Why Rates May Change at Calendar-Time Boundaries
A continuous-time model need not use the same transition intensities throughout its projection. For example, a hypothetical insurer may use one set of assumptions for the first two years of a projection and another set thereafter. The change may reflect a planned change in assumptions or a contract feature that takes effect at a specified calendar time. In either case, the model must state which rates apply in each time interval.
The key is to keep track of calendar time measured from the model’s time origin. A rate that changes at time \(t=2\) changes for everyone in the model at that calendar time, regardless of how long any particular person has spent in their current state. This is different from a rate that changes after an individual has spent a specified duration in a state.
This tutorial is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. It focuses on specifying and interpreting piecewise-constant intensity matrices, not on calculating finite-interval transition probabilities.
Defining the Time Intervals and Matrices
Suppose a model begins at calendar time \(t=0\), with a change at \(t=c\). A common interval convention is \([0,c)\) for the first matrix and \([c,\infty)\) for the second. The first matrix therefore applies at times strictly before \(c\); the second applies from \(c\) onward. For multiple boundaries, use consecutive intervals such as \([0,c_1)\), \([c_1,c_2)\), and \([c_2,\infty)\).
The interval convention must be clear, especially at a boundary. Assigning the boundary to the interval on its right is convenient and standard. The value assigned to the intensity at a single instant does not change the amount of time spent in either interval; what matters for the model is that the rates on each interval are specified consistently.
For each matrix, retain the same state order. The entry in row \(i\), column \(j\), for \(i\ne j\), is the intensity of transition from state \(i\) to state \(j\). A zero means that transition is not allowed in the model. The diagonal is determined by the other entries in its row, so that the row sums to zero.
All intensities must be nonnegative off the diagonal. A row with no possible exits has zero in every entry, including its diagonal; such a state is absorbing. These checks apply separately to every matrix in the schedule.
Worked Examples
Worked Example: A Two-State Model With One Change
A hypothetical life model has two states, Active and Dead, in that order. Time is measured in years from the start of coverage. The intensity of transition from Active to Dead is \(0.04\) per year for \(0\le t<5\), and \(0.06\) per year for \(t\ge5\). Dead is absorbing. Specify and interpret the interval matrices.
Method. Write the given Active-to-Dead intensity in the off-diagonal position in each interval. There are no exits from Dead, so its row consists of zeros. Then make each diagonal the negative sum of its row’s off-diagonal intensities.
For \(0\le t<5\), the Active row has one outgoing intensity, \(0.04\). Its diagonal is therefore \(-(0.04)=-0.04\). The Dead row has no outgoing intensities and has diagonal zero:
For \(t\ge5\), the Active-to-Dead intensity is \(0.06\). The Active diagonal is \(-(0.06)=-0.06\), while the Dead row remains all zeros:
In the first interval, \(0.04\) is the instantaneous intensity of death while Active; in the second, that intensity is \(0.06\). These are rates per year, not probabilities of dying over an entire year. Each row sums to zero, and the off-diagonal entries are nonnegative. Common mistake avoided: retaining the first matrix for someone who became Active before time 5. The boundary is calendar time \(t=5\), so the second matrix applies to everyone still in the model from that time onward.
Worked Example: Two Possible Exits Before and After a Boundary
A hypothetical model has three states, Active \(A\), receiving long-term care \(L\), and Dead \(D\), in that order. Time is measured in years from the start of the projection. Before calendar time 2, the intensities are \(A\to L=0.03\), \(A\to D=0.01\), \(L\to A=0.08\), and \(L\to D=0.12\), all per year. From time 2 onward, they are \(A\to L=0.05\), \(A\to D=0.015\), \(L\to A=0.04\), and \(L\to D=0.16\), all per year. Dead is absorbing. Construct both matrices.
Method. Use the stated order \(A,L,D\). Put each transition intensity in its row and destination column. In each transient state, add its outgoing intensities and take the negative of that total for the diagonal.
For \(0\le t<2\), the total exit intensity from \(A\) is \(0.03+0.01=0.04\), giving diagonal \(-0.04\). The total exit intensity from \(L\) is \(0.08+0.12=0.20\), giving diagonal \(-0.20\). Since \(D\) is absorbing, its row is all zeros:
For \(t\ge2\), the total exit intensity from \(A\) is \(0.05+0.015=0.065\), giving diagonal \(-0.065\). The total exit intensity from \(L\) is \(0.04+0.16=0.20\), giving diagonal \(-0.20\). The Dead row remains all zeros:
Check the row sums explicitly: in \(Q^{(1)}\), they are \(-0.04+0.03+0.01=0\), \(0.08-0.20+0.12=0\), and \(0\). In \(Q^{(2)}\), they are \(-0.065+0.05+0.015=0\), \(0.04-0.20+0.16=0\), and \(0\). Every off-diagonal entry is nonnegative. Common mistake avoided: calling this a two-state model because it has two non-dead states. It has three states in total: \(A\), \(L\), and \(D\).
Worked Example: Assigning the Matrix After Someone Enters a State
A hypothetical three-state model has state order \(1,2,3\), where state 3 is absorbing. Calendar time is measured in years from the model origin. Its intensity matrices are
A person enters state 2 at calendar time \(t=1.5\) and remains there until \(t=2.4\). State the outgoing intensities that apply before and after time 2, and check the relevant row sums.
Method. First identify the calendar interval, then read the row for state 2. From \(t=1.5\) up to but not including \(t=2\), \(Q^{(1)}\) applies. Its state-2 row is \((0.02,-0.07,0.05)\), so the intensities are \(q_{21}=0.02\) and \(q_{23}=0.05\) per year. The total exit intensity is \(0.02+0.05=0.07\), and the diagonal check is \(-0.07+0.02+0.05=0\).
At \(t=2\), the model switches to \(Q^{(2)}\), even though the person entered state 2 only half a year earlier. The state-2 row is now \((0.01,-0.11,0.10)\), so the intensities are \(q_{21}=0.01\) and \(q_{23}=0.10\) per year. The total exit intensity is \(0.01+0.10=0.11\), and the diagonal check is \(-0.11+0.01+0.10=0\). These rates continue to apply from \(t=2\) onward while the model remains in this interval. Common mistake avoided: treating the person’s entry time \(1.5\) as the start of a new intensity schedule. The stated boundary is at calendar time 2, not after a fixed duration in state 2.
Why the Diagonal Changes With the Outgoing Rates
The diagonal is not an independently chosen transition intensity. Its value follows from the requirement that, over a very short interval, the probabilities of remaining in the current state and moving to any other state add to one. This also explains why the diagonal must be recomputed whenever an outgoing intensity changes.
Suppose the process is in state \(i\) at time \(t\), and the outgoing intensities are constant in the interval containing \(t\). For a small positive duration \(h\), the probability of a transition from \(i\) to \(j\), for \(j\ne i\), is \(q_{ij}h+o(h)\). Here \(o(h)\) denotes a term that becomes negligible relative to \(h\) as \(h\) approaches zero. The probability of leaving \(i\) for any other state is therefore the sum of these exit probabilities:
The probability of remaining in \(i\) is consequently \(1-\left(\sum_{j\ne i}q_{ij}\right)h+o(h)\). By the definition of the diagonal intensity, that same probability is \(1+q_{ii}h+o(h)\). Equating the coefficients of \(h\) gives the required diagonal:
This argument applies within each constant-rate interval. When the schedule changes, the outgoing rates for the new interval determine a new diagonal. The formula does not turn an intensity into a finite-interval probability; it establishes the diagonal entry needed for a valid intensity matrix.
A Practical Construction and Checking Routine
When a problem gives rates that change at one or more calendar times, use the following sequence for each interval:
Write the boundaries in order and specify which interval includes each boundary, such as \([0,2)\) and \([2,\infty)\).
Use the same order for the rows and columns of every matrix.
Place each stated transition intensity in the row for its origin state and column for its destination state; place zero for a forbidden transition.
For each row, add its outgoing rates, negate the total, and verify that the row sums to zero.
Keep the units visible while constructing the schedule. If time is measured in years, a rate stated per month must be converted before it is combined with rates per year. Also check that the schedule defines a rate for every time in the projection; an omitted interval leaves the model unspecified there.
Common Errors and Why They Matter
- Using a duration in state instead of calendar time. A boundary at \(t=2\) applies at model time 2, not two years after each person enters a state.
- Keeping an old diagonal after changing off-diagonal rates. Recalculate the diagonal from that interval’s outgoing rates so the row still sums to zero.
- Calling a transition intensity a finite-interval probability. An intensity is an instantaneous rate. It is not, by itself, the probability of transitioning during an interval of a specified length.
- Changing the state order between matrices. A numerical entry can then appear in the wrong row or column even when the individual rates were read correctly.
- Omitting absorbing-state checks. An absorbing state has no exits, so its entire row, including its diagonal, is zero in every interval in which it remains absorbing.
A piecewise-constant schedule is useful whenever a model’s assumptions explicitly change at known projection times. Specifying it interval by interval makes both the change in assumptions and the direction of each possible transition transparent. The most important discipline is to read the calendar interval first and then read the applicable row of the matrix.
Check Your Understanding
For each problem, specify or interpret the rates using the calendar-time intervals given. Intensities are per year unless stated otherwise.
- A two-state model has states Working and Dead, with Dead absorbing. The Working-to-Dead intensity is \(0.025\) for \(0\le t<3\) and \(0.04\) for \(t\ge3\). Write both matrices. (Answer: \(\begin{pmatrix}-0.025&0.025\\0&0\end{pmatrix}\) for \(0\le t<3\); \(\begin{pmatrix}-0.04&0.04\\0&0\end{pmatrix}\) for \(t\ge3\), in the order Working, Dead.)
- A model has states \(A,B,D\), with \(D\) absorbing. For \(0\le t<1\), \(A\to B=0.02\), \(A\to D=0.03\), \(B\to A=0.04\), and \(B\to D=0.06\). Find the diagonals and verify the row sums. (Answer: \(q_{AA}=-0.05\), \(q_{BB}=-0.10\), \(q_{DD}=0\); all three row sums are zero.)
- In a model with boundary \(t=4\), a person enters state \(B\) at \(t=3\). The intensity \(B\to C\) is \(0.07\) before \(t=4\) and \(0.02\) from \(t=4\) onward. Which intensity applies at \(t=3.5\) and at \(t=4.2\)? (Answer: \(0.07\) per year at \(t=3.5\), and \(0.02\) per year at \(t=4.2\); the person’s state-entry time does not move the calendar boundary.)
- A state \(R\) has two possible exits with intensities \(0.015\) and \(0.025\) before \(t=2\), and \(0.03\) and \(0.01\) from \(t=2\) onward. Find its diagonal in each interval and check the total exit intensity. (Answer: \(-0.04\) before \(t=2\), with total exit intensity \(0.04\); \(-0.04\) from \(t=2\) onward, also with total exit intensity \(0.04\).)
- A proposed row for state \(M\) in one interval is \((-0.09,0.04,0.03)\), with the diagonal listed first. Is it valid as an intensity-matrix row? If not, correct it. (Answer: No. The off-diagonal entries sum to \(0.07\), so the diagonal must be \(-0.07\); the corrected row is \((-0.07,0.04,0.03)\).)