From a Rate Description to an Intensity Matrix
A transition intensity need not remain constant over time. A model might specify that a transition becomes more frequent as calendar time passes, or that its rate varies seasonally. To use such a description, first express each permitted transition rate as a function of time. Then place those functions in the appropriate off-diagonal entries of the intensity matrix and calculate each diagonal entry from the rates exiting that state.
Throughout this tutorial, \(t\) denotes calendar time measured from a stated origin, usually in years. It does not denote time since entering the current state. The model describes a time-inhomogeneous Markov process: conditional on the current state and calendar time, the transition rates are specified by the model. The time dependence means that the intensity matrix can change as \(t\) changes.
The intensity is a rate, not a probability. For a short interval of length \(h\), the probability of a transition from \(i\) to \(j\) is approximately \(q_{ij}(t)h\), with an error that becomes negligible relative to \(h\) as \(h\) approaches zero. A rate function alone does not give the exact probability of transition over a long interval.
How to Translate a Stated Rate Pattern
Before writing a matrix, identify four things: the states, the allowed transition routes, the time variable and its units, and the units and values of each stated rate. A rate described as increasing linearly must have both a starting value and a slope. A seasonal pattern needs a periodic function and a check that the resulting rate never becomes negative. Only after these pieces are clear should the functions be placed in the matrix.
State what \(t\) measures and over what range the model applies. Keep the units consistent when evaluating rates.
For each transition \(i\to j\), translate the verbal description into \(q_{ij}(t)\), including the rate’s units.
Put each permitted rate off the diagonal. Set each diagonal entry equal to the negative sum of the other entries in that row.
Confirm that rates are nonnegative on the stated time range, forbidden transitions are zero, and every row sums to zero.
Linear Intensities
Suppose a transition intensity begins at \(a\) when \(t=0\) and increases by \(b\) intensity units for each unit increase in time. Its linear form is \(a+bt\). If \(t\) is measured in years, then \(a\) is measured per year and \(b\) is measured per year per year. The model must be restricted to a range of \(t\) on which the intensity is nonnegative.
Worked Example: A Linearly Increasing Exit Rate
A hypothetical benefit model has states Active (A), Disabled (D), and Dead (X). Time \(t\) is measured in years after policy issue, and the model applies for \(0\le t\le5\). The intensity from A to D starts at 0.02 per year and increases by 0.01 per year for each year of calendar time. The intensity from A to X is constantly 0.03 per year. From D, the only possible transition is to X, at a constant intensity of 0.04 per year. State X is absorbing. Construct \(Q(t)\) and evaluate it at \(t=3\).
Method. Translate each stated rate into its function, then form each diagonal as the negative total exit intensity. The A-to-D intensity is \(q_{AD}(t)=0.02+0.01t\), while \(q_{AX}(t)=0.03\). The D-to-X intensity is \(q_{DX}(t)=0.04\).
The total exit intensity from A is \((0.02+0.01t)+0.03=0.05+0.01t\). The total exit intensity from D is 0.04. Since X has no exits, its row is all zeros. Using the state order A, D, X gives
At \(t=3\), the A-to-D rate is \(0.02+0.01(3)=0.05\), and the total A exit rate is \(0.05+0.01(3)=0.08\). Therefore
Check. The first row sums to \(-0.08+0.05+0.03=0\); the second sums to \(-0.04+0.04=0\); and the absorbing row sums to zero. The increasing rate ranges from 0.02 to 0.07 per year on the stated interval, so it remains nonnegative. This avoids the mistake of putting the positive total exit rate, rather than its negative, on the diagonal.
Seasonal Intensities
A periodic function can represent a rate that rises and falls over a repeating cycle. For example, a cosine term varies between \(-1\) and \(1\). In a function of the form \(c[1+a\cos(\omega t)]\), with \(c>0\), the rate is nonnegative for all \(t\) if \(0\le a\le1\). The period is \(2\pi/\omega\), when \(t\) and \(\omega\) use compatible units. The constant \(c\) sets the central level, while \(a\) controls the size of the fluctuation relative to that level.
Worked Example: A Seasonal Transition Rate
Consider a hypothetical model with states Active (A), Disabled (D), and Dead (X). Time \(t\) is measured in years. The intensity from A to D varies with a two-year cycle and is \(0.04[1+0.5\cos(\pi t)]\) per year. The intensity from A to X is 0.01 per year. From D, recovery to A occurs at 0.20 per year and transition to X at 0.06 per year. State X is absorbing. Write the matrix and evaluate it at \(t=1\).
Method. Use the stated seasonal function for the A-to-D transition and keep the other stated intensities constant. Since \(-1\le\cos(\pi t)\le1\), the seasonal intensity lies between \(0.04(1-0.5)=0.02\) and \(0.04(1+0.5)=0.06\) per year. It is therefore nonnegative for every \(t\).
Let \(\lambda(t)=0.04[1+0.5\cos(\pi t)]\). The total exit intensity from A is \(\lambda(t)+0.01\), and the total exit intensity from D is \(0.20+0.06=0.26\). With state order A, D, X,
At \(t=1\), \(\cos(\pi)=-1\), so \(\lambda(1)=0.04[1+0.5(-1)]=0.04(0.5)=0.02\) per year. Thus
Check. The active row sums to \(-0.03+0.02+0.01=0\), and the disabled row sums to \(0.20-0.26+0.06=0\). A common mistake is to treat the seasonal intensity as \(0.04\) at every time because 0.04 appears as its leading constant. That value is the central level, not the value at every point in the cycle.
Why the Diagonal Is the Negative Exit Rate
The diagonal rule remains necessary when rates depend on time. At each fixed time, the possible immediate outcomes from a state must account for all probability over a very short interval. The probability of remaining in the state is the amount left after the probabilities of leaving by each permitted route are taken into account. This gives the negative sum on the diagonal.
Here \(o(h)\) denotes a remainder whose ratio to \(h\) approaches zero as \(h\) approaches zero. Summing the probabilities of all possible states at time \(t+h\) must give one. For a finite-state model, summing these short-interval expressions for row \(i\) yields
The coefficient of \(h\) must therefore be zero, giving \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\). This argument applies separately at each time \(t\); the rates in the row are evaluated at that same time.
Keep Rate Units Consistent
A stated rate may be expressed per month while the model’s calendar-time variable is measured in years. These units must be reconciled before building the matrix. When a monthly instantaneous rate is given as a function of calendar time in years, multiply the rate by 12 to express the intensity per year. Do not multiply the time variable by 12 as well unless the stated function itself uses time measured in months.
Worked Example: Converting Monthly Rates to Annual Intensities
A hypothetical three-state model has states Active (A), Disabled (D), and Dead (X). Calendar time \(t\) is measured in years, with \(0\le t\le4\). The A-to-D intensity is reported as \(0.006+0.001t\) per month, where the formula’s \(t\) is in years. The A-to-X rate is 0.0015 per month. From D, the rate to A is 0.025 per month and the rate to X is 0.004 per month. Convert all intensities to annual units, write the matrix, and evaluate it at \(t=2\).
Method. Convert each monthly intensity to a per-year intensity by multiplying by 12. The A-to-D rate becomes \(12(0.006+0.001t)=0.072+0.012t\) per year. The A-to-X, D-to-A, and D-to-X rates become \(12(0.0015)=0.018\), \(12(0.025)=0.300\), and \(12(0.004)=0.048\) per year, respectively.
The total exit intensity from A is \(0.072+0.012t+0.018=0.090+0.012t\). The total exit intensity from D is \(0.300+0.048=0.348\). Therefore, in state order A, D, X,
At \(t=2\), the A-to-D intensity is \(0.072+0.012(2)=0.096\) per year, and the A exit total is \(0.090+0.012(2)=0.114\) per year. Hence
Check. The first row sums to \(-0.114+0.096+0.018=0\); the second sums to \(0.300-0.348+0.048=0\). A common unit error is to leave 0.096 in the matrix as a monthly intensity while treating the other entries as annual. The matrix would then mix units and would not represent a coherent set of rates.
What the Matrix Does—and Does Not—Say
The matrix \(Q(t)\) records the instantaneous transition rates at time \(t\). Its off-diagonal entries identify which transitions are possible and how their instantaneous rates vary; its diagonal entries summarize the total rate of leaving each state. It is not a matrix of finite-interval transition probabilities. In particular, an entry such as \(q_{AD}(t)=0.05\) per year is not itself a 5% probability of moving from A to D during a year.
A valid time-dependent specification has nonnegative off-diagonal entries over the interval where the model is used, zeros for forbidden transitions, and rows summing to zero. These checks catch many construction errors. They do not, by themselves, determine a probability of being in a state at a later time; that requires the appropriate transition-probability calculations.
Check Your Understanding
For each problem, write the requested rate or matrix and check the sign and row-sum conditions.
- A model has states 0, 1, and 2. The only exits from 0 have intensities \(q_{01}(t)=0.015+0.004t\) and \(q_{02}(t)=0.025\), with \(t\) in years. What is the total exit intensity from 0 at \(t=2\), and what is its diagonal matrix entry? (Answer: total exit intensity \(0.048\) per year; \(q_{00}(2)=-0.048\) per year.)
- A transition intensity is \(\lambda(t)=0.03[1+0.2\sin(\pi t)]\) per year. What is its range, and what is its value at \(t=0.5\)? (Answer: range \(0.024\) to \(0.036\) per year; \(\lambda(0.5)=0.036\) per year.)
- From state A, rates to B and C are 0.010 and 0.006 per month. Time is to be represented in years. What are the annual off-diagonal rates and the A diagonal entry? (Answer: \(q_{AB}=0.120\), \(q_{AC}=0.072\), and \(q_{AA}=-0.192\), all per year.)
- In state 0, \(q_{01}(t)=0.05-0.01t\) and \(q_{02}(t)=0.02\), in units per year, for \(0\le t\le4\). What is the total exit intensity and diagonal entry at \(t=3\), and does the first rate remain nonnegative throughout the model interval? (Answer: total \(0.04\) per year and diagonal \(-0.04\) per year at \(t=3\); yes, \(q_{01}(t)\) decreases to \(0.01\) per year at \(t=4\).)
- A three-state model has \(q_{01}(t)=0.02[1+0.5\cos(\pi t)]\), \(q_{02}=0.01\), and \(q_{12}=0.03\), with state 2 absorbing and \(t\) in years. Give the first row of \(Q(1)\) and check its sum. (Answer: \((-0.02,\ 0.01,\ 0.01)\); its row sum is zero.)