Integrating Age-Varying Forces of Mortality

Why Integrate an Age-Dependent Force?

A force of mortality describes the instantaneous rate of death at an attained age. When the force changes with age, the rate at the beginning of an interval is not enough to determine survival over the whole interval. The relevant quantity is the accumulated force over all ages traversed.

For a life aged \(x\), the attained age after \(s\) years is \(x+s\). Thus, survival for \(t\) years depends on the force values from age \(x\) through age \(x+t\), not only on the force at age \(x\). Integrating the force captures that changing exposure and gives the probability of surviving the interval.

Definition: If the force of mortality at attained age \(y\) is \(\mu_y\), then the integrated force from age \(x\) to age \(x+t\) is \(\int_0^t \mu_{x+s}\,ds\). Provided the force is nonnegative and integrable over this interval, the probability of survival is \({}_tp_x=\exp\left(-\int_0^t \mu_{x+s}\,ds\right)\), and the probability of death within the interval is \({}_tq_x=1-{}_tp_x\).

Why the Survival Formula Holds

The survival probability changes according to the force at the current attained age. In a small interval of length \(ds\), conditional on having survived to age \(x+s\), the probability of death is approximately \(\mu_{x+s}\,ds\). Equivalently, the survival function decreases at a rate equal to its current value multiplied by the force.

Write the differential relation.
Let \(p(s)={}_sp_x\). Then \(p(0)=1\) and \(\frac{dp(s)}{ds}=-\mu_{x+s}p(s)\).

Separate variables.
For positive survival probability, divide by \(p(s)\): \(\frac{d}{ds}\ln p(s)=-\mu_{x+s}\).

Integrate from 0 to \(t\).
Since \(p(0)=1\), \(\ln p(t)-\ln p(0)=-\int_0^t\mu_{x+s}\,ds\), so \(\ln p(t)=-\int_0^t\mu_{x+s}\,ds\).

Exponentiate.
This gives \({}_tp_x=\exp\left(-\int_0^t\mu_{x+s}\,ds\right)\). The death probability is its complement, \({}_tq_x=1-{}_tp_x\).

The force must be expressed using the same time scale as the integration variable. If ages and durations are measured in years, then \(\mu_y\) is a rate per year and \(ds\) is measured in years; their product is dimensionless, as required in the exponential. The formula applies when the force is integrable over the interval. A closed-form antiderivative is convenient but is not required for the formula itself.

Linear Forces of Mortality

A force that varies linearly with elapsed age from the starting age can be written as \(\mu_{x+s}=\alpha+\beta s\), where \(\alpha=\mu_x\) and \(\beta\) is the change in force per year of age. Integrating term by term gives a simple expression. The model is valid on the chosen interval only if the force remains nonnegative there.

\int_0^t(\alpha+\beta s)\,ds =\alpha t+\frac{\beta t^2}{2}, \qquad {}_tp_x=\exp\left(-\alpha t-\frac{\beta t^2}{2}\right).

Worked Example: Survival Under a Linearly Increasing Force

A hypothetical life is age 40. Over the next five years, assume the force at attained age \(y\) is \(\mu_y=0.010+0.002(y-40)\), measured per year. Find the five-year survival and death probabilities. The method is to express the force in terms of elapsed time and integrate it. This avoids treating the starting-age force as constant.

Step 1: Express the force over the interval. At elapsed time \(s\), attained age is \(40+s\), so \(\mu_{40+s}=0.010+0.002s\). The force rises from \(0.010\) at age 40 to \(0.020\) at age 45, so it is nonnegative throughout the interval.

Step 2: Calculate the integrated force.

\int_0^5(0.010+0.002s)\,ds =0.010(5)+0.002\left(\frac{5^2}{2}\right) =0.050+0.025 =0.075.

Step 3: Convert it to probabilities.

{}_5p_{40}=e^{-0.075}=0.927743\ldots\approx 0.9277, \qquad {}_5q_{40}=1-0.927743\ldots=0.072256\ldots\approx 0.0723.

Check. The average of the linearly increasing endpoint forces is \((0.010+0.020)/2=0.015\). Multiplying by five years gives integrated force \(0.015(5)=0.075\), agreeing with the integral. The probabilities sum to \(1.0000\) after rounding to four decimal places.

Gompertz Forces

A Gompertz force increases exponentially with attained age. A convenient parameterization relative to age \(x\) is \(\mu_{x+s}=\mu_x e^{\gamma s}\), where \(\mu_x>0\) and \(\gamma>0\). Equivalently, the force multiplies by \(e^\gamma\) for each additional year of age. Integrating the exponential gives a closed-form expression.

\int_0^t\mu_x e^{\gamma s}\,ds =\frac{\mu_x}{\gamma}\left(e^{\gamma t}-1\right), \qquad {}_tp_x =\exp\left[-\frac{\mu_x}{\gamma}\left(e^{\gamma t}-1\right)\right].

The parameter \(\gamma\) is measured per year when \(s\) is in years. The formula above assumes \(\gamma\ne 0\). If \(\gamma=0\), the force is constant at \(\mu_x\), and the integral is \(\mu_x t\), as in the constant-force model.

Worked Example: A Gompertz Force Over Ten Years

For a hypothetical life aged 50, assume \(\mu_{50+s}=0.020e^{0.1s}\) per year for the next ten years. Find the ten-year survival and death probabilities. Use the Gompertz integral, keeping \(\gamma=0.1\) per year and \(t=10\) years. This avoids substituting only the initial force into a constant-force formula.

Step 1: Integrate the force. Here \(\mu_{50}=0.020\), \(\gamma=0.1\), and \(\gamma t=1\).

\int_0^{10}0.020e^{0.1s}\,ds =\frac{0.020}{0.1}(e^1-1) =0.2(2.718282-1) =0.343656.

Step 2: Calculate survival and death probabilities.

{}_{10}p_{50}=e^{-0.343656}=0.7092, \qquad {}_{10}q_{50}=1-0.7092=0.2908.

Check. The force increases from \(0.020\) to \(0.020e\approx0.0544\) per year, so the integrated force over ten years should be greater than \(10(0.020)=0.200\), the value obtained by incorrectly holding the initial force fixed. The calculated integrated force, \(0.343656\), is greater than \(0.200\), as expected. The probabilities sum to \(1.0000\).

When Is a Closed-Form Integral Available?

The survival formula always calls for an integral of the force, but not every force has an antiderivative that can be expressed using elementary functions. Linear and exponential forces have familiar closed forms. Polynomial forces do as well, because each power of \(s\) can be integrated term by term. Some other functions, such as an exponential of a quadratic, generally do not have an elementary antiderivative; numerical integration can then estimate the accumulated force.

For numerical work, the integration interval and units still matter. A numerical method approximates the integral of the force, not the survival probability directly. After estimating the integral, exponentiate its negative to obtain survival, and subtract that value from one to obtain death probability. The following example uses Simpson’s rule with a stated number of subintervals, so the approximation can be reproduced.

Worked Example: Numerical Integration for a Non-Elementary Force

For a hypothetical life aged 30, suppose \(\mu_{30+s}=0.010e^{0.2s^2}\) per year for \(0\le s\le2\), with \(s\) in years. Estimate two-year survival using Simpson’s rule with four equal subintervals, and then find the death probability. The exponent \(0.2s^2\) makes this an example in which the integral has no elementary antiderivative.

Step 1: Set up Simpson’s rule. The width is \(h=2/4=0.5\). For \(f(s)=e^{0.2s^2}\), the values at \(s=0,0.5,1,1.5,2\), rounded to six decimal places, are \(1.000000,\ 1.051271,\ 1.221403,\ 1.568312,\ 2.225541\).

Step 2: Approximate the integrated force. Simpson’s rule is \( \int_0^2 f(s)\,ds\approx\frac{h}{3}[f(0)+f(2)+4(f(0.5)+f(1.5))+2f(1)]\). Including the factor \(0.010\) in the force gives:

\begin{aligned} \int_0^2 0.010e^{0.2s^2}\,ds &\approx 0.010\frac{0.5}{3} [1.000000+2.225541+4(1.051271+1.568312)+2(1.221403)]\\ &=0.010\frac{0.5}{3}(16.146679)\\ &\approx 0.026911. \end{aligned}

Step 3: Convert the approximation to probabilities.

{}_2p_{30}\approx e^{-0.026911}=0.9734, \qquad {}_2q_{30}\approx1-0.9734=0.0266.

Check. The force is at least \(0.010\) throughout the interval, so the integrated force must be at least \(0.020\). The estimate \(0.026911\) meets that check. The reported probabilities are rounded to four decimal places and sum to \(1.0000\).

Common Exam Trap: Freezing the Force at the Starting Age

The constant-force expression \(e^{-\mu_x t}\) applies only when the force remains constant over the interval. For an age-varying force, using the starting-age value throughout discards the change in mortality. In the linear example, that mistake would give \(e^{-0.010(5)}=e^{-0.050}=0.9512\) for survival. The correct integrated force is \(0.075\), giving survival \(e^{-0.075}=0.9277\). The incorrect result is too high because the force increases after age 40.

Before calculating, write the force at attained age \(x+s\), identify the interval \(0\le s\le t\), and integrate over that interval. Also check whether the proposed force stays nonnegative throughout; a linear expression that becomes negative is not a valid force on the portion where it is negative.

Key takeaway: For a changing force, integrate over elapsed time at attained age: \({}_tp_x=\exp(-\int_0^t\mu_{x+s}\,ds)\). Linear and Gompertz forces have closed-form integrals; other forces may require numerical integration. Always use the integrated force, not just the value at the starting age.

Check Your Understanding

For each problem, first identify the force over elapsed time and then integrate it before calculating survival or death probabilities.

A hypothetical life aged 60 has force \(\mu_{60+s}=0.015+0.001s\) per year for four years. Find four-year survival and death probabilities. (Answer: \({}_4p_{60}=0.9343,\ {}_4q_{60}=0.0657\).)
At age 45, a hypothetical life has Gompertz force \(\mu_{45+s}=0.012e^{0.08s}\) per year. Find five-year survival and death probabilities. (Answer: integrated force \(0.0738\); \({}_5p_{45}=0.9289,\ {}_5q_{45}=0.0711\).)
For a hypothetical life aged 70, let \(\mu_{70+s}=0.030-0.002s\) per year for eight years. Check that the force remains nonnegative, then find eight-year survival and death probabilities. (Answer: the force decreases to \(0.014\), so it remains nonnegative; integrated force \(0.176\); \({}_8p_{70}=0.8386,\ {}_8q_{70}=0.1614\).)
A hypothetical life aged 30 has force \(\mu_{30+s}=0.004\,2^{s/10}\) per year for ten years. Find the integrated force and ten-year survival probability. (Answer: integrated force \(0.04/\ln 2=0.0577\); \({}_{10}p_{30}=0.9439\).)
A hypothetical life has force \(\mu_{x+s}=0.010+0.002s\) per year, with \(s\) in years. How long until survival is \(0.9000\)? Use the positive solution and report the duration to two decimals. (Answer: solve \(0.010t+0.001t^2=-\ln(0.9)\), giving \(t\approx6.42\) years.)

What You'll Learn