Compare Continuous and Discrete Insurance on One Mortality Basis

Why Payment Timing Changes Insurance Values

A death benefit can be paid at the moment of death or at the end of the policy year in which death occurs. These contracts can have the same benefit amount, mortality basis, and interest basis, yet different expected present values because the payment dates differ. A payment made earlier has a larger present value.

To isolate this timing effect, use the same mortality model and interest assumption for both contracts. This tutorial compares a continuous benefit, paid at death, with a discrete benefit, paid at the end of the year of death. The formulas also provide a direct way to check whether a proposed value is consistent with its payment timing.

Definition: A continuous death benefit is paid at the time of death $T_x$. A year-end death benefit is paid at time $K_x+1$, where $K_x$ is the number of complete years lived after age $x$. For a term of $n$ years, each benefit is payable only if death occurs within those $n$ years. In this tutorial, $A_{x:\overline{n}|}$ denotes the $n$-year term-insurance EPV with payment at the end of the year of death, and $\bar{A}_{x:\overline{n}|}$ denotes the corresponding EPV with payment at the moment of death.

Valuation Formulas on a Shared Mortality Basis

Let $T_x$ be the complete future lifetime of a life aged $x$. In general, the continuous term-insurance EPV for a unit benefit is the expected discounted benefit, with discounting at force of interest $\delta$. If ${}_tp_x$ is the probability of survival for $t$ years and $\mu_{x+t}$ is the force of mortality at duration $t$, the density of death at time $t$ is ${}_tp_x\mu_{x+t}$. Thus

\bar{A}_{x:\overline{n}|} =\int_0^n e^{-\delta t}{}_tp_x\mu_{x+t}\,dt.

For a discrete benefit, death during policy year $k+1$ has probability ${}_kp_xq_{x+k}$, and the benefit is discounted for $k+1$ years. With effective annual interest rate $i$ and $v=1/(1+i)$, this gives

A_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}.

The two expressions use different benefit times. In the continuous formula, discounting is for the actual time $t$ of death. In the discrete formula, every death in policy year $k+1$ is discounted for the full $k+1$ years. Do not replace one timing convention with the other while keeping the same expression.

Constant-force model: If the force of mortality is the constant $\mu$ throughout the term, then ${}_tp_x=e^{-\mu t}$, and the probability of surviving one year is $p=e^{-\mu}$, with $q=1-p$. If interest is also expressed as a constant force $\delta$, then the equivalent effective annual rate is $i=e^\delta-1$, so $v=e^{-\delta}$. Under these assumptions, $$ \bar{A}_{x:\overline{n}|} =\frac{\mu}{\mu+\delta}\left(1-e^{-(\mu+\delta)n}\right), \qquad A_{x:\overline{n}|} =\sum_{k=0}^{n-1}e^{-\delta(k+1)}e^{-\mu k}(1-e^{-\mu}). $$ These constant-force formulas do not apply unchanged when the mortality force varies with time or age.

The continuous formula follows directly from the death-time density: at time $t$, the probability density of death is ${}_tp_x\mu$, and the present value of a unit benefit is $e^{-\delta t}$. Their product is integrated over all payable death times. Under constant forces, the integrand becomes $\mu e^{-(\mu+\delta)t}$, whose integral from $0$ to $n$ is the stated expression.

For the discrete expression, death in year $k+1$ requires survival for $k$ years and death during the following year, with probability $e^{-\mu k}(1-e^{-\mu})$. The benefit is then discounted by $e^{-\delta(k+1)}$. Summing these mutually exclusive possibilities gives the term-insurance EPV.

Worked Comparisons

Worked Example: Two-Year Term Under Equal Forces

A hypothetical two-year term policy pays $10,000 on death. Compare its continuous and year-end EPVs. Assume a constant force of mortality $\mu=\ln(1.25)$ and an effective annual interest rate $i=25\%$. State the payment timing in each calculation.

Method. Convert the annual interest rate to a force: $\delta=\ln(1+i)=\ln(1.25)$. The annual survival probability is $p=e^{-\mu}=0.8$, the annual death probability is $q=0.2$, and the discount factor is $v=1/1.25=0.8$.

For continuous payment, $\mu=\delta$, so $\mu/(\mu+\delta)=1/2$. Also, $e^{-(\mu+\delta)2}=e^{-4\ln(1.25)}=(0.8)^4=0.4096$. Therefore the unit-benefit EPV is

\bar{A}_{x:\overline{2}|} =\frac12(1-0.4096) =0.2952.

The EPV of the $10,000 benefit is $10{,}000(0.2952)=\$2{,}952.00$.

For year-end payment, the first-year death contribution is $vq=(0.8)(0.2)=0.16$. The second-year contribution is $v^2pq=(0.8)^2(0.8)(0.2)=0.1024$. Thus

A_{x:\overline{2}|}=0.16+0.1024=0.2624, \qquad 10{,}000(0.2624)=\$2{,}624.00.

The continuous value exceeds the year-end value by $\$2{,}952-\$2{,}624=\$328$. This difference is consistent with earlier payment under the continuous contract. Common mistake avoided: discounting every continuous death at the end of the policy year; the continuous calculation discounts to the actual time of death.

Worked Example: Two-Year Term With Different Benefits by Year

A hypothetical policy pays $6,000 for death in the first year and $9,000 for death in the second year. Calculate the EPV under continuous payment at death and under payment at the end of the year of death. Assume constant force of mortality $\mu=\ln(1.25)$ and effective annual interest $i=25\%$, so $\delta=\ln(1.25)$, $p=v=0.8$, and $q=0.2$.

Method. For continuous payment, integrate over each policy-year interval, using the benefit applicable to that interval. The integrated unit-benefit factors are one-half of the reduction in survival-and-discount probability over the interval. For year one, $1-e^{-2\mu}=1-(0.8)^2=0.36$, giving a factor of $0.5(0.36)=0.18$. For year two, the interval difference is $e^{-2\mu}-e^{-4\mu}=0.64-0.4096=0.2304$, giving a factor of $0.5(0.2304)=0.1152$.

The continuous EPV is therefore

6{,}000(0.18)+9{,}000(0.1152) =1{,}080+1{,}036.80 =\$2{,}116.80.

For year-end payment, the first-year probability-weighted discount factor is $vq=(0.8)(0.2)=0.16$. For year two it is $v^2pq=(0.8)^2(0.8)(0.2)=0.1024$. The EPV is

6{,}000(0.16)+9{,}000(0.1024) =960+921.60 =\$1{,}881.60.

The continuous value is $\$235.20$ greater. Common mistake avoided: applying the second-year benefit to all deaths, or assigning the first-year benefit to deaths in both years. Each benefit must be paired with the probability and payment timing for its own interval.

Worked Example: Whole-Life Values and the Geometric Sum

A hypothetical whole-life policy pays $15,000 on death. Assume a constant force of mortality $\mu=\ln(1.25)$ and effective annual interest $i=10\%$, so $\delta=\ln(1.10)$, $p=0.8$, $q=0.2$, and $v=10/11$. Find the continuous and year-end EPVs.

Continuous method. Let $\delta=\ln(1.10)$. Integrating over all possible death times gives the unit-benefit value

\bar{A}_x =\int_0^\infty \mu e^{-(\mu+\delta)t}\,dt =\frac{\mu}{\mu+\delta} =\frac{\ln(1.25)}{\ln(1.25)+\ln(1.10)} \approx 0.7007095.

The integral converges because $\mu+\delta>0$. The dollar EPV is $15{,}000(0.7007095)$, or $\$10{,}510.64$ to the nearest cent.

Year-end method. The first year's unit contribution is $vq=(10/11)(1/5)=2/11$. Each successive contribution is the preceding one multiplied by $vp=(10/11)(4/5)=8/11$. Therefore the infinite series is geometric:

A_x=\frac{2/11}{1-8/11}=\frac23.

The year-end dollar EPV is $15{,}000(2/3)=\$10{,}000.00$. The continuous value is $\$510.64$ greater. Common mistake avoided: using the year-end geometric series for the continuous benefit. The series values payments at integer times; continuous payment requires integrating over death times.

Comparing the Results Without Mixing Assumptions

A reliable comparison proceeds in a fixed order: identify the benefit amount and term, specify the mortality model, specify the interest basis, and then apply the formula for the stated payment time. If the force of mortality is constant, convert it to annual probabilities when using year-end payments. If the effective annual interest rate is given, use $v=1/(1+i)$ for those payments and $\delta=\ln(1+i)$ for continuous discounting on the equivalent constant-force interest basis.

The two values are not generally related by a universal fixed factor. Their difference depends on mortality, interest, term, and benefit pattern. A death near the start of a policy year produces a larger timing difference than one near its end, while a higher interest rate increases the cost of postponing payment. If the benefit varies by year, each interval or year must be valued with its own amount.

Specify the model.
State the mortality assumption and the effective annual interest rate or force of interest. Do not silently change either basis between the two values.

Match event probabilities to payment times.
For continuous insurance, use death-time density and discount to the actual death time. For year-end insurance, use ${}_kp_xq_{x+k}$ and discount for $k+1$ years.

Apply the benefit pattern and check.
Multiply each probability-weighted discount factor by the benefit payable for that time or year. Confirm that earlier payment has not been inadvertently discounted for longer.

A frequent exam trap is to multiply the annual death probability $q$ by $v^k$ rather than $v^{k+1}$ for a death in policy year $k+1$. The correct year-end present value includes $k+1$ years of discounting: $v^{k+1}{}_kp_xq_{x+k}$. Another trap is to use $q=1-e^{-\mu}$ as though it were a continuous death density; $q$ is a one-year probability, while the continuous density under constant force is $\mu e^{-\mu t}$.

Key takeaway: Under identical mortality and interest assumptions, continuous and year-end insurance differ only in the timing convention for the benefit, but that timing changes the EPV. State the event interval, payment date, and discount period explicitly before calculating. These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, keep the mortality basis, interest basis, and payment timing distinct.

A one-year term pays $5,000 continuously at death. Assume $\mu=\ln(1.25)$ and effective annual interest $i=10\%$. Calculate its EPV. (Answer: $\$955.51$)
A two-year term pays $7,000 for death in year one and $11,000 for death in year two, at the end of the year of death. Assume $\mu=\ln(1.25)$ and $i=10\%$. Calculate its EPV. (Answer: $\$2{,}727.27$)
A two-year term pays $12,000 continuously at death. Assume constant $\mu=\delta=\ln(1.25)$. Calculate its EPV. (Answer: $\$3{,}542.40$)
A whole-life policy pays $20,000 at the end of the year of death. Assume annual survival probability $p=0.9$ and effective annual interest $i=10\%$. Calculate its EPV. (Answer: $\$10{,}000.00$)
A three-year term pays $10,000 on death. Assume $\mu=\delta=\ln(1.25)$. Calculate both the continuous and year-end EPVs, then find their difference. (Answer: Continuous $\$3{,}689.28$; year-end $\$3{,}279.36$; difference $\$409.92$)

What You'll Learn