Why a Differential Equation Describes Remaining Probability
In a competing-decrement model, members of a group may leave through several distinct causes, such as death, withdrawal, or transition to another status. The probability of still being in the group changes over time because members who remain are exposed to each of those possible exits. The differential equation expresses this relationship locally: the rate at which the remaining probability falls depends both on how many are still at risk and on the combined intensity of their possible decrements.
The equation is useful because it links the cause-specific decrement assumptions to the probability of remaining in the group. Its initial condition matters just as much as its rate of change: for a cohort defined at entry, everyone is initially in the group, so the probability of remaining at duration zero is one.
The model concerns first exits from the group: once a decrement occurs, that member is no longer counted as remaining. Decrements are mutually exclusive causes of that first exit. The intensities may vary with duration. The differential equation below applies where the relevant probabilities and intensities have the regularity needed for the derivative to exist. It does not describe a model with re-entry unless that model is reformulated to account for return transitions.
Deriving the Survival Differential Equation
Over a short interval \([t,t+h)\), a member can remain in the group or leave by one of the competing causes. Conditional on being in the group at duration \(t\), the probability of leaving by cause \(j\) is \(\lambda_j(t)h+o(h)\). Because the causes are mutually exclusive first exits, their probabilities add to first order. The conditional probability of any decrement in the interval is therefore \(\Lambda(t)h+o(h)\).
To remain in the group until \(t+h\), the member must first have remained until \(t\), and then must avoid every decrement in the short interval. Multiplying these probabilities gives the local update:
Subtract \(R(t)\) from each side, divide by \(h\), and take the limit as \(h\) tends to zero. This yields the differential equation:
The negative sign reflects that decrements reduce the probability of still being in the group. The factor \(R(t)\) is essential: the total intensity is conditional on being in the group at time \(t\), so it applies only to the probability mass that has survived up to that time.
The Initial Condition
If the cohort is defined at the moment of entry and \(R(t)\) is the probability that a member of that cohort has had no decrement by duration \(t\), then every member is in the group at duration zero. Thus the initial condition is \(R(0)=1\). Together, the rate equation and this initial condition specify the survival problem.
The condition is about the choice of cohort and time origin, not about the value of any decrement intensity. If the question instead describes a population already observed at some later duration, the remaining probability should be defined conditionally from that starting point; it should not be assigned the entry-cohort initial condition without checking the definition. In the standard entry-cohort formulation used here, \(R(0)=1\).
List each mutually exclusive first-decrement cause and its intensity at duration \(t\).
Add the cause-specific intensities: \(\Lambda(t)=\sum_j\lambda_j(t)\).
Multiply the total intensity by the probability currently remaining, and use a negative sign: \(R'(t)=-\Lambda(t)R(t)\).
For a cohort observed from entry, set \(R(0)=1\).
Worked Applications
Worked Example: Local Change With Three Competing Causes
A hypothetical income-protection pool has three first-decrement causes: death, recovery from the covered condition, and withdrawal from the contract. At duration \(t=3\) years, the respective intensities are \(0.012\), \(0.018\), and \(0.005\) per year. Suppose the probability of still being in the pool at that duration is \(R(3)=0.72\). Find the instantaneous rate of change in \(R\) at year 3, then estimate the change over the next \(0.01\) year. State the method and identify the mistake the calculation avoids.
Method. Add the cause-specific intensities to obtain the total intensity, then apply the differential equation at the stated duration:
The instantaneous rate of change is
For \(h=0.01\) year, the first-order approximation to the change is \(R'(3)h\):
This is a local approximation, not an exact probability for the full interval. It uses the current rate of change over a short interval. The corresponding first-order conditional probability of any decrement during that interval is approximately \((0.035)(0.01)=0.00035\); multiplying by \(0.72\) gives the unconditional decrement probability of approximately \(0.000252\) for a member of the original cohort.
Check. The estimated remaining probability is lower than \(0.72\), as expected when the total intensity is positive. The common mistake avoided here is reporting \(-0.035\) as \(R'(3)\). That is the negative total intensity, but the probability is changing at rate \(-\Lambda(3)R(3)=-0.0252\), since only 72% of the original cohort remains at risk.
Worked Example: Applying the Entry Initial Condition
A hypothetical rehabilitation program tracks first exits from its active-care group. At entry, the instantaneous intensities of discharge to independent living, transfer to another facility, and death are \(0.024\), \(0.009\), and \(0.007\) per year, respectively. For a cohort of 8,000 entrants, state the initial remaining probability, find its initial derivative, and interpret the derivative in terms of the cohort. Do not project the cohort over a finite year.
Method. The probability is defined for the entry cohort, so at duration zero all members are in the active-care group:
The total intensity at entry is
The differential equation then gives
For the cohort of 8,000 entrants, the initial rate of change in the expected number remaining is \(8{,}000R'(0)\):
This is a rate at the start of observation, not a statement that exactly 320 members leave during the first year. Intensities and the number remaining may change over time, so a finite-period projection requires more than the initial derivative.
Check. The initial probability is one because the cohort consists of entrants observed at entry. The initial derivative is negative, and its cohort interpretation is a local expected-count rate. The common mistake avoided is using \(R(0)=8{,}000\); \(R\) is a probability, while 8,000 is the cohort size.
Worked Example: Checking a Proposed Remaining-Probability Function
For a hypothetical group, two cause-specific intensities at duration \(t\) years are \(\lambda_1(t)=0.010+0.005t\) and \(\lambda_2(t)=0.015\), with rates measured per year. An analyst proposes \(R(t)=1-0.020t-0.005t^2\) for \(0\le t\le2\). Check the initial condition and determine whether the proposed function satisfies the survival differential equation at \(t=1\). This is a local model check, not a calculation of a survival function from the intensities.
Method. First check the entry condition:
Thus the proposed function has the correct initial value. Next, add the intensities and evaluate them at year 1:
The proposed remaining probability at year 1 is
Differentiate the proposed function:
But the right-hand side required by the model at year 1 is
Since \(-0.030\ne-0.02925\), the proposed function does not satisfy the differential equation at year 1. Having the right initial value is necessary, but it is not sufficient: the function must also have the derivative required by the total intensity at each duration. The common mistake avoided is checking only \(R(0)=1\) and treating that as proof that a proposed function is valid.
Interpreting the Equation and Avoiding Traps
The differential equation is a local statement. At any duration \(t\), the cohort probability \(R(t)\) is exposed to the total intensity \(\Lambda(t)\) only through the members who have remained until \(t\). This explains the product \(R(t)\Lambda(t)\). It also explains why cause-specific intensities are added before the equation is applied: in a short interval, any of the mutually exclusive first-decrement causes removes a member from the same remaining group.
| Quantity | Meaning | Role in the equation |
|---|---|---|
| \(\lambda_j(t)\) | Instantaneous intensity for cause \(j\), conditional on remaining at \(t\) | One component of the total intensity |
| \(\Lambda(t)\) | Sum of the cause-specific intensities | Combined instantaneous exit rate among those still in the group |
| \(R(t)\) | Probability that an entry-cohort member has had no decrement by \(t\) | Scales the rate to the original cohort |
| \(R'(t)\) | Instantaneous rate of change in the remaining probability | Equals \(-\Lambda(t)R(t)\) |
An intensity has units of inverse time; it is not itself a probability over a year or any other fixed interval. For a sufficiently short interval of length \(h\), the conditional probability of a decrement is approximately \(\Lambda(t)h\), with a term of smaller order than \(h\) omitted. A finite-interval probability cannot generally be obtained by simply treating \(\Lambda(t)\) as that interval’s decrement probability.
- Do not omit \(R(t)\). The total intensity is conditional on still being in the group, so the derivative of the original cohort’s remaining probability includes the factor \(R(t)\).
- Do not use a positive sign. A decrement removes members from the group, so the remaining probability has a nonpositive derivative when intensities are nonnegative.
- Do not confuse a rate with a probability. The intensity has units per unit time; multiplying by a short time interval gives a first-order probability approximation.
- Do not change the time origin silently. \(R(0)=1\) is for a cohort observed from entry, with \(R\) defined as its remaining probability.
The equation is a foundation for further competing-decrement calculations: it states how the remaining probability must change given the cause-specific intensities. For this result, the essential tasks are to define the cohort and time origin clearly, add the intensities, and use the remaining probability as the multiplier. These checks are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use the competing first-decrement model and distinguish an instantaneous rate from a finite-interval probability.
- A hypothetical membership group has two cause-specific intensities of \(0.014\) and \(0.006\) per year at duration \(t=2\), when \(R(2)=0.80\). Find \(R'(2)\). (Answer: \(-0.016\) per year)
- For an entry cohort, the cause-specific intensities at duration zero are \(0.008\), \(0.017\), and \(0.005\) per year. State \(R(0)\) and calculate \(R'(0)\). (Answer: \(R(0)=1\), \(R'(0)=-0.030\) per year)
- At duration \(t=4\), a hypothetical group has three cause-specific intensities \(0.004\), \(0.011\), and \(0.005\) per year, and \(R(4)=0.65\). Find the total intensity and the instantaneous rate of change in remaining probability. (Answer: \(\Lambda(4)=0.020\) per year; \(R'(4)=-0.013\) per year)
- A hypothetical cohort has total intensity \(0.025\) per year at duration \(t=1\) and \(R(1)=0.90\). Using a first-order approximation, estimate \(R(1.02)\). (Answer: \(R'(1)=-0.0225\) per year and \(R(1.02)\approx0.89955\))
- A proposed function for an entry cohort has \(R(0)=0.98\), while its cause-specific intensities are nonnegative. Can it be the remaining probability for the entry cohort under the stated definition? Explain. (Answer: No. The entry-cohort initial condition is \(R(0)=1\), so the proposed function fails it.)