 Solution

Use one transition per time step in the printed discrete-time multi-state diagnosis model. Put $a={\delta }_E$, $b={\delta }_N$, and $r=1-b$. Let $I_k=S_{1k}\sim \mathrm{Poisson}(h_k(\theta ))$ be the new infection count. Let $U_k$ be the undiagnosed non-early count present in state 3 at step $k$, including those retained by its self-loop. Start with $I_0=U_0=0$.

For the realised random variables, define conditionallyusing independent individual transition choices, and setThe early split's two complementary counts are not independent conditional on $I_{k-1}$, nor are the late split's two counts conditional on $U_{k-1}$. Random state incidences must not be equated to their expectations.

Explicitly, the first four realised vectors areHere $E_k$ and $D_k$ have the conditional laws just given; $D_1=D_2=0$. The corresponding expressions solely in terms of the parameters are the mean flow vectors. Writing $h_j=h_j(\theta )$ and interpreting state 3 with its self-loop as $S_{3k}=U_k$, they areThese follow from $\mathbb{E}{U}_k=(1-a)h_{k-1}+r\mathbb{E}{U}_{k-1}$. They make the minimum one-step early delay and two-step late delay explicit. If “entering state 3” is reserved for first entry only, its incidence is $I_{k-1}-E_k$ with mean $(1-a)h_{k-1}$; $U_k$ is then a separate occupancy variable. The late-diagnosis expressions are unchanged. This distinction resolves the wording's use of incidence alongside a state-3 self-loop.