Geometric distribution


In probability theory and statistics, the geometric distribution is either of two discrete probability distributions:
Which of these one calls "the" geometric distribution is a matter of convention and convenience.
These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the former one ; however, to avoid ambiguity, it is considered wise to indicate which is intended, by mentioning the support explicitly.
The geometric distribution gives the probability that the first occurrence of success requires k independent trials, each with success probability p. If the probability of success on each trial is p, then the probability that the kth trial is the first success is
for k = 1, 2, 3,....
The above form of the geometric distribution is used for modeling the number of trials up to and including the first success. By contrast, the following form of the geometric distribution is used for modeling the number of failures until the first success:
for k = 0, 1, 2, 3, ....
In either case, the sequence of probabilities is a geometric sequence.
For example, suppose an ordinary die is thrown repeatedly until the first time a "1" appears. The probability distribution of the number of times it is thrown is supported on the infinite set and is a geometric distribution with p = 1/6.
The geometric distribution is denoted by Geo where 0 < p ≤ 1.

Definitions

Consider a sequence of trials, where each trial has only two possible outcomes. The probability of success is assumed to be the same for each trial. In such a sequence of trials, the geometric distribution is useful to model the number of failures before the first success. The distribution gives the probability that there are zero failures before the first success, one failure before the first success, two failures before the first success, and so on.

Assumptions: When is the geometric distribution an appropriate model?

The geometric distribution is an appropriate model if the following assumptions are true.
If these conditions are true, then the geometric random variable Y is the count of the number of failures before the first success. The possible number of failures before the first success is 0, 1, 2, 3, and so on. In the graphs above, this formulation is shown on the right.
An alternative formulation is that the geometric random variable X is the total number of trials up to and including the first success, and the number of failures is X − 1. In the graphs above, this formulation is shown on the left.

Probability Outcomes Examples

The general formula to calculate the probability of k failures before the first success, where the probability of success is p and the probability of failure is q = 1 − p, is
for k = 0, 1, 2, 3,....
E1) A doctor is seeking an anti-depressant for a newly diagnosed patient. Suppose that, of the available anti-depressant drugs, the probability that any particular drug will be effective for a particular patient is p = 0.6. What is the probability that the first drug found to be effective for this patient is the first drug tried, the second drug tried, and so on? What is the expected number of drugs that will be tried to find one that is effective?
The probability that the first drug works. There are zero failures before the first success. Y = 0 failures. The probability P is simply the probability that the first drug works.
The probability that the first drug fails, but the second drug works. There is one failure before the first success. Y= 1 failure. The probability for this sequence of events is P p which is given by
The probability that the first drug fails, the second drug fails, but the third drug works. There are two failures before the first success. Y = 2 failures. The probability for this sequence of events is P p P
E2) A newlywed couple plans to have children, and will continue until the first girl. What is the probability that there are zero boys before the first girl, one boy before the first girl, two boys before the first girl, and so on?
The probability of having a girl is p= 0.5 and the probability of having a boy is q = 1 − p = 0.5.
The probability of no boys before the first girl is
The probability of one boy before the first girl is
The probability of two boys before the first girl is
and so on.

Properties

Moments and cumulants

The expected value for the number of independent trials to get the first success, of a geometrically distributed random variable X is 1/p and the variance is /p2:
Similarly, the expected value of the geometrically distributed random variable Y = X − 1 is q/p = /p, and its variance is /p2:
Let μ = /p be the expected value of Y. Then the cumulants of the probability distribution of Y satisfy the recursion
Outline of proof: That the expected value is /p can be shown in the following way. Let Y be as above. Then

Expected Value Examples

E3) A patient is waiting for a suitable matching kidney donor for a transplant. If the probability that a randomly selected donor is a suitable match is p=0.1, what is the expected number of donors who will be tested before a matching donor is found?
With p = 0.1, the mean number of failures before the first success is E = /p =/0.1 = 9.
For the alternative formulation, where X is the number of trials up to and including the first success, the expected value is E = 1/p = 1/0.1 = 10.
For example 1 above, with p = 0.6, the mean number of failures before the first success is E = /p = /0.6 = 0.67.

General properties

Parameter estimation

For both variants of the geometric distribution, the parameter p can be estimated by equating the expected value with the sample mean. This is the method of moments, which in this case happens to yield maximum likelihood estimates of p.
Specifically, for the first variant let k = k1, ..., kn be a sample where ki ≥ 1 for i = 1, ..., n. Then p can be estimated as
In Bayesian inference, the Beta distribution is the conjugate prior distribution for the parameter p. If this parameter is given a Beta prior, then the posterior distribution is
The posterior mean E approaches the maximum likelihood estimate as α and β approach zero.
In the alternative case, let k1, ..., kn be a sample where ki ≥ 0 for i = 1, ..., n. Then p can be estimated as
The posterior distribution of p given a Beta prior is
Again the posterior mean E approaches the maximum likelihood estimate as α and β approach zero.
For either estimate of using Maximum Likelihood, the bias is equal to
which yields the bias-corrected maximum likelihood estimator

Computational methods

Geometric distribution using R

The R function dgeom calculates the probability that there are k failures before the first success, where the argument "prob" is the probability of success on each trial.
For example,
dgeom = 0.6
dgeom = 0.24
R uses the convention that k is the number of failures, so that the number of trials up to and including the first success is k + 1.
The following R code creates a graph of the geometric distribution from Y = 0 to 10, with p = 0.6.

Y=0:10
plot, type="h", ylim=c, main="Geometric distribution for p=0.6", ylab="P

Geometric distribution using Excel

The geometric distribution, for the number of failures before the first success, is a special case of the negative binomial distribution, for the number of failures before s successes.
The Excel function NEGBINOMDIST calculates the probability of k = number_f failures before s = number_s successes where p = probability_s is the probability of success on each trial. For the geometric distribution, let number_s = 1 success.
For example,
=NEGBINOMDIST = 0.6
=NEGBINOMDIST = 0.24
Like R, Excel uses the convention that k is the number of failures, so that the number of trials up to and including the first success is k + 1.