Free SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) Survival Models for Contingent Cash Flows Practice Questions

Survival models on SOA Exam ALTAM cover multi-state Markov models, transition intensities, and Kolmogorov forward and backward equations applied to insurance and pension benefit calculations.

127 questions 65 easy 41 medium 21 hard 2026 syllabus

Sample Questions

Question 1 Easy
The force of mortality μx+t\mu_{x+t} and the survival function tpx{}_tp_x are related by which of the following identities?
Solution
C is correct.

The fundamental relationship between the force of mortality and the survival function follows from the definition of the force of mortality as:
μx+t=−ddtln⁡tpx\mu_{x+t} = -\frac{d}{dt}\ln {}_tp_x
Integrating from 0 to tt and using 0px=1{}_0p_x = 1:
tpx=exp⁡ ⁣(−∫0tμx+s ds){}_tp_x = \exp\!\left(-\int_0^t \mu_{x+s}\,ds\right)
This is the general formula valid for any non-negative integrable force of mortality.
Question 2 Medium
For a Weibull survival model with survival function S(t)=e−λtγS(t) = e^{-\lambda t^{\gamma}}, where λ=0.002\lambda = 0.002 and γ=2\gamma = 2, derive the force of mortality μ(t)\mu(t) and evaluate it at t=10t = 10.
Solution
C is correct.

The force of mortality is μ(t)=−ddtln⁡S(t)\mu(t) = -\frac{d}{dt}\ln S(t). With S(t)=e−λtγS(t) = e^{-\lambda t^{\gamma}}, we get ln⁡S(t)=−λtγ\ln S(t) = -\lambda t^{\gamma}, so μ(t)=λγtγ−1\mu(t) = \lambda \gamma t^{\gamma-1}. With λ=0.002\lambda = 0.002 and γ=2\gamma = 2: μ(t)=0.002×2×t=0.004t\mu(t) = 0.002 \times 2 \times t = 0.004t. At t=10t = 10: μ(10)=0.004×10=0.04\mu(10) = 0.004 \times 10 = 0.04.
Question 3 Hard
For a Makeham mortality law with μx=A+Bcx\mu_x = A + Bc^x, the 10-year survival probability for a life aged 30 is given by 10p30=exp⁡ ⁣(−10A−Bc30(c10−1)ln⁡c){}_{10}p_{30} = \exp\!\left(-10A - \frac{Bc^{30}(c^{10}-1)}{\ln c}\right). With A=0.0005A = 0.0005, B=0.00005B = 0.00005, c=1.10c = 1.10, compute 10p30{}_{10}p_{30}.
Solution
D is correct.

The exact 10-year survival probability under Makeham's law is:

10p30=exp⁡ ⁣(−10A−Bc30(c10−1)ln⁡c){}_{10}p_{30} = \exp\!\left(-10A - \frac{Bc^{30}(c^{10}-1)}{\ln c}\right)

Computing each component:

10A=10×0.0005=0.00510A = 10 \times 0.0005 = 0.005

c10=1.110=2.593742c^{10} = 1.1^{10} = 2.593742, so c10−1=1.593742c^{10} - 1 = 1.593742

c30=1.130=17.449402c^{30} = 1.1^{30} = 17.449402

ln⁡c=ln⁡1.1=0.095310\ln c = \ln 1.1 = 0.095310

Gompertz term: 0.00005×17.449402×1.5937420.095310=0.001390490.095310=0.014589\frac{0.00005 \times 17.449402 \times 1.593742}{0.095310} = \frac{0.00139049}{0.095310} = 0.014589

Total exponent: −(0.005+0.014589)=−0.019589-(0.005 + 0.014589) = -0.019589

10p30=e−0.019589=0.9806{}_{10}p_{30} = e^{-0.019589} = 0.9806
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