Align Curtate and Complete Lifetime with Benefit Timing

Why Benefit Timing Determines the Lifetime Variable

A life-contingent payment depends not only on whether death occurs, but also on when the contract pays. If a benefit is paid at the end of the policy year of death, annual death probabilities naturally describe the event. If it is paid at the instant of death, the exact time to death is needed. Using the right mortality data with the wrong time convention can still produce the wrong probability or present value.

Let $T_x$ denote the complete future lifetime of a life aged $x$: the exact, possibly fractional, number of years from now until death. Let $K_x=\lfloor T_x\rfloor$ denote the curtate future lifetime, the number of complete years lived. Thus, if death occurs 2.4 years from now, $T_x=2.4$ and $K_x=2$. For a continuous lifetime distribution, the probability of death at an exact integer duration is zero; this convention avoids boundary complications.

Definition: The curtate future lifetime $K_x$ records the number of whole years lived, while the complete future lifetime $T_x$ records the exact time lived. For integer $k\geq 0$, $K_x=k$ corresponds to death during the interval $(k,k+1]$. A benefit paid at the end of that policy year is paid at time $k+1$, not at time $k$.

Match the Probability to the Payment Date

Write ${}_tp_x=P(T_x>t)$ for the probability that a life aged $x$ survives more than $t$ years, and ${}_tq_x=1-{}_tp_x$ for the probability of death by time $t$. The one-year death probability at age $x+k$ is $q_{x+k}$, and $p_{x+k}=1-q_{x+k}$.

An annual life table gives the probability of death during each policy year, not the exact death time within that year. The probability that the curtate lifetime equals $k$ is the probability of surviving $k$ years and then dying during the next year. A year-end payment uses the discount period $k+1$, because payment follows the year of death.

P(K_x=k)={}_kp_xq_{x+k} \qquad\text{and}\qquad \text{payment time}=k+1.

For a payment at the exact time of death, $T_x$ is the relevant variable. For example, an event described as death within 1.5 years is $T_x\leq 1.5$, and survival beyond 1.5 years is $T_x>1.5$. An annual life table alone does not determine such fractional-year probabilities unless an assumption about mortality within each year is supplied.

Key distinction: $K_x$ identifies the policy year of death; $T_x$ identifies the time of death. Use annual death probabilities for year-end payments. Use a complete-lifetime probability model when the event or payment time is fractional or continuous.

Why the Year-End Formula Works

The annual formula follows by conditioning on survival to the start of the year in which death occurs. To have $K_x=k$, the life must first survive $k$ complete years, an event with probability ${}_kp_x$. Given survival to age $x+k$, the probability of dying in the next year is $q_{x+k}$. Multiplying these conditional stages gives the probability of death in that year.

\begin{aligned} P(K_x=k) &=P(\text{survive }k\text{ years})\, P(\text{die in the next year}\mid\text{survive }k\text{ years})\\ &={}_kp_xq_{x+k}. \end{aligned}

The same event can be written as $k<T_x\leq k+1$. Its probability is ${}_kp_x-{}_{k+1}p_x$, which equals ${}_kp_xq_{x+k}$ because ${}_{k+1}p_x={}_kp_xp_{x+k}$. This equivalence is a useful check on an annual calculation.

Worked Example: A Benefit Paid at the End of the Second Year

A hypothetical policy pays $10,000 at the end of the policy year of death, provided death occurs during the second year. For a life aged $x$, suppose $q_x=0.10$ and $q_{x+1}=0.20$. The effective annual interest rate is $i=5\%$. Find the probability of payment and the expected present value of this second-year benefit. Use the annual, discrete mortality model given by the two one-year probabilities.

Method. A second-year death means $K_x=1$, equivalently $1<T_x\leq2$. First calculate survival through year one: ${}_1p_x=1-q_x=1-0.10=0.90$.

Conditional on surviving the first year, the probability of death during the second year is $q_{x+1}=0.20$. Therefore, $P(K_x=1)={}_1p_xq_{x+1}=0.90(0.20)=0.18$.

The benefit is paid at time $2$, not at time $1$. The discount factor is $v^2=(1.05)^{-2}=0.907029$. The expected present value of the benefit is $$10{,}000)(0.18)(0.907029)=$1{,}632.65$, rounded to the nearest cent.

Check. The probability of death in the second year is also ${}_1p_x-{}_2p_x=0.90-(0.90)(0.80)=0.18$. The calculation avoids the common error of discounting this year-end payment for only one year.

Fractional Durations Require a Mortality Assumption

An annual probability $q_{x+k}$ specifies the chance of death over a whole year, but not how deaths are distributed within that year. One common assumption for fractional-age calculations is the uniform distribution of deaths (UDD): conditional on being alive at the start of a year, deaths are spread uniformly over that year. Under UDD, for $0\leq r\leq1$,

{}_rp_{x+k}=1-rq_{x+k}, \qquad {}_rq_{x+k}=rq_{x+k}.

These relations apply within the specified policy year under UDD; they are not implied by annual data alone. To calculate survival or death probabilities across more than one year, multiply the survival probabilities for the successive portions of each year. This keeps the fractional duration attached to the correct age interval.

Worked Example: A Death Event by a Fractional Time

A hypothetical contract pays $4,000 immediately at death if death occurs within 1.4 years. Assume UDD within each policy year, with $q_x=0.30$ and $q_{x+1}=0.20$. The effective annual interest rate is $i=0\%$, so the requested expected payment is undiscounted. Find the probability of payment and its expected amount.

Method. The payment event is $T_x\leq1.4$, not simply a condition on the curtate lifetime. Death within the first year has probability $q_x=0.30$. For death between years one and 1.4, first survive year one, then die in the first 0.4 of the next year. By UDD, that probability is ${}_1p_x(0.4q_{x+1})=(0.70)(0.4)(0.20)=0.056$.

The two disjoint intervals give $P(T_x\leq1.4)=0.30+0.056=0.356$. The expected undiscounted payment is $$4{,}000(0.356)=$1{,}424$.

Check. The same survival probability is ${}_{1.4}p_x=(1-q_x)(1-0.4q_{x+1})=(0.70)(0.92)=0.644$; therefore, ${}_{1.4}q_x=1-0.644=0.356$. The calculation avoids treating all deaths during year two as though they occurred before 1.4 years.

Continuous Timing and Complete Lifetime

When a benefit is paid at the instant of death, its payment time is $T_x$. A continuous mortality model can supply the required distribution. The force of mortality $\mu_{x+t}$ describes the instantaneous death rate at duration $t$, conditional on survival to that duration. If the force is constant at $\mu$ over an interval, survival over $s$ years within that interval is $e^{-\mu s}$. For a payment made at time $t$, discount using $v^t=(1+i)^{-t}=e^{-\delta t}$, where $\delta=\ln(1+i)$ is the force of interest corresponding to effective annual rate $i$.

The distinction is not merely terminology. A probability of death during a year can be used to identify a year-end payment, but it cannot locate the payment within that year. With a continuous model, death density and discounting are both evaluated at the actual time of death.

Worked Example: Discounting a Payment at Death in the Second Year

A hypothetical policy pays $1,000 immediately at death if death occurs between times 1 and 2. Assume the force of mortality is constant at $\mu=0.10$ during the first year and $\mu=0.20$ during the second year. The force of interest is $\delta=0.05$, equivalent to effective annual rate $i=e^{0.05}-1=5.127\%$ (to three decimal places). Find the probability of a second-year death and the expected present value of the payment. Mortality is piecewise constant-force, and interest is continuously compounded at the stated force.

Method. Survival to time 1 is $e^{-0.10}=0.904837$. Conditional on survival to time 1, the probability of death during the next year is $1-e^{-0.20}=0.181269$. Thus, $P(1<T_x\leq2)=e^{-0.10}(1-e^{-0.20})=0.164019$, or 0.1640 to four decimals.

For a death at time $1+s$, where $0\leq s\leq1$, the probability density of death at that time, including survival through the first year, is $e^{-0.10}(0.20)e^{-0.20s}$. The discount factor at that time is $e^{-0.05(1+s)}$. Therefore, the expected present value is

\begin{aligned} 1000\int_0^1 e^{-0.10}(0.20)e^{-0.20s}e^{-0.05(1+s)}\,ds &=1000e^{-0.10}e^{-0.05}(0.20) \frac{1-e^{-0.25}}{0.25}\\ &=1000(0.904837)(0.951229)(0.20) \frac{1-0.778801}{0.25}\\ &=\$152.31. \end{aligned}

The intermediate integral uses $\int_0^1e^{-0.25s}\,ds=(1-e^{-0.25})/0.25$. The probability check is consistent with the same model: the density integrated without discounting gives $e^{-0.10}(1-e^{-0.20})=0.164019$. The calculation avoids assigning every second-year death a payment time of 2.

Common Timing Errors

The most reliable check is to translate the wording into a time event before calculating. “Death in the third policy year” means $2<T_x\leq3$, or $K_x=2$; “death by time 2.4” means $T_x\leq2.4$. These are not interchangeable events.

Discounting from the wrong date. If $K_x=k$ and payment is at year-end, discount for $k+1$ years. The lifetime count excludes the year of death, but the payment occurs at its end.
Reading $K_x\geq k$ as $K_x>k$. For integer $k$, $K_x\geq k$ means survival to time $k$, so its probability is ${}_kp_x$. In contrast, $K_x>k$ means surviving beyond year $k$, with probability ${}_{k+1}p_x$, under a continuous lifetime model.
Using annual mortality data for a fractional event without an assumption. The annual probability alone does not specify the probability of death in the first 0.4 of a year. State and apply a fractional-age assumption, such as UDD, or use a specified continuous mortality model.
Applying the fractional duration to the wrong age year. For a duration of 1.4 years, the first full year uses mortality at age $x$, and the additional 0.4 uses mortality during the year beginning at age $x+1$.

For instance, under the UDD assumptions in the fractional-time example, $P(T_x>1.4)=0.644$. Replacing that with ${}_2p_x=(0.70)(0.80)=0.56$ would incorrectly require survival through the entire second year. The latter is survival beyond two years, not beyond 1.4 years.

Translate the benefit wording.
Identify whether the event is a policy-year death, a death by a stated time, or an exact-time payment.

Select the lifetime variable.
Use $K_x$ for the year of death and $T_x$ for a fractional or continuous time.

Check the mortality model.
Use annual probabilities for whole policy years; state the fractional-year assumption or continuous force when needed.

Place the payment date before discounting.
A year-end benefit after death in year $k+1$ is discounted for $k+1$ years; an immediate benefit at death is discounted for $T_x$.

This alignment is essential in multi-step life-contract calculations. A correct mortality probability paired with an incorrect payment time gives an inconsistent value. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus: identify the lifetime event first, then match the probability model and payment timing to it.

Check Your Understanding

For each problem, identify the time convention before calculating.

A hypothetical policy pays at the end of the year of death. Given $q_x=0.12$ and $q_{x+1}=0.25$, find the probability that payment occurs at time 2. (Answer: 0.2200)
Assume UDD, $q_x=0.30$, and $q_{x+1}=0.20$. Find the probability of death by time 1.4. (Answer: 0.3560)
A year-end benefit is payable for death during the third policy year. Given ${}_2p_x=0.84$ and ${}_3p_x=0.756$, find the probability of payment. State the payment time. (Answer: 0.0840; time 3)
Assume a constant force of mortality $\mu=0.08$ during the first year and $\mu=0.12$ during the second year. Find the probability of death between times 1 and 1.5. (Answer: $e^{-0.08}(1-e^{-0.06})=0.0538$, to four decimals)
Under UDD, $q_x=0.10$ and $q_{x+1}=0.20$. A hypothetical $2,000 benefit is paid immediately if death occurs between times 1 and 1.5. At $i=0\%$, find its expected payment. (Answer: $180.00)

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