Free SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) Joint Life Insurance and Annuities Practice Questions

Joint life insurance and annuities on SOA Exam ALTAM cover joint-life and last-survivor statuses, common shock mortality models, reversionary annuities, and multiple life contingent cash flows.

120 questions 45 easy 56 medium 19 hard 2026 syllabus

Sample Questions

Question 1 Easy
For two independent lives with qx=0.05q_x = 0.05 and qy=0.04q_y = 0.04, calculate qxyq_{xy}, the probability the joint-life status fails within one year.
Solution
E is correct.

The joint-life status fails within one year if at least one of the two lives dies. Under independence: qxy=1−pxy=1−px⋅py=1−(1−qx)(1−qy)=1−(0.95)(0.96)=1−0.912=0.088q_{xy} = 1 - p_{xy} = 1 - p_x \cdot p_y = 1-(1-q_x)(1-q_y) = 1-(0.95)(0.96) = 1-0.912 = 0.088 Equivalently by inclusion-exclusion: qx+qy−qxqy=0.05+0.04−0.002=0.088q_x + q_y - q_x q_y = 0.05+0.04-0.002 = 0.088.
Question 2 Medium
The prospective reserve at time tt for a joint-life whole life insurance with net annual premium PxyP_{xy} is tVxy{}_tV_{xy}. Which of the following is the correct prospective formula?
Solution
C is correct.

The prospective reserve equals the expected present value of future benefits minus the expected present value of future premiums, both conditioned on both lives surviving to time tt with attained ages x+tx+t and y+ty+t: tVxy=Ax+t:y+t−Pxya¨x+t:y+t{}_tV_{xy} = A_{x+t:y+t} - P_{xy}\ddot{a}_{x+t:y+t} At t=0t=0: Axy−Pxya¨xy=0A_{xy} - P_{xy}\ddot{a}_{xy} = 0 by the equivalence principle. For t>0t>0, the attained ages change and the reserve grows.
Question 3 Hard
Given Ax=0.25A_x = 0.25, Ay=0.30A_y = 0.30, and Axy=0.40A_{xy} = 0.40, a student computes the last-survivor APV as Axˉyˉ=Ax+Ay−Axy=0.15A_{\bar{x}\bar{y}} = A_x + A_y - A_{xy} = 0.15. Which statement correctly evaluates this result?
Solution
D is correct.

A whole life insurance of 1 on a status pays when the status fails at time TT, so its APV E[vT]E[v^{T}] falls as the future lifetime lengthens. The joint-life status fails at the first death and the last-survivor status at the second, so on every outcome T(xy)=min⁡(T(x),T(y))T(xy) = \min(T(x), T(y)) and T(xˉyˉ)=max⁡(T(x),T(y))T(\bar{x}\bar{y}) = \max(T(x), T(y)). Discounting reverses the lifetime order:

Axˉyˉ≤min⁡(Ax,Ay)≤max⁡(Ax,Ay)≤AxyA_{\bar{x}\bar{y}} \leq \min(A_x, A_y) \leq \max(A_x, A_y) \leq A_{xy}

The given joint value fits this ordering: Axy=0.40≥0.30=max⁡(Ax,Ay)A_{xy} = 0.40 \geq 0.30 = \max(A_x, A_y).

On every outcome the pair T(xy),T(xˉyˉ)T(xy), T(\bar{x}\bar{y}) is the pair T(x),T(y)T(x), T(y) in sorted order, so vT(xy)+vT(xˉyˉ)=vT(x)+vT(y)v^{T(xy)} + v^{T(\bar{x}\bar{y})} = v^{T(x)} + v^{T(y)}. Taking expectations gives Axy+Axˉyˉ=Ax+AyA_{xy} + A_{\bar{x}\bar{y}} = A_x + A_y with no assumption about dependence between the lives, so

Axˉyˉ=0.25+0.30−0.40=0.15A_{\bar{x}\bar{y}} = 0.25 + 0.30 - 0.40 = 0.15

and 0.15≤min⁡(Ax,Ay)=0.250.15 \leq \min(A_x, A_y) = 0.25, as the ordering requires. The student's computation is valid.

Reading the ordering as Axˉyˉ≥max⁡(Ax,Ay)A_{\bar{x}\bar{y}} \geq \max(A_x, A_y) reverses it: the status that lasts longest pays latest and so has the smallest APV. The ratio AxAy/Axy=0.1875A_x A_y / A_{xy} = 0.1875 is not a last-survivor formula; the additive identity is exact. The same pathwise argument applies to term benefits and whole life benefits alike, and to dependent lives as well as independent ones.

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