Why Lifetime Probability Calculations Go Wrong
Many errors in lifetime calculations are not difficult algebra errors. They arise from using the wrong complement, dropping a conditioning event, or treating a rate or density as though it were a probability. The reliable approach is to identify the event first, express it with the survival function, and only then substitute a model.
Let \(T\) be a nonnegative continuous lifetime measured from a specified origin, and write \(S_0(t)=P(T>t)\) for its survival function. In the continuous models used here, \(S_0\) is differentiable where needed, the density is \(f_T(t)=-S_0'(t)\), and the force at time \(t\) is \(\mu_t=f_T(t)/S_0(t)\) when \(S_0(t)>0\). Time units must be consistent: a force stated per year must be integrated over time measured in years.
The survival function uses a strict inequality: \(S_0(t)=P(T>t)\). For a continuous lifetime, the probability of death at any single specified time is zero, so changing an interval endpoint from open to closed does not change its probability. The stated endpoint convention is still useful: it prevents confusion about which survival values belong in a calculation and remains precise even when an interval is described verbally.
Start With the Event, Not the Formula
For \(a<b\), the event \(a<T\leq b\) consists of those who survive beyond \(a\), but not beyond \(b\). Subtracting the latter group from the former gives the interval identity. For conditioning on survival to \(c\), the denominator is the probability of the condition, \(S_0(c)\); the numerator is the probability of both the requested event and that condition.
If \(c\leq a<b\), then every lifetime in \((a,b]\) already satisfies \(T>c\). Consequently,
This expression is not generally the same as \(1-S_0(b)/S_0(c)\). That latter expression is the probability of failure by \(b\), given survival to \(c\), and includes failures between \(c\) and \(a\). A useful check is to describe the event in words after calculating it: does the result cover exactly the requested interval, with exactly the requested conditioning?
Worked Example: Interval Probability, Conditional Survival, and Density
Suppose a hypothetical lifetime has constant force \(\mu_t=0.25\) per year for all \(t\geq0\), with \(S_0(0)=1\). Find (1) \(P(2<T\leq5)\), (2) \(P(T>5\mid T>2)\), and (3) the density at time 3. Give probabilities to four decimals and the density to four decimals per year.
Method. Under a constant force \(\mu\), survival is \(S_0(t)=e^{-\mu t}\). This follows from the force-survival relation \(S_0(t)=\exp(-\int_0^t\mu_s\,ds)\); with \(\mu_s=0.25\), the integral is \(0.25t\).
First calculate the needed survival values without rounding: \(S_0(2)=e^{-0.50}=0.60653066\) and \(S_0(5)=e^{-1.25}=0.28650480\). Therefore,
For survival beyond time 5 conditional on survival beyond time 2, divide the later survival probability by the conditioning probability:
The density is \(f_T(t)=\mu_tS_0(t)\). At time 3, \(S_0(3)=e^{-0.75}=0.47236655\), so
Check and trap. The conditional probability of failure between years 2 and 5 is the complement of surviving past year 5 given survival to year 2: \(1-0.47236655=0.52763345\), which is also \((S_0(2)-S_0(5))/S_0(2)\). A common wrong calculation treats the density \(f_T(3)=0.1181\) as the probability of death at year 3. A density has units per year and is not a point probability; for a continuous lifetime, \(P(T=3)=0\).
Force Is a Rate, Not a Death Probability
The force \(\mu_t\) describes the instantaneous rate of failure at time \(t\), conditional on survival to just before that time. It is not itself the probability of dying over a finite interval. Under a continuous lifetime model, integrating the force over an interval gives the cumulative force. Exponentiating its negative gives conditional survival over that interval; subtracting from 1 gives conditional failure.
To see why, the survival differential equation is \(S_0'(t)=-\mu_tS_0(t)\). Dividing by \(S_0(t)\) and integrating from \(t\) to \(t+h\) gives \(\ln S_0(t+h)-\ln S_0(t)=-\int_t^{t+h}\mu_s\,ds\). Thus the conditional survival probability is \(S_0(t+h)/S_0(t)=\exp(-\int_t^{t+h}\mu_s\,ds)\). Its complement is the stated conditional failure probability. For a small interval of length \(h\), this is approximately \(\mu_t h\) when the force is nearly constant; it is not generally equal to \(\mu_t h\) for a finite interval.
Worked Example: A Piecewise Force and a Conditional Interval
A hypothetical lifetime has force \(0.20\) per year for \(0\leq t<2\) and \(0.50\) per year for \(t\geq2\), with \(S_0(0)=1\). Find the unconditional probability \(P(1<T\leq4)\) and the probability \(P(1<T\leq4\mid T>1)\). Give probabilities to four decimals.
Method. Integrate the force separately over each interval it covers. From time 0 to 1 the cumulative force is \(0.20(1)=0.20\), so \(S_0(1)=e^{-0.20}=0.81873075\). From time 0 to 4 it is \(0.20(2)+0.50(2)=1.40\), so \(S_0(4)=e^{-1.40}=0.24659696\).
The unconditional interval probability is the difference in survival probabilities:
For the conditional probability, the numerator is the same interval probability, and the denominator is the probability of the condition, \(P(T>1)=S_0(1)\):
The cumulative force from time 1 to 4 is \(0.20(1)+0.50(2)=1.20\), confirming the conditional calculation directly as \(1-e^{-1.20}\). The unconditional probability also checks: \(0.81873075-0.24659696=0.57213379\).
Trap and correction. Using the later force \(0.50\) across all three years would give \(1-e^{-1.50}=0.7769\), which is wrong because the force from year 1 to year 2 is \(0.20\), not \(0.50\). Split the integral at the change point; a boundary itself has no effect in this continuous model.
Conditioning on Survival to an Earlier Time
A frequent conditioning error is to divide by survival to the start of the requested interval instead of survival to the time specified in the condition. These are different questions. If the condition is \(T>c\) and the event is \(a<T\leq b\), with \(c<a\), the conditional probability includes the possibility of surviving from \(c\) to \(a\) and then failing during the requested interval.
Worked Example: Do Not Replace the Conditioning Time
A hypothetical lifetime has force \(0.10\) per year for \(0\leq t<3\) and \(0.30\) per year for \(t\geq3\), with \(S_0(0)=1\). Calculate \(P(4<T\leq6\mid T>2)\). Also calculate the probability of failure between 4 and 6 conditional on survival to 4, to distinguish the two conditions. Give probabilities to four decimals.
Method. The cumulative force to time 2 is \(0.10(2)=0.20\), giving \(S_0(2)=e^{-0.20}\). To time 4 it is \(0.10(3)+0.30(1)=0.60\), so \(S_0(4)=e^{-0.60}\). To time 6 it is \(0.10(3)+0.30(3)=1.20\), so \(S_0(6)=e^{-1.20}\).
The requested interval probability conditional on survival to 2 is
For comparison, conditional on survival to 4, the probability of failure by 6 is \(1-S_0(6)/S_0(4)=1-e^{-0.60}=0.45118836\approx0.4512\). This is larger because it does not include the chance of surviving from time 2 to time 4. As a check, the requested answer can also be decomposed into survival from 2 to 4, \(e^{-0.40}=0.67032005\), times failure from 4 to 6 given survival to 4, \(0.45118836\); their product is \(0.30244061\).
Trap and correction. Reporting \(0.4512\) for the requested question silently changes the condition from \(T>2\) to \(T>4\). Keep the stated conditioning time in the denominator, \(S_0(2)\), and use the interval endpoints in the numerator.
Common Checks Before Finalizing a Calculation
A short checklist catches many errors before they reach a final answer:
- Translate the event. “Between \(a\) and \(b\)” means \(S_0(a)-S_0(b)\), not simply \(1-S_0(b)\).
- Use the correct condition. Given survival to \(c\), divide by \(S_0(c)\), not by survival to another time.
- Separate rates from probabilities. A force is an instantaneous rate; a density is probability per unit time. Neither is a finite-interval probability by itself.
- Respect model changes. For a piecewise force, split the integral at every change point and use the force appropriate to each segment.
- Retain precision. Carry unrounded values through subtraction and division, then round only the final result. This avoids a misleading last-decimal discrepancy.
These habits are useful whenever a lifetime model supplies a survival function, density, or force. They also support later calculations involving conditional survival and life-table quantities. The techniques here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each question, identify the event and conditioning time before calculating. Round probabilities to four decimals; report a density to four decimals per year.
- A hypothetical lifetime has constant force \(0.40\) per year from time 0 onward. Find \(P(1<T\leq3)\). (Answer: \(0.3691\))
- A lifetime has constant force \(0.15\) per year. Given survival to time 6, find the probability of failure during the next four years. (Answer: \(0.4512\))
- A hypothetical lifetime has force \(0.10\) per year before time 2 and \(0.40\) per year from time 2 onward. Find \(P(1<T\leq3\mid T>1)\). (Answer: \(0.3935\))
- A continuous lifetime has \(S_0(1)=0.90\), \(S_0(2)=0.80\), and \(S_0(5)=0.50\). Find \(P(2<T\leq5\mid T>1)\). (Answer: \(0.3333\))
- A hypothetical lifetime has constant force \(0.20\) per year. Find both \(f_T(4)\) and \(P(4<T\leq4.5\mid T>4)\). (Answer: \(f_T(4)=0.0899\) per year; probability \(=0.0952\))