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 22 easy 21 medium 5 hard 2026 syllabus

Sample Questions

Question 1 Easy
Five observations from a Uniform(0, θ\theta) distribution are: 2.1, 3.5, 1.8, 4.2, 2.9. What is the MLE of θ\theta?
Solution
A is correct.

For the Uniform(0, θ\theta) distribution, the MLE of θ\theta is the maximum order statistic: θ^=X(n)=max⁡(2.1,3.5,1.8,4.2,2.9)=4.2\hat{\theta} = X_{(n)} = \max(2.1, 3.5, 1.8, 4.2, 2.9) = 4.2.
Question 2 Medium
A complete sample of 25 exponential losses has total 10,000. The model has unknown mean θ\theta. Using maximum likelihood, estimate the variance of a loss from this model.
Solution
D is correct.

The exponential log-likelihood is −25ln⁡θ−10,000/θ-25\ln\theta-10{,}000/\theta, maximized at θ^=10,000/25=400\hat\theta=10{,}000/25=400. The model variance is θ2\theta^2. By invariance, its maximum likelihood estimate is θ^2=160,000\hat\theta^2=160{,}000. This is the variance of a loss, not the variance of an estimator.
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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