Editing Erlang distribution (section)

==Related distributions==

* The Erlang distribution is the distribution of the sum of ''k'' [[independent and identically distributed random variables]], each having an [[exponential distribution]]. The long-run rate at which events occur is the reciprocal of the expectation of <math>X,</math> that is, <math>\lambda/k.</math> The (age specific event) rate of the Erlang distribution is, for <math>k>1,</math> monotonic in <math>x,</math> increasing from 0 at <math>x=0,</math> to <math>\lambda</math> as <math>x</math> tends to infinity.<ref>Cox, D.R. (1967) ''Renewal Theory'', p20, Methuen.</ref> 
** That is: if <math> X_i \sim \operatorname{Exponential}(\lambda),</math> then <math display="block"> \sum_{i=1}^k{X_i} \sim \operatorname{Erlang}(k, \lambda)</math>
* Because of the factorial function in the denominator of the [[#Probability density function|PDF]] and [[#Cumulative distribution function (CDF)|CDF]], the Erlang distribution is only defined when the parameter ''k'' is a positive integer. In fact, this distribution is sometimes called the '''Erlang-''k'' distribution''' (e.g., an Erlang-2 distribution is an Erlang distribution with <math>k=2</math>). The [[gamma distribution]] generalizes the Erlang distribution by allowing ''k'' to be any positive real number, using the [[gamma function]] instead of the factorial function.
** That is: if k is an [[integer]] and <math> X \sim \operatorname{Gamma}(k, \lambda),</math> then <math> X \sim \operatorname{Erlang}(k, \lambda)</math>
*If <math> U \sim \operatorname{Exponential}(\lambda)</math> and <math> V \sim \operatorname{Erlang}(n, \lambda)</math> then <math> \frac{U}{V}+1 \sim \operatorname{Pareto}(1, n)</math>
*The Erlang distribution is a special case of the [[Pearson distribution|Pearson type III distribution]]{{citation needed|date=March 2016}}
*The Erlang distribution is related to the [[chi-squared distribution]]. If <math> X \sim \operatorname{Erlang}(k,\lambda),</math> then <math> 2\lambda X\sim \chi^2_{2k}.</math>{{citation needed|date=March 2016}}
*The Erlang distribution is related to the [[Poisson distribution]] by the [[Poisson process]]: If <math> S_n = \sum_{i=1}^n X_i</math> such that <math> X_i \sim \operatorname{Exponential}(\lambda),</math> then <math display="block"> S_n \sim \operatorname{Erlang}(n, \lambda)</math> and <math display="block"> \operatorname{Pr}(N(x) \leq n - 1) = \operatorname{Pr}(S_n > x) = 1 - F_X(x; n, \lambda) = \sum_{k=0}^{n-1} \frac{1}{k!}e^{-\lambda x} (\lambda x)^k.</math> Taking the differences over <math>n</math> gives the Poisson distribution.