# Strong Markov property for Poisson point process

The question is thoroughly contained in the title. I just say that I would only like to find a reference for this question. I have searched in some books, to no avail.

Here is what I mean exactly. Let's say we have a process $N$ on $\mathbb{R} _+\times \mathbb{R}^d$, $\mathbb{R} _+$ is time, with intensity being the Lebesgue measure on $\mathbb{R} _+\times \mathbb{R}^d$. Let $\mathscr{F} _t$ be the minimal $\sigma$ -algebra containing all random variable $N(Q \times U)$, where $Q \in \mathscr{B}([0;T])$, $U \in \mathscr{B}(\mathbb{R}^d)$. Or we may take minimal complete right-continuous $\sigma$ - algebra with this property. Let $\tau$ be a stopping time with respect to $(\mathscr{F} _t)$. It seems reasonable to conjecture, that the process $\bar N$ defined by

$$\bar N ([0;s] \times U) = N ([\tau;\tau + s] \times U), \ \ \ U \in \mathscr{B}(\mathbb{R}^d)$$

is a Poisson point process with the same intensity measure independent of $(\mathscr{F} _{\tau})$. I was not able however to find a reference to this statement.

In terms of random sets, it corresponds to the strong Markov property of the set $[0;\tau] \times \mathbb{R}^d$, which is not compact.

In the book by Rozanov, link, the strong Markov property is considered for compact stopping sets.

Update

I would like to add two things.

First, it looks like the strong Markov property for $\bar N$ follows from Theorem 20.9 here:

Theorem 20.9 Suppose $(X_t , P^x)$ is a Markov process with respect to $\{\mathscr{F}_t \}$, that Assumption 20.1 holds, and that $T$ is finite stopping time. If Y is bounded and measurable with respect to $\mathscr{F}_\infty$, then $$E^x[Y \circ \theta _T |\mathscr{F}_T ] = E^{X_T}Y, \ \ \ \ P^x-a.s.$$

Here $\{\mathscr{F}_t \}$ is the minimal right-continuous filtration, which contains all sets $N$ satisfying $P^x(N)=0$ for all $x$, and such that $(X_t)$ is adapted: $$\mathcal{F}_t^{00} = \sigma(X_s : s \leq t),$$ $$\mathcal{F}_t^{0} = \sigma \left(\mathcal{F}_t^{00} \cup \{ A \subset S : A \text{ is } \Bbb{P}^x\text{-null for all } x \in S\} \right) \quad \text{and} \quad \mathcal{F}_t = \mathcal{F}_{t+}^{0} = \bigcap_{\epsilon > 0} \mathcal{F}_{t+\epsilon}^{0},$$ $S$ is the state space and $P_t$ is associated with the process semigroup. Assumption 20.1:

Assumption 20.1 Suppose $P_t f$ is continuous on $S$ whenever $f$ is bounded and continuous on $S$.

Assumption 20.1 is satisfied, if we consider a Poisson point process as a canonical process in the space $D_S [0;T_1]$, $T_1 >0$, $S =\Gamma$. Here $\Gamma$ is the space of all simple counting measures over $\mathbb{R}^d$, equipped with the vague topology, i.e. the smallest topology such that for every $f \in C_K(\mathbb{R}^d)$ the mapping

$$\Gamma \ni \gamma \to \int f d \gamma$$ is continuous. With this topology, $Г$ is a Polish space, link. Then assumption 20.1 is equivalent to

\begin{equation} E g(\gamma _n \cup N _t) \to E g(\gamma \cup N_t), \ \ \ \text{whenever} \ \gamma _n \to \gamma \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{equation} where $g: \Gamma \to \mathbb{R}$ is a bounded and continuous function. Convergence $\gamma _n \to \gamma$ in the vague topology implies $\gamma _n \cup N _t \to \gamma \cup N_t$ a.s. Therefore, (1) follows by the bounded convergence Theorem.

Also, one should prove that $(N_t)$ is a Markov process under $\{\mathscr{F}_t \}$ (this is an assumption of Theorem 20.1).

Second, I have tried to prove the statement in the original question, using the idea suggested by Anthony Quas in his comment. To prove that $N$ is a strong Markov, enough to show that

1. for any $b>a>0$ and bounded open $U \subset \mathbb{R}^d$, $\bar N ((a;b),U)$ is an independent of $\mathscr{F} _\tau$ Poisson random variable with mean $(b-a)\lambda (U)$, $\lambda$ is the Lebesgue's measure on $\mathbb{R}^d$, and
2. for any $b_k>a_k>0$, $k=1,...,m$, and any bounded open $U_k \subset \mathbb{R}^d$, such that $((a_i;b_i) \times U_i) \cap( (a_j;b_j) \times U_j) = \varnothing$, $i \ne j$, random variables $\bar N ((a_k;b_k) \times U_k)$ are independent of each other.

To do so, let $\tau _n$ be the sequence of stopping times taking only countably many values, $\tau _n \downarrow \tau$, $\tau _n - \tau \leq \frac{1}{2^n}$. Then $N$ satisfies strong Markov property for $\tau _n$, and the processes $\bar N _n$ defined by $$\bar N _n ([0;s] \times U) = N ([\tau_n;\tau_n + s] \times U),$$ are Poisson point processes. To prove 1, note that $\bar N _n ((a;b) \times U) \to \bar N ((a;b) \times U)$ a.s. and all random variables $\bar N _n ((a;b) \times U)$ have the same distribution, therefore $\bar N ((a;b) \times U)$ is a Poisson random variable with mean $(b-a)\lambda (U)$. Random variables $\bar N _n ((a;b) \times U)$ are independent of $\mathscr{F}_{\tau}$, hence $\bar N ((a;b) \times U)$ is independent of $\mathscr{F}_{\tau}$, too. Similarly, 2 follows.

• Did you check Kallenberg's Foundation of Modern Probability, Theorem 12.14? – Jochen Wengenroth Jun 4 '14 at 14:12
• @Jochen Wengenroth It is not applicable here. The process $N$ is not pure jump type, since infinitely many points appear before any positive moment. It would do if we had a finite measure over $\mathbb{R} ^d$ as characteristic measure, instead of the Lebesgue measure. – Sinusx Jun 4 '14 at 14:20
• It looks as though your definition is not what you mean? Do you want $\bar N([0,s],U) := N([\tau,\tau+s],U)$? I don't see why you would subtract what happened before $\tau$. – Anthony Quas Jun 4 '14 at 18:50
• Have you tried defining $\tau_n=\min\{\frac kn\colon \frac kn\ge \tau\}$? This is another stopping time. Since it takes only countably many values, the Markov property is easy there. Since the stopping times $\tau_n$ converge to $\tau$, are you done? – Anthony Quas Jun 4 '14 at 18:55
• @Anthony Quas About your first comment, you are right, it was a typo. About proving this, I think I can deduce this property from strong Markov property for processes with finite characteristic measure. The thing is, I thought it was a natural question and it should have been mentioned somewhere, and since I could not find it, maybe there is a snag? – Sinusx Jun 4 '14 at 20:41