Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discrete prob. dist.

  1. Oct 28, 2009 #1
    how do i interpret this probability distribution:

    [tex]\sum_{k=r}^\infty \binom{k}{r}p^k(1-p)^{k-r}[/tex]

    where r is the number of successes, p is the probability, k trials.

    by looking at it, it seems like it's similar to a negative binomial distribution once you pull out a k/r. if you do some math after pulling out the k/r, it seems like it is the expected value of a geometric distribution. is this distribution saying that a negative binomial divided by the number of successes r means there is only one success, which is geometric?
  2. jcsd
  3. Oct 28, 2009 #2
    in other words, what does this sum equal?
  4. Oct 29, 2009 #3
    Do you mean to ask "what is a probabilistic interpretation of this summation formula?" - in which case there would be lots of different ways to write it as an expected value E[f(X)] = sum(f(k)*Prob[X=k]). One way you found was with negative binomial X and f(X)=X/r; the other way was E[g(Y)] for some function g where Y is geometric.

    However it sounds like you want to use E[f(X)]=E[g(Y)] to draw conclusions about how X and Y are related, but that's just not possible.

    Does that help?
  5. Oct 29, 2009 #4
    makes sense. thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook