Free SOA Exam FAM Introduction to Credibility Practice Questions (2026)

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

C is correct.

n_{0}^{agg} = n_{0}^{freq} \times (1 + 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

Z

in limited fluctuation credibility?

Solution

E is correct.

In limited fluctuation credibility,

Z

depends only on the sample size relative to the full credibility standard

n_{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

D is correct.

For Bühlmann credibility, we need the process variance

v

, the variance of hypothetical means

a

, and

k = v / a

.

Since each class is equally likely:

v = E [Var (X ∣Θ)] = \frac{1}{2} (3) + \frac{1}{2} (7) = 5

(For Poisson, variance = mean.)

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

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

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

With

n = 4

observations:

Z = \frac{n}{n + k} = \frac{4}{4 + 1.25} = \frac{4}{5.25} = \frac{16}{21} = 0.7619

Bühlmann estimate:

μ^= Z X ˉ + (1 - Z) μ = 0.7619 \times 6 + 0.2381 \times 5 = 4.5714 + 1.1905 = 5.7619

Rounded: 5.76.

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