Let $y_{1},…,y_{n}$ be real numbers and let $bar{y}$ be their average. Prove the following












1












$begingroup$


I am currently trying to solve the following problem which will ultimately help me finish a larger proof. Here is the equality that I am trying to calculate:



enter image description here



Here is my thought process so far. I will have to algebratically work with the sum on the left hand side of the equation, until all the $y_{i}^2$ terms can be collected into a single sum, and same with all the $y_{i}y_{j}$ terms (s.t $ineq j$. But I am getting stuck on where to begin here, factoring the square and using the linearity property of summations seems to be wrong when trying to work to the right side. Is there some algebraic trick that I'm not recognizing?






statistics summation







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am currently trying to solve the following problem which will ultimately help me finish a larger proof. Here is the equality that I am trying to calculate:



    enter image description here



    Here is my thought process so far. I will have to algebratically work with the sum on the left hand side of the equation, until all the $y_{i}^2$ terms can be collected into a single sum, and same with all the $y_{i}y_{j}$ terms (s.t $ineq j$. But I am getting stuck on where to begin here, factoring the square and using the linearity property of summations seems to be wrong when trying to work to the right side. Is there some algebraic trick that I'm not recognizing?






    statistics summation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am currently trying to solve the following problem which will ultimately help me finish a larger proof. Here is the equality that I am trying to calculate:



      enter image description here



      Here is my thought process so far. I will have to algebratically work with the sum on the left hand side of the equation, until all the $y_{i}^2$ terms can be collected into a single sum, and same with all the $y_{i}y_{j}$ terms (s.t $ineq j$. But I am getting stuck on where to begin here, factoring the square and using the linearity property of summations seems to be wrong when trying to work to the right side. Is there some algebraic trick that I'm not recognizing?






      statistics summation







      share|cite|improve this question









      $endgroup$




      I am currently trying to solve the following problem which will ultimately help me finish a larger proof. Here is the equality that I am trying to calculate:



      enter image description here



      Here is my thought process so far. I will have to algebratically work with the sum on the left hand side of the equation, until all the $y_{i}^2$ terms can be collected into a single sum, and same with all the $y_{i}y_{j}$ terms (s.t $ineq j$. But I am getting stuck on where to begin here, factoring the square and using the linearity property of summations seems to be wrong when trying to work to the right side. Is there some algebraic trick that I'm not recognizing?






      statistics summation




      statistics summation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 31 at 1:56









      mizkid45mizkid45

      274




      274






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Expanding the quadratic gives:



          $$begin{equation} begin{aligned}
          sum_i (y_i - bar{y})^2
          &= sum_i (y_i^2 - 2 bar{y} y_i + bar{y}^2) \[6pt]
          &= sum_i y_i^2 - 2 bar{y} sum_i y_i + n bar{y}^2 \[6pt]
          &= sum_i y_i^2 - 2n bar{y}^2 + n bar{y}^2 \[6pt]
          &= sum_i y_i^2 - n bar{y}^2. \[6pt]
          end{aligned} end{equation}$$



          Now, substituting the sample mean with its definition as a sum gives:



          $$begin{equation} begin{aligned}
          sum_i (y_i - bar{y})^2
          &= sum_i y_i^2 - n Big( frac{1}{n} sum_i y_i Big)^2 \[6pt]
          &= sum_i y_i^2 - frac{1}{n} Big( sum_i y_i Big)^2 \[6pt]
          &= sum_i y_i^2 - frac{1}{n} sum_i sum_j y_i y_j \[6pt]
          &= sum_i y_i^2 - frac{1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j \[6pt]
          &= frac{n-1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j. \[6pt]
          end{aligned} end{equation}$$






          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%2f3094388%2flet-y-1-y-n-be-real-numbers-and-let-bary-be-their-average-prov%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Expanding the quadratic gives:



            $$begin{equation} begin{aligned}
            sum_i (y_i - bar{y})^2
            &= sum_i (y_i^2 - 2 bar{y} y_i + bar{y}^2) \[6pt]
            &= sum_i y_i^2 - 2 bar{y} sum_i y_i + n bar{y}^2 \[6pt]
            &= sum_i y_i^2 - 2n bar{y}^2 + n bar{y}^2 \[6pt]
            &= sum_i y_i^2 - n bar{y}^2. \[6pt]
            end{aligned} end{equation}$$



            Now, substituting the sample mean with its definition as a sum gives:



            $$begin{equation} begin{aligned}
            sum_i (y_i - bar{y})^2
            &= sum_i y_i^2 - n Big( frac{1}{n} sum_i y_i Big)^2 \[6pt]
            &= sum_i y_i^2 - frac{1}{n} Big( sum_i y_i Big)^2 \[6pt]
            &= sum_i y_i^2 - frac{1}{n} sum_i sum_j y_i y_j \[6pt]
            &= sum_i y_i^2 - frac{1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j \[6pt]
            &= frac{n-1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j. \[6pt]
            end{aligned} end{equation}$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Expanding the quadratic gives:



              $$begin{equation} begin{aligned}
              sum_i (y_i - bar{y})^2
              &= sum_i (y_i^2 - 2 bar{y} y_i + bar{y}^2) \[6pt]
              &= sum_i y_i^2 - 2 bar{y} sum_i y_i + n bar{y}^2 \[6pt]
              &= sum_i y_i^2 - 2n bar{y}^2 + n bar{y}^2 \[6pt]
              &= sum_i y_i^2 - n bar{y}^2. \[6pt]
              end{aligned} end{equation}$$



              Now, substituting the sample mean with its definition as a sum gives:



              $$begin{equation} begin{aligned}
              sum_i (y_i - bar{y})^2
              &= sum_i y_i^2 - n Big( frac{1}{n} sum_i y_i Big)^2 \[6pt]
              &= sum_i y_i^2 - frac{1}{n} Big( sum_i y_i Big)^2 \[6pt]
              &= sum_i y_i^2 - frac{1}{n} sum_i sum_j y_i y_j \[6pt]
              &= sum_i y_i^2 - frac{1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j \[6pt]
              &= frac{n-1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j. \[6pt]
              end{aligned} end{equation}$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Expanding the quadratic gives:



                $$begin{equation} begin{aligned}
                sum_i (y_i - bar{y})^2
                &= sum_i (y_i^2 - 2 bar{y} y_i + bar{y}^2) \[6pt]
                &= sum_i y_i^2 - 2 bar{y} sum_i y_i + n bar{y}^2 \[6pt]
                &= sum_i y_i^2 - 2n bar{y}^2 + n bar{y}^2 \[6pt]
                &= sum_i y_i^2 - n bar{y}^2. \[6pt]
                end{aligned} end{equation}$$



                Now, substituting the sample mean with its definition as a sum gives:



                $$begin{equation} begin{aligned}
                sum_i (y_i - bar{y})^2
                &= sum_i y_i^2 - n Big( frac{1}{n} sum_i y_i Big)^2 \[6pt]
                &= sum_i y_i^2 - frac{1}{n} Big( sum_i y_i Big)^2 \[6pt]
                &= sum_i y_i^2 - frac{1}{n} sum_i sum_j y_i y_j \[6pt]
                &= sum_i y_i^2 - frac{1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j \[6pt]
                &= frac{n-1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j. \[6pt]
                end{aligned} end{equation}$$






                share|cite|improve this answer









                $endgroup$



                Expanding the quadratic gives:



                $$begin{equation} begin{aligned}
                sum_i (y_i - bar{y})^2
                &= sum_i (y_i^2 - 2 bar{y} y_i + bar{y}^2) \[6pt]
                &= sum_i y_i^2 - 2 bar{y} sum_i y_i + n bar{y}^2 \[6pt]
                &= sum_i y_i^2 - 2n bar{y}^2 + n bar{y}^2 \[6pt]
                &= sum_i y_i^2 - n bar{y}^2. \[6pt]
                end{aligned} end{equation}$$



                Now, substituting the sample mean with its definition as a sum gives:



                $$begin{equation} begin{aligned}
                sum_i (y_i - bar{y})^2
                &= sum_i y_i^2 - n Big( frac{1}{n} sum_i y_i Big)^2 \[6pt]
                &= sum_i y_i^2 - frac{1}{n} Big( sum_i y_i Big)^2 \[6pt]
                &= sum_i y_i^2 - frac{1}{n} sum_i sum_j y_i y_j \[6pt]
                &= sum_i y_i^2 - frac{1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j \[6pt]
                &= frac{n-1}{n} sum_i y_i^2 - frac{1}{n} sum_{i neq j} y_i y_j. \[6pt]
                end{aligned} end{equation}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 3:02









                BenBen

                1,900215




                1,900215






























                    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%2f3094388%2flet-y-1-y-n-be-real-numbers-and-let-bary-be-their-average-prov%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

                    MongoDB - Not Authorized To Execute Command

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                    Read more

                    How to fix TextFormField cause rebuild widget in Flutter

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                    Read more

                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                    Image
                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                    Read more