Introduction
A transition intensity can change because the environment changes over calendar time, or because a person has spent longer in a particular state. These are different effects. For example, a hypothetical insurer might observe that disability incidence rises with the time since a policy began, while the chance of leaving a care state rises with the length of the current care episode. The first clock is tied to a fixed time origin; the second restarts on entry to the state.
Keeping the clocks separate matters when specifying a multiple-state model. The intensity must be conditioned on the time information that actually determines the transition rate. If the model uses the wrong clock, it can assign the same transition behaviour to people whose relevant histories differ.
An intensity is a rate, not a probability over a finite interval. In the small-interval definition, conditional on being in state \(i\), the probability of a transition to \(j\) during the next interval of length \(h\) is \(q_{ij}h+o(h)\) as \(h\) approaches zero. If an intensity depends on duration, both the calendar time and the duration advance as the process remains in state \(i\). When the process enters a new state, its duration in that new state starts at zero.
Which Clock Determines the Intensity?
For a calendar-time model, the transition rate at time \(t\) is determined by \(t\), even if the process entered its current state at a different time. A rate might change over the policy’s lifetime because of an external schedule or a change in the conditions applying to all policies. Two people in the same state at the same calendar time therefore have the same specified intensity for a given transition, regardless of how long each has been in that state.
For a duration-dependent model, the relevant clock is time since entry to the current state. Two people in the same state at the same calendar time can have different intensities if their state durations differ. For example, a hypothetical long-term care model might assign a higher rate of recovery after a care episode has lasted longer. The intensity depends on duration \(s\), not merely on the date \(t\).
The two effects can also occur together. An intensity \(q_{ij}(t,s)\) can depend on both the date and the time in state. This is appropriate only when both clocks are relevant to the assumed transition mechanism. Do not interpret a dependence on \(t\) as a dependence on \(s\): they coincide only in special situations, such as a state entered at the model origin and not left since.
Worked Example: Reading a Calendar-Time Intensity
In a hypothetical two-state model, a policyholder is either Active or Disabled. There is only one possible transition from Active, to Disabled. Its intensity at calendar time \(t\), measured in years since issue, is \(q_{AD}(t)=0.02+0.01t\) per year. Find the intensity at \(t=2\), then find the probability of remaining Active from issue through time \(2\), assuming the policyholder starts Active and the stated transition is the only way to leave Active.
Method. Evaluate the intensity at the specified calendar time. For the finite-interval probability, accumulate the intensity over the full interval and exponentiate the negative accumulated intensity.
At \(t=2\), the instantaneous intensity is \(q_{AD}(2)=0.02+0.01(2)=0.04\) per year. This is a rate at that instant, not a 4% probability of becoming Disabled over a year.
Because there is only one way to leave Active, the total exit intensity is \(0.02+0.01t\). The accumulated intensity from \(0\) to \(2\) is \(\int_0^2(0.02+0.01t)\,dt=0.02(2)+0.01(2^2/2)=0.04+0.02=0.06\).
Therefore the probability of remaining Active is \(e^{-0.06}=0.9418\), to four decimal places. The probability of entering Disabled by time \(2\) is \(1-0.9418=0.0582\), to four decimal places. The probabilities sum to \(1.0000\), as required for these two exhaustive outcomes.
Common mistake avoided: using \(q_{AD}(2)=0.04\) as the two-year transition probability. The intensity changes over the interval, and an instantaneous rate is not a finite-interval probability; the accumulated intensity is \(0.06\), giving an exit probability of \(0.0582\).
From Exit Intensity to a Sojourn Probability
To calculate the probability of staying in a state, use the total intensity of leaving that state. If there are several possible destinations \(j\), the total exit intensity is the sum of the outgoing intensities. For instance, if state \(i\) has possible transitions to states \(j\) and \(k\), its total exit intensity is \(q_{ij}+q_{ik}\), with each intensity evaluated at the applicable time arguments.
Suppose a process is in state \(i\) at calendar time \(t\), with duration \(s\) in that state. Write \(q_i^{\mathrm{out}}(t,s)=\sum_{j\ne i}q_{ij}(t,s)\) for its total exit intensity. Conditional on no exit yet, advancing a small amount \(du\) changes the probability of remaining in the state by approximately the factor \(1-q_i^{\mathrm{out}}(t+u,s+u)\,du\). Multiplying these conditional survival factors over the interval and taking the limiting product gives the exponential of the negative integrated intensity.
This formula assumes the specified intensities describe the process while it remains in state \(i\), and that time and duration advance together during the sojourn. It gives the probability of no transition out of \(i\) during the interval. If there are multiple exit destinations, it does not by itself give the probability of each destination; those probabilities also depend on the separate transition intensities.
Worked Example: An Exit Rate That Depends on State Duration
A hypothetical care model has one possible exit from the Care state, to Recovery. The exit intensity depends only on duration \(s\) in Care and is \(q_{CR}(s)=0.10+0.05s\) per year, where \(s\) is measured in years. A person has just entered Care. Find the probability of remaining in Care for the next two years and the probability of recovery during that period.
Method. Since the duration at the start is zero and the only exit is to Recovery, integrate the duration-dependent exit intensity from duration \(0\) to duration \(2\). The survival probability is the exponential of the negative integral.
The accumulated exit intensity is \(\int_0^2(0.10+0.05s)\,ds=0.10(2)+0.05(2^2/2)=0.20+0.10=0.30\).
The probability of remaining in Care is \(e^{-0.30}=0.7408\), to four decimal places. The probability of recovery within two years is the complement, \(1-e^{-0.30}=0.2592\), to four decimal places. These probabilities add to \(1.0000\).
Common mistake avoided: using the starting intensity \(q_{CR}(0)=0.10\) as if it applied throughout both years. That would give \(1-e^{-0.10(2)}=0.1813\) for recovery, which is too low because the intensity increases as the duration in Care grows. Integrating the varying intensity gives the correct probability, \(0.2592\).
When Duration Must Be Distinguished
The practical test is whether people in the same current state and at the same calendar time can have different future transition rates because of how long they have already been in that state. If they can, a model that specifies rates using calendar time alone does not capture the stated behaviour. Duration must be recorded as relevant information, or otherwise represented in the model.
This distinction is separate from ordinary calendar-time inhomogeneity. A calendar-time intensity can vary over \(t\) and still assign the same rate to every person currently in state \(i\) at a given \(t\). A duration effect instead distinguishes, for example, a person who entered a state recently from someone who has remained there for years. A model can also include both effects if its assumptions call for both.
The state-duration clock restarts whenever the process enters a state. If someone leaves a state and later returns, duration in the returned state is measured from that new entry, not from the first time the person ever occupied it. Calendar time, by contrast, continues from the fixed model origin throughout.
Worked Example: Comparing Two Durations at the Same Date
At calendar time \(t=4\), two hypothetical individuals are both in a temporary Support state. One has been in Support for \(s=1\) year and the other for \(s=3\) years. Each has one possible exit, to Independent status, with duration-dependent intensity \(q(s)=0.08+0.04s\) per year. Find each individual’s probability of exiting Support during the next year.
Method. For each person, integrate the intensity along the next year of that person’s state duration. Calendar time is the same for both and does not appear in this intensity; their starting durations are different.
For the individual with \(s=1\), duration during the year runs from \(1\) to \(2\). The accumulated intensity is \(\int_0^1[0.08+0.04(1+u)]\,du=0.08+0.04(1+0.5)=0.14\). The exit probability is \(1-e^{-0.14}=0.1306\), to four decimal places.
For the individual with \(s=3\), duration runs from \(3\) to \(4\). The accumulated intensity is \(\int_0^1[0.08+0.04(3+u)]\,du=0.08+0.04(3+0.5)=0.22\). The exit probability is \(1-e^{-0.22}=0.1975\), to four decimal places.
The longer-duration individual has the larger probability of exiting within the next year. The difference arises even though both people are in the same state at the same calendar time. That is precisely the distinction the duration dependence represents.
Common mistake avoided: assigning both people the same intensity because they share the same state and calendar date. The stated intensity depends on duration, so the two accumulated intensities, \(0.14\) and \(0.22\), must be calculated separately.
Choosing and Checking the Time Scale
Before using a transition intensity, identify its time argument and units. If \(t\) is years since policy issue, a term involving \(t\) must use that same origin and be measured in compatible units. If \(s\) is years in the current state, it must start at zero on entry to that state. Mixing months and years changes the numerical intensity and can produce substantially incorrect probabilities.
A useful specification check is to ask what the model predicts for two individuals who share the current state and calendar time but have different state durations. A calendar-only intensity predicts equal instantaneous rates for them. A duration-dependent intensity may predict unequal rates. The answer should match the intended model assumptions rather than being chosen merely for algebraic convenience.
These distinctions are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In calculations, keep three quantities separate: the instantaneous transition intensity, its integral over an interval, and the resulting probability. The integrated intensity is dimensionless; exponentiating its negative gives the probability of no exit under the stated model.
Check Your Understanding
For each question, identify the relevant clock and use an integrated intensity for any finite-interval probability.
- A hypothetical two-state model has only one exit from Active, with calendar-time intensity \(q(t)=0.03+0.01t\) per year. A policyholder starts Active at \(t=0\). What is the probability of remaining Active through \(t=2\)? (Answer: \(e^{-0.08}=0.9231\), to four decimal places.)
- A state has one exit and duration-dependent intensity \(q(s)=0.20+0.10s\) per year. A person has just entered the state. What is the probability of exiting during the next year? (Answer: \(1-e^{-0.25}=0.2212\), to four decimal places.)
- A hypothetical state has one exit, with calendar-time intensity \(0.04\) per year until \(t=2\), then \(0.08\) per year. A person is in the state at \(t=1.5\). What is the probability of exiting by \(t=3\)? (Answer: \(1-e^{-0.10}=0.0952\), to four decimal places.)
- A state has one exit with duration-dependent intensity \(q(s)=0.05+0.03s\) per year. A person has already been in the state for \(s=2\) years. What is the probability of exiting in the next half-year? (Answer: \(1-e^{-0.05875}=0.0571\), to four decimal places.)