SOA Advanced Short-Term Actuarial Mathematics Examination Practice Test
ASTAM practice focuses on short-term actuarial loss models and the mathematical behavior of claim frequency, claim severity, and aggregate losses. The available question bank uses probability distributions, independence, conditional models, moments, and normal approximations to analyze insurance claims and portfolio risk.
Exam overview
About the SOA Advanced Short-Term Actuarial Mathematics Examination Exam
Problems require translating insurance scenarios into probability models and selecting calculations that answer specific risk questions. Skills include combining independent Poisson counts, applying Poisson thinning to classified claims, and using binomial or hypergeometric models for claim categories. Aggregate-loss calculations connect individual claim amounts and claim counts to expected loss and variance; other tasks interpret negative binomial parameters or apply a normal approximation to estimate a tail probability. Premium-principle questions test how a loading based on aggregate-loss variability changes the indicated premium. These methods support practical actuarial analysis of claim frequency, severity, portfolio totals, and risk margins. Practice with the underlying assumptions—such as independence, fixed claim amounts, and conditional classification—is important because a small change in the model can change the appropriate distribution or calculation.
Sample Practice Questions
Review a selection from this exam's published US Practice Tests practice bank.
Claim counts in two consecutive, non-overlapping six-month periods are independent Poisson random variables, each with mean 3. What is the probability of exactly 2 claims over the full year?
Show answer
Correct answer: CThe sum of independent Poisson counts is Poisson with mean 3+3=6. Thus P(N=2)=e⁻⁶6²/2!=(18)e⁻⁶. The correct value is 18e⁻⁶; option B is correct.
A portfolio contains 100 independent risks. Each risk has probability 0.02 of a claim, and each claim has a fixed amount of 500. What is the variance of aggregate loss?
Show answer
Correct answer: CAggregate loss is 500N, where N is Binomial(100,0.02). Its variance is 500² Var(N)=250,000×100×0.02×0.98=490,000,000. None of the listed options matches; this illustrates why the options should be checked carefully.
A claim count N is Poisson with mean 4. Conditional on N, each claim independently has probability 0.25 of being assigned to a particular coverage. What is the distribution of the number assigned to that coverage?
Show answer
Correct answer: AIndependent thinning of a Poisson count with retention probability 0.25 produces a Poisson count with mean 4×0.25=1.
Individual claim amounts have mean 500 and standard deviation 300. For 100 independent claims, what are the mean and standard deviation of aggregate claims?
Show answer
Correct answer: AThe aggregate mean is 100×500=50,000. Independence gives aggregate variance 100×300², so the standard deviation is √100×300=3,000.
An insurer uses a premium principle that adds a loading proportional to the standard deviation of aggregate loss S. If the loading factor is θ>0, which formula gives the premium under this principle?
Show answer
Correct answer: AThe standard-deviation premium principle adds θ times the standard deviation to the expected loss. Since the standard deviation is √Var(S), the premium is E[S]+θ√Var(S).
A portfolio has a fixed number of claims, each independently classified as large with probability 0.15. Given that 4 of 20 claims are large, what is the conditional distribution of the number of large claims among the first 8 claims?
Show answer
Correct answer: BConditioning on exactly 4 large claims among 20 fixes the total number of successes. The locations of those successes are uniformly distributed, so the count in a subset of 8 has a hypergeometric distribution with population size 20, 4 successes, and sample size 8.
A policyholder's annual claim count N has a Poisson distribution with mean 2. Each claim is independently classified as either type A with probability 0.3 or type B otherwise. What is the distribution of the type A claim count?
Show answer
Correct answer: AThinning a Poisson count by independently retaining each event with probability p produces a Poisson count with mean λp. Here the mean is 2×0.3=0.6.
How to use this practice test
The free US Practice Tests bank contains 50 ASTAM practice questions. Each session presents up to 20 questions selected at random and shows results after completion. Use a session to identify which models or calculations need more work, then revisit missed problems by writing down the assumptions and distribution used.
Preparation tips
- For aggregate-loss problems, distinguish fixed claim amounts from random severities and calculate the mean and variance using the stated independence assumptions.
- Practice Poisson addition and thinning, and verify how the resulting frequency distribution changes when claims are assigned independently to categories.
- Compare binomial and hypergeometric models: use the problem's sampling or conditioning setup to determine whether claim classifications are independent or made without replacement.
- For normal approximations, standardize the aggregate-loss threshold with the given mean and standard deviation before finding a tail probability.
- In premium-principle questions, identify whether the loading is based on expected loss, standard deviation, or another stated risk measure before substituting values.