Why Express a Term Reserve as a Difference?
A term policy pays a death benefit only if death occurs during a stated coverage period. A whole-life policy pays at death, whenever it occurs. This difference in coverage gives a useful valuation identity: the value of term insurance is the value of whole-life insurance minus the value of the whole-life coverage that begins after the term ends.
The same decomposition can be used at a policy duration to calculate a prospective reserve. It is especially important to keep the survival-and-interest factor separate from the whole-life value at the end of the term: surviving to that date is not the same event as dying during the last year of coverage.
Whole-Life Insurance as Term Plus Deferred Whole Life
Assume annual discrete benefits payable at the end of the year of death, with the same mortality basis and interest rate for all values. A whole-life benefit is paid either during the first \(n\) years or after survival through those \(n\) years. The first possibility has value \(A^1_{x:\overline{n}|}\). The second has value \({}_nE_xA_{x+n}\): the benefit value at age \(x+n\) must be discounted and weighted by the probability of surviving to that age.
The term \({}_nE_xA_{x+n}\) is the value of a deferred whole-life insurance: no benefit is payable during the deferral period, and whole-life coverage begins at its end. The identity is valid because the two benefit periods divide all possible death years into disjoint groups. It does not say that term insurance equals whole-life insurance minus an undiscounted whole-life value; the survival-and-interest factor is essential.
Worked Example: Separating Term and Deferred Whole-Life Values
A hypothetical policy pays $20,000 at the end of the year of death. It covers two years. The annual effective interest rate is \(i=0\), and the one-year death probabilities are \(q_x=0.25\), \(q_{x+1}=0.40\), and \(q_{x+2}=1\). Find the two-year term-insurance value and verify it using whole-life values.
Method. With \(i=0\), \(v=1\), so the relevant values are probabilities multiplied by the benefit. Survival through two years is \({}_2p_x=(1-0.25)(1-0.40)=0.75(0.60)=0.45\). At age \(x+2\), death in the following year is certain, so the whole-life value there is \(A_{x+2}=1\). Thus the deferred whole-life value is \({}_2E_xA_{x+2}=1(0.45)(1)=0.45\).
The whole-life insurance value at age \(x\) is \(A_x=1\), because death is certain by the end of the third year and there is no discounting. Therefore, \(A^1_{x:\overline{2}|}=1-0.45=0.55\), and the term benefit value is \(0.55(\$20{,}000)=\$11{,}000\).
Check by direct calculation. Death in the first year has probability \(0.25\). Death in the second year has probability \(0.75(0.40)=0.30\). The total probability of death during the term is \(0.25+0.30=0.55\), giving \(0.55(\$20{,}000)=\$11{,}000\), as required. The common mistake is to use the second-year death probability \(0.30\) as the probability of surviving two years. That would give a deferred value of \(0.30(\$20{,}000)=\$6{,}000\) and a term value of $14,000. The correct survival probability is \(0.45\), so the deferred value is $9,000 and the term value is $11,000.
From Term Values to Term Reserves
Consider an \(n\)-year term policy with a level annual net premium \(P\), payable in advance while the insured is alive and the premium-paying period has not ended. At issue, the premium is determined by the equivalence principle: the present value of expected premiums equals the present value of expected benefits. Here, “net” means that expenses are not included.
At duration \(t\), conditional on survival to that time, let \(m=n-t\) years of coverage remain. Suppose the reserve is measured just before the premium due at duration \(t\), for \(t<n\). The remaining premiums are due immediately and at each later policy anniversary through duration \(n-1\), provided the insured is alive then. Their value is a temporary life annuity-due value. The prospective reserve is therefore:
The timing convention matters. The annuity-due value includes the premium due at the reserve date. If the reserve is measured immediately after that premium has been paid, the premium at that instant is excluded. A formula using the wrong annuity timing will not match the contract’s cash flows.
The temporary annuity-due value also has a whole-life decomposition. The whole-life annuity includes payments at the start of every future policy year while the insured is alive. Its payments after the remaining \(m\)-year term have value \({}_mE_{x+t}\ddot{a}_{x+n}\). Hence \(\ddot{a}_{x+t:\overline{m}|}=\ddot{a}_{x+t}-{}_mE_{x+t}\ddot{a}_{x+n}\). Substitute this and the insurance decomposition into the prospective reserve formula.
This identity is an algebraic repackaging of the prospective reserve, not a change to the underlying cash flows. It applies when the same interest and mortality assumptions are used for all component values and the stated premium schedule is used. If premiums stop before the term ends, or benefit and premium timing differs from these assumptions, the corresponding annuity and insurance values must be adjusted.
Worked Reserve Calculation
Worked Example: A Reserve One Year Into a Two-Year Term
A hypothetical two-year term policy pays $10,000 at the end of the year of death. The effective annual interest rate is \(i=0.25\). The mortality assumptions are \(q_x=0.10\), \(q_{x+1}=0.40\), and \(q_{x+2}=1\). Annual net premiums are payable at durations 0 and 1 if the insured is alive. Find the net premium and the reserve at duration 1, measured just before the premium due then.
Step 1: Calculate the term-insurance value at issue. Here \(v=1/1.25=0.8\), with \(p_x=0.90\). The benefit is payable in year 1 with probability \(0.10\), or in year 2 with probability \(0.90(0.40)=0.36\). Thus \(A^1_{x:\overline{2}|}=0.10(0.8)+0.36(0.8^2)=0.08+0.2304=0.3104\).
Step 2: Calculate the premium annuity value. A premium is payable at issue with certainty, and the duration-1 premium is payable if the insured survives the first year: \(\ddot{a}_{x:\overline{2}|}=1+0.90(0.8)=1.72\). The net premium is \(P=\$10{,}000(0.3104)/1.72=\$3{,}104/1.72=\$1{,}804.651162\ldots\), or $1,804.65 to the nearest cent.
Step 3: Calculate the prospective reserve directly. Given survival to duration 1, one year of coverage remains. The value of the benefit is \(\$10{,}000(0.40)(0.8)=\$3{,}200\). The premium due at duration 1 is included because the reserve is measured just before it is paid. No later premium is due. Therefore, \({}_1V_x=\$3{,}200-\$1{,}804.651162\ldots=\$1{,}395.348837\ldots\), or $1,395.35 to the nearest cent.
Step 4: Verify using whole-life values. At age \(x+1\), \(A_{x+1}=0.40(0.8)+0.60(0.8^2)=0.704\), and \(\ddot{a}_{x+1}=1+0.60(0.8)=1.48\). At age \(x+2\), death is certain in one year, so \(A_{x+2}=0.8\) and \(\ddot{a}_{x+2}=1\). Also, \({}_1E_{x+1}=0.60(0.8)=0.48\). The whole-life values using the term premium are \(W_{x+1}(P)=\$7{,}040-(1.48)(\$1{,}804.651162\ldots)=\$4{,}369.116279\ldots\) and \(W_{x+2}(P)=\$8{,}000-\$1{,}804.651162\ldots=\$6{,}195.348837\ldots\). Thus \({}_1V_x=W_{x+1}(P)-0.48W_{x+2}(P)=\$4{,}369.116279\ldots-\$2{,}973.767442\ldots=\$1{,}395.348837\ldots\), agreeing with the direct calculation. The main mistake avoided is omitting the premium payable at duration 1 under the stated before-premium timing.
A Second Check on the Difference Formula
Worked Example: Verifying a Reserve from Whole-Life Components
A hypothetical three-year term policy has a $5,000 benefit and annual net premiums payable at durations 0, 1, and 2 while the insured is alive. Assume \(i=0\), with \(q_x=0.20\), \(q_{x+1}=0.30\), \(q_{x+2}=0.50\), and \(q_{x+3}=1\). Find the premium and the reserve at duration 1, just before that duration’s premium.
Premium at issue. The probability of death during the term is \(0.20+0.80(0.30)+0.80(0.70)(0.50)=0.20+0.24+0.28=0.72\). The term-insurance value is therefore \(0.72\). The expected number of premiums, valued at issue, is \(1+0.80+0.80(0.70)=1+0.80+0.56=2.36\). Thus \(P=\$5{,}000(0.72)/2.36=\$3{,}600/2.36=\$90{,}000/59=\$1{,}525.423728\ldots\), or $1,525.42 to the nearest cent.
Direct reserve. Given survival to duration 1, the probability of death during the remaining two years is \(0.30+0.70(0.50)=0.65\). The expected benefit value is \(0.65(\$5{,}000)=\$3{,}250\). The remaining premium annuity-due value is \(1+0.70=1.70\), since the current premium is due with certainty and the duration-2 premium is payable only if the insured survives the next year. The reserve is \(\$3{,}250-1.70(\$1{,}525.423728\ldots)=\$656.779661\ldots\), or $656.78 to the nearest cent.
Whole-life difference check. At age \(x+1\), \(A_{x+1}=1\), because death is certain by the end of the third year. The whole-life annuity-due value is \(\ddot{a}_{x+1}=1+0.70+0.70(0.50)=2.05\). At age \(x+3\), \(A_{x+3}=1\) and \(\ddot{a}_{x+3}=1\). The two-year survival factor from age \(x+1\) is \({}_2E_{x+1}=0.70(0.50)=0.35\). Therefore, \(W_{x+1}(P)=\$5{,}000-2.05(\$1{,}525.423728\ldots)=\$1{,}872.881356\ldots\), and \(W_{x+3}(P)=\$5{,}000-\$1{,}525.423728\ldots=\$3{,}474.576271\ldots\). Their difference is \(\$1{,}872.881356\ldots-0.35(\$3{,}474.576271\ldots)=\$656.779661\ldots\), the same reserve. The method checks both the deferred whole-life subtraction and the premium annuity’s survival weighting.
Common Exam Traps
- Using a death probability where a survival probability is required. The deferred whole-life component uses \({}_nE_x=v^n{}_np_x\), not the probability of death in year \(n\). The probability of death in a particular year is unconditional only after multiplying the probability of prior survival by the conditional death probability for that year.
- Forgetting discounting. A deferred whole-life value includes \(v^n\) as well as \({}_np_x\). Survival alone does not give the value at the earlier date.
- Mixing reserve timing. A before-premium reserve includes the premium due at the valuation date in the future-premium annuity. An after-premium reserve does not include that payment.
- Using a different premium in the whole-life components. In the reserve identity, both \(W_{x+t}(P)\) and \(W_{x+n}(P)\) use the term policy’s premium \(P\). They are components of an algebraic decomposition, not necessarily reserves for separately issued whole-life policies.
Check Your Understanding
For each problem, use annual discrete benefits and premiums. Unless stated otherwise, measure a reserve just before the premium due at that duration.
- At \(i=0\), a two-year term policy has \(q_x=0.20\), \(q_{x+1}=0.50\), and \(q_{x+2}=1\). Find the term-insurance value per unit benefit using the whole-life difference identity. (Answer: \(A^1_{x:\overline{2}|}=0.60\); deferred whole-life value \(=0.40\).)
- A hypothetical two-year policy pays $6,000 at the end of the year of death. Assume \(i=0\), \(q_x=0.10\), and \(q_{x+1}=0.30\). Find the annual net premium. (Answer: $1,168.42 to the nearest cent.)
- For the policy in Question 2, find the duration-1 reserve just before the premium due then. (Answer: $631.58 to the nearest cent.)
- A two-year policy has benefit $1,000, \(i=0.10\), \(q_x=0.20\), \(q_{x+1}=0.25\), and \(q_{x+2}=1\). Find the annual net premium using the term-insurance and annuity-due values. (Answer: $200.96 to the nearest cent.)
- For the policy in Question 4, find the duration-1 reserve just before the premium due then. (Answer: $26.32 to the nearest cent.)