Random line processes

August 2021

How do you draw lines at random in the plane, in such a way that the corresponding process does not depend on the origin?

Representing lines with points

Let \ell be 2d line – a one-dimensional subspace of R2\mathbb{R}^2.

Mathematically, it can be represented in many ways: for instance, if it is not vertical, there are two unique numbers α,β\alpha, \beta such that ={(x,αx+β):xR}\ell = \{ (x, \alpha x + \beta) : x \in \mathbb{R} \}. A less trivial way of representing lines is with orthogonality: if \ell does not go through the origin, there is a unique couple z=(r,θ)z=(r, \theta) with r0r \geqslant 0 and θ[0,2π[\theta \in [0, 2\pi[ such that \ell is the unique line passing through the point p=(rcos(θ),rsin(θ))p=(r\cos(\theta), r\sin(\theta)), and orthogonal to the line (0,p)(0, p). If \ell passes through zero, the same is true provided that θ[0,π[\theta \in [0, \pi[.

This correspondance

(θ,r)\ell \leftrightarrow (\theta, r)

is one-to-one between the set of lines L\mathscr{L}, and the "strip" A=]0,2π[×]0,[[0,π[×{0}\mathscr{A} = ]0, 2\pi [\times ]0, \infty [ \cup [0, \pi[\times \{0\}.

Random lines

A random line process is simply the set of lines obtained by sampling a random point process on the strip A\mathscr{A} and looking at the lines they represent.

For instance, a Poisson line process is nothing but the set of lines obtained by sampling a Poisson point process on A\mathscr{A}.

One could easily create many kinds of random processes by first sampling classical point process Φ\Phi in A\mathscr{A}, and then using the correspondance to get the line analog of this process. But...

Is it shift-invariant ?

Just as points in Rd\mathbb{R}^d, lines can be shifted. If xx is a point and \ell a line, +x\ell + x is simply the set {y+x:y}\{y+x : y \in \ell \}. A crucial question in geometric probability is to create random line processes LL that are shift-invariant: the distribution of LL, and the distribution of Lx={x:L}L - x = \{\ell - x : \ell \in L \} must be the same.

While such a condition is not too difficult to check for many interesting point processes[1], it is not so obvious for the kind of line processes I described above.

Indeed, when a line \ell represented by (θ,r)(\theta, r) is shifted by xx, its representation in A\mathscr{A} undergoes a non trivial transformation: if α\alpha is the argument of xx, then

x(θ,r) \ell - x \quad \leftrightarrow \quad (\theta', |r'|)

where r=r+xcos(θρ)r' = r + |x|\cos(\theta - \rho), and θ=θ\theta' = \theta if r>0r>0, otherwise it is equal to (θ+π)(\theta + \pi) modulo 2π2\pi.

Hence, if we note Tx:AAT_x : \mathscr{A} \to \mathscr{A} the group of transformations described above, we obtain the simple caracterization: a line process is shift-invariant if its representation in A\mathscr{A} is TxT_x-invariant for all xx. While this can be checked for what I described above as the Poisson line process, it also prevents other naive constructions from being stationary. For instance, sampling a Ginibre point process on C\mathbb{C}, then taking its restriction in the strip A\mathscr{A}does not yield a stationary line process.

The animated picture shows the effect of such transformations: on the right, the lines are slowly shifted by (t,0)(t,0) for tt between 2-2 and 22, and on the left, we see the corresponding A\mathscr{A}-representation being transformed by T(t,0)T_{(t,0)}. Since the shift is horizontal, horizontal lines merely move: this corresponds to the points on the lines θ=π/2,π/2\theta = \pi/2, -\pi/2 being invariant.

Is it isotropic ?

The same question can be asked with isotropy, that is: rotation invariance. When a line \ell is represented by (θ,r)(\theta, r) in A\mathscr{A}, rotating around the origin by an angle τ\tau is the same thing as replacing θ\theta by θ+τ\theta + \tau (modulo 2π2\pi). Consequently, a line process is isotropic if and only if its representation in A\mathscr{A} is invariant with respect to the transformations Rτ: (θ,r)(θ+τ,r)R_\tau : (\theta, r) \to (\theta + \tau, r) where the angle addition is taken modulo 2π2\pi.

A typical example of non-isotropic point process is obtained by choosing a finite set of directions, say (θ1,,θk)(\theta_1, \dotsc, \theta_k), and for each of them to sample a translation-invariant point process Φj\Phi_j on R\mathbb{R}. The union

{(θi,x):i[k],xΦi}\{ (\theta_i, x) : i \in [k], x \in \Phi_i \}

is translation-invariant, but certainly not isotropic.


[1] for instance, for determinantal point processes it is trivially verified if the kernel itself is shift-invariant.