Why Piecewise-Constant Force Models Are Useful
A lifetime model need not have the same force of mortality at every duration. A hypothetical warranty population, for example, might have one failure rate during an initial period, another after that period, and a third later. A piecewise-constant force gives a simple way to represent such changes without assuming that the survival function itself jumps when the rate changes.
The essential rule is to accumulate the force over time. Within each interval, the survival curve decreases exponentially at that interval’s rate. At a boundary between intervals, the new rate changes the curve’s slope, not its current survival value. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Definition and Construction
Suppose the force is \(\mu_1\) from time 0 to \(c_1\), \(\mu_2\) from \(c_1\) to \(c_2\), and so on. The force in the interval containing \(t\) determines how quickly survival declines there. However, survival at \(t\) also reflects the force accumulated in every earlier interval.
For a nonnegative force that is constant on successive intervals, the survival function is obtained by integrating the force from 0 to \(t\), then exponentiating the negative of that accumulated amount. The intervals can be written as half-open intervals, such as \([0,c_1)\) and \([c_1,c_2)\); the force’s value at an isolated boundary does not affect the integral or the survival probabilities.
If \(t\) lies in interval \(j\), let \(c_{j-1}\) be the beginning of that interval and let \(c_0=0\). Then the accumulated force consists of the completed earlier intervals plus the portion of interval \(j\) elapsed by time \(t\):
This expression applies when the force is \(\mu_k\) on \([c_{k-1},c_k)\) for each completed interval and is \(\mu_j\) on the current interval. If the last interval has no finite endpoint, its contribution is calculated only up to \(t\). A nonnegative force makes survival nonincreasing; if the force is positive and constant after the final change point, survival tends to zero as duration tends to infinity.
Why Survival Remains Continuous at a Change Point
The continuity at a rate-change point follows directly from accumulating force. On an interval with constant force \(\mu\), survival at elapsed time \(h\) after its starting point is the survival at that starting point multiplied by \(e^{-\mu h}\). As \(h\) decreases to zero, this multiplier tends to 1. Thus the survival value approaches its value at the start of the interval, even though the force may differ immediately before and after that time.
At time 0, \(S_0(0)=1\). At a later boundary, use the survival already accumulated up to that time.
Over an additional duration \(h\) in an interval with force \(\mu\), multiply the boundary survival by \(e^{-\mu h}\).
At a change point, retain the survival just reached. Only the force used for further durations changes.
More formally, suppose the force is \(\mu_1\) before \(c\) and \(\mu_2\) after \(c\). For \(t\geq c\), the accumulated force is \(\mu_1c+\mu_2(t-c)\). At the boundary, the value from the later interval is
which equals the value approached from the earlier interval. The same reasoning applies at every subsequent boundary. This also shows why one must not restart the survival calculation at 1 when a new force begins: doing so discards the survival already accumulated.
Worked Examples
Worked Example: Constructing Survival Across Two Change Points
A hypothetical insured device has a continuous lifetime \(T\), measured in years. Its force of failure is \(0.10\) per year for \(0\leq t<3\), \(0.20\) per year for \(3\leq t<5\), and \(0.05\) per year for \(t\geq5\). Construct \(S_0(t)\), find survival to durations 3, 5, and 7, and find the probability of surviving 3 years but not 5 years. Assume this force model applies throughout the stated durations.
Method. Integrate the force over each elapsed portion of time. At each boundary, carry forward the survival value from the preceding interval.
For \(0\leq t<3\), the accumulated force is \(0.10t\), so \(S_0(t)=e^{-0.10t}\). For \(3\leq t<5\), the accumulated force is \(0.10(3)+0.20(t-3)\). For \(t\geq5\), it is \(0.10(3)+0.20(2)+0.05(t-5)\). Therefore,
At \(t=3\), the first expression gives \(e^{-0.10(3)}=e^{-0.30}\). The second expression at the boundary gives \(e^{-0.30-0.20(0)}=e^{-0.30}\), so survival is continuous. At \(t=5\), the second expression gives \(e^{-0.30-0.20(2)}=e^{-0.70}\), and the third gives \(e^{-0.70-0.05(0)}=e^{-0.70}\).
The requested survival values are
The probability of surviving 3 years but not 5 years is \(P(3<T\leq5)=S_0(3)-S_0(5)\). Thus it is \(0.7408-0.4966=0.2442\), to four decimal places. The result is consistent with the interval: survival decreases from 0.7408 to 0.4966.
Common mistake avoided: The force \(0.20\) applies only to the two years from 3 to 5, not to the entire five-year duration.
Worked Example: Conditional Survival Across a Change Point
In a hypothetical population, a continuous lifetime \(T\) has force \(0.08\) per year for \(0\leq t<1\), \(0.12\) per year for \(1\leq t<4\), and \(0.04\) per year for \(t\geq4\). Given survival to year 2, calculate the probability of surviving to year 5. State the corresponding conditional probability of death by year 5.
Method. Conditional survival from year 2 to year 5 depends on the force accumulated only over \([2,5]\). This interval contains two years at force \(0.12\) and one year at force \(0.04\).
The conditional survival is the exponential of the negative accumulated force over those three years:
Therefore, the conditional probability of death by year 5, given survival to year 2, is the complement:
As a check using survival values from time 0, \(S_0(2)=e^{-0.08(1)-0.12(1)}=e^{-0.20}\), while \(S_0(5)=e^{-0.08-0.12(3)-0.04}=e^{-0.48}\). Their ratio is \(e^{-0.48}/e^{-0.20}=e^{-0.28}\), the same conditional survival. The calculation does not apply the force \(0.08\) after year 1, or the force \(0.12\) after year 4.
Common mistake avoided: Conditional survival begins at year 2, so the accumulated force from years 0 to 2 cancels from the ratio. It must not be counted again.
Worked Example: Building the Model and Checking a Conditional Probability
A hypothetical sensor has a continuous operating lifetime \(T\) in years. Its force of failure is \(0.05\) per year for \(0\leq t<2\), \(0.15\) per year for \(2\leq t<5\), and \(0.10\) per year for \(t\geq5\). Construct the survival function and calculate the probability that a sensor surviving to year 2 fails by year 4. Use four decimal places for probabilities.
Method. First accumulate the force from time 0 to express survival on each interval. For the conditional calculation, count only the force from year 2 through year 4.
For the first interval, the accumulated force is \(0.05t\). For the second interval, it is \(0.05(2)+0.15(t-2)\). For the final interval, it is \(0.05(2)+0.15(3)+0.10(t-5)\). Hence
At year 2, both adjoining expressions give \(e^{-0.10}\). At year 5, the middle expression gives \(e^{-0.10-0.15(3)}=e^{-0.55}\), which equals the final expression at \(t=5\). Thus the function carries survival continuously across both change points.
The interval from year 2 to year 4 lies entirely in the second interval, so the accumulated force conditional on reaching year 2 is \(0.15(4-2)=0.30\). The conditional survival to year 4 is \(e^{-0.30}=0.7408\). Therefore,
A check using the survival function gives \(S_0(2)=e^{-0.10}\) and \(S_0(4)=e^{-0.10-0.15(2)}=e^{-0.40}\). Their ratio is \(e^{-0.40}/e^{-0.10}=e^{-0.30}\), as required. The reported death probability is its complement, \(0.2592\), to four decimal places.
Common mistake avoided: The conditional probability is not calculated from time 0. The force accumulated before year 2 is already reflected in the condition that the sensor survived to year 2.
Reading the Model Correctly
A piecewise-constant force model has two different features that are easy to confuse. The force is the local rate governing the decline of survival at a particular duration. The survival value is cumulative: it reflects all force accumulated since time 0. At a boundary, the force may jump, but the survival value does not.
For two durations \(s<t\), conditional survival is the ratio \(S_0(t)/S_0(s)\), provided \(S_0(s)>0\). Equivalently, it is the exponential of the negative force accumulated between \(s\) and \(t\). This is why a conditional calculation across several intervals can be done by adding the rate-duration products only for the portions of those intervals between \(s\) and \(t\).
The most reliable check is to evaluate the expressions on both sides of every boundary. If they do not give the same survival value, a completed interval has probably been omitted, counted twice, or assigned the wrong duration. A second check is monotonicity: with nonnegative forces, survival must never increase as time advances.
Check Your Understanding
For each model, treat the force as applying on the stated half-open intervals and round probabilities to four decimal places.
- A lifetime has force \(0.07\) per year for \(0\leq t<2\) and \(0.11\) per year thereafter. What is the probability of surviving to year 3? (Answer: \(e^{-0.25}=0.7788\))
- A lifetime has force \(0.04\) per year for \(0\leq t<5\), \(0.10\) per year for \(5\leq t<8\), and \(0.03\) per year thereafter. Find \(S_0(8)\) and the probability of surviving from year 6 to year 9 given survival to year 6. (Answer: \(S_0(8)=e^{-0.50}=0.6065\); conditional survival \(=e^{-0.23}=0.7945\))
- A lifetime has force \(0.20\) per year for \(0\leq t<1\) and \(0.10\) per year thereafter. Given survival to year 0.5, find the probability of death by year 2. (Answer: \(1-e^{-0.20}=0.1813\))
- A lifetime has force \(0.06\) per year for \(0\leq t<2\), \(0.12\) per year for \(2\leq t<5\), and \(0.08\) per year thereafter. Find \(S_0(6)\) and the probability of death between years 1 and 4. (Answer: \(S_0(6)=e^{-0.56}=0.5712\); interval death probability \(=e^{-0.06}-e^{-0.36}=0.2441\))
- A lifetime has force \(0.05\) per year for \(0\leq t<4\), \(0.09\) per year for \(4\leq t<6\), and \(0.03\) per year thereafter. Given survival to year 3, find the probability of death by year 7. (Answer: \(1-e^{-0.26}=0.2289\))