Introduction
A lifetime model can describe the same random lifetime in several connected ways. The survival function gives the probability of lasting beyond a time; the distribution function gives the probability of having ended by that time; the density describes how probability is distributed over time; and the force describes the instantaneous rate of ending, given survival to that point.
Problems often provide one of these functions but ask for a probability or another function. The essential skill is to identify the event precisely, select the function that represents it, and keep unconditional and conditional probabilities distinct. This tutorial develops that workflow using continuous lifetime models. The material is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
For a continuous lifetime, the probability of an event at exactly one specified time is zero. Thus, \(P(T>t)\) and \(P(T\ge t)\) agree, as do probabilities using open or closed endpoints for a time interval. A density is not itself a probability at a point: probabilities over intervals are obtained by integrating the density or taking differences of distribution or survival values.
Translate Between Lifetime Functions
The basic conversions follow from the definitions. If the survival function is known, subtract it from one to obtain the distribution function. If it is differentiable, negate its derivative to obtain the density. The force is the density divided by the probability of having survived to that time.
These relationships require a continuous lifetime for the density and derivative formulas. A model with point masses, such as a probability of ending at exactly a fixed time, cannot be described everywhere by an ordinary density alone. Also, the force formula only applies where survival is positive; after the end of a lifetime’s possible range, the ratio \(f_T(t)/S_0(t)\) is undefined.
Worked Example: From a Piecewise Force to Interval Probabilities
A hypothetical device has lifetime \(T\), measured in years, with force \(\mu_t=0.20\) for \(0\le t<2\) and \(\mu_t=0.10\) for \(t\ge2\). Assume \(S_0(0)=1\). Find (a) \(S_0(t)\), (b) \(P(1<T\le3.5)\), and (c) the probability that the device fails by year 3 given that it lasts beyond year 1.
Method. Integrate the force over time to get cumulative force, then exponentiate its negative to obtain survival. This method avoids applying the second force rate before year 2.
(a) Survival function. Up to year 2, cumulative force is \(0.20t\), so \(S_0(t)=e^{-0.20t}\). After year 2, cumulative force includes \(0.20(2)=0.40\) from the first interval and \(0.10(t-2)\) from the second:
At \(t=2\), both expressions give \(e^{-0.40}\), so survival is continuous at the change in force.
(b) Unconditional interval probability. The event is survival beyond year 1 but not beyond year 3.5, so subtract the two survival probabilities. At year 3.5, cumulative force is \(0.40+0.10(1.5)=0.55\):
(c) Conditional failure probability. Given survival to year 1, failure by year 3 has probability \(1-S_0(3)/S_0(1)\). The cumulative force from year 1 to year 3 is \(0.20(1)+0.10(1)=0.30\):
Check. Both answers are between zero and one, and the conditional probability uses only the force accumulated after year 1. Probabilities are rounded to four decimal places.
Interval and Conditional Probabilities
For \(0\le a<b\), an unconditional interval probability is the difference between the probability of lasting beyond the start and the probability of lasting beyond the end. Conditioning on survival to \(a\) divides that interval probability by the probability of reaching \(a\). Equivalently, conditional survival to \(b\) is the ratio of the two survival probabilities.
The conditional formulas require \(S_0(a)>0\), so the conditioning event has positive probability. A useful justification is to write the conditional probability as a ratio of joint and conditioning probabilities. Since \(T>b\) implies \(T>a\) when \(b>a\), their intersection is simply \(T>b\):
Subtracting this conditional survival probability from one gives the conditional probability of ending by \(b\), after having survived to \(a\). This ratio is not the same as the unconditional probability \(S_0(a)-S_0(b)\): the latter has not been divided by the chance of reaching \(a\).
Worked Example: Recovering the Density and Force
A hypothetical research cohort is modeled by the survival function \(S_0(t)=(1+t/4)^{-2}\) for \(t\ge0\), where \(t\) is in years. Find \(F_T(t)\), \(f_T(t)\), and \(\mu_t\). Then calculate \(P(1.5<T\le4)\), \(P(2<T\le5\mid T>2)\), and the density at year 2.
Method. Use the survival-function identities directly, then use survival ratios for the conditional event. The common mistake avoided here is treating a density value as a probability.
Distribution function. Subtract the given survival function from one:
Density. Differentiate the survival function and change the sign:
Force. Divide the density by survival:
Unconditional interval probability. At \(t=1.5\), \(S_0(1.5)=(11/8)^{-2}=64/121=0.5289\); at \(t=4\), \(S_0(4)=1/4=0.2500\). Therefore,
Conditional interval probability. Here \(S_0(2)=(3/2)^{-2}=4/9\), and \(S_0(5)=(9/4)^{-2}=16/81\). Thus,
Density at year 2. Substitute \(t=2\) into the density:
Check. The interval probability is positive because survival decreases from year 1.5 to year 4. The density \(0.1481\) is a rate per year at that time, not a probability of exactly \(0.1481\) of ending at year 2. Probabilities are rounded to four decimal places.
Using a Time-Varying Force
A force can change continuously with time rather than staying constant on intervals. If \(\mu_t\) is nonnegative and integrable, and \(S_0(0)=1\), survival is obtained from the cumulative force:
This relationship lets us calculate survival ratios without first writing out the entire distribution function. In particular, conditional survival from \(a\) to \(b\) is the exponential of the negative force accumulated over that interval. The density at a time \(t\) is then the force at that time multiplied by survival to that time.
Worked Example: A Force That Increases With Time
A hypothetical component has force \(\mu_t=t/4\) per year for \(t\ge0\). Find the survival function and density, the probability that \(1<T\le2\), the probability of failure by year 2 given survival beyond year 1, and the density at year 2.
Method. Integrate the force from zero to \(t\), then apply the survival, interval, and conditional formulas. The main trap is using the unconditional interval probability as the conditional answer.
Survival function. The cumulative force is \(\int_0^t s/4\,ds=t^2/8\), so
Density. Since \(\mu_t=f_T(t)/S_0(t)\), multiply the force by survival:
Unconditional interval probability. Survival is \(S_0(1)=e^{-1/8}=0.8825\) at year 1 and \(S_0(2)=e^{-1/2}=0.6065\) at year 2. Therefore,
Conditional failure probability. The cumulative force from year 1 to year 2 is \((2^2-1^2)/8=3/8\). Hence,
Density at year 2. The force there is \(2/4=0.5\), and survival is \(e^{-1/2}=0.6065\):
Check. The conditional failure probability exceeds the unconditional probability over the same interval because the interval probability is divided by \(P(T>1)=0.8825<1\). The density is reported per year; probabilities are rounded to four decimal places.
A Reliable Calculation Sequence
When a problem combines lifetime functions, first identify what is supplied and what is requested. Then translate the event into survival values or an integral. This prevents a familiar expression from being used for the wrong event.
Mark the start and end times, and note whether the question is unconditional or conditioned on survival to a time.
Use \(F_T(b)-F_T(a)\) or \(S_0(a)-S_0(b)\) for an unconditional interval probability. For a conditional probability, divide by \(S_0(a)\).
From a differentiable survival function, obtain \(F_T=1-S_0\), \(f_T=-S_0'\), and \(\mu_t=f_T/S_0\) where survival is positive.
Survival must be nonincreasing, interval probabilities must be nonnegative, and conditional probabilities must lie between zero and one.
A particularly common exam trap is mixing an unconditional interval probability with a conditional one. For example, \(S_0(a)-S_0(b)\) answers \(P(a<T\le b)\); it does not answer \(P(a<T\le b\mid T>a)\). The correction is to divide by \(S_0(a)\), or equivalently use \(1-S_0(b)/S_0(a)\). Another trap is to use a force defined per year with times expressed in months without converting the units. The force and time must use compatible units in any integral or exponent.
Check Your Understanding
For each problem, state the event carefully before calculating. Give probabilities to four decimal places.
- A hypothetical sensor has constant force \(0.25\) per year. Given that it has lasted beyond year 2, what is the probability it fails by year 5? (Answer: 0.5276)
- A modeled lifetime has \(S_0(t)=(1+t/3)^{-1}\) for \(t\ge0\). What is the probability it fails by year 5, given that it lasts beyond year 2? (Answer: 0.3750)
- A component has force \(0.10\) per year for \(0\le t<3\) and \(0.20\) per year for \(t\ge3\). Find \(P(T>4\mid T>1)\). (Answer: 0.6703)
- A lifetime has force \(\mu_t=t/9\) for \(t\ge0\). Find its density at year 3 and the unconditional probability \(P(2<T\le4)\). (Answer: \(f_T(3)=0.2022\) per year; probability \(=0.3896\))
- A lifetime has \(S_0(t)=(4-t)^2/16\) for \(0\le t\le4\), and \(S_0(t)=0\) for \(t>4\). Find \(P(1<T\le3\mid T>1)\). (Answer: 0.8889)