A problem involving geometric distributions and conditional expectation












-1












$begingroup$


Take into account an infinity sequence of independent tosses of an unbiased die. Let $X$ be the number of tosses necessary to obtain a five and $Y$ the number of tosses necessary in order to obtain a six. Determine:



(a) $textbf{E}(X)$



(b) $textbf{E}(X mid Y = 1)$



(c) $textbf{E}(X mid Y=2)$



MY ATTEMPT



I am particularly having trouble to calculate the conditional expectation. As I far as I know, it should be something like
begin{align*}
textbf{E}(Xmid Y = y) = sum_{x} xcdot p_{Xmid Y}(x,y) = sum_{x} frac{xcdot p_{X,Y}(x,y)}{p_{Y}(y)}
end{align*}



But I don't know how to proceed. Could someone pave the way towards the right solution? Thanks in advance.






probability probability-theory conditional-expectation expected-value







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    Take into account an infinity sequence of independent tosses of an unbiased die. Let $X$ be the number of tosses necessary to obtain a five and $Y$ the number of tosses necessary in order to obtain a six. Determine:



    (a) $textbf{E}(X)$



    (b) $textbf{E}(X mid Y = 1)$



    (c) $textbf{E}(X mid Y=2)$



    MY ATTEMPT



    I am particularly having trouble to calculate the conditional expectation. As I far as I know, it should be something like
    begin{align*}
    textbf{E}(Xmid Y = y) = sum_{x} xcdot p_{Xmid Y}(x,y) = sum_{x} frac{xcdot p_{X,Y}(x,y)}{p_{Y}(y)}
    end{align*}



    But I don't know how to proceed. Could someone pave the way towards the right solution? Thanks in advance.






    probability probability-theory conditional-expectation expected-value







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Take into account an infinity sequence of independent tosses of an unbiased die. Let $X$ be the number of tosses necessary to obtain a five and $Y$ the number of tosses necessary in order to obtain a six. Determine:



      (a) $textbf{E}(X)$



      (b) $textbf{E}(X mid Y = 1)$



      (c) $textbf{E}(X mid Y=2)$



      MY ATTEMPT



      I am particularly having trouble to calculate the conditional expectation. As I far as I know, it should be something like
      begin{align*}
      textbf{E}(Xmid Y = y) = sum_{x} xcdot p_{Xmid Y}(x,y) = sum_{x} frac{xcdot p_{X,Y}(x,y)}{p_{Y}(y)}
      end{align*}



      But I don't know how to proceed. Could someone pave the way towards the right solution? Thanks in advance.






      probability probability-theory conditional-expectation expected-value







      share|cite|improve this question









      $endgroup$




      Take into account an infinity sequence of independent tosses of an unbiased die. Let $X$ be the number of tosses necessary to obtain a five and $Y$ the number of tosses necessary in order to obtain a six. Determine:



      (a) $textbf{E}(X)$



      (b) $textbf{E}(X mid Y = 1)$



      (c) $textbf{E}(X mid Y=2)$



      MY ATTEMPT



      I am particularly having trouble to calculate the conditional expectation. As I far as I know, it should be something like
      begin{align*}
      textbf{E}(Xmid Y = y) = sum_{x} xcdot p_{Xmid Y}(x,y) = sum_{x} frac{xcdot p_{X,Y}(x,y)}{p_{Y}(y)}
      end{align*}



      But I don't know how to proceed. Could someone pave the way towards the right solution? Thanks in advance.






      probability probability-theory conditional-expectation expected-value




      probability probability-theory conditional-expectation expected-value






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 26 at 19:51









      user1337user1337

      46310




      46310






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Here is a solution for (b). You should be able to get (c) similarly.



          Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in ${1,...,6}$ (the dice throws). Then $X=min{tge 0: Z_t = 5}$ and
          [
          E[X|Y=1] = sum_{k=1}^{infty} k frac{P(X=k,Y=1)}{P(Y=1)}
          ]
          and the prpobability of the denomiator is $1/6$.



          For $k=1$, the probability of the event ${X=1,Y=1}={Z_1=5,Z_1=6}$ is 0 since
          $Z_1$ takes only one value.



          For $kge 2$, the event ${X=k}$ can be rewritten as ${Z_1ne 5, ..., Z_{k-1}ne 5, Z_k=5}$
          and the intersection ${X=k,Y=1}$ can be rewritten as
          ${Z_1 =6, Z_2 ne 5,..., Z_{k-1}ne 5, Z_k=6}$. By independence, the probability of this event is $frac 1 6(frac 5 6)^{k-2}frac 1 6$ for $kge 2$. Hence
          [
          E[X|Y=1] = sum_{k=2}^{infty} k (1/6) (5/6)^{k-2}
          = frac 6 5 sum_{k=2}^{infty} k (1/6) (5/6)^{k-1}
          ]
          and $sum_{k=2}^{infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).






          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%2f3088684%2fa-problem-involving-geometric-distributions-and-conditional-expectation%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









            1












            $begingroup$

            Here is a solution for (b). You should be able to get (c) similarly.



            Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in ${1,...,6}$ (the dice throws). Then $X=min{tge 0: Z_t = 5}$ and
            [
            E[X|Y=1] = sum_{k=1}^{infty} k frac{P(X=k,Y=1)}{P(Y=1)}
            ]
            and the prpobability of the denomiator is $1/6$.



            For $k=1$, the probability of the event ${X=1,Y=1}={Z_1=5,Z_1=6}$ is 0 since
            $Z_1$ takes only one value.



            For $kge 2$, the event ${X=k}$ can be rewritten as ${Z_1ne 5, ..., Z_{k-1}ne 5, Z_k=5}$
            and the intersection ${X=k,Y=1}$ can be rewritten as
            ${Z_1 =6, Z_2 ne 5,..., Z_{k-1}ne 5, Z_k=6}$. By independence, the probability of this event is $frac 1 6(frac 5 6)^{k-2}frac 1 6$ for $kge 2$. Hence
            [
            E[X|Y=1] = sum_{k=2}^{infty} k (1/6) (5/6)^{k-2}
            = frac 6 5 sum_{k=2}^{infty} k (1/6) (5/6)^{k-1}
            ]
            and $sum_{k=2}^{infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Here is a solution for (b). You should be able to get (c) similarly.



              Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in ${1,...,6}$ (the dice throws). Then $X=min{tge 0: Z_t = 5}$ and
              [
              E[X|Y=1] = sum_{k=1}^{infty} k frac{P(X=k,Y=1)}{P(Y=1)}
              ]
              and the prpobability of the denomiator is $1/6$.



              For $k=1$, the probability of the event ${X=1,Y=1}={Z_1=5,Z_1=6}$ is 0 since
              $Z_1$ takes only one value.



              For $kge 2$, the event ${X=k}$ can be rewritten as ${Z_1ne 5, ..., Z_{k-1}ne 5, Z_k=5}$
              and the intersection ${X=k,Y=1}$ can be rewritten as
              ${Z_1 =6, Z_2 ne 5,..., Z_{k-1}ne 5, Z_k=6}$. By independence, the probability of this event is $frac 1 6(frac 5 6)^{k-2}frac 1 6$ for $kge 2$. Hence
              [
              E[X|Y=1] = sum_{k=2}^{infty} k (1/6) (5/6)^{k-2}
              = frac 6 5 sum_{k=2}^{infty} k (1/6) (5/6)^{k-1}
              ]
              and $sum_{k=2}^{infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Here is a solution for (b). You should be able to get (c) similarly.



                Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in ${1,...,6}$ (the dice throws). Then $X=min{tge 0: Z_t = 5}$ and
                [
                E[X|Y=1] = sum_{k=1}^{infty} k frac{P(X=k,Y=1)}{P(Y=1)}
                ]
                and the prpobability of the denomiator is $1/6$.



                For $k=1$, the probability of the event ${X=1,Y=1}={Z_1=5,Z_1=6}$ is 0 since
                $Z_1$ takes only one value.



                For $kge 2$, the event ${X=k}$ can be rewritten as ${Z_1ne 5, ..., Z_{k-1}ne 5, Z_k=5}$
                and the intersection ${X=k,Y=1}$ can be rewritten as
                ${Z_1 =6, Z_2 ne 5,..., Z_{k-1}ne 5, Z_k=6}$. By independence, the probability of this event is $frac 1 6(frac 5 6)^{k-2}frac 1 6$ for $kge 2$. Hence
                [
                E[X|Y=1] = sum_{k=2}^{infty} k (1/6) (5/6)^{k-2}
                = frac 6 5 sum_{k=2}^{infty} k (1/6) (5/6)^{k-1}
                ]
                and $sum_{k=2}^{infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).






                share|cite|improve this answer









                $endgroup$



                Here is a solution for (b). You should be able to get (c) similarly.



                Let $Z_1,Z_2,...,Z_n,...$ be a countable sequence of iid random variables valued in ${1,...,6}$ (the dice throws). Then $X=min{tge 0: Z_t = 5}$ and
                [
                E[X|Y=1] = sum_{k=1}^{infty} k frac{P(X=k,Y=1)}{P(Y=1)}
                ]
                and the prpobability of the denomiator is $1/6$.



                For $k=1$, the probability of the event ${X=1,Y=1}={Z_1=5,Z_1=6}$ is 0 since
                $Z_1$ takes only one value.



                For $kge 2$, the event ${X=k}$ can be rewritten as ${Z_1ne 5, ..., Z_{k-1}ne 5, Z_k=5}$
                and the intersection ${X=k,Y=1}$ can be rewritten as
                ${Z_1 =6, Z_2 ne 5,..., Z_{k-1}ne 5, Z_k=6}$. By independence, the probability of this event is $frac 1 6(frac 5 6)^{k-2}frac 1 6$ for $kge 2$. Hence
                [
                E[X|Y=1] = sum_{k=2}^{infty} k (1/6) (5/6)^{k-2}
                = frac 6 5 sum_{k=2}^{infty} k (1/6) (5/6)^{k-1}
                ]
                and $sum_{k=2}^{infty} k (1/6) (5/6)^{k-1} = E[X]$ already computed in (a).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 20:18









                jlewkjlewk

                1115




                1115






























                    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%2f3088684%2fa-problem-involving-geometric-distributions-and-conditional-expectation%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

                    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

                    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