Build the core actuarial skills assessed by the October 2026 Society of Actuaries FAM syllabus: transform insured losses, model aggregate claims, estimate parameters, apply credibility, develop reserves, indicate rates, value options, and calculate life-contingent benefits, premiums, and policy values. This free guide combines 20 original five-choice questions with 60 worked concepts across all ten syllabus domains. Every practice item supplies its required assumptions and data. Solutions emphasize timing, conditioning, parameterization, and the distinction between quantities that look similar but answer different questions. This is a representative preview, not a full-length examination. Its scope follows the supplied October 2026 syllabus as of October 16, 2026.
What this guide covers
Practice October 2026 SOA FAM with 20 original questions, worked solutions, and 60 focused actuarial concepts.
Short-Term Insurance and Reinsurance Coverages
1. Ordinary deductibles: payment per loss versus per payment
An ordinary deductible d transforms ground-up loss X into insurer payment Y = max(X − d, 0). Its unconditional mean is E[Y] = ∫d∞ S_X(x) dx, which includes zero payments for losses below the deductible. The mean payment given that a payment occurs is E[Y]/P(X > d), provided that probability is positive. These quantities answer different pricing questions: payment per original loss versus payment per paid claim. A deductible also reduces paid-claim frequency, so combining conditional severity with original frequency overstates aggregate cost.
Worked example: Suppose losses are exponential with mean $1,000 and the deductible is $500. Payment occurs with probability e^(−0.5), about 0.6065. Expected payment per loss is $606.53, while expected payment per paid claim remains $1,000 because of exponential memorylessness.
Common mistake: Using conditional payment severity with the original claim count overlooks losses eliminated entirely by the deductible and overstates expected payments.
Official syllabus reference: [1]
2. Coinsurance and payment limits depend on contract order
A payment formula must distinguish a limit on covered loss from a limit on insurer payment. If the insurer pays proportion c after deductible d and its payment is capped at m, then Y = min[c(X − d)₊, m]. If a covered-loss limit u applies before coinsurance, the formula instead becomes c min[(X − d)₊, u]. These arrangements can produce different maxima and attachment points. Write the transformation before integrating, estimating moments, or allocating reinsurance, because identical numerical labels do not establish identical coverage.
Worked example: For a $100,000 loss, a $10,000 deductible, and 80% insurer participation, the uncapped payment is $72,000. A $60,000 payment cap produces $60,000, whereas a $60,000 covered-loss cap before participation produces only $48,000.
Common mistake: Applying every limit before coinsurance silently changes a payment cap into a covered-loss cap and can materially understate coverage.
Official syllabus reference: [1]
3. Loss elimination ratios and deductible erosion under inflation
The loss elimination ratio for an ordinary deductible is LER(d) = E[min(X,d)]/E[X]. Thus expected insurer payment is E[X][1 − LER(d)]. If ground-up losses inflate by factor k while the deductible stays fixed, payment becomes (kX − d)₊ = k(X − d/k)₊. Inflation therefore scales losses and lowers the deductible in original-loss units. Payment inflation can exceed ground-up inflation because more losses cross the deductible. This calculation assumes unchanged loss frequencies and no additional policy limits or coverage changes.
Worked example: If X is uniform from $0 to $100, its mean is $50. With a $20 deductible, expected payment is $32. After 10% ground-up inflation, losses are uniform to $110 and expected payment becomes $90²/$220 = $36.82, a 15.1% increase.
Common mistake: Multiplying existing insurer payments by the inflation factor misses the additional payment caused by a fixed deductible losing relative value.
Official syllabus reference: [1]
4. Quota share versus surplus-share proportional reinsurance
Quota-share reinsurance cedes a fixed proportion of every covered policy payment, subject to treaty terms. Surplus-share reinsurance determines the ceded proportion from the policy's insured amount relative to the insurer's retained line and available treaty capacity. Once that proportion is established for a policy, covered losses are shared proportionally. Surplus share therefore permits different cession percentages across policies, while quota share ordinarily uses the same treaty percentage. These arrangements differ from excess-of-loss protection, whose recovery depends directly on the size of the covered loss.
Worked example: An insurer retains a $100,000 line on a $400,000 insured amount and has enough surplus treaty capacity for the remainder. The ceded share is 75%, so a covered $80,000 payment generates a $60,000 recovery, unlike a treaty with a fixed 30% quota share.
Common mistake: Treating surplus share as excess of loss confuses how the policy's cession percentage is established with how individual claims are paid.
Official syllabus reference: [1]
5. Per-loss layers versus aggregate stop-loss protection
A layer described as m excess of a pays min[(X − a)₊, m] on its defined subject loss. A per-loss treaty applies that calculation separately to each covered claim; aggregate stop-loss applies it to a specified period's aggregate subject loss. Generally, summing per-loss recoveries does not equal applying the same layer to total losses. Attachment, limit, covered loss basis, and any preceding reinsurance must be identified explicitly. Pricing requires the expected layer payment, while reserving must also reflect payments and recoveries already recognized.
Worked example: Two claims of $70,000 each produce $40,000 total recovery under $50,000 excess of $50,000 per loss: each contributes $20,000. A $50,000 excess of $100,000 annual aggregate treaty instead recovers $40,000 on their $140,000 combined amount.
Common mistake: Applying an aggregate attachment separately to each claim changes the treaty and can eliminate recoveries that aggregate protection was designed to provide.
Official syllabus reference: [1]
Severity, Frequency, and Aggregate Models
6. Scale, shape, and transformed severity families
A scale parameter changes monetary size without changing the standardized distribution's shape: if X = θW, quantiles scale by θ and kth moments by θᵏ. Shape parameters instead alter relative dispersion or tail behavior. For a Weibull survival function exp[−(x/θ)ᵏ], θ is scale and k is shape; k = 1 gives an exponential distribution. A gamma with shape one is also exponential. For lognormal X = exp(μ + σZ), multiplying losses shifts μ by the logarithm of the multiplier while leaving σ unchanged.
Worked example: A Weibull model with scale $2,000 and fixed shape has median θ(ln 2)^(1/k). Increasing scale by 15% increases its median and mean by 15%, but multiplies variance by 1.15² rather than by 1.15.
Common mistake: Assuming every parameter measured near the mean is a scale parameter ignores how shape changes alter tail probabilities and relative variability.
Official syllabus reference: [1]
7. Pareto tails and the existence of moments
For a Pareto Type II severity with survival S(x) = (θ/(θ + x))ᵅ, the positive kth moment exists only when α > k. In particular, the mean is θ/(α − 1) for α > 1, and variance is θ²α/[(α − 1)²(α − 2)] for α > 2. A finite mean does not imply finite variance. This distinction determines whether variance-based aggregate approximations and credibility calculations are available. Check the survival function's parameterization before borrowing formulas from a different Pareto convention.
Worked example: With θ = $3,000 and α = 1.5, expected severity is $6,000, but the second moment and variance are infinite. Expected aggregate loss can still be calculated for a finite-mean count model, although an ordinary finite-variance normal approximation lacks its required variance.
Common mistake: A plausible sample variance does not establish a finite population variance when the fitted tail shape implies the second moment diverges.
Official syllabus reference: [1]
8. The (a,b,0) frequency recursion
The (a,b,0) class satisfies p_k = (a + b/k)p_(k−1) for k ≥ 1, with p₀ and normalization completing the model. It includes Poisson, binomial, and negative binomial distributions. For Poisson mean λ, a = 0 and b = λ. For binomial parameters m and q, a = −q/(1−q) and b = (m+1)q/(1−q). Negative binomial models allow variance above the mean, whereas binomial variance is below the mean. Parameterization must be stated when interpreting negative binomial probabilities.
Worked example: For a Poisson count with mean 2, p₀ = e^(−2) and p₁ = 2p₀. The recursion gives p₂ = (2/2)p₁ = 2e^(−2). This computes successive probabilities without repeatedly evaluating factorials and also identifies the distribution's recursion coefficients.
Common mistake: Matching only the mean can select Poisson even when observed overdispersion makes a negative binomial model substantially more appropriate.
Official syllabus reference: [1]
9. Zero truncation versus zero modification
Zero truncation conditions a count on being positive, giving p_k^T = p_k/(1−p₀) for k ≥ 1 and zero probability at zero. Zero modification instead selects a new zero probability and rescales the positive probabilities while preserving their relative proportions. Members of the (a,b,1) class retain the recursion from k = 2 onward, allowing the initial probability structure to differ. These distinctions matter when a database omits claim-free policies or a portfolio has an unusually large or small zero-count proportion.
Worked example: A Poisson distribution with mean 1 has p₀ ≈ 0.3679. Among policies known to have at least one claim, the probability of exactly one claim is 0.3679/(1−0.3679) ≈ 0.5820, not the unconditional Poisson probability.
Common mistake: Fitting an ordinary count distribution to positive-only records treats missing zero-count observations as evidence that zero claims are intrinsically unlikely.
Official syllabus reference: [1]
10. Individual risk models retain policy-level differences
An individual risk model writes aggregate loss as a sum over a fixed set of policies, often using Y_i = I_iX_i with claim indicator I_i. If I_i is independent of severity X_i, E[Y_i] = q_iE[X_i] and Var(Y_i) = q_iVar(X_i) + q_i(1−q_i)(E[X_i])². Independent policy variances add; dependent policies require covariance terms. A collective model instead starts with a random total claim count and claim severities. Choosing between them depends on whether individual policy characteristics or pooled claim behavior are central.
Worked example: Two independent policies each pay $1,000 with probability 0.10 and otherwise pay nothing. Each has mean $100 and variance $90,000, so their aggregate mean is $200 and variance $180,000, using fixed policy-level Bernoulli outcomes.
Common mistake: Replacing a small fixed portfolio with a Poisson count without checking the approximation changes its count variance and aggregate loss distribution.
Official syllabus reference: [1]
11. Collective aggregate moments by conditioning
For S = Σᵢ₌₁ᴺX_i with independent identically distributed severities independent of N, E[S] = E[N]E[X]. Total variance gives Var(S) = E[N]Var(X) + Var(N)(E[X])². The first component represents severity variability at a given count, and the second represents changing numbers of claims. For Poisson N, this simplifies to Var(S) = λE[X²]. The simplification does not extend to arbitrary counts. Dependence between counts and severities, or dependence among severities, requires a different conditional analysis.
Worked example: Suppose annual counts have mean 10 and variance 15, while severity has mean $200 and variance $10,000. Aggregate mean is $2,000. Aggregate variance is 10 × 10,000 + 15 × 200² = $700,000 in squared-dollar units.
Common mistake: Using E[N]E[X²] for every collective model incorrectly substitutes the count mean for its variance outside the Poisson case.
Official syllabus reference: [1]
12. Convolution constructs aggregate probabilities
Convolution combines independent loss distributions by summing products of probabilities for combinations producing the same total. For discrete independent X and Y, P(X+Y=s) = Σ_x P(X=x)P(Y=s−x). Repeated convolution handles fixed claim counts. A collective model then mixes those conditional distributions over N, including S = 0 when N = 0. Specify the loss grid and preserve units when discretizing continuous severities. Convolution is exact for the stated discrete model, whereas discretization introduces an additional approximation that should not be confused with it.
Worked example: Two independent claims each equal $100 with probability 0.7 or $200 with probability 0.3. Their total equals $200, $300, or $400 with probabilities 0.49, 0.42, and 0.09. The middle probability includes both possible claim orderings.
Common mistake: Counting only one ordering of unequal severities omits valid combinations and causes the resulting aggregate probabilities to sum to less than one.
Official syllabus reference: [1]
13. Moment-matched aggregate approximations and stop-loss
A normal approximation matches aggregate mean m and variance s², while a lognormal match uses σ² = ln(1+s²/m²) and μ = ln(m)−σ²/2 for positive m. The normal can assign probability to negative losses; the lognormal cannot represent a positive atom at zero. For a normal approximation, expected stop-loss above d is sφ(k)+(m−d)[1−Φ(k)], where k=(d−m)/s. This is an approximation to E[(S−d)₊], not merely the probability of exceeding d. Tail accuracy depends on the underlying aggregate distribution.
Worked example: If aggregate losses are approximated by a normal distribution with mean $1,000 and standard deviation $200, stop-loss attachment at $1,000 gives expected payment 200φ(0), approximately $79.79. The exceedance probability is 0.50, but that probability alone does not value the coverage.
Common mistake: Multiplying an attachment by its exceedance probability does not calculate stop-loss cost, because payment depends on the amount above attachment.
Official syllabus reference: [1]
14. VaR, TVaR, and coherent risk-measure properties
Value at Risk is a loss quantile, while Tail Value at Risk averages quantiles above the selected probability level. With probability mass at VaR, TVaR requires only enough of that mass to complete the tail; it need not equal a simple conditional expectation. Coherence combines monotonicity, translation invariance, positive homogeneity, and subadditivity. TVaR is coherent for integrable losses under its quantile definition, whereas VaR can fail subadditivity. These properties describe how a measure responds to increased losses, cash additions, scaling, and portfolio aggregation.
Worked example: Two independent risks each lose $100 with probability 0.04 and otherwise lose zero. Each has 95% VaR of zero. Their sum has probability 0.9216 of zero and 95% VaR of $100, demonstrating that portfolio VaR can exceed the sum of individual VaRs.
Common mistake: Assuming diversification must reduce every numerical risk measure overlooks VaR's possible failure of subadditivity, especially for discrete rare-loss portfolios.
Official syllabus reference: [1]
Parametric Estimation
15. Complete-data likelihoods and parameter conventions
For independent complete observations, the likelihood multiplies their model densities or probability masses, viewed as a function of unknown parameters. Maximizing the log-likelihood often turns products into manageable sums. Under an exponential mean parameter θ, θ̂ is the sample mean; under a Poisson mean λ with equal exposure, λ̂ is the sample average count. Different exposures require exposure-adjusted means. Always check parameter constraints, boundary candidates, and whether a stationary point is a maximum. A rate parameter and its reciprocal mean parameter produce different numerical estimates.
Worked example: Three independent complete exponential losses are $200, $400, and $900. The mean-parameter estimate is θ̂ = $500. If the model is instead written with rate β, the estimate is β̂ = 1/500 per dollar; these describe the same fitted distribution.
Common mistake: Reporting a fitted exponential rate as though it were a mean reverses its interpretation and gives the estimate incorrect units.
Official syllabus reference: [1]
16. Grouped-data likelihoods use interval probabilities
When observations are recorded only as interval counts, the likelihood uses interval probabilities F(b_j)−F(a_j), raised to observed counts n_j. Multinomial constants can be omitted during maximization when they do not involve the model parameters. An open-ended final interval contributes a survival probability. Replacing each group by its midpoint generally changes the likelihood and may be especially misleading for wide or heavy-tailed intervals. Ensure the intervals match the sampling mechanism: complete grouped data and grouped data conditional on passing a threshold have different likelihoods.
Worked example: For exponential mean θ, suppose 60 losses are at most $1,000 and 40 exceed $1,000. The likelihood is proportional to [1−e^(−1000/θ)]⁶⁰[e^(−1000/θ)]⁴⁰. Maximization sets the fitted upper-group probability to 0.40, giving θ̂ = −1,000/ln(0.40).
Common mistake: Treating an open-ended group as its lower endpoint systematically understates large observations and is not equivalent to grouped maximum likelihood.
Official syllabus reference: [1]
17. Truncation conditions the sample; censoring limits information
A right-censored observation at c contributes S(c), because its exact value is unknown but exceeds c. A left-truncated observation x above d contributes f(x)/S(d), because only losses exceeding d enter the sample. When both apply, an upper-censored record contributes S(c)/S(d). For payment observations y = x−d, express the density at x = y+d with the truncation adjustment. Censoring retains a record with incomplete measurement; truncation removes ineligible records entirely. Their likelihood adjustments therefore cannot be exchanged or ignored.
Worked example: An insurer records only losses above $500 and caps recorded amounts at $5,000. An exact recorded ground-up loss of $1,200 contributes f(1,200)/S(500). A capped record contributes S(5,000)/S(500), because it establishes exceedance but not an exact $5,000 loss.
Common mistake: Calling every capped record an exact observation adds unsupported density information and generally biases estimates toward the recording limit.
Official syllabus reference: [1]
Introduction to Credibility
18. Full credibility from a probability tolerance
Limited-fluctuation credibility asks whether experience is sufficiently stable to fall within a selected relative tolerance r of its expectation with a selected probability. Under a normal approximation to Poisson counts, full credibility requires expected count λ ≥ (z/r)², where z is the chosen symmetric normal quantile. This threshold follows from standard deviation √λ and tolerance rλ. It is a modeling criterion, not a universal industry threshold. The approximation, selected confidence probability, and appropriate exposure basis must be stated before applying the numerical requirement.
Worked example: If a procedure selects z = 2 and tolerance r = 0.05, full credibility requires (2/0.05)² = 1,600 expected claims. Halving the tolerance to 0.025 quadruples the requirement to 6,400 because the count threshold depends on inverse squared tolerance.
Common mistake: Doubling the confidence tolerance does not double the required claim count; the normal-approximation count requirement changes with the square of the ratio.
Official syllabus reference: [1]
19. Severity variability and partial classical credibility
For compound Poisson aggregate losses, relative variance depends on severity variability as well as claim count. If severity coefficient of variation is CV, a full-credibility expected count is approximately (z/r)²(1+CV²). A common partial-credibility rule is Z = min[1,√(n/n₀)], followed by Z times experience plus (1−Z) times a suitable complement. The selected complement should estimate the same prospective quantity and exposure basis. Classical credibility measures fluctuation under chosen assumptions; it does not derive an optimal Bayesian weight from between-risk heterogeneity.
Worked example: With z = 2, tolerance 10%, and severity CV = 2, the full-credibility count is 400 × 5 = 2,000. If the applicable experience count is 500, Z = 0.50, so $300 experience blended with a $240 complement gives $270.
Common mistake: Using a frequency-only credibility standard for highly variable aggregate losses overlooks severity variation and gives experience more weight than the stated criterion supports.
Official syllabus reference: [1]
Pricing and Reserving for Short-Term Insurance Coverages
20. Development triangles and volume-weighted link ratios
A development triangle organizes cumulative claims by origin period and development age. A volume-weighted age-to-age factor divides the sum of later cumulative amounts by the sum of corresponding earlier amounts for matched periods. Do not include unmatched recent observations in the denominator. Paid and reported triangles measure different processes and require separately selected patterns. Calendar-period effects, operational changes, and changing business mix can disturb the assumed comparability of rows. Link ratios are estimates of development relationships, rather than proof that every origin period follows an identical trajectory.
Worked example: Two matched accident years have cumulative paid claims of $100 and $200 at 12 months, increasing to $150 and $260 at 24 months. The volume-weighted factor is 410/300 = 1.3667, rather than the unweighted average of 1.50 and 1.30.
Common mistake: Averaging individual development ratios without recognizing the selected weighting method gives small and large origin periods equal influence on the result.
Official syllabus reference: [1]
21. Chain-ladder separates ultimate claims from outstanding claims
Chain-ladder multiplies the latest cumulative claim amount by a selected cumulative development factor to estimate ultimate claims. For a paid triangle, outstanding claims equal estimated ultimate less paid to date. For a reported triangle, ultimate less reported estimates additional reporting development; total unpaid claims still require subtraction of paid, not reported. A tail factor may extend the observed development pattern beyond the last column. The method is responsive to current experience but can be unstable for immature periods and misleading when historical development no longer represents current claims handling.
Worked example: An accident year has $800,000 paid and a paid-to-ultimate factor of 1.50. Chain-ladder estimates ultimate claims of $1,200,000 and unpaid claims of $400,000. Reporting the ultimate amount as the reserve would include $800,000 already paid to claimants.
Common mistake: Subtracting reported claims from a reported-based ultimate produces unreported development, not the full unpaid claims reserve when case reserves remain outstanding.
Official syllabus reference: [1]
22. Expected-loss-ratio reserving uses an independent ultimate
The expected-loss-ratio method estimates ultimate claims as earned premium multiplied by a selected expected ultimate loss ratio. Outstanding claims are that ultimate estimate less paid claims. The selected ratio should match the premium's rate level, business mix, coverage, and loss cost period. Unlike chain-ladder, the ultimate estimate does not directly scale emerging claims. This can stabilize immature periods but can also delay recognition of a changed loss environment. The method's reserve can become implausible if paid claims exceed the independently selected ultimate, signaling assumptions needing reconsideration.
Worked example: Earned premium is $5 million, the selected expected loss ratio is 60%, and paid claims are $1.2 million. Expected ultimate claims are $3 million and the unpaid reserve is $1.8 million, regardless of the current paid development percentage.
Common mistake: Applying an expected loss ratio selected at old rates to premium at new rates can change expected claims without any underlying change in risk.
Official syllabus reference: [1]
23. Bornhuetter–Ferguson combines emergence with expected unobserved claims
Paid Bornhuetter–Ferguson estimates ultimate claims as paid claims plus independently expected ultimate claims times the expected unpaid proportion. If the paid cumulative development factor is F, that proportion is 1−1/F. Reported Bornhuetter–Ferguson instead estimates unreported development and then combines it with reported claims; obtaining total unpaid still requires subtracting paid. The method gives observed emergence increasing influence as development progresses. It differs from merely averaging chain-ladder and expected-loss-ratio ultimate estimates with arbitrary weights, although a development-based weighted representation can be derived.
Worked example: Expected ultimate claims are $2 million, actual paid claims are $700,000, and 25% is expected paid. The paid Bornhuetter–Ferguson reserve is $1.5 million and ultimate is $2.2 million. Chain-ladder would instead estimate ultimate at $2.8 million from current payments.
Common mistake: Multiplying actual paid claims by the expected unpaid percentage loses the independent expected-claims component that distinguishes Bornhuetter–Ferguson from development-only methods.
Official syllabus reference: [1]
24. Development and trend adjust different time dimensions
Development adjusts immature recorded losses toward their ultimate amount for the original exposure period. Trend moves loss costs from that historical period to the prospective cost period. A multiplicative annual trend g over h years gives factor (1+g)ʰ, with h determined by the relevant average dates. Frequency and severity may be trended separately when exposures and data support that approach. Development does not automatically account for future economic inflation, and trend does not complete unreported claims. Consistent treatment of claim adjustment expenses and coverage changes is also necessary.
Worked example: Reported historical losses of $500,000 have a development factor of 1.20 and require two years of 4% annual trend. Prospective losses are $500,000 × 1.20 × 1.04² = $648,960, before any separate exposure or coverage adjustment.
Common mistake: Using the valuation date as the start of every trend period can misstate the interval because historical losses represent an earlier average cost date.
Official syllabus reference: [1]
25. Premium on-level adjustments isolate rate changes
Premium on-leveling restates historical premium at a common rate level so historical loss ratios can support a prospective rate indication. The factor compares the selected current rate level with the average rate level represented in earned premium. An extension-of-exposures calculation rerates individual historical exposures where adequate detail exists; parallelogram methods approximate earning patterns under stated assumptions. On-leveling addresses rate changes, not claim inflation or exposure growth. If historical premium is understated relative to current rates, failing to on-level generally overstates the indicated increase from a loss-ratio analysis.
Worked example: Historical earned premium is $900,000 at average rate level 1.00, while current rates correspond to 1.08. On-level premium is $972,000. Adjusted losses of $650,000 therefore produce a current-level loss ratio of about 66.87%, rather than 72.22%.
Common mistake: Trending losses and increasing premium by the same inflation assumption does not substitute for reconstructing the actual rate levels represented in earned premium.
Official syllabus reference: [1]
26. Loss-cost and loss-ratio methods fund fixed and variable expenses
Under a loss-cost approach, a rate per exposure can be written R = (L+F)/(1−v−p), where L is prospective loss cost, F fixed expense per exposure, v variable expense proportion, and p target profit proportion. The denominator reflects expenses and profit measured as shares of premium. A loss-ratio indication compares required premium with current-level premium for matching exposures. Distinguish fixed dollars from premium-proportional costs and avoid counting an expense in both categories. Consistent assumptions should make the two methods reconcile for the same underlying portfolio.
Worked example: Expected losses are $400 per policy, fixed expenses $50, variable expenses 20% of premium, and profit 5%. Required rate is $450/0.75 = $600. If current rate is $570, the indicated increase is $600/$570 − 1, about 5.26%.
Common mistake: Adding 25% to losses and fixed expenses does not fund costs defined as 25% of premium; those costs require division by 0.75.
Official syllabus reference: [1]
Option Pricing Fundamentals
27. European option payoffs and put-call parity
A European call pays max(S_T−K,0), while a European put pays max(K−S_T,0). For matching strike and maturity on a non-dividend-paying asset, put-call parity is C−P = S₀−Ke^(−rT) under continuous interest. This follows because call plus a bond paying K has the same terminal payoff as put plus the asset. Use a discount factor consistent with the stated interest convention. Parity concerns current prices of replicating portfolios and does not equate a premium with its terminal payoff or investment profit.
Worked example: A stock costs $100, a one-year strike-$100 call costs $12, and the discount factor to maturity is 0.95. Parity gives P = 12 − 100 + 95 = $7. The put price follows without specifying a physical stock-return probability.
Common mistake: Subtracting the full strike rather than its present value mixes time-zero prices with a maturity cash flow and breaks parity.
Official syllabus reference: [1]
28. Binomial replication and risk-neutral probabilities
In a one-period binomial model, a non-dividend-paying asset moves from S₀ to uS₀ or dS₀. With risk-free accumulation R, the risk-neutral up probability is q = (R−d)/(u−d), provided d<R<u. A derivative price is its discounted risk-neutral expected payoff. Equivalently, hold Δ = (V_u−V_d)/[S₀(u−d)] units of the asset and a risk-free position to replicate both states. Multi-period valuation applies the same logic backward through the tree. Physical expected returns do not replace the no-arbitrage probability.
Worked example: An asset at $50 moves to $60 or $45, and R = 1.02. A call struck at $50 pays $10 or zero. Its hedge uses 10/15 = 2/3 shares, and q = (51−45)/15 = 0.40 gives price $4/1.02.
Common mistake: Using a forecasted up probability prices expected investment returns rather than the replicating payoff and can violate no-arbitrage valuation.
Official syllabus reference: [1]
29. Black–Scholes delta measures local share exposure
In the Black–Scholes model, a European call on a non-dividend-paying stock has value C = SΦ(d₁) − Ke^(−rT)Φ(d₂), where d₁ = [ln(S/K) + (r + σ²/2)T]/(σ√T) and d₂ = d₁ − σ√T. Its delta is Φ(d₁), the derivative of price with respect to stock price. A trader short m calls therefore buys mΦ(d₁) shares to offset small immediate stock movements. This hedge requires rebalancing because delta changes with stock price, time, and volatility. The same d₁ and d₂ must support both the quoted price and hedge.
Worked example: Let S = K = 100, r = 0, T = 1, and σ = 0.40. Then d₁ = 0.20 and d₂ = −0.20. Using Φ(0.20) = 0.57926, the call price is approximately 15.852. A trader short 100 calls buys approximately 57.926 shares for a local delta hedge.
Common mistake: Using Φ(d₂) as call delta confuses a pricing weight with local stock sensitivity; a static delta hedge also drifts as market inputs change.
Official syllabus reference: [1]
Long-Term Insurance Coverages and Retirement Financial Security Programs
30. Life and health promises, benefit triggers, and insurable interest
Long-term contracts are distinguished by the event triggering benefits and the duration of the promise. Term life pays on covered death during a specified period; whole life covers death throughout life; endowment combines death coverage with a survival payment. Long-term health coverage may pay on disability or other defined health events, requiring a trigger beyond simple mortality. Insurable interest describes a legitimate relationship to the insured risk that supports the contract's insurance purpose. Its precise legal application depends on jurisdiction and contract circumstances, rather than a universal examination rule.
Worked example: A borrower buys ten-year term coverage to protect a remaining mortgage balance. Death during the covered term triggers payment, while survival to the term's end does not. Replacing it with an endowment adds a survival benefit and changes the expected payment pattern.
Common mistake: Assuming every long-term contract pays only at death overlooks survival and health-state benefits and can lead to the wrong contingent-payment model.
Official syllabus reference: [1]
31. Pension promises and annuitization allocate different risks
A defined benefit pension specifies a benefit formula, often based on salary and service, so funding must support that promise despite investment and longevity experience. A defined contribution arrangement specifies contributions, leaving retirement resources dependent on account accumulation unless additional guarantees apply. A life annuity converts assets into payments contingent on survival and pools longevity risk across annuitants. Payment guarantees, survivor options, and inflation features change the promise and its cost. Distinguish the pension's accumulation design from the retirement payout product subsequently selected by a member.
Worked example: Two employees each retire with $300,000 in defined contribution accounts. One takes withdrawals and retains investment and longevity risk. The other purchases a fixed life annuity, transferring the covered longevity risk to the insurer while accepting the annuity's payment terms.
Common mistake: An annuity purchased from a defined contribution account changes the payout arrangement without retroactively changing the employer's contribution promise into defined benefit.
Official syllabus reference: [1]
Mortality Models
32. Conditional survival notation links lifetime distributions
For a life aged x, T_x is complete future lifetime and ₜp_x = P(T_x>t). If S₀ is survival from birth, ₜp_x = S₀(x+t)/S₀(x), provided S₀(x)>0. Death within t years has probability ₜq_x = 1−ₜp_x. Deferred death probability is ₜ|ᵤq_x = ₜp_x·ᵤq_(x+t). This conditioning distinguishes attained-age survival from unconditional survival since birth. It also supports life-table ratios such as ₜp_x = l_(x+t)/l_x when the table and fractional-age assumptions supply those values.
Worked example: If a life table has l₄₀=90,000, l₄₅=85,500, and l₄₇=82,935, survival from 40 to 45 is 0.95. Death between ages 45 and 47, viewed at age 40, has probability (85,500−82,935)/90,000 = 0.0285.
Common mistake: Using the probability of death after the deferred age without first allowing for survival to that age overstates the unconditional deferred event.
Official syllabus reference: [1]
33. Force of mortality is an instantaneous conditional rate
The force of mortality satisfies μ_(x+t) = −d ln(ₜp_x)/dt wherever differentiable. Therefore ₜp_x = exp[−∫₀ᵗ μ_(x+s)ds], and the future-lifetime density is f_T(t)=ₜp_x μ_(x+t). The force is a rate per unit time, not a probability, and its integral is cumulative hazard. For constant force μ, survival is e^(−μt) and one-year death probability is 1−e^(−μ). This distinction is important when converting annual mortality rates, integrating benefits, or applying an extra mortality force.
Worked example: A constant force of 0.03 per year gives two-year survival e^(−0.06), approximately 0.9418. Its one-year death probability is about 0.02955, slightly below 0.03. Multiplying the force directly by long durations would not produce a valid exact death probability.
Common mistake: Equating an annual death probability with the force ignores survival during the interval; the two coincide only as a small-interval approximation.
Official syllabus reference: [1]
34. Uniform distribution of deaths gives linear lives, not constant hazard
Under uniform distribution of deaths between integer ages x and x+1, l_(x+t) = l_x(1−tq_x) for 0≤t≤1. Thus survival from x+s to x+t is (1−tq_x)/(1−sq_x), and the fractional-age force is q_x/(1−tq_x). Although deaths are uniform relative to the initial cohort, the surviving population shrinks, so the force increases within the year. UDD interpolates a supplied annual table; it does not assert a constant mortality force. Crossing an integer age requires the next year's table rate.
Worked example: If q_x=0.12, survival from x+0.25 to x+0.75 is 0.91/0.97≈0.93814. The force at x+0.75 is 0.12/0.91≈0.13187 per year, illustrating why uniform deaths do not imply a flat conditional hazard.
Common mistake: Subtracting interval length times q_x from one without conditioning is valid only for an interval beginning at integer age x.
Official syllabus reference: [1]
35. Constant-force interpolation gives geometric survival
Under constant force between integer ages, set μ = −ln(p_x) within that year. Fractional survival from x to x+t is p_xᵗ, and from x+s to x+t is p_x^(t−s), provided both ages remain inside the interval. The corresponding lives interpolate geometrically rather than linearly. This differs from UDD while agreeing at integer endpoints. For an interval crossing a birthday, multiply the appropriate within-year factors using each annual survival probability. Keep the interpolation assumption explicit when valuing monthly or immediately payable benefits from annual data.
Worked example: If annual survival is 0.81, constant-force half-year survival is √0.81=0.90. Under UDD, survival from the integer age through half a year is 1−0.5×0.19=0.905. Both assumptions reproduce annual survival 0.81 but give different fractional values.
Common mistake: Applying one year's force across a birthday ignores the next year's mortality rate and extends the interpolation assumption beyond its stated interval.
Official syllabus reference: [1]
36. Select tables track selection age and duration
A select mortality rate q_[x]+t identifies both age at selection x and duration t since selection. Mortality can differ between lives at the same attained age because their selection durations differ. After the stated select period, ultimate rates indexed by attained age apply. A survival path must therefore move across the appropriate select row before entering ultimate mortality. Do not interpret the bracketed selection age as the current age at every duration. Select effects model the mortality consequences of selection; they do not automatically specify how that selection was performed.
Worked example: For a two-year select period, a life selected at age 45 uses q_[45] during year one, q_[45]+1 during year two, and ultimate q₄₇ during year three. A newly selected age-47 life can have a different first-year rate despite the same attained age.
Common mistake: Switching immediately to ultimate mortality after the first birthday discards the remaining select period and may misprice newly selected lives.
Official syllabus reference: [1]
37. Complete lifetime moments come from survival integrals
For nonnegative complete future lifetime T_x, E[T_x] = ∫₀∞ ₜp_x dt when the integral exists. Its second moment is E[T_x²] = 2∫₀∞ t·ₜp_x dt, and variance subtracts the squared mean. These formulas follow from tail integration and avoid differentiating survival into a density. If mortality changes at specified durations, split the integrals and preserve survival to each later segment. For a finite limiting age, stop integration at the maximum remaining lifetime rather than extending survival beyond the model's support.
Worked example: With constant force 0.04, complete lifetime is exponential: E[T]=25 years and E[T²]=2/0.04²=1,250 years squared. Variance is 1,250−25²=625, so the standard deviation also equals 25 years under this particular mortality model.
Common mistake: Integrating the density without a lifetime multiplier gives total probability, not expected lifetime; the survival integral already incorporates the required weighting.
Official syllabus reference: [1]
38. Curtate lifetime probabilities and discrete tail sums
The curtate future lifetime K_x counts complete future years lived, so K_x = floor(T_x). Its probability mass function is P(K_x = k) = ₖp_x q_(x+k), equivalently the difference between survival probabilities to times k and k + 1. For a finite expectation, E[K_x] = Σₖ₌₁∞ ₖp_x: each integer survival probability contributes one possible completed year. This discrete tail sum differs from the integral used for complete lifetime. Payment timing must remain explicit: annual insurance paid at the end of the death year uses time K_x + 1.
Worked example: A model gives survival probabilities at years 1, 2, 3, and 4 of 0.80, 0.50, 0.20, and 0. Thus E[K_x] = 0.80 + 0.50 + 0.20 = 1.50 years, while P(K_x = 2) = 0.50 − 0.20 = 0.30.
Common mistake: Replacing T_x with K_x for an immediate death benefit advances payment and overstates present value at positive interest.
Official syllabus reference: [1]
39. De Moivre, Gompertz, and Makeham imply different hazards
Under De Moivre's law with limiting age ω, future lifetime at age x is uniform on (0,ω−x), survival is (ω−x−t)/(ω−x), and force is 1/(ω−x−t). Gompertz instead uses an exponentially increasing age-related force Bcʸ, with c>1. Makeham adds an age-independent component A, giving μ_y=A+Bcʸ. Integrating the force produces the corresponding survival model. These laws differ in support and aging behavior, so formulas should be derived from the stated model rather than chosen merely because a life table is absent.
Worked example: Under De Moivre with limiting age 100, a life aged 60 has 40 years of maximum future lifetime and mean complete lifetime 20 years. Ten-year survival is 30/40=0.75, while the force at age 70 is 1/30 per year.
Common mistake: A finite limiting age is specific to models such as De Moivre and should not be imposed on an unbounded Gompertz model.
Official syllabus reference: [1]
Present Value Random Variables for Long-Term Insurance Coverages
40. Annual insurance present values follow the death year
For unit whole-life insurance payable at the end of the year of death, the present value random variable is Z=v^(K_x+1). Its expectation is A_x=Σₖ₌₀∞v^(k+1)ₖp_xq_(x+k). The extra one in the exponent places payment at the end of the death year rather than at its beginning. A benefit schedule can replace the unit amount with a year-specific benefit before summing. Identify the benefit trigger and payment time separately; the same death event can generate different present values under immediate and year-end settlement.
Worked example: If a life survives two complete years and dies during the third, K_x=2. A $50,000 benefit paid at death-year end has present value $50,000v³. Paying immediately at death instead requires the actual fractional lifetime, not merely the completed-year count.
Common mistake: Using v^K_x for an end-of-death-year benefit brings every payment forward one year and systematically overstates its present value.
Official syllabus reference: [1]
41. Endowment benefits partition death and survival outcomes
An n-year endowment insurance pays on death within n years or on survival to maturity. For a unit benefit paid at the end of the death year, its present value is v^(K_x+1) when K_x < n and v^n otherwise. Its actuarial present value therefore equals the sum of an n-year term insurance value and the pure endowment value v^n ₙp_x. These components occupy mutually exclusive outcomes. At positive interest, every payment occurs between years 1 and n, giving the useful consistency bounds v^n ≤ Z ≤ v.
Worked example: A two-year unit endowment has v = 0.90, q_x = 0.20, and q_(x+1) = 0.25. Term insurance contributes 0.20(0.90) + 0.80(0.25)(0.81) = 0.342. Survival contributes 0.80(0.75)(0.81) = 0.486. The total value is 0.828, exceeding the maturity discount factor 0.81.
Common mistake: Adding unrestricted whole-life insurance to a pure endowment double-counts benefits after maturity; use insurance restricted to deaths within the term.
Official syllabus reference: [1]
42. Annuity-due and annuity-immediate differ at payment dates
A unit whole-life annuity-due pays at times 0 through K_x and has present value Y=Σₖ₌₀ᴷvᵏ. Its expectation is ä_x=Σₖ₌₀∞vᵏₖp_x. A whole-life annuity-immediate begins at time one and pays only if the life survives each payment date, giving a_x=Σₖ₌₁∞vᵏₖp_x=ä_x−1. Temporary versions also impose a final payment boundary, so their difference requires attention to endpoints. The first due payment is certain for a life alive at valuation, while later payments remain contingent on survival.
Worked example: For a life dying during the third year, an annual annuity-due pays at times 0, 1, and 2, producing 1+v+v². The corresponding whole-life annuity-immediate pays at times 1 and 2, producing v+v², exactly one less.
Common mistake: Shifting an annuity's discount exponents without shifting its survival conditions can create a payment after death or remove a legitimate payment.
Official syllabus reference: [1]
43. Continuous insurance–annuity identities hold path by path
At constant force of interest δ>0, immediate unit whole-life insurance has present value Z=e^(−δT_x). A continuous whole-life annuity paying one per year has Y=∫₀ᵀe^(−δt)dt=(1−Z)/δ. Hence Z+δY=1 for every lifetime, and Ā_x+δā_x=1 after taking expectations. The annual counterpart is A_x+dä_x=1. These identities require matching whole-life benefits and timing conventions. Temporary or deferred arrangements need appropriate endpoint terms and cannot automatically use the unrestricted whole-life formula.
Worked example: At δ=0.05, suppose immediate unit insurance has expected value 0.40. The continuous whole-life annuity factor is (1−0.40)/0.05=12. A payment rate of $3,000 per year therefore has expected present value $36,000 under the same assumptions.
Common mistake: Substituting annual effective interest i for continuous force δ in the continuous identity mixes payment and discount conventions and changes the answer.
Official syllabus reference: [1]
44. Second moments use doubled discount force
For an insurance present value Z=e^(−δT), its second moment is E[e^(−2δT)], the corresponding unit insurance value evaluated at doubled force of interest. For annual death-year-end benefits, square v to obtain the second-moment discount factor; the equivalent effective interest is (1+i)²−1, not simply 2i. Variance is the second moment minus the squared first moment. Fixed benefits scale variance by their square. This technique applies directly to single discounted benefits, while multiple-payment annuities may require identities or cross-product terms.
Worked example: At effective interest 5%, an annual insurance second moment uses effective interest 1.05²−1=10.25%. If unit first and second moments are 0.30 and 0.12, a $10,000 benefit has variance 10,000²(0.12−0.09)=$3,000,000 in squared-dollar units.
Common mistake: Doubling effective interest rather than doubling the force gives an incorrect second-moment discount rate, even though the two may look numerically close.
Official syllabus reference: [1]
45. Covariance reflects shared lifetime dependence
Insurance and annuity present values on the same life are generally dependent. Under the continuous whole-life identity Y=(1−Z)/δ, Cov(Z,Y)=−Var(Z)/δ, and Var(Y)=Var(Z)/δ². Earlier death increases the discounted insurance benefit while reducing annuity payments, producing negative covariance. For a combined cash flow aZ+bY, variance includes 2abCov(Z,Y). Independence cannot be assumed merely because benefits have different names or payment schedules. Other benefit combinations require their own joint present-value reasoning, particularly when guarantees or limited terms interrupt the identity.
Worked example: If Var(Z)=0.04 and δ=0.05, the covariance between unit immediate insurance and the continuous annuity is −0.8. Omitting that covariance when valuing a package containing both benefits overstates its variance when both coefficients are positive.
Common mistake: Adding variances alone treats benefits on one life as independent and misses the offset created by their shared death time.
Official syllabus reference: [1]
46. Fractional payment frequency and claims acceleration
An m-thly annuity paying at annual rate one makes installments of 1/m, not one, at each payment date. Values require survival probabilities at fractional ages and a consistent interpolation assumption. Under UDD, immediate whole-life insurance and annual death-year-end insurance satisfy Ā_x=(i/δ)A_x for constant annual effective interest i and δ=ln(1+i). For benefits paid at the end of each 1/m-year death interval, the analogous factor is i/i^(m), where i^(m)=m[(1+i)^(1/m)−1]. These relationships require the stated UDD and payment assumptions.
Worked example: At i=5%, δ≈0.048790, so UDD converts annual insurance value 0.30 to immediate value approximately 0.30×0.05/0.048790=0.30744. Earlier settlement increases expected present value even though the death benefit amount and mortality table are unchanged.
Common mistake: Using twelve full annual payment amounts for a monthly annuity advertised at an annual rate multiplies the intended benefit by twelve.
Official syllabus reference: [1]
47. Deferred and temporary annuities require endpoint control
A temporary annuity pays only while alive within a limited payment interval. A deferred annuity additionally requires survival through a waiting period. For an n-year deferred whole-life annuity-due with its first payment at time n, expected value is vⁿₙp_x ä_(x+n). The survival factor is essential because the future annuity factor is conditional on being alive then. A temporary annuity-due can be obtained by subtracting the matching deferred tail from a whole-life annuity-due. Confirm the first and last payment dates before applying either decomposition.
Worked example: If ten-year survival is 0.80, v¹⁰=0.60, and the age-x+10 annuity-due factor is 12, the deferred factor is 0.60×0.80×12=5.76. A $5,000 annual benefit beginning at time ten therefore has expected present value $28,800.
Common mistake: Discounting a future annuity factor without multiplying by survival to commencement values it as though the beneficiary is guaranteed to reach that date.
Official syllabus reference: [1]
48. Present-value probability events reverse under discounting
For a positive immediate death benefit B at constant force δ, Z=Be^(−δT) decreases as lifetime increases. Therefore, for 0<z<B, P(Z>z)=P[T<ln(B/z)/δ]. A large insurance present value corresponds to early death. A continuous life annuity present value increases with lifetime, reversing the direction of its lifetime event. Deferred and term benefits can also create atoms at zero or at maturity values. Identify monotonicity, support, and any point masses before transforming a requested present-value probability into a mortality calculation.
Worked example: With B=$100,000, δ=0.04, and constant mortality force 0.02, Z>$50,000 means T<ln(2)/0.04. Its probability is 1−e^[−0.02ln(2)/0.04]=1−1/√2≈0.2929. It is an early-death event, despite being an upper present-value tail.
Common mistake: Keeping the inequality direction unchanged when solving a decreasing discount function selects late deaths instead of the requested large insurance values.
Official syllabus reference: [1]
49. Interest and mortality sensitivities depend on the benefit
Higher interest lowers present values of fixed nonnegative future payments when their timing distribution is unchanged. Higher mortality generally increases a pure whole-life death insurance value by bringing death payments forward, but decreases a pure life annuity value by shortening payments. A pure endowment decreases with mortality because survival becomes less likely. Mixed benefits, guarantees, and limited terms can produce different net effects. Separate changes in event probability from changes in payment timing, and do not transfer whole-life conclusions automatically to every term, endowment, or policy-value calculation.
Worked example: At constant force δ=0.04, raising mortality force from 0.02 to 0.03 increases immediate whole-life insurance from 0.02/0.06=0.3333 to 0.03/0.07=0.4286. The corresponding continuous annuity falls from 1/0.06=16.6667 to 1/0.07=14.2857.
Common mistake: Assuming higher mortality raises every insurance-related present value ignores survival benefits and the shorter payment duration of life annuities.
Official syllabus reference: [1]
Premium and Policy Value Calculation for Long-Term Insurance Coverages
50. Future loss random variables preserve cash-flow timing
The insurer's future loss random variable is present value of future benefits and expenses minus present value of future premiums. At issue, annual whole-life net loss can be written L₀=Bv^(K+1)−Pä_(K+1), with the annuity term representing premiums actually payable before death. At later durations, condition on survival and include only cash flows after the valuation point. A before-premium reserve includes the imminent premium as future income; an after-premium reserve does not. Correct signs and timing are necessary before calculating expectations, variances, premiums, or policy values.
Worked example: A policyholder pays $500 immediately and another $500 one year later, then dies during year two. With a $20,000 year-end death benefit, the realized issue loss excluding expenses is $20,000v²−$500−$500v. Other lifetime outcomes generate different premium and benefit present values.
Common mistake: Replacing random lifetime premium receipts by a fixed premium total changes both expected loss and its dependence on the death benefit.
Official syllabus reference: [1]
51. Net equivalence premiums match benefit and premium present values
The net equivalence principle chooses premium so expected future loss at issue is zero, excluding expenses and explicit profit. For level annual premiums, P=B A/ä, using the insurance factor for the promised benefit and the annuity factor for the actual premium-paying period. Limited-pay insurance uses a limited premium annuity even when coverage continues for life. Term insurance and endowment insurance require their own benefit factors. Equivalence balances expectations rather than guaranteeing a profit on each policy, and it does not by itself include a margin for portfolio variability.
Worked example: A $100,000 whole-life benefit has unit insurance value 0.25. Premiums are payable for at most ten years while alive, with premium annuity factor 8. The net annual premium is $25,000/8=$3,125, even if the lifetime annuity factor is much larger.
Common mistake: Using a whole-life premium annuity for a limited-pay contract assumes premiums continue after their contractual end and understates the required annual premium.
Official syllabus reference: [1]
52. Gross premiums distinguish issue, renewal, and percentage expenses
Gross equivalence adds expected expenses to benefits before balancing premium income. A fixed issue expense is paid once, while renewal expenses are weighted by the survival and discount factors at their actual payment dates. If renewal expenses begin with the second annual premium, their factor is the premium annuity-due factor minus one. Expenses expressed as a percentage of premium belong on both sides of the equation and usually reduce the effective premium-income coefficient. Benefit settlement expenses may instead follow death timing. Different expense bases should not be combined without checking their contingencies.
Worked example: Benefit present value is $8,000, premium annuity factor 10, issue expense $200, renewal expense $30, and every premium incurs 4% expense. Equivalence gives 0.96×10P=8,000+200+30×9, so annual gross premium is $882.29.
Common mistake: Multiplying renewal expenses by the full annuity factor when the issue date is excluded charges one additional renewal expense that the contract does not incur.
Official syllabus reference: [1]
53. Portfolio percentile premiums include lifetime-dependent variance
A portfolio percentile premium selects a target probability that aggregate future loss does not exceed an allowed amount. For n independent identical policies, a normal approximation uses aggregate mean nμ(P) and standard deviation √nσ(P). A zero-loss target leads to μ(P)+zσ(P)/√n=0. Both mean and standard deviation can depend on premium because lifetime premium receipts are random and correlated with benefits. Independence is essential to the √n scaling. The selected probability and approximation are scenario assumptions, not a universal required solvency standard.
Worked example: For 2,500 independent policies, suppose mean loss is $200−10P and loss standard deviation is approximately $500 over the relevant premium range. With selected z=2, the condition becomes 200−10P+2×500/50=0, giving P=$22 rather than the equivalence premium $20.
Common mistake: Dividing the individual standard deviation by portfolio size rather than its square root makes the risk margin disappear too quickly as business grows.
Official syllabus reference: [1]
54. Premiums for a specified expected present value of profit
Expected present value of profit is premium income minus benefits and expenses under the specified valuation basis. A target fixed profit H therefore requires E[PV premiums]−E[PV benefits and expenses]=H. If the target is a proportion of premium present value, it changes the premium-income coefficient rather than adding a fixed amount. State whether profit is measured at issue and whether expenses depend on premium. This principle controls expected profitability, while a percentile principle additionally addresses distributional uncertainty. The two criteria need not produce the same premium.
Worked example: Benefit and fixed-expense present value totals $9,000, premium annuity factor is 12, and percentage expenses are 10% of premiums. A target issue profit of $720 requires 0.90×12P−9,000=720, giving annual premium $900.
Common mistake: Adding a desired annual profit directly to an issue present-value target mixes payment periods and fails to discount the profit requirement consistently.
Official syllabus reference: [1]
55. Prospective and retrospective net policy values
A prospective net policy value is expected present value of future benefits minus future net premiums, conditional on survival to the valuation time. A retrospective value accumulates past net premiums less past expected benefit costs and allocates the balance to survivors. Under matching assumptions, premium basis, and timing conventions, the two approaches agree. Changing the valuation basis or using a premium calculated on another basis can break a naïvely applied equality. A policy value is a conditional amount per surviving policy, rather than the original cohort's undivided fund.
Worked example: At a before-premium valuation, remaining benefit present value is $18,000 and remaining premium annuity factor is 6. A net annual premium of $2,000 gives prospective value $18,000−$12,000=$6,000. The imminent premium belongs in the future annuity factor under this convention.
Common mistake: Comparing a retrospective after-premium fund with a prospective before-premium reserve creates an artificial difference caused by timing rather than actuarial inconsistency.
Official syllabus reference: [1]
56. Gross policy-value recursion allocates funds to deaths and survivors
For a before-premium gross reserve, a simple annual recursion is (V_t+P_t−e_t)(1+i)=q_(x+t)(B_(t+1)+e_death)+p_(x+t)V_(t+1), when premiums and renewal expenses occur at the year's start and death outgo at its end. The equation follows one year's funds into expected death payments and survivor reserves. More complicated timing requires corresponding discount or accumulation adjustments. A reserve immediately after receiving premium has a different recursion. Define each cash flow and valuation point before solving, especially when expenses or maturity benefits occur at boundaries.
Worked example: A policy begins the year with reserve $500, receives premium $100, pays expense $20, and earns 5%. With death probability 0.01 and death outgo $10,000, the survivor reserve satisfies 609=100+0.99V_next, giving V_next≈$514.14.
Common mistake: Leaving out the survival divisor treats the remaining fund per initial policy as though it were already an amount per survivor.
Official syllabus reference: [1]
57. Full preliminary term modifies net premium allocation
Modified net premium methods alter the allocation of valuation premiums while retaining an actuarial funding relationship. In full preliminary term for annual whole-life insurance, the first-year valuation premium funds one year of term coverage, α=vq_x for a unit benefit. Later valuation premiums correspond to a new whole-life policy issued one year older, β=A_(x+1)/ä_(x+1). The first anniversary reserve before the renewal premium is then zero under the matching net basis. This is a valuation-premium construction, not a change to the actual contractual premium or death coverage.
Worked example: For a $100,000 whole-life policy, suppose vq_x=0.002 and the age-x+1 whole-life net premium rate is 0.015. Full preliminary term uses a first-year valuation premium of $200 and later valuation premiums of $1,500, with zero reserve at the first anniversary before renewal.
Common mistake: Treating a modified valuation premium as the amount billed to the policyholder confuses reserve methodology with the contract's actual premium schedule.
Official syllabus reference: [1]
58. Reserve assumption changes affect benefits and premium offsets
A prospective reserve changes when mortality, interest, or expenses change because both future outgo and future premium income are revalued. Lower interest generally increases the present value of each positive cash flow, but the reserve is a difference between outgo and income, so its direction requires analysis. Mortality changes can increase death-benefit values while reducing premium annuities. Keep the contractual premium fixed unless the question explicitly recalculates a valuation premium. Repricing a new policy and revaluing an existing policy are different operations with different offsets.
Worked example: An existing policy has future benefit value $20,000, premium factor 8, and contractual annual premium $1,500, giving reserve $8,000. On a new basis, benefits become $22,000 and premium factor 8.5; reserve becomes $9,250, without changing the contractual premium.
Common mistake: Recomputing the contractual premium to restore issue equivalence during a reserve sensitivity calculation removes part of the effect being measured.
Official syllabus reference: [1]
59. Age rating and additive mortality forces model different extra risks
Age rating substitutes mortality at an older rated age, so an age addition k uses the relevant standard mortality trajectory beginning at x+k. A constant addition c to the force instead gives μ*_(x+t)=μ_(x+t)+c and survival ₜp*_x=e^(−ct)ₜp_x. Age rating changes mortality according to the age pattern; an additive force overlays the same instantaneous extra rate at every future duration. They need not generate equivalent benefits or premiums. Apply the chosen model consistently to both survival-contingent premiums and death-contingent benefits.
Worked example: If standard ten-year survival is 0.90, adding force 0.01 gives adjusted survival 0.90e^(−0.10)≈0.81435. An age rating must instead obtain ten-year survival from the rated starting age and can produce a different value even if initial mortality matches.
Common mistake: Matching two extra-risk models at the first year's mortality does not establish matching future survival, because their later-age mortality trajectories can diverge.
Official syllabus reference: [1]
60. Multiplying annual mortality rates differs from multiplying forces
A constant multiple of annual mortality rates specifies q*_y=kq_y at each applicable age, with probabilities remaining valid. Survival then follows products of 1−kq_y. Multiplying forces instead gives μ*_y=kμ_y and, over a period with unchanged scaling, survival (ₜp_y)ᵏ. These models agree only approximately when mortality probabilities are small. If annual rates are modified, a fractional-age assumption is still needed for within-year payments. A proposed multiplier producing q*>1 requires a separately specified valid model rather than silently accepting an impossible probability.
Worked example: With q_x=0.20 and multiplier 2, annual-rate multiplication gives q*_x=0.40. Doubling a constant within-year force instead gives q*_x=1−(1−0.20)²=0.36. The difference affects both death benefits and premiums contingent on surviving the year.
Common mistake: Doubling q and doubling the force are distinct assumptions; interchanging them becomes increasingly inaccurate as the original annual mortality probability grows.
Official syllabus reference: [1]