Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Credibility Practice Questions
Credibility theory on SOA Exam ASTAM covers Buhlmann and Buhlmann-Straub models, empirical Bayes estimation, and the mathematical connection between credibility weighting and linear regression.
145 questions44 easy63 medium38 hard2026 syllabus
Sample Questions
Question 1
Easy
In the context of Bayesian credibility, the prior distribution π(θ) represents:
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Correct Answer: B
Solution
B is correct.
In the Bayesian credibility (greatest accuracy) framework, the prior π(θ) represents the distribution of the risk parameter Θ across the heterogeneous population of risks. It captures how different risks are before any data on a specific risk is observed. When we select a risk at random, Θ is drawn from this prior distribution.
Question 2
Medium
The semiparametric empirical Bayes approach to credibility differs from the nonparametric approach in which key way?
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Correct Answer: C
Solution
C is correct.
In the semiparametric empirical Bayes approach, the conditional distribution of losses given the risk parameter, f(x∣θ), is assumed to belong to a parametric family (e.g., Poisson, normal). This allows the structural parameters v=E[σ2(Θ)] and a=Var[μ(Θ)] to be estimated using the known functional form of σ2(θ) and μ(θ). The prior π(θ) remains completely unspecified (nonparametric component).
Question 3
Hard
A portfolio of 5 risks is observed for one year each. The claim counts are 3, 7, 2, 5, and 8. Using the Buhlmann credibility model, estimate the structural parameters v and a from the data, assuming each risk has a Poisson process with its own unknown rate.
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Correct Answer: B
Solution
B is correct.
With r=5 risks each observed for n=1 year, the overall mean is Xˉ=(3+7+2+5+8)/5=25/5=5.0. For Poisson data with one observation per risk, the within-risk process variance estimate is v^=Xˉ=5.0 (using Poisson mean-variance equality). The between-risk variance is estimated by removing sampling noise from total variance: a^=S2−v^/n=S2−5.0 where S2=r−11∑(Xi−Xˉ)2=4(3−5)2+(7−5)2+(2−5)2+(5−5)2+(8−5)2=44+4+9+0+9=426=6.5.
With n=1 per risk: a^=S2−v^/1=6.5−5.0=1.5. k^=5.0/1.5≈3.33.
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