Free SOA Exam ASTAM Coverage Modifications Practice Questions (2026)

Sample Questions

Question 1 Easy

Losses are uniformly distributed on

[0, 5000]

. A policy limit of 3000 applies with no deductible. What is the expected payment per loss?

Solution

B is correct.

With a policy limit

u = 3000

and losses

X \sim U [0, 5000]

, the expected payment per loss is

E [X \land 3000] = \int_{0}^{3000} S (x) d x

where

S (x) = (5000 - x) /5000

. Evaluating:

(1/5000) [5000 x - x^{2} /2]_{0}^{3000} = (15, 000, 000 - 4, 500, 000) /5000 = 10, 500, 000/5000 = 2100

Question 2 Medium

Losses follow an exponential distribution with mean 1,000. A policy has an ordinary deductible of 500 and a maximum covered loss of 3,000. Calculate the expected payment per loss.

Solution

C is correct.

The expected payment per loss under an ordinary deductible

d

and maximum covered loss

u

is:

E [Y^{L}] = E [min (X, u)] - E [min (X, d)]

For an exponential distribution with mean

θ = 1000

, the limited expected value is

E [min (X, c)] = θ (1 - e^{- c / θ})

E [min (X, 3000)] = 1000 (1 - e^{- 3}) \approx 1000 (0.9502) = 950.2

E [min (X, 500)] = 1000 (1 - e^{- 0.5}) \approx 1000 (0.3935) = 393.5

E [Y^{L}] \approx 950.2 - 393.5 = 556.7 \approx 558

Question 3 Hard

Losses follow an exponential distribution with mean

θ = 2000

. A policy has an ordinary deductible of

d = 500

and a policy limit of

u = 3000

. Compute the variance of the payment per loss.

Solution

E is correct.

The payment per loss is

W = (X - 500)_{+} \land 3000 = min (max (X - 500, 0), 3000)

. To find

Var (W) = E [W^{2}] - (E [W])^{2}

, compute:

E [W] = E [X \land 3500] - E [X \land 500]

E [X \land c] = θ (1 - e^{- c / θ}) = 2000 (1 - e^{- c /2000}) .

E [X \land 3500] = 2000 (1 - e^{- 1.75}) \approx 2000 \times 0.8262 = 1652.4.

E [X \land 500] = 2000 (1 - e^{- 0.25}) \approx 2000 \times 0.2212 = 442.3.

E [W] \approx 1652.4 - 442.3 = 1210.1.

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