(statistics) A discrete probability distribution whose probability function is given by the equation P(x) = p (1 - p)x - 1 for x any positive integer, p(x) = 0 otherwise, when 0 ≤ p ≤ 1; the mean is 1/p.
| Sci-Tech Dictionary: geometric distribution |
(statistics) A discrete probability distribution whose probability function is given by the equation P(x) = p (1 - p)x - 1 for x any positive integer, p(x) = 0 otherwise, when 0 ≤ p ≤ 1; the mean is 1/p.
| 5min Related Video: Geometric distribution |
| Wikipedia: 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.
| Probability mass function |
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| Cumulative distribution function |
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| Parameters | success probability (real) |
success probability (real) |
|---|---|---|
| Support | ![]() |
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| Probability mass function (pmf) | ![]() |
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| Cumulative distribution function (cdf) | ![]() |
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| Mean | ![]() |
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| Median | (not unique if − log(2) / log(1 − p) is an integer) |
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| Mode | 1 | 0 |
| Variance | ![]() |
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| Skewness | ![]() |
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| Excess kurtosis | ![]() |
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| Entropy | ![]() |
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| Moment-generating function (mgf) | ![]() |
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| Characteristic function | ![]() |
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These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the former one (distribution of the number X); however, to avoid ambiguity, it is considered wise to indicate which is intended, by mentioning the range explicitly.
If the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is

for k = 1, 2, 3, ....
Equivalently, if the probability of success on each trial is p, then the probability that there are k failures before the first success is

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 { 1, 2, 3, ... } and is a geometric distribution with p = 1/6.
Contents |
The expected value of a geometrically distributed random variable X is 1/p and the variance is (1 − p)/p2:

Similarly, the expected value of the geometrically distributed random variable Y is (1 − p)/p, and its variance is (1 − p)/p2:

Let μ = (1 − p)/p be the expected value of Y. Then the cumulants κn of the probability distribution of Y satisfy the recursion

Outline of proof: That the expected value is (1 − p)/p can be shown (why summation and integration is exchanged is never shown) in the following way. Let Y be as above. Then
 \\
& {} =-p(1-p)\frac{d}{dp}\frac{1}{p}=\frac{1-p}{p}.
\end{align}](http://wpcontent.answers.com/math/2/b/0/2b083d465f3697a4440e45af9083edbb.png)
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
be a sample where
for
. 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(α, β) prior, then the posterior distribution is

The posterior mean E[p] approaches the maximum likelihood estimate
as α and β approach zero.
In the alternative case, let
be a sample where
for
. Then p can be estimated as

The posterior distribution of p given a Beta(α, β) prior is

Again the posterior mean E[p] approaches the maximum likelihood estimate
as α and β approach zero.





| 1 − | ∏ | (1 − p(m)). |
| m |

(where
is the floor (or greatest integer) function)
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