Constant Force Over an Age Interval

Why the Starting Age Matters

A force of mortality describes the instantaneous rate of death at an attained age. When that force is constant over an age interval, survival across the interval has a particularly direct form: the exponent depends on the length of time spent in the interval. The starting age identifies where the interval lies, but it is not itself the duration.

This distinction matters whenever a question asks about someone who has already survived to an age other than zero. For example, the probability that a life aged 58 survives another 2.5 years is a conditional probability over 2.5 years, not a probability of surviving from birth to age 60.5. The constant-force model makes these calculations straightforward, provided the interval and conditioning are read correctly.

Definition: Suppose a life is currently aged \(x\), and the force of mortality is constant at \(\mu\) throughout the age interval from \(x\) to \(x+t\). Then the conditional probability of surviving \(t\) years from age \(x\) is \({}_tp_x=e^{-\mu t}\), and the conditional probability of dying within those \(t\) years is \({}_tq_x=1-e^{-\mu t}\). Here, \(t\) is measured in years when \(\mu\) is expressed per year.

The assumption applies over the entire interval of interest. If the model specifies a constant force only from age 50 to age 55, the formula can be used for an interval wholly within those ages, such as ages 51.5 to 54. It cannot automatically be extended beyond age 55 without an additional mortality assumption.

From Force to Survival Over an Interval

The force \(\mu\) is a rate, not a probability. Under a constant force of \(0.03\) per year, for example, the probability of death in one year is \(1-e^{-0.03}\), not \(0.03\) exactly. For a fractional duration, the same formula uses that duration in years: six months is \(0.5\) years, and 18 months is \(1.5\) years.

The standard survival function \(S_0(y)\) gives the probability of surviving from birth to age \(y\). If \(S_0(x)>0\), conditional survival from age \(x\) to age \(x+t\) is the ratio of the survival probabilities to those ages:

{}_tp_x=\frac{S_0(x+t)}{S_0(x)}.

When the force is constant on this particular interval, this ratio equals \(e^{-\mu t}\). Thus, information about survival before age \(x\) is not needed to calculate the conditional probability from age \(x\). However, to calculate the probability from birth of surviving to \(x+t\), the probability \(S_0(x)\) is also needed.

Why the Exponential Formula Holds

For a life that has survived to age \(x\), let its conditional survival over a further duration \(t\) be denoted by \({}_tp_x\). The force of mortality at duration \(t\) is the instantaneous rate of death among lives that have survived to that duration. The survival function therefore satisfies the differential relationship

\frac{d}{dt}{}_tp_x=-\mu_{x+t}{}_tp_x.

On the interval under consideration, \(\mu_{x+t}=\mu\) is constant. Dividing by the positive survival probability and integrating from duration 0 to duration \(t\) gives

\int_0^t \frac{d}{ds}\log({}_sp_x)\,ds = -\int_0^t \mu\,ds \quad\Longrightarrow\quad \log({}_tp_x)-\log({}_0p_x)=-\mu t.

At duration zero, \({}_0p_x=1\), so \(\log({}_0p_x)=0\). Exponentiating yields \({}_tp_x=e^{-\mu t}\). The death probability is the complement of survival, giving \({}_tq_x=1-e^{-\mu t}\). This derivation also shows why the duration is \(t\): it is the length of the integration interval, not the attained age \(x+t\).

Worked Calculations

Worked Example: Survival From a Nonzero Age

A hypothetical insurance model assumes a constant force of mortality of \(0.032\) per year from age 40 through age 55. A life is known to be alive at age 47.5. Calculate the probability that the life survives to age 52, and the probability that the life dies before age 52. State the results to four decimal places.

Method and assumptions. The life is already alive at age 47.5, and the force is constant over the relevant interval. The duration from age 47.5 to age 52 is \(52-47.5=4.5\) years. The model applies because both ages lie within the specified constant-force interval.

Survival probability. Substitute \(\mu=0.032\) and \(t=4.5\):

{}_{4.5}p_{47.5}=e^{-0.032(4.5)} =e^{-0.144} =0.8659.

Death probability. Take the complement:

{}_{4.5}q_{47.5}=1-0.8659=0.1341.

The probabilities sum to \(0.8659+0.1341=1.0000\), as required. A common error would be to use 52 as the duration; 52 is the ending age, while the time spent in the interval is 4.5 years.

Worked Example: Recovering the Force From Annual Survival

A hypothetical mortality model has a constant force from ages 60 to 65. The probability that a life aged 60 survives one year is \(0.96\). Find the force per year, then calculate the probability that a life alive at age 61.25 dies before age 64.75. Give the force to four significant figures and the probability to four decimal places.

Method and assumptions. The one-year survival probability is given under the same constant-force model. First solve \(e^{-\mu}=0.96\) for the force. Then use the duration from age 61.25 to age 64.75, which is \(64.75-61.25=3.5\) years.

Recover the force.

e^{-\mu}=0.96 \quad\Longrightarrow\quad \mu=-\ln(0.96)=0.04082\text{ per year}.

Calculate the death probability. The interval lies within ages 60 to 65, so the constant-force assumption applies:

{}_{3.5}p_{61.25} =e^{-0.0408219945(3.5)} =0.8669, \qquad {}_{3.5}q_{61.25} =1-0.8669 =0.1331.

The result is consistent with survival being below one and death being its complement. The common mistake is to use 3.5 as the force, or to treat the annual death probability \(0.04\) as the force. The force is \(-\ln(0.96)\), and it must be multiplied by the duration in years.

Worked Example: Conditional and Unconditional Probabilities

In a hypothetical population, \(S_0(46)=0.84\). From age 46 through age 54, the force of mortality is constant at \(0.025\) per year. Calculate (a) the probability that a life alive at age 46 survives to age 49.2, (b) \(S_0(49.2)\), and (c) the probability that a newborn survives to age 46 and then dies before age 49.2. Give probabilities to four decimal places.

Method and assumptions. The interval from age 46 to age 49.2 has length \(49.2-46=3.2\) years. First calculate the conditional survival probability over those 3.2 years. To obtain probabilities from birth, multiply by \(S_0(46)=0.84\).

(a) Conditional survival from age 46.

{}_{3.2}p_{46} =e^{-0.025(3.2)} =e^{-0.08} =0.9231.

(b) Survival from birth to age 49.2. Use the survival-function ratio:

S_0(49.2) =S_0(46){}_{3.2}p_{46} =0.84e^{-0.08} =0.7754.

(c) Survival to age 46 followed by death before age 49.2. This is the probability of reaching age 46 multiplied by the conditional probability of dying over the next 3.2 years:

0.84\,{}_{3.2}q_{46} =0.84(1-e^{-0.08}) =0.0646.

As a check, the probabilities of surviving to age 49.2 and of surviving to age 46 but dying before age 49.2 sum to \(0.7754+0.0646=0.8400\), the probability of reaching age 46. The distinction to preserve is that \({}_{3.2}p_{46}\) is conditional on being alive at 46, whereas \(S_0(49.2)\) is measured from birth.

Common Mistakes and Model Checks

A frequent exam trap is to multiply the annual death probability by the number of years. If the one-year survival probability is \(0.96\) under a constant force, then the one-year death probability is \(0.04\). A tempting but incorrect calculation for two-year death is \(2(0.04)=0.08\). Under the constant-force model, two-year survival is \(0.96^2=0.9216\), so the correct two-year death probability is

$$ 1-0.96^2=1-0.9216=0.0784. $$

The linear calculation ignores that the second year's risk applies only to lives that survive the first year. Equivalently, the one-year probability \(q\) is not the force \(\mu\); under a constant force, \(q=1-e^{-\mu}\), so \(\mu=-\ln(1-q)\).

Before finalizing an interval calculation, check the following:

Duration: subtract the starting age from the ending age. Convert months to a fraction of a year if the force is per year.
Conditioning: decide whether the question concerns a life already alive at the starting age or a probability measured from birth.
Model range: confirm that the whole interval lies within the ages for which the force is stated to be constant.
Complement: for the same starting age and duration, survival and death probabilities must sum to one.

These calculations are useful whenever a mortality assumption is supplied as a force but a question asks for a probability over a specified age range. They also provide a direct check on the interpretation of annual survival information. Constant-force calculations over intervals beginning at nonzero ages are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated constant-force assumption over the full interval and keep the duration in years. Probabilities are rounded to four decimal places.

A hypothetical model has force \(0.025\) per year from ages 35 to 45. Find the survival and death probabilities for a life aged 38 surviving to age 42. (Answer: survival \(0.9048\); death \(0.0952\))
Under a constant force, one-year survival from age 60 is \(0.97\), and the force remains constant through age 64. Find the probability that a life alive at age 61.5 dies before age 63. (Answer: \(0.0447\))
A constant force of \(0.06\) per year applies from ages 70 to 74. Find the probability that a life alive at age 71.5 survives to age 74, and the corresponding death probability. (Answer: survival \(0.8607\); death \(0.1393\))
Suppose \(S_0(52)=0.91\), and the force is \(0.02\) per year from age 52 to age 59. Find the probability from birth of surviving to age 52 and then dying before age 56. (Answer: \(0.0700\))
A hypothetical model has constant force from ages 64 to 70. The three-year survival probability from age 64 is \(0.88\). Find the probability that a life alive at age 65.25 survives to age 69.75. (Answer: \(0.8255\))

What You'll Learn