How does Poisson thinning work?

Question

I'm trying to understand how "Poisson thinning" can be used to construct a nonhomogeneous Poisson process.

Here is what I got: Given a measure space $(E,\mathcal E,\alpha)$, a "Poisson random measure on $(E,\mathcal E)$ with intensity $\alpha$" is a transition kernel $\zeta$ from a probability space $(\Omega,\mathcal A,\operatorname P)$ to $(E,\mathcal E)$ such that

1. if $B\in\mathcal E$, then $$\zeta(B):=\zeta(\;\cdot\;,B)\sim\operatorname{Poi}(\alpha(B))\tag1;$$
2. if $n\in\mathbb N$ with $n\ge2$ and $B_1,\ldots,B_n$ are disjoint, then $\zeta(B_1),\ldots,\zeta(B_n)$ are independent.

Now, I guess the idea of "Poisson thinning" (but maybe it can be generalized) is the following: Let $N$ be a Poisson random measure on $$A:=\{(x,y)\in[0,\infty)^2:y\le f(x)\}$$ with intensity $\left.\lambda^{\otimes2}\right|_A$, where the latter is the restriction of the Lebesgue measure $\lambda^{\otimes2}$ on $\mathbb R^2$ to $A$ and $f:[0,\infty)\to[0,\infty)$ is Borel measurable. Using $N$, we now (somehow) want to construct a Poisson random measure $\zeta$ with intensity $\alpha:=\left.f\lambda\right|_{[0,\;\infty)}$, where the latter is the measure with density $f$ with respect to $\left.f\lambda\right|_{[0,\;\infty)}$.

How exactly is that done?

I've started working on this question as I was reading p. 253 of Non-Uniform Random Variate Generation. However, while Poisson thinning is described therein, I don't even understand their (confusing) definition of a (nonhomogeneous) Poisson process. Any help on this is highly appreciated!

Answer 1

I think there are two ways to define the thinning of a given random measure. Let me keep the more general description:

Let

• $(E,\mathcal E)$ be a measurable space;
• $\mu$ be a measure on $(E,\mathcal E)$;
• $p:E\to[0,\infty)$ be $\mathcal E$-measurable and $$\nu(B):=\int_Bp\;{\rm d}\mu\;\;\;\text{for }B\in\mathcal E.$$

The goal is to construct a random measure on $(E,\mathcal E)$ with intensity measure $\nu$ out of another given random measure.

Way 1:

If $\zeta$ is a random measure on $(E,\mathcal E)$ with intensity $\mu$, then $$\zeta_p(B):=\zeta(1_Bp)\;\;\;\text{for }B\in\mathcal E$$ is a random measure with intensity $\nu$.

Way 2:

If $$A:=\left\{(t,x)\in[0,\infty)\times E:t\le p(x)\right\},$$ $\lambda$ is the Lebesgue measure on $\mathbb R$ and $\zeta$ is a random measure on $(A,\left.(\mathcal B(\mathbb R)\otimes\mu)\right|_A)$ with intensity measure $\left.(\lambda\otimes\mu)\right|_A$, then $$\zeta_p(B):=\zeta(A\times(\mathbb R\times B))\;\;\;\text{for }B\in\mathcal E$$ is also a random measure with intensity $\nu$.

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I guess that way 2 is preferred in the book, cause it gives rise to a numerical simulation of a nonhomogeneous Poisson process, but any remark on this in the comments is highly appreciated.

Answer 2

A canonical definition of a Poisson random measure $N_\alpha$ with intensity measure $\alpha$ is exactly what you have written. You can turn a Poisson random measure $N_\alpha$ on $\mathbb{R}$ into a Poisson random process $P_\alpha$ on $\mathbb{R}$ with the same intensity measure by taking $$ P_\alpha(t) := N_\alpha([0,t]). $$ Poisson random process is a pure jump process and a jump size is a.s. $1$, and those moments of jumps determine the process. In the book the Poisson process on $\mathbb{R}$ is represented as a sequence of random variables $X_{1} < X_{2} < \dots < X_{i} <\dots$ that represent consecutive moments of jumps of $P_\alpha$.

The Poisson process on $A \subset\mathbb{R}^{n}$ is called homogenous if its intensity measure is $C \cdot\lambda^{n}|_{A}$, for $C > 0$ and $\lambda^{n}$ is a Lebesgue measure on $\mathbb{R}^{n}$. The Poisson process on $\mathbb{R}$ is called non-homogenous with rate $\lambda(t)$ if its intensity measure has density $\lambda(t)$ (total mass of which might not equal 1, but otherwise it is a density).

The Poisson thinning:
Using other methods we have already simulated a Poisson process $P_{\mu}$ with intensity rate $\mu(t) \ge \lambda(t)$. It means that we have an increasing sequence of moments of jumps $Z_{i}$ of $P_{\mu}$. The goal is to turn it into a Poisson process $P_{\lambda}$ with intensity rate $\lambda(t)$.

During step number $i$ we take moment $Z_{i}$, independently simulate a uniform random variable $U_{i}$ on $[0, 1]$, then we take that moment as a next moment of a jump for $P_{\lambda}$ only if $U_{i} \le \frac{\lambda(t)}{\mu(t)}$. Basically we discard all moments of jumps of $P_{\mu}$ for which $U$ is big. The moments of jumps we didn't discard form $P_{\lambda}$.

Why it works:
We use theorem 1.3. If needed, I can describe how.