Why Check Intensities Over the Whole Time Interval?
A transition intensity can vary with calendar time, age, or another time variable. Checking its value at just one instant is not enough: a function that is positive at the start of a model may become negative later, or it may be assigned to a transition that the model rules out. A time-dependent intensity must be checked over the entire period for which the model uses it.
The checks are local to the rate specification. They do not calculate transition probabilities. Instead, they determine whether the proposed rates are suitable inputs to a continuous-time multiple-state model. This tutorial is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
The time domain is part of the specification. For example, “\(0\le t\le 5\), with \(t\) in years” is materially different from “all \(t\ge 0\).” A function may be valid over the first domain but invalid over the second. For ordinary finite-state models, intensities should also be finite and integrable over finite modeled intervals; this ensures that the rates are usable across those intervals.
A Systematic Validity Check
For each state and each proposed outgoing intensity, first identify the time variable, its unit, and its modeled range. Then test the function over that range, not merely at a convenient sample time. For a continuous function on a closed finite interval, endpoint values and interior minima are especially useful: a negative minimum means the specification fails.
After checking the numerical functions, check the state structure. A nonnegative function is still invalid if it represents a transition that is forbidden by the model. Finally, verify the diagonal entry. It is not an independent rate to choose freely; it balances the outgoing rates in its row.
These checks apply at each modeled time. A rate may equal zero at an isolated time or at an interval endpoint without violating nonnegativity. A negative value at even one modeled time is not acceptable. For piecewise-defined functions, test each piece on its own interval and check the specified values at any boundaries.
Why the Diagonal Must Balance the Row
The diagonal-entry condition follows from the fact that the state probabilities over a short interval must sum to one. Given state \(i\) at time \(t\), the probability of moving to each different state \(j\) is \(q_{ij}(t)h+o(h)\). Adding those possible exits gives the probability of leaving \(i\) to first order in \(h\). Therefore the probability of remaining in \(i\) is one minus that total, to the same order.
By definition, the coefficient of \(h\) in the probability of remaining is the diagonal intensity. It is consequently the negative total outgoing intensity. This also explains why a diagonal entry is nonpositive, and why a row with no outgoing transitions has diagonal entry zero.
Worked Checks
Worked Example: Checking a Candidate Row and a Forbidden Route
A hypothetical three-state model has states \(A,B,C\), with time \(t\) measured in years over \(0\le t\le4\). The only permitted transition out of \(A\) is \(A\to B\). A proposed row specifies \(q_{AB}(t)=0.03+0.01t\), \(q_{AC}(t)=0.02-0.005t\), and \(q_{AA}(t)=-(0.05+0.005t)\), all in rates per year. Test the row and give the corrected entries.
Method. Check the range of each off-diagonal function, test whether its transition is permitted, and then check the row sum. The common mistake to avoid is treating nonnegativity and a zero row sum as sufficient for validity, without checking the model’s permitted routes.
For \(q_{AB}\), the function increases over the interval. Its endpoint values are \(q_{AB}(0)=0.03\) and \(q_{AB}(4)=0.03+0.04=0.07\). Thus it is nonnegative throughout \([0,4]\).
For \(q_{AC}\), the function decreases over the interval. Its endpoint values are \(q_{AC}(0)=0.02\) and \(q_{AC}(4)=0.02-0.02=0\). Thus it is also nonnegative throughout the interval. But \(A\to C\) is forbidden, so its intensity must be zero at every modeled time. Its nonnegative values do not make this transition permissible.
The proposed row sums to zero because \(0.03+0.01t+0.02-0.005t-(0.05+0.005t)=0\). Nevertheless, it is structurally invalid. Set \(q_{AC}(t)=0\). The only outgoing rate from \(A\) is then \(q_{AB}(t)\), so the diagonal must be its negative:
The corrected entries satisfy \(q_{AB}(t)\ge0\), \(q_{AC}(t)=0\), and \(q_{AA}(t)+q_{AB}(t)+q_{AC}(t)=0\) for every \(t\in[0,4]\). The correction retains the permitted transition and removes the forbidden one.
Worked Example: Finding Interior Minima Before Accepting Rates
In a hypothetical model, the permitted outgoing transitions from state \(X\) are \(X\to Y\) and \(X\to Z\). Time \(t\) is in years, with \(0\le t\le5\). The proposed intensities, in rates per year, are \(q_{XY}(t)=0.04+0.01(t-3)^2\) and \(q_{XZ}(t)=0.06-0.02t+0.003t^2\). Check both functions over the full interval and determine the diagonal entry.
Method. Use the minimum of each function on the interval to test nonnegativity, then sum the outgoing rates to obtain the diagonal. The common mistake to avoid is checking endpoints only when a function can have an interior minimum.
The first function is a positive constant plus a nonnegative square. Its minimum occurs at \(t=3\), where \(q_{XY}(3)=0.04\). Its endpoint values are \(q_{XY}(0)=0.04+0.09=0.13\) and \(q_{XY}(5)=0.04+0.04=0.08\). Therefore it is nonnegative throughout.
For the second function, differentiate to locate any interior turning point: \(q'_{XZ}(t)=-0.02+0.006t\). Setting this equal to zero gives \(t=0.02/0.006=10/3\), which lies inside the interval. Since the coefficient of \(t^2\) is positive, this turning point is a minimum. Its value is \(0.06-0.02(10/3)+0.003(100/9)=0.06-0.066666\ldots+0.033333\ldots=0.026666\ldots\), or \(2/75\). This is positive. The endpoint values, \(0.06\) at \(t=0\) and \(0.035\) at \(t=5\), are also positive.
Both transitions are permitted and both functions are nonnegative throughout the modeled interval. Adding the rates gives \(q_{XY}(t)+q_{XZ}(t)=0.19-0.08t+0.013t^2\). Hence the required diagonal is
The row sum is zero by construction, and the diagonal is negative because it is the negative of two positive outgoing intensities. The check is complete for this row.
Worked Example: Converting a Monthly Intensity to Annual Time
A hypothetical model gives the permitted transition intensity from state \(E\) to state \(F\) as \(\lambda(s)=0.002+0.0001s\) per month, where \(s\) is time in months and \(0\le s\le24\). The model is to use time \(t\) in years, over \(0\le t\le2\). Express and check the intensity on the annual time scale.
Method. Convert the time argument and the rate unit together. The common mistake to avoid is changing months to years in the argument but leaving a rate measured per month labelled as a rate per year.
The time variables are related by \(s=12t\). A rate of \(\lambda(s)\) per month must be multiplied by 12 to express the corresponding rate per year. Thus
Over \(0\le t\le2\), this is increasing, with values \(0.024\) per year at \(t=0\) and \(0.0528\) per year at \(t=2\). It is nonnegative throughout, and the transition is permitted. For a model with no transition back from \(F\) to \(E\), the corresponding reverse intensity must be zero. If a row contains no other outgoing route from \(E\), its diagonal is \(q_{EE}(t)=-q_{EF}(t)\).
As a check at \(t=2\), the original monthly rate is \(\lambda(24)=0.002+0.0024=0.0044\) per month. Multiplying by 12 gives \(0.0528\) per year, agreeing with the converted function. Calling \(0.0044\) a rate per year would miss the required conversion.
Common Mistakes and a Practical Checklist
A rate formula should not be accepted just because it looks plausible or gives a sensible value at the model’s starting time. A negative value means it cannot be a transition intensity there. A positive value on a forbidden route is also a model error, even if all row sums are correct.
A frequent domain error is to validate a function only over the period used in a calculation and then apply it beyond that period. For instance, \(0.04-0.003t\) is nonnegative on \(0\le t\le10\), but becomes negative for \(t>0.04/0.003=13.333\ldots\). It is therefore unsuitable if the model uses it for all \(t\ge0\). State the modeled interval before reaching a conclusion.
A reliable review follows this order:
- Write down what the time variable measures, its unit, and the interval on which the model uses the function.
- Check each off-diagonal function over that whole interval, including interior extrema and piecewise boundaries.
- Confirm that every nonzero off-diagonal entry represents a permitted transition.
- Set each diagonal entry to the negative sum of its row’s outgoing rates and verify the row sum is zero.
- Check that every rate and time argument uses consistent units.
These checks matter because a valid numerical function is only one part of a valid model. A row can fail through a negative rate, an incorrectly placed transition, a mismatched diagonal, or a unit error. Identifying which check fails makes the correction precise.
Check Your Understanding
For each problem, state whether the specification passes the requested checks over its stated interval.
- Time \(t\) is in years on \(0\le t\le3\). A permitted transition has \(q_{AB}(t)=0.01+0.004t\) per year, while \(A\to C\) is forbidden. A proposed \(q_{AC}(t)=0.005\) per year is included. Give the minimum of \(q_{AB}\), its value at \(t=3\), and the structural verdict on \(q_{AC}\). (Answer: \(q_{AB,\min}=0.010\) per year; \(q_{AB}(3)=0.022\) per year; the row is invalid because the forbidden \(A\to C\) rate must be zero.)
- On \(0\le t\le4\), with time in years, a proposed permitted intensity is \(q(t)=0.08-0.025t\) per year. Find where it reaches zero and decide whether it is valid over the full interval. (Answer: It reaches zero at \(t=3.2\) years and is negative thereafter, so it is invalid on \([0,4]\); \(q(4)=-0.02\) per year.)
- For \(0\le t\le6\), two permitted rates out of state \(R\) are \(q_{RS}(t)=0.03+0.002(t-4)^2\) and \(q_{RT}(t)=0.05-0.01t+0.001t^2\), in rates per year. Find each minimum and the diagonal at \(t=5\). (Answer: The minima are \(0.030\) at \(t=4\) and \(0.025\) at \(t=5\) per year; \(q_{RR}(5)=-(0.032+0.025)=-0.057\) per year.)
- A rate is given as \(\lambda(s)=0.003+0.0002s\) per month for \(0\le s\le36\), where \(s\) is in months. Express it using years \(t=s/12\), and evaluate it at \(t=3\). (Answer: \(q(t)=12\lambda(12t)=0.036+0.0288t\) per year; \(q(3)=0.1224\) per year.)
- A proposed permitted intensity is \(q(t)=0.04-0.003t\) per year, with \(t\) in years. Decide whether it is valid on \([0,10]\) and on \([0,20]\). (Answer: It is nonnegative on \([0,10]\), where \(q(10)=0.010\); it is invalid on \([0,20]\), since it reaches zero at \(t=13.333\ldots\) and is negative afterward.)