Increasing Benefits on a Term Contract

Why the Benefit’s Timing Matters

For a level term insurance benefit, the amount paid does not depend on when death occurs during the term. An increasing-benefit contract is different: a death later in the term produces a larger payment. Its value therefore depends on both the probability of death at each time and the benefit amount at that time. The essential step is to keep the benefit tied to the actual time of death throughout the valuation.

Consider a hypothetical contract covering a life aged $x$ for $n$ years. If death occurs at time $t$ after issue, where $0<t\le n$, suppose the benefit paid immediately is $B+ct$. Here $B$ is the initial benefit and $c$ is the annual rate of increase in benefit. The payment is made only if death occurs during the term; there is no maturity payment if the insured survives to time $n$.

Definition: A continuous $n$-year term insurance with a linearly increasing death benefit $B+ct$ pays $B+cT_x$ at the moment of death if $0<T_x\le n$, and pays zero otherwise. The parameter $c$ is measured in benefit currency per year. Unless stated otherwise, take $B\ge0$ and $c\ge0$.

Present Value of the Contract

Let $T_x$ be the future lifetime of a life aged $x$, and let the force of interest be constant at $\delta$. A payment at time $t$ has present-value factor $e^{-\delta t}$. If the future lifetime has a density, then the probability density of death at time $t$ is ${}_tp_x\mu_{x+t}$: the life must survive to $t$, then die in the next instant at force $\mu_{x+t}$.

\operatorname{APV} =E\!\left[e^{-\delta T_x}(B+cT_x)\mathbf{1}_{\{0<T_x\le n\}}\right] =\int_0^n (B+ct)e^{-\delta t}{}_tp_x\mu_{x+t}\,dt.

This expression applies to a death benefit payable immediately on death and with the stated time-dependent amount. It is not the value of a benefit paid at the end of the year of death, nor does it include a payment on survival to the end of the term. If the mortality model does not have a density, the expectation can instead be written as an integral with respect to the distribution of $T_x$.

The integral also explains the valuation method: at each possible death time, multiply the benefit at that time by its discount factor and the probability density of death there, then add these contributions over the covered interval. Linearity separates the constant part $B$ from the increasing part $ct$.

Constant Forces of Mortality and Interest

Suppose the force of mortality is a constant $\mu>0$ throughout the term, and the force of interest is a constant $\delta$. Then ${}_tp_x=e^{-\mu t}$ and $\mu_{x+t}=\mu$. Define $\lambda=\mu+\delta$. The term-insurance present value becomes

\operatorname{APV} =\mu\int_0^n (B+ct)e^{-\lambda t}\,dt.

For $\lambda\ne0$, evaluating the two integrals gives the working formula below. The formula allows a negative $\delta$, provided the interest model is valid; in that case the discount factor increases with time rather than decreases.

Result: Under constant mortality force $\mu$ and constant interest force $\delta$, with $\lambda=\mu+\delta\ne0$,

\operatorname{APV} =\mu\left[ B\frac{1-e^{-\lambda n}}{\lambda} +c\frac{1-e^{-\lambda n}(1+\lambda n)}{\lambda^2} \right].

When $\lambda=0$, use the limiting value

\operatorname{APV}=\mu\left(Bn+\frac{cn^2}{2}\right).

To justify the increasing-benefit term, integrate by parts. With $u=t$ and $dv=e^{-\lambda t}dt$, an antiderivative calculation over $0\le t\le n$ yields the required first moment of time. The constant-benefit term is the usual integral of an exponential.

\int_0^n t e^{-\lambda t}\,dt =\left[-\frac{t e^{-\lambda t}}{\lambda} -\frac{e^{-\lambda t}}{\lambda^2}\right]_0^n =\frac{1-e^{-\lambda n}(1+\lambda n)}{\lambda^2}.

If $\lambda=0$, the integrand $e^{-\lambda t}$ is 1. The integrals are then $\int_0^n 1\,dt=n$ and $\int_0^n t\,dt=n^2/2$, giving the stated limiting formula. This case can occur with a valid negative force of interest equal to $-\mu$, so it should not be dismissed solely because interest rates are usually positive.

Worked Example: Constant Forces Over a Five-Year Term

A hypothetical five-year term contract pays $100{,}000+10{,}000t$ dollars at death time $t$, measured in years since issue. Assume a constant mortality force $\mu=0.04$ and a constant interest force $\delta=0.06$, corresponding to effective annual interest $i=e^{0.06}-1\approx6.184\%$. Find the actuarial present value.

Method. Use the constant-force formula with $B=100{,}000$, $c=10{,}000$, $n=5$, and $\lambda=0.04+0.06=0.10$. First, $e^{-0.10(5)}=e^{-0.5}=0.60653066$.

The constant-benefit integral is $(1-0.60653066)/0.10=3.9346934$. The increasing-benefit integral is $[1-0.60653066(1+0.10(5))]/0.10^2 =[1-0.90979599]/0.01=9.020401$.

Substitute these values: $100{,}000(3.9346934)=393{,}469.34$, and $10{,}000(9.020401)=90{,}204.01$. Their sum is $483{,}673.35$; multiplying by $\mu=0.04$ gives $19{,}346.934$.

Answer: The APV is $19,346.93, rounded to the nearest cent. As a check, the maximum benefit is $100{,}000+10{,}000(5)=150{,}000$, and with nonnegative interest the APV must be no greater than that maximum.

Piecewise Mortality Forces

The same integral handles mortality that changes during the term. On an interval with constant mortality force $\mu_j$, survival from issue to time $t$ must still include the mortality experience in every earlier interval. If interest is constant at $\delta$, the combined exponential factor on that interval reflects both accumulated mortality and discounting. Splitting the valuation at every force-change time prevents an incorrect reset of survival.

Mark the intervals.
Write each interval over which the mortality force is constant, and identify the benefit $B+ct$ using time since issue.

Carry survival forward.
At each interval’s start, include the probability of surviving all earlier intervals.

Integrate and add.
Within each interval multiply benefit, discount, survival, and the current mortality force; integrate over that interval and sum.

Worked Example: A Mortality Force That Increases Midterm

A hypothetical four-year contract pays $50{,}000+5{,}000t$ dollars at death time $t$. The force of mortality is $0.10$ for $0\le t\le2$ and $0.20$ for $2<t\le4$. The force of interest is constant at $\delta=0.05$, so the effective annual rate is $e^{0.05}-1\approx5.127\%$. Find the APV.

Method. Split the integral at $t=2$. In the first interval, the combined exponent rate is $0.10+0.05=0.15$. Let $q_1=e^{-0.15(2)}=e^{-0.3}=0.74081822$. The two integrals are $(1-q_1)/0.15=1.7278785$ and $[1-q_1(1+0.15(2))]/0.15^2=1.6416139$.

The first-interval contribution is therefore $0.10[50{,}000(1.7278785)+5{,}000(1.6416139)]$. The bracket is $86{,}393.93+8{,}208.07=94{,}602.00$, so the contribution is $9{,}460.20$.

For the second interval, set $s=t-2$, so $0\le s\le2$. Survival to time 2 is $e^{-0.10(2)}=e^{-0.2}$; discounting to time 2 contributes $e^{-0.05(2)}=e^{-0.1}$. Their product is $e^{-0.3}$. The benefit becomes $60{,}000+5{,}000s$, and the combined rate within the interval is $0.20+0.05=0.25$.

Let $q_2=e^{-0.25(2)}=e^{-0.5}=0.60653066$. The integrals over $s$ are $(1-q_2)/0.25=1.5738774$ and $[1-q_2(1+0.25(2))]/0.25^2=1.4432642$. Thus the second contribution is $0.20e^{-0.3}[60{,}000(1.5738774)+5{,}000(1.4432642)]$. The bracket is $94{,}432.64+7{,}216.32=101{,}648.96$, giving $15{,}060.68$.

Adding the interval contributions gives $9{,}460.20+15{,}060.68=$$24,520.88, rounded to the nearest cent. The calculation retains the survival probability through the first two years instead of treating the second interval as a new contract.

Checks, Bounds, and Common Traps

A useful check is to compare an increasing-benefit contract with its maximum possible benefit, but the interest assumption matters. If $B,c\ge0$ and $\delta\ge0$, then $B+ct\le B+cn$ and $e^{-\delta t}\le1$. Consequently, $\operatorname{APV}\le(B+cn)P(T_x\le n)\le B+cn$.

Do not apply that bound without a nonnegative-interest assumption. More generally, the discount factor over the term is at most $\max_{0\le t\le n}e^{-\delta t}$, so $\operatorname{APV}\le(B+cn)\max_{0\le t\le n}e^{-\delta t}$. When $\delta<0$, this maximum is $e^{-\delta n}$, which exceeds 1. The APV can then exceed the maximum benefit because later payments have larger present values under negative interest.

A frequent exam error is to value every death using the end-of-term benefit. In Worked Example 1, that incorrect approach would use $150{,}000$ for every death time and produce $150{,}000(0.04)(3.9346934)=23{,}608.16$, rather than the correct $19{,}346.93$. The error replaces $B+ct$ by $B+cn$ inside the integral. Another common error is to use $c$ per month while $t$ is measured in years; the units of $c$ and $t$ must agree.

Key takeaway: For a continuously paid increasing term benefit, evaluate the benefit at the actual death time: integrate $(B+ct)$ against discounting and the death-time density only over the insured term. Under constant forces, use $\lambda=\mu+\delta$; under changing mortality, split the integral and carry survival through each interval. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, treat the benefit as payable immediately at death during the stated term, with no survival payment. Round dollar answers to the nearest cent.

A four-year contract pays $40{,}000+5{,}000t$ dollars. Assume constant mortality force $\mu=0.05$ and interest force $\delta=0.03$. Find the APV. (Answer: $8,466.71)
A two-year contract pays $8{,}000t$ dollars. Assume constant mortality force $\mu=0.10$ and interest force $\delta=0$. Find the APV. (Answer: $1,401.85)
A five-year contract pays $25{,}000+2{,}000t$ dollars under constant forces $\mu=0.06$ and $\delta=0.04$. Find the APV. (Answer: $6,984.49)
A two-year contract pays $10{,}000+2{,}000t$ dollars. The mortality force is $0.10$ in the first year and $0.20$ in the second year; the interest force is zero. Find the APV, carrying survival through the first year. (Answer: $3,171.99)
A two-year contract pays $100+10t$ dollars. Assume a constant mortality force $\mu=0.10$ and interest force $\delta=-0.10$, so $\lambda=0$. Find the APV. (Answer: $22.00)

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