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The first-order excitation amplitude is the Fourier transform of the temporal pulse envelope evaluated at the Rabi detuning. Two rectangular time windows therefore give exactly the mathematical structure of double-slit interference: each window gives a sinc function amplitude, and their separation gives a relative quantum phase $\delta T$. The transition probability is a sinc function squared envelope multiplied by ${\cos }^2(\delta T/2)$.

Pulse duration corresponds to slit width, pulse-start separation to slit-center separation, and Rabi detuning to the transverse wavevector sampled on the screen. The narrow fringes measure the long separation, while the broad envelope measures the short interaction duration. As in spatial interference, adding probabilities instead of coherent probability amplitudes would lose the fringes. Randomizing the relative pulse phase averages the cosine squared to $1/2$.

For a free-space electric dipole transition, the spontaneous emission rate in SI units isHere $|\mathbf{d}{|}^2$ includes the relevant dipole components; a specified single polarization requires its corresponding matrix element.

A complete scattering cycle returns the internal state to the ground state and changes the atomic momentum to $\mathbf{P}+\mathbf{p}-{\mathbf{p}}^{\prime }$. ThusFor photon-scattering recoil heating with isotropic emission of fixed momentum magnitude, $⟨{\mathbf{p}}^{\prime }⟩=0$, givingThe first term is the velocity-dependent work of absorption and can cool an atom if the photon opposes its motion. The last two are positive absorption and emission recoil energies. If the actual absorbed directions are also isotropic and independent of $\mathbf{P}$, the first term averages to zero. For approximately equal incoming and outgoing photon momenta $\hslash k$,Isotropic incident illumination alone would not guarantee isotropic absorption if Doppler-dependent excitation selected directions.

Source qualification for the trapped calculation. Set $m=M$ for the same atom. The printed Gaussian $\exp [-x^2/(2l_0{)}^2]$ has the wrong width for the specified harmonic oscillator: the printed ground and first-excited functions have squared norms $\sqrt{2}$ and $2\sqrt{2}$, respectively. The normalized oscillator states fixed by the given $l_0=\sqrt{\hslash /(m{\omega }_{\mathrm{HO}})}$ instead useAlso, one common complex phase multiplying both off-diagonal terms does not make the Hamiltonian a Hermitian operator. Use an absorption matrix element $(\hslash \Omega /2)e^{-ikx}$ and its Hermitian conjugate for emission; choosing the opposite propagation convention reverses the imaginary sideband phase but not a transition probability. The physical results below use these explicitly stated corrections. Parts (i) and (ii) also give the literal integrals of the printed functions.

Conditional motional heating and recoil averaging. With $\eta =k{l}_0/\sqrt{2}$, an oscillator eigenstate has $⟨K{⟩}_n=⟨V{⟩}_n=(n+1/2)\hslash {\omega }_{\mathrm{HO}}/2$. Therefore the absorption outcomes haveThe same statements hold for the specified emission outcomes $|e,0⟩\to |g,0⟩$ and $|e,0⟩\to |g,1⟩$. Internal excitation energy goes into, or comes from, the photon and is distinct from this external energy change. Emission into the first motional state makes the emitted photon correspondingly less energetic.

A sudden recoil from $|0⟩$ gives probabilities $P_n=e^{-{\eta }^2}{\eta }^{2n}/n!$, so $P_1\simeq {\eta }^2\ll 1$ while $P_0\simeq 1$. The rare first sideband transfers a whole trap quantum, much larger than $E_R$; the mean is neverthelessThe emission average is the same for a one-dimensional recoil of the same magnitude. Projecting isotropic three-dimensional emission onto one axis instead gives a factor $⟨{\cos }^2\theta ⟩=1/3$ for that axis.

A sudden unobserved kick initially raises kinetic energy by $E_R$ and leaves the position distribution, hence potential energy, unchanged. It creates a coherent superposition of motional states. The equal kinetic/potential increments above concern energy-eigenstate-conditioned outcomes, or time-averaged/dephased motion; these are different measurement conditions. There is no contradiction.

With $\Gamma >{\omega }_{\mathrm{HO}}$, motional sidebands are not cleanly resolved, and a short-lived excited atom has little time to move before decay. In the stronger sudden limit $\Gamma \gg {\omega }_{\mathrm{HO}}$ the recoil-kick picture is direct. For only the stated inequality, finite linewidth weights can change numerical factors; it does not fix an exact continuous-wave scattering rate. The average heating remains of recoil order per absorption or emission, and of order $2E_R$ when both absorption and emission kicks are counted at full magnitude. For the axial component of isotropic three-dimensional emission, the emission contribution instead carries the factor $1/3$ noted above. Strong confinement suppresses the chance of changing $n$, but does not make its averaged recoil energy disappear. This agrees with free-space recoil heating while replacing small continuous energy changes by rare quantized motional changes.

Include all four positions and a turn marker:The initial marker is $\mathrm{R}$. There are at most $2n^8$ such states before terminal compression. The two robots form one maximizing team in the adversarial search representation. At an R state their actions are joint legal moves, including waiting, with deterministic successor positions and marker M. At an M state the monster outcome is a pair of legal cardinal moves, producing marker R.

Check termination initially and after each simultaneous pair of moves. A state is a failure if $\{p,q\}\cap \{z_1,z_2\}\ne \varnothing$. Otherwise it is a success if $p=q$. Collision takes priority over meeting, so meeting on an occupied room fails. A safe meeting ends immediately, before a further monster response. Room swaps without a shared position at a checked boundary are neither capture nor meeting.

The monsters move randomly, so their nodes are chance nodes, not minimizing decisions. Under independent uniform legal moves, if monster $j$ currently has $d_j$ available destinations, each joint outcome has probability $1/(d_1{d}_2)$. If several commands produce the same state, add their probabilities. This specifies a stochastic game tree, equivalently a Markov decision process for the robot team. Uniform legal moves and independent monsters are explicit boundary/joint-law conventions; uniform attempted directions with blocked moves staying put is another convention, with different boundary transition probabilities but the same search method. For the one-cell grid, a monster with no legal move stays; all four initial positions then coincide and the instance immediately fails.

For a discrete memoryless channel with transition probabilities $M(y\mid x)$, the conditional probability of a length-$n$ output given a length-$n$ input factors asThe same transition probabilities apply at each use. Its channel capacity isThe unit is bits per channel use. For finite alphabets the maximum exists by continuity on the probability simplex; with more general alphabets one uses a supremum when appropriate.

An $(m,n)$ channel code consists of an encoder $c:{\mathcal{V}}_m\to {\mathcal{X}}^n$ and a decoder $d:{\mathcal{Y}}^n\to {\mathcal{V}}_m$. It represents each of $m$ messages by $n$ input symbols and recovers a message from the $n$ observed output symbols. Its code rate is $R=({\log }_{2}m)/n$ bits per use. For each message $w$, its conditional error probability is $\mathbb{P}(d(Y^n)\ne w\mid W=w)$; an average error also specifies the message probability distribution, while the maximum error takes the largest conditional error.

For every input probability mass function, conditional entropy is nonnegative, soTaking the maximum over input laws proves $C\le 1$ bit per channel use. This bound comes from the binary input alphabet itself; no specific transition probabilities or decoder are needed.

With no deletion states allowed, each match state must emit once in every aligned sequence. Thus the four gap-free alignment columns, $1,2,5,7$, become $M_1,M_2,M_3,M_4$. Columns $3$ and $4$ form one optional run, represented by a single looping insertion state $I_2$; column $6$ requires a separate insertion state $I_3$ after $M_3$. Two insertion states are sufficient and minimal in an ordered profile hidden Markov model: the two optional regions are separated by a compulsory emission and occupy different positions in the profile.

Estimate each transition probability by the number of observed uses of that transition divided by all transitions leaving its source state. The complete arrow diagram is given below; repeated labels denote the same state, so the second and third lines supply the branches of the first line. Start and End are silent:Three aligned sequences visit $I_2$, giving three exits to $M_3$, while one sequence emits twice there, giving one self-transition. Thus its exit and loop probabilities are $3/4$ and $1/4$, respectively. Two sequences visit $I_3$ and two bypass it. All absent transitions have probability zero.

The probability mass functions of the emitting states, estimated from all characters assigned to each state with no added counts, areIn particular, pooling the two optional columns into $I_2$ pools four emissions: two T, one A and one G. This is a generative Hidden Markov model, so it can generate other run lengths and combinations as well as the training sequences; reproducing each training column deterministically is not its purpose.

The sequence has five characters, while the four match states account for four compulsory emissions. Consequently exactly one insertion state emission is needed. An insertion at $I_3$ would have to emit T, which has probability zero there. The only nonzero hidden path is thereforeMultiply its transition probabilities and emission probabilities, including the probability of exiting $I_2$ immediately and bypassing $I_3$:Normally the forward algorithm sums over all compatible paths. Here all other paths have probability zero, so this single path gives the full sequence probability.

The printed “otherwise” branch requires correction. Its factor $3/4$ is the probability of ending at any of the three bases different from the initial base, not of ending at one specified different base. Assigning that probability to each of three alternatives would make a row of the transition matrix sum to $5/2-(3/2)e^{-\mu t}$, which exceeds one for $t>0$. This distinction is present in the source PDF, not introduced by the transcription.

For a consistent Jukes-Cantor model, let each ordered change from one nucleotide to a specified different nucleotide have instantaneous rate $q$. Its continuous-time Markov chain has diagonal generator entries $-3q$ and off-diagonal entries $q$. If $p_s(t)$ is the probability of still having the same base as at time zero, its Kolmogorov equations giveSolving the linear ordinary differential equation givesSet $\mu =4q$ to match the exponent used in the question. Symmetry then gives the properly normalized transition probabilitiesSumming the three different-base outcomes yieldsThe assumptions are a time-homogeneous Markov process, the same change rate between every distinct ordered base pair, identical rates across independent sites, and no insertions or deletions. The stationary distribution is uniform over the four bases; starting the process in that distribution is not required for this conditional calculation.

There is also a parameter-convention issue. The actual total mutation rate out of a base is $r=3q=3\mu /4$. Thus the exponent parameter $\mu$ in (b)–(d) is $4/3$ times the event rate used in the elementary counting interpretation of (a). Keeping this conversion explicit avoids changing the meaning of “mutation rate” halfway through the calculation.

Use $T_y=\inf \{n\ge 0:X_n=y\}$ and the positive return time $T_x^{+}=\inf \{n\ge 1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)={\sum }_v{c}_{uv}$; its transition matrix defines a reversible Markov chain, although the transition probabilities need not be symmetric.

The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a harmonic function on a graph away from the terminals: $h(u)={\sum }_v{p}_{uv}h(v)$. For this unit electric potential difference the electric current isThe first-step analysis therefore proves the return probability from effective resistance identity:Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.

For an infinite locally finite connected graph, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thuswith $1/\infty =0$. In particular recurrence is equivalent to infinite effective resistance to infinity.

After deleting an edge, consider any infinite component and choose a root $o$ away from that edge's endpoints. Its vertex degree $d(o)$ is unchanged. On every finite wired exhaustion the Rayleigh monotonicity principle makes the resistance no smaller; the limit of a sequence does too. A recurrent original network consequently leaves a recurrent component. Finite components give recurrent walks as well, with an isolated graph vertex viewed as an absorbing state. This proves that edge deletion preserves recurrence. The phrase “no less recurrent” refers to this recurrence comparison: at a deleted-edge endpoint the vertex degree factor also changes, so monotonicity of its first-return probability alone should not be inferred.

In the voter model on ${\mathbb{Z}}^2$, each site has a rate-one Poisson process clock; when it rings, that site copies one of its four neighbors, chosen uniformly. States are zero or one. For a cylinder function $f$, the local infinitesimal generator iswhere ${\eta }^{x\leftarrow y}$ agrees with $\eta$ except that the voter model opinion at $x$ is replaced by the voter model opinion at $y$. Only sites in the support of $f$ contribute, so this infinitesimal generator sum is finite.

The voter-model duality comes from independent Poisson processes of copying arrows of rate $1/4$ on each ordered neighboring pair. Trace the ancestors of specified sites backwards from time $t$: each ancestral line is a rate-one continuous-time random walk with uniform nearest-neighbor steps, and lines coalesce when they meet. The voter model opinions at time $t$ are the initial voter model opinions at those ancestral locations. This construction also explains the general duality theorem being used.

Two ancestral walks on ${\mathbb{Z}}^2$ meet almost surely. Until their meeting, their difference is a rate-two symmetric nearest-neighbor walk, and simple random walk in two dimensions is recurrent. If ${\tau }_{xy}$ is their meeting time, this givesfor every initial law $\nu$. If $\nu$ is invariant, the left side is the fixed probability $\nu (\eta (x)\ne \eta (y))$, so it is zero. This holds for every pair. Countability then implies that $\nu$ is concentrated on configurations with all voter model opinions equal. Both constant configurations are absorbing state. Thus the invariant measures of the two-dimensional voter model are exactlyNo translation-invariance assumption on $\nu$ was used.

For the last assertion, let the initial set of one-valued opinions $F$ be finite. Single-line duality givesEach transition probability tends to zero. One direct verification is the Fourier transform for the rate-one walk:The integrand has modulus at most one and tends to zero except at the single point $k=0$; dominated convergence proves the claim. For any finite observation set $K$, the union bound now givesTherefore every finite voter model opinion pattern converges in law to the all-zero pattern. These cylinder distributions determine weak convergence of probability measures in the compact product topology, proving finite-seed local extinction in the voter model and

Use the time-homogeneous continuous-time multi-state model displayed in part ii. Write $Z(t)$ for the state at time $t$, $Q=(q_{rs})$ for its generator matrix, andFor $r\ne s$, $q_{rs}={\lim }_{h\downarrow 0}{p}_{rs}(h)/h$ is the transition intensity, and $q_{rr}=-{\sum }_{s\ne r}{q}_{rs}$. Condition on the observed initial state, and treat the clinic observation process and administrative end of follow-up as noninformative. A panel-observed interval contributes its transition probability, not the density of a jump at the examination date; the transitions inside that interval need not have been observed.

For the first patient, the two clinic intervals contribute $p_{11}(2)p_{13}(2)$. The subsequent death is observed at elapsed time $u=0.2$ after the last clinic state, so its contribution is the conditional death-time densityThe equality uses the permitted directions: starting from state 3, only living states 3 and 4 can be occupied before death. To obtain this density directly, survival in a living state $s$ until time $u$, followed by death in $(u,u+du)$, has probability $p_{3s}(u)q_{s5}du+o(du)$; add over the unobserved state $s$. Hence the mixed panel and exact-death likelihood contribution isIt includes the possibility of an unrecorded transition $3\to 4\to 5$ between the last clinic visit and death. Keeping only $p_{33}(0.2)q_{35}$ would exclude that allowed history, while $p_{35}(0.2)$ would be the probability of death by that time rather than the density of death at that time. With dates recorded in whole days, an exact interval-likelihood treatment integrates this density over the relevant day; the density formulation is its fine-time approximation.

For the second patient, the three successive intervals with state 1 at both ends, then the interval ending in state 2, and the last interval remaining in state 2 give the panel-observed multi-state likelihoodThere is no additional death contribution after the final administrative observation. Each likelihood contribution already accounts for being alive in the observed state at that visit. For independent patients their joint conditional likelihood contribution is $L_1{L}_2$.

For an explicit form, put ${\lambda }_r=-q_{rr}$. Irreversibility givesandThese formulas follow by integrating the density of the one possible intermediate jump time; at equal exit rates the limiting expression is $q_{rs}u{e}^{-{\lambda }_{r}u}$. ThusUsing the displayed fitted rates as numerical values gives $g_{35}(0.2)\approx 0.289314$ per year, $L_1\approx 0.017596$ per year and $L_2\approx 0.016048$, providing a check of the probability-versus-density distinction.

A conservative Q-matrix has $q_{ij}\ge 0$ for $i\ne j$, finite exit rates $-q_{ii}={\sum }_{j\ne i}{q}_{ij}$, and hence row sums zero. It specifies the infinitesimal transition probabilities $P_{ij}(h)={\delta }_{ij}+h{q}_{ij}+o(h)$. For a finite state space,The first derivative at zero is the generator. On infinite state spaces one must additionally address nonexplosion and the domains needed for the forward and backward equations.

For patient $h$, let $Z_h(t)\in \{1,2,3\}$ be the fatigue state at time $t$ years since study entry. Let the covariate vector be $z=(s,d,a,g{)}^T$, representing the sex indicator, arthritis duration in years, HAQ disability score and haemoglobin measurement. Conditional on these covariates, the continuous-time multi-state model has the Markov property and generator matrixEach allowed transition intensity has its own log-linear transition intensity model, $q_{ij}(z)=\exp ({\eta }_{ij}+b_{ij}^{T}z)$. The zero entries exclude direct instantaneous jumps between mild and severe states, and the diagonals enforce zero row sums.

To express the estimates correctly, the printed intensities are evaluated at the sample covariate means $\bar{z}$, not at $z=0$. Order the allowed transitions as $12,21,23,32$. Their estimates arewith rows indexed by that transition order and columns by $s,d,a,g$ inEquivalently, the uncentred intercept is ${\hat{\eta }}_{ij}=\log {\hat{q}}_{ij}(\bar{z})-{\hat{b}}_{ij}^T\bar{z}$. The means are not printed, so the displayed rates cannot be treated as numerical zero-covariate intercepts.

The assumptions are independent patients conditional on their covariates; correctly observed fatigue states; no unmodelled history dependence once current state and covariates are known; positive adjacent-state intensities with the specified exponential covariate dependence and no interaction terms; and a noninformative observation and censoring mechanism for the conditional state-process likelihood. With fixed covariates the rates are homogeneous in time and holding times are exponential. Covariates updated at visits are treated as piecewise constant between observations, so each such interval has a constant generator, a piecewise-constant covariate approximation. In particular this is an approximation if a quantity such as duration changes continuously between visits.

For observations $Z_h(t_{hk})=s_{hk}$, conditioning on the initial observed state, the panel-observed multi-state likelihood is a product over patients and intervals ofwhere $z_{hk}$ is the covariate value held over that interval. Multiple unobserved jumps within an interval are integrated out by this transition matrix. For example, a mild state followed by a severe state can have positive likelihood despite $q_{13}=0$, because an unobserved moderate state can intervene. The panel observations within a patient are not independent; it is the Markov property that justifies multiplying conditional transition probabilities. The piecewise-constant handling of changing covariates is specified in the msm model documentation.

The fitted continuous-time multi-state model allows only adjacent fatigue-state jumps. Its four transition intensities, in reciprocal years, are shown here:

In state $2$, the total exit intensity is ${\lambda }_2=q_{21}+q_{23}=0.4684+0.3835=0.8519$. By the holding time law for a continuous-time Markov chain, the time until the next jump has an exponential distribution with this rate. ThereforeThe printed interval for $q_{22}=-{\lambda }_2$ gives the interval $(0.7219,1.005)$ for ${\lambda }_2$. Taking reciprocals reverses the endpoints, so the transformed $95\%$ confidence interval for the mean holding time isFor the next jump's destination, competing outgoing intensities giveIndeed the density of a first jump to $1$ at time $t$ is $q_{21}{e}^{-{\lambda }_{2}t}$; integrating gives the ratio above. This is a conditional next-jump probability, not a transition probability of being mild at a fixed later visit.

The second continuous-time Markov chain runs in the reverse direction at rate $\rho$, as in the right diagram. Its nonzero eigenvalues have real part $-3\rho /2$, and its transition probabilities tend to $1/3$ for each vertex. Independence therefore givesThus the limit exists for every pair of starting vertices and equals $\boxed{1/3}$.

One can also see the limit directly without spectral multiplication. For fixed starting vertices, $X_t-Y_t$ modulo three advances with each jump of either flea. Its jump count is the sum of independent Poisson processes with rates $1$ and $\rho$, hence is Poisson with rate $1+\rho$ (multiply their generating functions). The formula from part (a), with $t$ replaced by $(1+\rho )t$ and the appropriate initial residue, gives $p(t)$ and the same limit.

PutThe transition probabilities areThe first and second rows coincide because either an arrival to an empty queue or the single existing customer initiates the next service.

Use the standard infinite-population slotted ALOHA model with independent fresh arrivals with Poisson distribution $Y_t\sim \mathrm{Poisson}(\lambda )$, $\lambda >0$. Newly arrived packets attempt in the next slot. Given $N_t=n$ old backlogged packets, each independently attempts with fixed probability $f$, so the retransmission count has binomial distribution $R_t\sim \mathrm{Binomial}(n,f)$, independent of $Y_t$ and earlier random choices. Let $Z_t$ record whether $R_t+Y_t$ is zero, one, or at least two. A single transmission succeeds and departs; a collision makes every transmitted packet remain. ThereforeOnly the present backlog determines the law of the next independent choices, so $N_t$ is a time-homogeneous Markov chain.

For $0<f<1$, put $q=1-f$ and $b_n=nf{q}^{n-1}$, with $b_0=0$. Its nonzero transition probabilities areFor example, a single fresh arrival leaves the backlog unchanged only if no old packet transmits; if an old packet also transmits, that fresh arrival becomes a new backlogged packet. These cases explain the extra diagonal term and the suppressed one-step increase at $n=0$. At $f=1$ the same formulas hold with $q^0=1$, $b_1=1$ and $b_n=0$ for $n\ne 1$.

Here is an almost-sure proof of permanent collisions in constant-probability slotted ALOHA for $0<f<1$, rather than just a positive-drift heuristic. For $\theta >0$, the conditional exponential increment isAs $n\to \infty$ this tends to a number strictly below one. Choose $M\ge 1$ so the expression is at most one for $n\ge M$. Stopping $e^{-\theta N_t}$ when the chain first enters $\{0,\dots ,M-1\}$ gives a nonnegative supermartingale. Downward jumps have size at most one, so it hits at $M-1$. Optional sampling theorem for a supermartingale, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundThis is an irreducible Markov chain: every positive state can successively decrease to zero, and from zero Poisson batches reach every state at least two, with state one then reached from two. An irreducible recurrent Markov chain would hit that finite set with probability one from every state, contradicting the bound. Hence every state is a transient state, every finite set is visited only finitely often, and $N_t\to \infty$ almost surely.

The conditional success probability is $p(N_t)=e^{-\lambda }(b_{N_t}+\lambda q^{N_t})\to 0$. The centered success indicators form a martingale difference sequence with bounded increments. The strong law for martingales with bounded increments givesThe conditional means have Cesàro average zero, while the strong law of large numbers gives $t^{-1}{\sum }_{u<t}{Y}_u\to \lambda$. Therefore $N_t/t\to \lambda$ almost surely. Finallydecays exponentially along this linear backlog growth. Its sum over $t$ is almost surely finite. The Conditional Borel-Cantelli lemma then implies that only finitely many slots have zero or one transmission. Thus $\boxed{\mathbb{P}(\exists J:Z_t=\ast \text{ for every }t\ge J)=1}$. At $f=1$, a fresh batch of size at least two eventually occurs, after which at least two old packets retransmit in every slot, giving the same conclusion directly.

The qualification $f>0$ is essential: if $f=0$, then $Z_t$ depends only on the independent Poisson arrivals, and zero- or one-arrival slots occur infinitely often. Positive mean arrival rate alone, without the Poisson model or another assumption allowing collisions to develop, would also be insufficient: Bernoulli fresh arrivals and initial backlog zero never collide.

One alternative is binary exponential backoff: after $k$ collisions a packet selects a waiting time uniformly from $\{0,\dots ,2^k-1\}$ and tries when its counter expires. Its attempt behavior depends on its collision history, rather than a common constant $f$. Another possibility, if a backlog estimate ${\hat{N}}_t$ is available, is to set the attempt probability near $1/{\hat{N}}_t$, keeping the offered retransmission count near one. These descriptions illustrate adaptive access rules; they do not assert that every backoff variant is stable at arbitrary load.

Let $q_{rs}$ be the transition intensity from state $r$ to $s$, and let $Q$ be the transition intensity matrix, with $q_{rr}=-{\sum }_{s\ne r}{q}_{rs}$. For a continuous-time multi-state model with the time-homogeneous Markov property, the transition probability matrix is $P(u)=e^{uQ}$, with entry $p_{rs}(u)$. Condition on each recorded initial state. Panel visits contribute transition probabilities; an exact entry into the absorbing death state contributes a statistical probability densityThis mixed panel and exact-death likelihood sums over the living state just before death. It accounts for survival until the event; replacing its final factor by $p_{r3}(u)$ would count deaths throughout the interval.

Using the actual visit times recovered from the PDF, the three individual likelihood contributions areHere $g_{23}$ means $g_{r3}$ with $r=2$, not a transition probability. Subjects 8 and 9 supply no event-density factor after their last panel observation. Assume independent subjects and noninformative examination and censoring times; conditional on their observation schedule, its distribution supplies no additional statistical parameter-dependent factor. The patients' covariates can be incorporated by using their own $Q_i$ in these same expressions.

For the progressive structure used in the subsequent output, put $a=q_{12}$, $b=q_{13}$, $c=q_{23}$ and $\lambda =a+b$. The progressive illness-death model permits no recovery, soWhen $\lambda =c$, the continuous limit is $p_{12}(u)=au{e}^{-cu}$. Thus the likelihood contributions simplify toIntermediate unobserved disease transitions remain integrated into each panel transition probability.

The Markov property and the Chapman-Kolmogorov equation giveFor each intermediate state, one first uses the next-step transition probability and then the following one; the sampled history adds no information once $X_{2n}$ is known. Thus $W$ is a homogeneous Markov chain with transition matrix $P^2$ and initial law $\lambda$. In particular,If zero is not a state of $S$, this probability is zero.

Let $p_{rs}(t)=P(X(u+t)=s\mid X(u)=r)$ and $P(t)=e^{Qt}$, where time homogeneity removes dependence on $u$. A recorded state at the next clinic visit contributes a transition probability; an exactly observed death contributes a statistical probability density, not the probability of being dead at that time. If the last recorded living state is $r$, the mixed panel and exact-death likelihood factor after an interval $t$ isThis sums over the unobserved living state immediately before death.

Conditioning on the recorded initial states, the contribution of the three displayed patient histories isIn this irreversible illness-death model, $p_{21}=0$, $p_{11}(t)=e^{-(a+b)t}$ and $p_{22}(t)=e^{-ct}$, simplifying it toThe factor $c$ is essential: the death time is known exactly. Replacing the final statistical probability density by $p_{23}(34.6)$ would instead model interval observation of death and give a different likelihood.

The assumptions are independent patient histories with common rates; the Markov property; constant rates over calendar/follow-up time in this model; the stated absence of recovery and absorption at death; accurate state labels and death times; and an observation/follow-up mechanism that is noninformative for the latent process given the observed history. Clinic dates are conditioned on. The displayed living endpoints contribute only the shown observations, with noninformative right censoring if they are follow-up endpoints. Initial state probabilities are omitted by conditioning on them. Progression between visits can be unobserved, which is precisely why panel-observed multi-state likelihood uses the matrix exponential rather than assuming a transition occurs at a visit.