Cumulative Force and Exponential Survival

From Instantaneous Force to Survival Over Time

The force of mortality describes mortality at an instant. Survival over a span of time, however, depends on the force throughout that span. Integrating the force accounts for its accumulation and gives a direct formula for the probability of surviving to a specified time.

Let \(S_x(t)\) be the probability that a life aged \(x\) survives for at least \(t\) more years. At elapsed time \(u\), the life has attained age \(x+u\), and the force of mortality is \(\mu_{x+u}\). The earlier material on survival's differential equation gives \(S_x'(t)=-\mu_{x+t}S_x(t)\), with \(S_x(0)=1\). This tutorial integrates that equation.

Definition: The cumulative force of mortality from age \(x\) through \(t\) years of future exposure is the integral of the force over that period. It measures accumulated force, not a probability.

H_x(t)=\int_0^t \mu_{x+u}\,du

If time is measured in years, force has units of inverse years. The integral \(H_x(t)\) is dimensionless, as a probability must be when used in an exponential. The force may change with attained age; the integral includes its values throughout the interval, rather than only its value at the beginning or end.

Deriving Exponential Survival

Assume the survival function is differentiable and positive over the interval of interest, and the force is nonnegative and integrable there. The positivity assumption permits division by \(S_x(t)\). Starting with the differential equation, divide both sides by survival and integrate from time 0 to time \(t\):

Separate the survival term.
From \(S_x'(u)=-\mu_{x+u}S_x(u)\), divide by \(S_x(u)\): \(\frac{S_x'(u)}{S_x(u)}=-\mu_{x+u}\).

Integrate over elapsed time.
Integrating from 0 to \(t\) gives \(\int_0^t \frac{S_x'(u)}{S_x(u)}\,du=-\int_0^t\mu_{x+u}\,du\).

Apply the initial condition.
The left side is \(\ln S_x(t)-\ln S_x(0)\). Since \(S_x(0)=1\), \(\ln S_x(0)=0\), so \(\ln S_x(t)=-H_x(t)\).

Exponentiate.
Exponentiating both sides yields survival as the exponential of negative cumulative force.

Theorem: Under the stated conditions, the survival probability is

S_x(t)=\exp\left(-\int_0^t \mu_{x+u}\,du\right)=e^{-H_x(t)}

The negative sign is essential: nonnegative force makes cumulative force nondecreasing, so survival cannot increase with time. If the integrated force is finite, survival is positive. If the cumulative force tends to infinity, the formula gives survival tending to zero.

Using Cumulative Force for Interval Probabilities

The formula also gives probabilities over an interval that begins after time zero. For \(0\leq s\leq t\), survival from time \(s\) to time \(t\), conditional on surviving to \(s\), is the ratio \(S_x(t)/S_x(s)\). Substituting the exponential formula shows why the relevant accumulation is only the force between \(s\) and \(t\):

P(T_x>t\mid T_x>s) =\frac{S_x(t)}{S_x(s)} =\exp\left(-\int_s^t\mu_{x+u}\,du\right)

Here \(T_x\) denotes the future lifetime of a life aged \(x\). The conditional probability of dying by time \(t\), given survival to \(s\), is one minus this conditional survival probability. An unconditional probability over an interval is different: for example, \(P(s<T_x\leq t)=S_x(s)-S_x(t)\).

Worked Example: Survival Under a Linearly Increasing Force

A hypothetical model assigns a life aged \(x\) the force \(\mu_{x+u}=0.020+0.004u\) per year, where \(u\) is measured in years. Find the probability of surviving 5 years and the probability of dying within those 5 years.

Method. Integrate the force over the full five-year period, then apply the exponential survival formula. The force is nonnegative throughout the interval, and the model's time unit is years.

Step 1: Calculate cumulative force.

H_x(5)=\int_0^5(0.020+0.004u)\,du =0.020(5)+0.004\frac{5^2}{2} =0.100+0.050 =0.150

Step 2: Calculate survival. \(S_x(5)=e^{-0.150}=0.8607\), to four decimal places.

Step 3: Calculate the death probability. The probability of dying within five years is the complement of survival:

{}_5q_x=1-S_x(5)=1-0.8607=0.1393

Check and trap avoided. The probabilities sum to \(0.8607+0.1393=1.0000\). A common error is to report \(0.150\), the cumulative force, as the death probability. The correct probability is \(1-e^{-0.150}=0.1393\), not \(0.150\).

Worked Example: Integrating a Curved Force

For a different hypothetical life aged \(x\), suppose \(\mu_{x+u}=0.010+0.003u+0.0005u^2\) per year, with \(u\) in years. Calculate survival for 4 years and the corresponding probability of death within 4 years.

Method. Integrate each term separately. The quadratic term means the force does not change linearly, so substituting a single force value for the whole period would not give the cumulative force.

Step 1: Integrate from 0 to 4.

\begin{aligned} H_x(4) &=\int_0^4(0.010+0.003u+0.0005u^2)\,du\\ &=0.010(4)+0.003\frac{4^2}{2}+0.0005\frac{4^3}{3}\\ &=0.040+0.024+0.0106667\\ &=0.0746667 \end{aligned}

Step 2: Convert cumulative force to survival.

S_x(4)=e^{-0.0746667}=0.9281

Step 3: Take the complement. The probability of death within four years is \(1-0.9281=0.0719\), to four decimal places.

Check and trap avoided. Survival plus death probability is \(0.9281+0.0719=1.0000\). The integration uses the powers \(u\), \(u^2\), and \(u^3\) in the antiderivative; forgetting the division by 2 or 3 would overstate cumulative force and understate survival.

Worked Example: Conditional Survival Between Two Times

A hypothetical force model is \(\mu_{x+u}=0.040+0.010u\) per year. Find the probability that a life survives from year 2 through year 6, given that it has already survived to year 2. Also find the conditional probability of death during that interval.

Method. Use the integral from 2 to 6, not from 0 to 6, because the requested probabilities are conditional on survival to year 2.

Step 1: Integrate over the interval.

\begin{aligned} \int_2^6(0.040+0.010u)\,du &=0.040(6-2)+0.010\frac{6^2-2^2}{2}\\ &=0.160+0.160\\ &=0.320 \end{aligned}

Step 2: Calculate conditional survival. The probability is \(e^{-0.320}=0.7261\), to four decimal places.

Step 3: Calculate conditional death. The conditional probability of death during years 2 to 6 is \(1-0.7261=0.2739\).

Check and trap avoided. The probabilities sum to 1.0000. For comparison, integrating from 0 to 6 gives cumulative force \(0.040(6)+0.010(6^2/2)=0.420\), and unconditional survival to year 6 is \(e^{-0.420}=0.6570\). That is not the requested conditional survival; using it would include the earlier period before the conditioning event.

Interpreting the Integral Correctly

The cumulative force summarizes the total exposure to mortality over an interval. It is not a death probability, nor is it generally equal to the force at any one age multiplied by the interval length. That shortcut is valid only when the force is constant over the interval, an assumption that must be stated rather than silently imposed.

The exponential relationship also makes accumulation additive across adjacent intervals. For \(0\leq r\leq s\leq t\), the integral from \(r\) to \(t\) is the sum of the integrals from \(r\) to \(s\) and from \(s\) to \(t\). Correspondingly, survival over the full interval is the product of survival over its two parts:

\exp\left(-\int_r^t\mu_{x+u}\,du\right) = \exp\left(-\int_r^s\mu_{x+u}\,du\right) \exp\left(-\int_s^t\mu_{x+u}\,du\right)

This is useful as a consistency check and when a problem gives information in separate time intervals. Each integral must use the force at the attained ages corresponding to that interval. Keep elapsed time in the same units as the force specification: if the force is per year, integrate with time measured in years.

Key takeaway: Integrate the force over the exact time interval first. Exponentiating the negative integral gives survival; subtracting that survival from 1 gives the corresponding death probability. For conditional interval probabilities, begin the integral at the conditioning time.

These integration and interpretation skills are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the interval being considered before integrating. Give probabilities to four decimal places unless the answer is a time.

A hypothetical force is \(\mu_{x+u}=0.015+0.002u\) per year. Find the probability of surviving 6 years and the probability of dying within 6 years. (Answer: survival \(0.8816\); death \(0.1184\).)
A life has force \(\mu_{x+u}=0.020+0.001u\) per year. Given survival to year 2, find the probability of surviving to year 5. (Answer: \(0.9319\).)
For \(\mu_{x+u}=0.010+0.002u^2\) per year, find survival to year 3. (Answer: \(0.9531\).)
A hypothetical force is \(\mu_{x+u}=0.030+0.004u\) per year. Given survival to year 1, find the probability of dying before year 4. (Answer: \(0.1131\).)
A model has \(\mu_{x+u}=0.020+0.010u\) per year. At what positive time \(t\), to two decimal places in years, is survival equal to \(e^{-0.200}\)? (Answer: \(4.63\) years.)

What You'll Learn