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

$n_0^{\text{agg}} = n_0^{\text{freq}} \times (1 + \text{CV}^2) = 1{,}083 \times (1 + 1.5^2) = 1{,}083 \times (1 + 2.25) = 1{,}083 \times 3.25 = 3{,}519.75 \approx 3{,}520$.

(A) is $1.5 \times n_0$. (B) is $2 \times n_0$ (CV = 1). (C) is $2.5 \times n_0$. (D) is $5 \times n_0$ (CV = 2).

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.

(A) True: $Z$ is always between 0 and 1. (B) True: $Z = \sqrt{n/n_0}$ increases with $n$. (C) True: $Z = 1$ when $n \ge n_0$. (E) True by definition.

B 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:
$$v = E[\text{Var}(X|\Theta)] = \frac{1}{2}(3) + \frac{1}{2}(7) = 5$$
(For Poisson, variance = mean.)

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

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

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

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:
$$\hat{\mu} = Z \bar{X} + (1-Z)\mu = 0.7619 \times 6 + 0.2381 \times 5 = 4.5714 + 1.1905 = 5.7619$$

Rounded: 5.76.
Choice A is incorrect because 5.00 equals the overall mean $\mu$ and assigns zero credibility to the data.
Choice E is incorrect because 6.00 equals the sample mean $\bar{X}$ and assigns full credibility to the data.
Choice C is incorrect because 5.50 is the simple average of $\mu$ and $\bar{X}$, which implies $Z = 0.50$.
Choice D is incorrect because 6.24 results from computing $k = a/v = 0.80$ (inverting the ratio), giving $Z = 4/4.80 = 0.833$.