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.

146 questions 44 easy 63 medium 39 hard 2026 syllabus

Sample Questions

Question 1 Easy
In the context of Bayesian credibility, the prior distribution π(θ)\pi(\theta) represents:
Solution
B is correct.

In the Bayesian credibility (greatest accuracy) framework, the prior π(θ)\pi(\theta) represents the distribution of the risk parameter Θ\Theta 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\Theta 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?
Solution
C is correct.

In the semiparametric empirical Bayes approach, the conditional distribution of losses given the risk parameter, f(x∣θ)f(x \mid \theta), is assumed to belong to a parametric family (e.g., Poisson, normal). This allows the structural parameters v=E[σ2(Θ)]v = E[\sigma^2(\Theta)] and a=Var[μ(Θ)]a = \text{Var}[\mu(\Theta)] to be estimated using the known functional form of σ2(θ)\sigma^2(\theta) and μ(θ)\mu(\theta). The prior π(θ)\pi(\theta) 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 vv and aa from the data, assuming each risk has a Poisson process with its own unknown rate.
Solution
B is correct.

With r=5r = 5 risks each observed for n=1n = 1 year, the overall mean is Xˉ=(3+7+2+5+8)/5=25/5=5.0\bar{X} = (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\hat{v} = \bar{X} = 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\hat{a} = S^2 - \hat{v}/n = S^2 - 5.0
where S2=1r−1∑(Xi−Xˉ)2=(3−5)2+(7−5)2+(2−5)2+(5−5)2+(8−5)24=4+4+9+0+94=264=6.5S^2 = \frac{1}{r-1}\sum(X_i - \bar{X})^2 = \frac{(3-5)^2+(7-5)^2+(2-5)^2+(5-5)^2+(8-5)^2}{4} = \frac{4+4+9+0+9}{4} = \frac{26}{4} = 6.5.

With n=1n=1 per risk: a^=S2−v^/1=6.5−5.0=1.5\hat{a} = S^2 - \hat{v}/1 = 6.5 - 5.0 = 1.5. k^=5.0/1.5≈3.33\hat{k} = 5.0/1.5 \approx 3.33.

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