Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Severity, Frequency, and Aggregate Models Practice Questions

Practice severity, frequency, and aggregate loss models for Exam FAM. Questions test parametric distributions, compound models, and the recursive method for computing aggregate loss probabilities.

148 Questions
72 Easy
52 Medium
24 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
Which of the following is a property of a coherent risk measure?
Solution
Subadditivity is one of the four axioms of coherent risk measures (monotonicity, positive homogeneity, translation invariance, subadditivity).

Choice A is incorrect because VaR is not subadditive in general; there are well-known counterexamples where diversification increases VaR, violating the coherence requirement.
Choice B is incorrect because translation invariance gives ρ(X+c)=ρ(X)+c\rho(X+c) = \rho(X)+c, not ρ(c)=0\rho(c) = 0; the axiom describes how adding a deterministic amount shifts the risk measure.
Choice D is incorrect because coherent risk measures can handle heavy-tailed distributions; in fact, TVaR is specifically valued for its sensitivity to tail behavior.
Choice E is incorrect because TVaR (also called CTE or Expected Shortfall) is a coherent risk measure that satisfies all four axioms, including subadditivity.
Question 2 Medium
Which of the following statements about TVaR is FALSE?
Solution
TVaR0=E[X]\text{TVaR}_0 = E[X], while VaR0\text{VaR}_0 is the minimum possible value (often 0 for loss distributions). Since E[X]E[X] generally does not equal the minimum, (E) is FALSE.

(A) True: TVaR satisfies subadditivity. (B) True: TVaR \ge VaR always. (C) True for continuous distributions. (D) True: TVaR satisfies all four coherence axioms.
Question 3 Hard
Losses follow a Pareto distribution with α=3\alpha = 3 and θ=2,000\theta = 2{,}000. Calculate TVaR0.95(X)\text{TVaR}_{0.95}(X).

You are given: 0.051/3=0.36840.05^{1/3} = 0.3684.
Solution
Find VaR0.95\text{VaR}_{0.95}. Set S(x)=0.05S(x) = 0.05:

(2,000/(2,000+x))3=0.05(2{,}000/(2{,}000+x))^3 = 0.05, so 2,000/(2,000+x)=0.051/3=0.36842{,}000/(2{,}000+x) = 0.05^{1/3} = 0.3684.

x=2,000/0.36842,000=5,4292,000=3,429x = 2{,}000/0.3684 - 2{,}000 = 5{,}429 - 2{,}000 = 3{,}429.

VaR0.95=3,429\text{VaR}_{0.95} = 3{,}429.

For the Pareto distribution, the mean excess loss at dd is e(d)=(θ+d)/(α1)e(d) = (\theta + d)/(\alpha - 1):

e(3,429)=(2,000+3,429)/(31)=5,429/2=2,715e(3{,}429) = (2{,}000 + 3{,}429)/(3-1) = 5{,}429/2 = 2{,}715

TVaR0.95=VaR0.95+e(VaR0.95)=3,429+2,715=6,144\text{TVaR}_{0.95} = \text{VaR}_{0.95} + e(\text{VaR}_{0.95}) = 3{,}429 + 2{,}715 = 6{,}144

Equivalently: TVaRp=αα1VaRp+θα1=32(3,429)+2,0002=5,144+1,000=6,144\text{TVaR}_p = \frac{\alpha}{\alpha-1}\text{VaR}_p + \frac{\theta}{\alpha-1} = \frac{3}{2}(3{,}429) + \frac{2{,}000}{2} = 5{,}144 + 1{,}000 = 6{,}144.

The answer is VaR0.95=3,429\text{VaR}_{0.95} = 3{,}429 and TVaR0.95=6,144\text{TVaR}_{0.95} = 6{,}144.

Why other choices are wrong:
- Choice E (2,715) is the mean excess loss alone, not VaR.
- Choice C (4,429) and Choice D (6,429) use incorrect formulas for the Pareto quantile.
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