I would like to introduce some students to the notion of Poisson process beginning from the qualitative definition. What I mean is that I would like to introduce it as a special case of processes with independent increments and derive all properties from first principle. Some say this is the hard way, but it is probably the most rigorous as, otherwise, one has to guess the distribution, etc. prove that such a stochastic process exists, and so on.
So I am attacking the problem like this.
Let $\mathbb{F} = \{{\cal F}_t: t\ge 0\} $ be a (natural) filtration and let $\{\tau_n: n \in \mathbb{N}\} $ be a sequence of stopping times with respect to such filtration. Then $\{\tau_n: n \in \mathbb{N}\} $ generates a point process} if it satisfies the following requirements: $\tau_0 = 0; $ $\tau_n \le \tau_{n+1}; $ and if $\tau_n(\omega) < \infty, $ then $\tau_n(\omega) < \tau_{n+1}(\omega).$
The process $N=\{N_t: t \in \mathbb{R}^+\} $ defined by $$ N_t = \sum_{n\ge 1} 1_{\{t\ge \tau_n\}} $$ with values in $\overline{\mathbb{N}} $ is called the counting process associated to the sequence $\{\tau_n: n \in \mathbb{N}\}. $
A counting process $N $ is a Poisson process if it has:
(independent increments): for any $s, t, 0 \le s < t < \infty, $ $ N_t-N_s $ is independent of ${\cal F}_s; $
(stationary increments): for any $s, t, u, v, $ $0 \le s < t < \infty, $ $0 \le u , v < \infty, $ $t-s = v-u, $ then $N_t- N_s \stackrel{D}{=} N_v- N_u. $
In constructing all properties of the Poisson process from these "axioms", I come to a point (just a point, not a point process :-) ) where I need to prove that $E[N_1] < \infty. $ so that I can use it to justify an application of the dominated convergence theorem in one step and interchange limit and expectation.
Is there any simple way to justify the claim $E[N_1] < \infty$~?
I would be grateful for any suggestion. I think that Protter attacks this point using results for general Levy processes, but is there a way to prove that point without using any more structure? Come to think that if the process is not "exposive", meaning $\lim_n \tau_n = \infty $ it must be the case as there cannot be infinite jumps in the interval $[0, 1], $ but I am afraid that I introduce a circularity as I would first need to prove that a Poisson process is without explosions for which I need to prove that $P(N_t \ge 2) $ is $o(t) $ which is why I need to know that $E[N_1] < \infty $ in first place.
I would welcome any suggestions that allows to show that $E[N_1] < \infty $ without getting too much out of control.
Thank you.
For $t,s\geqslant0$, we have by independent increments $$\mathbb P(N_{t+s}=0) = \mathbb P(N_t=0)\mathbb P(N_{t+s}-N_t=0), $$ and by stationary increments $$\mathbb P(N_{t+s}-N_t=0) = \mathbb P(N_s=0), $$ so $$\mathbb P(N_{t+s}=0) = \mathbb P(N_t=0)\mathbb P(N_s=0). $$ Let $f(t) = \mathbb P(N_t=0)$. Then $$f(t+s)=f(t)f(s),\quad 0\leqslant f(t)\leqslant 1$$ so $f(t)= e^{-\lambda t}$ for some $\lambda>0$ (as otherwise $f$ is identically $0$, which leads to a contradiction). Therefore $$\mathbb P(N_{\tau_n+s}-N_{\tau_n}=0\mid \mathcal F_{\tau_n})=\mathbb P(N_s=0)=e^{-\lambda s}, $$ and since $$\{N_{\tau_n+s}-N_{\tau_n}=0\} = \{\tau_{n+1}-\tau_n>s\}, $$ it follows that $\tau_{n+1}-\tau_n\sim\textsf{Exp}(\lambda)$. From here we may use the renewal equation $$\mathbb E[N_t] = F+\mathbb E[N_t]\star F $$ where $F(t)=1-e^{-\lambda t}$ to show that $\mathbb E[N_t]=\lambda t<\infty$.