I can't seem to find how to derive this equation:
$$Pr(X(t)=n) = \frac1{n!}\left(\int_{0}^t \lambda(\tau) d\tau\right)^n\exp\left(-\int_{0}^t \lambda(\tau) d\tau\right)$$
Where $X(t)$ is the number of events over a time $t$.
The proof for a homogeneous rate $\lambda$ involves looking at the Binomial distribution where each trial is a small interval $\Delta$t. However, obviously that won't work in this case as the intervals won't have the same success rates.
In Homogeneous Poisson Process(HPP), the intensity parameter is a constant.
Non-Homogeneous Poisson Process(NHPP) is a generalization of Poisson Process, in the sense that, it allows the intensity parameter to be a function of time $t$. The dependence of intensity parameter on time implies that events are more likely to occur at certain times compared to other times.
For example, customers arriving into a bank are not uniform throughout the business hours; there will be some rush hours and some leisure hours. In a household, the number of units of electricity consumed will vary according to time. The vehicular traffic on the road will not be uniform throughout the day.
Due to the dependence of arrival rate on time, the process no longer has the stationarity property.
This generalization does not effect the form of the distribution of the process, but allows variation in the arrival rate to occur as determined by the function $\lambda(t).$
This can be observed from the following:
Let $G(s,t)$ denote the PGF of the NHPP. Then, by definition, we have \begin{eqnarray*} G(s,t+h) & = & E[s^{N(t+h)}] \\ & = & E[s^{N(t+h)-N(t)}\cdot s^{N(t)}]\\ & = & E[s^{N(t+h)-N(t)}] \cdot \underbrace{ E[s^{N(t)}]}_{G(s,t)}\\ & = & G(s,t) \cdot E[s^{N(t+h)-N(t)}]\\ & = & G(s,t)\cdot \left\{ \sum_{k=0}^{\infty} P\{N(t+h) - N(t)= k\}\cdot s^{k} \right\}\\ & = & G(s,t)\; \{\; P\{N(t+h)-N(t) = 0 \} \quad + \\ & & \quad\quad\quad P\{N(t+h)-N(t) = 1 \}\cdot s + \\ & & \quad\quad\quad \sum_{j=2}^{\infty}P\{N(t+h)-N(t) = j\} \cdot s^{j}\}\\ & = & G(s,t) \cdot \left[(1-\lambda(t)h + o(h)) + \lambda(t)h \cdot s + o(h)\right] \end{eqnarray*} Transposing $G(s,t)$ to the left hand side and dividing throughout by $h$ and letting $h\rightarrow 0$, yields $$\quad\quad\quad\quad G^{'}(s,t) = -(1-s)\cdot \lambda(t) \cdot G(s,t).$$ $$\frac{G^{'}(s,t)}{G(s,t)} = -(1-s)\cdot \lambda(t)$$ Integrating on both sides, we get $$\ln G(s,t) = -(1-s)\int_{0}^{t}\lambda(u)du + C(s)$$ $$G(s,t) = \; exp \left\{-(1-s)\int_{0}^{t}\lambda(u)du\right\}A(s)$$ But, $G(s,0) = 1 = A(s).$ Thus, $$G(s,t) = \;exp\left\{-(1-s)\int_{0}^{t}\lambda(u)du\right\}$$ which is the PGF of a Poisson process with mean $$E\left\{N(t)\right \}= \int_{0}^{t}\lambda(u)du$$