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Negative Binomial Question

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  • Negative Binomial Question

    Does anyone know what the expected value of the negative binomial distribution is? Originally I thought it was: r(1-p)/p but then I was reading "A first year in prob" by Sheldon and it said that the expected value was r/p. That made more sense because the negative is just like the geometric and the expected value of the geometric is 1/p. B/c the trials are all independent, just r/p made sense. But then I noticed several other books had r(1-p)/p. So I began to think maybe it was a typo in Sheldon. But then I saw there was a proof that it was r/p in sheldon. But then I found a website on the internet that also had a proof that it was equal to r(1-p)/p. They both looked legitimate. How can this be? Which is the real expected value of the negative binomial? Thanks, John
    Actuary.com - www.actuary.com
    Email - [email protected]

  • #2
    yah it is....nq/p or u can say r(1-p)/p
    Actuary.com - www.actuary.com
    Email - [email protected]

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    • #3
      Negative binomial

      Negative binomial distribution either counts how many times you try until you succeed r times, or how many times you fail until you succeed r times. Two possible approaches to basically the same process, but the resulting means are different.

      Yours,
      Krzys'
      Want to know how to pass actuarial exams? Go to: smartURL.it/pass

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      • #4
        Originally posted by krzysio
        Negative binomial distribution either counts how many times you try until you succeed r times, or how many times you fail until you succeed r times. Two possible approaches to basically the same process, but the resulting means are different.

        Yours,
        Krzys'
        Yup, I find this rather confusing when I fisrt looked it too.
        For the first case you mention :
        x is the total times you try until you succeed r times, the expected value will be r/p, (note, here the possible value for x is r, r+1, r+2 ... )
        the second case:
        x is the times you fail until you succeed r times, the expected value will be r(1-p)/p
        (note, here the possible value for x is 0, 1, 2,...)

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        • #5
          How about the mean of the geometric distribution?

          Hi,
          I read in two sources that the mean of Geometric Distribution is the reciprocal of the success rate, e.g. 1/p.
          But I also read in some solution that the the mean as 1-p/p.
          Which one is right?

          Thanks for help.
          My name is Yuchin.

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          • #6
            Originally posted by leerobb1966 View Post
            Hi,
            I read in two sources that the mean of Geometric Distribution is the reciprocal of the success rate, e.g. 1/p.
            But I also read in some solution that the the mean as 1-p/p.
            Which one is right?

            Thanks for help.
            You will benefit from reading this: http://en.wikipedia.org/wiki/Geometric_distribution and also using parantheses where appropriate. "1-p/p" means 1-(p/p) = 1-1 = 0", you obv. meant (1-p)/p

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            • #7
              Originally posted by NoMoreExams View Post
              You will benefit from reading this: http://en.wikipedia.org/wiki/Geometric_distribution and also using parantheses where appropriate. "1-p/p" means 1-(p/p) = 1-1 = 0", you obv. meant (1-p)/p
              Thank you so much!
              and I will be more careful with expressions.:tongue:
              My name is Yuchin.

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