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A frozen-age approximation holds an age covariate constant at the start of each observation interval, allowing homogeneous transition probabilities to enter a panel-observed multi-state likelihood. It approximates a model with continuously changing age. Reciprocal rate calibration of mean waiting times is exact for a fixed-age exponential distribution, but not for a continuously ageing Gompertz distribution.

An irreversible three-state continuous-time multi-state model permits only $1\to 2\to 3$, with state 3 absorbing. For constant rates $\lambda =q_{12}$ and $\mu =q_{23}$,At equal rates use $p_{12}(t)=\lambda t{e}^{-\lambda t}$. A panel-observed multi-state likelihood uses these transition probabilities, allowing unobserved intermediate visits between examinations.

The occupation time records the duration spent in a state of a continuous-time Markov chain. Conditional on $Z(0)=r$, its expected value is ${\int }_0^t{p}_{rs}(u)du$, where $p_{rs}$ is the transition probability.

Let $S_k={\sum }_{\ell =1}^k{Y}_{\ell }$, $S_0=0$, be the epochs of the renewal process. The definition uses the next epoch at or after $n$, so $X_n=0$ at an epoch. If $X_n=i\ge 1$, no renewal occurs before $n+i$ and necessarily $X_{n+1}=i-1$.

If $X_n=0$, time $n$ is a renewal epoch, and the next unused interarrival time is independent of the entire observed past with the original distribution. To justify the random index, condition on each possible number $k$ of renewals up to $n$: the event identifying $k$ depends only on $Y_1,\dots ,Y_k$, whereas $Y_{k+1}$ is independent of those variables. Summing over $k$ preserves that distribution and independence. Therefore the residual lifetime Markov chain has transition probabilitiesIts reachable state space is $E=\{j\ge 0:\mathbb{P}(Y_1>j)>0\}$, either an initial finite segment or all nonnegative integers. The conditional next-step law depends only on $X_n$, proving the Markov property. Initially $X_0=0$. The reset is $Y-1$, rather than $Y$, because one unit of time has elapsed by the next step.

The first departure from the no-symptoms state has an exponential distribution with rate $\lambda$. Spending the entire interval there means no departure at all, soThis is the holding time survival probability. The transition probability $p_{11}(t)$ only says that the patient is symptom-free at the endpoint, allowing an onset and recovery in between, and is therefore not the answer. The formula also covers $\lambda =0$, when state 1 is absorbing.

Visits can be missed or rescheduled because of illness, clinic availability or patient choice; follow-up may end through dropout or right censoring. The correct panel-observed multi-state likelihood uses each observed interval length ${\Delta }_{ik}$:Thus the four aggregated one-year counts are generally no longer sufficient; records must retain actual durations. Numerical maximum likelihood estimation uses the corresponding transition probabilities, or a matrix exponential of the generator when a larger model is fitted. One must not treat a two-year interval as a single one-year transition or as two known transitions through an unobserved intermediate state.

This conditional calculation is valid when visit scheduling and loss to follow-up are ignorable given the modeled information. If symptomatic patients are systematically examined earlier or leave the study differently, observation itself may be informative; the visit or dropout process may require a joint statistical model. Simply replacing one year by the actual interval does not correct informative observation.

Condition on each patient's initial observed state and assume independent patient trajectories and observation times that do not carry additional information about the unobserved path. The Markov property factors each trajectory's likelihood function into its successive one-year transition probabilities. ThusMore explicitly, with $q=\lambda +\mu >0$ and $E=e^{-q}$,Successive transitions within one patient need not be independent unconditionally: this is a product of conditional factors justified by the Markov property. If initial-state probabilities are modeled rather than conditioned on, their likelihood contributions must also be included. Zero-count terms contribute zero; a positive count attached to a zero model probability makes the log-likelihood equal to $-\infty$.

Define interaction-picture states by $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$ and $V_I(t)=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. Differentiation cancels the free Hamiltonian operator and yieldsAssume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in $|n⟩$, the first iteration of the integral equation givesfor $m\ne n$. The picture change only multiplies the measured amplitude by a phase. Squaring therefore givesFor the harmonic oscillator, $|1⟩=a^{†}|0⟩$ is normalized and $⟨1|(a+a^{†})|0⟩=1$. The energy difference is $\hslash \omega$, andHence the lowest nonzero transition probability isThe stronger remainder here follows from parity: every insertion of $a+a^{†}$ reverses oscillator parity, so the amplitude between $|0⟩$ and $|1⟩$ has only odd powers of $\lambda$. The general displayed $O({\lambda }^3)$ estimate remains valid.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

A random time $\tau$ is a stopping time for the natural filtration $({\mathcal{F}}_n)$ of a Markov chain when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. The Strong Markov property says that, conditionally on $\tau <\infty$ and $X_{\tau }=i$, the process after $\tau$ is a fresh copy of the chain started from $i$, independent of the pre-$\tau$ history.

Before capture define the half-separationFor $D_n=k\ge 2$, independence of the two moves gives the transition probabilitiesIgnore the holding steps. The resulting embedded chain is a biased random walk that moves left with probability $q$ and right with probability $1-q$. Its expected number of moves needed to descend one level is $1/(2q-1)$; this follows either from first-step analysis or from its drift $1-2q$. A non-holding move occurs with probability $1/2$ at each time, so its mean waiting time is two. Hence, for every $m\ge 2$, the expected time to go from separation $2m$ to $2(m-1)$ is

Write $q_i=-g_{ii}$ for the total rate of leaving $i$. The jump chain of a continuous-time Markov chain $X$ is the discrete-time chain $Y=(Y_n)$ recording the states occupied after successive jumps. Its transition probabilities arewhenever $q_i>0$.

The chain $X$ is irreducible if every state can be reached from every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the process returns to $i$ with probability one; an irreducible chain is recurrent when one, and hence every, state is recurrent.

The continuous-time Markov chain and its jump chain have exactly the same successive states. If $Y$ is transient, it visits each state only finitely often, so $X$ is transient. If $Y$ is recurrent, it visits a fixed state $i$ infinitely often. At those visits, $X$ has independent exponential holding times of rate $q_i<\infty$. Their sum diverges almost surely, so explosion cannot occur before all those returns. Hence $X$ returns to $i$ infinitely often and is recurrent. Thus, for an irreducible chain,