-to-one%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Сан-Квентин

                              Image
                              Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но
                              Read more

                              Алькесар

                              Image
                              Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове
                              Read more

                              Josef Freinademetz

                              Image
                              San Ujöp (Giuseppe/Josef) Freinademetz Josef Freinademetz in abiti cinesi   Religioso e missionario   Nascita 15 aprile 1852 Morte 28 gennaio 1908 Venerato da Chiesa cattolica Beatificazione 1975 Canonizzazione 5 ottobre 2003 Ricorrenza 28 gennaio Manuale Ujöp Freinademetz noto anche come Giuseppe o Josef Freinademetz (Oies, 15 aprile 1852 – Taikia, 28 gennaio 1908) è stato un presbitero austriaco, membro della Società del Verbo Divino, fu missionario in Cina; è venerato come santo dalla Chiesa cattolica. .mw-parser-output .citazione-table{margin-bottom:.5em;font-size:95%}.mw-parser-output .citazione-table td{padding:0 1.2em 0 2.4em}.mw-parser-output .citazione-lang{vertical-align:top}.mw-parser-output .citazione-lang td{width:50%}.mw-parser-output .citazione-lang td:first-child{padding:0 0 0 2.4em}.mw-parser-output .citazione-lang td:nth-child(2){padding:0 1.2em} «Io amo la Cina e i cinesi; voglio morire in mezzo a loro, e tra loro
                              Read more