Continuous Death Benefits Versus Year-End Benefits

Why the Payment Date Matters

A death benefit can be triggered by the same death but paid at different times. One contract pays at the moment of death; another pays at the end of the policy year in which death occurs. The benefit amount and mortality experience may be identical, yet the present values differ because the payments are discounted for different lengths of time.

To compare the contracts fairly, hold the death benefit and the coverage period fixed. Then distinguish the time of death from the payment date: a continuous benefit is discounted to the actual time of death, while a year-end benefit is discounted to the end of the death year. This timing distinction is especially important when the benefit amount changes from one policy year to the next.

Definition: Let $T_x$ be the future lifetime of a life aged $x$. A continuous death benefit is paid at time $T_x$. A year-end death benefit is paid at time $k+1$ if death occurs during policy year $k+1$, meaning $k<T_x\leq k+1$, for $k=0,1,2,\ldots$. If the benefit for that policy year is $B_k$, both contracts pay $B_k$; only the payment time differs.

Write $i$ for the effective annual interest rate, $v=1/(1+i)$, and $\delta=\ln(1+i)$ for the force of interest. Thus $v^t=e^{-\delta t}$ is the discount factor for a payment at time $t$. For the continuous formulas below, assume the future lifetime has an absolutely continuous distribution, with density $f_{T_x}(t)={}_tp_x\mu_{x+t}$. Here ${}_tp_x$ is the probability of surviving $t$ years and $\mu_{x+t}$ is the force of mortality at duration $t$. Under this assumption, death at an exact integer duration has probability zero.

The Two Present-Value Formulas

For an $n$-year term contract, the continuous benefit for deaths in policy year $k+1$ contributes the integral of the discounted benefit over that year. The year-end benefit instead uses the probability of dying in that year and discounts the entire amount to time $k+1$.

\operatorname{EPV}_{\mathrm{continuous}} = \sum_{k=0}^{n-1} B_k \int_k^{k+1} e^{-\delta t}\,{}_tp_x\mu_{x+t}\,dt

\operatorname{EPV}_{\mathrm{year\mbox{-}end}} = \sum_{k=0}^{n-1} B_k v^{k+1}\,{}_kp_x\,{}_1q_{x+k}

The annual death probability in the second formula is conditional on being alive at duration $k$: ${}_1q_{x+k}$ is the probability of dying within the following year. Multiplying by ${}_kp_x$ gives the unconditional probability that the insured survives to duration $k$ and then dies in that year. If $B_k$ is constant within the year, it can be taken outside the continuous integral. If the benefit varies continuously with time within a policy year, it must remain inside the integral as a time-dependent benefit.

For whole-life coverage, the same reasoning applies, with the sum extending over all policy years, provided the expected present value exists. For a term contract, do not include deaths after the term or a maturity payment unless the contract specifically provides one.

Why Immediate Payment Has a Larger Value

Suppose the two contracts pay the same nonnegative benefit for every death, interest is nonnegative, and death times have a continuous distribution. For a death in year $k+1$, the continuous payment occurs at $t\leq k+1$. Since $\delta\geq 0$, discounting to $t$ gives a factor at least as large as discounting to $k+1$.

Fix a possible time of death.
If $k<t\leq k+1$, both contracts pay $B_k$ for that death.

Compare the discount factors.
Because $\delta\geq 0$ and $t\leq k+1$, $e^{-\delta t}\geq e^{-\delta(k+1)}=v^{k+1}$.

Average over deaths.
Multiplying by the same nonnegative benefit and integrating over each death year preserves the inequality. Summing over covered years shows that the continuous benefit has at least as large an expected present value.

The difference for a particular death time is $B_k(e^{-\delta t}-e^{-\delta(k+1)})$. It is positive when $B_k>0$, $\delta>0$, and $t<k+1$. The expected values are equal if interest is zero and the benefits match, but they need not be equal if the benefit amounts differ. With a negative force of interest, the discount-factor comparison reverses, so the continuous payment need not have the larger value.

Theorem: Under a common mortality model and identical nonnegative benefits by policy year, if $\delta\geq0$, the expected present value of payment at death is at least the expected present value of payment at the end of the year of death. This comparison concerns payment timing only; it does not apply as stated when the benefit amounts or covered deaths differ.

Worked Comparisons

Worked Example: A Two-Year Level Benefit

A hypothetical two-year term policy pays $1,000 on death. Compare payment at the moment of death with payment at the end of the death year. The effective annual interest rate is $i=100\%$, so $v=1/2$ and $\delta=\ln 2$. Assume a constant force of mortality $\mu_{x+t}=\ln 2$ throughout the two years. Thus ${}_tp_x=e^{-t\ln 2}=2^{-t}$. Find both expected present values and check their ordering.

Method and setup. In each policy year, the force of mortality and the force of interest are both $\ln 2$. Therefore the continuous contribution for a $1,000 benefit in year $k+1$ is $1000\int_k^{k+1}e^{-2t\ln 2}\ln 2\,dt$. For the year-end contract, survival to duration $k$ is $2^{-k}$, and the conditional probability of death during the next year is $1-1/2=1/2$.

Continuous, year 1. The contribution is $1000\cdot\frac{1}{2}(1-2^{-2})=1000\cdot\frac{3}{8}=\$375.00$.

Continuous, year 2. The year-1 survival discount factor in the integral is $2^{-2}=1/4$, so the contribution is $1000\cdot\frac{1}{4}\cdot\frac{3}{8}=\$93.75$. The total is $\$375.00+\$93.75=\$468.75$.

Year-end, year 1. The death probability is $1/2$, and the payment discount factor is $v=1/2$. The contribution is $1000\cdot\frac{1}{2}\cdot\frac{1}{2}=\$250.00$.

Year-end, year 2. The probability of surviving year 1 and dying in year 2 is $(1/2)(1/2)=1/4$. The payment discount factor is $v^2=1/4$, giving $1000\cdot\frac{1}{4}\cdot\frac{1}{4}=\$62.50$. The total is $\$250.00+\$62.50=\$312.50$.

Check. The continuous value exceeds the year-end value by $\$468.75-\$312.50=\$156.25$, as expected under positive interest. The common error avoided here is discounting both benefits to the same date: the continuous benefit is discounted to the actual death time, not automatically to the end of the year.

Worked Example: Benefit Amounts That Change by Year

A hypothetical two-year policy pays $800 for a death in year 1 and $1,400 for a death in year 2. Compare continuous payment with year-end payment. The effective annual interest rate is $100\%$, and the force of mortality is $\ln 2$ at every duration. Find both expected present values.

Method. Use the same year-specific timing factors as in the previous calculation, but multiply each year’s factor by its own benefit. For a benefit $B_k$, the continuous factors are $3/8$ in year 1 and $(1/4)(3/8)=3/32$ in year 2. The year-end factors are $1/4$ in year 1 and $1/16$ in year 2.

Continuous value. The year-1 contribution is $800(3/8)=\$300.00$. The year-2 contribution is $1,400(3/32)=\$131.25$. Thus the total is $\$300.00+\$131.25=\$431.25$.

Year-end value. The year-1 contribution is $800(1/4)=\$200.00$. The year-2 contribution is $1,400(1/16)=\$87.50$. Thus the total is $\$200.00+\$87.50=\$287.50$.

Check. The difference is $\$431.25-\$287.50=\$143.75$, which is positive. The common error avoided is applying one benefit amount to both years or using the year-1 payment date for the year-2 benefit.

Worked Example: Finding an Equivalent Level Year-End Benefit

A hypothetical two-year term policy pays $1,000 continuously at death in year 1 and $600 continuously at death in year 2. Find the level benefit $C$ paid at the end of the year of death that has the same expected present value. The effective annual interest rate is $100\%$, and the force of mortality is $\ln 2$ throughout the term.

Method. First value the specified continuous benefits. Then calculate the present-value factor for a level year-end benefit and divide the continuous value by that factor.

Continuous value. The year-1 contribution is $1,000(3/8)=\$375.00$. The year-2 contribution is $600(3/32)=\$56.25$. The total is $\$375.00+\$56.25=\$431.25$.

Level year-end factor. The year-1 contribution per dollar of benefit is $(1/2)(1/2)=1/4$. The year-2 contribution per dollar is $(1/4)(1/4)=1/16$. The total factor is $1/4+1/16=5/16=0.3125$.

Solve and check. Equating expected present values gives $0.3125C=\$431.25$, so $C=\$431.25/0.3125=\$1,380.00$. Substitution gives $0.3125(\$1,380.00)=\$431.25$, as required. This equivalent year-end amount is greater than either annual benefit in this example because it must match the value of a schedule that pays sooner, on average.

Common Timing Traps

The most reliable way to avoid errors is to identify the payment time before applying a discount factor. A death in policy year $k+1$ does not mean the benefit is paid at duration $k$. It is paid at duration $T_x$ under continuous insurance and at duration $k+1$ under year-end insurance.

Using $v^k$ for a year-end payment. The year-end payment is made at $k+1$, so its factor is $v^{k+1}$, not $v^k$.
Using an annual death probability in the continuous formula. A continuous value requires integration over the death-time density. Annual probabilities alone do not identify when within the year death occurs.
Ignoring survival to the policy year. The unconditional probability of death in year $k+1$ is ${}_kp_x{}_1q_{x+k}$, not just ${}_1q_{x+k}$.
Assuming equal benefit amounts when they change by year. Apply $B_k$ separately in each year before comparing the totals.
Comparing unlike coverage. Both values must use the same term and the same death-benefit schedule. A maturity benefit or a different covered period changes the contract being valued.

Key takeaway: For the same death and benefit amount, payment at death uses the actual-time discount factor $e^{-\delta T_x}$; payment at the end of the death year uses $v^{k+1}$. Under nonnegative interest, the earlier payment has at least as large an expected present value.

Check Your Understanding

For each problem, compare the payment dates carefully and use the stated mortality and interest assumptions.

A hypothetical one-year policy pays $2,000 on death. The force of interest and the constant force of mortality are both $\ln 2$. Find the expected present values for payment continuously and at year-end. (Answer: continuous $750.00; year-end $500.00.)
A two-year policy pays $600 for death in year 1 and $1,600 for death in year 2. The effective annual interest rate is $100\%$, and the force of mortality is constantly $\ln 2$. Find both expected present values. (Answer: continuous $375.00; year-end $250.00.)
A three-year policy pays $1,000 on death, and the force of mortality is constantly $\ln 2$. If the effective annual interest rate is zero, find both expected present values. (Answer: continuous $875.00; year-end $875.00.)
A one-year continuous death benefit is $1,200. The force of interest and the constant force of mortality are both $\ln 2$. What level benefit paid at year-end has the same expected present value? (Answer: $1,800.00.)
For a two-year policy, the benefit is $1,000 in either year. The effective annual interest rate is $100\%$, and the constant force of mortality is $\ln 2$. Find the difference between the continuous and year-end expected present values. (Answer: $156.25.)

What You'll Learn