Destination-Specific Holding-Time Distributions

Why the Next Destination Belongs With the Waiting Time

When a person enters a state, two linked questions arise: how long will they remain there, and where will they go next? These questions need not be independent. For example, among people who leave a state quickly, one destination may be common; among those who remain longer, another may be more likely. A model that describes only the overall time to leave can miss this pattern.

A semi-Markov model represents the holding time and the next destination jointly at each state entry. This differs from specifying a single holding-time distribution and then assigning a destination separately: the latter construction imposes independence unless the model explicitly links the two. The focus here is one state with two possible exits, though the definitions extend to any finite set of destinations.

Definition: Suppose a process enters state $i$. Let $T_i$ be the time from entry until its next transition, and let $J_i$ be the destination state at that transition. The semi-Markov transition kernel for an allowed destination $j\ne i$ is $$ Q_{ij}(t)=P(T_i\le t,\ J_i=j\mid \text{entry into state }i), \qquad t\ge 0. $$ It is a joint, destination-specific cumulative distribution, not generally a conditional distribution given $J_i=j$. Its total over destinations is $P(T_i\le t)$.

The associated marginal holding-time survival function is $S_i(t)=P(T_i>t)$. Thus, when the possible destinations include all exits, $S_i(t)=1-\sum_{j\ne i}Q_{ij}(t)$. If a transition-time density exists, write $f_{ij}(t)$ for the joint density of $T_i=t$ and $J_i=j$, in the sense that $Q_{ij}(t)=\int_0^t f_{ij}(u)\,du$. The eventual probability of destination $j$ is $p_{ij}=P(J_i=j)=Q_{ij}(\infty)$, if the process leaves state $i$ with probability one.

If $p_{ij}>0$, the holding-time distribution conditional on destination $j$ has cumulative distribution $Q_{ij}(t)/p_{ij}$, and, when densities exist, density $f_{ij}(t)/p_{ij}$. This is different from $Q_{ij}(t)$, which includes the probability of choosing destination $j$ in the first place. Keeping those two quantities distinct is essential for interpretation.

From Competing Intensities to Joint Distributions

A useful continuous-time description specifies a duration-dependent cause-specific transition intensity for each possible exit. Let $\lambda_{ij}(u)$ be the instantaneous rate of transition from $i$ to $j$, at elapsed duration $u$ since entry, conditional on still being in $i$ just before $u$. In a small interval of length $du$, the probability of an $i$-to-$j$ transition is approximately $\lambda_{ij}(u)\,du$, conditional on having remained in $i$ until then. Assume these rates are nonnegative and locally integrable, there are no simultaneous exits, and the rates depend on the current state and elapsed time as specified by the model, not on omitted past history.

With two possible exits, say $j=1,2$, the total exit intensity at duration $u$ is $\lambda_{i1}(u)+\lambda_{i2}(u)$. The probability of still being in $i$ through duration $t$ is the exponential of minus the accumulated total intensity. The probability density of leaving at duration $t$ for a particular destination is survival up to $t$ multiplied by that destination’s intensity:

S_i(t)=\exp\left\{-\int_0^t \bigl[\lambda_{i1}(u)+\lambda_{i2}(u)\bigr]\,du\right\}, \qquad f_{ij}(t)=S_i(t)\lambda_{ij}(t).

This relation shows why a destination-specific distribution is not determined by its own intensity alone. Its density also depends on the chance of not having taken either exit earlier. Consequently, $Q_{ij}(t)=\int_0^t S_i(u)\lambda_{ij}(u)\,du$. If an exit eventually occurs with probability one, the destination probabilities are the integrated cause-specific densities, $\sum_j p_{ij}=1$.

Justification: The Total Exit Rate Determines Survival

The survival formula follows directly from the definition of the intensities. Over a short interval $[u,u+du)$, conditional on being in $i$ at $u$, the probability of either exit is approximately $[\lambda_{i1}(u)+\lambda_{i2}(u)]du$. Therefore the probability of surviving that interval is approximately one minus this quantity. In differential form,

\frac{dS_i(t)}{dt} =-S_i(t)\bigl[\lambda_{i1}(t)+\lambda_{i2}(t)\bigr], \qquad S_i(0)=1.

Dividing by $S_i(t)$ where it is positive and integrating from 0 to $t$ gives $\log S_i(t)=-\int_0^t[\lambda_{i1}(u)+\lambda_{i2}(u)]\,du$, hence the stated exponential survival. Conditional on surviving to $t$, the chance of the particular exit $j$ in the next small interval is approximately $\lambda_{ij}(t)dt$. Multiplying by the probability $S_i(t)$ of reaching that interval still in $i$ gives $f_{ij}(t)=S_i(t)\lambda_{ij}(t)$. This also verifies that summing the destination-specific densities recovers the overall exit density, $-S_i'(t)$.

Worked Example: Two Constant Exit Rates

Suppose a hypothetical process enters state $i$ and can move to states 1 or 2. Take constant intensities $\lambda_{i1}(t)=0.08$ per month and $\lambda_{i2}(t)=0.12$ per month. Assume these rates apply throughout the holding time, no other exit is possible, and the process eventually leaves $i$.

The total rate is $0.08+0.12=0.20$ per month, so $S_i(t)=e^{-0.20t}$, with $t$ measured in months. The joint densities are $f_{i1}(t)=0.08e^{-0.20t}$ and $f_{i2}(t)=0.12e^{-0.20t}$. Integrating over all possible holding times gives

p_{i1}=\int_0^\infty 0.08e^{-0.20t}\,dt =\frac{0.08}{0.20}=0.400, \qquad p_{i2}=\frac{0.12}{0.20}=0.600.

The conditional holding-time density given destination 1 is $f_{i1}(t)/p_{i1}=0.20e^{-0.20t}$, and the same result holds given destination 2. In this special constant-rate case, destination and holding time are independent: either destination has an exponential holding-time distribution with mean $1/0.20=5.00$ months. At 3 months, $S_i(3)=e^{-0.60}=0.5488$. The probability of having left for state 1 by then is $0.400(1-0.5488)=0.1805$; for state 2 it is $0.600(1-0.5488)=0.2707$. The two probabilities sum to $0.4512$, the probability of any exit by 3 months.

Worked Example: Duration-Dependent Competition

Now suppose the hypothetical intensities are $\lambda_{i1}(u)=0.10$ per month and $\lambda_{i2}(u)=0.05u$ per month, where $u$ is measured in months. The coefficient 0.05 therefore has units of per month squared. Both rates are nonnegative for $u\ge0$; the rate for exit 2 increases with time in the state.

The cumulative total intensity through $t$ is $0.10t+0.025t^2$. Hence $S_i(t)=\exp(-0.10t-0.025t^2)$, and the joint densities are $f_{i1}(t)=0.10S_i(t)$ and $f_{i2}(t)=0.05tS_i(t)$. At duration 2 months, the respective densities are $0.10e^{-0.30}=0.07408$ and $0.10e^{-0.30}=0.07408$, in units of probability per month. Thus, conditional on leaving at exactly that duration in the density sense, the two destinations have equal instantaneous shares. This statement concerns the relative densities at 2 months, not probabilities of an event at one exact time.

For a finite-window calculation, numerical integration gives $Q_{i1}(2)=\int_0^2 0.10e^{-0.10u-0.025u^2}\,du\approx0.1757$ and $Q_{i2}(2)=\int_0^2 0.05u e^{-0.10u-0.025u^2}\,du\approx0.0835$. For example, Simpson’s rule with step 0.5 month gives these approximations. Their sum is $0.2592$, agreeing with $1-S_i(2)=1-e^{-0.30}=0.2592$, up to rounding. Among exits occurring by 2 months, the estimated model probability of destination 2 is therefore $0.0835/0.2592\approx0.322$. The instantaneous destination-2 share grows with duration, but the cumulative share through 2 months is lower because early exits more often go to destination 1.

Conditional Holding Times Can Reveal Dependence

The following construction makes the association between time and destination especially clear. Suppose a hypothetical joint distribution assigns probability 0.60 to destination 1, with an exponential holding time of rate 0.40 per month conditional on that destination, and probability 0.40 to destination 2, with an exponential holding time of rate 0.10 per month conditional on that destination. This specifies the joint densities directly as the destination probabilities multiplied by their conditional densities.

Worked Example: Different Holding Times by Destination

For this model, $f_{i1}(t)=0.60(0.40)e^{-0.40t}=0.24e^{-0.40t}$ and $f_{i2}(t)=0.40(0.10)e^{-0.10t}=0.04e^{-0.10t}$. The densities integrate to 0.60 and 0.40, so the eventual destination probabilities sum to one. The overall survival function is the sum of the two conditional survival probabilities weighted by their destination probabilities: $S_i(t)=0.60e^{-0.40t}+0.40e^{-0.10t}$.

Conditional mean holding times are $1/0.40=2.50$ months for destination 1 and $1/0.10=10.0$ months for destination 2. By 2 months, the probability of exit to destination 1 is $0.60(1-e^{-0.80})=0.3304$, while the probability of exit to destination 2 is $0.40(1-e^{-0.20})=0.07251$. The probability of any exit by 2 months is their sum, $0.4029$; among those exits, the destination-1 fraction is $0.3304/0.4029=0.820$. This finite-time destination mix differs from the eventual mix of 0.60 and 0.40 because destination 1 tends to occur sooner.

The implied cause-specific intensities are $f_{ij}(t)/S_i(t)$. For example, at duration zero, the destination-1 intensity is $0.24$ per month and the destination-2 intensity is $0.04$ per month. Their ratio changes over time because the two conditional holding-time distributions differ. Thus the direct joint-distribution specification and the intensity specification are two ways to describe the same model, when the distributions are sufficiently regular.

Assumptions, Diagnostics, and Alternatives

The model is meaningful only relative to a clearly defined state-entry time and set of possible destinations. Its semi-Markov assumption is that, conditional on the current state and the model’s included information, the joint law of the next holding time and destination does not depend on the earlier path. Duration dependence is allowed: the exit rates can vary with time since entry. If the next transition also depends on omitted prior states, calendar time, or an individual’s changing covariates, then a state-specific kernel alone may not describe the process adequately.

Useful checks include comparing observed holding-time patterns across destinations, examining whether destination proportions change across duration ranges, and checking whether fitted cause-specific densities sum to the overall exit density. In intensity-based models, fitted intensities should be nonnegative, and the resulting survival function should be nonincreasing and remain between zero and one. Large discrepancies between predicted and observed exits in particular duration ranges can signal a poor duration form, omitted heterogeneity, or an incorrect state definition. With sparse data, apparent duration-by-destination patterns may be unstable; uncertainty matters as much as the fitted shape.

The main alternatives impose different structures. A Markov model with constant transition intensities implies exponential holding times and, for competing constant exits, independence of destination and holding time as in the first example. A model with one overall holding-time distribution and a separate destination probability can be simpler, but it assumes independence unless dependence is added. A fully specified semi-Markov kernel allows that dependence directly, while duration-dependent cause-specific intensities express it through the competing rates and shared survival. These approaches should not be treated as interchangeable without checking their assumptions.

A final limitation is that this description concerns a single transition after entry into a state. It does not by itself provide the probability of occupying each state at later calendar times along a multi-transition path. Also, if some individuals never leave state $i$, the destination probabilities can sum to less than one; such a model must include that possibility rather than silently renormalizing the observed exits. Censoring can prevent observing either a completed holding time or its destination, so empirical estimation requires appropriate treatment of censoring and its assumptions.

Common mistake: Treating $Q_{ij}(t)$ as the distribution of holding time conditional on destination $j$. It is instead the joint probability of a duration no longer than $t$ and destination $j$. Divide by $p_{ij}$ to obtain the conditional distribution, when $p_{ij}>0$.
Another pitfall: Using only $\lambda_{ij}(t)$ as the density of an exit to $j$. The correct density is $S_i(t)\lambda_{ij}(t)$, because the process must still be in $i$ at duration $t$.

Key takeaway: A destination-specific holding-time distribution is a joint description of when the process leaves a state and where it goes. Under competing duration-dependent intensities, each joint density is the survival from all exits multiplied by the intensity of the particular exit.

Check Your Understanding

Answer each question, then compare with the solutions below.

With constant competing rates 0.03 and 0.07 per month, what are the survival function and eventual destination probabilities?
In the two-constant-rate model, why is the joint density for destination 1 not just $\lambda_{i1}(t)$?
If $Q_{i2}(t)=0.24$ and $p_{i2}=0.40$, what is the probability that the holding time is at most $t$, conditional on destination 2?
At a given duration $t$, what does $f_{i1}(t)/[f_{i1}(t)+f_{i2}(t)]$ represent when the denominator is positive?

Solutions. (1) The total rate is $0.03+0.07=0.10$ per month, so $S_i(t)=e^{-0.10t}$. The destination probabilities are $0.03/0.10=0.300$ and $0.07/0.10=0.700$. (2) The process must survive in state $i$ until duration $t$; therefore the density is $S_i(t)\lambda_{i1}(t)$, not the intensity alone. (3) The conditional cumulative distribution is $Q_{i2}(t)/p_{i2}=0.24/0.40=0.600$. (4) It is the destination-1 share of the exit-time density at duration $t$, or equivalently the relative cause-specific intensity at that duration. It is not the probability of an event at exactly $t$.