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

Future Lifetime Distributions for Two Lives: First and Last Death

Updated 11 October 2026 · Fact-checked

For two lives (x) and (y), the time to first death is T_xy = min(T_x, T_y) and the time to last death is T_x̄ȳ = max(T_x, T_y). Assuming independence, t p_xy = t p_x × t p_y. Expectations follow from e_xy = Σ k p_xy, and e_x̄ȳ = e_x + e_y − e_xy.

Understand Future Lifetime Distributions for Two Lives

A joint life status (xy) is alive while both lives are alive. It fails at the first death. Its future lifetime is T_xy = min(T_x, T_y). A last survivor status (x̄ȳ) is alive while at least one life is alive. It fails at the second death. Its future lifetime is T_x̄ȳ = max(T_x, T_y).

Most exam questions assume the two future lifetimes T_x and T_y are independent. State this assumption in your answer. Under it, both lives survive t years with probability t p_x × t p_y. So the joint status survives t years with t p_xy = t p_x × t p_y, and the first death has happened by time t with probability t q_xy = 1 − t p_x × t p_y.

For the last survivor, both lives must be dead by time t. Under independence, t q_x̄ȳ = t q_x × t q_y. So t p_x̄ȳ = 1 − t q_x × t q_y. Expanding gives t p_x̄ȳ = t p_x + t p_y − t p_xy. This is inclusion-exclusion: at least one alive = x alive + y alive − both alive.

The joint status has force of mortality μ_x+t + μ_y+t at time t. This is because the first death can come from either life. The density of T_xy is therefore t p_xy × (μ_x+t + μ_y+t).

The expected future lifetime is the area under the survival curve. For the complete expectation, e°_xy = ∫ from 0 to ∞ of t p_xy dt. For the curtate expectation (whole years completed), e_xy = Σ from k=1 to ∞ of k p_xy. The same forms work for the last survivor with t p_x̄ȳ. Substituting t p_x̄ȳ = t p_x + t p_y − t p_xy gives e_x̄ȳ = e_x + e_y − e_xy. An even quicker argument: max + min = sum of the two lifetimes, so taking expectations gives the same result. The relation holds for curtate and complete expectations, and it does not need independence.

Key rules to remember

Time to first and last death
T_xy = min(T_x, T_y); T_x̄ȳ = max(T_x, T_y)
Also T_xy + T_x̄ȳ = T_x + T_y. The curtate versions are K_xy = min(K_x, K_y) and K_x̄ȳ = max(K_x, K_y).
Joint life survival probability
t p_xy = t p_x × t p_y
Needs independence. Then t q_xy = 1 − t p_x × t p_y.
Last survivor survival probability
t p_x̄ȳ = t p_x + t p_y − t p_xy
Holds for any joint distribution of (T_x, T_y). Under independence, t q_x̄ȳ = t q_x × t q_y.
Force of mortality of the joint status
μ_xy(t) = μ_x+t + μ_y+t
Under independence. Density of T_xy: f(t) = t p_xy × (μ_x+t + μ_y+t).
Density of the time to last death
f(t) = t p_x μ_x+t + t p_y μ_y+t − t p_xy (μ_x+t + μ_y+t)
Differentiate −t p_x̄ȳ with respect to t. Under independence, this equals the derivative of t q_x × t q_y.
Complete expectation of life
e°_xy = ∫₀^∞ t p_xy dt; e°_x̄ȳ = e°_x + e°_y − e°_xy
e°_x̄ȳ uses t p_x̄ȳ in the integral. The relation follows because t p_x̄ȳ = t p_x + t p_y − t p_xy.
Curtate expectation of life
e_xy = Σ (k = 1 to ∞) k p_xy; e_x̄ȳ = e_x + e_y − e_xy
e_x̄ȳ uses k p_x̄ȳ in the sum. Always start the sum at k = 1.
Probability of death in a given year
P(K_xy = k) = k p_xy − (k+1) p_xy
The same form works for the last survivor with k p_x̄ȳ. K is the curtate future lifetime.
Approximation linking curtate and complete
e°_xy ≈ e_xy + ½
Valid only under the assumption of uniform distribution of deaths within each year. It is an approximation, not an identity.

How to solve Future Lifetime Distributions for Two Lives questions

Use this order for any question on the distribution or expectation of the first or second death of two lives.

  1. 1Identify the status. Joint life (xy) means first death and the minimum. Last survivor (x̄ȳ) means second death and the maximum.
  2. 2State your assumption clearly. Write 'assume T_x and T_y are independent' unless the question gives a dependence.
  3. 3Write the survival function. Use t p_xy = t p_x × t p_y for the joint life. For the last survivor, use t p_x̄ȳ = t p_x + t p_y − t p_xy, or t q_x̄ȳ = t q_x × t q_y.
  4. 4If a density is needed, differentiate the distribution function. For T_xy, use f(t) = t p_xy (μ_x+t + μ_y+t).
  5. 5For an expectation, integrate t p over t (complete) or sum k p from k = 1 (curtate). Use the status asked about.
  6. 6For a last survivor expectation, find e_x, e_y and e_xy separately, then use e_x̄ȳ = e_x + e_y − e_xy.
  7. 7Check the units and bounds. Is the answer in years? Is e_xy ≤ min(e_x, e_y)? Is e_x̄ȳ ≥ max(e_x, e_y)?

Quickest way: Build the joint table first, then use e_x̄ȳ = e_x + e_y − e_xy

When to use it: Use this when you are given a table or a few values of k p_x and k p_y, or when both lives have constant (or Gompertz-type) mortality and you can write t p_xy directly.

  1. Write k p_x and k p_y in two columns.
  2. Multiply them to get k p_xy in a third column. This assumes independence.
  3. Sum each column from k = 1 to get e_x, e_y and e_xy.
  4. Compute e_x̄ȳ = e_x + e_y − e_xy.
  5. For constant forces μ_x and μ_y, skip the table. T_xy is exponential with force μ_x + μ_y, so e°_xy = 1 ÷ (μ_x + μ_y).
  6. Cross-check by summing k p_x̄ȳ = k p_x + k p_y − k p_xy if time allows.

Common mistakes in Future Lifetime Distributions for Two Lives

  • Using t p_xy = t p_x + t p_y or taking the average of the two lifetimes.

    Students mix up 'at least one alive' with 'both alive'.

    Fix: Joint life means both alive, so multiply (under independence). The addition-and-subtract form belongs to the last survivor only.

  • Writing t q_x̄ȳ = t q_x + t q_y.

    Students copy the pattern of the joint life formula without checking which event it describes.

    Fix: A last survivor status has failed only when both lives are dead, so multiply the death probabilities: t q_x̄ȳ = t q_x × t q_y (independence).

  • Starting the curtate sum at k = 0, or integrating from the wrong limits.

    Students remember that t p = 1 at t = 0 and include it.

    Fix: Curtate expectation is Σ from k = 1. Complete expectation integrates from 0 to ∞ (or to the limiting age).

  • Adding μ_x and μ_y for the last survivor status.

    The joint life rule μ_x+t + μ_y+t is applied to every status.

    Fix: The sum of forces applies to the joint life status only. For the last survivor, work from the survival probability and differentiate.

  • Applying e_x̄ȳ = e_x + e_y − e_xy but mixing curtate and complete values.

    Tables give e_x curtate, and a formula uses e°_xy.

    Fix: Use one type throughout. If you must convert, state the approximation e° ≈ e + ½ and apply it consistently.

  • Forgetting to state the independence assumption.

    Students treat it as obvious.

    Fix: Write it as the first line of your solution. Marks are usually awarded for stating the assumption.

Worked examples

Example 1

Two independent lives (x) and (y) have constant forces of mortality μ_x = 0.02 and μ_y = 0.03 per year. Calculate (a) e°_xy, (b) e°_x̄ȳ, (c) the probability that the first death occurs within 10 years, and (d) the probability that both lives are dead within 10 years.

Show the solution
  1. Assume T_x and T_y are independent. With constant forces, T_x and T_y are exponential, so t p_x = e^(−0.02t) and t p_y = e^(−0.03t).
  2. (a) t p_xy = e^(−0.05t). So e°_xy = ∫₀^∞ e^(−0.05t) dt = 1 ÷ 0.05 = 20 years.
  3. (b) e°_x = 1 ÷ 0.02 = 50 and e°_y = 1 ÷ 0.03 = 33.3333. Then e°_x̄ȳ = e°_x + e°_y − e°_xy = 50 + 33.3333 − 20 = 63.3333 years.
  4. (c) P(T_xy ≤ 10) = 1 − e^(−0.05 × 10) = 1 − e^(−0.5) = 1 − 0.60653 = 0.39347.
  5. (d) P(T_x̄ȳ ≤ 10) = 10 q_x × 10 q_y = (1 − e^(−0.2)) × (1 − e^(−0.3)) = 0.181269 × 0.259182 = 0.04698.

Answer: (a) 20 years; (b) 63.33 years; (c) 0.3935; (d) 0.0470.

Example 2

Lives (x) and (y) are independent. Life (x) has 1 p_x = 0.8, 2 p_x = 0.5 and 3 p_x = 0. Life (y) has 1 p_y = 0.9, 2 p_y = 0.6, 3 p_y = 0.2 and 4 p_y = 0. Calculate the curtate expectations e_xy and e_x̄ȳ, and the probability that the last death occurs in the third year (that is, K_x̄ȳ = 2).

Show the solution
  1. Assume independence. Then k p_xy = k p_x × k p_y.
  2. 1 p_xy = 0.8 × 0.9 = 0.72. 2 p_xy = 0.5 × 0.6 = 0.30. 3 p_xy = 0 × 0.2 = 0.
  3. e_xy = 0.72 + 0.30 + 0 = 1.02.
  4. e_x = 0.8 + 0.5 = 1.3 and e_y = 0.9 + 0.6 + 0.2 = 1.7.
  5. e_x̄ȳ = e_x + e_y − e_xy = 1.3 + 1.7 − 1.02 = 1.98.
  6. For the last death, find k p_x̄ȳ = k p_x + k p_y − k p_xy. 2 p_x̄ȳ = 0.5 + 0.6 − 0.30 = 0.80. 3 p_x̄ȳ = 0 + 0.2 − 0 = 0.20.
  7. P(K_x̄ȳ = 2) = 2 p_x̄ȳ − 3 p_x̄ȳ = 0.80 − 0.20 = 0.60.
  8. Check: 1 p_x̄ȳ = 0.8 + 0.9 − 0.72 = 0.98. Summing 0.98 + 0.80 + 0.20 gives 1.98, which matches.

Answer: e_xy = 1.02 years; e_x̄ȳ = 1.98 years; P(K_x̄ȳ = 2) = 0.60.

Exam tips

  • Write the independence assumption at the top of every answer. It earns a mark and shows you know when the product rule applies.
  • In MCQs, check the status first. Many wrong options are the right formula for the other status.
  • Use the sanity bounds: e_xy ≤ min(e_x, e_y) and e_x̄ȳ ≥ max(e_x, e_y). They catch arithmetic slips fast.
  • For derivation questions, give both routes if time allows: the survival probability route and the max + min = sum route. Show the final line e_x̄ȳ = e_x + e_y − e_xy.
  • In the computer-based paper, build columns for k p_x, k p_y and k p_xy, then sum. Label each column and state the assumptions in a cell comment or note.

Practice questions from Assurance and annuity functions involving two lives

Future Lifetime Distributions for Two Lives in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Future Lifetime Distributions for Two Lives: frequently asked questions

What is the difference between T_xy and T_x̄ȳ?

T_xy = min(T_x, T_y) is the time until the first of the two lives dies. T_x̄ȳ = max(T_x, T_y) is the time until the second death. Always T_xy ≤ T_x̄ȳ, and T_xy + T_x̄ȳ = T_x + T_y.

How do I derive e_x̄ȳ = e_x + e_y − e_xy?

Use k p_x̄ȳ = k p_x + k p_y − k p_xy for each k. Sum both sides from k = 1. The left side is e_x̄ȳ and the right side is e_x + e_y − e_xy. The same step with integrals gives the complete version.

Does e_x̄ȳ = e_x + e_y − e_xy need independence?

No. It follows from max + min = sum of the two lifetimes, which holds for any joint distribution. Independence is only needed to calculate e_xy itself from the individual survival probabilities.

What is the difference between curtate and complete expectation of life?

Complete expectation e° is the mean of the exact future lifetime T. Curtate expectation e is the mean of the whole number of years lived, K. The curtate version sums k p from k = 1, and the complete version integrates t p. Under uniform distribution of deaths, e° ≈ e + ½.