Why a Profit Signature Needs Diagnostics
A profit signature shows how profit emerges across policy durations. Its net present value (NPV) is useful only if each duration has been weighted by the right probability and assigned to the right time. A signature can look plausible while still using the probability of surviving to the end of a year instead of being in force at its start, shifting every duration by one, or leaving out profits when a policy terminates.
This tutorial sets out a consistent annual model for checking those errors. All products and data below are hypothetical. The model treats profit as measured at the end of each policy year and discounts that end-of-year amount to issue. If a cash flow occurs at another time, its timing must be reflected before it is included in the profit amount.
This convention separates three quantities that are easy to confuse: the probability of entering the year, the conditional outcome probabilities during the year, and the profit conditional on each outcome. If a profit vector is already expressed per policy issued, do not apply the in-force weights to it a second time. Conversely, a profit amount conditional on being in force at the start of a year is not yet a signature entry per policy issued.
The formula assumes one annual termination event at most, a single annual interest rate, and profits measured at year-end. The \(q_t\) values need not be constant. If termination can occur through several competing causes, the conditional outcomes and their probabilities must represent those causes consistently. If cash flows occur within the year, an annual end-of-year profit model may need adjustment rather than a mechanical use of this formula.
Deriving the Signature Weight
The signature formula follows by conditioning on whether a policy is still active at the start of the year and, if active, on its outcome during that year. This gives a direct diagnostic: every profit contribution needs a path from issue to the relevant policy-year outcome.
A policy issued at time 0 is in force at the start of year \(t\) with probability \(W_t\), the product of survival probabilities in years 1 through \(t-1\).
Conditional on being in force at the start of year \(t\), survival has probability \(p_t\) and termination has probability \(q_t\). The corresponding end-of-year profits are \(c_t\) and \(z_t\).
The conditional expected profit is \(p_tc_t+q_tz_t\). Multiplying by \(W_t\) gives the signature entry \(\sigma_t\) per policy issued.
Because \(\sigma_t\) is measured at time \(t\), its present value is \(v^t\sigma_t\), and summing these present values gives NPV.
The conditioning argument also explains why termination profits belong in the signature. A policy that terminates does not contribute the continuation profit \(c_t\), but it may produce a claim, surrender payment, release of an expense allowance, or another terminal result represented by \(z_t\). The model includes the net profit assigned to that outcome; it does not assume that termination profit is zero.
Worked Example: Checking In-Force Weights Across Durations
A hypothetical three-year contract earns an end-of-year profit of \(100\), \(120\), or \(80\) if it survives the relevant year. If it terminates in years 1, 2, or 3, its end-of-year profit is respectively \(-50\), \(-20\), or \(10\). Conditional termination probabilities are \(0.10\), \(0.20\), and \(0.25\). The effective annual interest rate is \(10\%\). Find the signature and NPV per policy issued.
Method. Calculate the probability of being in force at each year’s start, then combine the conditional continuation and termination profits. This avoids using an unweighted conditional profit as though every issued policy reached every duration.
The in-force weights are \(W_1=1\), \(W_2=1(0.90)=0.90\), and \(W_3=0.90(0.80)=0.72\). The year 1 expected profit conditional on being in force is \(0.90(100)+0.10(-50)=90-5=85\), so \(\sigma_1=1(85)=85\).
For year 2, the conditional expected profit is \(0.80(120)+0.20(-20)=96-4=92\). Only \(0.90\) of policies issued reach the start of year 2, so \(\sigma_2=0.90(92)=82.8\).
For year 3, the conditional expected profit is \(0.75(80)+0.25(10)=60+2.5=62.5\). The year 3 signature entry is \(\sigma_3=0.72(62.5)=45\). Thus the undiscounted signature is \((85,\ 82.8,\ 45)\).
The discount factors for the three year-ends are \(1/1.10\), \(1/1.10^2=1/1.21\), and \(1/1.10^3=1/1.331\). Therefore,
Rounded to the nearest cent, the NPV is \(179.51\). As a check, the signature entries include the start-of-year weights \(1\), \(0.90\), and \(0.72\), and each is discounted for one, two, and three years respectively. A common mistake is to discount the conditional profits \(85\), \(92\), and \(62.5\) without weighting years 2 and 3 for policies that have terminated. That would overstate the value by treating those policies as if they remained exposed to later profits.
Duration Indices and Payment Timing
Two separate indices appear in an annual signature. The probability weight for year \(t\) covers the years before \(t\), so it is \(W_t={}_{t-1}p\) when a constant annual survival notation is appropriate. The discount factor, in the end-of-year convention, covers \(t\) years, so it is \(v^t\). The difference is intentional: a policy must survive to the start of the year before earning that year’s profit, and that profit is then discounted from the end of the year.
A reliable audit labels both dates before calculating: when must the policy be in force? and when is the profit measured? An answer of “start of year \(t\)” calls for \(W_t\); “end of year \(t\)” calls for \(v^t\). If the assigned profit is instead at the start of year \(t\), the discount exponent is \(t-1\), not \(t\). The profit amount itself may also need to reflect interest earned between the cash-flow date and the measurement date.
Worked Example: Detecting a Shifted Probability or Discount Index
A hypothetical policy has a \(10\%\) probability of terminating in each of three years, conditional on being in force at that year’s start. It earns \(100\) at each year-end if it survives that year and zero if it terminates. The effective annual interest rate is \(10\%\). Calculate NPV, then identify the effects of two common index errors.
Method. Weight the year-\(t\) profit by the probability of reaching the start of year \(t\), and discount it from the end of that year. The weights are \(1\), \(0.90\), and \(0.90^2=0.81\). Because the termination profit is zero, the signature entries are \(100\), \(90\), and \(81\).
The correct NPV is
Rounded to the nearest cent, NPV is \(226.15\). If the analyst incorrectly weights year \(t\) by the probability of surviving to the end of that year, the weights become \(0.90\), \(0.81\), and \(0.729\). The resulting signature entries are \(90\), \(81\), and \(72.9\), and the incorrect NPV is \(90/1.10+81/1.21+72.9/1.331=203.5319\), or \(203.53\). The first year’s profit has been wrongly reduced even though every policy is in force at its start.
A different error is to discount all three end-of-year profits using \(v^{t-1}\), as if they were paid at the start of each year. That gives \(100+90/1.10+81/1.21=248.7603\), or \(248.76\). The first error misplaces the in-force probability; the second misplaces the payment time. Neither is corrected by the other.
Termination Profits Must Be Tested Explicitly
A useful diagnostic is to write the continuation and termination outcomes in separate columns before weighting them. This makes an omitted terminal amount visible. In a simplified model, \(z_t\) can be zero, but that must follow from the product cash flows and the stated profit definition; it should not be assumed merely because the policy does not continue.
Worked Example: Finding an Omitted Termination Profit
A hypothetical two-year contract has a \(10\%\) termination probability in each year. At each year-end, a policy that survives earns \(100\). A policy that terminates incurs an end-of-year profit of \(-200\). The effective annual interest rate is \(10\%\). Find the correct signature and NPV, and compare them with a calculation that omits the termination outcome.
Method. Calculate the conditional expected profit in each year using both outcomes, then multiply by the probability of reaching that year. This avoids counting only surviving policies while leaving the terminating policies’ profit out.
The probabilities of being in force at the start of years 1 and 2 are \(1\) and \(0.90\). In either year, conditional expected profit is \(0.90(100)+0.10(-200)=90-20=70\). Thus \(\sigma_1=1(70)=70\) and \(\sigma_2=0.90(70)=63\). Discounting gives
or \(115.70\) to the nearest cent. If the termination profits are omitted, only surviving policies contribute: the resulting entries are \(1(0.90)(100)=90\) and \(0.90(0.90)(100)=81\). The resulting NPV is \(90/1.10+81/1.21=148.7603\), or \(148.76\). The omission overstates NPV by \(33.06\), the present value of the missing expected termination losses. This comparison checks both the outcome weights and the terminal amounts.
A Timing Check When Cash Flows Occur Within the Year
The end-of-year model is a convention, not a claim that every premium, expense, or claim is literally paid at year-end. To use it consistently, cash flows occurring earlier in the year must be carried forward to the chosen profit measurement date, or valued at their actual dates in the NPV calculation. For example, a premium received at the start of the year earns interest until year-end if the profit vector is measured at year-end.
Consider a hypothetical year in which a premium of \(50\) is received at the start, an expense of \(10\) is paid at year-end, and a termination benefit of \(200\) is paid at year-end if the policy terminates. Let the termination probability be \(0.10\), and let \(i=10\%\). Accumulating the premium to year-end gives \(50(1.10)=55\). The continuation profit is \(55-10=45\); the termination profit is \(55-10-200=-155\). The conditional expected year-end profit is \(0.90(45)+0.10(-155)=40.5-15.5=25\). Discounting this expected amount one year gives \(25/1.10=22.7273\) per policy in force at the start of the year. Treating the premium as if it arrived at year-end would instead use \(50\), not \(55\), in the profit calculation and would not match the stated timing.
A Practical Audit Sequence
When a signature produces an unexpected NPV, check its construction in a fixed order. This helps distinguish a probability-weighting error from a timing error or an omitted outcome, rather than trying to correct the final NPV without locating the source.
- Identify the unit. Confirm whether each profit is conditional on being in force, conditional on an outcome, or already per policy issued.
- Check the in-force weight. Year \(t\)'s weight includes survival only through year \(t-1\), not through year \(t\).
- List each outcome. Include continuation and termination profits with their conditional probabilities.
- Label the measurement date. Match the discount exponent to when the profit is valued or paid.
- Reconcile the total. Recompute NPV as the sum of discounted signature entries and verify that no duration is counted twice or skipped.
This audit is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In exam calculations, stating the probability weight and timing convention alongside each duration entry makes the reasoning easier to check and reduces the chance that a plausible-looking signature hides a structural error.
Check Your Understanding
For each problem, construct the signature using the stated measurement timing before calculating the requested value.
- A one-year hypothetical contract has a \(20\%\) termination probability. Its end-of-year profit is \(60\) on survival and \(-40\) on termination. At \(i=5\%\), find NPV per policy issued. (Answer: \(38.10\))
- A two-year contract has a \(10\%\) termination probability in each year, conditional on being in force at the start of the year. End-of-year profit is \(100\) on survival and zero on termination. At \(i=0\%\), give the two signature entries and total NPV. (Answer: signature \((90,\ 81)\); NPV \(171\))
- A two-year hypothetical contract has \(q_1=0.25\) and \(q_2=0.20\). In year 1, end-of-year profit is \(40\) on survival and \(0\) on termination. In year 2, it is \(100\) on survival and \(20\) on termination. At \(i=10\%\), find NPV per policy issued. (Answer: \(79.34\))
- A three-year contract has a \(20\%\) annual termination probability, zero termination profit, and end-of-year continuation profit of \(50\) in each year. At \(i=10\%\), find its undiscounted signature and NPV. (Answer: signature \((50,\ 40,\ 32)\); NPV \(102.55\))
- A two-year hypothetical contract has a \(10\%\) annual termination probability. At each year-end, continuation profit is \(80\) and termination profit is \(-120\). At \(i=0\%\), find the signature and total NPV. (Answer: signature \((60,\ 54)\); NPV \(114\))