In a sequence of independent Bernoulli trials, the number of trials needed toΒ obtain πΒ successes has Negative Binomial (Pascal) Distribution
Negative Binomial Distribution has two parameters:
- π = number of needed successes
- π = probability of success
With π=1, it becomes a Geometric Distribution
A Negative Binomial Variable πΏ can be represented as a sum of π independent Geometric Variables (π1+ β¦ + ππ):
- πΏ = π1+ β¦ + ππ
Probability Mass Function
- π(π=π₯) = π{ the π₯π‘β Bernoulli trial results in the ππ‘βΒ success }
- π(π=π₯) = π{ (π-1) successes in the first (π₯-1) Bernoulli trials AND the last Bernoulli trial is success }
- π(π=π₯) =Β π{ (π-1) successes in the first (π₯-1) trials }Β π{ the last trial is success }
- π(π=π₯) = π(π=π-1; π,π=π₯-1) π(π=π₯) # where:
- π(π=π-1; π,π=π₯-1) - Binomial Distribution
- π(π=π₯) - Bernoulli Distribution
- π(π=π₯) =Β [(π₯-1) choose (π-1)]Β (1-π)π₯-πππ-1π
- π(π=π₯) = [(π₯-1) choose (π-1)] (1-π)π₯-πππ
- π(π=π₯) = [(π₯-1)!/[(π-1)!(π₯-π)!]] (1-π)π₯-πππ
Expectation
π[πΏ] = π/π
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the number of trials until the ππ‘β success is the summation of the following:
- Geometric number of trials π1until the first success
- Geometric number of trials π2until the next success
- β¦
- Geometric number of trials ππΒ until the next success
π[πΏ] = π[π1Β +Β β¦Β + ππ] = π/π
Variance
πππ(πΏ)Β = π(1 - π) / π2
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- πππ(πΏ) = πππ(π1 + β¦ + ππ) # πΏ is a summation of geometric random variables
- πππ(πΏ) = πππ(π1) + β¦ + πππ(ππ) # see variance properties:Β πππ(πΒ + π) =Β πππ(π)Β +Β πππ(π)Β +Β 2πΆππ£(π,π), covariance = 0
- πππ(πΏ) = [(1-π) / π2] + β¦ + [(1-π) / π2]
- πππ(πΏ) = π(1 - π) / π2
Subpages
- Negative Binomial Distribution vs Binomial Distribution
- Geometric Distribution vs Negative Binomial Distribution
TODO
Negative Binomial Distribution #1
- the number of trials need to obtain k successes
- is the sum of k i.i.d. geometric distributions
- P(X=x) = (x-1 choose k-1) p^k (1-p)
- E[X] Β = k/p
- Var[X] = k(1-p) / p
- M_X(t) = [ pe^t / (1 - qe^t) ]
Negative Binomial Distribution #2
- the number of failures before k successes
- P(X=x) = (x+k-1 choose x) (1-p)^x p
- E[X] Β = k(1-p)/p
- Var[X] = k(1-p)/p
- M_X(t) = [ p / (1 - qe^t) ]