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=1pxy=1pxpy=1(1qx)(1qy)=1(0.95)(0.96)=10.912=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+qyqxqy=0.05+0.040.002=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+tPxya¨x+t:y+t{}_tV_{xy} = A_{x+t:y+t} - P_{xy}\ddot{a}_{x+t:y+t} At t=0t=0: AxyPxya¨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+AyAxy=0.15A_{\bar{x}\bar{y}} = A_x + A_y - A_{xy} = 0.15. Which statement correctly evaluates this result?
Solution
A is correct.

The last-survivor status is alive whenever at least one of (x)(x) or (y)(y) is alive, so T(xˉyˉ)=max(T(x),T(y))T(x)T(\bar{x}\bar{y}) = \max(T(x), T(y)) \geq T(x) and T(xˉyˉ)T(y)T(\bar{x}\bar{y}) \geq T(y) almost surely. Since the whole life insurance APV is an increasing function of the future lifetime, AxˉyˉAxA_{\bar{x}\bar{y}} \geq A_x and AxˉyˉAyA_{\bar{x}\bar{y}} \geq A_y must hold under any dependence structure. With the given inputs, Axˉyˉ=0.15<0.25=Ax<0.30=AyA_{\bar{x}\bar{y}} = 0.15 < 0.25 = A_x < 0.30 = A_y, which is impossible. Furthermore, Axy=0.40>min(Ax,Ay)=0.25A_{xy} = 0.40 > \min(A_x, A_y) = 0.25 is itself a violation: since T(xy)T(x)T(xy) \leq T(x) always, we require AxyAxA_{xy} \geq A_x. The inconsistency arises purely from Axˉyˉ=0.15<AyA_{\bar{x}\bar{y}} = 0.15 < A_y.

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