Deriving Force from Density and Survival

Why Density and Survival Determine Force

A density describes how probability is distributed across possible future lifetimes. A survival function describes the probability that a life remains alive beyond a specified time. Their ratio answers a local question: among those still alive at a given time, how much probability density is concentrated at that time? That conditional, instantaneous rate is the force of mortality.

The denominator matters. The density $f_x(t)$ describes the future lifetime of a life aged $x$ at time zero, including the original group whether or not its members survive to time $t$. The force at time $t$, by contrast, is conditional on having survived to that time. Dividing by $S_x(t)$ makes that conditioning explicit.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, with survival function $S_x(t)=P(T_x>t)$ and density $f_x(t)$. For a continuous lifetime model with $S_x(t)>0$, the force of mortality at attained age $x+t$ is $$ \mu_{x+t}=\frac{f_x(t)}{S_x(t)}. $$ The density and survival function must describe the same future-lifetime distribution, and the ratio is used at times where the density and the conditional-probability limit are well defined.

The units are useful as a check. Survival probability is dimensionless, while a density has units of inverse time. Therefore the force has units of inverse time, such as per year when $t$ is measured in years. A force is not itself a probability. For a short interval of length $h$, the conditional probability of death is approximately $\mu_{x+t}h$, provided the force does not change too much over that interval.

Deriving the Density-to-Survival Ratio

Assume the future lifetime is continuous, has density $f_x$, and has survival $S_x(t)>0$. For $h>0$, death during the short interval after time $t$ means $t<T_x\leq t+h$. Conditional on surviving to $t$, its probability is

P(t<T_x\leq t+h\mid T_x>t) =\frac{P(t<T_x\leq t+h)}{P(T_x>t)} =\frac{\int_t^{t+h} f_x(s)\,ds}{S_x(t)}.

The numerator is the probability of being in that interval, and the denominator is the probability of reaching the start of the interval. Divide by the interval length and take the limit as the interval shrinks. If $f_x$ is continuous at $t$, the average density over $[t,t+h]$ approaches $f_x(t)$:

\begin{aligned} \mu_{x+t} &=\lim_{h\downarrow 0} \frac{P(t<T_x\leq t+h\mid T_x>t)}{h}\\ &=\lim_{h\downarrow 0} \frac{1}{S_x(t)} \left(\frac{1}{h}\int_t^{t+h} f_x(s)\,ds\right)\\ &=\frac{f_x(t)}{S_x(t)}. \end{aligned}

This is a conditional-rate result, not a rule for dividing by the probability of having died by time $t$. The denominator is survival to $t$, because the rate concerns the lives exposed to death at that time. If $S_x(t)=0$, there are no survivors at $t$ in the model, so this ratio does not define a force there. If the lifetime distribution has a point mass, the continuous-density argument also needs modification at that point; the formula is intended for the continuous setting stated above.

Calculating Force from Paired Functions

Worked Example: A Force at an Attained Age and a Short-Interval Probability

A hypothetical model for a life aged 60 has future-lifetime survival function $S_{60}(t)=(1+t/20)^{-2}$, for $t\geq0$, with time measured in years. Its density is $f_{60}(t)=\frac{1}{10}(1+t/20)^{-3}$. Calculate the force at $t=20$, identify the attained age, and find the conditional probability of death in the next 0.1 year. Compare that probability with the short-interval approximation.

Method and model check. The stated functions are paired: differentiating the survival function gives the stated density, and the survival starts at 1 and decreases to 0. The ratio applies because this is a continuous lifetime model and $S_{60}(20)>0$.

At $t=20$, the life has attained age $60+20=80$. Evaluate the survival and density:

S_{60}(20)=(1+20/20)^{-2}=2^{-2}=\frac14, \qquad f_{60}(20)=\frac{1}{10}(2)^{-3}=\frac{1}{80}.

Therefore the force is

\mu_{80}=\frac{1/80}{1/4}=\frac{4}{80}=0.0500\text{ per year}.

For the exact conditional probability over the next 0.1 year, use the survival ratio for the interval:

\begin{aligned} P(20<T_{60}\leq20.1\mid T_{60}>20) &=1-\frac{S_{60}(20.1)}{S_{60}(20)}\\ &=1-\left(\frac{2}{2.005}\right)^2 =1-\left(\frac{400}{401}\right)^2\\ &=\frac{801}{160801} \approx 0.004981. \end{aligned}

The local approximation is $\mu_{80}(0.1)=0.0500(0.1)=0.0050$, close to the exact probability. The approximation is not identical because the force can vary within the interval. Common mistake avoided: treating force as a probability for a year or using it without multiplying by the interval length. Here the force is per year, and the interval is 0.1 year.

Worked Example: Force from a Different Survival Curve

For a hypothetical life aged 40, suppose the future-lifetime survival function is $S_{40}(t)=(1+t/25)^{-3}$ for $t\geq0$. The corresponding density is $f_{40}(t)=\frac{3}{25}(1+t/25)^{-4}$. Find the force at $t=25$, then calculate the exact conditional probability of death in the following year.

Method. Use the density-to-survival ratio at the specified time. The model is continuous, and its survival probability is positive at $t=25$. First evaluate both functions:

S_{40}(25)=2^{-3}=\frac18, \qquad f_{40}(25)=\frac{3}{25}2^{-4}=\frac{3}{400}=0.0075.

The attained age is $40+25=65$, so

\mu_{65}=\frac{3/400}{1/8} =\frac{3}{50} =0.0600\text{ per year}.

For the exact one-year conditional death probability, compare survival at 26 with survival at 25:

\begin{aligned} P(25<T_{40}\leq26\mid T_{40}>25) &=1-\frac{S_{40}(26)}{S_{40}(25)}\\ &=1-\left(\frac{2}{2.04}\right)^3 =1-\left(\frac{50}{51}\right)^3\\ &=\frac{7651}{132651} \approx 0.0577. \end{aligned}

The force at the start of the year is 0.0600 per year, but the exact one-year probability is about 0.0577. They are different quantities: the first is an instantaneous rate, and the second accounts for the whole year. Common mistake avoided: reporting the force as the exact probability of death over a year.

Worked Example: Reading Force from Pointwise Data

A hypothetical continuous lifetime model gives $S_{55}(8)=0.75$ and $f_{55}(8)=0.018$ per year. Find the force at the attained age and estimate the conditional probability of death in the next quarter-year. Explain why the information is not enough to give the exact quarter-year probability.

Method. Since the density and survival refer to the same future-lifetime model and survival is positive, divide the density by survival:

\mu_{55+8}=\frac{0.018}{0.75}=0.0240\text{ per year}.

The attained age is 63. For a quarter-year, $h=0.25$ years, so the local approximation is

P(8<T_{55}\leq8.25\mid T_{55}>8) \approx \mu_{63}h =0.0240(0.25)=0.0060.

This is an approximation from the force at the start of the interval. The exact probability requires the survival at $8.25$, because it equals $1-S_{55}(8.25)/S_{55}(8)$. That value was not supplied, and a single pointwise density and survival value do not determine it. Common mistake avoided: treating a local rate as enough information to calculate an exact finite-interval probability.

Common Denominator and Time-Unit Errors

Several quick checks help prevent errors when applying the formula:

Use survival as the denominator. The required denominator is $S_x(t)$, not $1-S_x(t)$. The latter is the probability of having died by $t$, not the probability exposed to mortality at $t$.
Match the functions. $f_x(t)$ and $S_x(t)$ must describe the same life aged $x$, the same future-time origin, and the same time units.
Report attained age correctly. The force at future time $t$ for a life aged $x$ is $\mu_{x+t}$, not $\mu_x$.
Keep rates and probabilities distinct. A force has units of inverse time. Multiplying by a short interval gives a first-order probability approximation, not generally an exact probability.
Check whether the denominator is positive. A ratio with zero survival is not a defined conditional force for the stated population.

The ratio is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because it links standard lifetime distribution functions to an instantaneous mortality measure. In exam calculations, writing down the time point, attained age, units, and conditional denominator before substituting values makes the interpretation as important as the arithmetic.

Key takeaway: For a continuous future-lifetime model with positive survival, divide the density at time $t$ by survival to time $t$: $\mu_{x+t}=f_x(t)/S_x(t)$. The denominator conditions on being alive and exposed at that time.

Check Your Understanding

For each problem, state the force with its attained age and units where requested. Treat all models as continuous and time as measured in years unless stated otherwise.

A life aged 50 has $S_{50}(t)=(1+t/10)^{-1}$ and $f_{50}(t)=0.1(1+t/10)^{-2}$. Find the force at $t=10$. (Answer: $\mu_{60}=0.0500$ per year.)
For a hypothetical model, $S_{47}(3)=0.60$ and $f_{47}(3)=0.015$ per year. Find the force and attained age. (Answer: $\mu_{50}=0.0250$ per year.)
A model has $S_{30}(t)=(1+t/20)^{-2}$ and $f_{30}(t)=0.1(1+t/20)^{-3}$. Calculate the force at $t=10$. (Answer: $\mu_{40}=0.0667$ per year.)
At a specified time, a continuous model has $S_x(t)=0.80$ and $f_x(t)=0.040$ per year. Calculate the force. What incorrect result would come from dividing the density by the probability of having died by that time? (Answer: Correct force $=0.0500$ per year; incorrect ratio $=0.2000$ per year.)
A life aged 35 has $S_{35}(t)=(1+t/50)^{-2}$ and $f_{35}(t)=\frac{2}{50}(1+t/50)^{-3}$. Find the force at $t=50$ and the exact conditional probability of death in the next half-year. (Answer: $\mu_{85}=0.0200$ per year; probability $=401/40401\approx0.0099$.)

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