Payments Based on State Occupancy Duration

Why Occupancy Duration Matters

Some hypothetical disability income and long-term care contracts make a benefit payable only after a person has spent a specified period continuously in an eligible state. For example, a contract might pay a lump sum after two uninterrupted years of care, or begin a monthly payment only after a qualifying period in a disabled state. In these designs, being in the eligible state at a particular time is not enough: the benefit also depends on how long the current, uninterrupted spell in that state has lasted.

This condition changes the valuation problem. A standard state probability gives the chance that a person is in a state at a given time. It generally does not tell us whether the person has remained there continuously since the start of the qualifying spell. A transition out of the state breaks the spell; a later return starts a new spell.

Definition: The occupancy duration of an eligible state at time $t$ is the length of the current uninterrupted period spent in that state, measured up to time $t$. A duration-based benefit with threshold $d$ is payable only when the current spell has lasted at least $d$, subject to any other contract conditions. An exit from the eligible state resets the duration clock; a later return does not restore the earlier spell.

Probability of Satisfying a Duration Threshold

First consider a time-homogeneous continuous-time Markov model. Suppose the person starts in eligible state $J$. Let $\lambda_J$ be the total intensity of leaving $J$, meaning the sum of the intensities of all transitions from $J$ to other states. While in $J$, the waiting time until the next exit has an exponential distribution with rate $\lambda_J$. Thus the probability of remaining in $J$ without interruption for at least $t$ years is $e^{-\lambda_Jt}$.

This is a probability of no exit, not simply a probability of being in $J$ at time $t$. If the model permits leaving $J$ and returning, the ordinary probability of being in $J$ at time $t$ includes such returns, whereas uninterrupted occupancy does not.

Result: If the process starts in $J$, the total exit intensity from $J$ is constant at $\lambda_J$, and the force of interest is constant at $\delta$, then a benefit $B$ payable at time $d$ only if the person has remained continuously in $J$ through time $d$ has expected present value

\operatorname{EPV}=B e^{-(\lambda_J+\delta)d}.

If the exit intensity varies with time while the person remains in $J$, the no-exit probability through time $t$ is instead $ \exp\!\left(-\int_0^t \lambda_J(s)\,ds\right) $, where $\lambda_J(s)$ is the exit intensity at time $s$. The constant-intensity formula should not be used unless that assumption is appropriate.

Worked Example: A Lump Sum After Two Continuous Years

A hypothetical policyholder starts in eligible state $J$. The total intensity of leaving $J$ is constant at $0.12$ per year. A benefit of $\$8{,}000$ is payable at the end of two years only if the policyholder has remained in $J$ continuously from time zero. The force of interest is $0.03$ per year. Find the expected present value, rounded to the nearest cent.

Method. Multiply the benefit by both the probability of uninterrupted occupancy to time two and the discount factor to time two. The no-exit probability is $e^{-0.12(2)}=e^{-0.24}$, and the discount factor is $e^{-0.03(2)}=e^{-0.06}$.

Combining these factors gives

\operatorname{EPV} =8{,}000e^{-0.24}e^{-0.06} =8{,}000e^{-0.30} =8{,}000(0.7408182207) =\$5{,}926.5458.

Rounded to the nearest cent, the expected present value is $\$5{,}926.55$. This calculation avoids the common mistake of discounting the benefit but forgetting to multiply by the probability of satisfying the continuous-duration requirement.

Continuous Payments After the Threshold

A contract may pay continuously once the duration threshold has been reached, for as long as the person remains in the eligible state and before a specified end time. If the person starts in the eligible state at time zero, has constant exit intensity $\lambda_J$, and receives a continuous payment at rate $b$ per year from time $d$ to time $T$, then the payment at time $t$ is made only if the person has had no exit by $t$. At each such time, the probability of that condition is $e^{-\lambda_Jt}$.

\operatorname{EPV} =b\int_d^T e^{-\lambda_Jt}e^{-\delta t}\,dt =b\int_d^T e^{-(\lambda_J+\delta)t}\,dt =\frac{b}{\lambda_J+\delta} \left(e^{-(\lambda_J+\delta)d}-e^{-(\lambda_J+\delta)T}\right).

The lower integration limit is the threshold duration: there are no payments before $d$. The upper limit is the contract’s payment horizon. This expression assumes that the payment rate remains level, the interest force and exit intensity are constant, and no other event terminates payments.

Worked Example: Continuous Payments From Year One to Year Three

A hypothetical contract pays $\$600$ per year continuously from time 1 through time 3 while the policyholder remains in eligible state $J$. The policyholder starts in $J$, the constant exit intensity is $0.10$ per year, and the force of interest is $0.05$ per year. There are no payments before time 1. Find the expected present value, rounded to the nearest cent.

Method. At time $t$, a payment is made only if the policyholder has remained in $J$ without interruption through $t$. The expected discounted payment rate at that time is therefore $600e^{-(0.10+0.05)t}$. Integrate this rate over the actual payment interval, from 1 to 3:

\begin{aligned} \operatorname{EPV} &=600\int_1^3 e^{-0.15t}\,dt\\ &=\frac{600}{0.15}\left(e^{-0.15}-e^{-0.45}\right)\\ &=4{,}000(0.8607079764-0.6376281516)\\ &=4{,}000(0.2230798248)\\ &=\$892.3193. \end{aligned}

Rounded to the nearest cent, the expected present value is $\$892.32$. Integrating from time zero would incorrectly include payments before the qualifying duration has elapsed; integrating to time two would incorrectly shorten the stated payment period.

Annual Models and a Duration Clock

In an annual model, the same logic uses retention probabilities instead of continuous-time exit intensities. Suppose a person starts in eligible state $E$, and at the end of each year remains there with probability $r$. Assume that leaving $E$ ends the current qualifying spell and that the person cannot return, or that any return would not qualify as continuation of the original spell. The probability of uninterrupted occupancy through the end of year $k$ is $r^k$.

If a benefit $B_k$ is payable at the end of year $k$ only when occupancy has been uninterrupted since time zero, its expected present value at effective annual interest rate $i$ is $B_k r^k v^k$, where $v=1/(1+i)$. For several payment dates, add the expected present values at the eligible dates.

Worked Example: Payments at Two Annual Dates

In a hypothetical annual model, a policyholder starts in eligible state $E$. Each year, $E$ is retained with probability $0.90$; otherwise, the policyholder moves to absorbing state $X$. A benefit of $\$2{,}000$ is payable at the ends of years 2 and 3 only if occupancy of $E$ has been uninterrupted since time zero. The effective annual interest rate is $0.04$. Find the expected present value, rounded to the nearest cent.

Method. For a payment at year $k$, the probability of remaining in $E$ through that date is $0.90^k$, and the discount factor is $1.04^{-k}$. The two payments are valued separately:

\begin{aligned} \operatorname{EPV} &=2{,}000\left(\frac{0.90^2}{1.04^2}+\frac{0.90^3}{1.04^3}\right)\\ &=2{,}000\left(\frac{0.81}{1.0816}+\frac{0.729}{1.124864}\right)\\ &=2{,}000(0.7488905325+0.6480783468)\\ &=2{,}000(1.3969688793)\\ &=\$2{,}793.9378. \end{aligned}

Rounded to the nearest cent, the expected present value is $\$2{,}793.94$. The probability for the year-3 payment is $0.90^3$, not $0.90^2$: the policyholder must retain $E$ in all three annual transitions to satisfy the condition through year 3.

Why Ordinary State Probabilities Can Be Insufficient

For a benefit conditioned only on being in state $J$ at a payment date, an ordinary state probability may be sufficient. For a benefit conditioned on a continuous spell of length $d$, it generally is not. The distinction is important when a model allows a person to leave the eligible state and later return.

For example, suppose an annual model has two states, $A$ and $B$. From $A$, the person stays in $A$ for the year with probability $0.80$, or moves to $B$ with probability $0.20$. From $B$, the person returns to $A$ with probability $0.50$. Starting in $A$, the probability of being in $A$ at the end of year 2 is

$$ (0.80)(0.80)+(0.20)(0.50)=0.74. $$

But the probability of staying in $A$ continuously through both years is only

$$ (0.80)(0.80)=0.64. $$

The extra $0.10$ in the ordinary state probability comes from paths that leave $A$ and return. Those paths end in $A$, but they do not satisfy a two-year uninterrupted-occupancy requirement.

Key takeaway: Match the probability to the contract condition. Use an ordinary state occupancy probability when payment depends on the state at a particular time. Use a no-exit probability over the required spell when payment depends on uninterrupted occupancy. If exit and return are possible, the endpoint state alone does not record the duration condition.

Common Exam Traps

Using an endpoint probability for a continuous-spell condition. A person may leave and return before the payment date. Count only paths with no exit during the qualifying spell.
Starting payments too early. For a continuous benefit beginning after duration $d$, integrate from $d$, not from time zero.
Forgetting discounting or survival. A duration-conditioned payment generally requires both the probability of uninterrupted occupancy and the time-value factor.
Miscounting annual retention steps. A payment at the end of year $k$ requires $k$ successful annual retentions if the spell began at time zero.
Applying a constant-rate formula to changing assumptions. If the exit intensity varies, use its cumulative value over time rather than multiplying one rate by the whole duration.

Check Your Understanding

For each problem, value only payments that meet the stated continuous-occupancy condition. Monetary answers are rounded to the nearest cent.

A person starts in eligible state $A$, with constant exit intensity $0.06$ per year. A $\$10{,}000$ benefit is payable at time 3 only if the person has remained continuously in $A$. The force of interest is $0.02$. Find the expected present value. (Answer: $\$7{,}866.28$)
A policyholder starts in eligible state $E$ and retains it each year with probability $0.95$, with exit ending eligibility. A $\$1{,}000$ payment is made at the ends of years 2 and 4 only if $E$ has been occupied continuously. The effective annual interest rate is $0.05$. Find the expected present value. (Answer: $\$1{,}488.69$)
A hypothetical contract pays $\$400$ per year continuously from time 1.5 to time 4 while the person remains continuously in an eligible state. The person starts there; the constant exit intensity is $0.08$ and the force of interest is $0.04$. Find the expected present value. (Answer: $\$721.62$)
In an annual model, a person starts in $A$. Each year $A$ is retained with probability $0.80$, otherwise the person moves to $B$; from $B$, the person returns to $A$ with probability $0.50$. A $\$500$ benefit is payable at the end of year 2 only if the person has remained continuously in $A$. At effective annual interest $0.04$, find the expected present value. (Answer: $\$295.86$)
A person starts in an eligible state and must remain there continuously for three years to receive $\$6{,}000$ at time 3. The exit intensity is $0.04$ per year during the first year and $0.10$ per year during the next two years. The force of interest is $0.03$. Find the expected present value. (Answer: $\$4{,}313.54$)

What You'll Learn