1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interpret this probability distribution

  1. Oct 28, 2009 #1
    i took an exam today and was sorta stumped by this question. pls take a look, thx!

    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
    as in, what does this distribution equal?
  4. Oct 28, 2009 #3
    It's just summing up binomial random variables. We're summing over k - the number of trials. r is fixed so we are summing the probabilities that we have r successes for k = r,..., infinity trials.

    Hmm, let's take an analytic stab.

    Using the binomial theorem,

    [tex](p + (1 - p))^{k} = \sum_{i=0}^k \binom{k}{i}p^{k}(1-p)^{k-i}[/tex]


    [tex]1 = 1^{\infty} = (p + (1 - p))^{\infty} = \sum_{i=0}^\infty \binom{k}{i}p^{k}(1-p)^{k-i} = \sum_{i=0}^{r-1} \binom{k}{i}p^{k}(1-p)^{k-i} + \sum_{i=r}^\infty \binom{k}{i}p^{k}(1-p)^{k-i}[/tex]

    finally we get

    [tex]\sum_{i=r}^\infty \binom{k}{i}p^{k}(1-p)^{k-i} = 1 -\sum_{i=0}^{r-1} \binom{k}{i}p^{k}(1-p)^{k-i} = 1 - P({X < r})[/tex]

    Where [tex]X[/tex] is a binomial random variable with parameters k, p.
    Last edited: Oct 28, 2009
  5. Oct 28, 2009 #4
    i understand the first part of your explanation. where did the "Hmm" part come from? is this an alternative explanation?
  6. Oct 28, 2009 #5
    Yeah sorry I was wasn't finished editing it. Its confusing no doubt though :smile:. I guess it's the complement that we will have less than r successes in k trials, but i'm not sure.
  7. Oct 28, 2009 #6
    hm...i thought your first answer made sense. that this is equal to simply 1.
  8. Oct 28, 2009 #7
    Yeah i thought that made sense too but "doing the math" gave a different answer.

    Think about the same scenario again (a fair coin, look for 8 successes, k trials). Sum up the probabilities that we get 8 successes after k flips from k = 8 to infinity flips. Should this sum be 1? I have no idea so I'll resort to computation:

    8 trials:
    [tex]\sum_{i = 8}^8 \binom{8}{i}\left(\frac{1}{2}\right)^{8} = 0.003906[/tex]

    9 trials:
    [tex]\sum_{i = 8}^9 \binom{9}{i}\left(\frac{1}{2}\right)^{9} = 0.019531[/tex]

    20 trials:
    [tex]\sum_{i = 8}^{20} \binom{20}{i}\left(\frac{1}{2}\right)^{20} = 0.868412[/tex]

    50 trials:
    Sum is .999999

    Turns out it does come out to 1 and I was initially right.
  9. Oct 28, 2009 #8
    Well my math up there is right too. In the special case where r = k (the number of trials). P(X < r) is P(we have r successes in less than r trials) = 0. So the math gives the same result of 1. Hope that helped!
  10. Oct 28, 2009 #9
    let me sum it up analytically:

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

    where r = 0,1,2...n
  11. Oct 28, 2009 #10
    the first step is a direct result of the binomial theorem.
  12. Oct 29, 2009 #11
    Not exactly. You did not use the binomial theorem correctly. The summation must start at 0. Also n = [tex]\infty[/tex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Interpret this probability distribution
  1. Probability Distribution (Replies: 24)

  2. Probability Distribution (Replies: 13)