Poisson process is a Stochastic Processes proof

Question

Suppose that $\{X_t\}_{t>0}$ is a poisson process. Usually the books define it like $$X_t =\textit{"Number of arrivals in a subinterval $[0,t)$}"$$ and it assumes that for each $t\geq 0$ ,$X_t$ is a random variable. I would like to prove that in fact $\{X_t\}_{t>0}$ is a Stochastic Processes. I know that this implies proof for each $t\geq 0$ $X_t$ is a measurable function, $$X_t:(\Omega_1,\mathcal{F}_1)\rightarrow(\Omega_2,\mathcal{F}_2)$$ where $\Omega_1$ is the sample space of the random experiment and $\mathcal{F}_1$ its $\sigma-$algebra. I suppose that $\Omega_1= \{0,1,2,\ldots\} $ and $\mathcal{F}_1=\mathcal{P}(\Omega_1)$. To definate the sample space $\Omega_1$ first of all i have to definate the experiment. What it is? What the formal definition of $X_t$ is? i.e. the mathematic definition of $X_t(\omega)$ where $\omega\in\Omega_1$ . In addition to this, what would $\Omega_2$ and $\mathcal{F}_2$ be? or what the way to proof it is?

Answer 1

People very very rarely define probability spaces explicitly, instead relying on general theorems that prove that probability spaces with the properties they want actually exist. In the case of stochastic processes in continuous time this is actually more complicated than you might think. One typically relies on a theorem called Kolmogorov's extension theorem, which allows one to define only the so-called finite dimensional distributions in a suitably self-consistent manner in order to "lift" them into a distribution on function space (which is itself a rather complicated object).

That being said, in the case of a jump process with independent jump times and independent jump increments, there is a relatively simple way to do it. You simply need the jump time distribution and the jump distribution; in the case of a Poisson process, the former is the exponential distribution with rate $\lambda$ and the latter is just the trivial distribution that is always just $1$. Then let $T_i$ be iid random variables distributed according to the jump time distribution and let $J_i$ be iid random variables distributed according to the jump distribution. Then $X_t=\sum_{n : \sum_{i=1}^n T_i \leq t} J_n$.

This can be done in a straightforward manner as long as you have a probability space where two sequences of independent random variables can be defined. This is something you have probably taken for granted in the past, but it's not totally obvious how to do it; for example, a simple trick based on the probability integral transformation applied to a multivariate uniform distribution on $(0,1)$ will not work. Again this is something that you could use Kolmogorov's extension theorem to prove can be done (but at least it has been reduced to something that is plausible).