Actuarial Mathematics for Modelling · Assurance and annuity functions involving two lives

Assurances on Joint Life and Last Survivor Statuses

Updated 11 October 2026 · Fact-checked

A joint life assurance pays on the first death of two lives. A last survivor assurance pays on the second death. Value each as the expected present value of v raised to the time of the relevant death. For whole life, A_xy + A_x̄ȳ = A_x + A_y, so you can find one from the other.

Understand Assurances on Joint Life and Last Survivor Statuses

A status is a condition that is either met or not at each future time. The joint life status (xy) survives while both lives are alive. It fails at the first death. The last survivor status (x̄ȳ) survives while at least one life is alive. It fails at the second death.

Let T_x and T_y be the future lifetimes of the two lives. The joint life status fails at time T_xy = min(T_x, T_y). The last survivor status fails at T_x̄ȳ = max(T_x, T_y). A whole life assurance on a status pays 1 at the moment the status fails (or at the end of the year of failure, in the discrete case). Its present value is v^T, where T is the failure time of the status.

So A_xy is the EPV of 1 paid at the end of the year of first death, and A_x̄ȳ is the EPV of 1 paid at the end of the year of second death. The EPV of 1 at the continuous moment of death is written Ā_xy and Ā_x̄ȳ.

The key link comes from a simple fact. The set {T_xy, T_x̄ȳ} is the same as the set {T_x, T_y}. One of the two deaths is the first and the other is the second. So the two payments, one at first death and one at second death, are made at the same two times as the payments at the death of x and at the death of y. Therefore v^T_xy + v^T_x̄ȳ = v^Tx + v^Ty. Taking expectations gives A_xy + A_x̄ȳ = A_x + A_y. This holds for any joint distribution of the two lives, not only independent lives.

The same logic gives the survival probability link: the probability the last survivor status survives t years is tp_x̄ȳ = tp_x + tp_y − tp_xy. Here independence is not needed for this probability identity. But to compute tp_xy as tp_x × tp_y, you do need independence of the two lives.

Key rules to remember

Joint life status failure time: T_xy = min(T_x, T_y); The status (xy) fails on the first death.
Last survivor status failure time: T_x̄ȳ = max(T_x, T_y); The status (x̄ȳ) fails on the second death.
Whole life assurance on a status (discrete): A_xy = E[v^(K_xy + 1)] and A_x̄ȳ = E[v^(K_x̄ȳ + 1)]; K is the curtate future lifetime of the status. Payment is at the end of the year of failure.
Whole life assurance on a status (continuous): Ā_xy = E[v^T_xy] and Ā_x̄ȳ = E[v^T_x̄ȳ]; Payment at the moment the status fails.
Key relationship: A_xy + A_x̄ȳ = A_x + A_y; Also Ā_xy + Ā_x̄ȳ = Ā_x + Ā_y. Holds for any dependence between the lives, using the same basis for all four.
Last survivor from the others: A_x̄ȳ = A_x + A_y − A_xy; Use this to avoid computing second-death probabilities directly.
Survival probabilities: tp_xy = tp_x × tp_y (independent lives); tp_x̄ȳ = tp_x + tp_y − tp_xy; The first needs independence. The second always holds.
Term assurance on a status: Joint life: A¹_xy:n = Σ (k = 0 to n−1) v^(k+1) × kp_xy × q_(x+k:y+k). Last survivor: A¹_x̄ȳ:n = Σ (k = 0 to n−1) v^(k+1) × (kp_x̄ȳ − (k+1)p_x̄ȳ); The first formula is for the joint life status only. Here q_(x+k:y+k) = 1 − p_(x+k:y+k) is the probability the joint status fails in the year, so the first-death probability in year k+1 is kp_xy × q_(x+k:y+k). For the last survivor status, do not use this expression. Use the status's own survival probabilities: the probability it fails in year k+1 is kp_x̄ȳ − (k+1)p_x̄ȳ.
Recursion for joint life assurance: A_xy = v × q_xy + v × p_xy × A_(x+1:y+1); One-year recursion for the joint life status, with q_xy = 1 − p_xy.

How to solve Assurances on Joint Life and Last Survivor Statuses questions

Use this method for any question on assurances on two lives. It works for first death, second death and mixed cases.

1Identify the status. Payment on first death means joint life (xy). Payment on second death means last survivor (x̄ȳ).
2Note the timing of payment: end of year of death (discrete, A) or moment of death (continuous, Ā).
3Note the sum assured, any term or deferral, and whether the lives are independent.
4Decide which functions you are given. If you have A_x, A_y and A_xy, use A_x̄ȳ = A_x + A_y − A_xy. If you have A_x̄ȳ instead, rearrange to find A_xy.
5If you must build the value from probabilities, write the probability the status fails in each year and multiply by v^(k+1). Use tp_xy = tp_x × tp_y only for independent lives.
6Check that all functions use the same mortality table and interest rate. Mixed bases break the identity.
7Multiply by the sum assured, and give the answer in rupees where asked.
8Sense-check: A_xy ≥ A_x̄ȳ, since first death comes no later than second death.

Quickest way: Use the sum identity instead of the second-death probabilities

When to use it: Use it when the question gives or lets you find single-life assurances and one of the two-life assurances.

Write A_xy + A_x̄ȳ = A_x + A_y.
Insert the known values and solve for the unknown.
Multiply by the sum assured.
Check A_xy ≥ A_x̄ȳ. If A_xy is smaller, you have swapped them.

Common mistakes in Assurances on Joint Life and Last Survivor Statuses

Swapping which status gives first death and which gives second death.
The bar notation is easy to confuse. Students link 'last' with the larger-looking symbol and mix it up.
Fix: Write: no bar (xy) = both alive = pays on first death. Bar (x̄ȳ) = either alive = pays on second death. Then check A_xy ≥ A_x̄ȳ.
Writing A_x̄ȳ = A_x × A_y or A_xy = A_x × A_y.
Students copy the product rule used for survival probabilities of independent lives.
Fix: Assurance values are expectations of v^T, not probabilities. The product of v^Tx and v^Ty is not v^min. Use the sum identity instead.
Using tp_x̄ȳ = tp_x × tp_y.
Confusion between 'both survive' and 'at least one survives'.
Fix: Both survive is tp_xy. At least one survives is tp_x + tp_y − tp_xy.
Assuming the identity needs independent lives.
Students link all two-life formulas with independence.
Fix: A_xy + A_x̄ȳ = A_x + A_y holds for any dependence. Independence is only needed to compute tp_xy as a product.
Mixing discrete and continuous forms or different bases in the identity.
Values are taken from different tables or one is an A and another an Ā.
Fix: Use all A or all Ā, with the same mortality and interest rate for all four terms.
Forgetting to multiply by the sum assured at the end.
Functions are per unit of benefit, and the final step is easy to forget under time pressure.
Fix: Underline the sum assured in the question and multiply last.

Worked examples

Example 1

For two lives, A_x = 0.32, A_y = 0.41 and A_xy = 0.52, on the same basis. Find (a) A_x̄ȳ and (b) the EPV of a whole life assurance paying ₹10,00,000 at the end of the year of second death.

Show the solution

Check consistency first. First death is no later than either single death, so A_xy ≥ A_x and A_xy ≥ A_y. Here 0.52 ≥ 0.41 and 0.52 ≥ 0.32, which is fine.
(a) Use A_x̄ȳ = A_x + A_y − A_xy.
A_x̄ȳ = 0.32 + 0.41 − 0.52 = 0.21.
Check: second death is no earlier than either single death, so A_x̄ȳ ≤ min(A_x, A_y) = 0.32. And 0.21 ≤ 0.32, which is fine.
(b) EPV = 10,00,000 × 0.21 = ₹2,10,000.

Answer: (a) A_x̄ȳ = 0.21. (b) The EPV is ₹2,10,000.

Exam tips

Always check that A_xy ≥ max(A_x, A_y), that A_x̄ȳ ≤ min(A_x, A_y), and that A_xy ≥ A_x̄ȳ. It catches swapped values fast.
In written questions, state the identity and the assumptions (same basis, payment timing) before using it. Method marks depend on it.
Look for whether independence is stated. If it is, you may write tp_xy = tp_x × tp_y. If not, use the sum identity, which needs no assumption.
In Paper B, build the probabilities in a column for each year, then compute A_xy and A_x̄ȳ and test the identity as a check on your spreadsheet or R code.

Practice questions from Assurance and annuity functions involving two lives

Assurances on Joint Life and Last Survivor Statuses: frequently asked questions

What is the difference between A_xy and A_x̄ȳ?

A_xy is the EPV of a benefit paid on the first death of the two lives. A_x̄ȳ is the EPV of a benefit paid on the second death. A_xy is always at least as large as A_x̄ȳ, because the first death occurs no later than the second.

Do I need independent lives for A_xy + A_x̄ȳ = A_x + A_y?

No. The identity holds for any joint distribution of the two lifetimes. You only need independence when you build tp_xy as tp_x × tp_y.

How do I find A_x̄ȳ if I only know A_x, A_y and A_xy?

Rearrange the identity: A_x̄ȳ = A_x + A_y − A_xy. Make sure all three values use the same mortality, interest rate and payment timing.

Does the same relationship hold for continuous assurances?

Yes. Ā_xy + Ā_x̄ȳ = Ā_x + Ā_y, by the same argument: the two payment times are the same set of times, whichever way you label them.