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

Parametric estimation on SOA Exam FAM covers maximum likelihood estimation (MLE), method of moments, and goodness-of-fit testing for insurance loss distributions. These techniques are foundational to actuarial modeling.

48 Questions
24 Easy
19 Medium
5 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
Which of the following is a property of maximum likelihood estimators for large samples?
Solution
D is correct.

Key asymptotic properties of MLEs:

1. **Consistency**: θ^MLE→pθ\hat{\theta}_{\text{MLE}} \xrightarrow{p} \theta as n→∞n \to \infty.

2. **Asymptotic normality**: n(θ^−θ)→dN(0,1/I(θ))\sqrt{n}(\hat{\theta} - \theta) \xrightarrow{d} N(0, 1/I(\theta)) where I(θ)I(\theta) is the Fisher information.

3. **Asymptotic efficiency**: The MLE achieves the Cramér-Rao lower bound asymptotically, meaning it has the smallest asymptotic variance among consistent estimators.

4. **Invariance**: If θ^\hat{\theta} is the MLE of θ\theta, then g(θ^)g(\hat{\theta}) is the MLE of g(θ)g(\theta).

Important caveats:
For example, the MLE of σ2\sigma^2 in the normal model divides by nn, not n-1.
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
C is correct.

For right-censored exponential data, the MLE of θ is: θ^=total exposurenumber uncensored=2,400+2(500)8=3,4008=425\hat{θ} = \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.
Question 3 Hard
Loss amounts follow a Pareto distribution with pdf f(x)=αθα(x+θ)α+1f(x) = \frac{\alpha\theta^\alpha}{(x+\theta)^{\alpha+1}} for x>0x > 0. Both α\alpha and θ\theta are unknown. A sample of 3 observations is: 100, 400, 500. Write the system of equations that the MLEs must satisfy.
Solution
E is correct.

The log-likelihood is: ℓ(α,θ)=nln⁡α+nαln⁡θ−(α+1)∑ln⁡(xi+θ)\ell(\alpha, \theta) = n\ln\alpha + n\alpha\ln\theta - (\alpha+1)\sum\ln(x_i + \theta). Taking partial derivatives: ∂ℓ∂α=nα+nln⁡θ−∑ln⁡(xi+θ)=0\frac{\partial\ell}{\partial\alpha} = \frac{n}{\alpha} + n\ln\theta - \sum\ln(x_i + \theta) = 0. ∂ℓ∂θ=nαθ−(α+1)∑1xi+θ=0\frac{\partial\ell}{\partial\theta} = \frac{n\alpha}{\theta} - (\alpha+1)\sum\frac{1}{x_i + \theta} = 0. With n=3n = 3 and the given data: Equation 1: 3α+3ln⁡θ−[ln⁡(100+θ)+ln⁡(400+θ)+ln⁡(500+θ)]=0\frac{3}{\alpha} + 3\ln\theta - [\ln(100+\theta) + \ln(400+\theta) + \ln(500+\theta)] = 0. Equation 2: 3αθ−(α+1)[1100+θ+1400+θ+1500+θ]=0\frac{3\alpha}{\theta} - (\alpha+1)\left[\frac{1}{100+\theta} + \frac{1}{400+\theta} + \frac{1}{500+\theta}\right] = 0. This system must be solved simultaneously using numerical methods (e.g., Newton-Raphson). From Equation 1, we can express α\alpha as a function of θ\theta and substitute into Equation 2. From Equation 1: α=3∑ln⁡(xi+θ)−3ln⁡θ\alpha = \frac{3}{\sum\ln(x_i+\theta) - 3\ln\theta} (profile likelihood approach).
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