An exact Jordan signature in cumulative-count first-passage statistics

Abstract

For cumulative-count first-passage problems, the Laplace transform of the time ${\mathcal{T}}_N$ of the $N$th count is naturally expressed through powers of a one-count kernel. A Jordan defect of that kernel does not, however, automatically survive the complete sum over terminal phases. We give an exact three-state Markov-renewal example in which it does. At $s=1$, the one-count kernel has spectrum $\{1/2,1/8,1/8\}$, with a one-dimensional eigenspace at $1/8$, and for initial phase $1$, \[ \mathbb E_1[e^{-\mathcal T_N}] =\frac56\,2^{-N}+\frac{N+1}{6}\,8^{-N}. \] Thus the ordinary, unconditioned cumulative-count transform contains the explicit Jordan contribution $N/(68^N)$ at every threshold $N\ge 1$. For the same fixed stochastic process, the two eigenvalues meeting at $1/8$ unfold as \[ λ_\pm(s)=\frac18\pm\frac1{96}\sqrt{s-1} -\frac{131}{2304}(s-1)+O((s-1)^{3/2}), \] and the characteristic discriminant has a simple zero at $s=1$. We place the example beside two structures that suppress such a marginal signature: fixed post-count reset, which gives scalar renewal, and a common total holding rate, for which terminal summation scalarizes the matrix power. The model also admits a monitored-Lindblad realization, but the mechanism is entirely classical. The result provides a small exact benchmark for how generalized spectral modes can remain visible in finite-threshold first-passage statistics.

arXiv commentSubstantially revised and narrowed. Corrects the earlier exceptional-point interpretation, clarifies counted self-transitions and terminal-summed visibility, and adds an exact three-state Markov-renewal example with a nonzero N8^{-N} Jordan term and square-root eigenvalue ramification. The mechanism is classical; the Lindblad model is an embedding