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Exam FAM · Prospective Reserves · Tutorial 413 of 988

Continuous Whole-Life Reserves

Calculate continuous whole-life prospective reserves by valuing future death benefits and subtracting the value of premiums that remain payable.

Intermediate 10 min read

What You'll Learn

  • Define a continuous whole-life prospective reserve conditional on survival to the valuation time
  • Calculate the present value of a death benefit payable at the moment of death
  • Distinguish annual premiums due immediately before and after a premium date
  • Apply piecewise constant mortality forces and a constant force of interest
  • Check reserve calculations against the policy’s premium and timing assumptions

What a Continuous Whole-Life Reserve Measures

A whole-life policy pays a benefit whenever the insured dies, with no fixed end to the coverage. In a continuous whole-life insurance, the benefit is paid at the moment of death. A reserve at a given duration is the value, at that time, of the future policy benefits less the value of future premiums. The valuation is conditional on the insured being alive at that duration.

This tutorial considers level premiums payable annually in advance while the insured is alive, and a level death benefit payable immediately at death. That timing distinction matters: after an anniversary premium has been paid, it is no longer part of the future premium stream. The calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let \(B\) be a level benefit payable at the moment of death, and let \(P\) be the annual level premium payable at issue and at each subsequent policy anniversary while the insured is alive. The prospective reserve at duration \(k\), immediately after the premium due at that duration has been paid, is $$ {}_kV_x = B\bar{A}_{x+k}-P a_{x+k}. $$ Here \(\bar{A}_{x+k}\) is the expected present value of a unit benefit payable at death for a life aged \(x+k\), and \(a_{x+k}\) is the expected present value of annual premiums payable at times one year or more after the valuation time. If the valuation is immediately before the premium at duration \(k\), that premium is still outstanding, so the premium value is \(P\ddot{a}_{x+k}\), not \(Pa_{x+k}\).

The reserve is not a new premium or a separate payment to the policyholder. It is a prospective measure of the insurer’s net future obligations under the stated assumptions. The benefit amount, interest assumption, mortality assumption, premium amount, and premium timing must all be identified before applying the formula.

Continuous Insurance and Annual Premium Values

Assume a constant force of interest \(\delta\), so a payment at time \(t\) has discount factor \(e^{-\delta t}\). Let \({}_tp_x\) denote the probability that a life aged \(x\) survives for at least \(t\) years, and let \(\mu_{x+t}\) be the force of mortality at age \(x+t\). Conditional on survival to time \(t\), the probability of death in the next small interval of length \(dt\) is approximately \(\mu_{x+t}dt\). Thus the density of the future lifetime at time \(t\) is \({}_tp_x\mu_{x+t}\), and

$$ \bar{A}_x=\int_0^\infty e^{-\delta t}{}_tp_x\mu_{x+t}\,dt. $$

This integral is the expected present value of a unit benefit paid at the instant of death. It applies when the benefit is paid at death and the expected present value exists. It is not the formula for a benefit payable only at the end of the year of death.

For annual premiums in advance, the expected present value at issue of the premium stream is

$$ \ddot{a}_x=\sum_{j=0}^{\infty}e^{-\delta j}{}_jp_x. $$

The term for \(j=0\) is the premium paid at issue. If the net premium is determined by the equivalence principle, the expected present value of premiums equals the expected present value of benefits:

$$ P\ddot{a}_x=B\bar{A}_x, \qquad P=\frac{B\bar{A}_x}{\ddot{a}_x}. $$

At an integer duration \(k\), immediately after that duration’s premium has been paid, the remaining premium dates are \(k+1,k+2,\ldots\). Their value at time \(k\), conditional on survival, is \(Pa_{x+k}\), where

$$ a_{x+k}=\sum_{j=1}^{\infty}e^{-\delta j}{}_jp_{x+k}. $$

The benefit value uses mortality and interest from the valuation time onward. In particular, it is based on the mortality law at age \(x+k\), not on the insured’s mortality before duration \(k\).

Why the Prospective Formula Holds

The reserve definition follows directly from the two future cash-flow streams. Given survival to duration \(k\), a death at \(t\) more years in the future has probability density \({}_tp_{x+k}\mu_{x+k+t}\) and is discounted by \(e^{-\delta t}\). Integrating the discounted benefit over all possible future death times gives \(B\bar{A}_{x+k}\). Each future annual premium is due only if the insured is alive on its payment date; discounting and summing those survival-contingent payments gives \(Pa_{x+k}\) after the premium at duration \(k\) has been paid. Subtracting the latter value from the former gives the reserve.

At a time immediately before the anniversary premium, the premium due at that moment has not yet been paid. It belongs in the outstanding premium stream, changing the annuity factor from \(a_{x+k}\) to \(\ddot{a}_{x+k}\). This is a timing change, not a change in the policy’s mortality or benefit assumptions.

1
Specify the cash flows.
Identify the death benefit, when it is paid, the premium amount, and the premium dates still outstanding.
2
Value future benefits.
Use the mortality and interest assumptions after the valuation time to calculate \(B\bar{A}\).
3
Value future premiums.
Use \(Pa\) immediately after an annual premium date or \(P\ddot{a}\) immediately before one.
4
Subtract and check.
Calculate benefit value minus premium value, then verify that the timing and assumptions match the question.

Worked Calculations

Worked Example: Constant Mortality and Interest

A hypothetical whole-life policy pays \(B=\$8{,}000\) at the moment of death. The force of interest is \(\delta=\ln 2\), and the force of mortality is constantly \(\mu=\ln 2\) at all future ages. Premiums are annual in advance, beginning at issue. Find the net premium and the reserve immediately after the premium at duration 2.

Method. Under constant mortality force \(\mu\), survival is \({}_tp_x=e^{-\mu t}\), and the density of future lifetime is \(\mu e^{-\mu t}\). Therefore,

$$ \bar{A}_x=\int_0^\infty e^{-\delta t}\mu e^{-\mu t}\,dt =\frac{\mu}{\delta+\mu} =\frac{\ln 2}{2\ln 2} =\frac12. $$

For each whole year, \(v=e^{-\delta}=1/2\) and \({}_1p_x=e^{-\mu}=1/2\). The annual premium annuity-due value is

$$ \ddot{a}_x=\sum_{j=0}^{\infty}(v{}_1p_x)^j =\frac{1}{1-(1/2)(1/2)} =\frac43. $$

The equivalence premium is

$$ P=\frac{8{,}000(1/2)}{4/3} =8{,}000\left(\frac12\right)\left(\frac34\right) =\$3{,}000. $$

At duration 2, the mortality and interest assumptions are unchanged, so \(\bar{A}_{x+2}=1/2\). Premiums still outstanding are due at times 3, 4, and later; measured from duration 2, their annuity-immediate value is

$$ a_{x+2}=\sum_{j=1}^{\infty}\left(\frac14\right)^j =\frac{1/4}{1-1/4} =\frac13. $$

Thus the reserve immediately after the duration-2 premium is

$$ {}_2V_x=8{,}000\left(\frac12\right)-3{,}000\left(\frac13\right) =4{,}000-1{,}000 =\$3{,}000. $$

Check. The benefit value is \(\$4{,}000\), and the value of outstanding premiums is \(\$1{,}000\); their difference is the stated reserve. The calculation avoids counting the duration-2 premium a second time.

Worked Example: Mortality Force Increases After the First Year

A hypothetical policy pays \(B=\$12{,}000\) at the moment of death. The force of interest is \(\delta=\ln 2\). The force of mortality is \(\ln 2\) during the first policy year and \(\ln 4\) thereafter. Find the annual net premium and the reserve immediately after the premium at duration 1.

Method. In year 1, the discounted death density is \(e^{-(\ln 2)t}\ln 2e^{-(\ln 2)t}\). Its integral over that year is

$$ \int_0^1(\ln 2)e^{-2(\ln 2)t}\,dt =\frac38. $$

At duration 1, the conditional lifetime value of the unit death benefit, with mortality force \(\ln 4\) and interest force \(\ln 2\), is

$$ \bar{A}_{x+1}=\frac{\ln 4}{\ln 4+\ln 2}=\frac23. $$

The probability of surviving the first year is \(1/2\), and the discount factor to that year is also \(1/2\). The time-1-and-later contribution to the issue-age benefit value is therefore \((1/2)(1/2)(2/3)=1/6\), giving

$$ \bar{A}_x=\frac38+\frac16=\frac{13}{24}. $$

For annual premiums, the issue-date installment contributes 1 to \(\ddot{a}_x\), and the duration-1 installment contributes \(v{}_1p_x=(1/2)(1/2)=1/4\). From duration 2 onward, survival in each additional year is \(1/4\), so each successive discounted survival term has ratio \((1/2)(1/4)=1/8\). The tail beginning at duration 2 is

$$ \frac{e^{-2\delta}{}_2p_x}{1-1/8} =\frac{1/32}{1-1/8} =\frac1{28}, \qquad \ddot{a}_x=1+\frac14+\frac1{28}=\frac97. $$

Consequently,

$$ P=\frac{12{,}000(13/24)}{9/7} =6{,}500\left(\frac79\right) =\$5{,}055.56. $$

After the duration-1 premium has been paid, the remaining premium dates are durations 2 onward. At age \(x+1\), the one-year discounted survival factor is \((1/2)(1/4)=1/8\), so \(a_{x+1}=(1/8)/(1-1/8)=1/7\). The benefit value is \(\$12{,}000(2/3)=\$8{,}000\). Thus

$$ {}_1V_x =8{,}000-\frac{5{,}055.555\ldots}{7} =8{,}000-722.222\ldots =\$7{,}277.78. $$

Check. The issue-age premium equation uses \(\bar{A}_x=13/24\) and \(\ddot{a}_x=9/7\). The reserve uses the post-year-1 values and excludes the premium just paid.

Worked Example: Reserve After Two Years of Changing Mortality

A hypothetical policy pays \(B=\$8{,}880\) at the moment of death. The force of interest is \(\delta=\ln 2\). The force of mortality is \(\ln 2\) during the first two policy years and \(\ln 4\) thereafter. Annual premiums are payable in advance while the insured is alive. Find the net premium and the reserve immediately after the duration-2 premium.

Method. During each of the first two years, the discounted probability contribution for death in that year, measured from the beginning of that year, is \(3/8\). The first-year contribution is \(3/8\). The second-year contribution is multiplied by the discount-and-survival factor to duration 1, \((1/2)(1/2)=1/4\), and is therefore \(3/32\). At duration 2, the conditional value of the unit benefit is \(2/3\); the discount-and-survival factor to duration 2 is \((1/2)^2(1/2)^2=1/16\). Hence

$$ \bar{A}_x=\frac38+\frac3{32}+\frac1{16}\left(\frac23\right) =\frac{36+9+4}{96} =\frac{49}{96}. $$

The premium annuity-due has contributions 1 at issue and \(1/4\) at duration 1. Its duration-2 term is \(1/16\), and the terms thereafter have ratio \(1/8\). Thus

$$ \ddot{a}_x=1+\frac14+\frac{1/16}{1-1/8} =1+\frac14+\frac1{14} =\frac{37}{28}. $$

The net premium is

$$ P=\frac{8{,}880(49/96)}{37/28} =8{,}880\left(\frac{343}{888}\right) =\$3{,}430. $$

At duration 2, the future mortality force is constantly \(\ln 4\), so \(\bar{A}_{x+2}=2/3\). The annuity-immediate value of premiums still due is \(a_{x+2}=1/7\), as its first discounted survival factor is \(1/8\). Therefore,

$$ {}_2V_x=8{,}880\left(\frac23\right)-3{,}430\left(\frac17\right) =5{,}920-490 =\$5{,}430. $$

Check. The issue benefit value is \(8{,}880(49/96)=\$4{,}532.50\), while the premium value is \(3{,}430(37/28)=\$4{,}532.50\). They agree under the equivalence principle.

Timing Trap and Practical Checks

A frequent error is to use an annuity-due for premiums after the premium at the valuation duration has already been paid. For example, in the preceding example’s type of calculation, using \(\ddot{a}_{x+1}\) after the duration-1 installment would count that installment again. Under the mortality assumptions in the second example, \(\ddot{a}_{x+1}=1+1/7=8/7\). The incorrect calculation would be

$$ 8{,}000-5{,}055.56\left(\frac87\right) \approx \$2{,}222.22. $$

That is not the post-premium reserve: it includes a premium that has already been paid. The correct premium factor is \(a_{x+1}=1/7\), yielding \(\$7{,}277.78\). Before making the subtraction, mark whether the valuation is just before or just after the premium date.

Other useful checks are to confirm that mortality is conditional on survival to the valuation age, that the benefit is discounted to the moment of death, and that premiums are counted only while payable. A negative result is not automatically impossible under every contract design, but for these level-premium whole-life examples it should prompt a careful review of timing, assumptions, and arithmetic.

Key takeaway: For an annual-premium continuous whole-life policy, value the moment-of-death benefit using \(\bar{A}\), then subtract the value of premiums that remain unpaid. Immediately after an anniversary premium, use \(a\); immediately before it, use \(\ddot{a}\).

Check Your Understanding

For each problem, identify the premium timing before calculating. Amounts are hypothetical.

  1. A whole-life policy has benefit \(\$6{,}000\), constant force of interest \(\delta=\ln 2\), and constant force of mortality \(\mu=\ln 2\). Annual premiums begin at issue. Find the net premium and reserve immediately after the duration-1 premium. (Answer: \(P=\$2{,}250\), \({}_1V_x=\$2{,}250\).)
  2. A policy has benefit \(\$9{,}600\), with \(\delta=\ln 2\) and \(\mu=\ln 2\) at all future ages. Its annual net premium is determined by equivalence. Find the reserve immediately before the duration-1 premium. (Answer: \({}_1V_x=\$0.00\).)
  3. A policy has benefit \(\$6{,}000\), \(\delta=\ln 2\), and mortality force \(\ln 2\) during the first year and \(\ln 4\) thereafter. Find the annual net premium and reserve immediately after the duration-1 premium. (Answer: \(P=\$2{,}527.78\), \({}_1V_x=\$3{,}638.89\).)
  4. A policy has benefit \(\$4{,}440\), \(\delta=\ln 2\), and mortality force \(\ln 2\) for the first two years and \(\ln 4\) thereafter. Find the net premium and reserve immediately after the duration-2 premium. (Answer: \(P=\$1{,}715.00\), \({}_2V_x=\$2{,}715.00\).)
  5. A policy has benefit \(\$10{,}000\), \(\delta=\ln 2\), and constant mortality force \(\mu=\ln 2\). Find the net premium and reserve immediately after the duration-3 premium. (Answer: \(P=\$3{,}750.00\), \({}_3V_x=\$3{,}750.00\).)
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