Interpreting the Force of Mortality

Why the Force of Mortality Needs Careful Interpretation

Mortality quantities can describe different things even when they refer to the same age and time interval. A one-year death probability measures the chance of dying over a full year, conditional on being alive at its start. The force of mortality describes the instantaneous rate of death at a particular age. The density of future lifetime describes how probability is distributed across possible times of death. Keeping these meanings separate is essential when reading or using a survival model.

Let $T_x$ denote the future lifetime, in years, of a person alive at exact age $x$. The probability of surviving at least $t$ more years is denoted by ${}_tp_x=P(T_x>t)$. The force of mortality at attained age $x+t$ describes the immediate death rate at that age, conditional on survival to that age.

Definition: The force of mortality at age $x+t$, written $\mu_{x+t}$, is the limiting conditional probability of death over the next $h$ years, divided by $h$, as $h$ decreases to zero: $$ \mu_{x+t} = \lim_{h\downarrow 0} \frac{P(t<T_x\le t+h\mid T_x>t)}{h}. $$ It is an instantaneous rate, not a probability. When time is measured in years, its units are inverse years.

The condition $T_x>t$ matters: the force is conditional on the person having survived to time $t$, or attained age $x+t$. For a small positive interval $h$, the conditional probability of dying in that interval is approximately $\mu_{x+t}h$, provided the force does not change much over the interval. This is a local approximation, not an exact formula for a year-long probability in a general model.

Force, Interval Probability, and Density

Suppose the force varies continuously with attained age and there are no point masses of deaths at particular times. Survival over an interval is determined by the accumulated force over that interval. In particular, survival for $t$ years from age $x$ is the exponential of the negative accumulated force:

{}_tp_x = \exp\left(-\int_0^t \mu_{x+s}\,ds\right).

This relationship shows why a force is not itself a probability. The force at one instant describes the local rate, while survival or death over a nonzero interval depends on the force accumulated across that interval. If the force is constant at $\mu$ over a full year, the conditional probability of death during that year is $1-e^{-\mu}$, not $\mu$. The approximation $\mu$ is useful only when $\mu$ is small enough that higher-order effects are negligible.

The density of future lifetime, when it exists, is a different quantity again. It describes probability per unit of time at a particular future duration. The relationship between density, survival, and force follows by differentiating the survival probability.

Key relationship: If the future-lifetime distribution is continuous and the force is defined over the relevant ages, then $$ f_{T_x}(t)={}_tp_x\,\mu_{x+t}. $$ The density is the force at time $t$ multiplied by the probability of reaching time $t$ alive. It is not generally equal to the force alone.

To justify the relationship, write $G(t)={}_tp_x$. From the accumulated-force formula, $G(t)=\exp(-\int_0^t\mu_{x+s}\,ds)$. Differentiating gives $G'(t)=-\mu_{x+t}G(t)$. Since the distribution function of $T_x$ is $1-G(t)$, its density is the derivative $-G'(t)$. Therefore $f_{T_x}(t)=\mu_{x+t}{}_tp_x$, as claimed. This derivation also explains why density and force can differ: the density includes the chance of surviving to the time where the local force is being assessed.

Worked Interpretations

Worked Example: Constant Force and Three Different Quantities

A hypothetical mortality model assigns a constant force of mortality of $0.04$ per year to a person alive at age 60. Under this model, calculate (a) the force at age 60, (b) the probability of death within one year, and (c) the density of future lifetime at exactly one year. State what each result means.

Model and method. Assume the force remains constant at $0.04$ per year for all future durations. Under a constant force, survival for $t$ years is $e^{-0.04t}$.

(a) Force. The force at age 60 is given directly: $$ \mu_{60}=0.04\text{ per year}. $$ This is an instantaneous conditional rate, not a 4% probability of death over a year.

(b) One-year death probability. The probability of surviving one year is $$ {}_1p_{60}=e^{-0.04}=0.9607894. $$ Thus the probability of death within one year is $$ 1-{}_1p_{60}=1-e^{-0.04}=0.0392106. $$ To four decimal places, this is $0.0392$, or about $3.92\%$.

(c) Density at one year. The density is survival to time 1 multiplied by the force at that time: $$ f_{T_{60}}(1)=e^{-0.04}(0.04)=0.0384316\text{ per year}. $$ To four decimal places, the density is $0.0384$ per year. It describes probability per unit of time near one year; it is not the probability of death at exactly one year.

Check. The one-year probability $0.0392$ is less than the force value $0.0400$, as expected from $1-e^{-0.04}<0.04$. The density at one year is smaller still because it is multiplied by the probability of surviving to that time. A common mistake is to report $0.04$ as the one-year death probability; the exact probability under this model is $0.0392$.

Worked Example: An Age-Varying Force and a Short Interval

For a hypothetical life aged 50, suppose the force $t$ years after age 50 is $\mu_{50+t}=0.02+0.01t$ per year. Find the probability of surviving two years, the density of future lifetime at two years, and the conditional probability of death between durations 2 and 2.1.

Model and method. The force is measured per year, and $t$ is measured in years. First integrate the force to obtain survival. Then use the force-survival relationship for the density and integrate over the requested interval for the conditional probability.

Survival to duration 2. The accumulated force is $$ \int_0^2(0.02+0.01s)\,ds =0.02(2)+0.005(2^2) =0.04+0.02 =0.06. $$ Therefore, $$ {}_2p_{50}=e^{-0.06}=0.9417645. $$ The probability of death within two years is $1-0.9417645=0.0582355$, or $0.0582$ to four decimal places.

Density at duration 2. The force then is $$ \mu_{52}=0.02+0.01(2)=0.04\text{ per year}. $$ Thus, $$ f_{T_{50}}(2)={}_2p_{50}\mu_{52} =(0.9417645)(0.04) =0.0376706\text{ per year}. $$ To four decimal places, this is $0.0377$ per year.

Conditional death probability from duration 2 to 2.1. Given survival to duration 2, the accumulated force over the next $0.1$ year is $$ \int_2^{2.1}(0.02+0.01s)\,ds =0.02(0.1)+0.005(2.1^2-2^2) =0.002+0.00205 =0.00405. $$ The conditional probability of death in that interval is $$ 1-e^{-0.00405}=0.0040418, $$ or $0.0040$ to four decimal places.

Check. The conditional probability over the short interval is close to the force near age 52 multiplied by $0.1$ year, which is approximately $0.04(0.1)=0.004$. It is not the density at duration 2: the density $0.0377$ is unconditional and is expressed per year. Confusing these quantities loses both the conditioning and the time interval.

Worked Example: Same Current Force, Different Lifetime Densities

Compare two hypothetical mortality models for a life aged $x$. In Model A, the force is $0.02$ per year throughout the first two years. In Model B, it is $0.04$ during the first year and $0.02$ during the second year. In both models, assume the force at and after duration 2 is $0.02$ per year. Find the density of future lifetime at duration 2 under each model, and compare the conditional one-year death probability after duration 2.

Model and method. In each model, calculate survival to duration 2 from the accumulated force. Multiply that survival probability by the force at duration 2 to obtain the density. The conditional one-year probability after duration 2 depends on the force during that future year, not on the force before duration 2.

Model A. The accumulated force through duration 2 is $$ 0.02(2)=0.04, $$ so ${}_2p_x=e^{-0.04}=0.9607894$. The density at duration 2 is $$ f_{T_x}(2)=(0.9607894)(0.02)=0.0192158\text{ per year}. $$

Model B. The accumulated force through duration 2 is $$ 0.04(1)+0.02(1)=0.06, $$ so ${}_2p_x=e^{-0.06}=0.9417645$. The density at duration 2 is $$ f_{T_x}(2)=(0.9417645)(0.02)=0.0188353\text{ per year}. $$

Conditional one-year probability after duration 2. In both models, the force remains $0.02$ during the next year. Conditional on being alive at duration 2, the probability of death in that year is therefore $$ 1-e^{-0.02}=0.0198013, $$ or $0.0198$ to four decimal places in either model.

Check. Both models have the same force at duration 2 and the same conditional next-year death probability, but Model B has a lower density at duration 2 because fewer lives survive to that duration. A common mistake is to infer the density from the current force alone; the required factor ${}_2p_x$ accounts for survival up to that time.

Reading and Using the Quantities Correctly

The three quantities answer different questions, so a sound interpretation names both the time scale and the conditioning:

Force $\mu_{x+t}$: What is the instantaneous conditional death rate at attained age $x+t$, given survival to that age?
Interval probability: What is the chance of death over a stated interval, given survival at the interval's start? This depends on the force across the whole interval.
Density $f_{T_x}(t)$: How much unconditional probability per unit of time is concentrated near future duration $t$? It includes the probability of surviving to $t$.

For a sufficiently small interval of length $h$, the conditional death probability is approximately $\mu_{x+t}h$. The corresponding unconditional probability of death between durations $t$ and $t+h$ is approximately ${}_tp_x\mu_{x+t}h=f_{T_x}(t)h$. These approximations explain why a force and a density have similar roles locally but are not interchangeable: one is conditional, and the other incorporates survival from the starting age.

A rate also depends on its time unit. If time is measured in years, $\mu$ is per year and $h$ must be in years in the small-interval approximation. Substituting a number of months as though it were a number of years creates a factor-of-12 error. When the force changes over the interval, use the accumulated force rather than multiplying a single value by the interval length.

Key takeaway: The force of mortality is an instantaneous conditional rate. An interval death probability is obtained from the force over that interval, and the future-lifetime density equals the force multiplied by the probability of surviving to the corresponding duration.

Check Your Understanding

For each problem, distinguish the local rate from probabilities over intervals. Probabilities are rounded to four decimal places.

A hypothetical life aged $x$ is subject to constant force $0.03$ per year for one year. Give the force and the probability of death during that year. (Answer: force $0.0300$ per year; death probability $1-e^{-0.03}=0.0296$.)
For a life aged 40, let $\mu_{40+t}=0.01+0.02t$ per year. Find survival to duration 2, the force at duration 2, and the density at duration 2. (Answer: ${}_2p_{40}=e^{-0.06}=0.9418$; force $0.0500$ per year; density $0.0471$ per year.)
A model gives ${}_3p_x=0.8000$ and force $\mu_{x+3}=0.0400$ per year. Find the density at duration 3. (Answer: $f_{T_x}(3)=0.8000(0.0400)=0.0320$ per year.)
A hypothetical life has constant force $0.025$ per year over the next year. Find the conditional probability of death during the year, given survival at its start. (Answer: $1-e^{-0.025}=0.0247$.)
The force is $0.02$ per year during the first year and $0.05$ per year during the second year. Find the conditional probability of death during year 2 given survival to its start, and the unconditional probability of dying during year 2 from the original starting age. (Answer: conditional probability $1-e^{-0.05}=0.0488$; unconditional probability $e^{-0.02}(1-e^{-0.05})=0.0478$.)

What You'll Learn