Build the calculations and modeling judgments tested by CAS MAS-I: translate arrival rates into probabilities, value survival-contingent payments, estimate parameters from incomplete observations, and interpret regression output. This guide provides 20 original practice items and 60 worked concepts, allocated 5/5/10 questions and 15/15/30 concepts across the three published domains. It follows the supplied 2026 outline for use as of October 16, 2026; announced 2027 schedules do not replace that scope. Single-answer, multiple-selection, numerical-entry, and matching practice illustrate listed formats without claiming an official item mix. Citations establish syllabus relevance; technical explanations use established statistical knowledge. No R-code interpretation is required.
What this guide covers
Practice 20 original MAS-I items and study 60 concepts in probability, statistics, and extended linear models.
A. Probability Models (Stochastic Processes and Survival Models)
1. Poisson counts and exponential waiting times
A homogeneous Poisson process has independent, stationary increments and constant intensity λ. An interval of length t contains a Poisson(λt) count, so its mean and variance both equal λt. The waiting time to the next event is exponential with rate λ, and the waiting time to event k is gamma with shape k and rate λ. Match the rate's time unit to the interval before calculating. These relationships connect frequency questions with waiting-time questions, but depend on a constant rate rather than merely an average observed frequency.
Worked example: A portfolio generates claims at four per week under a homogeneous Poisson model. Over half a week the expected count is two, and the probability of no claims is e^(−2). The expected waiting time to the next claim is one quarter of a week.
Common mistake: A rate of four claims per week does not mean exactly four arrive each week or that waiting times are equally spaced.
Official syllabus reference: [2]
2. Integrated intensity in a nonhomogeneous process
For a nonhomogeneous Poisson process with deterministic intensity λ(t), define the cumulative intensity Λ(t) = ∫₀ᵗλ(u)du. The increment from a to b is Poisson with parameter Λ(b) − Λ(a), and disjoint increments remain independent. Arrival waiting times are generally not exponentially distributed on the original clock. Conditional on no event before a, the probability of waiting another s units is exp[−∫ₐᵃ⁺ˢλ(u)du]. Use the integral rather than a single endpoint rate when activity changes over time.
Worked example: Suppose reporting intensity is 1 + 2t claims per hour. Between hours one and two, integrated intensity is [t + t²]₁² = 4. The count therefore has mean and variance four, and its zero-count probability is e^(−4).
Common mistake: A changing deterministic intensity does not destroy independent increments; it removes stationarity of increments and constant-rate waiting-time formulas.
Official syllabus reference: [2]
3. Poisson thinning and superposition
Independent superposition adds Poisson process intensities. Conversely, independently assigning each event to a category with probability p produces a thinned Poisson process with intensity pλ(t). Complementary categories are independent processes under independent marking. If classification probability varies deterministically with time, the category intensity becomes p(t)λ(t). Independence of the original streams or marking mechanism is essential: shared random environmental effects can invalidate the simple conclusions. These rules allow one to model claim reporting channels, coverage categories, or severity bands without rebuilding the arrival process.
Worked example: Two independent claim streams have rates three and five per month. Their combined rate is eight. If each combined claim is independently referred for review with probability 0.25, reviewed claims form a Poisson process with rate two per month.
Common mistake: Splitting events by a fixed total quota is not independent thinning, and the resulting category processes need not be independent.
Official syllabus reference: [2]
4. Arrival times conditional on a Poisson count
Conditional on N(T) = n in a homogeneous Poisson process, the n unordered arrival times are independent Uniform(0, T) variables. Ordered arrivals therefore follow uniform order-statistic distributions. A subinterval occupying fraction p of the full interval contains a Binomial(n, p) count conditional on the total. For nonhomogeneous intensity, replace time fractions by integrated-intensity fractions. Conditioning changes independence: subinterval counts that were unconditionally independent must sum to the known total and are then negatively associated. This distinction often determines whether a binomial or Poisson calculation is appropriate.
Worked example: Exactly six claims occurred during a ten-day interval under a homogeneous Poisson model. Conditional on that total, the number occurring in the first two days is Binomial(6, 0.2), with expected count 1.2 and zero-count probability 0.8⁶.
Common mistake: Once the total count is fixed, counts in complementary subintervals cannot remain independent because one determines the other.
Official syllabus reference: [2]
5. Persistent heterogeneity in mixed Poisson models
A mixed Poisson model first draws a latent rate Λ, then generates a Poisson count conditional on that rate. Over time t, E[N(t)] = tE(Λ) and Var[N(t)] = tE(Λ) + t²Var(Λ). Persistent rate heterogeneity creates overdispersion and positive dependence between counts from disjoint intervals for the same risk. Conditional independence does not imply unconditional independence. A gamma-distributed rate yields a negative binomial marginal count. Distinguish a rate drawn once per risk from a newly independent rate drawn for every period.
Worked example: A risk's annual rate is two or six with equal probability. Its three-year expected count is twelve. Since the rate variance is four, the count variance is twelve plus nine times four, or forty-eight, despite conditional Poisson variability.
Common mistake: Replacing a random rate by its mean preserves the expected count but removes heterogeneity, understating variance and cross-period dependence.
Official syllabus reference: [2]
6. Compound Poisson reward totals
Let S = Σᵢ₌₁ᴺXᵢ, with N Poisson of mean m and independent identically distributed rewards independent of N. Then E(S) = mE(X) and Var(S) = mE(X²), not merely mVar(X). Total variance includes uncertainty about both claim sizes and the number of claims. The moment-generating function is M_S(t) = exp[m(M_X(t) − 1)] wherever the severity generating function exists. The aggregate distribution also has mass at zero from zero claims, with additional zero mass possible if rewards themselves can be zero.
Worked example: A month has Poisson mean five claims. Independent severities have mean $100 and variance $400. Expected aggregate loss is $500, while aggregate variance is 5(400 + 100²) = 52,000 dollars squared, including variability from the random count.
Common mistake: Multiplying severity variance by expected frequency omits the variation caused by a random number of claims with nonzero mean severity.
Official syllabus reference: [2]
7. Markov transition and first-step equations
A discrete-time Markov chain specifies next-state probabilities through a transition matrix P. If the current distribution is a row vector α, its distribution after n steps is αPⁿ. Multi-step probabilities sum over intermediate states through matrix multiplication. Expected hitting times can instead be found by first-step analysis: hᵢ = 1 + Σⱼpᵢⱼhⱼ outside the target set, with h = 0 inside it. A unique finite solution requires appropriate reachability and finite expected hitting time; an inaccessible target cannot be assigned a finite waiting time.
Worked example: From an active state, a policy remains active next year with probability 0.8 and becomes permanently closed with probability 0.2. Expected years to closure satisfy h = 1 + 0.8h, giving five years from the active state.
Common mistake: Squaring each transition probability separately does not give two-step probabilities; all possible intermediate states must be included through matrix multiplication.
Official syllabus reference: [2]
8. Stationary distributions and convergence conditions
A stationary distribution π satisfies πP = π with nonnegative entries summing to one. It describes a distribution unchanged by one transition, not necessarily the limit from every starting state. For a finite irreducible chain, the stationary distribution is unique; aperiodicity additionally ensures convergence of state probabilities to it. Periodic chains can oscillate even while possessing a stationary distribution. Reducible chains may have multiple stationary distributions. Solve the balance equations together with normalization, then separately examine the chain's structure before claiming a long-run limiting distribution.
Worked example: A two-state chain switches from A to B with probability 0.2 and from B to A with probability 0.3. Balance gives 0.2π_A = 0.3π_B, so π = (0.6, 0.4). Positive self-transition probabilities make this irreducible chain aperiodic.
Common mistake: Finding a solution to πP = π does not alone establish convergence from every starting state; periodicity and communicating classes matter.
Official syllabus reference: [2]
9. Brownian increments and geometric transformation
Standard Brownian motion has continuous paths, independent stationary increments, and B(t) − B(s) distributed normally with mean zero and variance t − s. An arithmetic model X(t) = x₀ + μt + σB(t) therefore has normal increments. A geometric model S(t) = S₀exp[(μ − σ²/2)t + σB(t)] stays positive and has lognormal levels, with E[S(t)] = S₀e^(μt). The variance correction in the exponent distinguishes the desired level-growth parameter from the drift of log levels. Brownian paths are continuous but not ordinary differentiable trajectories.
Worked example: An arithmetic reserve model has drift three monetary units per year and volatility two units per square-root year. Over a quarter year, its increment has mean 0.75 and variance one, because variance scales with elapsed time rather than its square.
Common mistake: Brownian standard deviation scales with the square root of elapsed time; treating it as proportional to time produces incorrect interval uncertainty.
Official syllabus reference: [2]
10. Limited expectations and deductible layers
For nonnegative X, the limited expected value at u is E[min(X, u)] = ∫₀ᵘS_X(x)dx. A payment layer with attachment d and maximum payment u is min[(X − d)₊, u], whose expectation equals ∫dᵈ⁺ᵘS_X(x)dx. Equivalently, subtract limited expectations at d + u and d. These are unconditional payment expectations and include zero payments below attachment. A per-payment expectation conditions on a positive payment and requires division by P(X > d), when that probability is positive.
Worked example: For an exponential loss with mean $1,000, a layer attaching at $500 and paying at most $2,000 has expected payment 1000[e^(−0.5) − e^(−2.5)], approximately $524.45. Losses below attachment remain part of the population with zero insurer payment.
Common mistake: A policy limit is a payment transformation, not truncation of the loss population; losses exceeding the limit still contribute capped payments.
Official syllabus reference: [2]
11. Survival functions, hazards, and cumulative hazards
For a continuous nonnegative lifetime T, S(t) = P(T > t), density f(t) = −S′(t), and hazard h(t) = f(t)/S(t) where survival is positive. The cumulative hazard H(t) = ∫₀ᵗh(u)du gives S(t) = exp[−H(t)] for an absolutely continuous lifetime starting with survival one. A hazard is an instantaneous conditional rate, not an interval probability. Calculate interval failure probabilities as 1 − exp[−∫ₐᵇh(u)du], conditional on survival to a. Constant hazard produces exponential survival.
Worked example: A lifetime has hazard h(t) = 0.02t per year. Its cumulative hazard by year five is 0.01(25) = 0.25, so five-year survival is e^(−0.25), approximately 0.7788, rather than one minus the endpoint hazard.
Common mistake: The hazard at an instant can exceed one because it is a rate; a probability over an interval cannot exceed one.
Official syllabus reference: [2]
12. Residual lifetime and aging behavior
Given survival to age a, remaining-life survival is P(T − a > t | T > a) = S(a + t)/S(a). The mean residual lifetime is the integral of this conditional survival over positive t, assuming finiteness. Exponential lifetimes are memoryless, so residual life has the original distribution. Other distributions generally retain age information. Increasing hazard often represents deterioration, whereas decreasing hazard can represent selection of more durable survivors. Do not reset the original lifetime distribution at age a unless the model actually has the memoryless property.
Worked example: A lifetime has S(t) = exp(−t²). Conditional survival for one additional year after surviving two years is S(3)/S(2) = e^(−5). This differs sharply from first-year survival e^(−1), reflecting the model's increasing hazard with age.
Common mistake: Survival to the current age changes the relevant population; using S(t) alone ignores conditioning unless the lifetime distribution is memoryless.
Official syllabus reference: [2]
13. Joint-life and last-survivor status
The joint-life lifetime is min(T₁, T₂), while the last-survivor lifetime is max(T₁, T₂). Joint-life survival requires both lives to survive; under independence it equals S₁(t)S₂(t). Last-survivor survival requires at least one survivor and equals S₁(t) + S₂(t) − P(T₁ > t, T₂ > t). Independence is needed to replace the joint probability by a product. These same formulas describe series and parallel systems respectively. Dependence affects both statuses, so marginal survival functions alone do not generally determine their values.
Worked example: At ten years, two independent lives have survival probabilities 0.8 and 0.7. Joint-life survival is 0.56, while last-survivor survival is 0.8 + 0.7 − 0.56 = 0.94. The latter includes outcomes where exactly one person survives.
Common mistake: Adding two survival probabilities double-counts the outcome where both survive; last-survivor calculations require subtracting the overlap once.
Official syllabus reference: [2]
14. Whole-life benefits and payment timing
A continuous whole-life insurance paying one at death has expected present value Ā = E[e^(−δT)] = ∫₀∞e^(−δt)f_T(t)dt under constant interest force δ. If death is exponential with mortality force μ, this becomes μ/(μ + δ). A benefit paid at the end of the year of death instead uses v^(K + 1), where K is curtate future lifetime. Payment timing changes discounting and cannot be switched silently. Benefit amounts multiply the unit value, while deferred or temporary benefits restrict the integration or summation range.
Worked example: With constant mortality force 0.03 and interest force 0.05, a $10,000 benefit paid immediately at death has expected present value 10,000 × 0.03/0.08 = $3,750. An end-of-year benefit requires a different discounting calculation.
Common mistake: Immediate payment at death and payment at the end of the death year are distinct contracts, even when mortality assumptions coincide.
Official syllabus reference: [2]
15. Life annuity integrals and insurance identities
A continuous life annuity paying one per year while alive has value ā = ∫₀∞e^(−δt)S(t)dt. With constant mortality force μ, ā = 1/(μ + δ). For the corresponding continuous whole-life insurance, Ā + δā = 1, under conditions ensuring the discounted survival boundary vanishes. A discrete annuity-due pays at the beginning of each period and uses survival probabilities at integer times, including the initial certain payment. Distinguish continuous payment rates from discrete payment amounts before applying an identity or interpreting the resulting units.
Worked example: A continuous annuity pays $2,000 per year, with mortality force 0.04 and interest force 0.06. Its value is 2000/0.10 = $20,000. The corresponding unit insurance value is 0.4, consistent with 0.4 + 0.06(10) = 1.
Common mistake: An annuity-due includes an immediate payment, whereas a continuous annuity starts accruing immediately without a separate full initial installment.
Official syllabus reference: [2]
B. Statistics
16. Sample moments and method-of-moments estimation
Sample moments replace population expectations with empirical averages. The first raw moment is X̄, and the second raw moment is n⁻¹ΣXᵢ². Method-of-moments estimation equates enough sample and theoretical moments to solve for unknown parameters. The empirical central second moment uses denominator n, whereas the usual unbiased sample variance uses n − 1. Moment estimators can be convenient without being sufficient or efficient. Check that the solution lies in the parameter space and that the required population moments exist before treating the equations as valid estimators.
Worked example: For an exponential distribution with mean θ, equating the theoretical mean to observations 2, 3, and 7 gives θ̂ = 4. Estimating the rate instead gives λ̂ = 1/4, illustrating how the parameterization changes the reported estimate.
Common mistake: The denominator n − 1 belongs to the unbiased variance correction, not automatically to every empirical moment used in estimation equations.
Official syllabus reference: [2]
17. Factorization and sufficient statistics
A statistic T is sufficient for a parameter when the conditional distribution of the full sample given T no longer depends on that parameter. The factorization theorem establishes sufficiency by writing the likelihood as gθ(T(x))h(x), with h parameter-free. Include support indicators, especially when endpoints depend on the parameter. Sufficiency concerns retained information, not unbiasedness or numerical closeness to the parameter. A one-to-one transformation of a sufficient statistic remains sufficient, and a sufficient statistic can be a vector when the model has several unknown parameters.
Worked example: For independent Poisson observations with common mean λ, the likelihood contains e^(−nλ)λ^(Σxᵢ) divided by the product of factorials. The parameter-dependent part uses only the sum, so the total count is sufficient for λ.
Common mistake: A statistic is not sufficient merely because it is unbiased; unbiasedness describes its expectation, while sufficiency describes retained parameter information.
Official syllabus reference: [2]
18. Likelihood maximization and invariance
Maximum likelihood selects parameter values maximizing the probability or density assigned to the observed data. Work with the log likelihood to simplify products, but check boundaries and parameter-dependent support rather than relying only on a zero derivative. MLE invariance means that a transformed parameter g(θ) has estimator g(θ̂), with suitable handling of nonunique maximizers. Invariance does not preserve unbiasedness. If a likelihood increases up to a support boundary, the maximum may occur there despite having no interior score root. Always distinguish the observed likelihood from a probability distribution over parameters.
Worked example: For exponential lifetimes with rate λ and no censoring, the log likelihood is n log λ − λΣtᵢ. Its maximizer is n/Σtᵢ. The estimated mean lifetime is therefore 1/λ̂ = Σtᵢ/n by invariance.
Common mistake: Solving the score equation finds candidates for an interior maximum; it does not replace support checks, boundary checks, or comparison of maxima.
Official syllabus reference: [2]
19. Estimator MSE and the Cramér–Rao boundary
Mean squared error decomposes as Var(θ̂) + Bias(θ̂)². An unbiased estimator can therefore have greater MSE than a biased competitor. Under regularity conditions, the Cramér–Rao lower bound for an unbiased estimator of θ is 1/I_n(θ), where I_n is sample Fisher information. Attaining that bound establishes efficiency within the relevant unbiased class. Parameter-dependent support can invalidate the usual regularity argument. Compare estimators using the requested criterion: minimum variance among unbiased estimators and minimum MSE across all estimators are different optimization problems.
Worked example: Estimator A has variance four and zero bias, so its MSE is four. Estimator B has variance one and bias one, so its MSE is two. Under squared-error loss, B performs better despite being biased.
Common mistake: Unbiasedness does not guarantee the smallest MSE, and a regular Cramér–Rao formula cannot be applied blindly to endpoint-parameter models.
Official syllabus reference: [2]
20. Consistency and asymptotic normality
Consistency means an estimator converges in probability to the true parameter as sample size grows. It does not require finite-sample unbiasedness. Under suitable regularity conditions, an MLE is approximately normal with variance given by inverse Fisher information, enabling approximate standard errors and intervals. This is a large-sample statement and can fail at boundaries, under nonidentifiability, or with irregular support. A useful sufficient route to consistency is bias tending to zero together with variance tending to zero. An estimator with persistent variance need not become accurate merely because its expectation is correct.
Worked example: The normal variance MLE divides squared deviations by n, so it is biased downward in finite samples. Its bias vanishes as n grows, and it converges to the population variance under the usual independent normal sampling assumptions.
Common mistake: An estimator can be consistent while biased in every finite sample; consistency concerns its limiting concentration rather than exact finite-sample expectation.
Official syllabus reference: [2]
21. Rejection regions, Type II error, and power
A Type I error rejects a true null hypothesis; its probability is controlled by the test's significance level under the stated null model. A Type II error fails to reject at a specified alternative, with probability β(θ). Power is 1 − β(θ), so it is a function across alternatives rather than one universal property of the test. Calculate each probability using the sampling distribution appropriate to that parameter value and the same rejection region. Lowering a rejection threshold typically increases both Type I error and power when other design features remain fixed.
Worked example: A known-variance mean test rejects when X̄ exceeds 12. Under an alternative where X̄ is normal with mean 13 and standard deviation one, Type II error is Φ(−1), approximately 0.1587, and power is approximately 0.8413.
Common mistake: A p-value is calculated under the null model; it is not the probability that the null hypothesis is true.
Official syllabus reference: [2]
22. Likelihood-ratio tests and rejection direction
A likelihood-ratio test compares the largest likelihood under the null restrictions with the largest likelihood in the full parameter space. The ratio lies between zero and one, and small values indicate that the restrictions fit poorly. Equivalently, large values of minus twice the log ratio favor rejection. In regular nested problems, this statistic has an asymptotic chi-square distribution with degrees of freedom equal to the number of independent restrictions. Exact simple-versus-simple likelihood ratios can instead identify most-powerful rejection regions through the Neyman–Pearson principle.
Worked example: A restricted likelihood maximum is e^(−52), and the unrestricted maximum is e^(−49). The deviance-style statistic is −2[−52 − (−49)] = 6. With one regular restriction, compare six with the supplied chi-square critical value.
Common mistake: A larger restricted-to-unrestricted likelihood ratio supports the restrictions more strongly; rejection uses small ratios or large negative twice-log ratios.
Official syllabus reference: [2]
23. Normal mean tests and Student t pivots
For independent normal observations with unknown variance, (X̄ − μ₀)/(S/√n) has a Student t distribution with n − 1 degrees of freedom under the null. Known population variance instead yields a normal pivot. A paired analysis applies the one-sample method to within-pair differences, preserving the pairing. For two independent samples, a pooled t test assumes common population variance; Welch's approach does not impose that equality. Choose the pivot from the sampling structure and assumptions, not solely from sample size or whether the observed standard deviations appear similar.
Worked example: Ten paired before-and-after differences have mean three and standard deviation two. Testing a zero mean difference gives t = 3/(2/√10), approximately 4.743, on nine degrees of freedom, because the ten differences form the analysis sample.
Common mistake: Paired observations should not be analyzed as two independent samples when the relevant uncertainty comes from within-pair differences.
Official syllabus reference: [2]
24. Chi-square and F tests for normal variances
For a normal sample, (n − 1)S²/σ² follows chi-square with n − 1 degrees of freedom. This pivot supports variance tests and intervals, whose endpoints are asymmetric because the reference distribution is skewed. For two independent normal samples, the ratio (S₁²/σ₁²)/(S₂²/σ₂²) follows an F distribution with the respective residual degrees of freedom. These exact results require normal populations and independence. Specify which variance occupies the numerator and use matching degrees of freedom; taking a reciprocal also swaps those degrees of freedom.
Worked example: A normal sample of size twelve has variance eighteen. Testing a population variance of nine gives chi-square statistic 11(18)/9 = 22 on eleven degrees of freedom. Whether this rejects depends on the stated alternative and supplied critical value.
Common mistake: Normal-theory variance tests are sensitive to distributional shape; their exact chi-square and F reference distributions do not hold for arbitrary populations.
Official syllabus reference: [2]
25. Frequency models and overdispersion
Poisson frequency has equal mean and variance. Binomial frequency has a fixed number of trials and variance below its mean when success probability is positive. A negative binomial model allows variance greater than the mean, often through gamma mixing of a Poisson rate. Under the parameterization E(N) = μ and Var(N) = μ + αμ², α measures extra-Poisson variation. Compare distributions using support and generating mechanisms as well as moments. Apparent overdispersion may reflect omitted heterogeneity, dependence, or excess zeros rather than identifying one unique frequency model.
Worked example: A portfolio's count model requires mean three and variance six. Under Var(N) = μ + αμ², α = (6 − 3)/9 = 1/3. This matches a negative binomial variance pattern but does not alone prove that distribution fits.
Common mistake: Variance exceeding the mean rules out a homogeneous Poisson marginal model, but it does not uniquely establish a negative binomial mechanism.
Official syllabus reference: [2]
26. Severity tails, parameterizations, and moments
Positive severity models differ in tail behavior and moment existence. Exponential severity has constant hazard and finite moments of all orders. Lognormal severity arises by exponentiating a normal variable and is right-skewed. For a Lomax model with survival S(x) = (1 + x/θ)^(−α), the mean is θ/(α − 1) only when α > 1, and variance is finite only when α > 2. State parameterizations explicitly because names such as Pareto can refer to different supports. A fitted distribution need not possess every moment used in aggregate calculations.
Worked example: A Lomax severity with scale $1,000 and shape 1.5 has mean $2,000, using 1000/(1.5 − 1). Its variance is infinite, so an aggregate variance calculation requiring a finite severity second moment is unavailable under this fitted model.
Common mistake: A distribution can have a finite mean and infinite variance; estimating an average does not establish that variance-based approximations are justified.
Official syllabus reference: [2]
27. Random-sum aggregate loss moments
For aggregate loss S = Σᵢ₌₁ᴺXᵢ with independent identically distributed severities independent of frequency, E(S) = E(N)E(X). Total variance gives Var(S) = E(N)Var(X) + Var(N)[E(X)]². The first term measures severity uncertainty conditional on claim count; the second measures frequency uncertainty. This formula accommodates binomial, negative binomial, and other counts, unlike the specialized compound Poisson simplification. Dependence between count and severity or among severities requires additional terms or a different conditional model. Finite second moments are necessary for a finite variance result.
Worked example: Frequency has mean ten and variance fifteen. Severity has mean $200 and standard deviation $50. Aggregate mean is $2,000, while aggregate variance is 10(2500) + 15(40,000) = 625,000 dollars squared under the stated independence assumptions.
Common mistake: Substituting expected frequency for frequency variance silently imposes the Poisson mean–variance equality and can substantially understate aggregate uncertainty.
Official syllabus reference: [2]
28. Order statistics through binomial counting
For an independent continuous sample, the event X_(k) ≤ x means at least k observations are at most x. Thus its distribution is obtained from a Binomial(n, F(x)) tail. The minimum has survival [1 − F(x)]ⁿ, and the maximum has distribution F(x)ⁿ. Transforming an order statistic by its continuous population distribution gives a Beta(k, n + 1 − k) variable. These formulas require independence and a common distribution. Discrete data introduce ties, so density formulas and continuous beta-transform claims require separate treatment.
Worked example: For five independent Uniform(0, 10) losses, the probability that the largest is at most eight is (0.8)⁵ = 0.32768. The maximum's expected value is 10 × 5/6, approximately 8.333, rather than the population mean five.
Common mistake: The kth order statistic is not the kth original observation; sorting changes its distribution and creates dependence among ordered observations.
Official syllabus reference: [2]
29. Censoring contributions to likelihood
Censoring retains a unit while revealing only partial information about its value. An exact continuous observation contributes f(x); right censoring at c contributes S(c); left censoring contributes F(c); interval censoring between a and b contributes F(b) − F(a). Under an independent, noninformative censoring mechanism, parameter-free censoring factors can be omitted from the likelihood. Replacing a censored value by its boundary falsely treats partial information as an exact outcome. Censoring indicators and exposure times therefore matter even when the missing tail value cannot be observed.
Worked example: Two exponential lifetimes fail at years two and three, while a third survives to censoring at year five. Their likelihood is proportional to λ²e^(−10λ), yielding rate estimate 0.2 per year from two failures over ten observed years.
Common mistake: A right-censored observation is not a failure at the censoring time and should not be discarded as if it supplied no information.
Official syllabus reference: [2]
30. Truncation and conditional observation likelihoods
Truncation excludes units outside the observation region entirely. With left truncation at d, an observed continuous value x > d contributes f(x)/S(d), conditioning on eligibility for observation. With right truncation at u, it contributes f(x)/F(u) for x < u. If thresholds differ by unit, each contribution uses its own normalizing probability. This differs from censoring, where excluded tail values remain represented by indicators. Ignoring truncation estimates the observed conditional population as though it were the original population and generally distorts severity parameters.
Worked example: Only exponential losses exceeding $500 enter a dataset. Their conditional density is λe^(−λx)/e^(−500λ), so the excesses x − 500 are exponential. For observed losses $700, $900, and $1,200, the rate estimate is 3/1300 per dollar.
Common mistake: Unrecorded below-threshold losses are truncated, whereas recorded units known only to lie below a threshold are left censored observations.
Official syllabus reference: [2]
C. Extended Linear Models
31. Exponential-family structure of a GLM
A generalized linear model combines an exponential-family response distribution, a linear predictor η = Xβ, and a link g(μ) = η. In the form exp{[yθ − b(θ)]/a(φ) + c(y, φ)}, the mean is b′(θ) and variance is b″(θ)a(φ). The canonical link connects μ to the natural parameter θ. Distribution and link perform different jobs: the distribution controls support and mean–variance behavior, while the link maps admissible means to a linear scale. A canonical link is convenient but is not mandatory.
Worked example: For a Poisson response, b(θ) = e^θ, so μ = e^θ and variance equals μ. The canonical link is log μ. This explains both positive fitted means and the Poisson model's mean-dependent variance in a single framework.
Common mistake: Choosing a log link does not by itself make a model Poisson; several response distributions can use the same link.
Official syllabus reference: [2]
32. Inverse links and coefficient scales
A link transforms the conditional mean, and prediction requires applying its inverse. Identity links yield additive mean effects; log links yield multiplicative mean effects; logit links yield multiplicative odds effects. A coefficient is interpreted on the linear-predictor scale before conversion to the response scale. For nonlinear inverse links, averaging linear predictors and then transforming generally differs from averaging transformed predictions. Check the admissible mean range: an identity-link probability model can leave zero to one, whereas inverse logit predictions remain within that interval for finite linear predictors.
Worked example: A log-link severity model has intercept log(500) and coefficient log(1.2) for a binary indicator. The baseline predicted mean is $500, and the indicated group has mean $600. The coefficient represents a twenty-percent mean increase, holding other predictors fixed.
Common mistake: Exponentiating a logistic coefficient produces an odds ratio, while exponentiating a log-link coefficient produces a conditional mean ratio.
Official syllabus reference: [2]
33. Variance functions and dispersion estimation
GLMs relate conditional variance to the mean through a variance function, often Var(Yᵢ) = φV(μᵢ)/wᵢ with known weights wᵢ. Poisson has V(μ) = μ and fixed dispersion one; gamma commonly has V(μ) = μ² with estimable dispersion. Pearson residual squares can estimate dispersion using their sum divided by residual degrees of freedom. Extra dispersion inflates coefficient covariance without necessarily changing a quasi-likelihood mean fit. Distinguish a full probability model from quasi-likelihood: ordinary likelihood-based AIC is not automatically available for a quasi model.
Worked example: A gamma model predicts means $100 and $200 with common dispersion 0.25 and unit weights. Their variances are 2,500 and 10,000 dollars squared. Both have coefficient of variation 0.5, demonstrating constant relative rather than constant absolute spread.
Common mistake: A dispersion estimate above one does not automatically identify the source of poor fit; mean misspecification and dependence can also inflate residual variation.
Official syllabus reference: [2]
34. Iteratively reweighted least squares
GLM coefficient estimation commonly uses iteratively reweighted least squares. At each iteration, current means determine a working response z = η + (y − μ)(dη/dμ) and weights proportional to (dμ/dη)²/Var(Y). Weighted least squares updates β, then means and weights are recalculated until convergence. These are computational working quantities, not a claim that the original responses have constant-variance normal errors. Link and variance choices jointly determine the weights. Convergence alone establishes neither adequate fit nor absence of separation or numerical instability.
Worked example: In a Poisson log-link model with current mean two, dη/dμ = 1/2. An observed count five gives working response η + (5 − 2)/2 = η + 1.5. The working weight is proportional to the current mean two.
Common mistake: IRLS weights usually change during fitting; treating initial weights as fixed can produce a different estimate from the intended GLM maximum.
Official syllabus reference: [2]
35. OLS geometry and estimability
Ordinary least squares minimizes squared residuals and, for a full-column-rank design, gives β̂ = (X′X)⁻¹X′y. The fitted vector is the projection Hy, and residuals are orthogonal to every design column. With an intercept, residuals sum to zero. Exact collinearity makes individual coefficients nonidentifiable without constraints, although some fitted values and coefficient combinations remain estimable. Near collinearity instead produces unstable coefficients and large standard errors. The projection interpretation explains both normal equations and why residual patterns can reveal structure absent from the model.
Worked example: A model includes an intercept and indicators for every category of a three-level factor. Each row's indicators sum to the intercept column, so the design is rank deficient. Dropping one category indicator establishes a reference group and restores identification.
Common mistake: A high overall fit does not make collinear coefficients individually precise; fitted values can be stable while separate slope estimates remain unstable.
Official syllabus reference: [2]
36. ANOVA decomposition and partial F tests
For nested normal linear models, a partial F test compares the reduction in residual sum of squares per added parameter with the full model's residual mean square. The statistic is [(SSE_R − SSE_F)/q]/[SSE_F/(n − p_F)], where q is the number of independent restrictions and p_F includes the intercept. Under the classical null assumptions it follows F(q, n − p_F). A one-coefficient partial F equals the corresponding t statistic squared. Sequential ANOVA contributions can depend on predictor entry order when predictors are not orthogonal.
Worked example: Adding three terms reduces SSE from 210 to 180 in a full model with six coefficients and sixty observations. The partial F statistic is (30/3)/(180/54) = 3. Compare it with the supplied F critical value for three and fifty-four degrees of freedom.
Common mistake: A significant joint F test does not imply that every added coefficient has a significant individual t test.
Official syllabus reference: [2]
37. Categorical contrasts and ordinal restrictions
An unordered factor with K levels typically needs K − 1 independent contrasts when an intercept is present. Reference coding interprets each coefficient relative to a baseline level; changing the baseline changes coefficients but not fitted values. An ordinal predictor has ordered levels, but coding them as consecutive numbers additionally assumes equal spacing and a linear effect on the link scale. Separate indicators relax that restriction. Continuous predictors similarly impose a specific functional relationship unless transformed or expanded. Choose coding to represent the substantive restrictions the model should actually test.
Worked example: Risk grades low, medium, and high can use two indicators without imposing equal effect increments. Coding them 1, 2, and 3 instead forces the medium-to-high link-scale change to equal the low-to-medium change, a stronger modeling assumption.
Common mistake: An ordered category label does not establish equal numerical distances between levels or justify a single linear coefficient without further assumptions.
Official syllabus reference: [2]
38. Interactions and conditional effects
An interaction allows one predictor's effect to depend on another predictor. In η = β₀ + β₁x + β₂z + β₃xz, the x slope on the link scale is β₁ + β₃z. Main effects are conditional effects when interacting predictors equal zero or their reference levels. Centering a continuous predictor can make those reference conditions meaningful without changing fitted values. Retaining relevant main effects with an interaction usually preserves a coherent hierarchical interpretation. With nonlinear links, response-scale effects can vary even without an explicit product term.
Worked example: A log-cost model has tenure coefficient −0.04 and tenure-by-commercial coefficient 0.03. Personal policies have annual mean multiplier e^(−0.04), while commercial policies have multiplier e^(−0.01). The interaction changes the tenure slope rather than adding a fixed dollar amount.
Common mistake: An interaction coefficient is a difference in link-scale effects, not automatically the full effect for the group carrying the interaction indicator.
Official syllabus reference: [2]
39. Offsets versus estimated control variables
An offset enters a linear predictor with a coefficient fixed at one. In a log-link count model, log exposure as an offset makes expected counts proportional to exposure while modeling the underlying rate. A control variable instead receives an estimated coefficient and adjusts comparisons for measured differences. Treating log exposure as a predictor estimates an exposure elasticity rather than imposing proportionality. Positive exposure is required for its logarithm. Offset placement must match the modeled response and link; an exposure offset appropriate for counts should not be applied mechanically to already normalized rates.
Worked example: A count model log μ = log e + β₀ + β₁x assigns the same predicted annual rate to two otherwise identical risks. If their exposures are one and three policy-years, their expected counts have ratio three regardless of the estimated coefficients.
Common mistake: An offset is not an ordinary predictor whose coefficient happens to be near one; its coefficient is fixed before fitting.
Official syllabus reference: [2]
40. Logistic probabilities and separation
Logistic regression models η = log[p/(1 − p)], with inverse p = e^η/(1 + e^η). A coefficient's exponential is a conditional odds ratio, while absolute probability changes depend on baseline risk. Complete separation occurs when a predictor combination perfectly separates binary outcomes; ordinary maximum likelihood coefficients may then diverge rather than having finite estimates. Large reported coefficients and unstable standard errors can signal this problem. A converged-looking output should therefore be assessed alongside data patterns and fitting warnings, particularly with sparse categories or rare events.
Worked example: A baseline event probability of 0.1 has odds 1/9. An odds ratio of two gives odds 2/9 and probability 2/11, approximately 0.182. Doubling the odds therefore does not double this probability to 0.2.
Common mistake: Very large logistic coefficients do not necessarily indicate a precisely estimated strong effect; sparse data or separation can make ordinary estimates unstable.
Official syllabus reference: [2]
41. Binomial denominators and classification thresholds
Grouped binomial data record successes Y out of m trials, with E(Y) = mp and Var(Y) = mp(1 − p) under independent equal-probability trials. A proportion alone omits its precision unless its denominator is retained. Predicted probabilities can then support classification through a chosen threshold. Lowering the threshold generally raises sensitivity and lowers specificity on a fixed dataset. Discrimination, calibration, and threshold-specific error costs are distinct: ranking risks well does not guarantee accurate probabilities, and a default threshold need not match the decision's relative costs.
Worked example: Two groups each show a claim proportion of 0.2, one from five policies and one from five hundred. Their point proportions match, but the larger denominator supplies much more binomial information. Treating both proportions as equally precise discards that difference.
Common mistake: A predicted probability threshold of one half is a decision convention, not a universal optimum for imbalanced outcomes or unequal error costs.
Official syllabus reference: [2]
42. Poisson fit, excess zeros, and residual checks
A Poisson GLM models nonnegative integer responses with variance equal to conditional mean. Its zero probability is exp(−μ), so fitted means imply a predicted number of zero observations by summing these probabilities. Compare zeros and residual spread conditionally, not merely against one overall mean. Unmodeled rate heterogeneity can create both extra zeros and overdispersion. A large Pearson statistic relative to residual degrees of freedom flags a discrepancy but does not uniquely diagnose its cause. Check the mean structure, exposure, dependence, and distributional assumptions before choosing a remedy.
Worked example: For fitted means one, two, and three, the expected number of zero-count observations is e^(−1) + e^(−2) + e^(−3), approximately 0.553. Replacing these means by their average two would produce a different expected zero count.
Common mistake: Many zeros do not alone establish a special zero-generating mechanism; low fitted means and heterogeneous Poisson rates can also produce many zeros.
Official syllabus reference: [2]
43. Gamma GLMs and lognormal retransformation
A gamma GLM is suitable for strictly positive continuous responses with variance proportional to mean squared under common dispersion. A log link directly models log E(Y | x). By contrast, normal regression on log Y models E(log Y | x); exponentiating its fitted value generally estimates a conditional median under lognormal errors rather than the conditional mean. With constant normal log-error variance σ², the conditional mean requires multiplication by exp(σ²/2). These are different modeling assumptions, even when both use logarithms. Exact zero severities require separate handling because gamma support is positive.
Worked example: A lognormal regression predicts log severity log(1000) with log-error variance 0.4. Its conditional median is $1,000, but its conditional mean is 1000e^(0.2), approximately $1,221.40. A gamma log-link prediction already targets the mean directly.
Common mistake: Exponentiating a fitted log response does not generally recover the original-scale mean because nonlinear transformations do not commute with expectation.
Official syllabus reference: [2]
44. Deviance and nested-model likelihood differences
Deviance compares a fitted model's likelihood with that of a saturated model that fits each observation separately. It is twice the saturated-minus-fitted log likelihood, with distribution-specific scaling conventions. For nested models fitted to the same observations and family, their deviance difference equals twice the improvement in maximized log likelihood when conventions match. Under suitable regularity and dispersion assumptions, compare that difference with chi-square degrees of freedom equal to the added parameters. Residual deviance itself is not universally chi-square, especially with sparse discrete observations or inadequate asymptotics.
Worked example: A reduced binomial model has deviance 82 and a larger nested model has deviance 75 after adding two coefficients. The improvement statistic is seven on two degrees of freedom under the usual approximation, not seventy-five on two degrees of freedom.
Common mistake: A deviance difference tests nested restrictions; comparing two residual deviances with different datasets does not create a valid likelihood-ratio test.
Official syllabus reference: [2]
45. AIC and BIC complexity penalties
AIC = −2ℓ + 2k and BIC = −2ℓ + k log n balance maximized likelihood against estimated parameter count. Smaller values are preferred within a comparable candidate set. Count all estimated likelihood parameters according to the model's convention, including dispersion when estimated. Compare models fitted to the same response observations with compatible likelihood constants and scales. BIC imposes a stronger per-parameter penalty than AIC when log n exceeds two. Neither criterion is a hypothesis-test p-value or an absolute goodness-of-fit certificate, and their different objectives can select different models.
Worked example: A model with log likelihood −80 and three estimated parameters has AIC 166. A larger model with log likelihood −78 and five parameters also has AIC 166. The likelihood gain exactly offsets the two-parameter AIC penalty, creating a tie.
Common mistake: Comparing AIC across different response transformations without accounting for the changed likelihood scale can produce an invalid model ranking.
Official syllabus reference: [2]
46. R-squared and its adjusted form
For OLS with an intercept, R² = 1 − SSE/SST measures the fraction of corrected sample variation explained by the fitted values. Adding predictors cannot lower training R². Adjusted R² instead uses residual degrees of freedom: 1 − [SSE/(n − p)]/[SST/(n − 1)], with p including the intercept. It can decrease when additional terms contribute little and can be negative. Neither measure substitutes for predictive validation. GLM pseudo-R² measures have different definitions, so they should not automatically inherit the variance-explained interpretation of ordinary OLS R².
Worked example: With n = 20, p = 4, SSE = 80, and SST = 100, R² is 0.20. Adjusted R² is 1 − (80/16)/(100/19) = 0.05, reflecting the degrees-of-freedom cost of fitting four coefficients.
Common mistake: An increase in training R² after adding predictors is automatic for nested OLS models and does not establish improved out-of-sample prediction.
Official syllabus reference: [2]
47. Residuals, leverage, and influence
A residual measures response discrepancy, leverage measures unusual predictor location, and influence measures how substantially fitting changes when an observation is removed. In OLS, leverage is hᵢᵢ from H = X(X′X)⁻¹X′, and raw residual variance is σ²(1 − hᵢᵢ). Standardized residuals account for this unequal residual variance. Cook's distance combines residual size and leverage to summarize influence. A high-leverage point can have a small residual because it pulls the fit toward itself. Diagnose these quantities together rather than equating one unusual feature with automatic deletion.
Worked example: A policy with an extreme exposure covariate may have high leverage even when its loss lies near the fitted line. Another policy near the predictor center can have a large residual but modest influence because it has little leverage.
Common mistake: A small residual does not prove an observation is uninfluential; a high-leverage observation can materially determine the fitted line.
Official syllabus reference: [2]
48. QQ plots and distributional tail patterns
A QQ plot compares ordered observed quantities with theoretical distribution quantiles. For normal-error diagnostics, approximate linearity supports the reference shape, while systematic tail departures suggest heavier tails, lighter tails, or skewness. With theoretical normal quantiles on the horizontal axis, heavy tails often place the lower tail below and upper tail above the reference line. A QQ plot does not test whether the conditional mean is correctly specified or whether variance depends on predictors. Residual dependence and fitted-model effects can also complicate interpretation, so evaluate the plotted residual definition and sample size.
Worked example: A residual QQ plot follows its reference line centrally but falls below it on the left and rises above it on the right. This pattern suggests heavier tails than the normal reference, motivating scrutiny of extreme residuals and normal-based inference.
Common mistake: A straight residual QQ plot does not establish constant variance or independence; it primarily addresses the distributional shape being compared.
Official syllabus reference: [2]
49. Added-variable and marginal model plots
An added-variable plot residualizes both the response and a selected predictor against the remaining design columns, then plots those residuals against each other. Its fitted slope equals the selected predictor's coefficient in the full OLS model. It reveals conditional association and potentially influential observations. A marginal model plot instead compares observed and model-implied response patterns against a selected variable, often using smooth curves. These plots answer different questions: marginal patterns can differ from partial relationships when predictors are correlated, so a raw response-versus-predictor slope is not a substitute for an adjusted effect.
Worked example: Territory and mileage are correlated. A raw loss-versus-mileage plot shows a steep trend, but the added-variable plot after adjusting both loss and mileage for territory shows a shallow slope. The full-model mileage coefficient reflects that adjusted relationship.
Common mistake: Residualizing only the response is insufficient for an added-variable plot; the selected predictor must also be adjusted for the other predictors.
Official syllabus reference: [2]
50. Histograms, boxplots, and conditional exploration
Histograms display distribution shape but depend on bin width and boundaries. Boxplots summarize median, quartiles, and convention-based whiskers; points beyond whiskers are flags rather than proven recording errors. Compare groups using both location and spread, and retain sample-size information where possible. An overall distribution can conceal distinct subgroup structures, while a marginal association can reverse after conditioning on another variable. Exploratory plots guide response-family, transformation, and interaction choices, but they do not alone establish a conditional model or a causal relationship.
Worked example: Two territories have equal median severity, but one boxplot has a much wider upper quartile and several large observations. A mean-only comparison misses this tail difference. Separate histograms can help determine whether a common positive-severity distribution is plausible.
Common mistake: Boxplot outlier labels are rule-based screening indicators, not evidence that observations are invalid or should automatically be removed before modeling.
Official syllabus reference: [2]
51. Bayes classification and nearest-neighbor flexibility
Under equal misclassification costs, the Bayes classifier chooses the class with greatest conditional probability at each predictor value. Its error is the lowest achievable using those predictors and true probabilities. K-nearest-neighbor classification estimates local class proportions instead. Small K produces flexible boundaries with lower bias and higher variance; large K smooths boundaries, increasing bias while reducing variance. Predictor scaling affects which observations are nearest. Training error generally favors excessive flexibility, so select K using validation rather than apparent fit. Unequal decision costs can change the optimal classification threshold.
Worked example: Among nine nearest standardized predictor neighbors, six have claims and three do not. A KNN classifier with equal costs predicts a claim. Using only the single nearest neighbor could reverse that prediction, illustrating the instability of a more flexible local rule.
Common mistake: The smallest training classification error does not identify the best K; very small neighborhoods can fit noise and generalize poorly.
Official syllabus reference: [2]
52. Cross-validation without preprocessing leakage
K-fold cross-validation repeatedly fits a model on K − 1 folds and evaluates it on the held-out fold, then aggregates validation losses. Every data-dependent preprocessing step, including scaling, imputation, and feature selection, must be learned within the training portion to evaluate the complete procedure. Otherwise held-out information leaks into fitting. Use the same folds to compare candidates and reserve an independent test set when an unbiased final assessment is required. Random folds assume an appropriate exchangeability structure; grouped or dependent observations may require a split that respects that structure.
Worked example: To tune ridge regression, split the data into folds first. Within each training fold, estimate predictor means and scales, fit each candidate penalty, and transform the held-out fold using those training estimates. Average its prediction errors across folds.
Common mistake: Selecting predictors on the full dataset before cross-validation leaks response information and can make the reported validation error too optimistic.
Official syllabus reference: [2]
53. Bootstrap uncertainty and resampling units
The nonparametric bootstrap approximates an estimator's sampling variability by repeatedly sampling n observations with replacement from an observed sample of size n and refitting the estimator. Variation across bootstrap estimates provides a standard-error estimate. The resampling unit must match the independence structure; independently resampling rows from clustered data breaks within-cluster dependence. A bootstrap sample contains repeated observations and omits some originals, so it is not an independent external test dataset. Bootstrap validity also depends on the statistic and sampling setting, especially for irregular boundary or extreme-value problems.
Worked example: An analyst estimates a slope from eighty independent policies, then draws one thousand bootstrap samples of eighty policies with replacement. The standard deviation of the one thousand refitted slopes estimates slope uncertainty under the empirical sampling distribution.
Common mistake: Drawing a sample without replacement of the original size simply rearranges the data and does not generate ordinary nonparametric bootstrap variability.
Official syllabus reference: [2]
54. Best subset and stepwise model search
Best subset selection evaluates all predictor subsets, whereas forward selection adds terms sequentially and backward elimination removes them from an initial larger model. Stepwise paths need not recover the globally best subset at each size. Training residual error always favors additional fitted terms, so select model size with a complexity criterion or validation. Search itself creates selection uncertainty, which ordinary final-model coefficient tests do not automatically reflect. Respect required interaction hierarchy and distinguish the number of predictors from the number of fitted coefficients when factors or basis expansions generate multiple columns.
Worked example: With six candidate numeric predictors, best subset search considers sixty-four subsets including the intercept-only model. Forward selection follows only one nested path. A predictor excluded early by forward search can still belong to the best two-predictor subset.
Common mistake: Forward selection does not guarantee the best model of each size because its earlier choices constrain all subsequent candidate models.
Official syllabus reference: [2]
55. Ridge penalties and standardization
Ridge minimizes residual sum of squares plus λ times the sum of squared slopes, normally leaving the intercept unpenalized. Standardizing predictors makes the penalty comparable across variables measured in different units. With λ = 0, a full-rank fit returns OLS; increasing λ shrinks the coefficient vector and stabilizes poorly identified directions. Ridge generally retains all predictors rather than selecting a sparse subset. Tune the penalty for predictive performance, and report coefficient interpretation on the intended scale after reversing standardization. The precise numerical λ depends on the objective's normalization convention.
Worked example: Mileage measured in miles and income measured in thousands of dollars have very different numerical scales. Standardizing both before ridge prevents the penalty from favoring a variable merely because its coefficient is numerically smaller under its chosen measurement units.
Common mistake: Ridge shrinkage is not invariant to arbitrary predictor rescaling unless the preprocessing or penalty explicitly compensates for those scale changes.
Official syllabus reference: [2]
56. Lasso sparsity and correlated alternatives
The lasso minimizes residual sum of squares plus λΣ|βⱼ|, usually excluding the intercept. The absolute-value penalty has corners that can place coefficient estimates exactly at zero, providing variable selection as well as shrinkage. Standardization remains important because the penalty acts on coefficient magnitudes. With strongly correlated predictors, the lasso may select one while dropping another, and the selected member can change across samples. A zero coefficient is a penalized modeling decision rather than proof of no relationship. Choose λ through an appropriate validation procedure rather than maximizing training fit.
Worked example: Two nearly duplicate exposure measures predict loss similarly. Across resamples, a lasso model may retain the first measure in some fits and the second in others. Similar predictive accuracy can coexist with unstable variable selection when information is redundant.
Common mistake: A lasso-selected predictor list should not be interpreted as a uniquely established set of causal drivers, especially when candidate predictors are highly correlated.
Official syllabus reference: [2]
57. Principal components versus partial least squares
Principal components regression first forms orthogonal directions explaining predictor variation, without using the response, then regresses the response on selected components. Partial least squares uses response information when constructing directions, seeking components associated with both predictors and outcome. Both reduce dimension and can stabilize correlated designs, but PCR may discard a low-variance direction that is highly predictive. Choose the number of components through validation. Standardization affects component geometry, and any response-informed component construction must occur inside training folds to avoid leakage during performance assessment.
Worked example: Two predictors vary mainly together, but the response depends on their small difference. The first principal component captures their shared variation and may predict poorly. Retaining the second component or using response-informed PLS can recover the relevant contrast.
Common mistake: Explaining most predictor variance does not guarantee explaining most response variation; principal components are constructed without knowing the outcome.
Official syllabus reference: [2]
58. Polynomial and step-function bases
Polynomial regression expands a predictor into powers while remaining linear in coefficients. Higher degree increases flexibility but can create unstable boundary behavior and severe collinearity in raw powers. Orthogonal polynomial bases change numerical representation without changing the fitted polynomial space at a fixed degree. Step functions instead divide the predictor range into intervals with separate levels, permitting discontinuities at cutpoints. Degree and cutpoints control complexity and should be selected deliberately. Both are basis expansions, so they can enter a linear model or a GLM with an appropriate link.
Worked example: A quadratic log-severity model uses age and age squared in the linear predictor. Its age effect on log mean is β₁ + 2β₂age, not a constant slope. A step model with age bands would instead impose constant predictions within each band.
Common mistake: A model containing squared predictors is still linear in its coefficients; nonlinear predictor transformations do not by themselves require nonlinear parameter estimation.
Official syllabus reference: [2]
59. Regression splines and smoothing penalties
Regression splines join low-degree polynomial pieces at knots while enforcing continuity conditions. Cubic splines usually keep the function and its first two derivatives continuous. Natural cubic splines add linear boundary behavior. Smoothing splines instead balance squared-error fit against a roughness penalty proportional to ∫[f″(x)]²dx. Larger roughness penalties favor smoother functions, approaching a straight line in the usual setting. Knot count, basis dimension, and smoothing penalty are related but distinct controls of complexity. Effective degrees of freedom summarize flexibility for a penalized smoother rather than simply counting displayed basis coefficients.
Worked example: A smoothing spline fitted to age and loss follows local fluctuations when its penalty is small. Increasing the penalty suppresses curvature, eventually producing an almost linear age trend. This changes flexibility without requiring the analyst to delete particular age observations.
Common mistake: More available spline basis functions do not necessarily mean a rougher final fit; a strong smoothing penalty can heavily restrict their effective contribution.
Official syllabus reference: [2]
60. Generalized additive models and interaction limits
A generalized additive model extends the GLM predictor to η = β₀ + f₁(x₁) + … + f_p(x_p), allowing smooth nonlinear effects while preserving additivity on the link scale. Centering constraints help identify individual smooths separately from the intercept. A log-link GAM combines smooth effects multiplicatively on the mean scale, while a logistic GAM remains additive in log odds. Additivity limits interactions: one predictor's link-scale effect does not vary with another unless an interaction component is explicitly included. Evaluate smoothness and predictive performance rather than interpreting every visible wiggle as structure.
Worked example: A log-frequency GAM includes smooth age and vehicle-value terms plus a log-exposure offset. Age can have a curved rate effect, but its log-rate curve is shared across vehicle values unless an age-by-value interaction is added to the model.
Common mistake: Allowing each predictor a flexible curve does not automatically model interactions; additive smooth effects still impose a specific separation of predictor contributions.
Official syllabus reference: [2]