Why Solve for Force Parameters?
A survival probability records the chance that a lifetime exceeds a specified duration. A force model describes how the instantaneous rate of failure changes over time. When the force is assumed constant, one survival probability can determine its value. When the force is piecewise constant, survival probabilities at interval boundaries can identify the force on each interval. The key is to use the probability over the correct time interval, rather than treating every observation as if it began at time zero.
This tutorial develops a practical method for fitting these models and checking the result. The examples use hypothetical lifetime data. Durations are measured in the time unit explicitly stated in each problem; force values are therefore expressed per that unit.
Solving for a Constant Force
Suppose the force is constant over the durations of interest and the lifetime begins at duration zero. If an observed survival probability at duration \(t>0\) is \(S(t)=s\), with \(0<s\leq 1\), then the exponential survival model gives one equation for the unknown force:
The logarithm is essential: it reverses the exponential. The value \(s=1\) gives \(\mu=0\); a value \(s=0\) cannot be produced at a finite duration by a finite constant force. If multiple survival probabilities are supplied, calculate the implied force from each one. They must agree if they are exact observations from the same constant-force model. If the data are rounded, small differences may be due to rounding, but a substantial disagreement signals that the model does not fit all the observations.
The duration \(t\) is measured from the model's origin. For example, \(S(5)\) means survival from duration zero through duration five; it is not automatically the probability of surviving five more units conditional on already being alive at duration two. The distinction matters whenever the force can change over time.
Worked Example: Checking a Constant-Force Fit
A hypothetical device lifetime is measured in years. A proposed constant-force model is fitted to the observed survival probability \(S(4)=0.6561\). Estimate the force, then check the model against a second observation \(S(7)=0.4782969\). Report the force to four significant figures.
Method. Solve the exponential survival equation using the first observation, then substitute the estimated force into the model at the second duration. This check avoids assuming that one fitted probability automatically validates a constant-force model.
First solve using \(t=4\):
Rounded to four significant figures, the estimated force is \(0.1054\) per year. Now use the unrounded estimate to predict survival at seven years:
The prediction equals the second observed probability, so the two observations are consistent with this constant-force model. As a further check, at four years the model gives \(e^{-(0.1053605157)(4)}=0.9^4=0.6561\), the value used to fit it.
Piecewise-Constant Forces
A piecewise-constant model divides time into intervals and assigns a constant force to each interval. For example, let the force be \(\mu_1\) from duration zero to \(c\), and \(\mu_2\) after \(c\). Survival to a time after \(c\) accumulates the effects of both intervals. The first interval contributes a factor \(e^{-\mu_1 c}\); the part of the second interval up to \(t\) contributes \(e^{-\mu_2(t-c)}\).
A useful way to fit the parameters is to divide survival probabilities at two boundaries. The ratio removes the survival accumulated before the later interval. More generally, if an interval has constant force \(\mu_j\) over duration \(b-a\), then
This ratio follows directly by subtracting the cumulative force accumulated by time \(a\) from that accumulated by time \(b\). Equivalently, the factors for all earlier intervals cancel in the quotient. The equation is valid when the force is constant throughout \([a,b]\) and the supplied survival probabilities use the same time origin. For exact data, \(0<S(b)/S(a)\leq1\) is necessary; a ratio above one would imply increasing survival and cannot be explained by a nonnegative force.
Worked Example: Estimating Two Forces from Boundary Probabilities
A hypothetical component lifetime is measured in months. Its force is constant at \(\mu_1\) during months 0 to 2 and at \(\mu_2\) during months 2 to 5. The observed probabilities are \(S(2)=0.729\) and \(S(5)=0.373248\). Estimate both forces and verify the fitted survival at month 5. Report each force to four significant figures.
Method. Use \(S(2)\) to fit the first interval, then use \(S(5)/S(2)\) to fit the second. The ratio is the key step: \(S(5)\) includes survival through both intervals, not just the second one.
Since \(0.729=0.9^3\), the force over the first two months is
Thus \(\mu_1\) is \(0.1580\) per month to four significant figures. The conditional survival factor across months 2 to 5 is obtained from the ratio:
That interval lasts \(5-2=3\) months, so
To verify, multiply the survival factors from both intervals:
This reproduces the observation exactly. The fitted forces, rounded to four significant figures, are \(0.1580\) and \(0.2231\) per month. The first value is not \(-\ln(0.729)\) itself: the first interval lasts two months, so the logarithm must be divided by two.
A General Fitting Procedure
For several intervals, write down the boundary times in increasing order and identify which force applies between each pair. Then use survival ratios interval by interval. This keeps the time units visible and makes it easier to catch a misplaced boundary or an incorrect duration.
Record each interval's start and end, its duration, and the force assumed to apply there.
Use survival probabilities from the same origin. Check that each observation is assigned to the correct cumulative interval.
For an interval from \(a\) to \(b\), calculate \(S(b)/S(a)\), then solve \(\mu=-\ln(S(b)/S(a))/(b-a)\).
Reconstruct survival at each supplied duration by multiplying the interval factors. Compare each reconstructed value with its observation.
The procedure also identifies whether the information is sufficient. One survival probability at the end of a single constant-force interval determines one force. If there are several unknown interval forces, there must be enough independent boundary information to solve for them. A probability at a time well inside a later interval may still be used, but its equation includes the accumulated force from every earlier interval.
Worked Example: Recovering Three Interval Forces and Verifying Them
A hypothetical study measures lifetimes in years. The force is constant on each of the intervals \([0,1)\), \([1,4)\), and \([4,6)\), with values \(\mu_1,\mu_2,\mu_3\). The observed survival probabilities are \(S(1)=0.8\), \(S(4)=0.512\), and \(S(6)=0.32768\). Find the three forces and verify all three observations. Report the forces to four significant figures.
Method. Recover each force from the survival ratio across its own interval. This avoids incorrectly assigning the cumulative survival loss from earlier years to a later interval.
The first interval lasts one year, so
where \(S(0)=1\). The second interval lasts \(4-1=3\) years. Its survival factor is \(S(4)/S(1)=0.512/0.8=0.64\), giving
The third interval lasts \(6-4=2\) years. Its factor is \(S(6)/S(4)=0.32768/0.512=0.64\), so
Now reconstruct the survival probabilities. At year 1, the factor is \(0.8\), so \(\widehat{S}(1)=0.8\). At year 4, the first two factors give
At year 6, include the third interval's factor:
All three observations are reproduced. The forces, to four significant figures, are \(0.2231\), \(0.1488\), and \(0.2231\) per year. The different durations explain why equal survival factors in the second and third intervals do not imply equal forces: the second interval lasts three years, while the third lasts two.
Consistency Checks and Common Exam Traps
A fitted model should be checked against every observation, not only the one used to determine its parameters. For a constant force, the implied values \(-\ln S(t)/t\) should agree across durations. For a piecewise-constant force, calculate the force from each interval ratio, then reconstruct the cumulative survival at each observed time. In either case, check that interval lengths and force units match: dividing a logarithm by years gives a force per year, while dividing by months gives a force per month.
- Using cumulative survival for a later interval. For an interval from \(a\) to \(b\), the relevant factor is \(S(b)/S(a)\), not \(S(b)\) alone.
- Using the wrong duration. Divide by \(b-a\), not by the endpoint \(b\), when fitting the force on an interval that starts at \(a>0\).
- Ignoring a model check. A parameter can fit one observation even when the assumed model is inconsistent with other supplied observations.
- Rounding too early. Keep full precision through the interval calculations and round only the final force or probability.
- Mixing time units. If durations are given in months, do not report a force per year unless you explicitly convert the unit.
For example, in the two-interval example, using the wrong calculation \(-\ln(0.373248)/3\) for the second force treats the survival probability at month 5 as if the second interval began at month 0. The correction is to remove the first interval's survival factor first:
The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus: identify the assumed force structure, translate survival observations into exponential equations, solve with logarithms, and verify the fitted probabilities.
Check Your Understanding
For each problem, state the time unit of the force and check whether the supplied probabilities fit the stated force structure.
- A hypothetical battery has constant force and survival \(S(3)=0.729\), where time is measured in years. Find the force and predict \(S(5)\). (Answer: \(\mu=0.1054\) per year; \(S(5)=0.5905\).)
- A hypothetical sensor has constant force. The observations are \(S(2)=0.81\) and \(S(4)=0.6561\), with time in months. Find the force from each observation and state whether they agree. (Answer: Both imply \(\mu=0.1054\) per month; the observations are consistent.)
- A hypothetical component has force \(\mu_1\) from year 0 to year 1 and \(\mu_2\) from year 1 to year 3. Given \(S(1)=0.7\) and \(S(3)=0.343\), find both forces. (Answer: \(\mu_1=0.3567\) per year and \(\mu_2=0.3567\) per year.)
- A constant-force model is proposed for hypothetical equipment, with \(S(2)=0.8\) and \(S(5)=0.5\), in years. Find the force implied by each observation and assess consistency. (Answer: \(0.1116\) per year from \(S(2)\), and \(0.1386\) per year from \(S(5)\); the observations are inconsistent with one constant force.)
- A hypothetical lifetime has constant force on \([0,1)\), \([1,3)\), and \([3,5)\), measured in years. Given \(S(1)=0.9\), \(S(3)=0.729\), and \(S(5)=0.531441\), find all three forces and verify \(S(5)\). (Answer: \(\mu_1=0.1054\), \(\mu_2=0.1054\), and \(\mu_3=0.1580\) per year; the interval factors are \(0.9\), \(0.81\), and \(0.729\), giving \(S(5)=0.9(0.81)(0.729)=0.531441\).)