Survivor Pensions Following Competing Active-Member Exits

Introduction

A survivor pension is payable only if the event that triggers it occurs. For an active pension-plan member, death before retirement may trigger a pension for a surviving spouse, while retirement or withdrawal may lead to different benefits—or no survivor pension of this kind. The expected present value must therefore reflect both the chance that death is the first active-member exit and the spouse’s chance of being alive to receive payments.

This tutorial values hypothetical continuous survivor pensions triggered by an active member’s death before retirement. It focuses on the death path: retirement and withdrawal reduce the chance of reaching that path, but are not treated as triggers for this pension. All examples assume the spouse is alive at the valuation date and that the member’s and spouse’s future lifetimes are independent.

Model and Valuation Formula

Definition: Suppose an active member can exit through death, retirement, or withdrawal, with constant forces $\lambda_d$, $\lambda_r$, and $\lambda_w$, respectively, during the coverage period $[0,n]$. The total active-member exit force is $q=\lambda_d+\lambda_r+\lambda_w$. Let the spouse’s constant force of mortality be $\mu$, and let the force of interest be $\delta$. If death at service duration $t$ triggers a continuous pension at rate $B(t)$ per year, payable immediately after death while the spouse is alive, the expected present value is

\operatorname{EPV} = \int_0^n \lambda_d e^{-qt} e^{-\mu t} e^{-\delta t} \left(\int_0^\infty B(t)e^{-\mu s}e^{-\delta s}\,ds\right)dt.

Here $t$ is the time from valuation to the member’s death, and $s$ is time from that death to a later pension payment. The factor $\lambda_d e^{-qt}$ is the density that the member’s first active exit is death at time $t$. The factor $e^{-\mu t}$ requires the spouse to survive from the valuation date to that time; the inner integral then allows for the spouse’s survival after the member dies. The discount factors account for both periods.

Since $B(t)$ is fixed once the member dies at time $t$, the inner integral is $B(t)/(\delta+\mu)$. Thus the formula simplifies to:

\operatorname{EPV} = \frac{\lambda_d}{\delta+\mu} \int_0^n B(t)e^{-(q+\mu+\delta)t}\,dt.

This form applies when the forces are constant, the spouse is alive initially, the lifetimes are independent, the pension begins immediately at the member’s death, and it is paid continuously while the spouse is alive. It does not apply unchanged if payments are deferred, are made at discrete dates, or depend on a spouse’s status at the member’s death in a way not represented by these assumptions.

Why the Formula Holds

The valuation can be justified by conditioning on the time and cause of the member’s first exit. Over a short interval from $t$ to $t+dt$, the probability that the member remains active until $t$ and then dies is approximately $e^{-qt}\lambda_d\,dt$. Independent spouse mortality gives probability $e^{-\mu t}$ that the spouse is alive at that time. The pension commencing at $t$ has value at time $t$ equal to $B(t)/(\delta+\mu)$: each future payment is discounted and weighted by the spouse’s probability of survival.

Discounting that value back to time zero contributes another factor $e^{-\delta t}$. Multiplying these terms and integrating over all possible death times from 0 to $n$ gives the valuation formula. In particular, retirement and withdrawal belong in $q$, even though neither cause triggers this survivor pension.

Find the death-at-$t$ density.
The member must remain active to $t$, with probability $e^{-qt}$, and then die with force $\lambda_d$. The density is $\lambda_d e^{-qt}$.

Require the spouse to be alive.
Under independence and constant spouse mortality, the probability of spouse survival to $t$ is $e^{-\mu t}$. After the member’s death, the expected present value of a continuous pension is $B(t)/(\delta+\mu)$.

Discount and integrate over death times.
Discount from $t$ to zero by $e^{-\delta t}$, multiply by the death density and spouse survival probability, and integrate from 0 to $n$.

Worked Valuations

Worked Example: Constant Pension with Three Competing Exits

A hypothetical plan covers an active member for five years. The constant forces are $\lambda_d=0.02$ for death, $\lambda_r=0.03$ for retirement, and $\lambda_w=0.01$ for withdrawal. Death before retirement or withdrawal triggers a continuous pension of $8,000 per year to the spouse, beginning immediately and continuing while the spouse is alive. The spouse’s force of mortality is $\mu=0.03$, and the force of interest is $\delta=0.05$. Find the expected present value. Assume independent lifetimes and that the spouse is alive at time zero.

Method. Use the constant-benefit formula, retaining all three active-member exit forces in the probability of remaining active. First,

q=0.02+0.03+0.01=0.06, \qquad q+\mu+\delta=0.06+0.03+0.05=0.14.

The value at the member’s death of the continuous pension while the spouse is alive is $8{,}000/(0.05+0.03)=100{,}000$. Therefore,

\begin{aligned} \operatorname{EPV} &=\frac{0.02}{0.05+0.03} \int_0^5 8{,}000e^{-0.14t}\,dt\\ &=\frac{8{,}000(0.02)}{0.08} \left(\frac{1-e^{-0.14(5)}}{0.14}\right)\\ &=2{,}000\left(\frac{1-e^{-0.70}}{0.14}\right). \end{aligned}

Since $e^{-0.70}=0.4965853$, the integral factor is $(1-0.4965853)/0.14=3.595819$. Hence the expected present value is $2{,}000(3.595819)=\$7{,}191.64$, rounded to the nearest cent.

Check. The pension’s value at death is $100,000, but death is not certain: the member may retire or withdraw first, and the spouse may die before or after the member. An expected value substantially below the at-death pension value is consistent with these contingencies.

Common mistake avoided. Using only the death force in active-member survival would omit retirement and withdrawal and overstate the probability of death before another exit.

Worked Example: Pension Amount Accruing with Service

A hypothetical plan covers a member until $n=\ln(2)/0.10$ years. If the member dies while active at service duration $t$, the spouse receives a continuous pension at rate $B(t)=1{,}000t$ dollars per year. The forces are $\lambda_d=0.02$, $\lambda_r=0.05$, and $\lambda_w=0.03$ for active-member death, retirement, and withdrawal. The spouse’s force of mortality is $\mu=0.03$, and the force of interest is $\delta=0.02$. Find the expected present value. Assume the spouse is alive initially and lifetimes are independent.

Method. Because the pension rate depends on service at death, retain $B(t)$ inside the integral. The total active exit force is $q=0.02+0.05+0.03=0.10$, so the combined exponent is $q+\mu+\delta=0.10+0.03+0.02=0.15$. Also, $n=\ln(2)/0.10=6.9314718$ years. Thus,

\operatorname{EPV} = \frac{0.02}{0.02+0.03} \int_0^n 1{,}000t e^{-0.15t}\,dt = 400\int_0^n t e^{-0.15t}\,dt.

Integration by parts gives $\int_0^n t e^{-kt}\,dt=[1-(1+kn)e^{-kn}]/k^2$, for $k>0$. Here $kn=0.15\ln(2)/0.10=1.0397208$, and $e^{-kn}=e^{-1.0397208}=0.3535534$. Therefore,

\begin{aligned} \int_0^n t e^{-0.15t}\,dt &= \frac{1-(1+0.15n)e^{-0.15n}}{0.15^2}\\ &= \frac{1-(1+1.0397208)(0.3535534)}{0.0225}\\ &= \frac{1-0.7211502}{0.0225} =12.39332. \end{aligned}

The expected present value is $400(12.39332)=\$4{,}957.33$, rounded to the nearest cent.

Check. The service-dependent pension is zero at $t=0$ and grows with the member’s service, so later deaths trigger larger pension rates. The calculation still weights those later deaths by the probability that the member remains active and the spouse remains alive.

Common mistake avoided. Replacing $B(t)=1{,}000t$ by a fixed amount loses the service dependence. The factor $t$ must remain in the integrand.

Worked Example: Death Force Changes Partway Through Coverage

A hypothetical plan covers a member for four years. During years 0 to 2, the forces of death, retirement, and withdrawal are $0.01$, $0.02$, and $0.02$. During years 2 to 4, they are $0.03$, $0.01$, and $0.01$. Death while active triggers a continuous pension of $6,000 per year to the spouse, beginning immediately and payable while the spouse is alive. The spouse’s force of mortality is $0.02$, and the force of interest is $0.04$. Assume the spouse is alive initially and lifetimes are independent. Find the expected present value.

Method. Apply the constant-force valuation separately over the two intervals. In each interval the total active exit force is $q=0.05$, so the combined exponent is $q+\mu+\delta=0.05+0.02+0.04=0.11$. The death force differs by interval, so it must be applied separately to each integral. The at-death pension value is $6{,}000/(0.04+0.02)=100{,}000$.

\begin{aligned} \operatorname{EPV} &=100{,}000\left[ 0.01\int_0^2 e^{-0.11t}\,dt + 0.03\int_2^4 e^{-0.11t}\,dt \right]\\ &=100{,}000\left[ \frac{0.01(1-e^{-0.22})}{0.11} + \frac{0.03(e^{-0.22}-e^{-0.44})}{0.11} \right]. \end{aligned}

Using $e^{-0.22}=0.8025188$ and $e^{-0.44}=0.6440364$, the first term in brackets is $0.01(0.1974812)/0.11=0.01795284$. The second is $0.03(0.1584824)/0.11=0.04322247$. Their sum is $0.06117531$, giving an expected present value of $\$6{,}117.53$, rounded to the nearest cent.

Check. The active exit force stays at $0.05$ in both intervals, so the survival exponent is unchanged at the boundary. The death intensity triples in the second interval, which is reflected by using $0.03$, not $0.01$, in that interval’s death integral.

Common mistake avoided. A constant death force cannot be used across the full four years when the stated force changes at year 2. Split the valuation at the change and use the appropriate cause-specific force on each interval.

Common Exam Trap: Counting the Wrong Exit Path

The survivor pension in these examples is triggered by death before retirement or withdrawal. Its value is not the value of a pension paid after any active-member exit. Retirement and withdrawal are competing exits: each removes the member from active status and therefore prevents this particular death-triggered benefit.

For example, in the first worked example, an incorrect calculation might use only the death force $0.02$ as the member’s exit force. That gives the wrong combined exponent $0.02+0.03+0.05=0.10$ and the incorrect value

\frac{8{,}000(0.02)}{0.08} \left(\frac{1-e^{-0.10(5)}}{0.10}\right) = 2{,}000\left(\frac{1-e^{-0.50}}{0.10}\right) = \$7{,}869.39.

The correction is to include all active-member exit forces in $q$, because the member must avoid death, retirement, and withdrawal until the death time. Thus $q=0.02+0.03+0.01=0.06$, and the correct combined exponent is $0.06+0.03+0.05=0.14$. The correct value is $\$7{,}191.64$, as calculated above.

A second timing check is equally important: spouse mortality applies both before the member’s death and during the pension. In the simplified formula, the former appears in the combined exponent $q+\mu+\delta$, while the latter appears in the pension-value denominator $\delta+\mu$. Omitting either effect overstates the value.

Key takeaway: For a death-triggered survivor pension, use the death force to identify the triggering cause, the sum of all active exit forces to model continued active status, and spouse survival both to the member’s death and throughout subsequent payments.

Check Your Understanding

For each problem, assume independent lifetimes, a spouse alive at time zero, and a continuous pension beginning immediately at the member’s death and payable while the spouse is alive. Unless stated otherwise, forces are constant and all money amounts are in dollars. Round final amounts to the nearest cent.

A member is covered for three years. The death, retirement, and withdrawal forces are $0.02$, $0.03$, and $0.01$; the spouse’s force of mortality is $0.02$; and $\delta=0.04$. Death triggers a level pension of $4,000 per year. Find its expected present value. (Answer: $3,359.15)
A member is covered for four years. The death force is $0.01$, the retirement force is $0.025$, and the withdrawal force is $0.015$. The spouse’s force of mortality is $0.02$, and $\delta=0.03$. The pension rate at death is $B(t)=500t$ dollars per year. Find the expected present value. (Answer: $615.52)
Coverage lasts two years. In each year the total active exit force is $0.04$; the death force is $0.02$ during the first year and $0.01$ during the second. The pension rate at death is $5,000 per year, the spouse’s force of mortality is $0.02$, and $\delta=0.02$. Find the expected present value. (Answer: $3,511.56)
A member is covered for five years, with death, retirement, and withdrawal forces $0.015$, $0.025$, and $0.01$. The spouse’s force of mortality and $\delta$ are both $0.02$ and $0.03$, respectively. Death triggers a level pension of $6,000 per year. Find the expected present value. (Answer: $7,082.45)
A member is covered for two years with constant death, retirement, and withdrawal forces $0.02$, $0.03$, and $0.01$. The spouse’s force of mortality is $0.03$, and $\delta=0.02$. The pension rate at death is $B(t)=1{,}000+200t$ dollars per year. Find the expected present value. (Answer: $856.47)

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