I would like to understand better the derivation of the inter-arrival time for a nonhomogeneous Poisson process. Can any one supply a link to a nice clear derivation? I'm only interested in the time to the first event so the survival function would be great but general case is fine. The only references I can find essentially state the result without a derivation.

Let $T_1, T_2, \dots$ denote the interarrival times of events of a nonhomogeneous Poisson process having intensity function $\lambda(t)$.

  1. Are the $T_i$ independent?
  2. Are the $T_i$ identically distributed?
  3. Find the distribution of $T_1$.

1 Answer 1


In general, the times $T_i$ are not independent and not identically distributed. So questions 1 and 2 are answered in the negative. It is not too difficult to find examples in which they are clearly dependent and/or clearly differently distributed. I leave such a demonstration for someone else but shall however try to derive the distribution of $T_1$.

First recall that if $\lambda(t)=\lambda$ is fixed (so in case of the usual homogeneous Poisson process), the probability that there is no point in an interval of length $s$ is $e^{-\lambda s}$. This is because we know that the number of points $N_s$ in an interval of length $s$ is Poisson distributed with parameter $\lambda s$ (constant rate $\times$ interval length): $$ \text{Prob}[N_s = n] = e^{-\lambda s} \frac{(\lambda s)^n}{n!} $$ so $$ \text{Prob}[N_s = 0] = e^{-\lambda s} $$

Secondly, consider the probability that no point occurs in the interval $[0,s]$ for the nonhomogeneous process. To obtain this probability, we are going to cut the interval in equal parts of length $\Delta$ and then let $\Delta\to 0$. Every part then becomes very (infinitesimally) small so that we can assume the rate $\lambda(t)$ is constant in that small part. If the rate is constant there (say equal to $\lambda$), the number of points in that part is Poisson distributed with parameter $\lambda \Delta$. So we have: \begin{align*} \text{Prob}\big[\text{no point in }[0,s]\big] & = \lim_{\Delta\to 0}\text{Prob}\big[\text{no point in }[0,\Delta], \ldots, \text{no point in }[s-\Delta,s]\big]\\ & = \lim_{\Delta\to 0}\prod_{i=0}^{s/\Delta-1}\text{Prob}\big[\text{no point in }[i\Delta,(i+1)\Delta]\big]\\ & = \lim_{\Delta\to 0}\prod_{i=0}^{s/\Delta-1} e^{-\lambda(i\Delta)\Delta}\\ & = \lim_{\Delta\to 0} \exp \Big[ - \sum_{i=0}^{s/\Delta-1} \lambda(i\Delta)\Delta) \Big]\\ & = \exp \Big[ - \lim_{\Delta\to 0} \sum_{i=0}^{s/\Delta-1} \lambda(i\Delta)\Delta \Big] = \exp \Big[ - \int_0^s \lambda(t) \text{d}t \Big] = e^{-\Lambda(0,s)} \end{align*} because the last limit is nothing but a Riemann sum that is the area under the graph of $\lambda(t)$ from $t=0$ to $t=s$. For convenience, let us call $\Lambda(t_1,t_2) =\int_{t_1}^{t_2} \lambda(t)\text{d}t$.

Finally, consider the probability density $f(t)$ function of $T_1$: \begin{align*} f(t)\text{d}t & = \text{Prob}[ t \leqslant T_1 < t+\text{d}t ]\\ & = \text{Prob}\big[\text{no point in }[0,t]\big] \cdot \text{Prob}\big[\text{1 point in }[t,t+\text{d}t[\big]\\ & = e^{-\Lambda(0,t)} \cdot e^{-\lambda(t)\text{d}t}\lambda(t)\text{d}t = e^{-\Lambda(0,t)} \lambda(t)\text{d}t \end{align*} so the density of $T_1$ is $$ f(t) = e^{-\Lambda(0,t)} \lambda(t) $$ and the distribution function $F(t)=\text{Prob}[T\leqslant t]$ follows by integration of $f(t)$ as $$ F(t) = 1- e^{-\Lambda(0,t)} $$

  • $\begingroup$ Thanks a lot! This explanation really helped out my problem $\endgroup$
    – GorillaInR
    Feb 13, 2018 at 0:23

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