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A clinic-observed interval ending in a recorded state contributes a transition probability. If the next observation is exact entry into an absorbing state but the preceding state is unobserved, its statistical probability density is ${\sum }_{s\text{ living}}{p}_{rs}(t)q_{sD}$, not $p_{rD}(t)$. Multiplying these conditional contributions yields a likelihood for mixed panel and exact-event observations.

Write $h_i={\mathbb{P}}_i(H^A<\infty )$. For $i\in A$, $h_i=1$. For $i\notin A$, conditioning on the first step and using the Markov property gives $h_i={\sum }_j{p}_{ij}{h}_j$. This establishes the required equations for the hitting probability.

To prove the hitting probability is the minimal nonnegative harmonic extension, let $h_i^{(n)}={\mathbb{P}}_i(H^A\le n)$. Then $h^{(0)}=1_A$ andIf $g$ is any nonnegative solution of the same boundary equations, $g\ge h^{(0)}$. Positivity of the transition probabilities implies by induction $g\ge h^{(n)}$ for every $n$. The events $\{H^A\le n\}$ increase to $\{H^A<\infty \}$, so $h^{(n)}\uparrow h$ and $g\ge h$. Therefore $h$ is the minimal nonnegative solution.

For the tournament won by two consecutive victories, a transient pair $ij$, with $i\ne j$ and third player $k$, moves with equal probabilities to the absorbing state $jj$ or the transient pair $jk$. Thus the two transient cycles arewith probability $1/2$ of absorption at each step. A full cycle without absorption has probability $1/8$, so absorption occurs almost surely.

Starting from $AC$, the successive absorption winners are $C,B,A$, with first-cycle probabilities $1/2,1/4,1/8$. Summing the geometric repetitions givesStarting from $BC$, the order is $C,A,B$, giving $(2/7,1/7,4/7)$.

Because the first game is between $A$ and $B$, the initial ordered pair of winners is $AA,AC,BB,BC$, each with probability $1/4$. Combining the immediate wins with the transient hitting probabilities,Their sum is one, consistent with almost-sure absorption.

There are three free transition intensities: progression, death from the initial state, and death from the advanced state. In the three-state illness-death model, state 3 is an absorbing state, and there is no recovery transition.

Writing $a=q_{12}$, $b=q_{13}$ and $c=q_{23}$, with all three nonnegative, the transition intensity matrix isEach diagonal entry is minus the sum of the row's outgoing transition intensities, rather than an additional unknown parameter. A continuous-time multi-state model with these rates describes the severity labels in the observations; an explicit cured state would require a richer state space.

Use three states of a continuous-time Markov chain: disease-free $1$, pre-clinical $2$, and clinical $3$. The transition intensities are $\lambda =q_{12}>0$ for onset and $\nu =q_{23}>0$ for progression. Clinical disease is an absorbing state. The mandatory pre-clinical phase excludes a direct $1\to 3$ jump, and this untreated progression model has no reverse transitions.

With row-vector probabilities and state order $(1,2,3)$, the generator matrix isIts rows sum to zero, so only the two off-diagonal transition intensities are unknown. This is a continuous-time multi-state model with sequential irreversible progression.

Assume the intended initial fortune satisfies $n\ge 2$ and put $X_0=n$. Conditional on current fortune $i>1$, the next gift is uniform on $\{1,\dots ,i-1\}$ and independent of the previous choices. The remaining fortune is therefore uniform on that same set. At fortune $1$ the process stays there. Consequently the future conditional distribution depends only on the current fortune, proving the Markov property, and the transition matrix on $\{1,\dots ,n\}$ isThe uniform decreasing Markov chain has state $1$ as an absorbing state, and until absorption the fortune strictly decreases, so the hitting time is at most $n-1$.

Let $e_i$ be the expected number of additional transitions to hit $1$, taking $e_1=0$. First-step analysis givesIn particular $e_2=1$. For $i\ge 3$, multiplying this recurrence by $i-1$ and the recurrence for $i-1$ by $i-2$, then subtracting, gives $(i-1)e_i=(i-1)e_{i-1}+1$. Hence $e_i-e_{i-1}=1/(i-1)$ for every $i\ge 2$, and telescoping yieldsThe PDF defines $\tau$ using times $i\ge 1$. If $n=1$ were included, that definition would give $\tau =1$ although the displayed empty sum is zero; the initial instruction to choose an integer between $1$ and $n-1$ also requires $n\ge 2$. The result above uses that intended assumption. A hitting time allowing time zero would have expectation zero from state $1$, but that is a different definition.

The three year states are transient states: from each there is positive probability of entering an absorbing state, and repeated trials make eventual departure certain because $q<1$. The singleton classes $\{L\}$ and $\{G\}$ are closed absorbing states, hence recurrent. Thus the chain has two closed recurrent classes and three transient singleton classes.

A sequential continuous-time Markov chain has transitions $1\to 2\to 3$ with rates $\lambda ,\nu$ and an absorbing state $3$. Its probability of ever leaving state $1$ by time $t$ is $1-e^{-\lambda t}$; its probability of occupying state $2$ is $\lambda (e^{-\lambda t}-e^{-\nu t})/(\nu -\lambda )$ for unequal rates.

For a finite-state continuous-time Markov chain, a transition intensity matrix has nonnegative off-diagonal rates $q_{rs}$ and $q_{rr}=-{\sum }_{s\ne r}{q}_{rs}$. Every row sums to zero. In the time-homogeneous case its transition probability matrix is the matrix exponential $P(t)=e^{tQ}$. An absorbing state has a zero row.

On $\{1,\dots ,n\}$, make $1$ an absorbing state and let each state $i>1$ move uniformly to $\{1,\dots ,i-1\}$. Its expected hitting time of $1$, allowing zero time when started at $1$, is $e_i=H_{i-1}$. First-step analysis gives $e_i=1+(i-1{)}^{-1}{\sum }_{j<i}{e}_j$; subtracting consecutive recurrences yields $e_i-e_{i-1}=1/(i-1)$ and hence the formula.