Statistics 3 - Geometric and negative binomial distributions Flashcards
What is the geometric distribution?
For successive independent trials, each with a constant probability of success, π, the number of trials needed to get one success, π, has the geometric distribution (written as πβΌGeo(π)), with the probability function:
P(π=π₯)=p(π₯)=π(1-π)Λ£β»ΒΉ Where π₯=1,2,3,β¦
How is the cumulative geometric distribution calculated?
The values of p(π₯) form a geometric distribution with first term π and common ratio, (1-π). As such the cumulative geometric distribution can be derived by considering the sum of the terms of the geometric series.
β
Ξ£p(r)=π(1-(1-π)Λ£)/1-(1-π)Λ£=π(1-(1-π)Λ£)/π=1-(1-π)Λ£
Κ³βΌΒΉ
Thus if πβΌGeo(π), then the cumulative geometric distribution is given by:
P(πβ€x)=1-(1-π)Λ£ Where π₯=1,2,3,β¦
How is the mean of a geometric distribution calculated?
If πβΌGeo(π), then:
The mean of π=E(π)=ΞΌ=1/π
How is the variance of a geometric distribution calculated?
If πβΌGeo(π), then:
The variance of π=Var(π)=ΟΒ²=1-π/πΒ²
What is the negative binomial distribution?
The binomial distribution, πβΌB(π,π), models the number of successes in a fixed number of trials, π. The probability of success in each trial, π, is constant and the trials are independent.
The negative binomial distribution, often written as πβΌNB(π,π) or πβΌNegative B(π,π), models the number of trials needed to achieve a fixed number of successes, π.
For successive trials, each with the constant probability of success, π, the number of trials needed to get π successes, π, has the negative binomial distribution with probability function:
P(π=π₯)=p(π₯)=(Λ£β»ΒΉα΅£ββ)πΚ³(1-π)Λ£β»Κ³ π₯=π,π+1,π+2,β¦
This is the probability of π-1 successes in π₯-1 trials multiplied by the probability of success in the π₯th trial
How is the mean of a negative binomial distribution calculated?
If πβΌNB(π,π), then:
The mean of π=E(π)=ΞΌ=π/π
How is the variance of a negative binomial distribution calculated?
If πβΌNB(π,π), then:
The variance of π=Var(π)=ΟΒ²=π(1-π)/πΒ²