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 Questions
44 Easy
63 Medium
38 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(xθ)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
The Greatest Accuracy Credibility (Buhlmann) estimate minimizes the mean squared error E[(ZXˉ+(1Z)μμ(Θ))2]E[(Z\bar{X}+(1-Z)\mu-\mu(\Theta))^2] over all linear estimates of the form c0+c1Xˉc_0 + c_1\bar{X}. Derive the optimal ZZ and explain why it equals n/(n+k)n/(n+k).
Solution
E is correct.

For a fixed risk Θ\Theta, the estimator μ^=ZXˉ+(1Z)μ\hat{\mu} = Z\bar{X}+(1-Z)\mu has MSE
E[(ZXˉ+(1Z)μμ(Θ))2].E\left[(Z\bar{X}+(1-Z)\mu-\mu(\Theta))^2\right].
Conditioning on Θ\Theta, E[XˉΘ]=μ(Θ)E[\bar{X}|\Theta] = \mu(\Theta) and Var(XˉΘ)=σ2(Θ)/n\text{Var}(\bar{X}|\Theta) = \sigma^2(\Theta)/n. The overall MSE decomposes as
MSE(Z)=Z2vn+(1Z)2a\text{MSE}(Z) = Z^2 \cdot \frac{v}{n} + (1-Z)^2 \cdot a
where v=E[σ2(Θ)]v = E[\sigma^2(\Theta)] and a=Var[μ(Θ)]a = \text{Var}[\mu(\Theta)]. Differentiating:
ddZMSE=2Zvn2(1Z)a=0    Zvn=(1Z)a\frac{d}{dZ}\text{MSE} = 2Z\frac{v}{n} - 2(1-Z)a = 0 \implies Z\frac{v}{n} = (1-Z)a
Z(vn+a)=a    Z=aa+v/n=nana+v=nn+v/a=nn+k.Z\left(\frac{v}{n}+a\right) = a \implies Z^* = \frac{a}{a+v/n} = \frac{na}{na+v} = \frac{n}{n+v/a} = \frac{n}{n+k}.

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