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

Severity, frequency, and aggregate loss models on SOA Exam FAM test parametric distributions (exponential, Pareto, lognormal), compound Poisson models, and the Panjer recursive method for computing aggregate loss probabilities.

152 questions 77 easy 53 medium 22 hard 2026 syllabus

Sample Questions

Question 1 Easy
Which of the following is a property of a coherent risk measure?
Solution
A is correct.

Subadditivity is one of the four axioms of coherent risk measures (monotonicity, positive homogeneity, translation invariance, subadditivity).
Question 2 Medium
Which of the following statements about TVaR is FALSE?
Solution
B is correct.

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, the statement that TVaR at level p=0p = 0 equals VaR at level p=0p = 0 is false.
Question 3 Hard
The number of claims NN follows a zero-truncated Poisson distribution with λ=1\lambda = 1. Calculate E[N]E[N] and Var(N)\text{Var}(N).

You are given: e−1=0.36788e^{-1} = 0.36788.
Solution
D is correct.

For a zero-truncated Poisson with parameter λ=1\lambda = 1: E[NT]=λ1−e−λ=11−0.36788=10.63212=1.582E[N^T] = \frac{\lambda}{1 - e^{-\lambda}} = \frac{1}{1 - 0.36788} = \frac{1}{0.63212} = 1.582. Var(NT)=E[NT](1+λ−E[NT])=1.582(1+1−1.582)=1.582×0.418=0.661\text{Var}(N^T) = E[N^T](1 + \lambda - E[N^T]) = 1.582(1 + 1 - 1.582) = 1.582 \times 0.418 = 0.661. Alternatively: Var(NT)=λ(1+λ)1−e−λ−(λ1−e−λ)2=20.63212−(1.582)2=3.164−2.503=0.661\text{Var}(N^T) = \frac{\lambda(1 + \lambda)}{1 - e^{-\lambda}} - \left(\frac{\lambda}{1 - e^{-\lambda}}\right)^2 = \frac{2}{0.63212} - (1.582)^2 = 3.164 - 2.503 = 0.661. The answer is E[NT]=1.582E[N^T] = 1.582, Var(NT)=0.661\text{Var}(N^T) = 0.661.
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