Geometric distribution

Geometric distribution
Probability distribution Probability function of the geometric distribution (variant B) for (blue), (green) and (red) ${\displaystyle p=0.2}$ ${\displaystyle p=0.5}$ ${\displaystyle p=0.8}$
Distribution function
parameter	p ∈ (0,1) - single success probability
Expected value	${\displaystyle {\frac {1}{p}}}$(A) or (B) ${\displaystyle {\frac {1-p}{p}}}$
Variance	${\displaystyle {\frac {1-p}{p^{2}}}}$
Crookedness	${\displaystyle {\frac {2-p}{\sqrt {1-p}}}}$
Bulge	${\displaystyle 9+{\frac {p^{2}}{1-p}}}$

The geometric distribution is a probability distribution in stochastics that is univariate and belongs to the discrete probability distributions . It is derived from independent Bernoulli experiments and defined in two variants:

option A

the probability distribution of the number of Bernoulli attempts necessary to be successful. This distribution is defined on the set . ${\displaystyle X}$ ${\displaystyle \mathbb {N} }$

Variant B

the probability distribution of the number of failed attempts before the first success. This distribution is defined on the set . ${\displaystyle Y}$ ${\displaystyle \mathbb {N} _{0}}$

The two variants are related . Which of these is called “geometric distribution” is either determined in advance or the one that is more appropriate is chosen. ${\displaystyle X=Y+1}$

The geometric distribution is used:

• in the analysis of the waiting times until a certain event occurs.
  • in determining the service life of devices and components, d. H. waiting until the first failure
• when determining the number of frequent events between consecutive rare events such as errors:
  • Determination of the reliability of devices ( MTBF )
  • Determination of Risk in Actuarial Science
  • Determination of the error rate in the data transmission, for example the number of successfully transmitted TCP packets between two packets with retransmission

Definition of the geometric distribution

A discrete random variable or with the parameter (probability of success), (probability of failure) is sufficient for the geometric distribution if: ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle p}$ ${\displaystyle q=1-p}$ ${\displaystyle G(p)}$

option A

The following applies to the probability that one needs exactly attempts in order to achieve the first success ${\displaystyle n}$

${\displaystyle \operatorname {P} (X=n)=p(1-p)^{n-1}=pq^{n-1}\quad (n=1,2,\dotsc )}$

Variant B

The following applies to the probability of having failed attempts before the first success ${\displaystyle n}$

${\displaystyle \operatorname {P} (Y=n)=p(1-p)^{n}=pq^{n}\quad (n=0,1,2,\dotsc )}$

In both cases the values ​​for the probabilities form a geometric sequence .

The geometric distribution thus has the following distribution functions

option A

${\displaystyle F(n)=\operatorname {P} (X\leq n)=p\sum _{i=1}^{n}q^{i-1}=p\sum _{i=0}^{n-1}q^{i}=p{\frac {1-q^{n}}{1-q}}=1-q^{n}=1-(1-p)^{n}}$

Variant B

${\displaystyle F(n)=\operatorname {P} (Y\leq n)=p\sum _{i=0}^{n}q^{i}=p{\frac {1-q^{n+1}}{1-q}}=1-q^{n+1}=1-(1-p)^{n+1}}$

properties

Expected value

The expected values ​​of the two geometric distributions are

option A

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

Variant B

${\displaystyle \operatorname {E} (Y)=\operatorname {E} (X)-1={\frac {1-p}{p}}}$.

The expectation value can be derived in different ways:

• ${\displaystyle \operatorname {E} (X)=p\sum _{k=1}^{\infty }k\,(1-p)^{k-1}=p\sum _{k=0}^{\infty }\,{\frac {\mathrm {d} }{\mathrm {d} p}}\left(-(1-p)^{k}\right)=-p{\frac {\mathrm {d} }{\mathrm {d} p}}\left(\sum _{k=0}^{\infty }\,(1-p)^{k}\right)=-p{\frac {\mathrm {d} }{\mathrm {d} p}}\left({\frac {1}{p}}\right)={\frac {1}{p}}}$.

• ${\displaystyle \operatorname {E} (X)=\sum _{k=1}^{\infty }kp(1-p)^{k-1}=\sum _{k=0}^{\infty }(k+1)p(1-p)^{k}=\sum _{k=0}^{\infty }kp(1-p)^{k}+\sum _{k=1}^{\infty }p(1-p)^{k-1}=(1-p)\operatorname {E} (X)+1}$

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

Where is because is the probability function . ${\displaystyle \sum _{k=1}^{\infty }p(1-p)^{k-1}=1}$ ${\displaystyle p(1-p)^{k-1}}$

• The expected value can be broken down by case distinction. The first experiment is likely to be successful, i.e. it will be realized with 1. Likely the first experiment is unsuccessful, but the expected value for the number of still following experiments is due to the memoryless again . So it applies ${\displaystyle \operatorname {E} (X)}$ ${\displaystyle p}$ ${\displaystyle X}$ ${\displaystyle 1-p}$ ${\displaystyle \operatorname {E} (X)}$

${\displaystyle \operatorname {E} (X)=p\cdot 1+(1-p)\cdot (1+\operatorname {E} (X))=1+(1-p)\cdot \operatorname {E} (X)}$, so . ${\displaystyle \operatorname {E} (X)={\frac {1}{p}}}$

• If you carry out experiments, the expected value for the number of successful experiments is . Therefore, the expected interval between two successful experiments (including one successful experiment) is . ${\displaystyle n}$ ${\displaystyle n\cdot p}$ ${\displaystyle {\tfrac {n}{n\cdot p}}}$ ${\displaystyle \operatorname {E} (X)={\tfrac {1}{p}}}$

Variance

The variances of the two geometric distributions are

${\displaystyle \operatorname {Var} (X)=\operatorname {Var} (Y)={\frac {1-p}{p^{2}}}={\frac {1}{p^{2}}}-{\frac {1}{p}}}$.

The derivation can be made via

${\displaystyle \operatorname {Var} (X)}$	${\displaystyle =\operatorname {E} (X^{2})-\operatorname {E} (X)^{2}=p\sum _{k=1}^{\infty }k^{2}(1-p)^{k-1}-{\frac {1}{p^{2}}}}$
	${\displaystyle =p\sum _{k=1}^{\infty }k(k+1)(1-p)^{k-1}-p\sum _{k=1}^{\infty }k(1-p)^{k-1}-{\frac {1}{p^{2}}}}$
	${\displaystyle =p{\frac {\mathrm {d} ^{2}}{\mathrm {d} p^{2}}}\sum _{k=1}^{\infty }(1-p)^{k+1}+p{\frac {\mathrm {d} }{\mathrm {d} p}}\sum _{k=1}^{\infty }(1-p)^{k}-{\frac {1}{p^{2}}}}$
	${\displaystyle =p{\frac {\mathrm {d} ^{2}}{\mathrm {d} p^{2}}}\left(\sum _{k=0}^{\infty }(1-p)^{k}\cdot (1-p)^{2}\right)+p{\frac {\mathrm {d} }{\mathrm {d} p}}\left(\sum _{k=0}^{\infty }(1-p)^{k}\cdot (1-p)\right)-{\frac {1}{p^{2}}}}$
	${\displaystyle =p{\frac {\mathrm {d} ^{2}}{\mathrm {d} p^{2}}}\left({\frac {1}{1-(1-p)}}\cdot (1-p)^{2}\right)+p{\frac {\mathrm {d} }{\mathrm {d} p}}\left({\frac {1}{1-(1-p)}}\cdot (1-p)\right)-{\frac {1}{p^{2}}}}$
	${\displaystyle =p{\frac {\mathrm {d} ^{2}}{\mathrm {d} p^{2}}}\left({\frac {(1-p)^{2}}{p}}\right)+p{\frac {\mathrm {d} }{\mathrm {d} p}}\left({\frac {1-p}{p}}\right)-{\frac {1}{p^{2}}}}$
	${\displaystyle =p\cdot {\frac {2}{p^{3}}}-p\cdot {\frac {1}{p^{2}}}-{\frac {1}{p^{2}}}={\frac {2}{p^{2}}}-{\frac {1}{p}}-{\frac {1}{p^{2}}}={\frac {1}{p^{2}}}-{\frac {1}{p}}}$.

Memory loss

The geometric distribution is a memoryless distribution, i.e. i.e., it applies to

option A

${\displaystyle \operatorname {P} (X=n+k\,|\,X>n)=\operatorname {P} (X=k)\quad n,k=1,2,\dotsc }$

Variant B

${\displaystyle \operatorname {P} (Y=n+k\,|\,Y\geq n)=\operatorname {P} (Y=k)\quad n,k=0,1,2,\dotsc }$

So if a geometrically distributed random variable is known to be greater than the value (variant A) or at least has the value (variant B), the probability that it exceeds this value by is just as great as that an identical random variable takes on the value at all . ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle k}$ ${\displaystyle k}$

The lack of memory is a defining quality; the geometric distribution is therefore the only possible memoryless discrete distribution. Its constant counterpart here is the exponential distribution .

Relation to reproductivity

The sum of independent geometrically distributed random variables with the same parameter is not geometrically distributed , but negatively binomially distributed . Thus the family of geometric probability distributions is not reproductive . ${\displaystyle \textstyle X=\sum _{i=1}^{k}X_{i}}$ ${\displaystyle X_{1},\dotsc ,X_{k}}$ ${\displaystyle p}$

Crookedness

The skew results for both variants as follows:

${\displaystyle \operatorname {v} (X)=\operatorname {v} (Y)={\frac {2-p}{\sqrt {1-p}}}}$.

Bulge

The curvature can also be shown closed for both variants as

${\displaystyle \beta _{2}=9+{\frac {p^{2}}{1-p}}}$.

That’s the excess

${\displaystyle \gamma =6+{\frac {p^{2}}{1-p}}}$.

mode

option A

In variant A, mode is 1.

Variant B

With variant B the mode is 0.

Median

option A

In variant A, the median is

${\displaystyle {\tilde {m}}=\left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!}$.

Here is the Gauss bracket . The median is not necessarily unique. ${\displaystyle \lceil \cdot \rceil }$

Variant B

Here is the median

${\displaystyle {\tilde {m}}=\left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil -1\!}$.

It doesn't have to be unique either.

entropy

The entropy of both variants is

${\displaystyle \mathrm {H} ={\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}}$.

Characteristic function

The characteristic function has the form

option A

${\displaystyle \varphi _{X}(s)={\frac {pe^{is}}{1-(1-p)e^{is}}}}$.

Variant B

${\displaystyle \varphi _{Y}(s)={\frac {p}{1-(1-p)e^{is}}}}$.

Moment generating function

The moment generating function of the geometric distribution is

option A

${\displaystyle M_{X}(s)={\frac {pe^{s}}{1-(1-p)e^{s}}}}$

Variant B

${\displaystyle M_{Y}(s)={\frac {p}{1-(1-p)e^{s}}}}$.

Probability generating function

The probability generating function of the geometric distribution is

option A

${\displaystyle m_{X}(t)={\frac {pt}{1-(1-p)t}}}$

Variant B

${\displaystyle m_{Y}(t)={\frac {p}{1-(1-p)t}}}$.

Relationships with other distributions

Relationship to the negative binomial distribution

Generalization to several successes
A generalization of the geometric distribution is the negative binomial distribution , which indicates the probability that attempts are necessary for successes or (in an alternative representation) that the -th success occurs after failures have already occurred. ${\displaystyle r}$ ${\displaystyle n}$ ${\displaystyle r}$ ${\displaystyle k=n-r}$

Conversely, the geometric distribution is a negative binomial distribution with . The geometric distribution thus applies to the convolution . ${\displaystyle r=1}$ ${\displaystyle \operatorname {Geom} (p)*\operatorname {Geom} (p)=\operatorname {NegBin} (2,p)}$

Relationship to the exponential distribution

Convergence of the geometrical distribution
For a sequence of geometrically distributed random variables with parameters hold with a positive constant . Then the sequence for large converges to an exponentially distributed random variable with parameters . ${\displaystyle X_{1},X_{2},X_{3},\dotsc }$ ${\displaystyle p_{1},p_{2},p_{3},\dotsc }$ ${\displaystyle \lim _{n\to \infty }np_{n}=\lambda }$ ${\displaystyle \lambda }$ ${\displaystyle {\tfrac {X_{n}}{n}}}$ ${\displaystyle n}$ ${\displaystyle \lambda }$

In analogy to the discrete geometric distribution, the continuous exponential distribution determines the waiting time until the first occurrence of a rare Poisson-distributed event. The exponential distribution is therefore the continuous analogue of the discrete geometric distribution.

Relationship to the composite Poisson distribution

The geometric distribution in variant B arises as a special case of the composite Poisson distribution in combination with the logarithmic distribution . Select and as the parameter . This means that the geometric distribution is also infinitely divisible . ${\displaystyle p_{\text{log}}=1-p_{\text{geom}}}$ ${\displaystyle -\lambda =\ln(1-p_{\text{log}})}$

Relationship to the urn model

The geometric distribution can be derived from the urn model if is. Then the geometric distribution arises when pulling with replacement from an urn with balls, of which are marked. It is then the waiting time for the first success. ${\displaystyle p={\frac {p_{1}}{p_{2}}}\in \mathbb {Q} }$ ${\displaystyle p_{2}}$ ${\displaystyle p_{1}}$

Random numbers

Random numbers for the geometric distribution are usually generated using the inversion method. This method is particularly suitable for geometric distribution, since the individual probabilities of simple recursion are sufficient. The inversion method can only be carried out here with rational operations ( addition , multiplication ) and without calculating and storing the distribution function beforehand, which guarantees a fast algorithm for the simulation. ${\displaystyle \operatorname {P} (X=k+1)=(1-p)\operatorname {P} (X=k)}$

Web links

• University of Konstanz - Interactive Animation (Java)
• Interactive worksheets for the geometric distribution on MathsPrisma
• StatWiki Detailed derivation of the torque generating function