Constant Force and Exponential Lifetime

From Constant Force to an Exponential Lifetime

A constant force of mortality means that the instantaneous rate of death does not change with elapsed time. Under this assumption, the probability of surviving any fixed duration depends on the length of that duration and the constant rate. The resulting lifetime distribution is exponential.

The model is useful because a single parameter describes survival, death, and interval probabilities. Its simplicity also makes it important to check the model’s assumptions: a constant force over all durations is a specific assumption, not a general description of mortality.

Definition: Let $T$ be a continuous lifetime measured from time zero, and let $S_0(t)=P(T>t)$ be its survival function. The force of mortality at elapsed time $t$ is denoted by $\mu_t$. A constant-force model assumes $\mu_t=\mu$ for every $t\geq 0$, where $\mu>0$ is measured in inverse units of time. The lifetime $T$ then has an exponential distribution with rate $\mu$.

The time unit is part of the model. If $\mu$ is stated per year, durations in the exponent must be measured in years. If the rate is per day, durations must be measured in days. The product $\mu t$ is dimensionless.

Deriving the Survival Function, CDF, and Density

For a lifetime with force $\mu_t$, the survival function satisfies the differential equation $S_0'(t)=-\mu_t S_0(t)$, with $S_0(0)=1$. Under a constant force, this becomes $S_0'(t)=-\mu S_0(t)$. Separating variables and applying the initial condition gives the survival function.

\frac{S_0'(t)}{S_0(t)}=-\mu \quad\Longrightarrow\quad \ln S_0(t)-\ln S_0(0)=-\mu t \quad\Longrightarrow\quad S_0(t)=e^{-\mu t}, \qquad t\geq 0.

This derivation explains why the survival probability is exponential: over each small duration, the loss of survival probability is proportional to both the current probability of being alive and the constant force. The initial value $S_0(0)=1$ fixes the multiplicative constant.

Theorem: If $T$ has a constant force of mortality $\mu>0$ for all $t\geq 0$, then for $t\geq 0$, $$ S_0(t)=P(T>t)=e^{-\mu t}, \qquad F_T(t)=P(T\leq t)=1-e^{-\mu t}, \qquad f_T(t)=\mu e^{-\mu t}. $$ For $t<0$, $F_T(t)=0$. The density formula applies for $t\geq 0$; outside that range the density is zero.

The CDF follows from $F_T(t)=1-S_0(t)$. Differentiating it gives the density:

f_T(t)=\frac{d}{dt}\left(1-e^{-\mu t}\right)=\mu e^{-\mu t}, \qquad t\geq 0.

The density is not itself a probability at a single time. For example, the probability of death in an interval is found by integrating the density over that interval, or equivalently by subtracting survival probabilities. Because this model is continuous, $P(T=t)=0$ for any particular $t$. Thus, for $0\leq a<b$,

P(a<T\leq b)=F_T(b)-F_T(a) =S_0(a)-S_0(b) =e^{-\mu a}-e^{-\mu b}.

If the probability is conditional on survival to time $a$, divide the interval probability by $P(T>a)=S_0(a)$. This gives

P(a<T\leq b\mid T>a) =\frac{S_0(a)-S_0(b)}{S_0(a)} =1-e^{-\mu(b-a)}.

The unconditional and conditional probabilities answer different questions. The first concerns the original population at time zero; the second concerns only those who have survived to time $a$.

Applying the Model

For each calculation, first identify the constant force and its time unit, then express every duration in that same unit. Use survival directly for “survive beyond” questions, the CDF for “die by” questions, and a difference of survival probabilities for death within an interval.

Worked Example: Survival and Death by a Fixed Year

A hypothetical coverage model assumes a lifetime $T$ has a constant force of mortality $\mu=0.04$ per year. Calculate (a) the probability of surviving 10 years and (b) the probability of dying after year 5 and by year 12. Give probabilities to four decimal places.

Model and method. The assumed distribution is exponential with rate $\mu=0.04$ per year. Both durations are already in years, so no time conversion is needed. Use $S_0(t)=e^{-0.04t}$, and obtain the interval probability by subtracting the survival probabilities at its endpoints.

(a) Survive 10 years. Substitute $t=10$:

P(T>10)=S_0(10)=e^{-(0.04)(10)}=e^{-0.40}=0.670320\ldots

Rounded to four decimal places, the probability is $0.6703$.

(b) Die after year 5 and by year 12. The required event is $5<T\leq 12$. Evaluate both survival probabilities:

S_0(5)=e^{-(0.04)(5)}=e^{-0.20}=0.818731\ldots, \qquad S_0(12)=e^{-(0.04)(12)}=e^{-0.48}=0.618783\ldots

Subtract the probability of surviving past year 12 from the probability of surviving past year 5:

P(5<T\leq 12)=S_0(5)-S_0(12) =0.818731\ldots-0.618783\ldots =0.199947\ldots

Rounded to four decimal places, the probability is $0.1999$. The method avoids treating the interval probability as $S_0(12)$, which is the probability of surviving beyond the interval, not dying within it. The results are consistent: the probability of dying in the interval is positive and less than the probability of surviving past year 5.

Worked Example: Death in a Future Interval Given Survival

A hypothetical model for a continuous lifetime assumes an exponential distribution with constant force $\mu=0.025$ per month. Given that a person has survived 6 months, calculate the probability of dying after month 6 and by month 18. Also calculate the unconditional probability of surviving beyond month 18. Give probabilities to four decimal places.

Model and method. The force is $0.025$ per month and every duration is measured in months. For the conditional probability, divide the interval probability by the probability of surviving to month 6. For the unconditional survival probability, use the survival function from time zero.

Conditional death probability. First calculate the relevant survival probabilities:

S_0(6)=e^{-(0.025)(6)}=e^{-0.15}=0.860708\ldots, \qquad S_0(18)=e^{-(0.025)(18)}=e^{-0.45}=0.637628\ldots

The unconditional probability of surviving to month 6 and then dying by month 18 is $S_0(6)-S_0(18)$. Conditioning on survival to month 6 gives

P(6<T\leq 18\mid T>6) =\frac{S_0(6)-S_0(18)}{S_0(6)} =\frac{0.860708\ldots-0.637628\ldots}{0.860708\ldots} =0.259182\ldots

Rounded to four decimal places, the conditional probability is $0.2592$.

Unconditional survival beyond month 18. This probability is measured from time zero:

P(T>18)=S_0(18)=0.637628\ldots

Rounded to four decimal places, it is $0.6376$. The conditional calculation is not the same as the unconditional interval probability, which is $0.860708\ldots-0.637628\ldots=0.223080\ldots$. The method avoids omitting the conditioning denominator. As a check, the conditional interval probability is larger than the unconditional interval probability because the latter also includes the chance of not surviving to month 6.

Worked Example: Keeping the Time Unit Consistent

For a hypothetical equipment-insurance model, the lifetime $T$ until a specified event is assumed to be exponential with constant force $\mu=0.001$ per day. Calculate the probability of surviving 30 days and the probability of the event occurring between day 30 and day 90, conditional on survival to day 30. Give probabilities to four decimal places.

Model and method. The rate is per day, so use 30 and 90 as day durations. First evaluate survival at each time. Then use the ratio of the interval probability to the probability of reaching day 30.

Survive 30 days.

P(T>30)=e^{-(0.001)(30)}=e^{-0.03}=0.970446\ldots

Rounded to four decimal places, this is $0.9704$.

Event between day 30 and day 90, given survival to day 30. The survival probabilities are

S_0(30)=e^{-0.03}=0.970446\ldots, \qquad S_0(90)=e^{-(0.001)(90)}=e^{-0.09}=0.913931\ldots

Thus

P(30<T\leq 90\mid T>30) =\frac{S_0(30)-S_0(90)}{S_0(30)} =\frac{0.970446\ldots-0.913931\ldots}{0.970446\ldots} =0.058235\ldots

Rounded to four decimal places, the conditional probability is $0.0582$. This method avoids inserting a duration measured in months or years into a rate stated per day. The check is that the conditional probability is between zero and one, as required.

Common Mistakes and Model Checks

A constant force produces a valid exponential lifetime model when it is positive and applies over the durations being modeled. If the force changes with time, $e^{-\mu t}$ with one fixed $\mu$ does not generally describe survival over the whole period. A value $\mu=0$ would give $S_0(t)=1$ for every finite $t$, so it does not define a proper exponential lifetime with eventual death.

A frequent error is to calculate an unconditional interval probability when the question asks for a probability conditional on survival to the interval’s start. For example, under a constant force $\mu$, the incorrect calculation for $P(a<T\leq b\mid T>a)$ is $e^{-\mu a}-e^{-\mu b}$. That expression is $P(a<T\leq b)$, without conditioning. The correction is to divide by $S_0(a)$:

P(a<T\leq b\mid T>a) =\frac{e^{-\mu a}-e^{-\mu b}}{e^{-\mu a}} =1-e^{-\mu(b-a)}.

Another common error is confusing “survive to time $t$” with “die by time $t$.” These are complementary events: $P(T>t)=e^{-\mu t}$, whereas $P(T\leq t)=1-e^{-\mu t}$. Before calculating, translate the wording into an event and keep the rate and duration in matching units.

Key takeaway: Under a constant force $\mu>0$, survival to time $t$ is $e^{-\mu t}$, death by time $t$ is $1-e^{-\mu t}$, and death in $(a,b]$ is $e^{-\mu a}-e^{-\mu b}$. For a conditional interval probability, divide by the probability of surviving to the interval’s start.

These formulas are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. Their reliable use depends on stating the constant-force assumption, matching the units of $\mu$ and time, and distinguishing conditional from unconditional events.

Check Your Understanding

For each problem, assume a continuous lifetime and use the constant-force model stated. Give probabilities to four decimal places where appropriate.

A hypothetical lifetime has a constant force of $0.03$ per year. Find the probability of surviving 20 years and the probability of dying by year 20. (Answer: $0.5488$ and $0.4512$, respectively.)
A model has constant force $0.08$ per month. Find the unconditional probability that death occurs after month 3 and by month 9. (Answer: $e^{-0.24}-e^{-0.72}=0.2999$.)
A lifetime has constant force $0.012$ per week. Given survival to week 4, find the probability of death after week 4 and by week 9. (Answer: $1-e^{-0.06}=0.0582$.)
A hypothetical model has constant force $0.0005$ per day. Using 365 days in a year, find the probability of surviving one year. (Answer: $e^{-(0.0005)(365)}=0.8332$.)
A lifetime has constant force $0.04$ per year. Given survival to year 8, find the probability of death after year 8 and by year 15. (Answer: $1-e^{-(0.04)(7)}=0.2442$.)

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