Urns, balls, and simplifying an infinite series












3












$begingroup$


Consider a large number of urns and a large number of people each randomly placing $n sim Poisson(lambda)$ balls in urns (people do not place multiple balls in the same urn). Urns receiving multiple balls pick a ball at random. If an individual's balls are chosen by multiple urns, then the individual randomly chooses an urn and we call that a match. Letting z denote the probability that person i is chosen by urn j, conditional on placing a ball in j, I conjecture that the probability that an urn that has chosen a ball forms a match with that ball is given by:



$$displaystylesum_{ngeq 1} left( frac{lambda^ne^{-lambda}}{n!} right)left[sum_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]$$



Simplifying the expression inside of the brackets yields



$$ frac{(1-z)^{n-1}}{n} sum_{k=0}^{n-1}
{n choose k+1}left( frac{z}{1-z} right)^k $$



This is close to a familiar looking series for the inner brackets, but I'm not sure how to proceed. Any suggestions or references?






probability sequences-and-series







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Consider a large number of urns and a large number of people each randomly placing $n sim Poisson(lambda)$ balls in urns (people do not place multiple balls in the same urn). Urns receiving multiple balls pick a ball at random. If an individual's balls are chosen by multiple urns, then the individual randomly chooses an urn and we call that a match. Letting z denote the probability that person i is chosen by urn j, conditional on placing a ball in j, I conjecture that the probability that an urn that has chosen a ball forms a match with that ball is given by:



    $$displaystylesum_{ngeq 1} left( frac{lambda^ne^{-lambda}}{n!} right)left[sum_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]$$



    Simplifying the expression inside of the brackets yields



    $$ frac{(1-z)^{n-1}}{n} sum_{k=0}^{n-1}
    {n choose k+1}left( frac{z}{1-z} right)^k $$



    This is close to a familiar looking series for the inner brackets, but I'm not sure how to proceed. Any suggestions or references?






    probability sequences-and-series







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Consider a large number of urns and a large number of people each randomly placing $n sim Poisson(lambda)$ balls in urns (people do not place multiple balls in the same urn). Urns receiving multiple balls pick a ball at random. If an individual's balls are chosen by multiple urns, then the individual randomly chooses an urn and we call that a match. Letting z denote the probability that person i is chosen by urn j, conditional on placing a ball in j, I conjecture that the probability that an urn that has chosen a ball forms a match with that ball is given by:



      $$displaystylesum_{ngeq 1} left( frac{lambda^ne^{-lambda}}{n!} right)left[sum_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]$$



      Simplifying the expression inside of the brackets yields



      $$ frac{(1-z)^{n-1}}{n} sum_{k=0}^{n-1}
      {n choose k+1}left( frac{z}{1-z} right)^k $$



      This is close to a familiar looking series for the inner brackets, but I'm not sure how to proceed. Any suggestions or references?






      probability sequences-and-series







      share|cite|improve this question











      $endgroup$




      Consider a large number of urns and a large number of people each randomly placing $n sim Poisson(lambda)$ balls in urns (people do not place multiple balls in the same urn). Urns receiving multiple balls pick a ball at random. If an individual's balls are chosen by multiple urns, then the individual randomly chooses an urn and we call that a match. Letting z denote the probability that person i is chosen by urn j, conditional on placing a ball in j, I conjecture that the probability that an urn that has chosen a ball forms a match with that ball is given by:



      $$displaystylesum_{ngeq 1} left( frac{lambda^ne^{-lambda}}{n!} right)left[sum_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]$$



      Simplifying the expression inside of the brackets yields



      $$ frac{(1-z)^{n-1}}{n} sum_{k=0}^{n-1}
      {n choose k+1}left( frac{z}{1-z} right)^k $$



      This is close to a familiar looking series for the inner brackets, but I'm not sure how to proceed. Any suggestions or references?






      probability sequences-and-series




      probability sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 0:11









      Math1000

      19.3k31745




      19.3k31745










      asked Jan 25 at 23:17









      JohnJohn

      255




      255






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Simplifying the expression inside of the brackets:



          $left[sumlimits_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]tag1$



          Using the fact that $frac{1}{k+1}=intlimits_0^1 t^kdt$ and replace the order the sum and integral we get:



          $intlimits_0^1sumlimits_{k=0}^{n-1} {n-1choose k} (zt)^k(1-z)^{n-1-k}dttag2$



          According to Binomial formula we can write:



          $intlimits_0^1 (zt+1-z)^{n-1}dt=[frac{(zt+1-z)^n}{zn}]_{t=0}^1=frac{1}{zn}-frac{(1-z)^n}{zn}tag3$



          So the result of the sum inside the brakets is:



          $frac{1}{zn}-frac{(1-z)^n}{zn}tag4$



          Regarding the whole expression:



          $sumlimits_{n=1}^infty frac{lambda^ne^{-lambda}}{n!}big(frac{1}{zn}-frac{(1-z)^n}{zn}big)tag5$



          Using the same trick as in (2) we have:



          $frac{e^{-lambda}}{z}intlimits_0^1 frac{1}{t}sumlimits_{n=1}^infty frac{lambda^n}{n!}big(t^n-t^n(1-z)^nbig)tag6$



          Known that $sumlimits_{n=1}^infty frac{x^n}{n!}=sumlimits_{n=0}^infty frac{x^n}{n!}-1;$ we can form the(5) in the following way:



          $frac{e^{-lambda}}{z}intlimits_0^1 big(frac{e^{lambda t}}{t}-frac{e^{lambda t(1-z)}}{t}big)dt tag7$



          This integral can be solved by using the integral functions.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 5:40






          • 1




            $begingroup$
            @Claude Leibovici, Thank you, and regards.
            $endgroup$
            – JV.Stalker
            Jan 26 at 6:35










          • $begingroup$
            @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
            $endgroup$
            – John
            Jan 26 at 16:04










          • $begingroup$
            @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
            $endgroup$
            – JV.Stalker
            Jan 27 at 5:15











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Simplifying the expression inside of the brackets:



          $left[sumlimits_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]tag1$



          Using the fact that $frac{1}{k+1}=intlimits_0^1 t^kdt$ and replace the order the sum and integral we get:



          $intlimits_0^1sumlimits_{k=0}^{n-1} {n-1choose k} (zt)^k(1-z)^{n-1-k}dttag2$



          According to Binomial formula we can write:



          $intlimits_0^1 (zt+1-z)^{n-1}dt=[frac{(zt+1-z)^n}{zn}]_{t=0}^1=frac{1}{zn}-frac{(1-z)^n}{zn}tag3$



          So the result of the sum inside the brakets is:



          $frac{1}{zn}-frac{(1-z)^n}{zn}tag4$



          Regarding the whole expression:



          $sumlimits_{n=1}^infty frac{lambda^ne^{-lambda}}{n!}big(frac{1}{zn}-frac{(1-z)^n}{zn}big)tag5$



          Using the same trick as in (2) we have:



          $frac{e^{-lambda}}{z}intlimits_0^1 frac{1}{t}sumlimits_{n=1}^infty frac{lambda^n}{n!}big(t^n-t^n(1-z)^nbig)tag6$



          Known that $sumlimits_{n=1}^infty frac{x^n}{n!}=sumlimits_{n=0}^infty frac{x^n}{n!}-1;$ we can form the(5) in the following way:



          $frac{e^{-lambda}}{z}intlimits_0^1 big(frac{e^{lambda t}}{t}-frac{e^{lambda t(1-z)}}{t}big)dt tag7$



          This integral can be solved by using the integral functions.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 5:40






          • 1




            $begingroup$
            @Claude Leibovici, Thank you, and regards.
            $endgroup$
            – JV.Stalker
            Jan 26 at 6:35










          • $begingroup$
            @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
            $endgroup$
            – John
            Jan 26 at 16:04










          • $begingroup$
            @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
            $endgroup$
            – JV.Stalker
            Jan 27 at 5:15
















          2












          $begingroup$

          Simplifying the expression inside of the brackets:



          $left[sumlimits_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]tag1$



          Using the fact that $frac{1}{k+1}=intlimits_0^1 t^kdt$ and replace the order the sum and integral we get:



          $intlimits_0^1sumlimits_{k=0}^{n-1} {n-1choose k} (zt)^k(1-z)^{n-1-k}dttag2$



          According to Binomial formula we can write:



          $intlimits_0^1 (zt+1-z)^{n-1}dt=[frac{(zt+1-z)^n}{zn}]_{t=0}^1=frac{1}{zn}-frac{(1-z)^n}{zn}tag3$



          So the result of the sum inside the brakets is:



          $frac{1}{zn}-frac{(1-z)^n}{zn}tag4$



          Regarding the whole expression:



          $sumlimits_{n=1}^infty frac{lambda^ne^{-lambda}}{n!}big(frac{1}{zn}-frac{(1-z)^n}{zn}big)tag5$



          Using the same trick as in (2) we have:



          $frac{e^{-lambda}}{z}intlimits_0^1 frac{1}{t}sumlimits_{n=1}^infty frac{lambda^n}{n!}big(t^n-t^n(1-z)^nbig)tag6$



          Known that $sumlimits_{n=1}^infty frac{x^n}{n!}=sumlimits_{n=0}^infty frac{x^n}{n!}-1;$ we can form the(5) in the following way:



          $frac{e^{-lambda}}{z}intlimits_0^1 big(frac{e^{lambda t}}{t}-frac{e^{lambda t(1-z)}}{t}big)dt tag7$



          This integral can be solved by using the integral functions.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 5:40






          • 1




            $begingroup$
            @Claude Leibovici, Thank you, and regards.
            $endgroup$
            – JV.Stalker
            Jan 26 at 6:35










          • $begingroup$
            @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
            $endgroup$
            – John
            Jan 26 at 16:04










          • $begingroup$
            @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
            $endgroup$
            – JV.Stalker
            Jan 27 at 5:15














          2












          2








          2





          $begingroup$

          Simplifying the expression inside of the brackets:



          $left[sumlimits_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]tag1$



          Using the fact that $frac{1}{k+1}=intlimits_0^1 t^kdt$ and replace the order the sum and integral we get:



          $intlimits_0^1sumlimits_{k=0}^{n-1} {n-1choose k} (zt)^k(1-z)^{n-1-k}dttag2$



          According to Binomial formula we can write:



          $intlimits_0^1 (zt+1-z)^{n-1}dt=[frac{(zt+1-z)^n}{zn}]_{t=0}^1=frac{1}{zn}-frac{(1-z)^n}{zn}tag3$



          So the result of the sum inside the brakets is:



          $frac{1}{zn}-frac{(1-z)^n}{zn}tag4$



          Regarding the whole expression:



          $sumlimits_{n=1}^infty frac{lambda^ne^{-lambda}}{n!}big(frac{1}{zn}-frac{(1-z)^n}{zn}big)tag5$



          Using the same trick as in (2) we have:



          $frac{e^{-lambda}}{z}intlimits_0^1 frac{1}{t}sumlimits_{n=1}^infty frac{lambda^n}{n!}big(t^n-t^n(1-z)^nbig)tag6$



          Known that $sumlimits_{n=1}^infty frac{x^n}{n!}=sumlimits_{n=0}^infty frac{x^n}{n!}-1;$ we can form the(5) in the following way:



          $frac{e^{-lambda}}{z}intlimits_0^1 big(frac{e^{lambda t}}{t}-frac{e^{lambda t(1-z)}}{t}big)dt tag7$



          This integral can be solved by using the integral functions.






          share|cite|improve this answer









          $endgroup$



          Simplifying the expression inside of the brackets:



          $left[sumlimits_{k=0}^{n-1} {n-1choose k} z^k(1-z)^{n-1-k}left(frac{1}{k+1} right) right]tag1$



          Using the fact that $frac{1}{k+1}=intlimits_0^1 t^kdt$ and replace the order the sum and integral we get:



          $intlimits_0^1sumlimits_{k=0}^{n-1} {n-1choose k} (zt)^k(1-z)^{n-1-k}dttag2$



          According to Binomial formula we can write:



          $intlimits_0^1 (zt+1-z)^{n-1}dt=[frac{(zt+1-z)^n}{zn}]_{t=0}^1=frac{1}{zn}-frac{(1-z)^n}{zn}tag3$



          So the result of the sum inside the brakets is:



          $frac{1}{zn}-frac{(1-z)^n}{zn}tag4$



          Regarding the whole expression:



          $sumlimits_{n=1}^infty frac{lambda^ne^{-lambda}}{n!}big(frac{1}{zn}-frac{(1-z)^n}{zn}big)tag5$



          Using the same trick as in (2) we have:



          $frac{e^{-lambda}}{z}intlimits_0^1 frac{1}{t}sumlimits_{n=1}^infty frac{lambda^n}{n!}big(t^n-t^n(1-z)^nbig)tag6$



          Known that $sumlimits_{n=1}^infty frac{x^n}{n!}=sumlimits_{n=0}^infty frac{x^n}{n!}-1;$ we can form the(5) in the following way:



          $frac{e^{-lambda}}{z}intlimits_0^1 big(frac{e^{lambda t}}{t}-frac{e^{lambda t(1-z)}}{t}big)dt tag7$



          This integral can be solved by using the integral functions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 26 at 5:02









          JV.StalkerJV.Stalker

          93149




          93149








          • 1




            $begingroup$
            This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 5:40






          • 1




            $begingroup$
            @Claude Leibovici, Thank you, and regards.
            $endgroup$
            – JV.Stalker
            Jan 26 at 6:35










          • $begingroup$
            @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
            $endgroup$
            – John
            Jan 26 at 16:04










          • $begingroup$
            @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
            $endgroup$
            – JV.Stalker
            Jan 27 at 5:15














          • 1




            $begingroup$
            This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
            $endgroup$
            – Claude Leibovici
            Jan 26 at 5:40






          • 1




            $begingroup$
            @Claude Leibovici, Thank you, and regards.
            $endgroup$
            – JV.Stalker
            Jan 26 at 6:35










          • $begingroup$
            @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
            $endgroup$
            – John
            Jan 26 at 16:04










          • $begingroup$
            @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
            $endgroup$
            – JV.Stalker
            Jan 27 at 5:15








          1




          1




          $begingroup$
          This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
          $endgroup$
          – Claude Leibovici
          Jan 26 at 5:40




          $begingroup$
          This is a nice solution, for sure. I did not work the problem but I wonder if, once the expression inside brackets has been fully simplified (this eliminates the summation over $k$), we could not do the same using the complete and incomplte gamma functions. By the way, $+1$. Cheers.
          $endgroup$
          – Claude Leibovici
          Jan 26 at 5:40




          1




          1




          $begingroup$
          @Claude Leibovici, Thank you, and regards.
          $endgroup$
          – JV.Stalker
          Jan 26 at 6:35




          $begingroup$
          @Claude Leibovici, Thank you, and regards.
          $endgroup$
          – JV.Stalker
          Jan 26 at 6:35












          $begingroup$
          @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
          $endgroup$
          – John
          Jan 26 at 16:04




          $begingroup$
          @JV.Stalker Thanks so much. Can you explain what you mean when you write "this integral can be solved by using the integral functions"? Also, I presume this cannot be expressed in terms of elementary functions, but is there an approximation that can? Thanks again.
          $endgroup$
          – John
          Jan 26 at 16:04












          $begingroup$
          @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
          $endgroup$
          – JV.Stalker
          Jan 27 at 5:15




          $begingroup$
          @John, Sorry I leave out the "exponential" expression from my sentence. Accordace with Wolfram Alpha the last integral equal to: $frac{e^{-lambda}}{z}big(Ei(lambda t)-Ei(lambda t(z-1))big)|_0^1=frac{e^{-lambda}}{z}Big(ln(1-z)+Gamma(0,lambda(z-1))-Gamma(0,lambda)Big)$ what was seen by Claude Leibovici in advance. The approximation could be a nother exercise, I start to learn this topic. Best regards,
          $endgroup$
          – JV.Stalker
          Jan 27 at 5:15


















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