Poisson point process in polar coordinates

Question

Let $D = \mathbb{R^+} \times (\mathbb{R}\backslash \{0\})$
Let $\mu(dt \times dx)$ be a $\sigma$-finite measure on the Borel $\sigma$-algebra $\sigma(D)$.
Let $M(dt \times dx)$ be the Poisson random measure with intensity measure $\mu$, i.e. for $B \in \sigma(D)$, $M(B)$ is a Poisson random variable with intensity $\int_B \mu (dt \times dx)$.
(or more generally $E \int f(t,x)dM = \int f(t,x)d \mu$ etc.)

Question:
Is it possible to express $M$ in polar coordinates?
What is the transformed intensity measure?
What I mean is this:
Taking $E = \mathbb{R^+} \times ((-\pi/2, \pi/2) \backslash \{0\})$, how do I define a Poisson random measure $Q$ on $\sigma(E)$, so that for a set $B \in \sigma(D) \cap \sigma(E)$, $M(B)$ and $Q(B)$ have the same distribution. Let $q(dt \times dx)$ be the intensity measure of $Q$. What is the expression for $q$ in terms of $\mu$?

Finally, if $\mu(dt, dx) = dt \times m(dx)$ (i.e., $\mu$ is homogeneous in time, does $q$ have the same property? ie can $q$ be written as $q(dt \times d \theta) = d t \times r(d \theta)$?

References?
Many thanks in advance!!

Answer 1

This kind of thing is studied at length in Random Measures, Point Processes, and Stochastic Geometry by Baccelli, Blaszczyszyn, and Karray. The book is made freely available in pdf form by the authors, and you can find it by searching for the title.

Theorem. Given two locally compact second-countable Hausdorff spaces $G, G'$ equipped with their Borel $\sigma$-algebras and $g:G \to G'$ measurable such that $g^{-1}(B)$ is relatively compact for all relatively compact Borel $B \in \mathcal{B}(G')$, for any Poisson point process $\Phi$ on $G$ of intensity measure $\Lambda$, one has that $\Phi\circ g^{-1}$ is a Poisson point process on $G'$ of intensity measure $\Lambda' = \Lambda \circ g^{-1}$.

See the mentioned book for proof.

In particular, this can be applied to a polar change of coordinates between, e.g., $G = \mathbb{R}^2$ with the nonnegative $x$-axis removed and $G' = (0,\infty) \times (0, 2\pi)$. I.e. $g^{-1}(r,\theta) = (r\cos \theta, r \sin \theta)$. Assuming $\Lambda$ has a density $\lambda$ on $\mathbb{R}^2$, for any measurable $f: G \to \mathbb{R}^+$ one has $$\begin{align} \int f(r\cos\theta, r\sin\theta)\,\Lambda \circ g^{-1}(dr,d\theta)) &= \int f(x,y)\,\Lambda(dx, dy) \\ &=\int f(x,y) \lambda(x,y) \,dx\,dy \\ &= \int f(r \cos \theta, r \sin \theta) \lambda(r\cos\theta, r\sin\theta)r \,dr\,d\theta. \end{align} $$ In this case we see the intensity measure $$\Lambda'(dr,d\theta) = \Lambda\circ g^{-1}(dr,d\theta) = \lambda(r\cos\theta,r\sin\theta) r \,dr\,d\theta.$$ Thus if $\Lambda(dx, dy) = dx \times m(y)dy$, we find that $$\Lambda'(dr,d\theta) = m(r\sin\theta)r \,dr\,d\theta$$ so it will not in general be homogeneous in "time" (the $r$ coordinate). In order to get time homogeneity, one needs that $\Lambda$ is radial.