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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]

    so

    [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].
     
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