Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 101 / 5 / a

From the masses of the Poisson distribution, its probability generating function isAt $z=0$, the term $z^0$ is interpreted as one, so this also gives $\mathbb{P}(Z=0)=e^{-\lambda }$.

One constructive definition of a rate-$\lambda$ Poisson process uses independent and identically distributed exponential random variables $E_j$ of rate $\lambda$. Set $T_0=0$, $T_k={\sum }_{j=1}^k{E}_j$ andThe strong law of large numbers gives $T_k/k\to 1/\lambda$ almost surely, so the maximum is finite at every finite $t$. The process starts at zero, has right-continuous nondecreasing integer paths, and jumps by one at the arrival times.

For $k\ge 1$, exactly $k$ arrivals by $t$ means $T_k\le t<T_{k+1}$. Integrating the first $k$ exponential gaps and the survival probability of the next gap givesThe last factor is the volume of the $k$-dimensional simplex, obtained for example by iterated integration. For $k=0$, directly $\mathbb{P}(N_t=0)=\mathbb{P}(E_1>t)=e^{-\lambda t}$. ThusAt a fixed time $s$, conditional on the counting history, the residual time until the next arrival has rate-$\lambda$ exponential distribution, independently of that history, by the memoryless property. After that arrival all subsequent gaps are fresh independent exponentials. Thus the process of future counts $N_{s+u}-N_s$ has the same law as $N_u$ and is independent of the past. This proves stationary independent increments, the equivalent usual defining properties of the Poisson process, and gives