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Erlang Distribution


Given a Poisson distribution with a rate of change lambda, the distribution function D(x) giving the waiting times until the hth Poisson event is

D(x)=1-sum_(k=0)^(h-1)e^(-lambdax)((lambdax)^k)/(k!)
(1)
=1-(Gamma(h,xlambda))/(Gamma(h))
(2)

for x in [0,infty), where Gamma(x) is a complete gamma function, and Gamma(a,x) an incomplete gamma function. With h explicitly an integer, this distribution is known as the Erlang distribution, and has probability function

 P(x)=(lambda(lambdax)^(h-1))/((h-1)!)e^(-lambdax).
(3)

It is closely related to the gamma distribution, which is obtained by letting alpha=h (not necessarily an integer) and defining theta=1/lambda. When h=1, it simplifies to the exponential distribution.

Evans et al. (2000, p. 71) write the distribution using the variables b=1/lambda and c=h.


See also

Exponential Distribution, Gamma Distribution, Poisson Distribution

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References

Evans, M.; Hastings, N.; and Peacock, B. "Erlang Distribution." Ch. 12 in Statistical Distributions, 3rd ed. New York: Wiley, pp. 71-73, 2000.

Referenced on Wolfram|Alpha

Erlang Distribution

Cite this as:

Weisstein, Eric W. "Erlang Distribution." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ErlangDistribution.html

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