Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Parametric Estimation Practice Questions

Work through parametric estimation problems for Exam FAM. Questions cover maximum likelihood estimation, method of moments, and goodness-of-fit testing for loss distributions.

54 Questions
27 Easy
19 Medium
8 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
An estimator θ^\hat{\theta} is consistent if, as n→∞n \to \infty:
Solution
Consistency means convergence in probability: θ^→pθ\hat{\theta} \xrightarrow{p} \theta as n→∞n \to \infty. This requires both that the bias vanishes and the variance vanishes, but the definition is convergence in probability.

(A) describes unbiasedness, not consistency. (B) is necessary but not sufficient by itself. (D) describes efficiency. (E) is unbiasedness, which is a different property.
Question 2 Medium
A sample of 10 losses from an exponential distribution has 8 exact observations summing to 2,400 and 2 right-censored observations at 500 each. Calculate the MLE of θ\theta.
Solution
For right-censored exponential data, the MLE of θ\theta is:
θ^=total exposurenumber uncensored=2,400+2(500)8=3,4008=425\hat{\theta} = \frac{\text{total exposure}}{\text{number uncensored}} = \frac{2{,}400 + 2(500)}{8} = \frac{3{,}400}{8} = 425

Total exposure includes both exact observations and censored exposure times.

(A) is 2,400/82{,}400/8 (ignoring censored data). (B) is 3,400/103{,}400/10. (D) is 2,400/52{,}400/5. (E) is 3,400/6.183{,}400/6.18.
Question 3 Hard
Loss amounts follow a lognormal distribution. A sample of n=20n = 20 yields ∑ln⁡xi=140\sum \ln x_i = 140 and ∑(ln⁡xi)2=1,020\sum (\ln x_i)^2 = 1{,}020. Using the Fisher information for μ\mu, calculate the width of a 95% confidence interval for μ\mu.
Solution
MLEs.
μ^=∑ln⁡xin=14020=7.0\hat{\mu} = \frac{\sum \ln x_i}{n} = \frac{140}{20} = 7.0
σ^2=∑(ln⁡xi)2n−μ^2=1,02020−49=51−49=2.0\hat{\sigma}^2 = \frac{\sum (\ln x_i)^2}{n} - \hat{\mu}^2 = \frac{1{,}020}{20} - 49 = 51 - 49 = 2.0

Fisher information and Cramer-Rao bound.
For Yi=ln⁡Xi∼N(μ,σ2)Y_i = \ln X_i \sim N(\mu, \sigma^2), the Fisher information for μ\mu is n/σ2n/\sigma^2.
Var(μ^)=σ2n=2.020=0.10\text{Var}(\hat{\mu}) = \frac{\sigma^2}{n} = \frac{2.0}{20} = 0.10
SE=0.10=0.31623\text{SE} = \sqrt{0.10} = 0.31623

95% CI width.
Width=2×1.96×0.31623=2×0.61981=1.2396≈1.240\text{Width} = 2 \times 1.96 \times 0.31623 = 2 \times 0.61981 = 1.2396 \approx 1.240

The answer is 1.240.

Why other choices fail:
- Choice A (0.620): The half-width only: 1.96×0.316=0.6201.96 \times 0.316 = 0.620.
- Choice B (1.386): Uses s2=20×2.0/19=2.105s^2 = 20 \times 2.0/19 = 2.105, giving SE = 0.3245, width = 2×1.96×0.3245=1.2722 \times 1.96 \times 0.3245 = 1.272. Not 1.386.
- Choice C (0.876): Intermediate error.
- Choice D (1.568): Inflated variance estimate.
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