Erlang distribution

The Erlang distribution is a continuous probability distribution , a generalization of the exponential distribution and a special case of the gamma distribution . It was developed by Agner Krarup Erlang for the statistical modeling of the interval lengths between telephone calls.

The Erlang distribution is used in queuing theory to record the distribution of the time span between events of a Poisson process , for example the arrival of customers, and in quality assurance to describe lifetimes . In call centers , this distribution is used for personnel deployment planning in order to determine the number of agents required based on the expected call volume in the time interval.

The Erlang distribution density provides the distribution of the probability that the th event will occur after the time or space has elapsed, if one expects events per unit interval (see derivation ). It describes a chain of successive events. The most likely distance to the -th event ( mode ) is smaller than the mean value ( expected value ) because shorter event intervals occur more frequently. If the distances of the respective individual events , sorted according to size, are filled into a histogram , this shows an exponential distribution accordingly. ${\displaystyle x}$ ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle n}$ ${\displaystyle n}$

Density of the Erlang distribution, ${\displaystyle \lambda =1}$

definition

The Erlang distribution with the parameters (a positive real number ) and (a natural number ) is a special gamma distribution that is determined by the density function ${\displaystyle \operatorname {Erl} (\lambda ,n)}$ ${\displaystyle \lambda >0}$ ${\displaystyle n\geq 1}$

${\displaystyle f(x)={\begin{cases}\displaystyle {\frac {\lambda ^{n}x^{n-1}}{(n-1)!}}\,\mathrm {e} ^{-\lambda x}&x\geq 0\\0&x<0\end{cases}}}$

and which differs from the general gamma distribution through the restriction to natural numbers in the second parameter.

For an Erlang-distributed random variable , the probability that it is within the interval is given by the distribution function ${\displaystyle X}$ ${\displaystyle X}$ ${\displaystyle 0\leq X\leq x}$

${\displaystyle F(x)={\begin{cases}\displaystyle {\frac {\lambda ^{n}}{(n-1)!}}\int _{0}^{x}t^{n-1}\mathrm {e} ^{-\lambda t}\,\mathrm {d} t={\frac {\gamma (n,\lambda x)}{(n-1)!}}=1-{\frac {\Gamma (n,\lambda x)}{(n-1)!}}=1-\mathrm {e} ^{-\lambda x}\sum _{i=0}^{n-1}{\frac {(\lambda x)^{i}}{i!}}&x\geq 0\\0&x<0\end{cases}}}$

given, where or denotes the incomplete gamma function . ${\displaystyle \gamma }$ ${\displaystyle \Gamma }$

Derivation and Interpretation

The Erlang distribution can be interpreted as the probability density of receiving the th event after a certain time . Let the events be Poisson distributed . ${\displaystyle t}$ ${\displaystyle n}$

Let us consider the probability that the -th event is in the time interval . This is obviously the probability that events are in the interval multiplied by the probability that exactly one event is in. Since the events are Poisson distributed and independent at disjoint intervals, this is: ${\displaystyle n}$ ${\displaystyle [t,t+\Delta t]}$ ${\displaystyle n-1}$ ${\displaystyle [0,t]}$ ${\displaystyle [t,t+\Delta t]}$

${\displaystyle {\frac {(\lambda \cdot t)^{n-1}}{(n-1)!}}\mathrm {e} ^{-\lambda \cdot t}\cdot \lambda \cdot \Delta t\cdot \mathrm {e} ^{-\lambda \cdot \Delta t}}$.

This is first order : ${\displaystyle \Delta t}$

${\displaystyle \lambda \cdot {\frac {(\lambda \cdot t)^{n-1}}{(n-1)!}}\mathrm {e} ^{-\lambda \cdot t}\cdot \Delta t}$,

so that the Erlang distribution results as:

${\displaystyle \lambda \cdot {\frac {(\lambda \cdot t)^{n-1}}{(n-1)!}}\mathrm {e} ^{-\lambda \cdot t}}$.

properties

Since an Erlang-distributed random variable is the sum of independently and identical to parameter exponentially distributed random variables , the following properties result. ${\displaystyle X}$ ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle X_{1},\dotsc ,X_{n}}$

Expected value

The Erlang distribution has the expected value

${\displaystyle \operatorname {E} (X)=\operatorname {E} \left(\sum _{k=1}^{n}X_{k}\right)=\sum _{k=1}^{n}\operatorname {E} (X_{k})={\frac {n}{\lambda }}.}$

Variance

The variance results analogously to

${\displaystyle \operatorname {Var} (X)=\operatorname {Var} \left(\sum _{k=1}^{n}X_{k}\right)=\sum _{k=1}^{n}\operatorname {Var} (X_{k})={\frac {n}{\lambda ^{2}}}.}$

mode

The mode, the maximum density, is included

${\displaystyle {\frac {n-1}{\lambda }}.}$

Characteristic function

From the characteristic function of an exponentially distributed random variable one obtains that of an Erlang-distributed random variable:

${\displaystyle \varphi _{X}(t)=\left({\frac {\lambda }{\lambda -it}}\right)^{n}.}$

Moment generating function

The same results for the torque generating function

${\displaystyle M_{X}(t)=\left({\frac {\lambda }{\lambda -t}}\right)^{n}.}$

entropy

The entropy of the Erlang distribution is

${\displaystyle H(X)=(1-n)\psi (n)+\ln \left({\frac {\Gamma (n)}{\lambda }}\right)+n}$

where ψ ( p ) denotes the digamma function .

Relationships with other distributions

Relationship to the exponential distribution

• The Erlang distribution is a generalization of the exponential distribution, because it merges into it . ${\displaystyle \operatorname {Erl} (\lambda ,n)}$ ${\displaystyle n=1}$ ${\displaystyle \operatorname {Erl} (\lambda ,1)=\operatorname {Exp} (\lambda )}$
• Let there be many random variables , all exponentially distributed with the same parameter and stochastically independent . Then the random variable is Erlang-distributed with the parameters and . ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle Y_{i}\ (i=1,\dotsc ,n)}$ ${\displaystyle X=Y_{1}+Y_{2}+\dotsb +Y_{n}}$ ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle (n\in \mathbb {N} ,\lambda \geq 0)}$

Relationship to the Poisson distribution

• For a Poisson process , the random number of events up to a defined point in time is determined by means of the Poisson distribution , the random time up to the -th event is Erlang-distributed. In this case , this Erlang distribution changes into an exponential distribution, with which the time to the first random event and the time between two successive events can be determined. ${\displaystyle \operatorname {Poi} (\lambda )}$ ${\displaystyle n}$ ${\displaystyle n=1}$
• The Erlang distribution is the conjugate distribution to the Poisson distribution.

Relationship to constant uniform distribution

An Erlang distribution can be generated as a convolution of evenly continuously distributed functions : ${\displaystyle n}$ ${\displaystyle X(0,1)}$

${\displaystyle \operatorname {Erl} (\lambda ,n)\sim -{\frac {1}{\lambda }}\ln {\left(\prod _{i=1}^{n}x_{i}\right)}.}$

Relationship to the gamma distribution

The Erlang distribution with the parameter and degrees of freedom corresponds to a gamma distribution with natural shape parameter (and inverse scale parameter ). ${\displaystyle \lambda }$ ${\displaystyle n}$ ${\displaystyle p=n}$ ${\displaystyle b=\lambda }$

Web links

• Erlang distributions: Explanation and illustration of connections to other distributions

Individual evidence

1. ↑ Frodesen, Skjeggestad, Tofte: Probability and Statistics in Particle Physics , Universitetsforlaget, Bergen Oslo Tromsø p. 98.