How do ratemaking and reserving calculations differ?

Ratemaking estimates a price for future coverage; reserving estimates the remaining cost of claims already incurred. Suppose projected losses are 80 and fixed expenses are 20 per exposure, with a target profit of 5% of premium. The indicated premium is (80 + 20) / 0.95 = 105.26. Separately, if estimated ultimate claims are 150 and 90 has been paid, the estimated unpaid claims are 60. The two results answer different questions and depend on different assumptions.

Do I need to study all three Exam 6 jurisdictions?

No. Exam 6U, 6C and 6I are alternatives. Study the jurisdiction applicable to your pathway, and keep its regulation, financial reporting and solvency framework separate from the others.

What is the difference between the PCPA examination and project?

The examination tests predictive analytics concepts and interpretation through questions. The separate project asks you to make and explain practical choices about data preparation, modeling, validation and communication. For example, if a model will predict next year's claims, validate it on later periods and exclude information that would only become available after the prediction date.

ACAS Study Guide: Methods and Applications

Start with concept 1 Choose a topic

ACAS candidates move between probability, finance, statistics, insurance pricing, reserving, reporting and predictive analytics. This guide connects those subjects through the decisions an actuary makes with imperfect data. You will calculate event probabilities and present values, select models for claim counts and censored losses, turn claim experience into rate indications, and develop observed claims into estimates of ultimate cost. It also shows how to check whether a result is credible: compare assumptions with data, inspect model diagnostics, test sensitivity and explain uncertainty. U.S., Canadian and international reporting examples stay distinct, so a rule from one jurisdiction does not silently carry into another. Each worked calculation states its assumptions and interprets the result.

Probability (Exam 1 / SOA P)

1. Count the possible outcomes

Count the possible outcomes by deciding whether the result records a group or a sequence. If five distinct claims are available and two reviewers each receive one claim, assigning claim A first and claim B second differs from the reverse assignment. There are five choices followed by four, giving 20 ordered outcomes. If the task only identifies two claims for review, those assignments describe the same pair. Divide by 2!, the number of orders within each pair, to get 10 selections. More generally, selecting r objects from n distinct objects gives n!/[r!(n−r)!]; arranging them gives n!/(n−r)!. Before assigning equal probabilities, check that your listed outcomes are equally likely. The Probability Exam syllabus includes combinatorics and sample spaces.

Worked example: Five distinct claims are available. How many ways can you choose two claims for a single unordered review batch? A. 5; B. 10; C. 20; D. 25; E. 120. Choosing in sequence gives 5 × 4 = 20, but each pair appears in two orders. Thus 20 ÷ 2 = 10, choice B. If reviewer order mattered, 20 would be correct.

Watch the distinction: Dividing by 2! when two positions have different meanings loses valid outcomes. First ask whether swapping the selected claims changes the recorded result.

Reference: [2]

2. Update probabilities with evidence

A flag changes the probability of a large claim, but the flag's detection rate is not the probability that a flagged claim is large. Define L as large and F as flagged. Bayes' rule gives P(L|F) = P(F|L)P(L)/P(F). The denominator must include every route to a flag: P(F) = P(F|L)P(L) + P(F|not L)P(not L). This is the law of total probability applied to the two claim groups. A useful diagnostic is to imagine 1,000 claims: convert each conditional rate into counts, then compare large flagged claims with all flagged claims. The original prevalence matters greatly. Even a flag that catches most large claims may produce many false alarms when large claims are uncommon. Keep the condition after the vertical bar attached to the population used for that percentage.

Worked example: Suppose 10% of claims are large. A flag catches 80% of large claims and 20% of other claims. What is P(large|flag)? A. 8%; B. 20%; C. 30.8%; D. 80%; E. 90%. Large flagged claims make up 0.10 × 0.80 = 0.08 of all claims; other flagged claims make up 0.90 × 0.20 = 0.18. Therefore 0.08/(0.08 + 0.18) = 4/13 ≈ 30.8%, choice C.

Watch the distinction: The 80% detection rate is P(flag|large), not P(large|flag). Reversing the condition ignores the much larger group of ordinary claims that can also be flagged.

Reference: [2]

3. Model a claim count

Model a claim count by matching the distribution to how claims arise. A Poisson count is useful when events occur independently over a stated exposure interval at an approximately constant rate, with the chance of multiple events in a very short interval negligible. Its parameter λ is both the mean and variance of the count, measured in expected claims for that interval. The probability of n claims is e^(−λ)λ^n/n!, so zero claims have probability e^(−λ). If exposure doubles while the event rate stays constant, λ doubles; the probability of zero does not simply halve. Check for clustering, changing exposure, and excess variance before relying on the model. A fixed number of independent policies with only one possible claim each may instead lead to a binomial count model.

Worked example: A hypothetical portfolio has a Poisson mean of 2 claims during one year. What is the probability of zero claims? A. 0.0677; B. 0.1353; C. 0.2707; D. 0.5000; E. 2.0000. Using P(N = 0) = e^(−2)2^0/0! gives e^(−2) ≈ 0.1353, choice B. Its expected annual count is 2 claims; that mean does not guarantee two claims in any particular year.

Watch the distinction: A mean of two claims is not a probability of two claims. The Poisson mean sets the whole count distribution, including a positive chance of no claims.

Reference: [2]

4. Turn losses into insurer payments

Turn a covered loss into an insurer payment by reading each contract feature as an operation on a clearly defined amount. Under the stated convention, subtract the per-loss deductible from the loss, stopping at zero; multiply the remainder by the insurer's coinsurance share; then cap the resulting insurer payment at the payment limit. The formula is Y = min[c(L−d)_+, u], where L is loss, d is deductible, c is insurer share, u is the payment limit, and (x)_+ means max(x,0). This order matters because a limit on insurer payment differs from a limit on covered loss before coinsurance. Sketch the payment function across low and high losses: it is zero below the deductible, rises at slope c, then becomes flat at u. For a real contract, identify precisely what the deductible and limit apply to before calculating.

Worked example: A $9,000 covered loss has a $1,000 per-loss deductible, an 80% insurer share after deductible, and a $5,000 insurer payment limit. What is paid? A. $4,000; B. $5,000; C. $6,400; D. $7,200; E. $8,000. The pre-limit payment is 0.80 × ($9,000 − $1,000) = $6,400. Capping it at $5,000 gives choice B; the insurer retains no obligation above that stated cap.

Watch the distinction: Applying the $5,000 cap to the original loss and then subtracting the deductible changes the contract's stated order. Name the amount each provision modifies first.

Reference: [2]

5. Read dependence from a joint distribution

Read dependence from the joint distribution before calculating correlation. For binary indicators X and Y, E[X] and E[Y] are their probabilities of equaling one, while E[XY] is the probability both equal one. Covariance is E[XY] − E[X]E[Y]. It retains the product of the variables' units and indicates whether joint high values occur more or less often than independence predicts. Correlation divides covariance by the two standard deviations, making a dimensionless measure between −1 and 1 when variances are positive. For Bernoulli variables with success probability p, variance is p(1−p). Independence implies zero covariance when moments exist, but zero covariance alone generally does not establish independence. Check that the supplied joint probabilities form a valid table; for binary variables, the remaining cells follow from the marginals and must be nonnegative.

Worked example: Let X and Y be binary, each equal to one with probability 0.5, and let P(X = 1,Y = 1) = 0.4. What is their correlation? A. −0.6; B. 0; C. 0.15; D. 0.4; E. 0.6. Covariance is 0.4 − 0.5² = 0.15. Each standard deviation is √(0.5 × 0.5) = 0.5, so correlation is 0.15/(0.5 × 0.5) = 0.6, choice E.

Watch the distinction: Covariance of 0.15 is not correlation of 0.15. Divide by both standard deviations, and verify neither variance is zero before interpreting correlation.

Reference: [2]

6. Approximate a portfolio total

Approximate a portfolio total by aggregating its mean and variance first. For 100 independent losses with common mean μ and standard deviation σ, the sum has mean 100μ and variance 100σ², so its standard deviation is 10σ. The central limit theorem says that, under suitable conditions, a standardized sum of many independent, identically distributed losses is approximately standard normal. This supports a probability estimate even when each loss is not normal. Standardize a threshold x as (x−total mean)/total standard deviation, then use the stated normal probability. The approximation can be poor for a small portfolio or losses dominated by rare, very large amounts; independence is also essential to the simple variance calculation. Compare the threshold with the total mean before calculating to catch an incorrect sign or scale.

Worked example: One hundred independent losses each have mean $100 and standard deviation $20. Estimate P(total > $10,200), given P(Z > 1) = 0.1587. A. 0.0228; B. 0.1587; C. 0.5000; D. 0.8413; E. 0.9772. Total mean is 100 × $100 = $10,000; total standard deviation is √100 × $20 = $200. The threshold is one standard deviation above the mean, giving choice B by normal approximation.

Watch the distinction: Multiplying the individual standard deviation by 100 gives $2,000 and overstates spread. For independent losses, add variances first, then take the square root.

Reference: [2]

Financial Mathematics (Exam 2 / SOA FM)

7. Convert interest conventions

Convert interest conventions to a common period before comparing returns. A nominal annual rate j compounded m times yearly applies j/m interest each compounding period. One dollar therefore grows to (1+j/m)^m dollars after a year, so the effective annual rate is (1+j/m)^m−1. This is greater than j when j is positive and m exceeds one because later interest earns interest. The effective rate measures actual one-year accumulation under the stated compounding assumption; it does not account for fees, default, or inflation. Conversely, an effective annual rate i corresponds to nominal j = m[(1+i)^(1/m)−1] at that compounding frequency. The Financial Mathematics Exam syllabus covers these conversions. Always label whether a quoted percentage is nominal or effective.

Worked example: An investment quotes a hypothetical 12% nominal annual rate compounded monthly. What is its effective annual rate? A. 1.00%; B. 12.00%; C. 12.36%; D. 12.68%; E. 24.00%. Monthly interest is 0.12/12 = 0.01. A $1 balance becomes $1.01^12 ≈ $1.126825, so the effective annual rate is 0.126825, or about 12.68%, choice D. This assumes reinvestment at the same monthly rate.

Watch the distinction: Multiplying the monthly rate by twelve returns the nominal 12%, not the effective annual yield. Compounding requires raising the monthly accumulation factor to the twelfth power.

Reference: [3]

8. Value a payment stream

Value a payment stream by putting each cash flow on a timeline and moving it to the valuation date. With annual effective interest i, a payment at the end of year t has present value C_t/(1+i)^t. An annuity-immediate begins at time 1; an otherwise identical annuity-due begins at time 0. Moving every payment one year earlier multiplies the entire present value by 1+i, provided the interest rate is constant and the number of payments is unchanged. You can calculate term by term or use the level-annuity formula C[1−(1+i)^(−n)]/i. The timeline guards against a common off-by-one error, especially when a first payment is described as 'today' or 'one period from today.' State the valuation date and whether cash flows are certain; these formulas do not incorporate claim or survival probabilities.

Worked example: Three $100 payments arrive at the end of years 1, 2, and 3, with a 10% annual effective rate. Their value today is $100/1.1 + $100/1.1² + $100/1.1³ = $248.69. What if they arrive at times 0, 1, and 2? A. $225.90; B. $248.69; C. $260.00; D. $273.55; E. $300.00. Multiply by 1.1 to get $273.55, choice D.

Watch the distinction: An annuity-due does not add a fourth payment. It shifts the same payments one period earlier, increasing present value by one accumulation factor.

Reference: [3]

9. Split a loan payment

Split a loan payment by applying interest to the balance outstanding just before payment. If B is that balance, i the effective rate for the payment period, and P the payment, interest due is iB. The rest, P−iB, reduces principal, leaving B(1+i)−P. This recursion lets you build an amortization schedule and check that each row's ending balance equals the next row's beginning balance. When P is smaller than interest due, the balance grows despite a payment; when P equals interest, principal does not fall. For an outstanding balance at a later date, you can accumulate the original loan and subtract accumulated payments, or discount remaining scheduled payments if their terms are known. Keep payment and interest periods consistent, and separate any fees from contractual interest.

Worked example: A loan balance is $1,000 immediately before a period with 5% effective interest, followed by a $300 payment. What is the new balance? A. $700; B. $750; C. $800; D. $950; E. $1,050. Interest is 0.05 × $1,000 = $50. The payment reduces principal by $300 − $50 = $250, leaving $1,000 − $250 = $750, choice B.

Watch the distinction: Subtracting the full $300 from principal overlooks the $50 interest earned during the period. Allocate the payment to interest first, then reduce the balance.

Reference: [3]

10. Price a coupon bond

Price a coupon bond as the present value of its promised coupons plus its redemption payment. Use the investor's yield per coupon period, expressed on the same basis as coupon timing. For a two-year annual-pay bond, the first coupon is discounted one year and the final coupon plus redemption two years. A coupon rate above the required yield produces a price above par when redemption equals par; below-yield coupons produce a discount. The comparison follows from valuing par at yield-rate coupons: those coupons would price exactly at par under the same assumptions. The quoted coupon rate is based on face value, not market price, while yield is the discount rate that equates price and cash flows. This calculation assumes the scheduled payments occur; credit risk or an issuer call would require additional treatment.

Worked example: A two-year bond has $1,000 face value, pays $60 at each year-end, and redeems for $1,000 at year 2. At a 5% annual effective yield, what is its price? A. $980.00; B. $1,000.00; C. $1,018.59; D. $1,060.00; E. $1,120.00. Price = $60/1.05 + $1,060/1.05² = $57.14 + $961.45 = $1,018.59, choice C. Its 6% coupon rate exceeds yield.

Watch the distinction: Discounting the final $60 coupon but forgetting the $1,000 redemption severely understates price. List every dated cash flow before applying discount factors.

Reference: [3]

11. Use spot and forward rates

Use spot rates to value payments at their own maturities, and forward rates to describe the interest implied between maturities. With annual effective one-year spot s1 and two-year spot s2, investing one dollar for two years at s2 gives (1+s2)^2. An equivalent path invests one year at s1, then the second year at forward rate f1,2, giving (1+s1)(1+f1,2). Equating the two accumulation factors yields f1,2 = (1+s2)^2/(1+s1)−1. This is an implied rate from today's term structure, not a guaranteed prediction of the rate available next year. Ensure both spot rates use the same annual effective convention. With a different compounding basis, convert first; simply subtracting percentages discards compounding and may give a materially different result.

Worked example: The one-year annual effective spot rate is 4%, and the two-year annual effective spot rate is 5%. What second-year forward rate is implied? A. 1.00%; B. 5.00%; C. 5.50%; D. 6.01%; E. 10.00%. Two-year accumulation is 1.05² = 1.1025. Dividing by first-year accumulation 1.04 gives 1.060096, so f1,2 ≈ 6.01%, choice D.

Watch the distinction: Taking 5% − 4% gives the difference between quoted spot rates, not the second-year forward rate. Equate full-period accumulation factors instead.

Reference: [3]

12. Measure interest-rate sensitivity

Measure first-order interest-rate sensitivity with modified duration. For a small parallel yield change Δy, the approximate proportional price change is −D_modΔy, using yield expressed as a decimal. Thus positive-duration assets generally lose value when yield rises. Duration gives the slope of the price-yield relationship at the starting yield; curvature means the estimate deteriorates for larger changes. In asset-liability management, equal present values make current assets cover current liabilities, while matched dollar durations make their first-order responses to a small common yield shift similar. That helps immunize the surplus against that specified shift, but it does not address every yield-curve movement, cash-flow change, or credit event. Check that asset and liability durations refer to the same valuation date and rate convention before comparing them.

Worked example: An asset is worth $10,000 with modified duration 4. A one-percentage-point yield rise means Δy = 0.01. Approximate its new value. A. $9,600; B. $9,900; C. $10,000; D. $10,400; E. $14,000. Estimated change is −4 × 0.01 × $10,000 = −$400, so value is about $9,600, choice A. This linear estimate omits convexity and assumes the relevant yield shift applies to its cash flows.

Watch the distinction: Matching asset and liability present values balances them today but does not match their reactions to rates. Compare dollar durations as well as present values.

Reference: [3]

MAS-I: Modern Actuarial Statistics I

13. Count arrivals over time

Count arrivals over time with a Poisson process when claims occur independently and the expected number is proportional to exposure length. A homogeneous rate λ claims per year implies an interval of t years has count N(t) distributed Poisson(λt). Hence E[N(t)] = Var[N(t)] = λt and P(N(t)=0) = e^(−λt). For disjoint intervals, counts are independent under this model, which lets you combine exposure periods or calculate waiting-time probabilities. If claim intensity changes with season or exposure, replace λt with the integrated rate over the interval; the constant-rate formula then needs adjustment. Diagnose the process using counts by interval: systematic seasonality or variance well above the mean can signal that a simple homogeneous Poisson process is inadequate. The MAS-I content outline includes Poisson process modelling.

Worked example: A hypothetical homogeneous claim process has rate 3 claims per year. Over half a year, expected claims are 3 × 0.5 = 1.5, and the Poisson zero-count probability is e^(−1.5) ≈ 0.2231. Thus about 22.3% of comparable half-year intervals would have no claims under this model. Using e^(−3) would answer a full-year question instead.

Watch the distinction: The annual rate is not automatically the mean for every observation window. Multiply claims per year by the interval length in years before evaluating the Poisson probability.

Reference: [4]

14. Connect survival and hazard

Connect survival and hazard by distinguishing survival from time zero from survival over a later interval. If T is the time until an event, S(t)=P(T>t). For a case known to survive to time a, the probability of surviving to b>a is P(T>b|T>a)=S(b)/S(a). The hazard h(t) describes instantaneous event intensity among those still event-free at t; when differentiable, h(t)=−S'(t)/S(t). A constant hazard h gives exponential survival S(t)=e^(−ht), so survival across any interval depends only on its length. The hazard is a rate per unit time, not the probability of an event at one exact instant. Check units in an exponent: if h is per year, t must be in years. In a changing-hazard model, use the integral of hazard between a and b.

Worked example: Suppose S(t)=e^(−0.02t), with t measured in years. A claimant has remained event-free through year 5. The probability of remaining event-free through year 8 is S(8)/S(5) = e^(−0.16)/e^(−0.10) = e^(−0.06) ≈ 0.9418. The three-year conditional event probability is 1 − 0.9418 = 0.0582, or about 5.82%, under the stated model.

Watch the distinction: Using S(8) alone answers survival from time zero. Once survival through year 5 is known, divide by S(5) to condition on that information.

Reference: [4]

15. Estimate a population mean

Estimate a population mean with the sample average, then measure uncertainty about that average separately from variation in one future observation. For observations x1 through xn, the mean is their sum divided by n. Estimate population variance with s² = Σ(xi−x̄)²/(n−1) when observations are an independent representative sample and the usual unbiased sample-variance estimate is intended. The estimated standard deviation of an individual observation is s; the estimated standard error of the mean is s/√n. The reduction reflects averaging independent noise. It does not erase sampling bias, dependence, or a change in the claim mix. With only a few observations, the standard error itself is unstable, so avoid treating it as a precise confidence statement. If values are censored or exposure weights differ, a plain average may not estimate the desired quantity.

Worked example: Three observed claim severities are $100, $200, and $300. Their mean is ($100+$200+$300)/3 = $200. Squared deviations total $10,000 + $0 + $10,000 = $20,000, so sample variance is $20,000/(3−1) = $10,000 squared dollars and sample standard deviation is $100. Estimated standard error of the mean is $100/√3 ≈ $57.74, assuming independent, comparable observations.

Watch the distinction: The $100 sample standard deviation describes variation among claims; it is not the estimated standard error of their mean. Divide by √3 for the latter.

Reference: [4]

16. Account for censored losses

Account for censoring according to what the data reveal. An exact severity x contributes its probability density f(x) to a continuous-data likelihood. A right-censored record at c states only that the loss exceeds c, so it contributes the survival probability S(c)=P(X>c). Treating c as an exact loss invents a value and tends to pull fitted severities downward. For an exponential model with mean θ, f(x)=e^(−x/θ)/θ and S(c)=e^(−c/θ). These contributions have different units: a density is per dollar and a survival probability is dimensionless, but their product is valid as the likelihood of the observed data pattern under a fixed observation scheme. Distinguish censoring from truncation: with truncation, some losses never enter the dataset, so inclusion probabilities must also be addressed.

Worked example: Assume exponential severity with mean θ=$1,000. One exact $500 loss contributes f(500)=e^(−0.5)/$1,000 ≈ 0.0006065 per dollar. Another record says only that its loss exceeded $500; it contributes S(500)=e^(−0.5) ≈ 0.6065. Their likelihood contribution is the product of these two terms. Assigning $500 to the second record would incorrectly treat its lower bound as an observed severity.

Watch the distinction: A censored value records a boundary, not an exact claim amount. Use the probability of exceeding that boundary, because every larger severity remains compatible with the observation.

Reference: [4]

17. Specify a frequency GLM

Specify a frequency GLM by matching a count response to its exposure. A Poisson model with log link writes log E[N_i] = log e_i + β0 + β1X_i, where e_i is exposure in policy-years. The log exposure is an offset: its coefficient is fixed at one, so doubling otherwise identical exposure doubles expected count while leaving expected claims per policy-year unchanged. Exponentiating β1 gives the frequency ratio for a one-unit change in X at equal exposure. This interpretation is multiplicative, even though coefficients add on the log scale. Poisson variance equals its mean, an assumption worth checking through residuals or dispersion diagnostics. An offset needs positive exposure and the same exposure definition across records. The model estimates associations conditional on included predictors; a coefficient alone does not establish a causal effect.

Worked example: Suppose log E[N] = log(exposure) − 2 + 0.4X, with exposure in policy-years and X either zero or one. At one policy-year and X=0, expected count is e^(−2) ≈ 0.1353. At X=1 it is e^(−1.6) ≈ 0.2019. The frequency ratio is e^0.4 ≈ 1.49, so the model predicts about 49% higher claim frequency at equal exposure.

Watch the distinction: A coefficient of 0.4 on a log link is not an increase of 0.4 claims per policy-year. Exponentiate it to interpret the multiplicative frequency change.

Reference: [4]

18. Diagnose an extended linear model

Diagnose an extended linear model with several kinds of evidence. AIC combines model fit and a penalty for parameter count; among models fitted by comparable likelihoods to the same response data, lower AIC is preferred on that criterion. Its absolute value has no standalone quality label. Residual plots answer a different question: they can reveal curvature, changing spread, outliers, or dependence that a summary statistic conceals. A systematic pattern against fitted values suggests the chosen mean function, variance function, link, or omitted predictor needs investigation. Check the plot's residual definition, because raw, Pearson, and deviance residuals have different scales. A smaller AIC does not validate assumptions or guarantee performance on future data. Compare out-of-sample predictions where the decision calls for prediction, then examine whether remaining residual patterns are practically important.

Worked example: Two models use the same response records and comparable likelihoods. Model A has AIC 420; Model B has AIC 412. B improves AIC by 8 points, so it is preferred on that measure. Yet B's residual spread rises steadily with fitted values. That pattern calls for reviewing its variance specification or influential high-mean cases before selecting it. AIC comparison and residual diagnosis answer separate questions.

Watch the distinction: Choosing AIC 412 and stopping overlooks visible model failure. AIC ranks the compared fits with a complexity penalty; it does not certify their residual behavior.

Reference: [4]

MAS-II: Modern Actuarial Statistics II

19. Blend experience with credibility

Blend class experience with a broader complement when the class estimate is informative but noisy. A credibility estimate has the form Z × class estimate + (1−Z) × complement, with 0≤Z≤1. The complement is the estimate used for the portion of information the class experience cannot reliably supply; it might be an appropriate broader group or prior expectation adjusted to the same cost basis. As class information becomes more credible, Z generally rises, but the method used to determine Z matters. Limited-fluctuation, Bühlmann, and Bayesian approaches use different assumptions and calculations. Check that both inputs refer to the same exposure unit, coverage, and cost period. The MAS-II content outline covers credibility methods; the Exam 5 ratemaking guide shows a related pricing use.

Worked example: A class has observed loss cost $120 per exposure, a suitable broader complement is $100 per exposure, and credibility Z is stipulated as 0.30. The blended estimate is 0.30 × $120 + 0.70 × $100 = $36 + $70 = $106 per exposure. It is $6 above the complement because the class result exceeds it, but only 30% of the $20 difference is reflected.

Watch the distinction: Using Z for both inputs gives weights that sum to 0.60 here and distorts the estimate. The complement receives weight 1−Z, after both figures are put on a comparable basis.

Reference: [5]

20. Separate shared and individual effects

Separate shared and individual effects when observations belong to groups. A random-intercept mixed model can be written Y_ij = β0 + βX_ij + b_j + ε_ij, where b_j is a group-specific deviation and ε_ij is individual noise. Two policies in the same branch share b_j, so their outcomes are correlated if the branch-effect variance is positive and individual errors are independent. The fixed part describes the population-level relationship; the random part describes variation among branches. A branch prediction combines the fixed prediction with its estimated branch deviation, often shrunk toward zero when the branch has little data. Ignoring the shared component can make standard errors too small and overstate the amount of independent information. Check whether grouping reflects the actual data-generating structure, and whether residual patterns remain within and across groups.

Worked example: Suppose a model's fixed prediction is $100 per policy and branch A's estimated random intercept is +$5. Its conditional branch-A prediction is $100+$5=$105 per policy. Two branch-A policies each receive that $5 adjustment; across branches, their shared contribution to covariance is the branch-intercept variance. The $5 estimate itself is a branch-specific prediction, not the covariance or a guarantee about either individual outcome.

Watch the distinction: Treating two policies from the same branch as fully independent discards their common branch effect. Their individual errors may differ, while the random intercept remains shared.

Reference: [5]

21. Choose a local or tree prediction

A nearest-neighbor method predicts from observations close to the new policy in predictor space. Decide how to scale variables before measuring distance: an unscaled dollar amount could overwhelm a binary coverage indicator. For severity, an unweighted three-neighbor prediction is the mean of the three observed amounts. A regression tree instead divides policies through successive predictor tests and predicts from all training observations in the resulting terminal node. Neighbors can change with every new policy; a tree gives the same prediction to all policies in one node. Small neighborhoods and deep trees can both chase noise, so choose neighborhood size or tree complexity using held-out performance. The MAS-II content outline includes both methods.

Worked example: Suppose the three nearest comparable claims have severities of $80, $100, and $150. With equal neighbor weights, the prediction is ($80 + $100 + $150) / 3 = $110. A fitted tree places the same new claim in a terminal node whose training claims average $120, so its tree prediction is $120. Neither figure is automatically preferable; compare errors on claims excluded from fitting.

Watch the distinction: A tree prediction is not the mean of the three closest observations. It uses the observations in the assigned terminal node, which may contain cases at quite different distances.

Reference: [5]

22. Reduce and group predictors

Principal components analysis replaces correlated numeric predictors with directions formed from weighted sums. Standardize variables first when their units or scales differ materially. A component score is a coordinate along its loading vector; the first principal component captures the greatest available variance, with later components constrained to new directions. High variance does not guarantee strong prediction of claim outcomes, so test whether retaining components helps the intended model. Clustering answers another question: which observations are similar enough to group together? A method such as k-means assigns observations to centers and updates those centers to reduce within-group distances. Its output is a group label rather than a new continuous coordinate. Both procedures depend on scaling, and neither makes the resulting structure inherently meaningful to insurance decisions.

Worked example: Let standardized predictors be x1 = 1 and x2 = −1. The component with loading vector (1/√2, 1/√2) gives score (1 − 1)/√2 = 0. The perpendicular component (1/√2, −1/√2) gives score (1 + 1)/√2 = √2, about 1.414. A clustering method would instead compare the point (1, −1) with its centers and assign a group.

Watch the distinction: A zero first-component score does not mean the observation contains no information. Opposing predictor values can cancel along one direction while producing a substantial score along another.

Reference: [5]

23. Compare ranking performance

Rank policies by a model score, then ask what share of actual claims falls among the highest-scored policies. Top-decile lift compares that share with the 10% of policies selected. A lift above one means the selected group contains more claims than a random tenth would contain, assuming policies have comparable exposure. If exposure differs, compare claim rates or use an exposure-aware grouping and denominator; otherwise long-duration policies may dominate. Lift evaluates prioritization for uses such as investigation or outreach. It does not show whether predicted counts match observed counts. Check calibration separately by comparing predicted and actual totals within score bands. Also inspect whether a high lift is stable on held-out data, since sorting and evaluating the same training observations can exaggerate performance.

Worked example: A portfolio has 1,000 equally exposed policies and 200 observed claims. The highest-scored 100 policies contain 50 claims, or 25% of the total. Top-decile lift is 25% / 10% = 2.5. That group has 0.50 claims per policy versus 0.20 overall. The model ranks these policies usefully, but this calculation cannot tell whether it predicted 50 claims or 80.

Watch the distinction: Lift of 2.5 does not mean each selected policy will have exactly 2.5 claims. It compares an observed concentration of claims with the share of policies selected.

Reference: [5]

24. Forecast a stationary series

A mean-reverting AR(1) model expresses the next value as a long-run mean plus a fraction of the latest deviation from that mean and an unpredictable innovation. With coefficient 0.6, a positive deviation shrinks by 40% at each forecast step. Setting future innovations to their expected value of zero gives point forecasts; actual observations will still vary. A coefficient whose absolute value is below one supports the usual stationary AR(1) construction with constant innovation variance. A deterministic trend differs: it adds a time-dependent level that need not revert after a shock. Before relying on an AR forecast, examine the series for trend or seasonality and check residual dependence. The MAS-I study guide can refresh the model diagnostic habits used here.

Worked example: Assume a stationary AR(1) claim-count index with mean 10, coefficient 0.6, and latest observed value 15. The one-step forecast is 10 + 0.6(15 − 10) = 13 index units. Apply the same relation to that forecast: 10 + 0.6(13 − 10) = 11.8 units two steps ahead. Both forecasts move toward 10; neither is a guaranteed future observation.

Watch the distinction: Repeatedly adding the latest five-unit deviation treats a temporary shock as a permanent trend. Under this AR(1) model, the deviation is multiplied by 0.6 each step.

Reference: [5]

Exam 5: Basic Ratemaking and Estimating Claim Liabilities

25. Build a future rate indication

Work in a consistent unit, such as dollars per earned exposure. Project losses and loss adjustment expenses to the period when the proposed rate will apply, then add fixed expenses expressed per exposure. Variable expenses and the selected profit or contingency provision are percentages of premium, so they consume a share of the premium you seek. If P is indicated premium, L projected loss cost, F fixed expense, and v the combined variable provision, the balance equation is P = L + F + vP. Rearranging gives P = (L + F)/(1 − v), provided v is below one. Check what each provision includes to avoid double-counting. Compare the resulting premium with the current average premium to obtain an indicated percentage change, then evaluate implementation constraints separately.

Worked example: Suppose projected loss and adjustment expense cost is $80 per exposure, fixed expense is $10 per exposure, and variable expense plus profit is 25% of premium. P = ($80 + $10)/(1 − 0.25) = $120 per exposure. Check: $80 losses + $10 fixed expense + 0.25 × $120, or $30, equals $120. If current premium is $100, the indicated increase is ($120/$100) − 1 = 20%.

Watch the distinction: Adding 25% of the $90 cost gives $112.50, which falls short. The 25% provision is a share of the indicated premium, not a markup on fixed costs.

Reference: [6]

26. Align historical premium and loss

An observed loss ratio can mix premiums charged at different rate levels with losses from an earlier cost period. On-level premium estimates what the historical exposures would have earned under one chosen rate level. Apply rate changes according to when policies were written and earned; a single multiplier is only justified when the stated data support it. Develop losses to an appropriate basis if needed, then trend them from their historical cost period to the future period covered by the proposed rates. A loss trend should reflect both severity and frequency effects when relevant. Divide projected losses by on-level premium only after these timing and coverage bases are aligned. For additional ratemaking context, use the Exam 5 study guide.

Worked example: Assume $1,000,000 of earned premium requires a stated 10% on-level adjustment, giving $1,000,000 × 1.10 = $1,100,000. Assume $700,000 of losses is already on the desired development basis and needs one 5% trend adjustment: $700,000 × 1.05 = $735,000. The aligned projected loss ratio is $735,000/$1,100,000 = 66.82%. The assumptions specify the full timing adjustments for this illustration.

Watch the distinction: Applying the premium rate change to losses confuses price with claim cost. On-leveling restates premium; loss trend moves expected claim costs to the future cost period.

Reference: [6]

27. Refine rates by class

A class relativity compares expected cost per common exposure unit with a selected base class. If expected costs differ, a relativity can reflect that difference while the base rate sets the overall premium level. First estimate credible class costs and consider whether differences are stable and appropriate for the classification. Then choose a reference factor, commonly 1, and divide each other class cost by the reference cost. Finally, weight class premiums by projected exposure to check the portfolio average against the overall indication. Changing relativities while holding the base rate fixed can shift total premium when the class mix is uneven. A revenue-neutral refinement therefore requires solving for a new base rate or another balancing adjustment. Monitor the effects on individual policy premiums as well as aggregate revenue.

Worked example: Class A has expected loss cost $80 per exposure and class B $120, so with A at 1.00, B is $120/$80 = 1.50. Suppose projected exposures are 75 in A and 25 in B, and the target average premium is $120. Weighted factor = 0.75(1) + 0.25(1.5) = 1.125. Set base rate to $120/1.125 = $106.67; B's premium is about $160.00.

Watch the distinction: A class B factor of 1.5 does not itself produce the desired overall rate level. The base rate must account for projected class mix and the selected average premium.

Reference: [6]

28. Develop claims to ultimate

A cumulative reported-loss triangle shows how losses for each accident cohort emerge as that cohort matures. An age-to-ultimate factor estimates the portion still unreported or not yet reflected in reported values. Multiply reported losses at the stated age by the selected factor to estimate ultimate losses. Subtract paid losses, rather than reported losses, when estimating the total unpaid amount. The difference between ultimate and reported is future reported development; the difference between reported and paid is currently reported but unpaid. Their sum equals ultimate less paid. Select factors from relevant historical patterns, and examine changes in claims handling, coverage, inflation, and large losses before applying them. A factor derived from older cohorts can be misleading if reporting speed has changed.

Worked example: At 12 months, a hypothetical accident cohort has $600,000 cumulative reported losses, including $500,000 paid. With a selected 12-to-ultimate factor of 1.5, estimated ultimate is $600,000 × 1.5 = $900,000. Future reported development is $300,000, and current reported unpaid is $100,000. Total estimated unpaid is $300,000 + $100,000 = $400,000, which also equals $900,000 − $500,000.

Watch the distinction: Subtracting reported losses from ultimate yields only expected future reported development. Total unpaid claims also include amounts already reported but not yet paid.

Reference: [6]

29. Blend observed and expected losses

The Bornhuetter-Ferguson method combines reported experience with an independently selected expected ultimate. It treats the unreported share as an expected proportion of that prior ultimate rather than projecting every early fluctuation in reported losses. If p is the expected reported proportion, estimated ultimate equals reported losses plus (1 − p) times expected ultimate. A development factor can imply p as its reciprocal when the bases are consistent. This approach is useful when an immature cohort has volatile or sparse reported experience, but its result depends on the quality of the expected loss estimate. Check whether the prior reflects current exposure, rate level, coverage, and cost conditions. As maturity increases, the expected unreported share falls, so observed experience contributes more of the ultimate estimate.

Worked example: Suppose expected ultimate losses are $1,000,000 and 40% is expected to be reported by the valuation date. If $500,000 is actually reported, the expected unreported amount is (1 − 0.40) × $1,000,000 = $600,000. Bornhuetter-Ferguson ultimate is $500,000 + $600,000 = $1,100,000. If $350,000 has been paid, the implied unpaid amount is $1,100,000 − $350,000 = $750,000.

Watch the distinction: Multiplying actual reported losses by the inverse reporting percentage gives a development estimate, not the stated Bornhuetter-Ferguson estimate. The latter applies the unreported percentage to expected ultimate.

Reference: [6]

30. Test reserve reasonableness

An ultimate loss estimate becomes easier to challenge when translated into frequency, severity, loss ratio, or cost per exposure. Divide projected ultimate losses by expected ultimate claim count to obtain implied average severity, ensuring the loss and count definitions match. Compare that severity with a prior period after considering inflation, mix, limits, and claim maturity. A large increase may be genuine, but it may also signal changed case reserving, faster settlement of small claims, altered claim coding, or a shifted count definition. If small claims close sooner, remaining open claims can look more severe even without an equivalent rise in ultimate cost. Use several diagnostics together: claim-count development, paid versus reported patterns, large-claim effects, and actual versus expected emergence. Document why the selected reserve remains plausible.

Worked example: A hypothetical cohort has indicated ultimate losses of $1,200,000 and 600 expected ultimate claims. Implied severity is $1,200,000/600 = $2,000 per claim. Comparable prior severity was $1,500, so the increase is ($2,000/$1,500) − 1 = 33.33%. Investigate changes in claim mix, inflation, and settlement practice before accepting the indication; the calculation alone does not identify the cause.

Watch the distinction: A higher implied severity is not proof that every claim became costlier. Changes in which claims are counted or settled can move the average without the same underlying cost change.

Reference: [6]

Exam 6U: United States Regulation and Financial Reporting

31. Locate the U.S. regulator’s role

U.S. insurance regulation is primarily carried out by states, with federal law and federal agencies affecting particular parts of the market. For a rate question, start with the jurisdiction where the policy is issued and identify that state's insurance authority and applicable filing framework. For solvency, identify the insurer's domestic regulator and the other states in which it operates. The National Association of Insurance Commissioners coordinates state regulators and develops model materials, data tools, and financial reporting practices; adoption and enforcement depend on the relevant authorities. A model law is therefore not the same thing as an enacted state requirement. The Exam 6U content outline includes these regulatory distinctions.

Worked example: A hypothetical insurer proposes a new auto rate for policies issued in State A. The review question goes to State A's relevant insurance authority under its applicable rate-filing rules. The insurer may use an NAIC model as background, but NAIC publication alone does not approve the filing. If the same insurer faces a capital concern, its domestic state's solvency supervision is a separate question from the proposed rate.

Watch the distinction: Do not treat the NAIC as the office that approves an individual insurer's state rate. Its coordinating work may influence state rules, but the applicable state authority makes the regulatory decision.

Reference: [7]

32. Trace a public insurance program

Government and industry programs can make coverage available when the voluntary market does not supply it on workable terms, or they can spread a concentrated risk across participants. Analyze a program in sequence: who qualifies, who issues or services coverage, where premium or other funding comes from, who pays claims initially, and who ultimately retains the cost. Member assessments, policyholder charges, public funding, and reinsurance are economically different funding sources. An allocation percentage describes how a stated pool cost is shared; it does not by itself describe premiums collected or recoveries received. Compare the program with voluntary coverage to see whether risks are assigned, shared, or backed by a separate layer. Assess whether the arrangement meets its access and stability aims without assuming all participants bear equal risk.

Worked example: A hypothetical residual auto pool records $20 million of losses and allocates them to two insurers at 60% and 40%. Their gross allocated losses are 0.60 × $20 million = $12 million and 0.40 × $20 million = $8 million. These amounts sum to $20 million. No premiums, reinsurance, or public recoveries were specified, so the example cannot determine either insurer's final net cost.

Watch the distinction: A 60% participation share does not mean an insurer writes 60% of voluntary-market policies. It specifies the share of this pool's stated losses under the hypothetical allocation.

Reference: [7]

33. Read U.S. statutory results

Statutory accounting measures an insurer's financial position for regulatory reporting, with particular attention to obligations and assets available to meet them. Statutory surplus is admitted assets less recognized liabilities. An asset can have economic value yet be excluded or limited for statutory purposes, so use admitted assets in this subtraction. Underwriting results affect surplus through recognized premiums, losses, and expenses, alongside investment results and other changes. Distinguish a balance-sheet stock of surplus at a date from an income statement's flow over a period. An increase in a recognized unpaid-claim liability, with all other items fixed, lowers both the period result and ending surplus. Other reporting bases may recognize or measure items differently; do not carry a figure across bases without checking its definition.

Worked example: Assume a U.S. insurer reports $500 million of admitted assets and $420 million of liabilities. Statutory surplus is $500 million − $420 million = $80 million. If an updated claim estimate raises recognized liabilities by $5 million and nothing else changes, liabilities become $425 million and surplus becomes $75 million. This arithmetic describes the stated balance-sheet effect, not the insurer's overall solvency.

Watch the distinction: Subtracting liabilities from total assets without checking admission rules may overstate statutory surplus. The statutory calculation uses admitted assets under its reporting basis.

Reference: [7]

34. Interpret solvency indicators

Risk-based capital relates an insurer's available capital measure to capital calculated for specified risks. Always identify the numerator and denominator before interpreting a ratio; a percentage without those definitions is ambiguous. For a ratio expressed relative to authorized control level RBC, divide total adjusted capital by that level and multiply by 100%. The result shows a multiple of the stated denominator. It is a screening measure, not a complete assessment of future claim payments. Reserve uncertainty, catastrophe concentration, asset risk, reinsurance collectibility, and business plans may require further analysis. Compare reporting dates on a consistent basis and ask what drove a change in the ratio. Avoid inferring a regulatory response from an isolated constructed example when no applicable action criteria have been provided.

Worked example: In a constructed illustration, total adjusted capital is $90 million and authorized control level RBC is $30 million. The ratio is ($90 million/$30 million) × 100% = 300%, meaning capital is three times that denominator. If capital fell to $75 million with the denominator unchanged, the ratio would be 250%. The decline warrants investigation of its cause; these figures alone do not establish an action outcome.

Watch the distinction: A 300% ratio does not mean 300% of all future claims are funded. Its denominator is a defined RBC measure, not total potential losses.

Reference: [7]

35. Explain the appointed actuary’s conclusion

The U.S. appointed actuary assesses the reasonableness of carried unpaid claim and claim-adjustment-expense amounts for the applicable reporting opinion. Start with the booked amounts and the actuarial estimates on consistent gross, ceded, and net bases. Review methods, data quality, development assumptions, and the effect of material subsequent information. An actuarial range describes outcomes supported by selected assumptions; simply falling inside it does not settle every opinion question. Materiality, reserve composition, and company-specific risks matter. The risk of material adverse deviation concerns whether uncertainty could produce a materially worse outcome under the relevant reporting framework. Communicate the basis of the conclusion and significant uncertainty clearly, distinguishing an estimate of unpaid claims from a guarantee of final settlement cost.

Worked example: Suppose a hypothetical insurer carries $100 million of unpaid claims and a supported actuarial range is $95 million to $110 million. The carried amount is $5 million above the low end and $10 million below the high end. That location is useful but insufficient for an opinion. The actuary also examines methods, materiality, unusual exposures, and company-specific adverse-deviation risks before reaching and explaining a conclusion.

Watch the distinction: Choosing the midpoint of a range does not automatically make the booked reserve adequate. A range depends on assumptions, and an opinion also considers materiality and relevant risks.

Reference: [7]

36. Assess a reinsurance transaction

For an excess-of-loss cover, identify the insurer's retention and the reinsurer's maximum payment per covered claim. For a covered loss L, ceded loss is the smaller of the positive amount above retention and the cover limit; net loss is gross loss less ceded loss. That arithmetic describes claim allocation, while financial reporting asks a further question: does the contract transfer significant insurance risk, or does its substance function mainly as financing? Terms such as experience accounts, repayment features, and cancellation provisions can change the economic result. A nominal ceded amount is also a receivable subject to collection risk. Evaluate the contract's full cash-flow pattern, accounting treatment, and effect on reported capital and earnings rather than judging it from one large claim.

Worked example: A covered claim is $9 million under a hypothetical $5 million excess of $3 million contract. The amount above retention is $9 million − $3 million = $6 million, capped at $5 million. Ceded loss is therefore $5 million and the insurer's net claim is $9 million − $5 million = $4 million. The calculation assumes full coverage and collection; accounting treatment requires examining the entire contract.

Watch the distinction: The $5 million cover limit is not the insurer's retention. Here the insurer bears the first $3 million plus $1 million above the exhausted reinsurance layer.

Reference: [7]

Exam 6C: Canada Regulation and Financial Reporting

37. Map Canadian insurance oversight

In Canada, separate prudential supervision of an insurer's financial strength from provincial or territorial rules governing insurance products, rates, and market conduct. For a federally regulated insurer, the federal prudential supervisor examines matters such as capital and risk management. Provincial and territorial authorities have their own responsibilities for insurers operating in their jurisdictions, including consumer-facing requirements. The exact rate-filing treatment depends on the province and product, so locate the applicable jurisdiction before describing a review. This division matters when a company proposes a rate change while also facing solvency pressure: a sound pricing indication does not answer the capital question, and capital adequacy does not itself approve a rate. Identify the entity, policy location, and regulatory question before selecting the authority or reporting measure.

Worked example: A hypothetical federally regulated insurer proposes an auto rate change for policies in Province A. Its financial resilience raises a federal prudential question. Whether and how the proposed rate is reviewed raises a Province A insurance-regulation question. An actuary would prepare the rate evidence for the applicable provincial process and analyze solvency separately. No specific provincial filing rule is assumed in this example.

Watch the distinction: Federal supervision of the insurer's capital does not mean a federal authority approves every provincial auto rate. Rate and consumer-protection requirements must be checked in the relevant province.

Reference: [8]

38. Evaluate a Canadian insurance program

A residual market or compensation arrangement should be understood through its purpose and its cash flows. First identify who qualifies and what coverage or protection the program supplies. Then trace whether policies remain with a servicing insurer, move into a shared pool, or trigger compensation after an insurer failure. These designs allocate different risks and should not be treated as interchangeable. For a shared-risk pool, calculate each member's portion from the stated allocation basis, then consider premiums, expenses, recoveries, and assessments before describing net cost. Ask whether the program improves access or protects claimants, and whether its funding and allocation remain workable under stress. A simple share calculation demonstrates distribution of losses, not the full economics or legal duties of any named Canadian program.

Worked example: A hypothetical Canadian pool has $15 million of eligible claims and member shares of 50%, 30%, and 20%. Allocations are $15 million × 0.50 = $7.5 million, × 0.30 = $4.5 million, and × 0.20 = $3 million. The shares total 100% and the allocated claims total $15 million. If the pool expands access to coverage, these allocations show how members share its stated claim cost before other cash flows.

Watch the distinction: Do not confuse a residual pool with a claimant compensation scheme after insurer insolvency. Both can involve industry funding, but they address different coverage and failure problems.

Reference: [8]

39. Read IFRS 17 insurance liabilities

In Canadian IFRS 17 reporting, the liability for remaining coverage concerns obligations for insured events that have not yet occurred within the covered service. The liability for incurred claims concerns events already incurred, including claims that have not yet been reported. This distinction follows the event and service period, not simply whether cash has been paid. Each component may reflect measurement elements appropriate to its contracts and accounting approach; do not treat a simple illustrative sum as a complete valuation recipe. Reinsurance contracts held are measured and presented separately from insurance contracts issued, even when recoveries reduce the insurer's economic exposure. The Exam 6C content outline includes IFRS 17 reporting and Canadian solvency topics.

Worked example: A hypothetical portfolio reports $40 million of liability for remaining coverage and $18 million of liability for incurred claims. Together those stated components are $40 million + $18 million = $58 million before considering any separately presented reinsurance asset. The $18 million relates to events already incurred, whether reported or not. The $40 million relates to coverage still to be provided under the stated measurement.

Watch the distinction: An unpaid claim is not automatically part of remaining coverage merely because payment is future. Once the insured event has occurred, its obligation belongs with incurred claims.

Reference: [8]

40. Relate capital to risk

The Canadian Minimum Capital Test compares available capital with capital required for a property and casualty insurer under the applicable framework. A ratio is available capital divided by required capital, expressed as a percentage. State the reporting entity and capital definitions before calculating; importing a U.S. RBC denominator would make the result misleading. A higher ratio generally provides more measured headroom relative to the stated requirement, but it compresses varied risks into one figure. Stress tests and financial condition analysis explore what could happen if claims, catastrophes, investments, or reinsurance recoveries differ from expectations. Examine both the size of a stressed capital decline and the assumptions behind it. A single ratio cannot reveal the timing of losses, concentration of exposures, or management's feasible response.

Worked example: In a purely hypothetical MCT-style illustration, available capital is $120 million and required capital is $80 million. The ratio is ($120 million/$80 million) × 100% = 150%. If a separately modeled adverse scenario reduces available capital by $30 million while required capital is assumed unchanged, the illustrative ratio becomes $90 million/$80 million = 112.5%. That sensitivity shows why the initial ratio is only one view of risk.

Watch the distinction: Do not read 150% as a universal pass threshold or as the U.S. RBC ratio. It is this example's available capital divided by its stated Canadian required capital.

Reference: [8]

41. Assess the appointed actuary’s work

For a Canadian property and casualty insurer, the appointed actuary’s work connects liability valuation with the insurer’s financial condition. Start by checking the data and methods used to estimate insurance obligations, then examine assumptions about claim development, discounting, and uncertainty. Materiality helps determine which differences could affect a reader’s decision; it does not excuse an unsupported estimate. Peer review provides an independent challenge to important judgments, including model choices and the clarity of the report. Financial condition testing asks what could happen under adverse scenarios, so it adds a forward-looking view to the valuation date estimate. A useful reporting sequence is to state the result, explain its principal sensitivities, test plausible stresses, and describe how management actions might change the outcome. The Exam 6C content outline identifies these Canadian valuation and reporting responsibilities.

Worked example: Suppose an adverse scenario reduces a Canadian insurer’s projected capital from $120 million to $85 million. The decline is $120 million − $85 million = $35 million, or about 29.2% of the starting amount. The actuary should explain the assumed loss shock, its timing, and any credible management response. The $85 million result is scenario dependent; it is not a revised valuation of every insurance liability.

Watch the distinction: A peer-reviewed point estimate does not settle the financial condition question. Review challenges the valuation, while scenario testing shows how future adverse events could change capital and risk.

Reference: [8]

42. Show reinsurance in Canadian reporting

Under IFRS 17, analyze insurance contracts issued and reinsurance contracts held as distinct positions. The insurer remains responsible for valid claims under its issued policies even when a reinsurer is expected to reimburse part of them. First estimate the gross insurance cash flows and related liability. Then measure the reinsurance position using the held contract’s terms, coverage period, timing, and expected recoveries, including the effect of possible reinsurer nonperformance where relevant. A simple gross-minus-recovery calculation can describe economic exposure, but it does not replace separate presentation and measurement. This distinction also matters when the reinsurance contract covers only certain events or attaches above a retention: the recovery will not necessarily move in step with gross claims. Reconcile gross, ceded, and net figures to explain the risk position without concealing the underlying obligation.

Worked example: Assume an insurer owes $10 million for incurred claims and expects a $4 million recovery under qualifying reinsurance. Show the $10 million gross claim obligation and the $4 million reinsurance recovery position separately. The simplified economic net is $10 million − $4 million = $6 million. If collectability or coverage changes, revise the recovery assessment; the gross claim obligation does not disappear.

Watch the distinction: Subtracting the expected recovery and reporting only a $6 million claim liability hides both the insurer’s gross obligation and its dependence on the reinsurer. Separate measurement makes that dependence visible.

Reference: [8]

Exam 6I: International Regulation and Financial Reporting

43. Compare international supervision

For China, Hong Kong, Malaysia, and Singapore, begin with the same supervisory questions, then answer them under the selected market’s framework. Identify the local insurance supervisor and the legal entity seeking authorization. Ask what business it may write, how products or rates are overseen, what conduct rules govern distribution and claims, and how solvency is measured and monitored. China’s National Financial Regulatory Administration, Hong Kong’s Insurance Authority, Bank Negara Malaysia, and the Monetary Authority of Singapore serve different national or local systems; their requirements should not be treated as interchangeable. Group ownership adds another question: which supervisor examines the local insurer, and how are group risks shared or constrained? Organize a comparison by supervisory function rather than copying one market’s capital ratio or filing process into another. The international Exam 6I outline specifies these four markets.

Worked example: A fictional insurer plans to sell property cover in Hong Kong. Identify the Hong Kong Insurance Authority as the local supervisor, then examine authorization, distribution conduct, any applicable product requirements, and solvency for the Hong Kong entity. If the parent operates in Singapore, Singapore’s supervisory framework may matter to the group, but it does not substitute for the local Hong Kong assessment.

Watch the distinction: A shared term such as risk-based capital does not imply identical rules across these markets. Compare the purpose and calculation of each local measure before comparing results.

Reference: [9]

44. Trace catastrophe risk sharing

Catastrophe financing is easiest to understand as a set of layers. The primary insurer pays covered policyholders and retains losses within its own retention. Reinsurance can reimburse a defined portion above that retention, subject to attachment points, limits, exclusions, and contract conditions. A public or industry program may cover a further layer, but only if the stated event and eligibility conditions are met. Draw the loss amount on one scale and mark each layer’s start and end; this prevents counting the same dollar twice. Then distinguish who initially settles claims from who ultimately bears their cost. The insurer may face liquidity pressure while waiting for reimbursement even when much of the final cost is transferred. Compare structures by coverage availability, incentives, funding, and the possibility that a severe event exhausts one or more layers.

Worked example: Assume a hypothetical $100 million covered catastrophe. Primary insurers retain the first $20 million. Reinsurers cover losses from $20 million through $70 million, a $50 million layer. A stated public program covers $70 million through $100 million, or $30 million. The allocations sum to $20 million + $50 million + $30 million = $100 million, assuming every layer’s conditions are satisfied.

Watch the distinction: Calling reinsurance the payer of policyholder claims confuses claim settlement with ultimate risk bearing. The primary insurer can remain liable to policyholders while recovering covered amounts from other layers.

Reference: [9]

45. Stress-test solvency

A solvency assessment compares resources available to absorb losses with the risks the insurer could face. A current capital measure summarizes a valuation date, while stress testing asks how results change under a specified shock. Own risk and solvency assessment, or ORSA, brings that forward-looking analysis into decisions about strategy, risk appetite, and capital planning. State whether a stress changes expected losses, assets, liabilities, or several of them; otherwise a capital movement may be counted twice. Next consider timing, concentrations, reinsurance recoverability, and management actions. A ratio calculated under one jurisdiction’s framework is meaningful within that framework, but its numerator and required-capital denominator may differ elsewhere. A sound conclusion therefore describes the stressed position and the assumptions needed to sustain it, rather than declaring an insurer safe from one unqualified number.

Worked example: Suppose a hypothetical catastrophe shock raises expected losses from $40 million to $70 million, an incremental $30 million. If that increase is entirely unanticipated and reduces an illustrative $100 million capital amount dollar for dollar, stressed capital is $70 million. The insurer would compare that position with its applicable capital needs, liquidity, recoveries, and possible responses. The calculation alone does not establish a regulatory outcome.

Watch the distinction: Do not subtract the full $70 million again if the original $40 million expected loss is already reflected in capital. The incremental effect in this example is $30 million.

Reference: [9]

46. Interpret IFRS 17 performance

IFRS 17 separates the result from providing insurance service from the effects of financing insurance contracts. Insurance revenue reflects services provided during the period; insurance service expenses include the costs associated with that service. Their difference is the insurance service result before separately reported insurance finance effects. This separation helps you ask whether a change arose from claims and coverage performance or from discount rates and the passage of time. The premium allocation approach, when eligible, simplifies measurement of the liability for remaining coverage, often making it practical for shorter coverage patterns. It does not remove the need to assess liabilities for claims that have already occurred. Read a reported result in sequence: identify the contract group and period, calculate the service margin shown by the relevant line items, then consider financing and reinsurance effects before interpreting total profit.

Worked example: A fictional insurer reports $100 million of insurance revenue and $82 million of insurance service expenses for a period. The insurance service result is $100 million − $82 million = $18 million. If a separate insurance finance expense were $3 million and there were no other effects, the combined amount would be $15 million. The $18 million still describes service performance before that financing effect.

Watch the distinction: Treating insurance revenue as cash premium received confuses service recognized during a period with cash collection. Their timing can differ, even when the same contracts ultimately generate both amounts.

Reference: [9]

47. Evaluate international reinsurance

A reinsurance contract can transfer insured loss volatility while introducing other exposures. Evaluate what losses the contract actually covers, how much risk the reinsurer assumes, and whether contract terms limit reimbursement. Then assess counterparty risk: a valid receivable may be worth less if the reinsurer cannot pay in full. A commutation settles future obligations for an agreed amount, ending the parties’ rights and duties under the commuted portion. Its value depends on the settlement price versus expected future recoveries, uncertainty, timing, and the risk released. For a financial health assessment, follow the sequence from gross claim to contractual ceded amount, expected collectible amount, and retained economic cost. This shows why a large nominal ceded balance can offer less protection than it appears to, especially after a catastrophe affecting several counterparties.

Worked example: Assume a $12 million reinsurance recoverable on covered claims, but only $9 million is expected to be collectible after a counterparty shortfall. The shortfall is $12 million − $9 million = $3 million. Relative to full collection, the insurer’s expected net economic claim cost rises by $3 million. A proposed commutation must be assessed against the $9 million collectible estimate and remaining uncertainty.

Watch the distinction: A contract’s stated ceded amount is not automatically cash available to pay claims. Coverage disputes, credit quality, settlement timing, and commutation can all change the amount ultimately realized.

Reference: [9]

48. Apply actuarial professional judgment

Professional judgment is needed when credible actuarial methods produce different financial estimates. Start by explaining each method’s data, assumptions, and sensitivity to recent experience. Investigate whether differences reflect claim development, inflation, exposure mix, discounting, or data quality. Model governance adds checks on intended use, validation, changes, and ownership of limitations; a reproducible calculation is only one part of that control. The actuary should communicate a selected result and its uncertainty in terms decision makers can act on, while meeting the professional standards and local reporting duties that apply to the assignment. Across jurisdictions, the same analytical discipline can be used without assuming that reporting forms or legal responsibilities are identical. A midpoint can be a useful comparison, but it needs an independent reason before it becomes a selected estimate.

Worked example: Two defensible reserve methods produce $95 million and $105 million. Their midpoint is ($95 million + $105 million)/2 = $100 million. Suppose the higher method responds to recent claim inflation while the lower method relies on older development. Explain that $10 million spread and test the inflation assumption before selection. Reporting $100 million solely because it is halfway between the methods would leave the central judgment unsupported.

Watch the distinction: Averaging two estimates does not automatically reduce bias. If both methods share a flawed data source, their midpoint inherits that weakness; if assumptions differ, explain which better fits the exposure.

Reference: [9]

PCPA Examination: Predictive Analytics

49. Match data to the decision

Before building a property and casualty prediction, state the decision it will support. That determines the target, the observation unit, and which information can legitimately be used. For claim frequency, a policy period can be one row, claim count the target, and earned policy-years the exposure. Include predictors known by the decision date, such as selected policy characteristics; fields created after a claim may leak the answer into the model. A claim-level table serves a different purpose because policies with no claims have no row there. Check that joins preserve the intended policy periods and that exposure is measured consistently. Overall frequency is total claims divided by total earned exposure, not the simple average of row-level rates when exposures differ. This setup makes both model interpretation and validation align with the business decision.

Worked example: Two policies contribute 0.5 and 1.0 earned policy-years and have 1 and 0 claims. Together they have 1 claim over 1.5 policy-years, so observed frequency is 1/1.5 = 0.667 claims per policy-year. Their individual rates are 2 and 0; averaging those rates gives 1, which overweights the half-year policy.

Watch the distinction: A table containing only claims cannot directly estimate policy claim frequency. It excludes the zero-claim policies and their exposure, so the denominator needed for frequency is missing.

Reference: [10]

50. Interpret transformations and missingness

A transformation changes the question a model can answer. Taking a logarithm compresses large positive values; grouping a continuous predictor trades detail for a simpler pattern. Before transforming, inspect the distribution, missing values, and the meaning of special codes. A recorded zero may be a real value, an unknown value, or an administrative placeholder, and those cases should not be merged casually. For missing predictors, compare defensible treatments such as imputation, an explicit missingness indicator, or exclusion when justified. Fit any imputation rule on training data and apply the same rule to validation data so the holdout does not inform model construction. Review whether missingness itself reflects underwriting or claim processes that could change over time. Interpret coefficients after treatment: their meaning depends on the transformed scale and reference category.

Worked example: In a fictional vehicle dataset, age zero means a new vehicle, while 12% of records have unknown age. Replacing unknown ages with zero would combine that 12% with genuinely new vehicles. Instead, use an explicit unknown category or a training-set imputation plus missingness indicator. Compare held-out performance and inspect the resulting age effect before selecting a treatment.

Watch the distinction: Imputing missing values from the full dataset before splitting leaks validation information into training. Even a simple average should be estimated within the training partition and then reused.

Reference: [10]

51. Choose a GLM response

Choose a generalized linear model by asking what one observation’s response represents. A count response records the number of claims and commonly uses a Poisson log-link model with earned exposure as an offset; check whether observed variability challenges its variance assumption. A binary response records whether any claim occurred and calls for a model of probability, commonly with a logit link. Severity is the size of a claim conditional on a claim existing, so a positive-response model such as Gamma with a log link can be appropriate. Policies with no claim have no observed claim severity of zero; they belong in the frequency analysis. Aggregate cost combines occurrence and size and may require a model able to represent both zero outcomes and positive amounts, such as a suitable Tweedie specification. Let the business target decide which response is relevant.

Worked example: Suppose three policies have counts 0, 1, and 2 over differing earned periods. A count GLM uses those counts and a log-exposure offset. Converting them to 0, 1, and 1 instead changes the target to any-claim probability and loses the distinction between one and two claims. A severity model would use the positive claim amounts for the claiming policies.

Watch the distinction: Entering zero-claim policies as zero-dollar claims in a positive severity model changes the response definition. Zero policies contribute to frequency or aggregate cost, while severity describes claim size when a claim occurs.

Reference: [10]

52. Interpret an offset and coefficient

In a log-link frequency GLM, an exposure offset represents observation time with a coefficient fixed at one. Write the mean count as exposure × exp(intercept + predictor effects). Halving exposure therefore halves predicted count for an otherwise identical risk, while predicted frequency per policy-year remains unchanged. A fitted predictor coefficient acts differently: increasing that predictor by one unit multiplies frequency by exp(coefficient), assuming other variables are fixed. An offset is appropriate when expected count should scale proportionally with exposure; check whether that assumption is sensible for the coverage and data. Interpret effects on the response scale, not as additions to the raw count. If you want a fuller review of GLM interpretation alongside other statistics, the MAS-II study guide provides related context.

Worked example: Suppose a one-year policy has predicted count 0.2. An identical half-year policy receives 0.5 × 0.2 = 0.1 expected claims. A predictor coefficient of log(1.25) multiplies either count by 1.25: the one-year prediction becomes 0.25 and the half-year prediction becomes 0.125. Both still imply the same annual frequency of 0.25 claims per policy-year.

Watch the distinction: Treating log(exposure) as a freely estimated predictor lets the model learn a scaling power. An offset fixes proportional scaling; the predictor coefficient estimates a separate risk effect.

Reference: [10]

53. Read validation and diagnostics

Training fit tells you how closely a model describes data used to estimate it; validation performance tests predictions on data withheld from that fitting. A more flexible model can lower training deviance by following noise, then perform worse on new observations. Compare candidates on the same holdout rows, response definition, exposure treatment, and deviance convention. After choosing a promising model, examine diagnostic patterns: residuals by fitted value or segment, unusually influential observations, unstable coefficients, and calibration across risk groups. A single overall score can hide a poor fit in an important segment. Repeated splits or cross-validation can reduce dependence on one lucky partition, provided related records remain grouped appropriately. Prefer the model that supports the decision with credible held-out performance and understandable limitations, not merely the lowest training statistic.

Worked example: Model A has training deviance 900 and validation deviance 980. Model B has training deviance 820 and validation deviance 1,100, calculated on the same respective datasets. B improves training deviance by 80 but worsens validation deviance by 120. On these figures A is the stronger predictive choice. Next inspect segment residuals before relying on A for a business decision.

Watch the distinction: Lower training deviance does not establish better prediction. If added complexity learns sample noise, the validation result can worsen even while the fitted training result improves.

Reference: [10]

54. Translate model output

A useful model explanation names the unit, time period, comparison, and decision. A predicted claim frequency is an expected count per exposure unit, not a guarantee that an individual policy will claim. Translate a difference into both absolute and relative terms, then ask whether the model is sufficiently calibrated in the relevant segment. Pricing needs more than frequency: expected severity, expenses, reinsurance, uncertainty, and practical suitability can affect the final decision. A ranking model may identify higher-risk policies without predicting their claim levels accurately, so state which property has been validated. Communicate uncertainty through observed variation, validation error, or a clearly described sensitivity rather than inventing precision. The intended audience should be able to tell what the estimate supports, what data population it covers, and what further assumptions a proposed action requires.

Worked example: A model estimates 0.12 claims per policy-year for segment A and 0.18 for segment B. The difference is 0.06 claims per policy-year; the relative increase is 0.06/0.12 = 50%. If both segments had the same expected severity, B would have higher expected claim cost. A price decision still requires the severity estimate, expenses, and a check of model performance in both segments.

Watch the distinction: Calling segment B ‘50% more expensive’ from frequency alone assumes equal claim severity and other costs. The model supports a frequency comparison; the premium conclusion needs additional components.

Reference: [10]

PCPA Project: Applied Predictive Analytics

55. Define an original project question

An applied predictive analytics project begins with a decision stated narrowly enough to test. Identify who will use the result, what action they might take, and when the prediction must be available. Define one observation row and the target over a fixed outcome period. For a frequency question, earned policy-years provide the exposure measure and a claim count provides the response. List candidate predictors available before the prediction date, then specify how success will be judged on withheld data. Held-out Poisson deviance can compare count predictions, while segment calibration can reveal a business weakness hidden by an average score. Write down data boundaries and exclusions so the estimate has a clear population. The PCPA project study guide can help organize this decision sequence.

Worked example: For a fictional auto portfolio, define one row per policy-year, claim count during that year as the target, and earned years as exposure. A policy active for six months contributes 0.5 earned years. Split policies before fitting and compare candidate count models by held-out Poisson deviance on the same validation records. The resulting model predicts future frequency for similar policies, within the chosen data period and coverage definition.

Watch the distinction: Beginning with whichever columns are easiest to model can produce an accurate-looking answer to the wrong question. Specify the decision, target period, observation unit, and available predictors first.

Reference: [10]

56. Audit and prepare project data

Data preparation should leave an auditable path from raw records to modeling rows. Check identifiers and dates before removing apparent duplicates: two identical-looking transactions might be legitimate, while a repeated policy-period record could double-count exposure and claims. Test basic constraints, such as nonnegative earned exposure and plausible dates, then investigate extreme values against source records instead of deleting them automatically. Summarize missingness by field and segment; a field missing mostly for one product may reflect a process change. Record each correction, exclusion, and imputation rule with the number of affected rows. Reconcile totals after cleaning, especially policy count, earned exposure, claim count, and claim amount. This protects the model from errors that a fitting algorithm may absorb without warning and makes later validation results interpretable.

Worked example: A toy file has 1,002 rows, including two exact duplicate policy-period records confirmed by identifier and dates. Removing the extra copies leaves 1,000 rows. If each duplicate contains 1 earned policy-year and 1 claim, the raw totals were overstated by 2 policy-years and 2 claims. Recalculate all totals after removal, and retain a record of why those rows were excluded.

Watch the distinction: Do not delete every repeated identifier. One policy may legitimately appear in multiple periods or transactions; duplicate detection must use the intended observation unit and source context.

Reference: [10]

57. Fit a defensible baseline

A baseline model gives later refinements a fair comparison. For claim frequency, begin with count as the response, a log link, and log earned exposure as an offset. The intercept then represents an expected annual rate for the reference risk, and predicted count scales with time insured. Check that every modeled row has positive exposure and that the target and exposure refer to the same coverage period. Fit the baseline on training data, inspect predicted versus observed counts by broad groups, and calculate a held-out performance measure. Consider overdispersion and influential records before interpreting coefficients too strongly. Add predictors only when they improve the decision or address a clear diagnostic issue. A simple, transparent baseline makes it possible to tell whether added complexity contributes useful information.

Worked example: Suppose a hypothetical intercept-only Poisson baseline predicts 40 claims across 200 earned policy-years. Its fitted frequency is 40/200 = 0.2 claims per policy-year. A policy with 0.5 earned years receives predicted count 0.5 × 0.2 = 0.1. The offset supplies that scaling; it does not imply the half-year policy has half the underlying annual risk.

Watch the distinction: Using earned years as an ordinary predictor may estimate an arbitrary exposure effect. The offset fixes the intended proportional relationship between time insured and expected claim count.

Reference: [10]

58. Validate by the decision unit

A validation split must represent the decision the model will face. If the task is to predict outcomes for new policies, all rows from one policy should stay in one partition. Otherwise a later row in validation can share stable characteristics, history, or identifiers with a row already used for training. That leakage can make the model appear more accurate than it will be on genuinely unseen policies. If the task is to forecast a future period for existing policies, a time-based split may better match deployment, with predictors limited to information available before that period. Decide the split rule before comparing models, then apply the same validation records to every candidate. Check that claim outcomes and exposure remain balanced enough for a meaningful comparison, especially in small or rare-event segments.

Worked example: A policy appears in both 2024 and 2025 data. For a new-policy prediction exercise, assign both rows to training or both to validation. If its 2024 row is in training and 2025 row is in validation, the model may benefit from information about the same policy. For a future-year forecast, instead train on earlier periods and ensure 2025 predictors contain no later outcomes.

Watch the distinction: A random row split is not automatically independent validation. Repeated policies, households, or claims can cross the split and reveal information about supposedly unseen observations.

Reference: [10]

59. Refine the model with evidence

Refinement should answer a diagnostic or business question, then survive a fair comparison. Add a predictor because it represents a plausible risk difference or corrects a visible pattern, and compare the revised model with the baseline on the same held-out observations. Examine coefficient direction and stability across folds or periods; a dramatic coefficient that changes sign may signal sparse data, collinearity, or an unstable interaction. Plot residuals or observed-to-predicted ratios by segment to see whether the added term fixes the intended weakness. A lower validation deviance is useful evidence, but inspect whether gains are concentrated in a tiny group or depend on outliers. Retain a simpler model when a complex term improves training fit but harms validation or makes decisions fragile.

Worked example: A fictional baseline has validation deviance 510. Adding a territory factor lowers it to 480, an improvement of 30 on the same records. Adding a large interaction instead raises deviance to 530, a deterioration of 20 versus baseline. Retain territory for coefficient and segment review; reject the interaction on this evidence. Check whether territory’s improvement persists across other validation periods.

Watch the distinction: A large training-fit gain does not justify keeping an interaction when held-out results deteriorate. Compare models on the same validation population and inspect stability before deciding.

Reference: [10]

60. Explain the resulting decision

Finish an applied analysis by connecting model output to a specific action and its assumptions. State what was predicted, for whom, over what period, and how well predictions performed on held-out data. Separate predictive evidence from an intervention effect: a model may identify high-cost policies, but it does not by itself prove that an action will reduce their claims. Estimate the proposed action’s benefit and cost on a consistent per-policy or portfolio basis, then show a sensitivity that could reverse the decision. Technical readers need model choices, diagnostics, and uncertainty; business readers need the expected impact and conditions under which it holds. Recommend a decision only within the population and data conditions that were evaluated, with a way to monitor actual results after implementation.

Worked example: Suppose a hypothetical intervention costs $20 per selected policy and is assumed to avoid $35 of claims per selected policy. Estimated net benefit is $35 − $20 = $15 per policy, or $15,000 for 1,000 selected policies. If realized avoidance is only $15, net benefit becomes −$5 per policy. The recommendation therefore depends on validating both selection performance and the intervention effect.

Watch the distinction: A model that predicts high claims does not establish that an intervention prevents them. Treat avoided claims as a separate assumption requiring evidence, then test the decision if that effect is smaller.

Reference: [10]

Official references and further reading

All study guides

Choose your exam version

SOA P

SOA FM

CAS MAS-I

CAS MAS-II

CAS Exam 5

CAS Exam 6U

CAS Exam 6C

CAS Exam 6 International

CAS PCPA Exam

CAS PCPA Project

Probability (Exam 1 / SOA P)

1. Count the possible outcomes

2. Update probabilities with evidence

3. Model a claim count

4. Turn losses into insurer payments

5. Read dependence from a joint distribution

6. Approximate a portfolio total

Financial Mathematics (Exam 2 / SOA FM)

7. Convert interest conventions

8. Value a payment stream

9. Split a loan payment

10. Price a coupon bond

11. Use spot and forward rates

12. Measure interest-rate sensitivity

MAS-I: Modern Actuarial Statistics I

13. Count arrivals over time

14. Connect survival and hazard

15. Estimate a population mean

16. Account for censored losses

17. Specify a frequency GLM

18. Diagnose an extended linear model

MAS-II: Modern Actuarial Statistics II

19. Blend experience with credibility

20. Separate shared and individual effects

21. Choose a local or tree prediction

22. Reduce and group predictors

23. Compare ranking performance

24. Forecast a stationary series

Exam 5: Basic Ratemaking and Estimating Claim Liabilities

25. Build a future rate indication

26. Align historical premium and loss

27. Refine rates by class

28. Develop claims to ultimate

29. Blend observed and expected losses

30. Test reserve reasonableness

Exam 6U: United States Regulation and Financial Reporting

31. Locate the U.S. regulator’s role

32. Trace a public insurance program

33. Read U.S. statutory results

34. Interpret solvency indicators

35. Explain the appointed actuary’s conclusion

36. Assess a reinsurance transaction

Exam 6C: Canada Regulation and Financial Reporting

37. Map Canadian insurance oversight

38. Evaluate a Canadian insurance program

39. Read IFRS 17 insurance liabilities

40. Relate capital to risk

41. Assess the appointed actuary’s work

42. Show reinsurance in Canadian reporting

Exam 6I: International Regulation and Financial Reporting

43. Compare international supervision

44. Trace catastrophe risk sharing

45. Stress-test solvency

46. Interpret IFRS 17 performance

47. Evaluate international reinsurance

48. Apply actuarial professional judgment

PCPA Examination: Predictive Analytics

49. Match data to the decision

50. Interpret transformations and missingness

51. Choose a GLM response

52. Interpret an offset and coefficient

53. Read validation and diagnostics

54. Translate model output

PCPA Project: Applied Predictive Analytics

55. Define an original project question

56. Audit and prepare project data

57. Fit a defensible baseline

58. Validate by the decision unit

59. Refine the model with evidence