Deriving Discrete Cause-Contingent EPV

Why Partition by Exit Year and Cause?

A pension member may leave active service through several competing causes. A hypothetical plan might pay one lump sum for retirement, another for withdrawal, and a survivor benefit following death in service. If each benefit is paid at the end of the year in which the first exit occurs, its expected present value depends on two probabilities: the chance the member is still active when that year begins, and the conditional chance of exiting from the relevant cause during that year.

The key is to separate those events without counting the same member in more than one year. The member can receive a first-exit benefit for only one cause and one year. This tutorial derives that separation and applies it to discrete annual valuations. The treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Consider a member active at time 0 and annual periods $k=1,2,\ldots,n$. Let $p_k$ be the conditional probability of remaining active throughout year $k$, given active status at the start of that year. For each modeled exit cause $j$, let $q_k^{(j)}$ be the conditional probability of a first exit from cause $j$ during year $k$, given active status at the start of that year. Let $B_k^{(j)}$ be the benefit payable at the end of year $k$ if cause $j$ is the first exit cause in that year.

All probabilities in a given year are conditional on being active at its start. If the modeled causes do not include every possible exit, $p_k$ must still allow for all causes that end active membership. Thus, the modeled cause probabilities and continuation probability satisfy

p_k+\sum_{j\in\mathcal{J}}q_k^{(j)}+r_k=1,

where $\mathcal{J}$ is the set of modeled benefit causes and $r_k$ is the conditional probability of any unmodeled exit that ends active membership. If all exit causes are modeled, $r_k=0$. An unmodeled exit can have no benefit in the valuation, but it cannot be treated as active continuation. If the model is intentionally restricted to listed causes, state the assumption that every outcome outside those causes is continued active membership.

Assume a constant effective annual interest rate $i$, with $v=1/(1+i)$. Benefits are paid at the end of the exit year, whether or not the exit occurs at an earlier point within that year. The benefit amounts may vary by year and cause, but are fixed conditional on that year and cause for purposes of this formula.

Deriving the Discrete Cause-Contingent EPV

The probability of being active at the start of year $k$ is the product of the probabilities of continuing through each previous year. Conditional probabilities are used successively: to reach year 2 the member must continue through year 1; to reach year 3 the member must continue through years 1 and 2. Therefore,

{}_{k-1}P_{\mathrm{active}} = \prod_{r=1}^{k-1}p_r, \qquad \prod_{r=1}^{0}p_r=1.

To receive benefit $B_k^{(j)}$, the member must be active at the beginning of year $k$, then exit from cause $j$ during that year. The probability of this joint event is the probability of reaching year $k$ multiplied by the conditional cause probability $q_k^{(j)}$. The benefit is paid at time $k$, so its present value is $v^k B_k^{(j)}$.

Reach the year.
The probability of being active at the start of year $k$ is $\prod_{r=1}^{k-1}p_r$.

Select the first-exit cause.
Given active status at the start of year $k$, the probability of cause $j$ is $q_k^{(j)}$. The joint event probability is $\left(\prod_{r=1}^{k-1}p_r\right)q_k^{(j)}$.

Discount and sum.
Discount the end-of-year benefit by $v^k$, then sum over all modeled causes and years.

Theorem: Under the annual first-exit model and end-of-exit-year payment timing described above, the expected present value at time 0 is

\operatorname{EPV} = \sum_{k=1}^{n}v^k \left(\prod_{r=1}^{k-1}p_r\right) \sum_{j\in\mathcal{J}}q_k^{(j)}B_k^{(j)}.

The sum includes benefits only for modeled causes. All exits that end active membership, including unmodeled exits, must nevertheless be reflected in each $p_k$.

This expression follows directly by partitioning outcomes into mutually exclusive events: first exit in year $k$ from cause $j$. On each such event, the present value is $v^kB_k^{(j)}$; otherwise that event contributes nothing. The expectation of the benefit present value is therefore the sum of each event’s present value multiplied by its probability. Because a first exit can occur in only one year and from only one cause, no benefit event is counted twice.

Worked Applications

Worked Example: Two Benefit Causes Over Three Years

A hypothetical pension arrangement pays $5,000 on retirement and $2,000 on withdrawal, at the end of the year of first exit. The annual effective interest rate is $i=5\%$. Conditional on being active at the start of each year, the retirement and withdrawal probabilities are, respectively, $(0.06,0.04)$ in year 1, $(0.12,0.08)$ in year 2, and $(0.18,0.12)$ in year 3. Assume these are the only exit causes. Calculate the EPV.

Method. Find each year’s continuation probability, then multiply each year’s expected conditional benefit by the probability of reaching that year and by its discount factor.

Year 1. The continuation probability is $p_1=1-0.06-0.04=0.90$. The conditional expected benefit is $0.06(5000)+0.04(2000)=300+80=380$. The year 1 contribution is $v(380)=380/1.05=361.9048$.

Year 2. The continuation probability is $p_2=1-0.12-0.08=0.80$, so the probability of reaching year 2 is $p_1=0.90$. The conditional expected benefit is $0.12(5000)+0.08(2000)=600+160=760$. The year 2 contribution is $v^2(0.90)(760)=684/1.05^2=620.4082$.

Year 3. The continuation probability is $p_3=1-0.18-0.12=0.70$. The probability of reaching year 3 is $p_1p_2=(0.90)(0.80)=0.72$. The conditional expected benefit is $0.18(5000)+0.12(2000)=900+240=1140$. The year 3 contribution is $v^3(0.72)(1140)=820.8/1.05^3=709.0379$.

Adding the contributions gives $361.9048+620.4082+709.0379=1691.3509$. The EPV is $1,691.35, rounded to the nearest cent.

Check and mistake avoided. The reach probabilities are $1$, $0.90$, and $0.90(0.80)=0.72$, so they decrease as exits accumulate. Using the year 3 cause probabilities without first multiplying by $0.72$ would value year 3 as though every member were still active.

Worked Example: Unmodeled Exits Still Reduce Later Benefits

A hypothetical plan pays $3,000 for withdrawal and $10,000 for death in service at the end of the exit year. The effective annual interest rate is $4\%$. In year 1, the conditional probabilities of withdrawal, death, and disability exit are $0.10$, $0.02$, and $0.03$; disability has no benefit in this valuation. In year 2, the corresponding probabilities are $0.08$, $0.04$, and $0.02$. Calculate the EPV for the two modeled benefits.

Method. Include disability in the calculation of active continuation, but omit it from the benefit sum because no disability benefit is being valued here.

In year 1, $p_1=1-0.10-0.02-0.03=0.85$. The conditional expected modeled benefit is $0.10(3000)+0.02(10000)=300+200=500$. Thus the year 1 contribution is $500/1.04=480.7692$.

In year 2, $p_2=1-0.08-0.04-0.02=0.86$, although $p_2$ is not needed for a benefit payable only through year 2. The probability of reaching year 2 is $p_1=0.85$. Its conditional expected modeled benefit is $0.08(4000)+0.04(12000)=320+480=800$, using the year 2 benefit amounts specified here: $4,000 for withdrawal and $12,000 for death. The year 2 contribution is $0.85(800)/1.04^2=680/1.0816=628.6982$.

The total is $480.7692+628.6982=1109.4674$, so the EPV is $1,109.47, rounded to the nearest cent.

Check and mistake avoided. Year 1 has an unmodeled disability exit probability of $0.03$, so its continuation probability is $0.85$, not $1-0.10-0.02=0.88$. Omitting disability from the benefit sum does not make disabled members active in year 2.

Worked Example: Solving for a Later-Year Benefit

A hypothetical two-year arrangement has no interest. In year 1, the conditional probabilities of cause A and cause B are $0.10$ and $0.15$; both are exits, but only cause A pays a benefit, of $1,000. In year 2, conditional on being active at its start, the probability of cause A is $0.20$, and cause A pays an unknown benefit $X$. Cause B pays nothing in either year. The EPV is $475. Find $X$.

Method. Since the interest rate is $0\%$, each discount factor is 1. First calculate the year 1 continuation probability, then write the EPV as the sum of the cause A contributions.

Year 1 continuation is $p_1=1-0.10-0.15=0.75$. The year 1 benefit contribution is $0.10(1000)=100$. The probability-weighted year 2 benefit contribution is $0.75(0.20)X=0.15X$. Therefore,

475=100+0.15X,\qquad X=\frac{475-100}{0.15}=\frac{375}{0.15}=2500.

Thus the required year 2 cause A benefit is $2,500. The check is $100+0.15(2500)=100+375=475$, as required.

Check and mistake avoided. Using $0.90$ as the probability of reaching year 2 would ignore year 1 cause B, which also ends active membership. That incorrect calculation would give $475=100+0.90(0.20)X$, or $X\approx $2,083.33$, and would understate the benefit required.

Timing, Assumptions, and Common Traps

The exponent on the discount factor is determined by the payment date, not merely by the year in which the exit occurs. In this model, a benefit paid at the end of year $k$ is discounted by $v^k$. A benefit payable immediately at the moment of exit, or at a different fixed date, requires a different timing treatment; the end-of-year formula should not be used without adjustment.

Item	Correct treatment	Common error
Conditional exit probability	Apply $q_k^{(j)}$ only after accounting for the probability of reaching year $k$.	Treat $q_k^{(j)}$ as an unconditional probability from time 0.
Unmodeled exit	Include it in the calculation of $p_k$ if it ends active membership.	Count an unmodeled exit as continued active service.
Payment at year-end	Discount by $v^k$ for a payment at the end of year $k$.	Use $v^{k-1}$, as if the payment were at the start of that year.
Cause-specific benefit	Multiply the benefit by the probability of its own cause.	Apply the combined exit probability to every cause’s benefit.

The year-specific probabilities must be on a consistent conditional basis: each $q_k^{(j)}$ and $p_k$ is conditional on active status at the start of year $k$. The formula assumes the annual transition probabilities and benefit amounts are specified for the relevant year and status. If the benefit depends on another feature, such as service at exit, the appropriate amount must be used for that year and cause before summing.

Key takeaway: For a cause-contingent benefit paid at the end of the first-exit year, multiply its conditional cause probability by the probability of reaching that year, multiply by the benefit, and discount by $v^k$. Continuation probabilities must reflect every exit that ends active membership, whether or not that exit has a modeled benefit.

Check Your Understanding

For each problem, use the annual first-exit model and the stated payment timing.

A two-year model has $i=0$. In year 1, cause A and cause B probabilities are $0.10$ and $0.20$; their benefits are $1,000 and $2,000. In year 2, their probabilities are $0.20$ and $0.10$; their benefits are $3,000 and $0. Find the EPV. (Answer: $920)
In one year, two exhaustive exit causes have probabilities $0.12$ and $0.08$, with end-of-year benefits of $8,000 and $5,000. If $i=5\%$, find the EPV, rounded to the nearest cent. (Answer: $1,295.24)
In year 1, retirement, death, withdrawal, and unmodeled disability exits have conditional probabilities $0.05, 0.02, 0.08,$ and $0.10$. In year 2, assume the member is active at its start and the probabilities of retirement, death, and withdrawal are $0.10, 0.05,$ and $0.15$, with no other exits. With $i=0$, year 1 benefits are $4,000, $10,000, and $1,000 for retirement, death, and withdrawal; year 2 benefits are $5,000, $12,000, and $2,000. Find the EPV. (Answer: $1,530)
With $i=0$, year 1 probabilities for cause A and cause B are $0.10$ and $0.10$; cause A pays $2,000 and cause B pays nothing. In year 2, cause A has conditional probability $0.25$ and pays $X$; cause B has probability $0.10$ and pays nothing. If the EPV is $600, find $X$. (Answer: $2,000)
In one year, modeled causes have probabilities $0.07$ and $0.03$, while an unmodeled exit ending active membership has probability $0.04$. What is $p_1$, and what total probability should be assigned to continued active membership? (Answer: $p_1=0.86$; the continuation probability is $0.86$, not $0.90$)

Item	Correct treatment	Common error
Conditional exit probability	Apply \(q_k^{(j)}\) only after accounting for the probability of reaching year \(k\).	Treat \(q_k^{(j)}\) as an unconditional probability from time 0.
Unmodeled exit	Include it in the calculation of \(p_k\) if it ends active membership.	Count an unmodeled exit as continued active service.
Payment at year-end	Discount by \(v^k\) for a payment at the end of year \(k\).	Use \(v^{k-1}\), as if the payment were at the start of that year.
Cause-specific benefit	Multiply the benefit by the probability of its own cause.	Apply the combined exit probability to every cause’s benefit.

What You'll Learn