Free SOA Exam FAM Mortality Models Practice Questions (2026)

Sample Questions

Question 1 Easy

In a life table, which of the following relationships is correct?

Solution

B is correct.

In a life table:

l_{x}

= expected number alive at exact age

x

;

d_{x}

= expected number of deaths between ages

x

and x+1. Therefore:

d_{x} = l_{x} - l_{x + 1}

Equivalently:

d_{x} = l_{x} \cdot q_{x}

(number alive times probability of death).

Question 2 Medium

For a constant force of mortality

μ = 0.02

, calculate

e_{x}

, the curtate expectation of life.

Solution

B is correct.

The curtate expectation of life is:

e_{x} = \sum_{k = 1}^{\infty}_{k} p_{x}

With constant force

μ

_{k} p_{x} = e^{- μ k} = (e^{- μ})^{k}

.

Let

p = e^{- μ} = e^{- 0.02} = 0.98020

.

This is a geometric series:

e_{x} = \sum_{k = 1}^{\infty} p^{k} = \frac{p}{1 - p} = \frac{0.98020}{0.01980} = 49.505 \approx 49.5

Note: The complete expectation is

e ˚_{x} = 1/ μ = 50

, while the curtate expectation is slightly less because it only counts complete years survived. Under UDD,

e ˚_{x} \approx e_{x} + 0.5

, consistent with

50 \approx 49.5 + 0.5

Question 3 Hard

For a Gompertz model

μ_{x} = B c^{x}

, you are given:

_{10} p_{50} = 0.920

and

_{10} p_{60} = 0.850

Calculate

c^{10}

Solution

A is correct.

For Gompertz, the survival probability is:

_{10} p_{x} = exp (- \frac{B c ^{x} ( c ^{10} - 1 )}{ln c})

Taking logarithms of each:

ln (_{10} p_{50}) = - \frac{B c ^{50} ( c ^{10} - 1 )}{ln c}

ln (_{10} p_{60}) = - \frac{B c ^{60} ( c ^{10} - 1 )}{ln c}

Dividing:

\frac{ln ( _{10} p _{60} )}{ln ( _{10} p _{50} )} = \frac{c ^{60}}{c ^{50}} = c^{10}

ln (0.920) = - 0.08338

ln (0.850) = - 0.16252

c^{10} = \frac{- 0.16252}{- 0.08338} = 1.9492 \approx 1.949

The answer is 1.949.

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