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Semi-Markov and Non-Markov Models · Tutorial 323 of 699

Residual Sojourn Time After State Survival

Learn how to condition a sojourn-time distribution on surviving in a state to a landmark, and how the resulting residual time differs from a fresh sojourn.

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What You'll Learn

  • Derive the residual survival function and distribution from the original sojourn survival function.
  • Calculate destination-specific residual probabilities from a semi-Markov transition kernel.
  • Explain how duration-dependent hazards change after survival to a landmark.
  • Identify when residual and fresh sojourn distributions agree.
  • Recognize how unknown sojourn age or latent heterogeneity affects residual-time calculations.
  • Check assumptions and avoid common landmark-conditioning errors.

Why survival to a landmark changes the waiting-time question

Suppose a person enters a clinical state, such as living with an illness, and is still in that state at a later landmark time. The question “how much longer will this person remain here?” is not generally answered by the original distribution of time spent in the state. The person has already spent time there without exiting. That survival is information about the sojourn, and the remaining-time distribution must be conditioned on it.

The distinction matters whenever the chance of leaving a state depends on how long the current sojourn has lasted. A person who has remained in a state for a long time may have a different near-term transition risk from someone just entering it. This tutorial derives that conditional distribution, including its destination-specific form.

Definition: Consider a sojourn that begins when an individual enters state \(i\). Let \(T_i\) be the time from entry until the individual leaves state \(i\), and let \(J\) be the destination state at exit. Define the sojourn survival function \(S_i(t)=P(T_i>t)\), for \(t\geq 0\). For a landmark \(a\) with \(S_i(a)>0\), the residual sojourn time is \(R_a=T_i-a\), conditional on \(T_i>a\). It is the additional time in state \(i\), given that the individual is still there at elapsed sojourn time \(a\).

The landmark \(a\) is measured from entry into the current sojourn, not automatically from the study’s calendar-time origin. The conditioning event \(T_i>a\) means that the current visit to state \(i\) has lasted beyond \(a\); it does not describe a person who has already left and later returned.

Deriving the residual distribution

For \(r\geq 0\), the event \(R_a>r\), given \(T_i>a\), is exactly the event that the sojourn lasts beyond \(a+r\), given that it lasts beyond \(a\). Therefore,

$$ P(R_a>r\mid T_i>a) =\frac{P(T_i>a+r)}{P(T_i>a)} =\frac{S_i(a+r)}{S_i(a)}. $$

This is the residual survival function. Subtracting it from one gives the conditional cumulative distribution function:

$$ P(R_a\leq r\mid T_i>a) =1-\frac{S_i(a+r)}{S_i(a)} =\frac{S_i(a)-S_i(a+r)}{S_i(a)}. $$

The numerator in the last expression is the probability of exiting between elapsed times \(a\) and \(a+r\). Dividing by \(S_i(a)\) conditions on having reached the landmark without exiting. If the sojourn has a continuous density \(f_i(t)\), the residual density is \(f_i(a+r)/S_i(a)\) for \(r>0\). These formulas also allow distributions with atoms when probabilities are interpreted using the stated strict survival event and the corresponding interval events.

Key result: The residual distribution is obtained by shifting the original survival function to the landmark and renormalizing by survival to that landmark. It is not generally obtained by restarting the original sojourn clock at zero.

When exit can lead to several destinations, the same conditioning applies to the transition kernel. Write \(Q_{ij}(t)=P(T_i\leq t,J=j)\), the probability of exiting state \(i\) by elapsed time \(t\) and entering destination \(j\). For a sojourn that has survived to \(a\),

$$ P(R_a\leq r,J=j\mid T_i>a) =\frac{Q_{ij}(a+r)-Q_{ij}(a)}{S_i(a)}. $$

Summing over possible destinations gives the total residual distribution, because the events of exiting to distinct destinations are disjoint. If exit occurs eventually with probability one, the probability that the eventual destination is \(j\), conditional on surviving to \(a\), is \([Q_{ij}(\infty)-Q_{ij}(a)]/S_i(a)\). Conditioning on survival may therefore change not only the expected remaining time but also the distribution of the eventual destination.

What happens to the hazard?

For a continuous sojourn distribution, define the sojourn hazard at elapsed duration \(t\) as the instantaneous exit rate among individuals who have remained in the state until \(t\):

$$ h_i(t)=\lim_{\Delta t\downarrow 0} \frac{P(t<T_i\leq t+\Delta t\mid T_i>t)}{\Delta t}. $$

At residual time \(u\) after surviving to landmark \(a\), the individual has accumulated total sojourn duration \(a+u\). Thus the residual hazard, at residual duration \(u\), equals \(h_i(a+u)\), not \(h_i(u)\). The accumulated time in the state is still relevant; conditioning has not reset it.

$$ h_{R_a}(u)=h_i(a+u). $$

This identity follows directly from the conditional probability defining the hazard: being at residual time \(u\) means surviving in the original sojourn to \(a+u\), and an exit in the next short interval is an exit at original duration \(a+u\). In a model with destination-specific cause hazards, each destination’s hazard is evaluated at that same accumulated duration.

A useful special case is the exponential distribution, for which \(S_i(t)=e^{-\lambda t}\), with constant rate \(\lambda>0\). Then

$$ \frac{S_i(a+r)}{S_i(a)} =\frac{e^{-\lambda(a+r)}}{e^{-\lambda a}} =e^{-\lambda r} =S_i(r). $$

The residual survival is identical to the fresh sojourn survival. This is the memoryless property. For a general continuous proper distribution, equality of residual and fresh distributions at every landmark requires the memoryless property and characterizes the exponential distribution. Equality at one particular landmark alone does not establish that the whole distribution is exponential.

Worked Example: Residual time under a Weibull sojourn model

Suppose, hypothetically, that a sojourn has survival function \(S(t)=\exp[-(t/4)^2]\), where \(t\) and the scale 4 are measured in years. Find the probability that someone who has already remained in the state for 3 years stays at least 2 additional years.

The condition for applying the residual formula is \(S(3)>0\), which holds because the specified survival function is positive at every finite time. The requested probability is

$$ P(R_3>2\mid T>3) =\frac{S(5)}{S(3)} =\frac{\exp[-(5/4)^2]}{\exp[-(3/4)^2]} =\exp[-(25-9)/16] =e^{-1} =0.3679. $$

Thus, among individuals still in the state after 3 years, about 36.8% are predicted to remain there for more than 2 additional years under this hypothetical model. By comparison, a newly entered individual has probability \(S(2)=e^{-0.25}=0.7788\) of remaining for more than 2 years. The two probabilities differ because the landmark group has already survived to a duration at which the hazard is higher.

Indeed, differentiating the cumulative hazard \((t/4)^2\) gives \(h(t)=t/8\) per year. The residual hazard at \(u\) years after the landmark is \((3+u)/8\) per year, whereas a fresh sojourn’s hazard at duration \(u\) is \(u/8\) per year. The formula and the hazard both show why restarting the clock would be inappropriate here.

Destination-specific residual risks

In a multi-state setting, an exit from state \(i\) can have different destinations. The residual waiting-time calculation must retain which destination occurs; summing over destinations too early can hide changes in destination composition after landmark survival. The following example uses a simple kernel to make that distinction explicit.

Worked Example: Residual exits to two destinations

Consider a hypothetical sojourn with survival \(S(t)=e^{-0.1t^2}\), where time is measured in years. Suppose each exit goes to destination 1 with probability 0.6 and destination 2 with probability 0.4, independently of the sojourn duration. This specifies \(Q_{i1}(t)=0.6[1-S(t)]\) and \(Q_{i2}(t)=0.4[1-S(t)]\). Find, among people still in the state at \(a=2\) years, the probability of exiting within the next year to each destination.

The landmark condition is valid because \(S(2)=e^{-0.4}=0.6703>0\). The total conditional probability of an exit in the next year is

$$ \frac{S(2)-S(3)}{S(2)} =1-\frac{e^{-0.9}}{e^{-0.4}} =1-e^{-0.5} =0.3935. $$

For destination 1, the kernel increment is \(0.6[S(2)-S(3)]\). Therefore,

$$ P(R_2\leq 1,J=1\mid T>2) =\frac{0.6[S(2)-S(3)]}{S(2)} =0.6(0.3935) =0.2361. $$

Similarly, the probability of exiting within one year to destination 2 is \(0.4(0.3935)=0.1574\). The two probabilities sum to \(0.3935\), the total probability of an exit in that interval. Given that an exit occurs within the year, the destination probabilities are \(0.2361/0.3935=0.6000\) and \(0.1574/0.3935=0.4000\). In this particular model, landmark survival does not change destination proportions because destination and duration were assumed independent. That conclusion would not generally follow without this assumption.

Survival selection and unknown sojourn age

Landmark conditioning can change the composition of a group even when the model contains latent types. For example, people with longer typical sojourns are more likely to remain in the state until a later landmark. The next calculation shows how that selection changes residual survival.

Worked Example: A mixture of two sojourn types

Suppose, hypothetically, that half of entrants belong to a type with exponential exit rate \(0.2\) per year and half to a type with rate \(1.0\) per year. Type is unobserved, but remains fixed during a sojourn. Find the probability of staying at least one more year after surviving to year 2.

Before conditioning, the survival function is the mixture \(S(t)=0.5e^{-0.2t}+0.5e^{-t}\). At year 2, the contributions to survival are \(0.5e^{-0.4}=0.3352\) for the slower type and \(0.5e^{-2}=0.06767\) for the faster type. The total is \(S(2)=0.4029\). Hence the conditional type weights among survivors are \(0.3352/0.4029=0.8320\) and \(0.06767/0.4029=0.1680\), respectively.

The one-year residual survival is therefore the weighted average of the two types’ one-year exponential survivals:

$$ P(R_2>1\mid T>2) =0.8320e^{-0.2}+0.1680e^{-1} =0.7430. $$

For a fresh entrant, the one-year survival probability is instead \(0.5e^{-0.2}+0.5e^{-1}=0.5933\). The increase to 0.7430 among landmark survivors occurs because the slower-exiting type is overrepresented among those who have already survived two years. Although each type separately has the exponential memoryless property, the marginal mixture does not: conditioning changes its type weights.

If the current sojourn age is unknown, one cannot generally substitute the calendar landmark into \(S_i(a+r)/S_i(a)\). Let \(A\) denote the elapsed age of the current sojourn at the landmark. If the distribution of \(A\) among individuals known to be in state \(i\) is available, then the residual survival is a mixture of age-specific residual survivals:

$$ P(R>r\mid \text{in state }i\text{ at landmark}) =\int \frac{S_i(a+r)}{S_i(a)} \,dP(A\leq a\mid \text{in state }i\text{ at landmark}). $$

This expression presumes that, conditional on age and the model information used, the sojourn distribution is \(S_i\). If people in the state have different unobserved types, the relevant conditional distribution must also account for those types. The age mixture and the type mixture are distinct sources of selection; neither is resolved by simply treating everyone as a new entrant at the landmark.

Assumptions, checks, and alternatives

The residual formulas are probability identities once the sojourn distribution is specified. Their model-based use nevertheless depends on defining a coherent sojourn and conditioning event. In particular:

  • The time origin is entry into the current visit to state \(i\), and \(T_i\) measures time to its exit.
  • The landmark is an elapsed sojourn duration \(a\), and \(S_i(a)>0\); otherwise the conditional distribution is undefined.
  • The survival or transition kernel used applies to the population and conditioning information of interest. If relevant types or covariates differ, their distribution among landmark survivors must be represented.
  • For destination-specific calculations, the kernel \(Q_{ij}\) must describe the joint event of exit time and destination; destination independence must not be assumed unless justified.

A practical check is to verify that the residual survival at \(r=0\) is 1, is nonincreasing in \(r\), and lies between 0 and 1. Destination-specific probabilities should be nonnegative and sum to the total residual exit probability. If a model’s residual survival increases with \(r\), exceeds 1, or fails these sum checks, there is likely an algebra, coding, or kernel-definition error. Also check whether \(a\) is duration since entry or calendar time: confusing the two is a common source of incorrect predictions.

The main alternative is to use a fresh sojourn distribution \(S_i(r)\) at the landmark. That is correct when the landmark individual is a new entrant, or when the sojourn distribution is memoryless in the relevant model. It is generally wrong for a person who has already survived a nonzero duration under a duration-dependent model. A state-transition probability over a calendar-time interval answers a different question: it may account for multiple transitions and paths, whereas the residual formula concerns the remaining time in one current sojourn. Use the quantity matching the question.

QuestionRelevant quantityKey distinction
How long does a new entrant remain in state \(i\)?\(S_i(r)\)Clock starts at entry.
How much longer does a survivor at duration \(a\) remain?\(S_i(a+r)/S_i(a)\)Conditions on survival to \(a\).
What is the probability of occupying a state at a future calendar time?A transition probabilityCan include multiple transitions, not just one residual sojourn.

The residual distribution is limited by the quality and scope of the assumed sojourn model. It does not by itself adjust for an unknown age distribution, unmeasured heterogeneity, or a change in the process at the landmark. In applied work, the landmark population and the meaning of state entry should be stated explicitly. A fitted sojourn distribution that ignores selection among survivors may produce a mathematically valid conditional calculation for the wrong target population.

Common mistake: “The person is still in state \(i\), so their remaining time has the same distribution as the time for a new entrant.” This silently resets the sojourn clock. The correction is to condition on survival to elapsed age \(a\): use \(S_i(a+r)/S_i(a)\), and use the corresponding kernel increment for destination-specific probabilities. The fresh distribution agrees only under additional conditions, such as the exponential memoryless model.

Check Your Understanding

Answer each question using the conditional residual-time definitions. Short solutions follow the questions.

  1. If \(S(t)=e^{-0.3t}\), what is the probability of remaining at least 2 additional time units after surviving to \(a=5\)?
  2. If \(S(t)=e^{-0.05t^2}\), calculate the probability of exiting within the next 2 time units after surviving to \(a=2\).
  3. State the destination-specific residual probability in terms of \(Q_{ij}\), \(S_i\), \(a\), and \(r\). What extra assumption would allow destination proportions to remain unchanged after landmark survival?
  4. Why can a mixture of exponential sojourn distributions fail to have the memoryless property, even though each component is memoryless?

Solutions. (1) The exponential residual survival is unchanged, so the probability is \(e^{-0.3(2)}=e^{-0.6}=0.5488\). (2) The exit probability is \(1-S(4)/S(2)=1-e^{-0.05(16-4)}=1-e^{-0.6}=0.4512\). (3) It is \([Q_{ij}(a+r)-Q_{ij}(a)]/S_i(a)\). Destination proportions remain unchanged in the example when destination is independent of sojourn duration. (4) Survival to the landmark changes the mixture weights, generally overrepresenting types with longer sojourns, so the marginal residual distribution is not the original mixture distribution.

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