Poisson distribution (PDF): $$f(k;x,\lambda) = \frac{(\lambda x)^k e^{-\lambda x}}{k!}\quad\textrm{for }x,\lambda\ge 0$$

Erlang distribution (PDF): $$f(x;k,\lambda) = \frac{\lambda^k x^{k-1} e^{-\lambda x}}{(k-1)!}\quad\textrm{for }x,\lambda\ge 0$$

where $$\lambda$$ is the arrival rate of a Poisson process, and $$k$$ is the number of arrival events. The Poisson distribution tells the probability of the number of arrivals over a given amount of time $$x$$. The Erlang distribution tells the probability of the time interval over a given number of arrivals $$k$$. The difference is that, in Erlang distribution, it is assumed to have an arrival at time $$t=0$$ (which is not counted toward $$k$$) and $$t=x$$ (which is counted toward $$k$$), respectively, whereas Poisson does not.

## Derivation

Poisson is derived using the law of rare events: Assume $$B(n,p)$$ denote the binomial distribution of $$n$$ Bernoulli events with “success” probability $$p$$. Assume the rate of occurrence of “success” is $$\lambda$$, then $$\lambda x = np$$ where $$x$$ is the time to finish $$n$$ Bernoulli events. Define $$X_n$$ to be the number of “success” events occurred. So

\begin{aligned} \lim_{n\to\infty}\Pr[X_n = k] &= \lim_{n\to\infty}\binom{n}{k}p^k(1-p)^{n-k} \\ &= \lim_{n\to\infty}\frac{n!}{k!(n-k)!}(\frac{\lambda x}{n})^k(1-\frac{\lambda x}{n})^{n-k} \\ &= \lim_{n\to\infty}(\frac{n!}{n^k(n-k)!})\frac{(\lambda x)^k}{k!}(1-\frac{\lambda x}{n})^n(1-\frac{\lambda x}{n})^{-k} \\ &= \lim_{n\to\infty}(\frac{n(n-1)\cdots(n-(k-1))}{n^k})\frac{(\lambda x)^k}{k!}(1-\frac{\lambda x}{n})^n(1-\frac{\lambda x}{n})^{-k} \\ &= 1\cdot\frac{(\lambda x)^k}{k!}\cdot e^{-\lambda x}\cdot 1 \\ &= \frac{(\lambda x)^k}{k!}e^{-\lambda x} \end{aligned}

The formula for Erlang distribution is derived by induction. The case of $$f(x;1,\lambda)$$ is trivial because this is the exponential distribution, i.e. the time for the next arrival is $$x$$.

\begin{aligned} f(x;1,\lambda) &= \lambda e^{-\lambda x} \\ &= \frac{\lambda^1 x^0 e^{-\lambda x}}{0!} \end{aligned}

Consider $$f(x;k-1,\lambda)=\frac{\lambda^{k-1} x^{k-2} e^{-\lambda x}}{(k-2)!}$$, then $$f(x;k,\lambda)$$ is defined by the convolution:

\begin{align} f(x;k,\lambda) &= \int_0^x f(u;k-1,\lambda)\cdot\lambda e^{-\lambda(x-u)} du \\ &= \int_0^x \frac{\lambda^{k-1} u^{k-2} e^{-\lambda u}}{(k-2)!}\cdot\lambda e^{-\lambda(x-u)} du \\ &= \frac{\lambda^k}{(k-2)!} \int_0^x u^{k-2} e^{-\lambda u} e^{-\lambda(x-u)} du \\ &= \frac{\lambda^k e^{-\lambda x}}{(k-2)!} \int_0^x u^{k-2} du \\ &= \frac{\lambda^k e^{-\lambda x}}{(k-2)!} \frac{x^{k-1}}{k-1} \\ &= \frac{\lambda^k x^{k-1} e^{-\lambda x}}{(k-1)!} \end{align}