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If a customer arrives according to a Possion process with rate $\lambda$, how can I show that the time interval $X$ taken to receive $k$ customers is an Erlang-$k$ random variable with parameters $n$ and $\lambda$?

User1865345
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Alex_DeLarge
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    https://en.wikipedia.org/wiki/Erlang_distribution – user2974951 Dec 14 '22 at 11:12
  • Ok, Poisson process imply that inter-arrival times are exponentially distribute, and so the waiting time until we see k arrives is the sum of all the inter-arrival times up to k, which is the sum of k exponential random variables, that is an Erlang distribution. I'm looking for help to put this into formulas... – Alex_DeLarge Dec 14 '22 at 11:54
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    That Wikipedia page says If $S_n = \sum\limits_{i=1}^n X_i$ such that $X_i \sim \operatorname{Exponential}(\lambda)$ [i.i.d.], then $S_n \sim \operatorname{Erlang}(n, \lambda)$. Is that what you are looking for, or do you want an induction on the convolution of $\operatorname{Exponential}(\lambda)$ and $\operatorname{Erlang}(n-1, \lambda)$ using the density functions? – Henry Dec 14 '22 at 12:53
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    This result is developed from basic principles at https://stats.stackexchange.com/questions/214421. – whuber Dec 14 '22 at 15:12

2 Answers2

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You can use the following approaches:

  • The waiting time for a single event is exponential. The waiting time untill $k$ events is the sum of $k$ independent exponential distributed variables and is Erlang distributed. (See for instance here: How is the Erlang pdf derived?)

  • Alternatively you can use the correspondence between the probability that $k$ or more events happen before time $t$ (modelled by the Poisson distribution) and the probability that that the time to see $k$ events takes $t$ or less time (which is Erlang distributed).

    For the Poisson distribution we have for the number events $K$ given the time $t$: $$P(K \geq k; \theta = \lambda t) = 1-F(k;\lambda t) = 1-\frac{\Gamma( k , \lambda t)}{(k-1)!} = \frac{\gamma( k , \lambda t)}{(k-1)!}$$ where $\Gamma$ and $\gamma$ are respectively the upper and lower incomplete Gamma functions.

    For the Erlang distribution we have for the time $T$ given the number of events $k$: $$P(t \leq T; k, \lambda) = F(t;k,\lambda) = \frac{\gamma( k, \lambda t)}{(k-1)!}$$

    This argument also occurs here in the opposite direction (deriving the Poisson distribution, starting with exponential waiting times).

Both these two approaches relate the Erlang distribution to Poisson/exponential, but it doesn't derive the Poisson/exponential distributions. A more rigorous derivation/approach is given here: What distribution to use to model time before a train arrives? or here: Waiting time of Poisson process follows Gamma distribution

Sextus Empiricus
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With your help and suggestions in comments, I think I can answer to my original question. Let'start with this:

  • If $X$ and $Y$ are independent random variables with probability density functions $f$ and $g$, then $X + Y$ has density $f ∗ g$.

Let $T_1, . . . , T_n$ be independent, exponential random variables with parameter $\lambda$. Thus, each of the $T_i$ has probability density function (for $t \geq 0$):

$$f(t)=\lambda e^{-\lambda t}$$

Using the previous formula, $T_1 + T_2$ has density: $$(f * f)(t) = \int_{0}^tf(t-s)f(s)ds = \lambda^2te^{-\lambda t} = f_2(t)$$

Continuing we have $T_1 + T_2 + T_3$ with density: $$(f * f * f)(t) = (f*f_2)(t)= \frac{\lambda^3t^2}{2}e^{-\lambda t} = f_3(t) $$

Now we can prove by induction that $T_1 + ...+ T_n$ has density: $$(\underbrace{f * ... * f}_{\textrm{n times}})(t) = (f*f_{n-1})(t) = \frac{\lambda^nt^{n-1}}{(n-1)!}e^{-\lambda t}$$ which is, by definition, the probability density function of the Erlang-k distribution.

I think it's all correct, thank you all!

Alex_DeLarge
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