Why Incomplete and Rounded Inputs Need Care
Integrated life-contract problems may give mortality in several forms: a life-table count at each age, a one-year death probability, or a survival probability over multiple years. They may also report values only to a fixed number of decimal places. A calculation can be arithmetically correct and still be unreliable if a rounded figure is treated as exact or if the reconstructed probabilities do not agree with the other information.
The practical approach is to identify the underlying quantities, reconstruct only what the contract calculation needs, retain unrounded values during the calculation, and check the result against the original data. The examples use annual discrete contracts: benefits are paid at the end of the year of death, and premiums are paid at the beginning of each year while the insured is alive. These timing assumptions are stated explicitly because changing them changes the present values.
Reconstruct Before Valuing
A missing probability is often recoverable from a life-table count or a multi-year survival probability. For example, if \({}_2p_x\) and \({}_1q_x\) are known, then the next year's survival probability must satisfy \({}_2p_x=(1-{}_1q_x){}_1p_{x+1}\). Hence
This reconstruction is valid only if \(1-{}_1q_x>0\), and the resulting probability must lie between 0 and 1. A value outside that range signals inconsistent inputs, a misread time interval, or a calculation error.
The same discipline applies to life-table counts. If \(\ell_x\) and \(\ell_{x+2}\) are given but \(\ell_{x+1}\) is missing, the intermediate count cannot be uniquely determined without an additional input, such as \({}_1q_x\). Once it is reconstructed, calculate the next-year probability from the adjacent counts, not from an unrelated rounded value unless the data are stated to share that basis.
Worked Example: Reconstructing a Two-Year Term Premium
A hypothetical two-year term policy pays $10,000 at the end of the year of death if death occurs in the next two years. Annual premiums are due at times 0 and 1 while the insured is alive. A life table gives \(\ell_{50}=10{,}000\), \(\ell_{51}=9{,}500\), and \(\ell_{52}=8{,}550\). The effective annual interest rate is \(i=4\%\). Find the net annual premium and the prospective reserve at time 1 immediately after the second premium is paid.
Method and assumptions. Use annual discrete benefits and premiums as specified. First recover the one-year probabilities from adjacent life-table counts; then equate the expected present values of benefits and premiums. Let \(v=1/(1.04)=25/26\).
For age 50, \({}_1q_{50}=(10{,}000-9{,}500)/10{,}000=0.05\), so \({}_1p_{50}=0.95\). For age 51, \({}_1q_{51}=(9{,}500-8{,}550)/9{,}500=0.10\), so \({}_1p_{51}=0.90\). The probability of death in the second policy year is \({}_1p_{50}{}_1q_{51}=0.95(0.10)=0.095\).
The benefit is paid at time 1 for death in the first year, or at time 2 for death in the second year. Its expected present value is
A premium is paid at time 0 with certainty and at time 1 only if the insured survives the first year. The premium annuity factor is \(1+v(0.95)=1+0.9134615385=1.9134615385\). Therefore the net premium is
At time 1 immediately after the premium has been paid, only the second policy year remains. Conditional on being alive at age 51, the death probability in that year is 0.10. The prospective reserve is therefore \(10{,}000v(0.10)=\$961.54\). The result is consistent with the policy's remaining one-year term benefit and contains no future premium, since both scheduled premiums have now been paid.
Check. The reconstructed two-year survival probability is \(0.95(0.90)=0.855\), which agrees with \(\ell_{52}/\ell_{50}=8{,}550/10{,}000=0.855\). This independent check confirms that the adjacent probabilities match the original life-table counts. The common mistake avoided is using \(\ell_{52}/\ell_{50}\) as a one-year survival probability.
What Rounded Inputs Actually Tell You
A displayed rounded value is not generally an exact input. If a probability is reported as 0.050 to the nearest 0.001, its underlying value is ordinarily in the interval from 0.0495 up to, but not including, 0.0505. Reconstructed quantities inherit that uncertainty. Carrying extra digits in a calculator prevents additional rounding error; it does not remove uncertainty already present in the data.
For a contract calculation, it is often useful to compute a central estimate using the displayed values, then state that it is approximate. If the question requires a reliable range, propagate the input intervals through the reconstruction and valuation. Do not present cents in a way that implies the underlying inputs were known to the nearest cent.
Worked Example: A Rounded Death Rate and a Partial Survival Input
A hypothetical two-year term policy pays $1,000 at the end of the year of death. Premiums are annual in advance while the insured is alive. The effective annual interest rate is \(i=4\%\). The first-year death probability is reported as 0.050 to the nearest 0.001, and two-year survival is reported as 0.855 to the nearest 0.001. Estimate the second-year death probability and net annual premium, and assess the precision of the result.
Method and assumptions. Use the displayed values as central estimates first. Write \(v=25/26\), take \({}_1q_x=0.050\), and take \({}_2p_x=0.855\). The reconstructed second-year survival probability is \({}_1p_{x+1}=0.855/0.950=0.900\), so the central estimate of \({}_1q_{x+1}\) is 0.100.
The probability of death in the second policy year is \(0.950(0.100)=0.095\). Thus the expected present value of benefits is
The expected present value of premiums per dollar of annual premium is \(1+v(0.950)=1.9134615385\). Hence the central estimate of the net premium is \(135.909763/1.9134615385=\$71.03\).
That estimate should not be mistaken for a premium known to the nearest cent. Under ordinary rounding to the nearest 0.001, the first-year death probability lies from 0.0495 to 0.0505, and two-year survival lies from 0.8545 to 0.8555. Substituting the extreme combinations into \({}_1q_{x+1}=1-{}_2p_x/(1-{}_1q_x)\) gives an approximate range from 0.099947 to 0.100053. The central estimate 0.100 is plausible, but it is not exact.
Using the central estimate, the reserve at time 1 immediately after the second premium is \(1{,}000v(0.100)=\$96.15\). The reconstructed range for the second-year death probability alone corresponds to a reserve of approximately $96.10 to $96.20. This range illustrates why retaining calculator digits cannot justify reporting more precision than the mortality inputs support.
Check and common mistake. The central values satisfy \(0.950(0.900)=0.855\), as required. A common mistake is to treat both reported figures as exact, report a second-year rate of exactly 0.100, and imply a cent-accurate premium. The corrected interpretation is that $71.03 is a central estimate based on rounded inputs.
Missing Life-Table Entries and Contract Values
When one life-table entry is missing, use any supplied adjacent probability to reconstruct it, then use the completed table consistently. This matters because a premium or reserve may depend on more than one age. A useful contract check is to calculate the probability of reaching each year in two ways when possible: by multiplying successive survival probabilities and by taking the ratio of the corresponding life-table counts.
Worked Example: Filling a Missing Count for an Endowment
A hypothetical two-year endowment pays $5,000 at time 2 if the insured is alive then, or at the end of the year of death if death occurs earlier. Annual premiums are due at times 0 and 1 while the insured is alive. The effective annual interest rate is \(i=5\%\). The life table gives \(\ell_{60}=10{,}000\), \(\ell_{62}=9{,}215\), and \({}_1q_{60}=0.03\), but does not give \(\ell_{61}\). Find the net annual premium and the reserve at time 1 immediately after the second premium.
Method and assumptions. Reconstruct the missing count from the first-year death probability, then determine the second-year probability. The endowment pays at time 2 for everyone who survives to time 1, whether death occurs in the second year or the insured survives to the endowment date.
Since \({}_1p_{60}=1-0.03=0.97\), \(\ell_{61}=\ell_{60}{}_1p_{60}=10{,}000(0.97)=9{,}700\). Then \({}_1q_{61}=(9{,}700-9{,}215)/9{,}700=485/9{,}700=0.05\), and \({}_1p_{61}=0.95\). The two-year survival check is \(0.97(0.95)=0.9215\), equal to \(9{,}215/10{,}000\).
Set \(v=1/1.05=20/21\). The first-year death benefit is paid at time 1 with probability 0.03. For every policy reaching time 1, either a second-year death benefit or a survival benefit is paid at time 2. Together those outcomes have probability 0.97. Thus
The premium factor is \(1+v(0.97)=1+0.9238095238=1.9238095238\), so \(P=4{,}541.950113/1.9238095238=\$2{,}360.91\). At time 1 immediately after the second premium, the policy has one year remaining and pays $5,000 at time 2 for certain, conditional on being in force at time 1. Its prospective reserve is \(5{,}000v=\$4{,}761.90\).
Check and common mistake. The completed life table reproduces the given two-year count ratio, and the endowment benefit at time 2 has probability 0.97, not \(0.97(0.95)\): the benefit is paid whether the insured dies in the second year or survives it. Omitting the second-year death benefit would undervalue the contract.
Precision and Consistency Checks
A short sequence of checks catches many errors before they reach a final premium or reserve:
Record whether each mortality figure is a count, a one-year probability, or a multi-year probability, and note its stated rounding precision.
Use adjacent life-table counts or the product relation for survival over consecutive years. Check that every reconstructed probability lies between 0 and 1.
Apply the stated benefit and premium pattern to the appropriate survival and death probabilities; keep intermediate values unrounded.
Compare implied multi-year survival with the supplied count ratio or probability. Round the final amount as requested, but label results based on rounded inputs as estimates.
An especially useful consistency test is \({}_2p_x={}_1p_x{}_1p_{x+1}\). If a supplied two-year survival probability conflicts materially with the product of the one-year values, do not silently choose whichever value makes the arithmetic convenient. Recheck age indexing, the time span represented by each quantity, and whether rounding can plausibly explain the discrepancy.
Check Your Understanding
For each problem, retain unrounded working values and use the stated timing assumptions.
- A life table has \(\ell_{40}=8{,}400\) and \(\ell_{41}=7{,}980\). Find \({}_1q_{40}\). (Answer: \(0.050\))
- For a life aged \(x\), \({}_1q_x=0.04\) and \({}_2p_x=0.912\). Reconstruct \({}_1q_{x+1}\). (Answer: \(0.050\))
- A one-year term policy pays $2,000 at the end of the year of death, with \({}_1q_x=0.08\) and \(i=5\%\). A single net premium is paid at issue. Find the premium. (Answer: $152.38)
- A two-year term policy pays $10,000 at the end of the year of death. Premiums are due at times 0 and 1 while alive. Given \(\ell_x=5{,}000\), \(\ell_{x+1}=4{,}800\), \(\ell_{x+2}=4{,}560\), and \(i=4\%\), find the net annual premium. (Answer: $430.77)
- A table gives \(\ell_x=1{,}000\), \(\ell_{x+1}=960\), and separately reports \({}_1q_x=0.05\). Is the information consistent, using the life-table count as exact? (Answer: No; the counts imply \({}_1q_x=(1{,}000-960)/1{,}000=0.040\), not 0.050.)