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For an isotropic Poisson line process of intensity $cdpd\theta$, its pair intersections form a locally finite point process. Coincident intersections and parallel pairs have probability zero. Its intensity measure is $\pi c^2$ times area. Indeed the Poisson factorial moment measure and the change of variables from signed distances to intersection location giveThe factor $1/2$ counts unordered line pairs. Shared lines create dependence between intersections; this process is not a Poisson point process.

Identify a line with its unique parameter pair in $E=(0,\infty )\times [0,2\pi )$, usingA Poisson line process means a Poisson point process on the Borel parameter space $E$: for each measurable $B$ of finite mean measure, the number of its parameter points is Poisson with mean $\mu (B)$, and counts in finitely many disjoint sets are independent. Infinite-measure sets are handled by a countable finite-measure exhaustion. Here $\mu (dp,d\theta )=dpd\theta$ is sigma-finite and nonatomic, so there are countably many distinct lines almost surely. It is not an assertion that line counts are Poisson with respect to planar area.

Fix $r>0$ and let $D_r=\{z:|z|<r\}$. Exactly the lines with $0<p<r$ hit this disk, apart from the zero-probability tangent case. Thus the count of disk-hitting lines isEvery intersection in $D_r$ uses two of these lines. Writing $N_r=\#(\Phi \cap D_r)$, we haveThis also proves local finiteness of the intersection point process. Initially it gives a non-strict bound $\mathbb{P}(N_r\ge 1)\le 1-(1+2\pi r)e^{-2\pi r}$.

To establish strictness, choose the two parameter boxesEach has positive measure, and both lie among the disk-hitting lines. If one line from each box intersected at $z\in D_r$, thenBut their distance parameters differ by more than $r/4$, a contradiction. The event of exactly one line in each box and no other disk-hitting lines has probabilityOn this event $M_r=2$ but $N_r=0$. ThereforeThe following original diagram illustrates why two disk-hitting lines need not produce an intersection inside the disk.

To decide whether $\Phi$ is Poisson, compute its intensity measure and compare its void probability. Use the equivalent signed line coordinates $L(t,\theta )$ with $t\in \mathbb{R}$ and $0\le \theta <\pi$. If $t>0$ this is the original pair $(t,\theta )$; if $t<0$ it is $(-t,\theta +\pi )$. The measure is $dtd\theta$, with no extra factor of two. Translation by $a$ sends $t$ to $t+a\cdot n_{\theta }$, preserving this measure. Thus both the line and intersection processes are stationary.

Almost surely no two lines are parallel and no three are concurrent. In each finite-intensity parameter region, conditional line parameters have continuous joint densities. Parallel pairs and concurrent triples satisfy lower-dimensional equations and have probability zero; a countable exhaustion proves the assertion for all lines. Hence different unordered pairs have distinct intersections almost surely. For a bounded Borel set $A$, the second Poisson factorial moment measure identity givesFor nonparallel normals, their intersection $z$ satisfies $t=z\cdot n_{\theta }$, $s=z\cdot n_{\varphi }$. The absolute Jacobian determinant of this transformation is $|\sin (\varphi -\theta )|$. Consequentlysince the inner integral is $2$ for every $\theta$. In particular $\mathbb{E}{N}_r={\pi }^2{r}^2$.

If $\Phi$ were a Poisson point process, the count $N_r$ would therefore be Poisson with that mean, giving $\mathbb{P}(N_r=0)=e^{-{\pi }^2{r}^2}$. But the earlier necessary-two-lines argument givesAt $r=2/\pi$ this lower bound is $5e^{-4}$, whereas the supposed Poisson void probability is $e^{-4}$. This contradiction proves $\Phi$ is not a Poisson point process. Its points share parent lines, and the explicit void/mean comparison demonstrates the resulting dependence rather than merely asserting it.