Generalised Binomial Series











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I am reading an old paper "1971A Generalized Negative Binomial Distribution" by G. C. Jain and P. C. Consul. The paper mainly start from
$$(1+z)^n=sum_{x=0}^{infty}frac{n}{n+xbeta}binom{n+xbeta}{x}z^x(1+z)^{-beta x},quad where quad |frac{beta z}{1+z}|<1.$$



I can understand this. But I cannot understand why (the last two lines of the paper)
$$sum_{x=0}^{infty}frac{n}{n+beta x}binom{n+beta x}{x}=1.$$



Expecting some hints from you.



Thanks in advance.










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  • Does that old paper have an author?
    – Lord Shark the Unknown
    Nov 7 at 4:21










  • Yes, G. C. Jain and P. C. Consul.
    – gouwangzhangdong
    Nov 7 at 7:29






  • 2




    Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
    – Batominovski
    Nov 22 at 20:46












  • Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
    – Zvi
    Nov 23 at 17:57

















up vote
1
down vote

favorite












I am reading an old paper "1971A Generalized Negative Binomial Distribution" by G. C. Jain and P. C. Consul. The paper mainly start from
$$(1+z)^n=sum_{x=0}^{infty}frac{n}{n+xbeta}binom{n+xbeta}{x}z^x(1+z)^{-beta x},quad where quad |frac{beta z}{1+z}|<1.$$



I can understand this. But I cannot understand why (the last two lines of the paper)
$$sum_{x=0}^{infty}frac{n}{n+beta x}binom{n+beta x}{x}=1.$$



Expecting some hints from you.



Thanks in advance.










share|cite|improve this question
























  • Does that old paper have an author?
    – Lord Shark the Unknown
    Nov 7 at 4:21










  • Yes, G. C. Jain and P. C. Consul.
    – gouwangzhangdong
    Nov 7 at 7:29






  • 2




    Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
    – Batominovski
    Nov 22 at 20:46












  • Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
    – Zvi
    Nov 23 at 17:57















up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am reading an old paper "1971A Generalized Negative Binomial Distribution" by G. C. Jain and P. C. Consul. The paper mainly start from
$$(1+z)^n=sum_{x=0}^{infty}frac{n}{n+xbeta}binom{n+xbeta}{x}z^x(1+z)^{-beta x},quad where quad |frac{beta z}{1+z}|<1.$$



I can understand this. But I cannot understand why (the last two lines of the paper)
$$sum_{x=0}^{infty}frac{n}{n+beta x}binom{n+beta x}{x}=1.$$



Expecting some hints from you.



Thanks in advance.










share|cite|improve this question















I am reading an old paper "1971A Generalized Negative Binomial Distribution" by G. C. Jain and P. C. Consul. The paper mainly start from
$$(1+z)^n=sum_{x=0}^{infty}frac{n}{n+xbeta}binom{n+xbeta}{x}z^x(1+z)^{-beta x},quad where quad |frac{beta z}{1+z}|<1.$$



I can understand this. But I cannot understand why (the last two lines of the paper)
$$sum_{x=0}^{infty}frac{n}{n+beta x}binom{n+beta x}{x}=1.$$



Expecting some hints from you.



Thanks in advance.







summation negative-binomial






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edited Nov 7 at 7:29

























asked Nov 7 at 4:20









gouwangzhangdong

517




517












  • Does that old paper have an author?
    – Lord Shark the Unknown
    Nov 7 at 4:21










  • Yes, G. C. Jain and P. C. Consul.
    – gouwangzhangdong
    Nov 7 at 7:29






  • 2




    Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
    – Batominovski
    Nov 22 at 20:46












  • Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
    – Zvi
    Nov 23 at 17:57




















  • Does that old paper have an author?
    – Lord Shark the Unknown
    Nov 7 at 4:21










  • Yes, G. C. Jain and P. C. Consul.
    – gouwangzhangdong
    Nov 7 at 7:29






  • 2




    Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
    – Batominovski
    Nov 22 at 20:46












  • Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
    – Zvi
    Nov 23 at 17:57


















Does that old paper have an author?
– Lord Shark the Unknown
Nov 7 at 4:21




Does that old paper have an author?
– Lord Shark the Unknown
Nov 7 at 4:21












Yes, G. C. Jain and P. C. Consul.
– gouwangzhangdong
Nov 7 at 7:29




Yes, G. C. Jain and P. C. Consul.
– gouwangzhangdong
Nov 7 at 7:29




2




2




Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
– Batominovski
Nov 22 at 20:46






Is there any condition on $beta$? The last sum is not equal to $1$ when $n$ is a positive integer and $beta:=0$ (that is, the sum $sumlimits_{x=0}^n,dfrac{n}{n+beta,x},displaystyle binom{n+beta,x}{x}$ equals $2^n$ when $beta:=0$).
– Batominovski
Nov 22 at 20:46














Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
– Zvi
Nov 23 at 17:57






Which equation number as in the paper is the equation in your question? I see (2.2) is the first equation in your question, what about your second equation? This is the paper, right?
– Zvi
Nov 23 at 17:57












1 Answer
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I think this result provided by the paper is wrong. Simplify run the numerical calculation in any mathematical software will show this is not a probability distribution function at all.



The range of $beta$, $beta=0$ or $beta geq 1$, could be found in "1980The Generalized Negative Binomial Distribution and its Characterization by ZeroRegression". There are several subsequent papers pointing out the parameter setting in 1971 G. C. Jain and P. C. Consul is not correct.



Let me know if you have other comments.






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    up vote
    0
    down vote













    I think this result provided by the paper is wrong. Simplify run the numerical calculation in any mathematical software will show this is not a probability distribution function at all.



    The range of $beta$, $beta=0$ or $beta geq 1$, could be found in "1980The Generalized Negative Binomial Distribution and its Characterization by ZeroRegression". There are several subsequent papers pointing out the parameter setting in 1971 G. C. Jain and P. C. Consul is not correct.



    Let me know if you have other comments.






    share|cite|improve this answer

























      up vote
      0
      down vote













      I think this result provided by the paper is wrong. Simplify run the numerical calculation in any mathematical software will show this is not a probability distribution function at all.



      The range of $beta$, $beta=0$ or $beta geq 1$, could be found in "1980The Generalized Negative Binomial Distribution and its Characterization by ZeroRegression". There are several subsequent papers pointing out the parameter setting in 1971 G. C. Jain and P. C. Consul is not correct.



      Let me know if you have other comments.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I think this result provided by the paper is wrong. Simplify run the numerical calculation in any mathematical software will show this is not a probability distribution function at all.



        The range of $beta$, $beta=0$ or $beta geq 1$, could be found in "1980The Generalized Negative Binomial Distribution and its Characterization by ZeroRegression". There are several subsequent papers pointing out the parameter setting in 1971 G. C. Jain and P. C. Consul is not correct.



        Let me know if you have other comments.






        share|cite|improve this answer












        I think this result provided by the paper is wrong. Simplify run the numerical calculation in any mathematical software will show this is not a probability distribution function at all.



        The range of $beta$, $beta=0$ or $beta geq 1$, could be found in "1980The Generalized Negative Binomial Distribution and its Characterization by ZeroRegression". There are several subsequent papers pointing out the parameter setting in 1971 G. C. Jain and P. C. Consul is not correct.



        Let me know if you have other comments.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 23:28









        gouwangzhangdong

        517




        517






























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