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!!



Poisson point process in polar coordinates
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.


