We showed earlier that the time until the first event in a Poisson process with rate \(\lambda\) events per unit time has an exponential distribution with parameter \(\lambda\). We will now find the distribution of the time until the second event in the process.
Time until second event
If \(X\) is the time until the second event in a Poisson process with rate \(\lambda\), it has probability density function
\[ f(x) \;\;=\;\; \begin{cases} \lambda^2 x e^{-\lambda x} & x \gt 0 \\[0.3em] 0 & \text{otherwise} \end{cases} \]If the second event in the Poisson process occurs after time \(x\), there must have been either 0 or 1 event before time \(x\). If we define \(Y_x\) to be the number of events before time \(x\),
\[ Y_x \;\;\sim\;\; \PoissonDistn(\lambda x) \]Therefore
\[ \begin{align} F(x) \;\;=\;\; 1 - P(Y_x = 0) - P(Y_x = 1) \;\;&=\;\; 1 - \frac {(\lambda x)^0 e^{-\lambda x}}{0!} - \frac {(\lambda x)^1 e^{-\lambda x}}{1!} \\[0.2em] &=\;\; 1 - e^{-\lambda x} - \lambda x e^{-\lambda x} \end{align} \]We can find the probability density function by differentiating this,
\[ \begin{align} f(x) \;\;=\;\; F'(x) \;\;&=\;\; \lambda e^{-\lambda x} - \left(\lambda e^{-\lambda x} - \lambda^2 x e^{-\lambda x} \right) \\ &=\;\; \lambda^2 x e^{-\lambda x} \end{align} \]The pdf of the time until the third event in the Poisson process can be found in a similar way.
Time until third event
If \(X\) is the time until the third event in a Poisson process with rate \(\lambda\), it has probability density function
\[ f(x) \;\;=\;\; \begin{cases} \dfrac{\lambda^3}{2!} x^2 e^{-\lambda x} & x \gt 0 \\[0.5em] 0 & \text{otherwise} \end{cases} \]If the 3rd event in the Poisson process occurs after time \(x\), there must have been either 0, 1 or 2 events before time \(x\). Therefore
\[ \begin{align} F(x) \;\;&=\;\; 1 - P(Y_x = 0) - P(Y_x = 1) - P(Y_x = 2) \\[0.4em] &=\;\; 1 - \frac {(\lambda x)^0 e^{-\lambda x}}{0!} - \frac {(\lambda x)^1 e^{-\lambda x}}{1!} - \frac {(\lambda x)^2 e^{-\lambda x}}{2!} \\[0.2em] &=\;\; 1 - e^{-\lambda x} - \lambda x e^{-\lambda x} - \frac{\lambda^2}2 x^2 e^{-\lambda x} \end{align} \]Again,
\[ \begin{align} f(x) \;\;&=\;\; F'(x) \\ &=\;\; \lambda e^{-\lambda x} - \left(\lambda e^{-\lambda x} - \lambda^2 x e^{-\lambda x} \right) - \left(\lambda^2 x e^{-\lambda x} - \frac{\lambda^3}{2} x^2 e^{-\lambda x} \right) \\ &=\;\; \frac{\lambda^3}{2} x^2 e^{-\lambda x} \end{align} \]Generalising this, the time until the \(k\)'th event in a Poisson process has a distribution called an Erlang distribution with parameters \(\lambda\) and \(k\). Its probability density function is shown below.
Time until k'th event
The time until the \(k\)'th event in a Poisson process with rate \(\lambda\), has a distribution called an Erlang distribution
\[ X \;\; \sim \; \; \ErlangDistn(k, \lambda) \]with probability density function
\[ f(x) \;\;=\;\; \begin{cases} \dfrac{\lambda^k}{(k-1)!} x^{k-1} e^{-\lambda x} & x \gt 0 \\[0.5em] 0 & \text{otherwise} \end{cases} \]The full proof is not given here, but note how the terms cancelled in the proof for \(k = 3\). Similar cancelling occurs when \(k\) is higher, giving the result.
The Erlang distribution rarely arises in practical applications, but a generalisation of it called a gamma distribution is often encountered. We will therefore not show the distribution's pdf graphically until we look at gamma distributions later.