= Solution
Assume individual incubation delays are <independent> of the <Inhomogeneous Poisson process> of infections and of one another. This marking assumption is needed in addition to specifying a marginal incubation density.
In the discrete cohort model let $N_i\sim\operatorname{Poisson}(h_i)$ be the <independent> infection counts. Mark each infection by its eventual onset interval. For cohort $i$, the categories have probabilities $q_{ik}$ for observed intervals, with the remaining probability assigned to onsets outside the window. <Poisson thinning> makes the counts $N_{ik}$ in these categories mutually <independent> <Poisson random variables> with means $h_iq_{ik}$. One direct proof is their <probability generating function>:
$$
\mathbb E\!\left[\prod_k z_k^{N_{ik}}\right]
=\exp\!\left[h_i\sum_kq_{ik}(z_k-1)\right]
=\prod_k\exp[h_iq_{ik}(z_k-1)].
$$
Different cohorts are <independent>. Summing their counts therefore gives \b[independent Poisson onset counts in disjoint intervals]:
$$
\boxed{Y_k=\sum_iN_{ik}\sim\operatorname{Poisson}(\mu_k),\qquad
\mu_k=\sum_i h_iq_{ik},\qquad Y_j\perp Y_k\ (j\ne k).}
$$
The same argument applies exactly in continuous time by the <Independent marking theorem for Poisson point processes> and mapping each marked infection to its onset time. The endpoint model approximates its means; its independent-Poisson conclusion is exact within that discrete model. A fixed cohort size would instead induce negatively correlated onset-bin counts, so the <Poisson process> infection assumption matters.
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