Multiple Life Models: Joint Life and Last Survivor

Joint Life Status

The joint life status (xy) fails at the first death among two lives, (x) and (y). This status is fundamental to pension and insurance products that provide benefits while both members of a couple are alive. The survival function for the joint life status, assuming independence of the two lives, is t_p_{xy} = t_p_x * t_p_y. The force of mortality for the joint status is the sum of the individual forces: mu_{xy}(t) = mu_x(t) + mu_y(t). This additive property of forces of mortality under independence is one of the most useful results in multiple life theory.

Insurance and annuity values for joint life statuses follow the same structural formulas as single life values. The whole life insurance on joint life status, A_{xy}, pays a benefit at the moment of the first death. The annuity a-ddot_{xy} makes payments as long as both lives survive. The fundamental relationship A_{xy} = 1 minus d * a-ddot_{xy} continues to hold, connecting insurance and annuity values through the discount rate. Joint life annuities appear frequently in pension plans that provide benefits ceasing at the first death of a married couple.

Last Survivor Status

The last survivor status fails at the second (final) death. It is relevant for insurance products that provide coverage until both lives have died, or for annuities that continue paying as long as at least one member of a couple survives. The key identity linking the last survivor future lifetime to the joint life and individual future lifetimes is T_{last survivor} = T_x + T_y minus T_{joint life}.

This identity translates directly to actuarial present values. For annuities: a-ddot_{last survivor} = a-ddot_x + a-ddot_y minus a-ddot_{xy}. For insurance: A_{last survivor} = A_x + A_y minus A_{xy}. These formulas allow you to compute last survivor values from single life and joint life values without needing separate last survivor survival functions. This is extremely practical because joint life survival probabilities are straightforward to compute from individual survival probabilities under independence, whereas last survivor probabilities would require more complex calculations directly.

Contingent Insurance

Contingent insurance benefits depend on the order of death. The notation A_{xy}^1 denotes insurance that pays a benefit at the death of (x) only if (x) dies before (y). Similarly, A_{xy}^{bar 1} (or A with the "1" over y) pays at the death of (y) only if (y) dies first. The fundamental decomposition is A_{xy} = A_{xy}^1 + A_{xy}^{bar 1}, because the joint life insurance pays at the first death regardless of which life dies first.

Computing contingent insurance values requires integrating over the joint density of the two future lifetimes. Under independence, A_{xy}^1 = integral from 0 to infinity of v^t * t_p_{xy} * mu_x(t) dt. The integrand reflects that (x) must survive to time t (both must survive, giving t_p_{xy}), and then (x) dies at rate mu_x(t). This integral can be evaluated analytically for simple mortality laws such as constant force (exponential lifetime) or computed numerically for more complex mortality assumptions.

Worked Example with Constant Force

Suppose (x) has constant force of mortality mu_x = 0.04 and (y) has constant force mu_y = 0.03, with force of interest delta = 0.05. Under constant force, t_p_x = e^(-0.04t), t_p_y = e^(-0.03t), and t_p_{xy} = e^(-0.07t). The joint life insurance is A_{xy} = (mu_x + mu_y) / (mu_x + mu_y + delta) = 0.07 / 0.12 = 0.5833.

The contingent insurance on (x) dying first is A_{xy}^1 = mu_x / (mu_x + mu_y + delta) = 0.04 / 0.12 = 0.3333. Similarly, A_{xy}^{bar 1} = mu_y / (mu_x + mu_y + delta) = 0.03 / 0.12 = 0.25. Notice that 0.3333 + 0.25 = 0.5833 = A_{xy}, confirming the decomposition. The joint life annuity is a-bar_{xy} = 1 / (mu_x + mu_y + delta) = 1 / 0.12 = 8.3333.

Reversionary Annuities

A reversionary annuity pays benefits to one life after the death of another. The reversionary annuity to (y) after the death of (x) is denoted a_{x|y} and equals a-ddot_y minus a-ddot_{xy}. This represents the difference between an annuity that pays as long as (y) is alive and one that pays only while both are alive. The difference captures payments made to (y) after (x) has died.

Reversionary annuities are common in pension plans that provide a survivor benefit to the spouse after the retiree dies. For example, a "joint and 50% survivor" pension pays the full benefit while both spouses are alive and 50% of the benefit to the surviving spouse. The present value of this benefit equals the full annuity on joint life status plus 50% of the reversionary annuity to the surviving spouse.

Common Pitfalls

Students frequently make errors with the independence assumption. The formulas t_p_{xy} = t_p_x * t_p_y and mu_{xy} = mu_x + mu_y hold only when the two lives are independent. If the problem states a common shock model or other dependence structure, these simple multiplicative or additive relationships do not apply. Always check whether independence is stated or implied.

Another common mistake is confusing the joint life status with the last survivor status. Remember that joint life fails at the first death (both must survive), while last survivor fails at the last death (at least one must survive). Mixing these up reverses the meaning of every formula. Also be careful with contingent insurance notation: the superscript "1" goes over the life whose death triggers the payment, not over the life that must survive.

Study Tips

Exam LTAM tests multiple life models extensively, particularly the identities connecting joint life, last survivor, and individual life values. Memorize the core identities: the last survivor decomposition, the contingent insurance decomposition, and the reversionary annuity formula. Practice with constant force of mortality problems first, since they yield clean closed-form answers that build intuition. Then move to problems using life tables, where numerical computation is required. Be prepared for questions that combine multiple life models with premium and reserve calculations, as these multi-step problems are common on the exam.