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. $\endgroup$ – Dilip Sarwate Dec 4 '12 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 Oct 5 '16 at 4:26

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! $\endgroup$ – Miguel Anguita Dec 29 '19 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/… $\endgroup$ – Sextus Empiricus Dec 29 '19 at 20:06
  • $\begingroup$ I know the definition. But not how to do the specific f.ch of Erlang one. $\endgroup$ – Miguel Anguita Dec 29 '19 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 $ $\endgroup$ – Sextus Empiricus Dec 30 '19 at 7:59
  • $\begingroup$ Yeah, great, but how would you integrate it in an easy form? $\endgroup$ – Miguel Anguita Dec 31 '19 at 9:51

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