Weights of simple moving average are not adding up to one












3














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





























    3














    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



























      3












      3








      3







      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















      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














          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





























            4














            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





















              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














              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


























                7














                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
























                  7












                  7








                  7






                  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












                  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

                  32.1k22267




                  32.1k22267























                      4














                      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


























                        4














                        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
























                          4












                          4








                          4






                          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












                          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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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

                              Plaza Victoria

                              Image
                              Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                              Read more

                              Puebla de Zaragoza

                              Image
                              Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                              Read more

                              Musa

                              Image
                              Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                              Read more