Whole-Life Values as Limits of Term Values

Why a Term Value Approaches a Whole-Life Value

A continuous term insurance pays a benefit if death occurs during a stated term. A whole-life insurance pays at death whenever death occurs. If the term is lengthened while the benefit and other assumptions stay fixed, more possible death times are covered. The term value therefore approaches the whole-life value, but the difference between them is not generally negligible at a finite duration.

That difference is the value of the benefit for death after the term. It depends on two things: the probability of surviving to the end of the term and the value, at that time, of the remaining whole-life cover. Making that tail explicit is useful for checking calculations and for deciding whether a term duration is long enough for a particular approximation.

Definition: Suppose a life aged \(x\) has future lifetime \(T_x\), and a unit benefit is paid immediately at death. With a constant force of interest \(\delta\), the continuous whole-life insurance value is \(\bar A_x=E[e^{-\delta T_x}]\). The \(n\)-year continuous term insurance value, denoted \(\bar A^1_{x:\overline{n}|}\), is \(E[e^{-\delta T_x}\mathbf{1}(T_x\le n)]\). The value omitted by stopping cover at time \(n\) is \(E[e^{-\delta T_x}\mathbf{1}(T_x>n)]\).

Here the benefit amount is one dollar; for a level benefit of \(B\) dollars, multiply each value by \(B\). The formulas below assume a continuous lifetime model with an integrable death-time density and a finite whole-life value. Unless stated otherwise, interest is nonnegative. The same split between term and tail holds whenever the relevant expected present values are finite.

Write \({}_tp_x=P(T_x>t)\) for the probability of surviving \(t\) years, and \(\mu_{x+t}\) for the force of mortality at duration \(t\). The density of death at duration \(t\) is \({}_tp_x\mu_{x+t}\), so the two values can also be written as integrals.

\bar A^1_{x:\overline{n}|} =\int_0^n e^{-\delta t}{}_tp_x\mu_{x+t}\,dt, \qquad \bar A_x =\int_0^\infty e^{-\delta t}{}_tp_x\mu_{x+t}\,dt.

The Tail Identity and the Limit

Split the whole-life integral at duration \(n\). The first part is exactly the term value. In the second part, measure time from \(n\): a death at duration \(n+s\) has discount factor \(e^{-\delta n}e^{-\delta s}\). Conditional on survival to \(n\), the remaining lifetime follows the future-lifetime model at age \(x+n\). This gives the tail identity.

\begin{aligned} \bar A_x &=\int_0^n e^{-\delta t}{}_tp_x\mu_{x+t}\,dt +\int_n^\infty e^{-\delta t}{}_tp_x\mu_{x+t}\,dt\\ &=\bar A^1_{x:\overline{n}|} +e^{-\delta n}{}_np_x\,\bar A_{x+n}. \end{aligned}

The factor \({}_np_x\) is essential: the future whole-life value \(\bar A_{x+n}\) is conditional on being alive at duration \(n\). The factor \(e^{-\delta n}\) discounts that conditional value back to time 0. Thus the omitted tail is not simply the future whole-life value; it is that value weighted by survival and discounting to the term boundary.

Theorem: For a unit benefit payable immediately at death, the difference between the whole-life value and the \(n\)-year term value is \(\bar A_x-\bar A^1_{x:\overline{n}|}=e^{-\delta n}{}_np_x\bar A_{x+n}\). As the term duration increases, \(\bar A^1_{x:\overline{n}|}\) is nondecreasing and converges to \(\bar A_x\), provided the whole-life value is finite.

The convergence follows directly from the interpretation of the value. For each possible death time, extending the term cannot remove a payment that was previously covered. More formally, if \(m>n\), then the term integral over \([0,m]\) contains the term integral over \([0,n]\), plus a nonnegative integral over \((n,m]\). Also, as \(n\) increases without bound, the covered intervals increase to all finite death times. The monotone convergence theorem then gives the whole-life expected present value as the limit.

\bar A^1_{x:\overline{n}|}\le \bar A^1_{x:\overline{m}|}\le \bar A_x \quad (m>n), \qquad \lim_{n\to\infty}\bar A^1_{x:\overline{n}|}=\bar A_x.

There is also a useful bound when \(\delta\ge 0\). After duration \(n\), the discount factor for a death payment is at most \(e^{-\delta n}\), and the probability of surviving to that duration is \({}_np_x\). Since the benefit is one, the tail cannot exceed their product. This bound can be useful when the future whole-life value is not known, although the tail identity is usually more informative.

0\le \bar A_x-\bar A^1_{x:\overline{n}|} =e^{-\delta n}{}_np_x\bar A_{x+n} \le e^{-\delta n}{}_np_x.

Constant-Force Mortality

Under a constant force of mortality \(\mu>0\), survival for \(t\) years is \(e^{-\mu t}\), and the death-time density is \(\mu e^{-\mu t}\). With constant force of interest \(\delta\ge 0\), the whole-life and term values have closed forms. These formulas make the tail and the rate of convergence especially clear.

\bar A_x=\frac{\mu}{\mu+\delta}, \qquad \bar A^1_{x:\overline{n}|} =\frac{\mu}{\mu+\delta}\left(1-e^{-(\mu+\delta)n}\right), \qquad \bar A_x-\bar A^1_{x:\overline{n}|} =\frac{\mu}{\mu+\delta}e^{-(\mu+\delta)n}.

For example, the term formula follows by integrating the death-payment density discounted to time 0: \(\int_0^n \mu e^{-(\mu+\delta)t}\,dt\). The tail has the same exponential rate because both survival to the term boundary and discounting to that boundary contribute to its size.

Worked Example: Calculating the Term Value and Its Tail

A hypothetical continuous insurance pays one dollar immediately on death. Assume a constant annual force of mortality \(\mu=0.04\), a constant annual force of interest \(\delta=0.06\), and a term of \(n=10\) years. Find the whole-life value, the 10-year term value, and the omitted tail. Give values to four decimal places.

Method. Under constant forces, use \(\mu+\delta=0.04+0.06=0.10\). First calculate the whole-life value:

\bar A_x=\frac{0.04}{0.10}=0.4000.

The survival probability to year 10 is \(e^{-0.04(10)}=e^{-0.4}=0.6703\), and the discount factor to year 10 is \(e^{-0.06(10)}=e^{-0.6}=0.5488\). Their product is \(e^{-1}=0.3679\). The tail identity gives:

\bar A_x-\bar A^1_{x:\overline{10}|} =0.4000e^{-1} =0.1471518\ldots \approx 0.1472.

Subtract the tail from the whole-life value:

\bar A^1_{x:\overline{10}|} =0.4000-0.1471518\ldots =0.2528482\ldots \approx 0.2528.

As a check, the direct term formula gives \(0.4(1-e^{-1})=0.2528482\ldots\), the same result. The common mistake to avoid is treating the 10-year term value as the whole-life value: the omitted tail, \(0.1472\), is still part of the whole-life value.

Choosing a Duration From a Tail Tolerance

The tail identity can be rearranged to find a duration at which the omitted value falls below a specified tolerance. Under constant forces, if the whole-life value is \(C=\mu/(\mu+\delta)\), then the tail after \(n\) years is \(Ce^{-(\mu+\delta)n}\). For a strict requirement that the tail be less than \(\varepsilon\), solve the inequality before rounding the duration. If durations must be whole years, choose the smallest integer that satisfies the strict inequality.

Worked Example: Finding the Shortest Integer Term

A hypothetical benefit of one dollar is payable continuously at death. Assume constant annual forces \(\mu=0.06\) and \(\delta=0.04\). Find the shortest whole-number term duration for which the omitted tail is strictly less than \(0.01\). Also check the chosen duration and the preceding duration.

Method. The whole-life value is \(0.06/(0.06+0.04)=0.6\), so the tail after \(n\) years is \(0.6e^{-0.10n}\). Solve:

0.6e^{-0.10n}<0.01 \quad\Longleftrightarrow\quad e^{-0.10n}<\frac{1}{60} \quad\Longleftrightarrow\quad n>\frac{\ln 60}{0.10}=40.9434.

The shortest whole-number duration greater than \(40.9434\) is 41 years. At 41 years, the tail is \(0.6e^{-4.1}=0.0099436\ldots\), which is less than \(0.01\). At 40 years, it is \(0.6e^{-4}=0.0109894\ldots\), which is not less than \(0.01\). Therefore, 41 years is the shortest qualifying integer duration. The term value at 41 years is \(0.6-0.0099436\ldots=0.5900564\ldots\), or \(0.5901\) to four decimal places.

The strict inequality matters: rounding \(40.9434\) down would not meet the requirement. This method also avoids rounding an intermediate value of the tail before deciding which integer duration qualifies.

When Mortality Changes After the Term

The identity does not require a constant force of mortality. It only requires that the future whole-life value be evaluated under the appropriate mortality model after survival to the term boundary. A convenient approach for a piecewise-constant force is to calculate the value during the covered period, then calculate the discounted survival probability to the boundary and multiply by the future whole-life value there.

Worked Example: A Change in Mortality Force at the Term Boundary

For a hypothetical life, assume the force of mortality is \(0.02\) per year for the first 5 years and \(0.05\) per year thereafter. The force of interest is a constant \(\delta=0.03\). A unit benefit is payable immediately at death. Find the 5-year term value, the tail beyond year 5, and the whole-life value. Give values to four decimal places.

Method. During the first 5 years, the combined force in the discounted death-payment integral is \(0.02+0.03=0.05\). Thus:

\bar A^1_{x:\overline{5}|} =\int_0^5 0.02e^{-0.05t}\,dt =\frac{0.02}{0.05}(1-e^{-0.25}) =0.4(1-0.7788008) =0.0884797\ldots \approx 0.0885.

The probability of surviving 5 years is \(e^{-0.02(5)}=e^{-0.10}\). The discount factor to year 5 is \(e^{-0.03(5)}=e^{-0.15}\). Their product is \(e^{-0.25}=0.7788008\). After year 5, the future whole-life value uses mortality force \(0.05\) and interest force \(0.03\):

\bar A_{x+5}=\frac{0.05}{0.05+0.03}=0.625.

Therefore, the tail is \(e^{-0.25}(0.625)=0.4867505\ldots\), or \(0.4868\) to four decimal places. Adding the term and tail values gives the whole-life value: \(0.0884797\ldots+0.4867505\ldots=0.5752302\ldots\), or \(0.5752\). As a check, the two components sum to the whole-life value by the tail identity.

A common mistake is to omit the probability of surviving the first 5 years and calculate the tail as \(e^{-0.15}(0.625)=0.5379\). That calculation discounts the future value but treats survival to year 5 as certain. The correct tail includes both survival and discounting: \(e^{-0.10}e^{-0.15}(0.625)=0.4868\).

Interpretation and Exam Traps

The tail identity is valuable as both a calculation tool and a check. It separates the present value of cover before the term boundary from the present value of cover after it. The term value is always no greater than the whole-life value; their difference is nonnegative. A negative difference indicates a calculation or assumption error.

When comparing values across durations, keep the benefit amount and assumptions fixed. A longer term includes all death times covered by a shorter term, plus additional death times, so its value cannot decrease. If the term value is presented as a proportion of the whole-life value, the proportion increases toward one but need not equal one at any finite duration.

Do not omit survival to the term boundary. The future value at age \(x+n\) is conditional on reaching that age, so it must be multiplied by \({}_np_x\).
Do not discount the future value twice. The factor \(e^{-\delta n}\) moves the value at duration \(n\) back to time 0; any discounting within \(\bar A_{x+n}\) is already measured from that future time.
Keep force and time units consistent. A force stated per year must be multiplied by a duration in years in the exponential.
Do not round before testing a strict threshold. Solve the inequality at full precision, then select the required integer duration and verify it.

The relationships in this tutorial are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In valuation work, they provide a direct way to quantify the effect of truncating continuous whole-life coverage rather than assuming that a long term automatically has the whole-life value.

Check Your Understanding

For each problem, state the model assumptions and use the tail identity or its constant-force form. Values are rounded only at the final step.

A hypothetical unit benefit is payable continuously at death under constant annual forces \(\mu=0.04\) and \(\delta=0.01\). For a 5-year term, find the whole-life value, term value, and omitted tail. Give values to four decimal places. (Answer: whole life \(0.8000\); term \(0.1770\); tail \(0.6230\).)
A hypothetical unit benefit is payable continuously at death with constant annual forces \(\mu=0.05\) and \(\delta=0.05\). Find the 6-year term value and the omitted tail, to four decimal places. (Answer: term \(0.2256\); tail \(0.2744\).)
Under constant annual forces \(\mu=0.02\) and \(\delta=0.03\), find the shortest whole-number term for which the tail is at most \(0.05\). Verify that the chosen duration meets the condition and the preceding year does not. (Answer: 42 years; tail at 42 years \(0.0490\), and at 41 years \(0.0515\).)
A hypothetical life has force of mortality \(0.01\) per year for the first 4 years and \(0.04\) per year thereafter. The force of interest is \(\delta=0.02\). For a unit benefit payable continuously at death, find the 4-year term value, the tail after year 4, and the whole-life value, to four decimal places. (Answer: term \(0.0377\); tail \(0.5913\); whole life \(0.6290\).)

What You'll Learn