= Solution
From the masses of the <Poisson distribution>, its <probability generating function> is
$$
\boxed{\mathbb E[z^Z]=\sum_{k=0}^\infty z^ke^{-\lambda}\frac{\lambda^k}{k!}
=e^{\lambda(z-1)}\qquad(0\leq z\leq1).}
$$
At $z=0$, 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 $T_0=0$, $T_k=\sum_{j=1}^kE_j$ and
$$
N_t=\max\{k\geq0:T_k\leq t\}.
$$
The <strong law of large numbers> gives $T_k/k\to1/\lambda$ <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 $k\geq1$, exactly $k$ arrivals by $t$ means $T_k\leq t<T_{k+1}$. Integrating the first $k$ exponential gaps and the survival probability of the next gap gives
$$
\begin{aligned}
\mathbb P(N_t=k)
&=\int_{\substack{e_1,\ldots,e_k>0\\e_1+\cdots+e_k\leq t}}
\lambda^ke^{-\lambda(e_1+\cdots+e_k)}
e^{-\lambda(t-e_1-\cdots-e_k)}\,de_1\cdots de_k\\
&=e^{-\lambda t}\lambda^k\frac{t^k}{k!}.
\end{aligned}
$$
The last factor is the volume of the $k$-dimensional simplex, obtained for example by iterated integration. For $k=0$, directly $\mathbb P(N_t=0)=\mathbb P(E_1>t)=e^{-\lambda t}$. Thus
$$
\boxed{N_t\sim\operatorname{Poisson}(\lambda t).}
$$
At 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
$$
\boxed{N_t-N_s\sim\operatorname{Poisson}(\lambda(t-s)),\qquad
\mathbb E(N_t-N_s)=\lambda(t-s)\quad(0\leq s<t).}
$$
Back to article page
