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

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 $x$.

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: