The Conditional-Probability Limit for Force of Mortality

Why Condition on Survival?

A mortality rate at duration $t$ describes risk among lives that have reached duration $t$. The relevant group is therefore not everyone alive at the start of observation, but those who have survived to $t$. The force of mortality makes this idea precise by taking a limit of the conditional probability of death over a short interval after $t$.

Let $T_x$ denote the future lifetime of a life currently aged $x$, and let $S_x(t)=P(T_x>t)$ be its survival function. Thus, $T_x>t$ means that the life survives beyond duration $t$. We consider a positive interval $h$ after that duration and ask for the chance of dying by $t+h$, given survival to $t$.

Definition: If $S_x(t)>0$ and the limit exists, the force of mortality at age $x+t$ is $$ \mu_{x+t} = \lim_{h\downarrow 0} \frac{P(t<T_x\le t+h\mid T_x>t)}{h}. $$ Here $h$ is measured in the same time unit as $t$. The force has units of inverse time, such as per year.

The condition $S_x(t)>0$ ensures that survival to $t$ is an event with positive probability, so the conditional probability is defined. The limit is taken through positive values of $h$: it describes the risk immediately after duration $t$. For the usual continuous lifetime model, there is no point mass at an exact duration that would need separate treatment at the interval endpoints.

Deriving the Conditional-Probability Expression

The event $t<T_x\le t+h$ is the event of surviving to $t$ but not surviving beyond $t+h$. Since it is contained in the event $T_x>t$, its probability conditional on survival to $t$ is the probability of that interval event divided by $P(T_x>t)$.

\begin{aligned} P(t<T_x\le t+h\mid T_x>t) &=\frac{P(t<T_x\le t+h)}{P(T_x>t)}\\ &=\frac{S_x(t)-S_x(t+h)}{S_x(t)}\\ &=1-\frac{S_x(t+h)}{S_x(t)}. \end{aligned}

The second line follows because $P(t<T_x\le t+h)=P(T_x>t)-P(T_x>t+h)$. Dividing this interval probability by $S_x(t)$ gives the conditional probability among those who survived to $t$. Therefore, the force can equivalently be written as the limit of this conditional survival loss per unit time:

\mu_{x+t} = \lim_{h\downarrow 0} \frac{1-S_x(t+h)/S_x(t)}{h}.

This is a limiting statement, not an assertion that the probability over any chosen finite interval is exactly $\mu_{x+t}h$. For sufficiently small $h$, the quotient may be close to its limit; for a finite $h$, its value depends on how the survival probability changes throughout that interval.

Worked Examples

Worked Example: A Uniform Future Lifetime

Hypothetical model. A life aged $x$ has future lifetime $T_x$ uniformly distributed between 0 and 4 years. Its survival function is $S_x(s)=(4-s)/4$ for $0\le s\le4$, and is zero after 4. Find the force at duration $t=1$, using the conditional-probability limit. Also find the conditional probability of death in the next 0.2 years and compare it with the force times 0.2.

Method. First condition on survival to year 1, then form the interval quotient and take its limit. The common mistake to avoid is using the unconditional probability of death in the interval without dividing by the probability of reaching year 1.

At duration 1, $S_x(1)=(4-1)/4=3/4$. For $0<h<3$, $S_x(1+h)=(3-h)/4$. Thus the conditional probability of death over the next $h$ years is

\begin{aligned} P(1<T_x\le1+h\mid T_x>1) &=\frac{S_x(1)-S_x(1+h)}{S_x(1)}\\ &=\frac{3/4-(3-h)/4}{3/4}\\ &=\frac{h/4}{3/4} =\frac{h}{3}. \end{aligned}

Divide by $h$ and let $h$ decrease to zero. The quotient is $1/3$ for every $h$ in the stated range, so $\mu_{x+1}=1/3$ per year.

For the specified interval $h=0.2$, the conditional probability is $0.2/3=1/15=0.066666\ldots$, or $0.0667$ to four decimal places. The force times the interval is also $(1/3)(0.2)=1/15$. The match occurs here because the conditional probability is exactly proportional to $h$ in this model and over this range; it is not a general rule. As a check on the conditioning, the unconditional probability of dying between years 1 and 1.2 is $S_x(1)-S_x(1.2)=0.05$, which differs from the conditional probability $0.0667$.

Worked Example: A Nonlinear Survival Function

Hypothetical model. Suppose a life aged $x$ has survival function $S_x(s)=1/(1+s/2)^2$ for $s\ge0$, with $s$ measured in years. Calculate the force at duration $t=2$. Then calculate the conditional probability of death in the next 0.1 years and compare its quotient by 0.1 with the limiting force.

Method. Use the ratio of the survival probability at $t+h$ to the survival probability at $t$, then simplify the difference quotient before taking the limit. The common mistake to avoid is substituting the finite value $h=0.1$ into the limit and reporting that quotient as the force.

The survival probabilities needed are

S_x(2)=\frac{1}{(1+2/2)^2}=\frac14, \qquad S_x(2+h)=\frac{1}{(2+h/2)^2}=\frac{4}{(4+h)^2}.

Therefore, the conditional death probability divided by $h$ is

\begin{aligned} \frac{P(2<T_x\le2+h\mid T_x>2)}{h} &=\frac{1- S_x(2+h)/S_x(2)}{h}\\ &=\frac{1-16/(4+h)^2}{h}\\ &=\frac{(4+h)^2-16}{h(4+h)^2}\\ &=\frac{8+h}{(4+h)^2}. \end{aligned}

Taking $h\downarrow0$ gives $\mu_{x+2}=8/16=0.5$ per year.

For $h=0.1$, the quotient is $8.1/(4.1)^2=8.1/16.81\approx0.4819$ per year. The conditional probability itself is $0.1(0.4819)\approx0.0482$, to four decimal places. By contrast, the limiting force multiplied by 0.1 is $0.5(0.1)=0.05$. The values are close but not equal: the force is the limit as the interval shrinks, while 0.1 years is a finite interval. This also checks the time unit: the quotient and force are per year, while the probability is dimensionless.

Worked Example: A Change in the Survival Curve at the Starting Duration

Hypothetical model. A survival curve is specified by $S_x(s)=1-0.1s$ for $0\le s\le2$, $S_x(s)=0.8-0.2(s-2)$ for $2<s\le5$, $S_x(s)=0.2(6-s)$ for $5<s\le6$, and zero for $s>6$. Find the force at duration $t=2$. Then find the conditional probability of death over the next 0.5 years. All durations are in years.

Method. Since the definition uses $h>0$, evaluate survival just to the right of duration 2. The common mistake to avoid is using the slope from before duration 2, even though the conditional interval begins at duration 2 and proceeds forward.

At the starting duration, $S_x(2)=1-0.1(2)=0.8$. For $0<h\le3$, the applicable part of the curve is the segment after duration 2, so $S_x(2+h)=0.8-0.2h$. Consequently,

\begin{aligned} P(2<T_x\le2+h\mid T_x>2) &=\frac{0.8-(0.8-0.2h)}{0.8}\\ &=\frac{0.2h}{0.8} =0.25h. \end{aligned}

After dividing by $h$, the quotient is $0.25$ for the stated positive $h$, so $\mu_{x+2}=0.25$ per year. Over 0.5 years, the conditional probability is $0.25(0.5)=0.125$, or $0.1250$ to four decimal places.

The survival curve decreases at a rate of 0.1 on the segment before duration 2, but that earlier segment does not determine this right-hand limit. Using that earlier rate would give $0.1/0.8=0.125$ per year, which is not the force immediately after duration 2. The calculation is checked by substituting $h=0.5$: $S_x(2.5)=0.8-0.2(0.5)=0.7$, and the conditional probability is $(0.8-0.7)/0.8=0.125$.

Interpreting the Limit Correctly

The numerator in the definition is a conditional probability, so it lies between 0 and 1. Dividing by $h$ measures that probability per unit of elapsed time. As $h$ shrinks, the quotient may approach a finite value: that value is the force at the age reached at duration $t$. If the limit does not exist, the force is not defined by this limit at that point, even if conditional probabilities over finite intervals remain meaningful.

A force is not itself a probability. It can exceed 1 per year without violating the probability rules, because its units are inverse time. Nor should it be confused with the probability of death over one year. The conditional probability over a specified interval must be calculated from the survival model, or approximated using the force when the interval is sufficiently short and the approximation is suitable.

The conditioning also determines the correct population for the calculation. A death probability over $(t,t+h]$ among all lives at duration zero is an unconditional probability. The force concerns that same interval among only those who survived to $t$. When $S_x(t)$ is substantially below 1, omitting this denominator can cause a material error.

Key takeaway: To obtain the force at duration $t$, calculate the probability of dying in $(t,t+h]$ conditional on $T_x>t$, divide by the positive interval length $h$, and take the limit as $h\downarrow0$. Keep the conditioning, direction of the interval, and time units explicit.

Check Your Understanding

For each hypothetical model, use the conditional-probability limit. Probabilities are reported to four decimal places when needed.

A future lifetime has survival function $S_x(s)=(5-s)/5$ for $0\le s\le5$, and zero thereafter. Find the force at duration 1 and the conditional probability of death in the next 0.4 years. (Answer: force $0.2500$ per year; probability $0.1000$.)
For $s\ge0$, let $S_x(s)=1/(1+s)$, with $s$ in years. Find the force at duration 1, then calculate the conditional probability of death over the next 0.2 years. (Answer: force $0.5000$ per year; probability $0.0909$.)
At duration $t=3$, suppose $S_x(3)=0.70$ and, for sufficiently small positive $h$ measured in years, $S_x(3+h)=0.70-0.14h$. Find the force at duration 3 and the conditional probability of death over the next 0.25 years, using this segment of the curve. (Answer: force $0.2000$ per year; probability $0.0500$.)
At duration 2, a model has $S_x(2)=0.64$ and $S_x(2+h)=0.64(1-0.30h)$ for sufficiently small positive $h$, with $h$ in years. Find the force at duration 2 and the conditional probability over a 0.05-year interval, assuming the stated expression applies over that interval. (Answer: force $0.3000$ per year; probability $0.0150$.)
A life has $S_x(4)=0.50$, while for sufficiently small positive $h$, $S_x(4+h)=0.50-0.10h$, with durations in years. A candidate uses $0.10$ as the force at duration 4. Identify the error and calculate the correct force. (Answer: the probability change must be conditioned on survival to duration 4; the force is $0.10/0.50=0.2000$ per year.)

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