A particle system with no clock of its own

Countably many particles follow F_j(t) = R_j ⊙ Z^(j)(t): independent copies of a single mark process Z, each rescaled componentwise by an atom R_j of a Poisson measure with the homogeneous intensity ∏ r_i⁻²dr_i. Taking logarithms turns scaling into shifting, and the scalar case shown here is a cloud of copies of log Z displaced by Poisson points of intensity e^−xdx. The paper's point: the system can be stationary although the mark is not, and stationarity is a property of the intensity ν_Z, not of Z. Every curve below carries a visible origin — its own shape is anchored at one instant — and for a stationary choice the cloud has the same law after a time shift. From E. Hashorva, Stationary multihomogeneous Poisson particle systems.

the system on a log scale in a window that slides with time: each curve is log F_j(t) = log R_j + log Z^(j)(t), the upper envelope is dark, particles above the level u are orange

particles above u at each time (bars) against the exact mean e^−u E Z(t) (orange). A stationary system has a flat mean; a flat mean alone does not establish stationarity of the whole path intensity.

Mark

Level u = -2.0 Hurst index H = 0.70 (Brown–Resnick only)

At least 1200 Poisson atoms log R_j = −log Γ_j; only the top curves are drawn, all retained curves are counted. For cosine marks, sampling continues until Γ exceeds 3e^4.5. Since Z ≤ 3, no omitted atom can exceed any selectable level: these grid-point counts have the exact law. The Brown–Resnick cloud uses 1200 atoms and is a truncated approximation; its envelope and counts can be too low.

The mixture surprise

A finite mixture of marks with shift-invariant intensities has a shift-invariant intensity. The converse is the paper's question, and for continuous log-Gaussian marks on a connected index group the answer is yes: every component particle system must itself be stationary. Drop connectedness and it fails. Example 4.2 takes T = ℤ/2ℤ, so a trajectory is just a pair (F(0), F(1)), and stationarity means the cloud’s distribution is invariant under reflection about the diagonal. With log Z_k(1) ∼ N(0,2) and N(−2,2) neither component is — their two expected level counts differ by a factor e — and yet the mixture {ℒ(Z₁) + e ℒ(Z₂)}/(1+e) is invariant in law under that reflection.

Colour shows the logarithm of the bin count. The stationary mixture has a symmetric intensity, while each sampled cloud fluctuates around it. Counted are the particles above the level v at time 0 (right of the vertical line) and at time 1 (above the horizontal line); the upper-right region contains particles above at both times. The two means are e^−v and e^−vE Z(1), in ratio e, e⁻¹, and exactly 1. The finite display and both level counts are sampled without a lower-tail truncation, using Gaussian exponential tilting.