Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Coverage Modifications Practice Questions

Coverage modifications on SOA Exam ASTAM cover the mathematical treatment of ordinary and franchise deductibles, policy limits, coinsurance provisions, and the effect of inflation on loss distributions.

117 questions 43 easy 49 medium 25 hard 2026 syllabus

Sample Questions

Question 1 Easy
Losses are uniformly distributed on [0, 5000][0,\, 5000]. A policy limit of 3000 applies with no deductible. What is the expected payment per loss?
Solution
B is correct.

With a policy limit u=3000u = 3000 and losses X∼U[0,5000]X \sim U[0, 5000], the expected payment per loss is E[X∧3000]=∫03000S(x) dxE[X \wedge 3000] = \int_0^{3000} S(x)\, dx where S(x)=(5000−x)/5000S(x) = (5000 - x)/5000. Evaluating: (1/5000)[5000x−x2/2]03000=(15,000,000−4,500,000)/5000=10,500,000/5000=2100(1/5000)[5000x - x^2/2]_0^{3000} = (15{,}000{,}000 - 4{,}500{,}000)/5000 = 10{,}500{,}000/5000 = 2100.
Question 2 Medium
Losses follow an exponential distribution with mean 1,000. A policy has an ordinary deductible of 500 and a maximum covered loss of 3,000. Calculate the expected payment per loss.
Solution
C is correct.

The expected payment per loss under an ordinary deductible dd and maximum covered loss uu is:
E[YL]=E[min⁡(X,u)]−E[min⁡(X,d)]E[Y^L] = E[\min(X, u)] - E[\min(X, d)]
For an exponential distribution with mean θ=1000\theta = 1000, the limited expected value is E[min⁡(X,c)]=θ(1−e−c/θ)E[\min(X, c)] = \theta(1 - e^{-c/\theta}).
E[min⁡(X,3000)]=1000(1−e−3)≈1000(0.9502)=950.2E[\min(X, 3000)] = 1000(1 - e^{-3}) \approx 1000(0.9502) = 950.2
E[min⁡(X,500)]=1000(1−e−0.5)≈1000(0.3935)=393.5E[\min(X, 500)] = 1000(1 - e^{-0.5}) \approx 1000(0.3935) = 393.5
E[YL]≈950.2−393.5=556.7≈558E[Y^L] \approx 950.2 - 393.5 = 556.7 \approx 558
Question 3 Hard
Losses follow an exponential distribution with mean θ=2000\theta = 2000. A policy has an ordinary deductible of d=500d = 500 and a policy limit of u=3000u = 3000. Compute the variance of the payment per loss.
Solution
E is correct.

The payment per loss is W=(X−500)+∧3000=min⁡(max⁡(X−500,0),3000)W = (X-500)_+ \wedge 3000 = \min(\max(X-500,0), 3000). To find Var(W)=E[W2]−(E[W])2\text{Var}(W) = E[W^2] - (E[W])^2, compute: E[W]=E[X∧3500]−E[X∧500]E[W] = E[X \wedge 3500] - E[X \wedge 500]. E[X∧c]=θ(1−e−c/θ)=2000(1−e−c/2000).E[X \wedge c] = \theta(1-e^{-c/\theta}) = 2000(1-e^{-c/2000}). E[X∧3500]=2000(1−e−1.75)≈2000×0.8262=1652.4.E[X \wedge 3500] = 2000(1-e^{-1.75}) \approx 2000 \times 0.8262 = 1652.4. E[X∧500]=2000(1−e−0.25)≈2000×0.2212=442.3.E[X \wedge 500] = 2000(1-e^{-0.25}) \approx 2000 \times 0.2212 = 442.3. E[W]≈1652.4−442.3=1210.1.E[W] \approx 1652.4 - 442.3 = 1210.1.
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