Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Severity Models Practice Questions

Severity models on SOA Exam ASTAM cover parametric loss distributions (Pareto, lognormal, Weibull), tail behavior analysis, mixture distributions, and transformations used to fit insurance claim data.

120 Questions
39 Easy
50 Medium
31 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
A loss XX follows a lognormal distribution with parameters μ=7\mu = 7 and σ=1.2\sigma = 1.2. What is E[X]E[X]?
Solution
D is correct.

If ln⁡X∼N(μ,σ2)\ln X \sim N(\mu, \sigma^2), then the mean of XX is:
E[X]=eμ+σ2/2.E[X] = e^{\mu + \sigma^2/2}.
With μ=7\mu = 7 and σ=1.2\sigma = 1.2:
E[X]=e7+(1.2)2/2=e7+1.44/2=e7+0.72=e7.72≈2249.4.E[X] = e^{7 + (1.2)^2/2} = e^{7 + 1.44/2} = e^{7 + 0.72} = e^{7.72} \approx 2249.4.
Question 2 Medium
Which of the following correctly describes the relationship between a distribution's hazard rate h(x)h(x) and its tail classification?
Solution
B is correct.

Tail classification by hazard rate behavior:
- Increasing hazard rate (IHR): lighter than exponential — tail decays faster than e−λxe^{-\lambda x} for some λ\lambda.
- Constant hazard rate: exponential distribution — light-tailed, with MGF M(t)=(1−θt)−1M(t) = (1 - \theta t)^{-1} defined for t<1/θt < 1/\theta.
- Decreasing hazard rate (DHR): heavier than exponential — survival function decays slower than any pure exponential.

The exponential sits at the boundary between IHR (lighter) and DHR (heavier) distributions. It is classified as light-tailed because its MGF exists in a neighborhood of zero.
Question 3 Hard
Loss severity XX follows a lognormal distribution with μ=6\mu = 6 and σ=1.5\sigma = 1.5. A policy imposes an ordinary deductible of d=500d = 500. What is the expected per-payment amount (mean excess loss at d=500d = 500)?
Solution
A is correct.

For the lognormal with parameters μ\mu and σ\sigma, define z=(ln⁡d−μ)/σz = (\ln d - \mu)/\sigma. The mean excess loss at dd is: e(d)=[E[X]⋅Φ(σ−z)]/[1−Φ(z)]−de(d) = [E[X] \cdot \Phi(\sigma - z)] / [1 - \Phi(z)] - d, where E[X]=eμ+σ2/2E[X] = e^{\mu + \sigma^2/2}. With μ=6\mu = 6, σ=1.5\sigma = 1.5, d=500d = 500: E[X]=e6+1.125=e7.125≈1248.8E[X] = e^{6 + 1.125} = e^{7.125} \approx 1248.8. z=(ln⁡500−6)/1.5=(6.2146−6)/1.5=0.2146/1.5≈0.143z = (\ln 500 - 6)/1.5 = (6.2146 - 6)/1.5 = 0.2146/1.5 \approx 0.143. Φ(0.143)≈0.557\Phi(0.143) \approx 0.557, 1−Φ(0.143)≈0.4431 - \Phi(0.143) \approx 0.443. Φ(1.5−0.143)=Φ(1.357)≈0.913\Phi(1.5 - 0.143) = \Phi(1.357) \approx 0.913. e(500)=(1248.8×0.913)/0.443−500≈1139.8/0.443−500≈2572−500=2072e(500) = (1248.8 \times 0.913)/0.443 - 500 \approx 1139.8/0.443 - 500 \approx 2572 - 500 = 2072.
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