Start With the Valuation Condition
At a policy duration, the key question is not simply “what cash flows are in the contract?” It is which cash flows can still occur, given what is known at that duration? If the insured is known to be alive at duration \(k\), past premiums and past benefits are no longer future cash flows. The remaining death benefit is conditional on survival to the start of a future year followed by death during that year. A future premium, by contrast, is conditional on the insured being alive on its due date.
These events have different probabilities. Confusing them can produce an incorrect prospective value even when the contract benefits and premium amount have been read correctly. This tutorial focuses on identifying the outstanding cash flows and assigning those probabilities; it does not calculate complete policy reserves.
Assume annual premiums are payable in advance while the insured is alive: at policy times \(0,1,2,\ldots\). We place the valuation immediately after any premium due at time \(k\). Thus, a future premium at time \(k+j\) is payable only if the insured survives to that date, and its conditional probability is \({}_jp_{x+k}\). In a contract with premiums payable only through time \(n-1\), there is no premium due at time \(n\).
Match Each Contract to Its Remaining Cash Flows
The benefit pattern determines which future events produce payments. Whole-life insurance has a death benefit whenever death occurs in the future. Term insurance has a death benefit only during the remaining coverage period. An endowment insurance pays on death during the remaining term and pays a survival benefit at maturity. At a duration within the term, the insured’s survival to that duration is already part of the conditioning; it must not be multiplied into future probabilities again.
| Contract | Remaining benefit events at duration \(k\) | Conditional probability |
|---|---|---|
| Whole life | Death in each future year \(j=1,2,\ldots\) | \({}_{j-1}p_{x+k}q_{x+k+j-1}\) |
| \(n\)-year term, \(k<n\) | Death in each remaining year \(j=1,\ldots,n-k\) | \({}_{j-1}p_{x+k}q_{x+k+j-1}\) |
| \(n\)-year endowment, \(k<n\) | Death in each remaining year, or survival to maturity | Death: \({}_{j-1}p_{x+k}q_{x+k+j-1}\); maturity: \({}_{n-k}p_{x+k}\) |
Premiums require a separate count. For whole life with annual premiums payable in advance indefinitely, future installments occur at times \(k+j\), \(j=1,2,\ldots\), and each is conditional on survival to that date. For an \(n\)-year contract with premiums due at times \(0\) through \(n-1\), after the premium at time \(k\) has just been paid, the remaining installments are at \(k+1,\ldots,n-1\). Their indices are \(j=1,\ldots,n-k-1\). The last death benefit can therefore occur after the last premium date.
Why the Conditional Death Probability Has This Form
For death to occur during future year \(j\), two events must happen in sequence: the insured must survive the first \(j-1\) years from the valuation time, and then die during the following year. Conditional on being alive at duration \(k\), the probability of the first event is \({}_{j-1}p_{x+k}\). Given survival to the start of that year, the probability of death during it is \(q_{x+k+j-1}\). Multiplication of these conditional probabilities gives
This also shows why the premium probability is different. A premium at time \(k+j\) is due if the insured reaches that date, so the relevant event is survival for all \(j\) years, with probability \({}_jp_{x+k}\). A premium is not paid merely because the insured survived to the beginning of its year.
As a useful check, the death-in-year probabilities over \(r\) years and the probability of surviving all \(r\) years sum to one:
To see this, write each death probability as the probability of survival to the start of the year minus the probability of survival to its end: \({}_{j-1}p_{x+k}q_{x+k+j-1}={}_{j-1}p_{x+k}-{}_jp_{x+k}\). The sum telescopes to \(1-{}_rp_{x+k}\), and adding \({}_rp_{x+k}\) gives 1.
Worked Example: Outstanding Whole-Life Cash Flows
A hypothetical whole-life policy was issued at age \(x\). At duration 2, immediately after the premium due then, the insured is alive. The death benefit is $100,000, and the annual premium is $3,000, payable in advance for life. Assume \(i=5\%\). For the next three years, assume the one-year death probabilities at the successive attained ages are \(q_{x+2}=0.02\), \(q_{x+3}=0.03\), and \(q_{x+4}=0.04\). Identify the benefit and premium probabilities for those years.
Method. Start from age \(x+2\), the age at the valuation duration. For a death benefit, multiply survival to the beginning of the year by that year’s \(q\). For a premium due at the end of the year, use survival to that premium date.
Future year 1. Death during this year has probability \(q_{x+2}=0.02\). Survival to the premium date is \({}_1p_{x+2}=1-0.02=0.98\).
Future year 2. Survival to its start is \(0.98\). Death during it has probability \(0.98(0.03)=0.0294\). Survival to its premium date is \({}_2p_{x+2}=0.98(0.97)=0.9506\).
Future year 3. Survival to its start is \(0.9506\). Death during it has probability \(0.9506(0.04)=0.038024\). Survival to its premium date is \({}_3p_{x+2}=0.9506(0.96)=0.912576\).
Thus, in the first three future years, the benefit amounts are $100,000 at the ends of those years with respective probabilities \(0.02\), \(0.0294\), and \(0.038024\). Premiums of $3,000 are due at those dates with respective probabilities \(0.98\), \(0.9506\), and \(0.912576\). The discount factors at \(i=5\%\) are \(v=1/1.05\), \(v^2=1/1.05^2\), and \(v^3=1/1.05^3\). These first three years do not exhaust whole-life coverage or premiums; the same method continues for every later year. The probabilities are consistent because the three death probabilities sum to \(0.087424\), and survival through year 3 is \(0.912576\), giving 1.
Common mistake avoided: using \(0.98\) as the year-2 death probability ignores the additional death condition in year 2. The correct probability is survival to that year, \(0.98\), multiplied by its mortality probability, \(0.03\), giving \(0.0294\).
Worked Example: Remaining Benefits and Premiums on Term Insurance
A hypothetical 10-year term policy has a $80,000 death benefit and annual premiums of $1,200 due in advance at times 0 through 9 while the insured is alive. At duration 4, immediately after the premium then due, the insured is alive. Assume \(i=4\%\), and suppose the death probabilities for the six remaining years are \(0.02, 0.03, 0.04, 0.05, 0.06,\) and \(0.07\), in order. Find the probability attached to each remaining death benefit and each remaining premium.
There are \(10-4=6\) years of coverage remaining. For each benefit, multiply the probability of reaching the start of its year by the mortality probability for that year. The survival probability to the premium date is updated by multiplying by the one-year survival probability.
| Future year \(j\) | Survival to start of year | Death-benefit probability | Survival to end of year | Premium due? |
|---|---|---|---|---|
| 1 | 1 | \(1(0.02)=0.02\) | \(1(0.98)=0.98\) | Yes, probability \(0.98\) |
| 2 | 0.98 | \(0.98(0.03)=0.0294\) | \(0.98(0.97)=0.9506\) | Yes, probability \(0.9506\) |
| 3 | 0.9506 | \(0.9506(0.04)=0.038024\) | \(0.9506(0.96)=0.912576\) | Yes, probability \(0.912576\) |
| 4 | 0.912576 | \(0.912576(0.05)=0.0456288\) | \(0.912576(0.95)=0.8669472\) | Yes, probability \(0.8669472\) |
| 5 | 0.8669472 | \(0.8669472(0.06)=0.052016832\) | \(0.8669472(0.94)=0.814930368\) | Yes, probability \(0.814930368\) |
| 6 | 0.814930368 | \(0.814930368(0.07)=0.05704512576\) | \(0.814930368(0.93)=0.75788524224\) | No premium at maturity |
The benefit may be paid at the end of each of the six years. The remaining premiums are at times 5, 6, 7, 8, and 9, so their probabilities are the survival probabilities at the ends of future years 1 through 5. No premium is due at time 10: the in-advance premium schedule ends at time 9. The benefit probabilities sum to \(1-0.75788524224=0.24211475776\), the probability of death within the remaining term. Survival to maturity has probability \(0.75788524224\), so the two outcomes total 1.
Common mistake avoided: charging a sixth remaining premium at time 10 would extend the premium schedule beyond its stated final date. The six remaining years refer to benefit coverage, not six future premium dates.
Worked Example: Separating Death and Maturity Benefits
A hypothetical five-year endowment policy pays $50,000 on death during the term or $50,000 at the end of year 5 if the insured survives. Premiums of $2,000 are payable in advance at times 0 through 4 while alive. At duration 2, just after that date’s premium, the insured is alive. Assume \(i=5\%\), with \(q_{x+2}=0.10\), \(q_{x+3}=0.20\), and \(q_{x+4}=0.25\). Identify the remaining benefit and premium events and their probabilities.
Death benefits. The first remaining year’s death probability is \(0.10\). Survival through that year is \(1-0.10=0.90\). The second year’s death probability is \(0.90(0.20)=0.18\), leaving survival probability \(0.90(0.80)=0.72\). The third year’s death probability is \(0.72(0.25)=0.18\). These death benefits are mutually exclusive: only one can be paid.
Maturity benefit. The insured must survive all three remaining years. Its conditional probability is \(0.72(1-0.25)=0.72(0.75)=0.54\). Thus the maturity payment is distinct from the three death payments; it is payable only if the insured is alive at time 5.
Premiums. Premiums remain due at times 3 and 4, corresponding to future years 1 and 2. Their probabilities are \(0.90\) and \(0.72\), respectively. There is no premium at time 5 because the stated premium schedule ends at time 4.
As a check, the death-benefit probabilities sum to \(0.10+0.18+0.18=0.46\). Adding the maturity probability gives \(0.46+0.54=1\). The future premium probabilities are not part of that check: they describe separate payment events, not possible benefit outcomes.
Common mistake avoided: using \(0.72\) as the maturity probability stops after the second remaining year. Maturity is three years away, so the third year’s survival factor \(0.75\) must also be included.
Common Exam Traps
- Using the wrong starting age. At duration \(k\), condition from attained age \(x+k\), not from issue age \(x\). Past survival is already known.
- Using survival where death is required. A death benefit in future year \(j\) requires \({}_{j-1}p_{x+k}q_{x+k+j-1}\), not \({}_jp_{x+k}\).
- Using death where survival is required. A premium due at a future date requires survival to that date, with probability \({}_jp_{x+k}\).
- Extending benefits beyond the term. Term and endowment death benefits stop at maturity; only whole-life death coverage continues indefinitely.
- Counting a premium at maturity without checking the schedule. With annual premiums in advance through time \(n-1\), the maturity date \(n\) is not a premium date.
- Adding overlapping endowment events. Death before maturity and survival to maturity are mutually exclusive. Their probabilities should sum to one over the remaining term.
Careful event labeling is the foundation for later prospective-value calculations. Write down the valuation condition, mark each future payment date, and state the event that makes each cash flow payable before substituting probabilities or discount factors. This discipline is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, condition on survival to the stated duration and use annual premiums payable in advance while alive.
- A whole-life policy is valued at duration 1, just after its premium. The next year’s mortality probability is \(q_{x+1}=0.04\). What probabilities apply to the death benefit payable at the end of that year and the next premium? (Answer: death benefit \(0.0400\); next premium \(0.9600\).)
- A 8-year term policy is valued at duration 5 just after the premium. The next three mortality probabilities are \(0.10, 0.20,\) and \(0.25\). Give the three death-benefit probabilities and identify the number of premiums still due. (Answer: death probabilities \(0.1000,\ 0.1800,\ 0.1800\); two premiums remain, at times 6 and 7.)
- A 4-year endowment is valued at duration 2 just after the premium. The remaining mortality probabilities are \(0.05\) and \(0.10\). Find the two death-benefit probabilities, the maturity probability, and the number of premiums remaining. (Answer: death probabilities \(0.0500\) and \(0.0950\); maturity probability \(0.8550\); one premium remains.)
- A term policy has three years of coverage remaining. Conditional on survival to the valuation date, survival probabilities to the ends of years 1, 2, and 3 are \(0.90,\ 0.75,\) and \(0.60\). Find the death-benefit probability in each year and verify the total probability of death during the term. (Answer: \(0.1000,\ 0.1500,\ 0.1500\); total death probability \(0.4000\), with \(0.6000\) survival to maturity.)
- A six-year endowment is valued at duration 3 just after that date’s premium. It has annual premiums in advance at times 0 through 5. How many years of death coverage remain, how many premiums remain, and at what future times are those premiums due? (Answer: three years of death coverage remain; two premiums remain, due at times 4 and 5.)