Fractional-Year Death Timing Under Constant Force

Why Condition on Death During the Year?

A one-year mortality rate gives the probability of death during a year, but it does not by itself say how deaths are distributed within that year. Under the constant-force assumption, deaths are not uniformly distributed across the year: the probability density of death time decreases as time since the start of the year increases. This tutorial derives the distribution of death timing given that death occurs during the year, then applies it to interval probabilities and quantiles.

Let $T_x$ denote the complete future lifetime of a life aged $x$, measured in years. Define $U=T_x\mid T_x\le 1$: $U$ is the time from age $x$ to death, conditional on death before the end of the year. The event $T_x\le 1$ is equivalently death during the year of age $x$, apart from endpoints that have probability zero in a continuous lifetime model.

Definition: Under a constant force of mortality $\mu_x=\mu>0$ during the year of age $x$, survival for $t$ years within that year is ${}_tp_x=e^{-\mu t}$, for $0\le t\le 1$. Thus $p_x={}_1p_x=e^{-\mu}$, and $q_x={}_1q_x=1-e^{-\mu}$. The conditional fractional-year death time $U$ has distribution $$ F_U(t)=P(U\le t)=\frac{1-e^{-\mu t}}{q_x} =\frac{1-p_x^t}{q_x},\qquad 0\le t\le1. $$ Its density is $f_U(t)=\mu e^{-\mu t}/q_x$ for $0<t<1$.

The distribution is conditional: its total probability over the year is one. Without conditioning, the probability of death by time $t$ is $1-e^{-\mu t}$, which reaches $q_x$, not one, at $t=1$. If $q_x=0$, conditioning on death during the year is undefined because the conditioning event has probability zero.

Deriving the Conditional Distribution

The result follows directly from conditional probability. For $0\le t\le1$, the event $U\le t$ means that death occurs by time $t$, given that death occurs by time 1. Since death by time $t$ is a subset of death by time 1, the conditional probability is the ratio of those two probabilities.

\begin{aligned} P(U\le t) &=P(T_x\le t\mid T_x\le1)\\ &=\frac{P(T_x\le t)}{P(T_x\le1)}\\ &=\frac{1-e^{-\mu t}}{1-e^{-\mu}}\\ &=\frac{1-e^{-\mu t}}{q_x}. \end{aligned}

Differentiating this conditional distribution function with respect to $t$ gives the density. It is decreasing because $e^{-\mu t}$ decreases with $t$. Thus, conditional on death occurring during the year, earlier times have higher density than later times. This does not mean that an exact time has positive probability: for a continuous distribution, the probability of death at any single specified time is zero.

f_U(t)=\frac{d}{dt}F_U(t)=\frac{\mu e^{-\mu t}}{q_x},\qquad 0<t<1.

The conditional probability that death occurs after time $a$ but no later than time $b$, where $0\le a<b\le1$, is obtained by subtracting the distribution function at the endpoints:

P(a<U\le b)=\frac{e^{-\mu a}-e^{-\mu b}}{q_x} =\frac{p_x^a-p_x^b}{q_x}.

Likewise, the conditional probability of surviving beyond time $t$ within the year is

P(U>t)=\frac{e^{-\mu t}-e^{-\mu}}{q_x} =\frac{p_x^t-p_x}{q_x},\qquad 0\le t\le1.

For a conditional death-time percentile, let $t_r$ be the time by which a fraction $r$ of deaths during the year have occurred, where $0<r<1$. Set $F_U(t_r)=r$ and solve:

r=\frac{1-e^{-\mu t_r}}{q_x} \quad\Longrightarrow\quad t_r=-\frac{\ln(1-rq_x)}{\mu} =\frac{\ln(1-rq_x)}{\ln(1-q_x)}.

The final equality uses $\mu=-\ln(1-q_x)$. Both logarithms in the ratio are negative, so $t_r$ is positive. The quantile is measured in years from the start of the age year.

Worked Applications

Worked Example: Probability of Death in Part of the Year

A hypothetical life aged $x$ has one-year death probability $q_x=0.20$. Assume a constant force of mortality throughout the year. Find the probability that death occurs between 0.4 and 0.9 years after age $x$, conditional on death during the year.

Step 1: Identify the annual survival probability and force. The annual survival probability is $p_x=1-q_x=0.80$. Under constant force, $\mu=-\ln(p_x)=-\ln(0.80)=0.2231436$ per year.

Step 2: Use the conditional interval formula. Here $a=0.4$, $b=0.9$, and $q_x=0.20$, so

\begin{aligned} P(0.4<U\le0.9) &=\frac{0.80^{0.4}-0.80^{0.9}}{0.20}\\ &=\frac{0.914610-0.818052}{0.20}\\ &=\frac{0.096558}{0.20}\\ &=0.4828. \end{aligned}

The probability is 0.4828, or 48.28% to two decimal places. As a check, the numerator is the unconditional probability of death in that interval, 0.096558; dividing by $q_x=0.20$ conditions on death during the full year. The method avoids treating a constant force as if it implied a uniform distribution of death times.

Worked Example: Finding a Conditional Median

A hypothetical age year has $q_x=0.50$, with a constant force of mortality during the year. Find the median time to death conditional on death during that year. Also calculate the conditional density at the median.

Step 1: Calculate the force. Since $p_x=1-0.50=0.50$, the force is $\mu=-\ln(0.50)=0.6931472$ per year.

Step 2: Set the conditional percentile to one-half. The median has $r=0.50$, so

\begin{aligned} t_{0.5} &=-\frac{\ln(1-0.5(0.50))}{0.6931472}\\ &=-\frac{\ln(0.75)}{0.6931472}\\ &=\frac{0.2876821}{0.6931472}\\ &=0.4150\text{ years}. \end{aligned}

Step 3: Evaluate the conditional density there. From the quantile equation, $e^{-\mu t_{0.5}}=1-rq_x=0.75$. Therefore

f_U(t_{0.5})=\frac{0.6931472(0.75)}{0.50} =1.0397\text{ per year}.

The median is 0.4150 years after the start of the year, and the density there is 1.0397 per year. A density is not a probability and may exceed one; probabilities come from integrating the density over an interval. The quantile expression is positive because it can be written as $\ln(0.75)/\ln(0.50)$, a ratio of two negative numbers.

Worked Example: Starting From Life-Table Counts

A hypothetical life table has $\ell_x=1{,}000$ and $\ell_{x+1}=700$. Assume the force of mortality is constant during this age year. Find (a) the probability of death in the second half of the year conditional on death during the year, and (b) the conditional median death time.

Step 1: Convert the counts to the annual death probability. The number dying during the year is $d_x=\ell_x-\ell_{x+1}=1{,}000-700=300$. Hence $q_x=d_x/\ell_x=300/1{,}000=0.30$, and $p_x=0.70$.

Step 2: Calculate the probability for the second half. The interval is $0.5<U\le1$. Using the conditional interval formula,

\begin{aligned} P(0.5<U\le1) &=\frac{0.70^{0.5}-0.70^1}{0.30}\\ &=\frac{0.836660-0.700000}{0.30}\\ &=\frac{0.136660}{0.30}\\ &=0.4555. \end{aligned}

Step 3: Find the conditional median. The force is $\mu=-\ln(0.70)=0.3566749$ per year. Thus

t_{0.5} =\frac{\ln(1-0.5(0.30))}{\ln(0.70)} =\frac{\ln(0.85)}{\ln(0.70)} =\frac{-0.1625189}{-0.3566749} =0.4557\text{ years}.

The second-half probability is 0.4555, and the median is 0.4557 years. As a check, the first-half conditional probability is $1-0.4555=0.5445$, which is greater than one-half, consistent with a median before 0.5 years.

Interpretation and Exam Traps

Under a constant force, the conditional distribution is a truncated exponential distribution, restricted to the first year and normalized by the probability of death during that year. It differs from a uniform distribution: the density $f_U(t)$ slopes downward, while under the uniform-deaths assumption the conditional death time is uniform on the year. The assumptions therefore give different answers to within-year timing questions even when they use the same $q_x$.

The constant-force formulas apply only when the force is constant over the interval being modeled. If mortality varies within the year, then $e^{-\mu t}$ with one fixed $\mu$ is not the appropriate survival expression. A force that is specified as constant in separate subintervals must instead be handled using the survival accumulated over each part; do not apply one annual constant force across a boundary where it changes.

Common exam trap: The conditional probability is not just the unconditional probability of death in the requested interval. It must be divided by $q_x$. A second trap is reversing the signs in the quantile formula: $-\ln(1-rq_x)/\mu$ is positive, whereas $-\ln(1-rq_x)/\ln(1-q_x)$ is negative. Since $\mu=-\ln(1-q_x)$, the correct equivalent expression is $\ln(1-rq_x)/\ln(1-q_x)$.

A useful final check is to evaluate the conditional distribution at its endpoints: $F_U(0)=0$ and $F_U(1)=(1-e^{-\mu})/q_x=1$. For any $0\le a<b\le1$, the interval probability must be between zero and one. These checks often reveal a missing conditioning denominator, a reversed interval, or an incorrect sign.

The methods here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In a calculation, state the annual $q_x$, the constant-force assumption, and whether the requested probability is conditional. Then use the conditional distribution consistently, keeping times in years and rounding only at the end.

Check Your Understanding

For each problem, assume a constant force during the stated age year and condition on death during that year unless stated otherwise. Round probabilities and times to four decimal places.

A hypothetical age year has $q_x=0.10$. Find $P(0.2<U\le0.6)$. (Answer: 0.4041)
A hypothetical age year has $q_x=0.40$. Find the conditional median time to death in years. (Answer: 0.4368 years)
A hypothetical life table has $\ell_x=800$ and $\ell_{x+1}=600$. Find the conditional probability that death occurs after 0.5 years. (Answer: 0.4641)
If $q_x=0.20$, find the conditional 25th percentile of death time in years. (Answer: 0.2299 years)
If $q_x=0.30$, find $P(0.2<U\le0.8)$. (Answer: 0.5980)

What You'll Learn