Free SOA Exam ASTAM Aggregate Models Practice Questions (2026)

Sample Questions

Question 1 Easy

Which of the following distributions belongs to the

(a, b, 0)

class with

a > 0

and

b \geq 0

Solution

D is correct.

The

(a, b, 0)

class consists of counting distributions whose probabilities satisfy

\frac{p _{k}}{p _{k - 1}} = a + \frac{b}{k}, k = 1, 2, 3, \dots

with

p_{0} > 0

.

For the negative binomial with parameters

r > 0

and

β > 0

p_{k} = (k r + k - 1) (\frac{β}{1 + β})^{k} (\frac{1}{1 + β})^{r},

\frac{p _{k}}{p _{k - 1}} = \frac{r + k - 1}{k} \cdot \frac{β}{1 + β} = \frac{β}{1 + β} + \frac{( r - 1 ) β / ( 1 + β )}{k} .

Hence

a = \frac{β}{1 + β} > 0, b = \frac{( r - 1 ) β}{1 + β} .

With

r \geq 1

we get

b \geq 0

, and

b = 0

exactly in the geometric case

r = 1

. So the negative binomial is the

(a, b, 0)

member with

a > 0

and

b \geq 0

.

For comparison, the Poisson gives

a = 0

b = λ

, and the binomial gives

a = - q / (1 - q) < 0

b = (m + 1) q / (1 - q)

; both are in the class but neither has

a > 0

Question 2 Medium

An exponential severity with mean

θ = 500

is discretized using the method of rounding with span

h = 200

. Compute the masses at grid points 0, 200, and 400.

Solution

B is correct.

Under the method of rounding with span

h = 200

, grid point

k \cdot h

collects all probability mass within

[(k - 0.5) h, (k + 0.5) h)

. For

k = 0

: the interval is

(- 100, 100)

, but since support is

[0, \infty)

, this is

[0, 100)

p_{0} = F (100) - F (0) = (1 - e^{- 100/500}) - 0 = 1 - e^{- 0.2} \approx 0.1813

. For

k = 1

(grid point 200): interval

[100, 300)

p_{200} = F (300) - F (100) = e^{- 0.2} - e^{- 0.6} \approx 0.8187 - 0.5488 = 0.2700

. For

k = 2

(grid point 400): interval

[300, 500)

p_{400} = F (500) - F (300) = e^{- 0.6} - e^{- 1.0} \approx 0.5488 - 0.3679 = 0.1809

Question 3 Hard

For a compound Poisson

S

with

λ = 2

and severity

P (X = 1) = P (X = 2) = 0.5

, use the Panjer recursion to compute

g_{0}

g_{1}

, and

g_{2}

, then find

P (S \leq 2)

Solution

E is correct.

For compound Poisson with

a = 0

b = λ = 2

, and

f_{1} = f_{2} = 0.5

g_{0} = e^{- λ} = e^{- 2}

. For

s = 1

g_{1} = \frac{b}{1} f_{1} g_{0} = \frac{2}{1} (0.5) (e^{- 2}) = e^{- 2} .

For

s = 2

g_{2} = \frac{b}{2} f_{1} g_{1} + \frac{2 b}{2} f_{2} g_{0} = \frac{2}{2} (0.5) (e^{- 2}) + \frac{2 \cdot 2}{2} (0.5) (e^{- 2}) = 0.5 e^{- 2} + e^{- 2} = 1.5 e^{- 2} .

Therefore

P (S \leq 2) = e^{- 2} (1 + 1 + 1.5) = 3.5 e^{- 2} \approx 0.4726

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