Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Introduction to Credibility Practice Questions

Credibility theory on SOA Exam FAM introduces limited fluctuation credibility, greatest accuracy credibility (Bayesian), and the Buhlmann credibility model for blending individual experience with class data.

40 Questions
27 Easy
7 Medium
6 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
If the full credibility standard for frequency is 1,083, what is the full credibility standard for aggregate losses when severity has a coefficient of variation of 1.5?
Solution
D is correct.

n0agg=n0freq×(1+CV2)=1,083×(1+1.52)=1,083×(1+2.25)=1,083×3.25=3,519.75≈3,520n_0^{\text{agg}} = n_0^{\text{freq}} \times (1 + \text{CV}^2) = 1{,}083 \times (1 + 1.5^2) = 1{,}083 \times (1 + 2.25) = 1{,}083 \times 3.25 = 3{,}519.75 \approx 3{,}520.
Question 2 Medium
Which of the following is NOT a property of the credibility factor ZZ in limited fluctuation credibility?
Solution
E is correct.

In limited fluctuation credibility, ZZ depends only on the sample size relative to the full credibility standard n0n_0. It does NOT depend on the prior distribution. The prior distribution is relevant in Buhlmann (greatest accuracy) credibility, not limited fluctuation.
Question 3 Hard
An insurer classifies risks into two equally likely classes:

- **Class 1:** Claims follow a Poisson distribution with mean 3
- **Class 2:** Claims follow a Poisson distribution with mean 7

A risk is selected at random and observed for 4 periods, producing a sample mean of 6 claims per period.

Using Bühlmann credibility, calculate the expected number of claims for this risk in the next period.
Solution
C is correct.

For Bühlmann credibility, we need the process variance vv, the variance of hypothetical means aa, and k=v/ak = v/a.

Since each class is equally likely:
v=E[Var(X∣Θ)]=12(3)+12(7)=5v = E[\text{Var}(X|\Theta)] = \frac{1}{2}(3) + \frac{1}{2}(7) = 5
(For Poisson, variance = mean.)

μ=E[E(X∣Θ)]=12(3)+12(7)=5\mu = E[E(X|\Theta)] = \frac{1}{2}(3) + \frac{1}{2}(7) = 5

a=Var[E(X∣Θ)]=12(3−5)2+12(7−5)2=12(4)+12(4)=4a = \text{Var}[E(X|\Theta)] = \frac{1}{2}(3-5)^2 + \frac{1}{2}(7-5)^2 = \frac{1}{2}(4) + \frac{1}{2}(4) = 4

k=va=54=1.25k = \frac{v}{a} = \frac{5}{4} = 1.25

With n=4n = 4 observations:
Z=nn+k=44+1.25=45.25=1621=0.7619Z = \frac{n}{n+k} = \frac{4}{4+1.25} = \frac{4}{5.25} = \frac{16}{21} = 0.7619

Bühlmann estimate:
μ^=ZXˉ+(1−Z)μ=0.7619×6+0.2381×5=4.5714+1.1905=5.7619\hat{\mu} = Z \bar{X} + (1-Z)\mu = 0.7619 \times 6 + 0.2381 \times 5 = 4.5714 + 1.1905 = 5.7619

Rounded: 5.76.
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