Free SOA Exam FAM Parametric Estimation Practice Questions (2026)

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} p θ

n \to \infty

.

2. **Asymptotic normality**:

n (θ^- θ) d N (0, 1/ I (θ))

where

I (θ)

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^

is the MLE of

θ

, then

g (θ^)

is the MLE of

g (θ)

.

Important caveats:
For example, the MLE of

σ^{2}

in the normal model divides by

n

, 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

Solution

D is correct.

For right-censored exponential data, the MLE of θ is:

θ^= \frac{total exposure}{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) = \frac{α θ ^{α}}{( x + θ ) ^{α + 1}}

for

x > 0

. Both

α

and

θ

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) = n ln α + n α ln θ - (α + 1) \sum ln (x_{i} + θ)

. Taking partial derivatives:

\frac{\partial ℓ}{\partial α} = \frac{n}{α} + n ln θ - \sum ln (x_{i} + θ) = 0

\frac{\partial ℓ}{\partial θ} = \frac{n α}{θ} - (α + 1) \sum \frac{1}{x _{i} + θ} = 0

. With

n = 3

and the given data: Equation 1:

\frac{3}{α} + 3 ln θ - [ln (100 + θ) + ln (400 + θ) + ln (500 + θ)] = 0

. Equation 2:

\frac{3 α}{θ} - (α + 1) [\frac{1}{100 + θ} + \frac{1}{400 + θ} + \frac{1}{500 + θ}] = 0

. This system must be solved simultaneously using numerical methods (e.g., Newton-Raphson). From Equation 1, we can express

α

as a function of

θ

and substitute into Equation 2. From Equation 1:

α = \frac{3}{\sum l n ( x _{i} + θ ) - 3 l n θ}

(profile likelihood approach).

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