Why Piecewise Force Calculations Need Care
A lifetime model may use different mortality rates over different durations. For example, a hypothetical warranty program might model the time until a covered component fails using one force during its initial service period and another force later. To calculate a probability at a given time, it is not enough to multiply the force at that time by the whole duration. Each rate applies only over its own interval.
This tutorial works through how to accumulate those intervals and then calculate survival, density and death probabilities. The lifetime \(T\) is measured from time zero, and all rates and durations below use years. The force at duration \(t\) is denoted by \(\mu_t\). A force is an instantaneous rate, not a probability of death over a whole year.
A Reliable Calculation Method
For a piecewise-constant force, draw or list the time intervals first. On each interval, multiply its force by only the portion of that interval that has elapsed. Add these products to obtain cumulative force. Then calculate survival by exponentiating the negative cumulative force. This order helps prevent a common error: applying the rate in force at the endpoint to the entire elapsed duration.
Record the force and the duration spent in each interval up to the time of interest.
For an interval of length \(r\) with constant force \(c\), its contribution to cumulative force is \(cr\).
Use \(S_0(t)=e^{-M(t)}\), and use the survival difference for an interval death probability.
At a time inside an interval, multiply the force on that interval by survival at that time.
Why Exponential Survival Still Applies
The exponential survival formula follows from the differential equation for survival under a force. Over a short time interval, a force \(\mu_t\) produces an instantaneous rate of decrease in survival, so \(S_0'(t)=-\mu_tS_0(t)\). Since survival is positive, divide by \(S_0(t)\) and integrate from zero to \(t\):
Because \(S_0(0)=1\), its logarithm is zero, giving \(S_0(t)=e^{-M(t)}\). This derivation does not require the force to be constant. For a piecewise-constant force, the integral is simply the sum of the force-duration products on the intervals. Differentiating survival at times where the force is defined gives \(f_T(t)=-S_0'(t)=\mu_tS_0(t)\).
A force may jump at a breakpoint. The cumulative force, and therefore survival, remains continuous because a single point contributes zero duration to the integral. The density may have different one-sided limits on either side of the breakpoint. That jump does not create a point mass: the probability of death at exactly one time is zero for a continuous lifetime.
Worked Example: Survival, Density and an Interval Probability
A hypothetical lifetime has force \(0.02\) per year for \(0\leq t<3\), \(0.05\) per year for \(3\leq t<6\), and \(0.03\) per year for \(t\geq6\). Find survival and density at \(t=8\), and find the probability of death between durations 2 and 5. Give probabilities to four decimals and density to four decimals.
Method. Accumulate force interval by interval, then use the survival and density formulas. The key mistake to avoid is applying the force at year 8 to all eight years.
Cumulative force through year 8. The first interval contributes \(3(0.02)=0.06\). The second contributes \((6-3)(0.05)=0.15\). The final interval contributes \((8-6)(0.03)=0.06\). Thus \(M(8)=0.06+0.15+0.06=0.27\).
Survival at year 8. Therefore \(S_0(8)=e^{-0.27}=0.7633795\ldots\), or \(0.7634\) to four decimals.
Density at year 8. Year 8 is in the interval with force \(0.03\), so \(f_T(8)=0.03S_0(8)=0.03(0.7633795\ldots)=0.0229014\ldots\), or \(0.0229\) to four decimals.
Death between years 2 and 5. The cumulative force through year 2 is \(M(2)=2(0.02)=0.04\). Through year 5 it is \(M(5)=3(0.02)+2(0.05)=0.06+0.10=0.16\). Hence
The interval death probability is \(0.1086\) to four decimals. As a check, survival falls from about \(0.9608\) at year 2 to \(0.8521\) at year 5, so the probability of death between those times is their positive difference.
Conditional Death Probabilities Across a Change
When the question conditions on survival to a duration \(a\), the relevant probability is not the unconditional probability of death by a later time. For \(b>a\), divide the probability of surviving to \(a\) and dying by \(b\) by the probability of surviving to \(a\). Equivalently, use the increase in cumulative force after \(a\):
The accumulated force before \(a\) cancels. Only the force accumulated from \(a\) to \(b\) is needed for this conditional probability. This is a calculation from the survival model; it does not assert that a piecewise-force lifetime is memoryless.
Worked Example: Conditional Death After Surviving to a Change Period
A hypothetical lifetime has force \(0.01\) per year for the first 3 years, \(0.04\) per year from year 3 to year 5, and \(0.02\) per year thereafter. Given survival to year 4, find the probability of death by year 7. Give the probability to four decimals.
Method. Calculate the force accumulated from year 4 to year 7, then apply the conditional survival formula. The common mistake to avoid is using the cumulative force from year 0 as though the question were unconditional.
From year 4 to year 5, one year is spent at force \(0.04\), contributing \(1(0.04)=0.04\). From year 5 to year 7, two years are spent at force \(0.02\), contributing \(2(0.02)=0.04\). Thus \(M(7)-M(4)=0.04+0.04=0.08\).
The conditional probability of death by year 7 is
To four decimals, the answer is \(0.0769\). As a check, the conditional probability of surviving from year 4 to year 7 is \(e^{-0.08}=0.9231\), and the death and survival probabilities add to 1. Using \(1-e^{-0.15}\) would incorrectly include the force accumulated before year 4.
What Happens at a Force Breakpoint?
At a breakpoint, survival is found from all force accumulated up to that time. The density on either side uses the corresponding force on that side. If the force changes from \(c_1\) to \(c_2\) at time \(r\), then the one-sided density limits are \(c_1S_0(r)\) and \(c_2S_0(r)\). Survival has the same value on both sides, but the density limits need not match.
For an interval probability that crosses a breakpoint, do not try to assign a death probability to the breakpoint itself. A continuous lifetime has no point mass there. Instead, calculate survival at the interval's two endpoints and subtract, or add the separate probabilities across the subintervals.
Worked Example: A Density Jump and a Death Probability Across Two Changes
A hypothetical lifetime has force \(0.015\) per year for \(0\leq t<2\), \(0.025\) per year for \(2\leq t<5\), and \(0.01\) per year for \(t\geq5\). Find the survival probability at year 5, the one-sided density limits there, and the probability of death between years 3 and 8. Give probabilities and densities to four decimals.
Method. First accumulate force to each required endpoint. For the density limits at year 5, use the force on the interval immediately before and immediately after that time. The mistake to avoid is treating a jump in density as a point mass.
Through year 5, the first interval contributes \(2(0.015)=0.030\), and the second contributes \(3(0.025)=0.075\). Therefore \(M(5)=0.030+0.075=0.105\), and \(S_0(5)=e^{-0.105}=0.9003245\ldots\), or \(0.9003\) to four decimals.
Immediately before year 5, the force is \(0.025\). The left-hand density limit is \(0.025(0.9003245\ldots)=0.0225081\ldots\), or \(0.0225\). Immediately after year 5, the force is \(0.01\). The right-hand density limit is \(0.01(0.9003245\ldots)=0.0090032\ldots\), or \(0.0090\).
For the interval probability, cumulative force through year 3 is \(M(3)=2(0.015)+1(0.025)=0.055\). Through year 8 it is \(M(8)=2(0.015)+3(0.025)+3(0.01)=0.030+0.075+0.030=0.135\). Thus
The answer is \(0.0728\) to four decimals. The check is that survival at year 3 exceeds survival at year 8, and their difference is positive. The different density limits at year 5 do not change the fact that the probability of death at exactly year 5 is zero.
Common Mistakes to Avoid
- Using the last force for the whole duration. If the force changes, calculate each force-duration contribution separately before exponentiating.
- Subtracting the wrong survivals. For \(P(a<T\leq b)\), subtract later survival from earlier survival: \(S_0(a)-S_0(b)\), not the reverse.
- Ignoring the condition in a conditional probability. Given survival to \(a\), use only the cumulative force from \(a\) to \(b\), not from time zero to \(b\).
- Confusing density and probability. \(f_T(t)\) is a density, not the probability of death at exactly \(t\). A probability over an interval comes from a survival difference or an integral of the density.
- Assuming the density must be continuous. A force jump can make the density jump. The survival function remains continuous, and there is no point mass at the breakpoint.
Check Your Understanding
For each problem, identify the force intervals before calculating. Probabilities and densities in the answers are rounded to four decimals.
- A lifetime has force \(0.03\) per year for the first 2 years and \(0.01\) thereafter. Find survival at year 6, density at year 6, and the probability of death between years 3 and 6. (Answer: \(S_0(6)=0.9048\), \(f_T(6)=0.0090\), and \(P(3<T\leq6)=0.0276\).)
- A lifetime has force \(0.02\) per year for the first 4 years and \(0.04\) thereafter. Given survival to year 5, find the probability of death by year 8. (Answer: \(1-e^{-0.12}=0.1131\).)
- A lifetime has force \(0.01\) per year for the first 3 years, \(0.03\) per year from year 3 to year 6, and \(0.02\) thereafter. Find the probability of death between years 2 and 8. (Answer: \(e^{-0.02}-e^{-0.16}=0.1281\).)
- A lifetime has force \(0.015\) per year for the first year, \(0.025\) per year from year 1 to year 4, and \(0.005\) thereafter. Find survival and density at year 6, and the probability of death between years 3 and 6. (Answer: \(S_0(6)=0.9048\), \(f_T(6)=0.0045\), and \(P(3<T\leq6)=0.0322\).)
- A lifetime has force \(0.02\) per year for the first 3 years and \(0.06\) thereafter. Given survival to year 2, find the probability of death between years 2 and 5. (Answer: \(1-e^{-0.14}=0.1306\).)