Introduction
A continuous-time state model describes how an insured, policy, or other modeled unit moves among a set of states. Its transition intensities may be fixed over time, or they may change as time passes. This tutorial focuses on the fixed-rate case: a time-homogeneous intensity matrix. Recognizing this structure makes the model easier to specify and helps prevent errors when rates are recorded in different time units.
The key point is that time homogeneity concerns the rates attached to transitions, not the probabilities of being in each state. Under a time-homogeneous model, a given transition has the same intensity at every time while the process occupies its starting state. The resulting intensity matrix is therefore constant. Its entries still need to respect the chosen state order and the row-sum convention.
What “Constant” Means in the Model
An intensity is a rate, not a finite-interval probability. In the convention used here, conditional on being in state \(i\) at time \(t\), the probability of a transition to a distinct state \(j\) in a short interval of length \(h\) is \(q_{ij}h+o(h)\) as \(h\) approaches zero. The time unit for \(q_{ij}\) must match the unit for \(h\). For example, a rate per year is multiplied by an interval measured in years.
Time-homogeneous does not mean that all states have the same transition rates. It means each particular transition has a rate that does not vary with time. Rates from one state can differ from rates out of another state. Transitions that the model forbids have zero direct intensity, and an absorbing state has no outgoing transitions.
To construct the matrix, first fix the state order, then enter each specified rate in its off-diagonal position. Use zero for a forbidden direct transition. Finally, set each diagonal entry to the negative sum of the other entries in its row. This final step ensures that the rate of leaving a state is the sum of its competing outgoing intensities.
For example, decide whether rows and columns are ordered as Active, Disabled, Dead before filling any entries.
Express every rate using the same unit, such as per year, before placing it in the matrix.
Place each fixed transition rate at its origin-row and destination-column intersection; enter zero for a forbidden direct transition.
For each row, make its diagonal the negative sum of the off-diagonal rates. Verify that every row sums to zero.
Worked Examples
Worked Example: Constructing a Constant Matrix for Three States
Hypothetical model. A policyholder is classified as Active (A), Disabled (D), or Dead (M). Assume a time-homogeneous continuous-time Markov model, with rates measured per year. The fixed rates are A to D, \(0.04\); A to M, \(0.01\); D to A, \(0.20\); and D to M, \(0.08\). Death is absorbing. Construct the intensity matrix in the order A, D, M, and check the exits from each state.
Method. Place each stated rate in the row for its starting state and the column for its destination. The direct transition A to M is allowed and has rate \(0.01\); no other direct transitions are specified, so they receive zero.
For the Active row, the total outgoing intensity is \(0.04+0.01=0.05\) per year, giving diagonal entry \(-0.05\). For the Disabled row, the total is \(0.20+0.08=0.28\) per year, giving \(-0.28\). Since Dead is absorbing, all off-diagonal entries in its row are zero and its diagonal is zero.
Check. The row sums are \(-0.05+0.04+0.01=0\), \(0.20-0.28+0.08=0\), and \(0+0+0=0\). The total exit intensities are \(0.05\) from Active, \(0.28\) from Disabled, and zero from Dead. The matrix is constant because the assumed rates do not vary with time. A common mistake is to put \(+0.05\) on the Active diagonal; the diagonal must be the negative total exit intensity.
Worked Example: Converting Monthly Rates Before Building the Matrix
Hypothetical model. A service contract has states Covered (C), Under Review (R), and Terminated (T). Assume a time-homogeneous Markov model with rates initially specified per month. The fixed rates are C to R, \(0.006\) per month; C to T, \(0.001\) per month; R to C, \(0.010\) per month; and R to T, \(0.004\) per month. Terminated is absorbing. Construct the matrix using rates per year.
Method. There are 12 months in a year, so multiply each monthly rate by 12. Thus C to R is \(12(0.006)=0.072\) per year, C to T is \(12(0.001)=0.012\) per year, R to C is \(12(0.010)=0.120\) per year, and R to T is \(12(0.004)=0.048\) per year.
The C exit intensity is \(0.072+0.012=0.084\) per year, so the C diagonal is \(-0.084\). The R exit intensity is \(0.120+0.048=0.168\) per year, so the R diagonal is \(-0.168\). The T row is zero because T is absorbing.
Check. The row sums are \(-0.084+0.072+0.012=0\), \(-0.168+0.120+0.048=0\), and \(0\). The units are consistent throughout. A common mistake is to multiply the matrix by 12 after its diagonal has already been formed without explaining the unit change. Converting every off-diagonal rate first and then recomputing the diagonals makes the units and row sums transparent.
Worked Example: A Forbidden Direct Transition Is Not a Zero End-State Probability
Hypothetical model. A system has three states, A, B, and C, and constant rates measured per year. The direct transition A to C is forbidden. The only route from A to C is through B; the rates are A to B, \(0.020\) per year, and B to C, \(0.030\) per year. Assume no other route into C is relevant to the leading-order calculation. For \(h=0.001\) year, identify the direct A-to-C intensity and first-order probability term, then estimate the leading contribution to the probability of being in C at the end of the interval, given that the system starts in A.
Method. Because the direct transition is forbidden, \(q_{AC}=0\), and the direct, first-order term for an A-to-C transition is \(q_{AC}h=0\). But ending in C is possible after two transitions, A to B and then B to C, within the interval. It is therefore incorrect to conclude that the probability of being in C at the end is exactly zero.
To obtain the leading contribution from the two-step route, suppose the first transition occurs at time \(s\), where \(0\le s\le h\). To leading order, the contribution from a first transition in an interval of length \(ds\) around \(s\) is \(q_{AB}ds\). The remaining time for the B-to-C transition is \(h-s\), so its first-order contribution is \(q_{BC}(h-s)\). Integrating these contributions gives
Substitute the given values: \(q_{AB}q_{BC}=(0.020)(0.030)=0.0006\) per year squared, and \(h^2=(0.001)^2=0.000001\) year squared. Therefore the leading contribution is \(0.0006(0.000001)/2=0.0000000003\), a dimensionless probability contribution.
Check and trap. The direct intensity and its first-order term are zero, as required. The two-step contribution is positive and of order \(h^2\), not \(h\). The common incorrect calculation says “A to C is forbidden, so the probability of being in C after \(h\) is zero.” The correction is to distinguish no direct transition from no possible route to the destination during the interval. Higher-order paths may contribute additional terms; the computed value is the leading contribution from the specified two-step route.
Why the Matrix Is Constant: A Short Justification
The definition of time homogeneity applies separately to each off-diagonal transition. For a short interval of length \(h\), the intensity definition gives a first-order transition probability \(q_{ij}h\), conditional on being in state \(i\). If the model is time-homogeneous, \(q_{ij}\) has the same value at every starting time. Therefore the off-diagonal entries used to describe the model do not need to be recalculated as calendar time changes.
The diagonal follows from the same short-interval reasoning. Starting in state \(i\), the first-order probability of leaving for any distinct state is the sum of the competing first-order terms:
The first-order probability of remaining in state \(i\) is consequently \(1-h\sum_{j\ne i}q_{ij}+o(h)\). The coefficient of \(h\) for remaining in the same state is the diagonal intensity, so it is \(-\sum_{j\ne i}q_{ij}\). This justifies the zero row sum and shows why changing the diagonal independently of its outgoing rates would not describe the stated model.
Checks and Exam Traps
When asked to identify or construct a time-homogeneous intensity matrix, state the assumption clearly: the specified transition rates are constant over time, and the model uses a fixed state set with the Markov property. Then check the units, state order, and every row. The following distinctions are especially important:
- Constant intensity is not a constant transition probability. The rate is fixed, but a finite-interval probability is not simply the rate itself and may include paths involving multiple transitions.
- Zero direct rate does not rule out an indirect route. A forbidden direct transition has no first-order term, but a sequence of allowed transitions can produce a positive end-of-interval probability of higher order.
- Rates require time units. A rate per month cannot be combined directly with a rate per year in one matrix unless they are converted to a common unit.
- The diagonal is not an independent rate to choose. It is the negative sum of the other entries in its row, including zero for an absorbing state's diagonal.
- Homogeneity applies transition by transition. It does not require the rates from different states, or for different destinations, to be equal.
These checks matter because the matrix is the compact statement of the model's transition structure and rates. This skill is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus: a correctly constructed constant matrix makes the assumptions explicit and avoids confusing rates with probabilities.
Check Your Understanding
For each question, use the stated state order and time units. Answers are provided at the end of each item.
- A two-state hypothetical model has states Open and Closed. The only transition is Open to Closed at a fixed rate of \(0.07\) per year, and Closed is absorbing. Write the intensity matrix in the order Open, Closed. (Answer: \(\begin{pmatrix}-0.07&0.07\\0&0\end{pmatrix}\) per year.)
- A hypothetical model has states A, B, and C, in that order. Its fixed rates per year are A to B, \(0.03\), B to A, \(0.02\), and B to C, \(0.05\); C is absorbing. What is the total exit intensity from B, and what is the B diagonal entry? (Answer: \(0.07\) per year; \(-0.07\) per year.)
- A contract model has a fixed transition rate of \(0.004\) per month from state X to state Y and no other exit from X. Express the X-to-Y intensity and X diagonal using rates per year. (Answer: \(0.048\) and \(-0.048\) per year.)
- In a hypothetical model, A to C is forbidden, while A to B and B to C have constant rates \(0.04\) and \(0.05\) per year. For \(h=0.002\) year, find the direct A-to-C first-order term and the leading two-step contribution to ending in C, given a start in A. (Answer: direct first-order term \(0\); leading two-step contribution \((0.04)(0.05)(0.002)^2/2=0.000000004\).)
- A proposed row of an intensity matrix has off-diagonal entries \(0.015\), \(0\), and \(0.025\) per year. Find the diagonal entry and explain what a zero in the middle off-diagonal position means. (Answer: the diagonal is \(-0.040\) per year; the zero means there is no direct transition to that destination, not necessarily that the destination cannot be reached indirectly during a finite interval.)