Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$












0












$begingroup$



Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
with parameter $p$. Let $N$ be a Poisson random variable with with
parameter $lambda$ which is independet of the $X_i.$



(a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



(b) Use (a) to identify the distribution of $Z$.




The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



$$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



$$G_Z(s)=(1-p+ps)^N.$$



I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






probability probability-distributions generating-functions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
    with parameter $p$. Let $N$ be a Poisson random variable with with
    parameter $lambda$ which is independet of the $X_i.$



    (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



    (b) Use (a) to identify the distribution of $Z$.




    The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



    $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



    and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



    $$G_Z(s)=(1-p+ps)^N.$$



    I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






    probability probability-distributions generating-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
      with parameter $p$. Let $N$ be a Poisson random variable with with
      parameter $lambda$ which is independet of the $X_i.$



      (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



      (b) Use (a) to identify the distribution of $Z$.




      The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



      $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



      and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



      $$G_Z(s)=(1-p+ps)^N.$$



      I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






      probability probability-distributions generating-functions







      share|cite|improve this question









      $endgroup$





      Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
      with parameter $p$. Let $N$ be a Poisson random variable with with
      parameter $lambda$ which is independet of the $X_i.$



      (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



      (b) Use (a) to identify the distribution of $Z$.




      The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



      $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



      and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



      $$G_Z(s)=(1-p+ps)^N.$$



      I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






      probability probability-distributions generating-functions




      probability probability-distributions generating-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 17:21









      ParsevalParseval

      3,0001719




      3,0001719






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            $endgroup$
            – Parseval
            Jan 4 at 17:30










          • $begingroup$
            @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            $endgroup$
            – angryavian
            Jan 4 at 17:37










          • $begingroup$
            Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            $endgroup$
            – Parseval
            Jan 4 at 17:55








          • 1




            $begingroup$
            @Parseval No, the way I wrote it was a bit misleading, sorry.
            $endgroup$
            – angryavian
            Jan 4 at 18:04






          • 1




            $begingroup$
            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            $endgroup$
            – Just_to_Answer
            Jan 4 at 20:02











          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%2f3061853%2ffind-the-probability-generating-function-of-z-sum-i-1nx-i%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












          $begingroup$

          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            $endgroup$
            – Parseval
            Jan 4 at 17:30










          • $begingroup$
            @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            $endgroup$
            – angryavian
            Jan 4 at 17:37










          • $begingroup$
            Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            $endgroup$
            – Parseval
            Jan 4 at 17:55








          • 1




            $begingroup$
            @Parseval No, the way I wrote it was a bit misleading, sorry.
            $endgroup$
            – angryavian
            Jan 4 at 18:04






          • 1




            $begingroup$
            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            $endgroup$
            – Just_to_Answer
            Jan 4 at 20:02
















          2












          $begingroup$

          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            $endgroup$
            – Parseval
            Jan 4 at 17:30










          • $begingroup$
            @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            $endgroup$
            – angryavian
            Jan 4 at 17:37










          • $begingroup$
            Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            $endgroup$
            – Parseval
            Jan 4 at 17:55








          • 1




            $begingroup$
            @Parseval No, the way I wrote it was a bit misleading, sorry.
            $endgroup$
            – angryavian
            Jan 4 at 18:04






          • 1




            $begingroup$
            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            $endgroup$
            – Just_to_Answer
            Jan 4 at 20:02














          2












          2








          2





          $begingroup$

          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer











          $endgroup$



          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 18:03

























          answered Jan 4 at 17:27









          angryavianangryavian

          42.1k23381




          42.1k23381












          • $begingroup$
            Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            $endgroup$
            – Parseval
            Jan 4 at 17:30










          • $begingroup$
            @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            $endgroup$
            – angryavian
            Jan 4 at 17:37










          • $begingroup$
            Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            $endgroup$
            – Parseval
            Jan 4 at 17:55








          • 1




            $begingroup$
            @Parseval No, the way I wrote it was a bit misleading, sorry.
            $endgroup$
            – angryavian
            Jan 4 at 18:04






          • 1




            $begingroup$
            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            $endgroup$
            – Just_to_Answer
            Jan 4 at 20:02


















          • $begingroup$
            Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            $endgroup$
            – Parseval
            Jan 4 at 17:30










          • $begingroup$
            @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            $endgroup$
            – angryavian
            Jan 4 at 17:37










          • $begingroup$
            Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            $endgroup$
            – Parseval
            Jan 4 at 17:55








          • 1




            $begingroup$
            @Parseval No, the way I wrote it was a bit misleading, sorry.
            $endgroup$
            – angryavian
            Jan 4 at 18:04






          • 1




            $begingroup$
            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            $endgroup$
            – Just_to_Answer
            Jan 4 at 20:02
















          $begingroup$
          Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
          $endgroup$
          – Parseval
          Jan 4 at 17:30




          $begingroup$
          Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
          $endgroup$
          – Parseval
          Jan 4 at 17:30












          $begingroup$
          @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
          $endgroup$
          – angryavian
          Jan 4 at 17:37




          $begingroup$
          @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
          $endgroup$
          – angryavian
          Jan 4 at 17:37












          $begingroup$
          Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
          $endgroup$
          – Parseval
          Jan 4 at 17:55






          $begingroup$
          Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
          $endgroup$
          – Parseval
          Jan 4 at 17:55






          1




          1




          $begingroup$
          @Parseval No, the way I wrote it was a bit misleading, sorry.
          $endgroup$
          – angryavian
          Jan 4 at 18:04




          $begingroup$
          @Parseval No, the way I wrote it was a bit misleading, sorry.
          $endgroup$
          – angryavian
          Jan 4 at 18:04




          1




          1




          $begingroup$
          @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
          $endgroup$
          – Just_to_Answer
          Jan 4 at 20:02




          $begingroup$
          @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
          $endgroup$
          – Just_to_Answer
          Jan 4 at 20:02


















          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%2f3061853%2ffind-the-probability-generating-function-of-z-sum-i-1nx-i%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more