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Parameter space for the negative binomial distribution

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data
    For the negative binomial distribution, with r known, describe the natural parameter space


    2. Relevant equations
    the pmf for the negative binomial distribution with parameters r and p can be
    1) [itex]P(X=x|r,p)= \binom {x-1}{r-1}p^{r}(1-p)^{x-r} [/itex] where [itex]x=r,r+1,... [/itex], or
    2) [itex]P(Y=y|r,p)= \binom {y+r-1}{y}p^{r}(1-p)^{y} [/itex] where [itex]y=0,1,... [/itex].

    A distribution, like the one above where r is known, is a member of the exponential family of distributions. An exponential distribution is one that can be expressed as...

    [itex]h(x)c^{*}(\eta) exp(\sum_{i=1}^{k} \eta_i t_i(x)) [/itex]

    The parameter space are the values of [itex]\eta [/itex] such that [itex]\sum_A h(x) exp(\sum_{i=1}^{k} \eta_i t_i(x)) < \infty [/itex] where [itex]A [/itex] is the support of the pmf.
    3. The attempt at a solution
    Rewriting the 2nd pmf for the negative binomial distribution, as an exponential distribution, I have

    [itex]h(y) = \binom {y+r-1}{y} [/itex], [itex]c(p) = p^{r} \cdot I_(0,1)(p) [/itex], [itex]t_1(y)=y [/itex], and [itex]w_1(p) = ln(1-p) [/itex].

    Then I let [itex]\eta = w_1(p) [/itex], and find the values for [itex]\eta [/itex] where the sum converges.

    I have [itex]\sum_{y=0}^{\infty} \binom{r+y-1}{y}(e^{\eta})^{y} [/itex], and I don't recognize this sum as anything that converges.

    Any help would be appreciated.
     
  2. jcsd
  3. Aug 13, 2014 #2

    Ray Vickson

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    [tex]\frac{1}{(1-z)^r} \equiv (1-z)^{-r} = \sum_{k=0}^{\infty} \binom{r+k-1}{k} z^k.[/tex]
    The reason for the name "negative binomial distribution" is that it comes from "negative binomial coefficients" ##\binom{-r}{k} = (-1)^k \binom{r+k-1}{k}## associated with the "negative binomial" series ##(1-z)^{-r}##. You ought to be able to show that the series converges for ##|z| < 1## because it just generalizes the series for ##1/(1-z)##.
     
  4. Aug 13, 2014 #3
    That look's like a winner.

    I had thought of using the 1st pmf for the negative binomial distribution. Then I'd have the exponential distribution as:

    [itex]h(x) = \binom {x-1}{r-1} \cdot I_{(r,r+1,...)}(x) [/itex], [itex] c(p) = p^{r} \cdot I_{(0,1)} [/itex], [itex]t_1(x) = x-r [/itex], and [itex]w_1(p) = ln(1-p) [/itex].

    Then letting [itex]w_1(p) = \eta [/itex] and solving the series for [itex]\eta [/itex]:

    [itex]\sum_X \binom {x-r}{r-1} e^{(x-r)\eta} = \sum_X \binom {x-r}{r-1} (e^{\eta})^{x-r} [/itex],

    then I recognized this series: [itex] \sum_{x=r}^{\infty} \binom {x-1}{r-1}w^{x-r} = (1-w)^{r} [/itex].

    So I have [itex] (1-e^{\eta})^r [/itex].

    I'm not sure, but I think the constraint is that [itex] |e^{\eta}| < 1 [/itex].

    Is this right?
     
  5. Aug 13, 2014 #4

    Ray Vickson

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    I cannot make any sense out of what you write above.

    Your previous post had a simple, explicit question, and I answered it. If you put ##z = e^{\eta}## you will get the summation you had in your first post, and I already pointed out what restrictions apply to ##z##.
     
  6. Aug 13, 2014 #5
    Thanks for your first reply. I thanked you for it.

    Maybe my reply was lengthy and ambiguous. I asked you about the constraints for w in this series,

    [itex]\sum_{x=r}^{\infty} \binom {x-1}{r-1} w^{x-r} = (1-w)^{r} [/itex], which I found at this webpage.

    I think it's the same series you gave, with a change in the index, letting [itex]x=y+r [/itex], I have

    [itex] \sum_{y=0}^{\infty} \binom {r+y-1}{y} z^{y} = \sum_{x=r}^{\infty} \binom {x-1}{x-r}z^{x-r} = \sum_{x=r}^{\infty} \binom {x-1}{r-1}z^{x-r} [/itex]

    My question is this: Is the webpage wrong?
     
  7. Aug 13, 2014 #6
    Looking at the negative binomial distribution as a particular case of the general binomial series, I see the formula on the webpage was wrong.

    Note to self: do not trust all websites.
     
    Last edited: Aug 13, 2014
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