Integrated Case Checks for Probability and Cash-Flow Consistency

Why Integrated Case Checks Matter

A multi-state valuation can contain individually plausible probabilities and benefit amounts yet still produce an incorrect value. A transition row may fail to sum to one; a death benefit may be applied to people who remain alive; or a payment made at the end of a year may be discounted as though it were paid at the beginning. An effective check therefore examines the probabilities, benefit triggers, and payment dates together.

This tutorial develops a repeatable audit for hypothetical disability, long-term-care, and pension cases. The aim is not to replace the valuation method, but to test whether its parts fit together. The checks are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: A state probability is the probability that a life is in a specified state at a specified time. A transition probability is the probability of moving from one state to another over a stated period, conditional on the starting state. An exit-cause probability is the probability of leaving a state during that period through one particular cause, conditional on being in the state at its start.

These quantities answer different questions. For example, the probability of being retired at year-end can include both people newly retired during the year and people already retired who remain so. It is not, by itself, the probability of retirement as an exit cause from the active state.

A Consistency Framework

For a discrete-time model, list all possible destinations from each starting state. If state $j$ is reached from state $i$ with probability $p_{ij}$, then a complete transition row must satisfy

\sum_j p_{ij}=1.

If the starting-state probabilities are $\pi_i$, the probability of destination state $j$ at the next time is $\sum_i \pi_i p_{ij}$. The resulting destination probabilities must sum to one when the initial probabilities sum to one and all destinations, including absorbing states such as death, are included.

For continuous-time models, the corresponding check is that the outgoing transition rates from a state are properly specified. If the only exits from a state are to distinct states with constant rates $\lambda_1,\ldots,\lambda_m$, the total exit rate is $\lambda=\sum_k\lambda_k$. The probability of still being in the state after time $t$ is $e^{-\lambda t}$, and the probability of exit through cause $k$ by time $t$ is $(\lambda_k/\lambda)(1-e^{-\lambda t})$.

Key check: A benefit is payable only on the event stated in its definition. A benefit payable while disabled depends on disability occupancy; a recovery refund depends on a recovery transition; and a death benefit depends on a death transition. Do not substitute one probability for another merely because both involve the same state.

The timing check is equally direct. A payment at time $t$ is discounted from time $t$, not from the time the policy entered a state or the time the valuation began. With annual effective interest rate $i$, its present-value factor is $v^t=(1+i)^{-t}$. Under force of interest $\delta$, the factor is $e^{-\delta t}$.

Deriving the Continuous-Time Exit and Payment Check

Suppose a person starts in a care state. Recovery and death are the only exits, with constant transition rates $\lambda_R$ and $\lambda_D$, respectively. Let $\lambda=\lambda_R+\lambda_D$. Over a short interval of length $dt$, the probability of remaining in care is approximately $1-\lambda dt$. Thus the occupancy probability $p_C(t)$ satisfies $p_C(t+dt)\approx p_C(t)(1-\lambda dt)$, which gives the differential equation $p_C'(t)=-\lambda p_C(t)$, with $p_C(0)=1$. Its solution is $p_C(t)=e^{-\lambda t}$.

Conditional on still being in care at time $t$, the probability of recovery during the next short interval is approximately $\lambda_Rdt$; the analogous death probability is $\lambda_Ddt$. Therefore the unconditional probabilities of those transitions in that interval are approximately $e^{-\lambda t}\lambda_Rdt$ and $e^{-\lambda t}\lambda_Ddt$. This explains both the cause allocation and the expected-value integrals.

P(\text{recovery by }T)=\int_0^T \lambda_R e^{-\lambda t}\,dt =\frac{\lambda_R}{\lambda}(1-e^{-\lambda T}),

\operatorname{EPV}(\text{death benefit }B) =\int_0^T B\lambda_D e^{-(\lambda+\delta)t}\,dt.

The second formula assumes a death benefit $B$ is paid immediately at death, a constant force of interest $\delta$, and constant transition rates over the valuation period. It does not apply unchanged to a benefit paid only at year-end, to rates that vary with time, or to a death benefit payable only after another condition is met.

Worked Example: Auditing Continuous Care Benefits and Exit Payments

A hypothetical insured starts in a care state for a one-year period. Recovery occurs at constant rate $0.20$ per year and death at constant rate $0.10$ per year. The contract pays continuously at $\$12{,}000$ per year while the insured is in care, $\$2{,}000$ immediately on recovery, and $\$10{,}000$ immediately on death. The force of interest is $\delta=0.05$. Find the exit probabilities by cause and the expected present value of all three benefits.

Method and assumptions. The rates are constant, recovery and death are the only exits, and transition benefits are paid at the transition time. The total exit rate is $0.20+0.10=0.30$. The probability of remaining in care for one year is

e^{-0.30}=0.7408182.

The probability of leaving care during the year is $1-0.7408182=0.2591818$. Allocating that exit probability by rate gives recovery probability $(0.20/0.30)(0.2591818)=0.1727879$ and death probability $(0.10/0.30)(0.2591818)=0.0863939$. Check: $0.7408182+0.1727879+0.0863939=1.0000000$.

For the present value, let

J=\int_0^1 e^{-(0.30+0.05)t}\,dt =\frac{1-e^{-0.35}}{0.35} =\frac{1-0.7046881}{0.35} =0.8437483.

The care payment has expected present value $12{,}000J=\$10{,}124.98$. The recovery payment has expected present value $2{,}000(0.20)J=400J=\$337.50$. The death payment has expected present value $10{,}000(0.10)J=1{,}000J=\$843.75$. Thus the total is $\$10{,}124.98+\$337.50+\$843.75=\$11{,}306.23$, rounded to the nearest cent.

As a cross-check, the integrand combines the care rate and the transition-benefit rates: $(12{,}000+400+1{,}000)J=13{,}400(0.8437483)=\$11{,}306.23$. The component calculation and combined calculation agree. A common mistake is to use the death probability for the continuous care payment; care payments instead depend on time spent in care.

Worked State and Cash-Flow Audits

The next example shows why starting-state populations must be propagated separately. The final state probability for retirement is assembled from all routes into retirement; it is not just the active members’ retirement exit probability.

Worked Example: Checking a One-Year Pension Transition Table

A hypothetical pension group begins with probability $0.80$ active and $0.20$ retired. During the year, an active member has probabilities $0.78$ of remaining active, $0.12$ of retiring, $0.06$ of withdrawing, and $0.04$ of dying. A retired member has probability $0.90$ of remaining retired and $0.10$ of dying. At year-end, the plan pays $\$6{,}000$ to each person then retired, including those newly retired, $\$4{,}000$ to each withdrawing active member, and $\$20{,}000$ on each death. Find the year-end state probabilities and the present value of the expected payments at $i=0.04$.

Method. First check each transition row: $0.78+0.12+0.06+0.04=1$ for active members, and $0.90+0.10=1$ for retired members. Then multiply each transition probability by the probability in its starting state.

The year-end active probability is $0.80(0.78)=0.624$. The year-end retired probability is $0.80(0.12)+0.20(0.90)=0.096+0.180=0.276$. The withdrawal probability is $0.80(0.06)=0.048$. The death probability is $0.80(0.04)+0.20(0.10)=0.032+0.020=0.052$. These outcomes check: $0.624+0.276+0.048+0.052=1.000$.

The expected year-end pension payment is $0.276(\$6{,}000)=\$1{,}656$. The expected withdrawal refund is $0.048(\$4{,}000)=\$192$, and the expected death payment is $0.052(\$20{,}000)=\$1{,}040$. Total expected payment is $\$1{,}656+\$192+\$1{,}040=\$2{,}888$. All payments occur at the end of year one, so

\operatorname{EPV}=\frac{\$2{,}888}{1.04}=\$2{,}776.92.

The answer is rounded to the nearest cent. The active group contributes $0.032$ to deaths and the retired group contributes $0.020$; the total death probability is $0.052$, not twice either contribution. A further trap is counting the $\$6{,}000$ pension for every active member who retires during the year as well as for every person retired at year-end. The contract pays once per eligible person at year-end, so the correct probability is $0.276$, not $0.096+0.276$.

In a multi-period case, perform the probability check at every valuation date before using the probabilities to weight benefits. The following example keeps the transition rules fixed for two years and distinguishes a new-entry payment from an ongoing state payment.

Worked Example: Reconciling a Two-Year Disability Valuation

A hypothetical insured starts healthy. At each year-end, a healthy person is healthy with probability $0.80$, newly disabled with probability $0.15$, and dead with probability $0.05$. A disabled person is healthy with probability $0.20$, disabled with probability $0.70$, and dead with probability $0.10$. Death is absorbing. At each year-end, the contract pays $\$10{,}000$ if the insured is disabled then, $\$3{,}000$ on a healthy-to-disabled transition during that year, and $\$15{,}000$ on a death transition during that year. Payments are not made after death. Find the expected present value for two years at $i=0.05$.

Method. Propagate state probabilities year by year and calculate each payment from its own trigger. Both transition rows sum to one: $0.80+0.15+0.05=1$ and $0.20+0.70+0.10=1$.

At year one, the probabilities of healthy, disabled, and dead are $0.80$, $0.15$, and $0.05$. At year two, the healthy probability is $0.80(0.80)+0.15(0.20)=0.640+0.030=0.670$. The disabled probability is $0.80(0.15)+0.15(0.70)=0.120+0.105=0.225$. The dead probability is $0.80(0.05)+0.15(0.10)+0.05=0.040+0.015+0.050=0.105$. The year-two probabilities sum to $0.670+0.225+0.105=1.000$.

At year one, the expected disability-state payment is $0.15(\$10{,}000)=\$1{,}500$. The entry payment is $0.15(\$3{,}000)=\$450$, and the death payment is $0.05(\$15{,}000)=\$750$. Total expected payment at year one is $\$2{,}700$.

For year two, the disability-state payment is $0.225(\$10{,}000)=\$2{,}250$. The new-entry probability is the probability of being healthy at year one times the healthy-to-disabled transition probability: $0.80(0.15)=0.120$. Its payment is $0.120(\$3{,}000)=\$360$. The year-two death-transition probability is $0.80(0.05)+0.15(0.10)=0.040+0.015=0.055$, giving a payment of $0.055(\$15{,}000)=\$825$. Thus year-two expected payment is $\$2{,}250+\$360+\$825=\$3{,}435$.

Discount each total from its own year-end date:

\operatorname{EPV}=\frac{\$2{,}700}{1.05}+\frac{\$3{,}435}{1.05^2} =\$2{,}571.4286+\$3{,}115.6463 =\$5{,}687.07.

The result is rounded to the nearest cent. The death check also reconciles: the cumulative probability of death by year two is $0.050+0.055=0.105$, exactly the year-two dead-state probability. A common error is to use the year-two healthy probability $0.670$ to calculate new disability entries; entries during year two start from the year-one healthy probability $0.800$.

Common Audit Traps

Mixing state probabilities and exit probabilities. A year-end state can contain both entrants and people who remained in that state. Identify the starting state for every transition probability.
Leaving out a destination. A row that omits death, withdrawal, recovery, or another modeled destination may not sum to one. Do not silently assign the remainder to survival unless that interpretation is part of the assumptions.
Double-counting mutually exclusive outcomes. A person who dies during the year is not also in the year-end living state. Apply each benefit to its stated event and check whether two triggers can occur together.
Using a state probability for a transition benefit. The probability of being disabled at year-end is generally not the probability of becoming disabled during the year.
Discounting from the wrong date. A payment at the end of year two uses $v^2$, not $v$. An immediate transition payment in a continuous-time model requires discounting at the transition time inside the integral.
Changing the payment convention mid-calculation. A continuous annuity while in a state and a lump sum at a transition have different probability and timing structures.

Practical takeaway: For each valuation date, first reconcile the probability mass; then identify the exact state or transition that triggers each cash flow; finally discount from the contractual payment date. A total expected present value is credible only when all three checks agree.

Check Your Understanding

For each problem, check the probability totals and apply the stated payment timing.

A person starts in care. Recovery and death occur at constant rates $0.12$ and $0.08$ per year, with no other exits. Find the probabilities of remaining in care, recovering, and dying by the end of one year. Give probabilities to four decimals. (Answer: remaining $0.8187$, recovery $0.1088$, death $0.0725$.)
A pension group starts with probabilities $0.60$ active and $0.40$ retired. Active transitions are active $0.75$, retirement $0.15$, withdrawal $0.06$, and death $0.04$. Retired transitions are retired $0.92$ and death $0.08$. Find the year-end probabilities in active, retired, withdrawal, and death outcomes. (Answer: active $0.450$, retired $0.458$, withdrawal $0.036$, death $0.056$.)
A single benefit of $\$4{,}200$ is payable exactly at the end of year two, with annual effective interest $i=0.03$. Find its present value to the nearest cent. (Answer: $\$3{,}958.90$.)
A healthy person has annual transitions healthy $0.70$, disabled $0.20$, death $0.10$; a disabled person has transitions healthy $0.25$, disabled $0.65$, death $0.10$. The person starts healthy. Find the probability of death during year two and the cumulative probability of death by the end of year two. (Answer: year-two death $0.090$; cumulative death $0.190$.)
A person starts disabled for a one-year period. At year-end the probabilities are disabled $0.65$, recovered $0.30$, and dead $0.05$. The contract pays $\$9{,}000$ if disabled at year-end, $\$1{,}000$ on recovery, and $\$12{,}000$ on death, all at year-end. At $i=0.04$, find the expected present value to the nearest cent. (Answer: $\$6{,}490.38$.)

What You'll Learn