Erlang distribution

Erlang-distributed random variates can be generated from uniformly distributed random numbers (${\displaystyle U\in [0,1]}$) using the following formula:[6]

${\displaystyle E(k,\lambda )=-{\frac {1}{\lambda }}\ln \prod _{i=1}^{k}U_{i}=-{\frac {1}{\lambda }}\sum _{i=1}^{k}\ln U_{i}}$

• If ${\displaystyle X\sim \operatorname {Erlang} (k,\lambda )}$ then ${\displaystyle a\cdot X\sim \operatorname {Erlang} \left(k,{\frac {\lambda }{a}}\right)}$ with ${\displaystyle a\in \mathbb {R} }$
• If ${\displaystyle X\sim \operatorname {Erlang} (k_{1},\lambda )}$ and ${\displaystyle Y\sim \operatorname {Erlang} (k_{2},\lambda )}$ then ${\displaystyle X+Y\sim \operatorname {Erlang} (k_{1}+k_{2},\lambda )}$ if ${\displaystyle X,Y}$ are independent

• 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 ${\displaystyle X,}$ that is, ${\displaystyle \lambda /k.}$ The (age specific event) rate of the Erlang distribution is, for ${\displaystyle k>1,}$ monotonic in ${\displaystyle x,}$ increasing from 0 at ${\displaystyle x=0,}$ to ${\displaystyle \lambda }$ as ${\displaystyle x}$ tends to infinity.[12]
  • That is: if ${\displaystyle X_{i}\sim \operatorname {Exponential} (\lambda ),}$ then
    $${\displaystyle \sum _{i=1}^{k}{X_{i}}\sim \operatorname {Erlang} (k,\lambda )}$$

    
• Because of the factorial function in the denominator of the PDF and 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 ${\displaystyle k=2}$). 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 ${\displaystyle X\sim \operatorname {Gamma} (k,\lambda ),}$ then ${\displaystyle X\sim \operatorname {Erlang} (k,\lambda )}$
• If ${\displaystyle U\sim \operatorname {Exponential} (\lambda )}$ and ${\displaystyle V\sim \operatorname {Erlang} (n,\lambda )}$ then ${\displaystyle {\frac {U}{V}}+1\sim \operatorname {Pareto} (1,n)}$
• The Erlang distribution is a special case of the Pearson type III distribution
• The Erlang distribution is related to the chi-squared distribution. If ${\displaystyle X\sim \operatorname {Erlang} (k,\lambda ),}$ then ${\displaystyle 2\lambda X\sim \chi _{2k}^{2}.}$
• The Erlang distribution is related to the Poisson distribution by the Poisson process: If ${\displaystyle S_{n}=\sum _{i=1}^{n}X_{i}}$ such that ${\displaystyle X_{i}\sim \operatorname {Exponential} (\lambda ),}$ then
  $${\displaystyle S_{n}\sim \operatorname {Erlang} (n,\lambda )}$$

  
  and
  $${\displaystyle \operatorname {Pr} (N(x)\leq n-1)=\operatorname {Pr} (S_{n}>x)=1-F_{X}(x;n,\lambda )=\sum _{k=0}^{n-1}{\frac {1}{k!}}e^{-\lambda x}(\lambda x)^{k}.}$$

  
  Taking the differences over ${\displaystyle n}$ gives the Poisson distribution.

• Coxian distribution
• Engset calculation
• Erlang B formula
• Erlang unit
• Phase-type distribution
• Traffic generation model

• Ian Angus "An Introduction to Erlang B and Erlang C", Telemanagement #187 (PDF Document - Has terms and formulae plus short biography)
• Stuart Harris "Erlang Calculations vs. Simulation"

• Erlang Distribution
• Resource Dimensioning Using Erlang-B and Erlang-C