If each arrival is exponentially distributed, then the $k$th arrival time is Erlang distributed. The Erlang PDF is: $$ f_{Y_k}(y) = \lambda e^{-\lambda y} \frac{(\lambda y)^{k-1}}{(k-1)!} $$ How is this derived?

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    $\begingroup$ You need each arrival to be exponentially distributed (and independent of past arrivals). If arrival times are discrete and geometrically distributed with parameter $p$, the $k$-th arrival time is a negative binomial or Pascal random variable with parameters $(k,p)$. Both derivations can be done with a $(k-1)$-fold convolution, via moment-generating functions or characteristic functions, etc. $\endgroup$ Commented Dec 4, 2012 at 15:18
  • $\begingroup$ Here's a derivation given by convolution: math.unl.edu/~scohn1/428s05/queue3.pdf Essentially the Erlang distribution is the result of convolving the exponential distribution with itself k-1 times. Note that convolution as the sum of random variables is particularly important to understanding what's going on here. $\endgroup$
    – hedgepig
    Commented Oct 5, 2016 at 4:26

1 Answer 1


You can compute this by means of a convolution of $X \sim Exp(\lambda)$ and $Y_k \sim Erlang(k,\lambda)$ resulting in the distribution of $Y_{k+1} = X + Y_k$. (see also Why is the sum of two random variables a convolution?)

$$\begin{array}{rcccl} f_{Y_{k+1}}(s) &=& \int_0^s \underbrace{ \vphantom{ \frac{(\lambda t)^{k-1}}{(k-1)!}} \lambda e^{-\lambda (s-t)}}_{\text{exponential density}} \cdot \underbrace{ \lambda^k e^{-\lambda t} \frac{t^{k-1}}{(k-1)!}}_{\text{Erlang density with $k$}} dt && \\ & =& \lambda^{k+1} e^{-\lambda s} \int_{0}^s \frac{t^{k-1}}{(k-1)!} dt &=& \underbrace{ \vphantom{ \frac{(\lambda t)^{k-1}}{(k-1)!}} \lambda^{k+1} e^{-\lambda s} \frac{s^{k}}{k!}}_{\text{Erlang density with $k+1$}} \end{array}$$

That a sum of $k$ identical and independent exponential distributed variables is Erlang distributed will follow by induction.

Alternatively you can use Fourier transform or the characteristic function to quickly observe that characteristic function of the Erlang distribution $\varphi(t)_{\text{Erlang}} = \left(1 - \frac{it}{\lambda} \right)^{-k}$ is a product of the characteristic function of the exponential distribution $\varphi(t)_{exponential} = \left(1 - \frac{it}{\lambda} \right)^{-1}$.

  • $\begingroup$ Sorry for writing in this post, but how would you prove the characteristic function of the Erlang? I have exponential one but I don't know exactly how to do the Erlang one. Thanks! $\endgroup$
    – makux_gcf
    Commented Dec 29, 2019 at 18:23
  • $\begingroup$ @MiguelAnguita I am not sure what you are missing in understanding how to proof/compute the characteristic function. Anyway, here's the definition: en.m.wikipedia.org/wiki/… $\endgroup$ Commented Dec 29, 2019 at 20:06
  • $\begingroup$ I know the definition. But not how to do the specific f.ch of Erlang one. $\endgroup$
    – makux_gcf
    Commented Dec 29, 2019 at 20:13
  • $\begingroup$ @MiguelAnguita $$\phi (t) = \int_{-\infty}^{\infty} e^{itx} \lambda^k e^{-\lambda x} \frac{x^{k-1}}{(k-1)!} dx$$ which relates to integrating $\int e^{(it-\lambda) x}x^{k-1} dx $ $\endgroup$ Commented Dec 30, 2019 at 7:59
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    $\begingroup$ @MiguelAnguita you can do this with repeated integration by parts (note I should have written limits $0$ to $\infty$ instead of $-\infty$ to $\infty$). You could also do it more in one step and figure out the coefficients of $$F(x) = \sum_{n=0}^{k-1} a_n e^{(it-\lambda)x} x^n$$ such that it's derivative is $$F(x)' = \sum_{n=0}^{k-1} a_n ((it-\lambda) e^{(it-\lambda)x} x^n+ n e^{(it-\lambda)x} x^{n-1})$$ which should give you a recursive formula for $a_n$ $\endgroup$ Commented Dec 31, 2019 at 18:04

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