Geometric distribution

The geometric distribution is an appropriate model if the following assumptions are true.[3]

• The phenomenon being modelled is a sequence of independent trials.
• There are only two possible outcomes for each trial, often designated success or failure.
• The probability of success, p, is the same for every trial.

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.

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

${\displaystyle \Pr(Y=k)=q^{k}\,p.}$

for k = 0, 1, 2, 3, ...

E1) A doctor is seeking an antidepressant 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 Pr(zero failures before first success) is simply the probability that the first drug works.

${\displaystyle \Pr(Y=0)=q^{0}\,p\ =0.4^{0}\times 0.6=1\times 0.6=0.6.}$

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 Pr(first drug fails) ${\displaystyle \times }$ p(second drug succeeds), which is given by

${\displaystyle \Pr(Y=1)=q^{1}\,p\ =0.4^{1}\times 0.6=0.4\times 0.6=0.24.}$

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 Pr(first drug fails) ${\displaystyle \times }$ p(second drug fails) ${\displaystyle \times }$ Pr(third drug is success)

${\displaystyle \Pr(Y=2)=q^{2}\,p,=0.4^{2}\times 0.6=0.096.}$

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 (success) is p= 0.5 and the probability of having a boy (failure) is q = 1 − p = 0.5.

The probability of no boys before the first girl is

${\displaystyle \Pr(Y=0)=q^{0}\,p\ =0.5^{0}\times 0.5=1\times 0.5=0.5.}$

The probability of one boy before the first girl is

${\displaystyle \Pr(Y=1)=q^{1}\,p\ =0.5^{1}\times 0.5=0.5\times 0.5=0.25.}$

The probability of two boys before the first girl is

${\displaystyle \Pr(Y=2)=q^{2}\,p\ =0.5^{2}\times 0.5=0.125.}$

and so on.

The expected value for the number of independent trials to get the first success, and the variance of a geometrically distributed random variable X is:

${\displaystyle \operatorname {E} (X)={\frac {1}{p}},\qquad \operatorname {var} (X)={\frac {1-p}{p^{2}}}.}$

Similarly, the expected value and variance of the geometrically distributed random variable Y = X - 1 (See definition of distribution ${\displaystyle \Pr(Y=k)}$) is:

${\displaystyle \operatorname {E} (Y)=\operatorname {E} (X-1)=\operatorname {E} (X)-1={\frac {1-p}{p}},\qquad \operatorname {var} (Y)={\frac {1-p}{p^{2}}}.}$

Consider the expected value ${\displaystyle \mathrm {E} (X)}$ of X as above, i.e. the average number of trials until a success. On the first trial we either succeed with probability ${\displaystyle p}$, or we fail with probability ${\displaystyle 1-p}$. If we fail the remaining mean number of trials until a success is identical to the original mean. This follows from the fact that all trials are independent. From this we get the formula:

${\displaystyle \mathrm {E} (X)=p\cdot 1+(1-p)\cdot (1+\mathrm {E} (X)),}$

which if solved for ${\displaystyle \mathrm {E} (X)}$ gives:

${\displaystyle \mathrm {E} (X)={\frac {1}{p}}.}$

That the expected value of Y as above is (1 − p)/p can be shown in the following way:

${\displaystyle {\begin{aligned}\mathrm {E} (Y)&{}=\sum _{k=0}^{\infty }(1-p)^{k}p\cdot k\\&{}=p\sum _{k=0}^{\infty }(1-p)^{k}k\\&{}=p(1-p)\sum _{k=0}^{\infty }(1-p)^{k-1}\cdot k\\&{}=p(1-p)\left[{\frac {d}{dp}}\left(-\sum _{k=0}^{\infty }(1-p)^{k}\right)\right]\\&{}=p(1-p){\frac {d}{dp}}\left(-{\frac {1}{p}}\right)={\frac {1-p}{p}}.\end{aligned}}}$

The interchange of summation and differentiation is justified by the fact that convergent power series converge uniformly on compact subsets of the set of points where they converge.

Let μ = (1 − p)/p be the expected value of Y. Then the cumulants ${\displaystyle \kappa _{n}}$ of the probability distribution of Y satisfy the recursion

${\displaystyle \kappa _{n+1}=\mu (\mu +1){\frac {d\kappa _{n}}{d\mu }}.}$

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(Y) = (1 − p)/p =(1 − 0.1)/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(X) = 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(Y) = (1 − p)/p = (1 − 0.6)/0.6 = 0.67.

The moments for the number of failures before the first success are given by

${\displaystyle {\begin{aligned}\mathrm {E} (Y^{n})&{}=\sum _{k=0}^{\infty }(1-p)^{k}p\cdot k^{n}\\&{}=p\operatorname {Li} _{-n}(1-p)&({\text{for }}n\neq 0)\end{aligned}}}$

where ${\displaystyle \operatorname {Li} _{-n}(1-p)}$ is the polylogarithm function.

• The probability-generating functions of X and Y are, respectively,

${\displaystyle {\begin{aligned}G_{X}(s)&={\frac {s\,p}{1-s\,(1-p)}},\\[10pt]G_{Y}(s)&={\frac {p}{1-s\,(1-p)}},\quad |s|<(1-p)^{-1}.\end{aligned}}}$

• Like its continuous analogue (the exponential distribution), the geometric distribution is memoryless. That is, the following holds for every m and n.

${\displaystyle \Pr\{X>m+n|X>n\}=\Pr\{X>m\}}$

The geometric distribution supported on {1, 2, 3, ... } is the only memoryless discrete distribution. Note that the geometric distribution supported on {0, 1, 2, ... } is not memoryless.

• Among all discrete probability distributions supported on {1, 2, 3, ... } with given expected value μ, the geometric distribution X with parameter p = 1/μ is the one with the largest entropy.[4]
• The geometric distribution of the number Y of failures before the first success is infinitely divisible, i.e., for any positive integer n, there exist independent identically distributed random variables Y₁, ..., Y_n whose sum has the same distribution that Y has. These will not be geometrically distributed unless n = 1; they follow a negative binomial distribution.
• The decimal digits of the geometrically distributed random variable Y are a sequence of independent (and not identically distributed) random variables. For example, the hundreds digit D has this probability distribution:

${\displaystyle \Pr(D=d)={q^{100d} \over 1+q^{100}+q^{200}+\cdots +q^{900}},}$

where q = 1 − p, and similarly for the other digits, and, more generally, similarly for numeral systems with other bases than 10. When the base is 2, this shows that a geometrically distributed random variable can be written as a sum of independent random variables whose probability distributions are indecomposable.

• Golomb coding is the optimal prefix code for the geometric discrete distribution.[5]
• The sum of two independent Geo(p) distributed random variables is not a geometric distribution. [1]

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.[9][10]

Specifically, for the first variant let k = k₁, ..., k_n be a sample where k_i ≥ 1 for i = 1, ..., n. Then p can be estimated as

${\displaystyle {\widehat {p}}=\left({\frac {1}{n}}\sum _{i=1}^{n}k_{i}\right)^{-1}={\frac {n}{\sum _{i=1}^{n}k_{i}}}.\!}$

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

${\displaystyle p\sim \mathrm {Beta} \left(\alpha +n,\ \beta +\sum _{i=1}^{n}(k_{i}-1)\right).\!}$

The posterior mean E[p] approaches the maximum likelihood estimate ${\displaystyle {\widehat {p}}}$ as α and β approach zero.

In the alternative case, let k₁, ..., k_n be a sample where k_i ≥ 0 for i = 1, ..., n. Then p can be estimated as

${\displaystyle {\widehat {p}}=\left(1+{\frac {1}{n}}\sum _{i=1}^{n}k_{i}\right)^{-1}={\frac {n}{\sum _{i=1}^{n}k_{i}+n}}.\!}$

The posterior distribution of p given a Beta(α, β) prior is[11]

${\displaystyle p\sim \mathrm {Beta} \left(\alpha +n,\ \beta +\sum _{i=1}^{n}k_{i}\right).\!}$

Again the posterior mean E[p] approaches the maximum likelihood estimate ${\displaystyle {\widehat {p}}}$ as α and β approach zero.

For either estimate of ${\displaystyle {\widehat {p}}}$ using Maximum Likelihood, the bias is equal to

${\displaystyle b\equiv \operatorname {E} {\bigg [}\;({\hat {p}}_{\mathrm {mle} }-p)\;{\bigg ]}={\frac {p\,(1-p)}{n}}}$

which yields the bias-corrected maximum likelihood estimator

${\displaystyle {\hat {p\,}}_{\text{mle}}^{*}={\hat {p\,}}_{\text{mle}}-{\hat {b\,}}}$

The R function dgeom(k, prob) 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(0,0.6) = 0.6

dgeom(1,0.6) = 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(Y, dgeom(Y,0.6), type="h", ylim=c(0,1), main="Geometric distribution for p=0.6", ylab="Pr(Y=Y)", xlab="Y=Number of failures before first success")

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(number_f, number_s, probability_s) 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.[12]

For example,

=NEGBINOMDIST(0, 1, 0.6) = 0.6

=NEGBINOMDIST(1, 1, 0.6) = 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.