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The central exposed to risk E^c_x differs from the initial exposed to risk E_x principally because:
A
B
C
D
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2026 Statistics
Key Facts: IAI CS2 Exam
5
Syllabus Areas
IAI CS2 syllabus
25%
Survival Models Weight
IAI CS2 syllabus
25%
Stochastic Processes Weight
IAI CS2 syllabus
3h15m
Paper A Duration
IAI CS2 syllabus
1h45m
Paper B Duration
IAI CS2 syllabus
R
Paper B Software
IAI CS2 syllabus
IAI Subject CS2 (Risk Modelling and Survival Analysis) is assessed through a 3-hour-15-minute paper-based Paper A and a 1-hour-45-minute computer-based Paper B in R, combined for a single result. The current syllabus, which mirrors IFoA CS2, weights Stochastic Processes and Survival Models most heavily at about 25% each, Random Variables/Distributions for Risk Modelling and Time Series at about 20% each, and Machine Learning at about 10%. IAI sets the pass mark and the INR exam fee separately for each diet rather than publishing fixed values.
Sample IAI CS2 Practice Questions
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1In the Kaplan-Meier (product-limit) estimator of the survival function S(t), what role does a censored observation play at the moment of censoring?
A.It causes the survival estimate to drop by 1/n at the censoring time
B.It is removed from the dataset entirely and ignored in all calculations
C.It reduces the risk set for subsequent event times but does not itself cause a drop in the survival estimate
D.It is treated identically to a death at that time
Explanation: A right-censored observation contributes information up to the censoring time, so it remains in the risk set until censoring and then leaves it. Because no event (death) occurs, the product-limit factor (1 - d_j/n_j) is unchanged at that instant, so S(t) does not drop at a censoring time; it only steps down at observed event times.
2A study follows 10 lives. Deaths occur at t=2 (1 death) and t=5 (1 death), with one life censored at t=3. Using Kaplan-Meier, what is the estimated survival probability S(5)?
A.0.9000
B.0.8000
C.0.7000
D.0.7875
Explanation: At t=2 the risk set is 10 with 1 death, giving factor 9/10 = 0.9. The censoring at t=3 lowers the risk set but causes no drop. At t=5 the risk set is 10 - 1 death - 1 censored = 8 with 1 death, giving factor 7/8 = 0.875. Thus S(5) = 0.9 x 0.875 = 0.7875.
3The Nelson-Aalen estimator estimates which quantity directly?
A.The survival function S(t)
B.The cumulative hazard function H(t)
C.The probability density function f(t)
D.The force of mortality at a single point only
Explanation: The Nelson-Aalen estimator is a step function that estimates the cumulative (integrated) hazard H(t) = sum over event times of d_j/n_j. The survival function can then be recovered via S(t) = exp(-H(t)), but the Nelson-Aalen estimator itself targets the cumulative hazard.
4In the Cox proportional hazards model, the hazard is written lambda(t; z) = lambda_0(t) exp(beta^T z). What does the term lambda_0(t) represent?
A.The baseline hazard, common to all individuals when all covariates are zero
B.The cumulative hazard for the highest-risk individual
C.A covariate-specific multiplier that varies by individual
D.The probability of survival to time t
Explanation: lambda_0(t) is the baseline hazard, the hazard for an individual whose covariates are all zero. The exponential factor exp(beta^T z) scales this baseline up or down according to each individual's covariates, which is why the model is 'proportional' hazards.
5A key advantage of the Cox proportional hazards model over fully parametric survival models is that it:
A.Requires the baseline hazard to be exponential
B.Cannot handle censored data
C.Estimates the regression coefficients without specifying the form of the baseline hazard
D.Always produces a constant hazard over time
Explanation: The Cox model is semi-parametric: it estimates the covariate coefficients beta via partial likelihood without assuming any functional form for lambda_0(t). This flexibility is its main appeal, since the proportional-hazards structure is preserved regardless of the baseline shape.
6In the two-state (alive-dead) Markov model of mortality, the maximum likelihood estimate of the constant force of mortality mu is given by:
A.Number of deaths divided by the initial number of lives
B.Total waiting time divided by number of deaths
C.The square root of the number of deaths
D.Number of deaths divided by total waiting time (central exposed to risk)
Explanation: Under the two-state model with constant force mu, deaths follow a Poisson-type likelihood and the MLE is mu-hat = D / E^c_x, where D is the observed number of deaths and E^c_x is the central exposed to risk (total time lives are observed alive). This is the classic occurrence-exposure rate.
7The central exposed to risk E^c_x differs from the initial exposed to risk E_x principally because:
A.Central exposed to risk ignores all deaths during the year
B.Initial exposed to risk is always larger by exactly the number of new entrants
C.Central exposed to risk counts time observed, ending at death/exit, while initial exposed to risk extends a deceased life's contribution to the end of the rate interval
D.They are identical and the distinction is purely notational
Explanation: Central exposed to risk measures the actual person-time lives spend under observation, terminating at death. Initial exposed to risk adds, for each death, the remaining fraction of the rate interval, since under the q_x convention a life is assumed exposed for the whole year of age in which it dies. Hence E_x = E^c_x + (deaths' remaining time).
8Graduation of crude mortality rates is performed primarily to:
A.Increase the number of deaths in each age group
B.Remove all censored observations from the dataset
C.Produce a smooth set of rates that progress regularly with age while remaining adherent to the data
D.Convert forces of mortality into survival probabilities
Explanation: Graduation smooths crude (raw) rates so that mortality progresses smoothly by age, reflecting the belief that the underlying rates vary regularly. The graduated rates must balance smoothness against adherence (goodness of fit) to the observed crude rates.
9Which statistical test is commonly used after graduation to check whether the graduated rates adhere to the crude data overall by examining standardised deviations?
A.The chi-squared test
B.The Kaplan-Meier log-rank test
C.The Durbin-Watson test
D.The Dickey-Fuller test
Explanation: The chi-squared test sums the squared standardised deviations z_x = (actual - expected)/sqrt(expected) across age groups and compares the total to a chi-squared distribution. It is a primary overall goodness-of-fit test for a graduation, though it must be supplemented by tests for runs, bias and individual large deviations.
10The 'signs test' applied to graduation residuals primarily detects:
A.Overall bias (an excess of positive or negative deviations)
B.The presence of a unit root
C.Heteroscedasticity in the deviations
D.The exact magnitude of the largest deviation
Explanation: The signs test counts the number of positive deviations and compares it to a Binomial(n, 0.5) distribution. A significant imbalance indicates overall bias in the graduation, meaning the graduated rates are systematically above or below the crude rates.
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