Reconciling Exit Benefits and Survival Benefits

Why Reconcile Exit and Survival Benefits?

A decrement calculation does not end when the cause-specific exit probabilities have been found. If a benefit is payable on one or more exits, and another benefit is payable to those who remain in the group, the expected value must account for every possible outcome exactly once. A useful check is to ask whether the outcomes form a complete partition: either one of the specified exits occurs, or the individual survives the period.

The payment amounts and timing matter as much as the probabilities. A benefit paid at the end of a year is discounted for a full year; a benefit paid at the exact time of exit is discounted to that time. The formulas below make those assumptions explicit so that the calculation and its check use the same basis.

Definition: Consider one year for a person who is in a specified decrement group at its start. Let $q_j$ be the probability of exit during that year from cause $j$, conditional on being in the group at the start, and let $p$ be the probability of remaining in the group through the end of the year. If the listed causes exhaust the ways to exit, then $$ p+\sum_{j=1}^{m}q_j=1. $$ A benefit of amount $B_j$ is paid for exit from cause $j$, and a benefit of amount $B_S$ is paid for survival to year-end. All amounts and payment times must be stated as part of the model.

The One-Year Expected-Value Check

First suppose every benefit is paid at the end of the year, whether triggered by an exit or by survival. Let the effective annual interest rate be $i$, with $v=1/(1+i)$. The events “exit from cause 1,” through “exit from cause $m$,” and “survive the year” are mutually exclusive and collectively exhaustive. Therefore the expected present value, measured at the start of the year, is

\operatorname{EPV} =v\left(\sum_{j=1}^{m}q_jB_j+pB_S\right).

This is also a direct reconciliation: the probability-weighted amounts inside the parentheses include one and only one payment for each possible outcome. If there is no survival benefit, set $B_S=0$. If a survival payment is payable only under additional conditions, such as survival in a particular state, its probability must be the probability of that more specific event rather than the overall survival probability.

Why the formula holds: Let $I_j$ indicate that cause $j$ occurs during the year, and let $I_S$ indicate survival through the year. Exactly one of these indicators is 1, so the end-of-year payment is $Z=\sum_j B_j I_j+B_S I_S$. By linearity of expectation, $E[Z]=\sum_j B_jP(I_j=1)+B_SP(I_S=1)=\sum_jq_jB_j+pB_S$. Since the whole payment is made at time 1, its present value is $vZ$, giving the stated formula.

Worked Example: Reconciling Two Exit Benefits and a Survival Benefit

A hypothetical one-year contract covers a person in a decrement group. During the year, the probability of exit from cause A is $0.08$, the probability of exit from cause B is $0.12$, and the person receives a survival benefit if still in the group at year-end. Cause A pays $12,000, cause B pays $5,000, and survival pays $1,000. All payments are at year-end. The effective annual interest rate is $i=5\%$. Find the expected present value and check the probability partition.

Method. First infer the survival probability from the complete partition, then weight each outcome’s payment by its probability and discount once.

The stated exit probabilities sum to $0.08+0.12=0.20$. Thus

p=1-0.08-0.12=0.80,\qquad 0.08+0.12+0.80=1.00.

The probability-weighted end-of-year benefit is

(0.08)(\$12{,}000)+(0.12)(\$5{,}000)+(0.80)(\$1{,}000) =\$960+\$600+\$800=\$2{,}360.

The discount factor is $v=1/1.05=0.95238095\ldots$. Therefore

\operatorname{EPV}=\frac{\$2{,}360}{1.05}=\$2{,}247.6190\ldots

Rounded to the nearest cent, the expected present value is $2,247.62. The check is that the three event probabilities sum to 1 and the three weighted payments total $2,360 before discounting.

Common mistake avoided: The survival benefit is not another exit benefit. Omitting it would give only $1,560 before discounting; counting the survival event as both an exit and survival would count more than one outcome for the same person.

More Than One Year: Weight Each Year by Its Entry Probability

For a sequence of years, the cause probabilities for a particular year are conditional on being in the decrement group at the start of that year. A person who exits in an earlier year cannot receive a later benefit under this model. Consequently, a year’s expected payment must be multiplied by the probability of reaching that year still in the group.

For example, suppose year $t$ has conditional cause probabilities $q_{tj}$, survival probability $p_t$, and end-of-year payments $B_{tj}$ on cause $j$ and $B_{tS}$ on survival. If $\pi_t$ is the probability of being in the group at the start of year $t$, then the year-$t$ contribution to the present value at time 0 is

v^t\pi_t\left(\sum_j q_{tj}B_{tj}+p_tB_{tS}\right).

Here $\pi_1=1$, and, when the only way to leave the group is through the listed decrements, $\pi_{t+1}=\pi_t p_t$. This weighting separates two questions: conditional on reaching the year, what is the expected payment in that year; and what is the probability of reaching it?

Worked Example: A Survival Payment in Each of Two Years

A hypothetical two-year benefit pays at each year-end. In year 1, conditional on being in the group at the start of the year, the probabilities of exit from A and B are $0.10$ and $0.05$. In year 2, conditional on reaching that year, the corresponding probabilities are $0.08$ and $0.12$. If a person survives a year, the payment for that year is $500; an exit instead pays $10,000 for cause A or $6,000 for cause B. The effective annual interest rate is $i=4\%$. Find the expected present value over the two years.

Method. Calculate the expected payment conditional on entering each year, multiply the second-year amount by the probability of reaching year 2, and discount each year’s payment to time 0.

In year 1, the survival probability is $p_1=1-0.10-0.05=0.85$. The year-1 expected payment conditional on being in the group at time 0 is

(0.10)(\$10{,}000)+(0.05)(\$6{,}000)+(0.85)(\$500) =\$1{,}000+\$300+\$425=\$1{,}725.

In year 2, the conditional survival probability is $p_2=1-0.08-0.12=0.80$. Conditional on reaching year 2, its expected payment is

(0.08)(\$10{,}000)+(0.12)(\$6{,}000)+(0.80)(\$500) =\$800+\$720+\$400=\$1{,}920.

The probability of reaching year 2 is $0.85$, so the unconditional expected year-2 payment is $(0.85)(\$1{,}920)=\$1{,}632$. The discount factors are $v=1/1.04$ and $v^2=1/1.04^2=1/1.0816$. Thus

\operatorname{EPV} =\frac{\$1{,}725}{1.04}+\frac{\$1{,}632}{1.0816} =\$1{,}658.653846\ldots+\$1{,}508.875740\ldots =\$3{,}167.529586\ldots

Rounded to the nearest cent, the expected present value is $3,167.53. The probability of receiving a year-2 payment is $0.85$, not 1, because only those who survive year 1 reach the second year.

Common mistake avoided: Applying year-2 conditional probabilities directly to the original cohort overstates the year-2 contribution. The year-2 probabilities apply only to people who entered year 2 in the group.

When Exit Benefits Are Paid at the Exit Time

The end-of-year formula does not apply unchanged when a benefit is paid immediately at the random time of exit. Under a continuous-time competing-risks model with constant cause-specific forces $\mu_1,\ldots,\mu_m$ throughout the year, write $\mu=\sum_j\mu_j$. Assume the person begins in the group, exits are the only ways to leave it, and the benefit for cause $j$ is paid at the exact exit time. The probability of still being in the group at time $t$ is $e^{-\mu t}$; the probability of exit from cause $j$ during a small interval of length $dt$, conditional on still being present at $t$, is approximately $\mu_jdt$. Hence the unconditional cause-$j$ exit density at time $t$ is $\mu_j e^{-\mu t}$.

Let the force of interest be $\delta$, and let a survival benefit $B_S$ be paid at time 1 if the person remains in the group. Integrating the discounted exit payments over possible exit times and adding the discounted survival payment gives

\operatorname{EPV} =\sum_{j=1}^{m}B_j\int_0^1 e^{-\delta t}\mu_j e^{-\mu t}\,dt +B_Se^{-\delta}e^{-\mu}.

The integral combines the probability density of exit at time $t$ with the discount factor $e^{-\delta t}$. This expression relies on constant forces over the interval and exit benefits paid at the exact exit time; it is not the end-of-year benefit formula.

Worked Example: Exact-Time Exit Payments and a Year-End Survival Payment

A hypothetical one-year contract has two constant cause-specific forces: $\mu_A=0.10$ and $\mu_B=0.20$ per year. It pays $10,000 immediately on an A exit, $4,000 immediately on a B exit, and $2,000 at year-end if the person survives in the group. The effective annual interest rate is $i=0\%$, so $\delta=0$. Find the expected present value.

Method. Use the continuous exit densities for the immediate payments, then add the year-end survival benefit using the probability of remaining in the group through the year.

The total force is $\mu=0.10+0.20=0.30$. The survival probability is

p=e^{-0.30}=0.7408182207.

For completeness, the one-year exit probabilities are

q_A=\frac{0.10}{0.30}(1-e^{-0.30})=0.0863939264,\qquad q_B=\frac{0.20}{0.30}(1-e^{-0.30})=0.1727878529.

They reconcile with survival: $0.0863939264+0.1727878529+0.7408182207=1.0000000000$, subject to rounding.

With $\delta=0$, the common integral for the exit payments is

\int_0^1 e^{-0.30t}\,dt =\frac{1-e^{-0.30}}{0.30} =0.8639392643.

The two cause-specific present values are therefore

(10{,}000)(0.10)(0.8639392643)=\$863.9392643,

(4{,}000)(0.20)(0.8639392643)=\$691.1514114.

The survival payment has expected value $(\$2{,}000)(0.7408182207)=\$1{,}481.6364414$. Adding all three disjoint outcome contributions gives

\operatorname{EPV} =\$863.9392643+\$691.1514114+\$1{,}481.6364414 =\$3{,}036.7271171.

Rounded to the nearest cent, the expected present value is $3,036.73. Because the interest rate is zero, discounting does not alter the amounts; the different exit and survival probabilities still determine the expected value.

Common mistake avoided: Multiplying each exit probability by its benefit gives an undiscounted expectation only when the exit payments are treated as year-end payments. Here exits are paid at their exact times. The integral accounts for when each exit occurs.

Common Reconciliation Traps

A reliable check is to write down the outcome set before calculating. For an annual multiple-decrement model, verify that the conditional cause-specific probabilities and survival probability sum to 1. For multiple years, also verify that each later-year contribution is weighted by the probability of reaching that year. These checks concern the structure of the expected-value calculation; they do not repair probabilities or benefit definitions that were specified incorrectly.

Leaving out survival. If survival triggers a payment, it is part of the expected value even though no decrement occurs.
Counting survival twice. Each person must belong to exactly one outcome category for the period.
Mixing conditional and unconditional probabilities. A year-specific exit probability is generally conditional on being in the group at that year’s start. Multiply by the probability of reaching the year to obtain its unconditional contribution from the original cohort.
Discounting at the wrong time. A year-end payment uses a full-year discount factor; an immediate exit payment requires discounting to the exit time.
Confusing forces with probabilities. A cause-specific force is not itself a one-year exit probability. Under constant competing forces, the probability for cause $j$ over a year is $\frac{\mu_j}{\mu}(1-e^{-\mu})$, not simply $\mu_j$.

The partition is useful because it gives two independent ways to review a result: confirm that the probabilities account for all outcomes, and confirm that each outcome’s payment is weighted and discounted according to its trigger and timing. This discipline applies whether a calculation is written as a single expectation or as separate cause-contingent and survival contributions. The approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the relevant outcome partition before calculating. Answers are rounded to the nearest cent where needed.

A one-year benefit pays $9,000 on cause A, $3,000 on cause B, and $600 on survival to year-end. The respective cause probabilities are $0.06$ and $0.14$. If $i=3\%$ and all payments occur at year-end, what is the expected present value? (Answer: $1,398.06)
In a one-year model, $q_A=0.10$, $q_B=0.20$, and $i=0\%$. Payments are $5,000 on A, $2,500 on B, and $1,000 on survival, all at year-end. Find the expected present value. (Answer: $1,700.00)
A two-year contract pays at each year-end. In year 1, conditional exit probabilities are $0.05$ for A and $0.15$ for B; in year 2 they are $0.10$ for each cause. Each exit pays $4,000 for A or $2,000 for B, while survival pays $500 for that year. With $i=0\%$, find the expected value of all payments from time 0. (Answer: $1,700.00)
Under constant forces $\mu_A=0.20$ and $\mu_B=0.10$, with $i=0\%$, benefits of $3,000 and $6,000 are paid at the exact times of A and B exits. A $1,000 benefit is paid at year-end on survival. Find the expected present value. (Answer: $1,777.55)
In a one-year end-of-year model, $q_A=0.12$, $q_B=0.08$, and survival pays $1,000. The exit benefits are $10,000 for A and $4,000 for B, and $i=5\%$. What is the expected present value? Also find the mean exit benefit conditional on an exit occurring. (Answer: expected present value $2,209.52; conditional mean exit benefit $7,600.00)

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