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For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval for the rate transforms to one for the mean by reversing and inverting its endpoints. The monotonic transformation preserves the coverage event; it does not give a prediction interval for one realized holding time.

If a positive rate $q$ has a confidence interval $[L,U]$, monotonicity transforms it into $[1/U,1/L]$ for the reciprocal mean $1/q$. This preserves the coverage event exactly and reverses the endpoints. It applies directly to mean exponential holding times.

With initial exit rates $a$ to the advanced living state and $b$ directly to death, and death rate $c>0$ from the advanced state, the expected time to death from state 1 is $1/(a+b)+a/((a+b)c)$. The first term is initial holding time; the second weights the advanced-state holding time by the probability of reaching it.

For a finite-state homogeneous continuous-time Markov chain with generator matrix $Q$, integrate the indicator of occupying state $j$ over $[0,T]$. Interchanging the bounded time integral and expected value gives the displayed integral of transition probabilities. It counts all visits, not only the first holding time. Summing over states gives $T$. A zero eigenvalue of $Q$ does not cause a problem for this integral, although it prevents casually writing it with an inverse matrix $Q^{-1}$.

Start a Markov chain at a nonabsorbing state $i$. If each departure from $i$ has probability $p>0$ of absorption before the next return, the Strong Markov property makes the number of visits to $i$, including the initial one, a positive-integer geometric random variable of parameter $p$. For a continuous-time Markov chain whose holding times at $i$ are independent rate-${\lambda }_i$ exponentials, independently of the embedded jump chain, the total occupation time at $i$ is a sum of $K$ such holding times and has expected value $1/(p{\lambda }_i)$.

For $P_{ii}<1$, observe a discrete-time Markov chain only when its state changes. The probability that its next different state is $j$ is ${\sum }_{r\ge 0}{P}_{ii}^r{P}_{ij}=P_{ij}/(1-P_{ii})$. Its holding times are finite geometric waiting times. This construction deletes self-transitions and differs from the existing continuous-time jump-chain concept.

A state's holding time in a time-homogeneous continuous-time Markov chain is exponentially distributed with rate $-q_{rr}>0$. Its mean is $-1/q_{rr}$; the probability that its next jump is to state $s$ is $q_{rs}/(-q_{rr})$. An absorbing state has infinite holding time.

For a continuous-time Markov chain with absorbing endpoints $0,N$ and rates $q_{j,j\pm 1}={\lambda }_j/2$, the jump chain is a simple symmetric random walk. On leaving $i$, absorption before returning has probability $p_i=\frac{1}{2}(1/i+1/(N-i))$, by gambler's ruin on each side. The geometric return count before absorption and independent rate-${\lambda }_i$ holding times give ${\mathbb{E}}_{i}I=1/(p_i{\lambda }_i)$. The rates at other states affect elapsed excursion times but not their return probabilities.

For a homogeneous continuous-time Markov chain, the holding time in a living state $r$ before the next transition has survival probability $e^{-{\lambda }_{r}t}$, where ${\lambda }_r={\sum }_{s\ne r}{q}_{rs}=-q_{rr}$. Indeed, conditional on still occupying that state, the probability of leaving in the next $dt$ is ${\lambda }_{r}dt+o(dt)$, so its survival function satisfies $S^{\prime }(t)=-{\lambda }_{r}S(t)$ and $S(0)=1$. Integrating gives the mean holding time from a transition intensity matrix:This is the mean time until any allowed exit from that state, not necessarily the patient's remaining lifetime.

For confidence limits, use the supplied standard error $s_r$ of the corresponding diagonal generator estimate. Changing its sign does not change its standard error, so it is also the standard error of ${\hat{\lambda }}_r$. The delta method, with derivative $d(1/\lambda )/d\lambda =-1/{\lambda }^2$, givesThe resulting approximate 95% Wald confidence intervals areThe diagonal standard errors already incorporate the covariance among outgoing rates. They should not be replaced by the square root of a sum of squared off-diagonal standard errors, which would silently assume those rate estimates independent.

Alternatively construct a positive rate confidence interval and apply the confidence interval for an inverse rate transformation. Here every normal rate interval has a positive lower endpoint, givingThese approximate intervals are, respectively, $(10.863,12.830)$, $(4.497,7.161)$, $(2.397,3.117)$ and $(1.548,2.631)$ years. They are asymmetric because inversion is nonlinear; positive log-rate or likelihood-based intervals could also be inverted. The symmetric delta intervals and the inverted-rate intervals are two legitimate first-order approaches and need not have numerically identical endpoints.

The death state has zero exit rate, so its holding time is infinite. There is no finite mean-sojourn estimate or ordinary finite confidence interval for this structurally absorbing state. All intervals above concern the mean parameter, not a prediction interval for one patient's realized holding time.

A renewal process counts arrivals $N(t)=\max \{j:A_j\le t\}$, where $A_0=0,A_j=S_1+\cdots +S_j$ and the holding times are independent, identically distributed positive random variables. Their sum tends to infinity almost surely, so the count is finite on bounded time intervals. Use $N(s)=X_s$ in the question's notation. The index $J=N(s)+1=\min \{j:A_j>s\}$ is a stopping index for the holding-time sequence. Conditional on $J=j$ and the first $j$ holding times, every later $S_{j+k}$ retains its original probability distribution. Summing over $j$ provesThe straddling time $T_1=S_J$ is different. Let $\overline{F}(t)=P(S_1>t)$ and let $U$ be the renewal measure $U(B)={\sum }_{j\ge 0}P(A_j\in B)$. Independence of the next holding time from its start givesFor two numbers in $[0,1]$, their minimum is at least their product. Since $\overline{F}(\max (t,a))=\min (\overline{F}(t),\overline{F}(a))$, subtracting $\overline{F}(t)$ times the second integral provesThese formulae handle holding-time atoms with the stated renewal convention $A_j\le s<A_{j+1}$; sums may be interchanged by nonnegativity.

For rate-$\lambda$ exponentials, the renewal process is Poisson. The elapsed age $B_s=s-A_{N(s)}$ has probability distribution $\min (E,s)$ with $E\sim \mathrm{Exp}(\lambda )$: for $0\le b<s$, $P(B_s>b)$ is the probability of no arrivals in $(s-b,s]$, namely $e^{-\lambda b}$, with an atom $e^{-\lambda s}$ at $s$. The forward residual $R_s$ is an independent rate-$\lambda$ exponential by independent future increments. Thus $T_1=B_s+R_s$. Conditioning on the age gives the exact survival lawFor $s,t>0$ the extra term is strictly positive, proving strict inequality. For fixed $t$ and $s\to \infty$, the limit is $e^{-\lambda t}(1+\lambda t)$, which equals the survival probability of the sum of two independent rate-$\lambda$ exponentials. This is the finite-time form of the stochastic length bias of a renewal interval.

A Markov jump process, considered up to its explosion time, is an adapted càdlàg piecewise-constant process whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. Describe its rates by an off-diagonal kernel $q(x,dy)$ with finite total rate $q(x,E)$. In state $x$ the holding time has the exponential distribution with this rate, followed by a destination distributed as $q(x,dy)/q(x,E)$; a zero-rate state is absorbing. Its Markov jump-process generator acts byThe Markov property must hold relative to the specified filtration, not merely the natural filtration: future-revealing information could change the jump compensator.

Let $\mu$ count jumps, marked by their destination states. For this jump measure of a Markov jump process, the predictable compensator isThe compensated random-measure theorem says that a predictable $H(s,y)$ with $\mathbb{E}{\int }_0^t\int |H|d\nu <\infty$ gives a true martingaleThe same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $1_A{1}_{(s,t]}H$, for $A\in {\mathcal{F}}_s$, shows that each increment has zero conditional expectation. Taking $H(s,y)=f(y)-f(X_{s-})$ gives the generator local martingale $f(X_t)-f(X_0)-{\int }_0^{t}Qf(X_s)ds$. This states the integrability conditions in the compensated jump-measure local martingale result explicitly.

For the birth-process claim first take the customary fixed initial state $X_0=i$. Write $N_t=X_t-i$, ${\Lambda }_t={\int }_0^t\lambda (X_s)ds$ and $a=e^{\theta }-1$. The compensator of $N$ is $\Lambda$, whose integrand may equivalently use $X_{s-}$ because jump times have zero Lebesgue measure. A birth multiplies $e^{\theta X}$ by $e^{\theta }$, so the jump product rule givesFor an explicit localization, stop at ${\tau }_n=\inf \{t:X_t\ge n\}\wedge n$ with $n>i$. Up to that time the statewise rates are bounded by the maximum of the finitely many rates below $n$, the jump count is bounded, and $M$ and its compensated-integral integrand are bounded on the finite stopped horizon. The compensated-measure theorem therefore makes each stopped process a true martingale. The times increase to the explosion lifetime (or infinity if there is no explosion), proving $M$ is a local martingale on $[0,\zeta )$ for every real $\theta$.

Now suppose $\lambda (j)\le C$ uniformly. For $C>0$, propose candidate births at the times of a rate-$C$ Poisson process, accepting each with probability $\lambda (X_{s-})/C$ using fresh independent uniform marks. While the state is fixed, Poisson thinning gives a holding time with the required exponential distribution of rate $\lambda$; accepted births have exactly the prescribed law. Thus there can be no explosion, and the true birth count $N_T$ is stochastically dominated by a Poisson random variable of mean $CT$. In particular,The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,This is one integrable dominating variable for the entire stopped family. Conditional dominated convergence removes the localization and provesso the exponential martingale of a pure birth process is a genuine martingale for uniformly bounded rates. Only uniform integrability on each finite horizon is asserted; it need not be uniformly integrable over infinite time.

For a random initial state, the same proof requires $\mathbb{E}{e}^{\theta X_0}<\infty$. The source does not explicitly specify the initial distribution, so that qualification cannot be dropped: take constant rate one and an independent initial law $\mathbb{P}(X_0=n)=6/({\pi }^2{n}^2)$. For $\theta >0$, already $\mathbb{E}{M}_0=\infty$, precluding even the integrability in the definition of a martingale. The fixed-initial-state interpretation above supplies the intended complete proof; bounded rates alone do not remedy an arbitrary heavy-tailed initial law.

For independent identically distributed positive holding times $S_i$, define $T_0=0$, $T_n={\sum }_{i=1}^n{S}_i$ and the renewal count $X_t=\max \{n:T_n\le t\}$. With $0<\mu =E{S}_1<\infty$, the renewal strong law is $X_t/t\to 1/\mu$ almost surely. The elementary renewal theorem is $m(t)/t\to 1/\mu$, where $m(t)=E{X}_t$. More generally the rate is zero when the mean holding time is infinite, under the usual nonexplosion assumptions.

One complete cycle consists of ten stops of mean one minute and ten journeys of mean four minutes, so its mean duration is fifty minutes. Over the eight-hour shift, the elementary renewal approximation gives $480/50=9.6$ completed circuits on average.

For the end-of-shift calculation use the long-run occupation probabilities of the twenty successive phases. Each stop phase has probability $1/50$, and each travel phase probability $4/50$. Alighting stop $j$ is therefore selected with probability $1/50+4/50=1/10$, from either stopping there or travelling toward it. If that stop is the station, no further stops are passed. If it is stop $j\ge 2$, the intermediate stops are $j+1,\dots ,10$, numbering $10-j$. Thus the requested approximate mean isThe approximation treats eight hours as long compared with a cycle and ignores the initial-phase transient. Exponential holding times justify the twenty-phase Markov description; independent cycle times justify the renewal rate.

For $0<i<N$, the total rate out of $i$ is ${\lambda }_i$. The holding time of the continuous-time Markov chain and its next state satisfyThe holding time is independent of the direction of this jump: the conditional probability of destination $j$ is the rate $q_{ij}$ divided by the total rate ${\lambda }_i$, independently of the elapsed holding duration. This is the exponential-clock construction of a continuous-time Markov chain with the given infinitesimal generator matrix.

At $i=0$ or $i=N$, all rates out are zero. These are absorbing states; the path remains at its initial endpoint and $J_i=\infty$ almost surely. There is no next visited state, so $X_{J_i}$ is not a finite-time jump value at either endpoint. If one separately uses $X_{\infty }$ for the limiting state, that limit is the same endpoint.

The embedded jump chain in the interior is a simple symmetric random walk on $\{0,\dots ,N\}$, stopped at the endpoints. From any interior state it has probability at least $2^{-N}$ of hitting a boundary in the next $N$ jumps, by following successive leftward choices. Thus its number of jumps before absorption is finite almost surely. Each of this finite number of holding times is finite almost surely, so absorption also occurs in finite continuous time. The finite set of positive interior rates causes no explosion; in particular their maximum bounds the total jump rate. When $N=1$, the chain starts at an already absorbing endpoint.

The rates alter the clock but not the sequence of visited states. The gambler's ruin calculation from part (a), translated to an initial state $i$ between endpoints $0,N$, therefore givesThis includes probability one at zero and probability zero at $N$.

Use the positive-integer convention for the geometric distribution:Conditional on $K=k$, independence gives $\mathbb{E}[S\mid K=k]=k/\lambda$. Taking expectations givesThis is the convention required below because a chain starting at $i$ makes at least one visit to $i$. If instead $K$ counts failures on $\{0,1,\dots \}$, its mean is $(1-p)/p$, and the corresponding answer for that different random sum is $(1-p)/(p\lambda )$.

For the chain of part (b), let $K_i$ count visits to $i$, including the initial visit, before absorption. After a departure to $i-1$, gambler's ruin on $\{0,\dots ,i\}$ gives probability $1/i$ of reaching zero before returning to $i$. After a departure to $i+1$, the analogous probability of reaching $N$ before returning is $1/(N-i)$. The departure is equally likely in either direction, so the probability of absorption before a return isAt every return to $i$, the Strong Markov property resets these alternatives. Hence the geometric return count before absorption gives $K_i\sim \mathrm{Geom}(p_i)$ on positive integers. One may construct the chain from its embedded jump chain and a separate sequence of fresh independent exponential clocks. Then its successive holding times at $i$ are independent and identically distributed rate-${\lambda }_i$ exponential random variables, independently of $K_i$, andUsing the first calculation gives the occupation time at an interior state of a variable-rate symmetric walk:For $N=2,i=1$, $p_i=1$ and the only visit is the initial one; the degenerate positive-integer geometric distribution at one handles this case. Other states' rates affect the time of each excursion, but not its probability of return, which explains why only ${\lambda }_i$ appears in this expectation.

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 command uses the model without covariates, starts in mild fatigue at time zero, and calculates expected accumulated years in each state between times $0$ and $10$. Let $Z(t)$ be the fatigue state and $Q$ its fitted generator matrix. For the occupation time of a continuous-time Markov chain,the expected finite-horizon occupation time isThe first equality follows by interchanging a bounded time integral and expected value, and the second uses the homogeneous continuous-time Markov chain transition matrix. The reported expectations are 7.014800 years mild, 1.724686 years moderate, and 1.260514 years severe. They sum to ten because the person occupies exactly one state at every time. All repeated entries are counted: these are not the lengths of single holding times, the durations until first reaching each state, or the probabilities of occupying the states at year ten. The interpretation of the time limits is also given in the package 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.

Convolve three rate-$\lambda$ exponential distributions. The first two have convolution density ${\lambda }^{2}x{e}^{-\lambda x}{1}_{x>0}$; convolving once more givesThus $X$ is an Erlang distribution of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yieldsFor $t\ge \lambda$ the moment-generating function is infinite.

Realize each holding time as a consecutive block of three exponential holding times from one Poisson process $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb{E}{R}_t=p_1(t)+2p_2(t)$. Taking expectations givesThis proves the formula without assuming that the renewal count itself is Poisson.

The fitted progressive illness-death model has three allowed arrows: $1\to 2$ with estimated transition intensity $0.1849$, $1\to 3$ with $0.01935$, and $2\to 3$ with $0.06143$, all in years${}^{-1}$. State 3 is an absorbing state, and state 2 has no return arrow.

Once in state 2, the holding time has an exponential distribution with rate $q_{23}$, so the mean holding time from a transition intensity matrix isApply a confidence interval for an inverse rate to the printed rate interval $(0.03552,0.1063)$: since inversion reverses order, the approximate 95% interval for the mean isThe time-homogeneous Markov property makes the future depend only on the current state; the exponential distribution also has the memoryless property. For a person currently in state 2,Equivalently, $p_{22}(2)=p_{22}(1{)}^2\simeq 0.9404163^2$. Thus the fitted two-year death probability is approximately 11.6%.