Continuous Payments While Occupying a Benefit State

Why Occupancy Determines a Continuous Benefit

Some long-term benefits accrue continuously while a policyholder is in a designated state. The amount paid over a short interval depends on whether the policyholder occupies that state during the interval, not merely on whether a transition into it has occurred. Valuation therefore requires the probability of occupying the payment state at each time.

This tutorial develops the expected present value (EPV) of such payments. It focuses on the integration step: obtaining a state-occupancy probability from the transition model, discounting it, and integrating over the payment term. The resulting method is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let $X_t$ denote the policyholder’s state at time $t$, and let $B$ be a payment state. The occupancy probability at time $t$ is $p_B(t)=P(X_t=B)$. If a benefit accrues at rate $b(t)$ dollars per year only while the policyholder occupies $B$, the expected present value over $[0,T]$, under constant force of interest $\delta$, is $$ \operatorname{EPV}=\int_0^T e^{-\delta t}b(t)p_B(t)\,dt. $$ For a constant annual benefit rate $b$, this becomes $b\int_0^T e^{-\delta t}p_B(t)\,dt$. For an unbounded term, take $T=\infty$ if the integral converges.

The effective annual interest rate $i$ and force of interest are related by $\delta=\ln(1+i)$. Thus the accumulation over $t$ years is $e^{\delta t}=(1+i)^t$, and the discount factor is $e^{-\delta t}$. Continuous benefit rates are expressed per year, so time and transition intensities must also be measured in years and per year.

The state model specifies how occupancy probabilities change over time. For a time-homogeneous continuous-time Markov chain with row-vector state probabilities, initial distribution $\boldsymbol{\pi}(0)$, and generator matrix $Q$, the state probabilities are $\boldsymbol{\pi}(t)=\boldsymbol{\pi}(0)e^{Qt}$. The generator’s off-diagonal entries are transition intensities, and each diagonal entry is the negative sum of the other entries in its row. In applications below, the needed occupancy probability is derived directly from the transition structure.

Why the Discounted Occupancy Integral Holds

Let the instantaneous benefit rate be $b(t)$ while the policyholder is in $B$, and zero otherwise. The random present value of the benefit is the integral of the discounted payment rate multiplied by an indicator of occupancy. Taking its expectation gives the valuation formula.

Z=\int_0^T e^{-\delta t}b(t)\mathbf{1}_{\{X_t=B\}}\,dt \quad\Longrightarrow\quad E[Z]=\int_0^T e^{-\delta t}b(t)E[\mathbf{1}_{\{X_t=B\}}]\,dt =\int_0^T e^{-\delta t}b(t)p_B(t)\,dt.

Here $E[\mathbf{1}_{\{X_t=B\}}]=P(X_t=B)$. Interchanging expectation and integration is justified for a nonnegative benefit rate by the theorem for integrating nonnegative functions; on a finite term, bounded rates also ensure the expected present value is finite. The formula counts the payment accumulated during time spent in $B$. A transition lump sum is different: it is valued using the probability density or intensity of the relevant transition, as covered in the earlier material on benefits paid at state transitions.

Key distinction: A continuous benefit is valued using the probability of occupying the payment state, $p_B(t)$. An entry intensity multiplied by the probability of being in another state describes a flow of transitions, not the amount of time spent receiving the benefit.

Worked Valuations

Each example assumes constant transition intensities over the stated term. An absorbing state is one that, once entered, cannot be left. Unless a problem specifies otherwise, there are no payments in absorbing nonpayment states.

Worked Example: Sequential Entry into a Benefit State

A hypothetical contractholder starts in state $A$. The process moves from $A$ to benefit state $B$ at intensity $0.20$ per year, then from $B$ to absorbing nonpayment state $C$ at intensity $0.10$ per year. There are no other transitions. A benefit of $9,000 per year is payable continuously while in $B$, with no time limit. The effective annual interest rate is $i=e^{0.10}-1=0.1052$ to four significant figures, so $\delta=0.10$. Find the EPV. The method is to derive $p_B(t)$, then integrate its discounted value over the full term.

Starting from $A$, occupancy of $B$ at time $t$ requires entry from $A$ before $t$ and no exit from $B$ by $t$. Equivalently, solving the state-probability equation $p_B'(t)=0.20p_A(t)-0.10p_B(t)$, with $p_A(t)=e^{-0.20t}$ and $p_B(0)=0$, gives

p_B(t)=2\left(e^{-0.10t}-e^{-0.20t}\right).

The occupancy probability starts at zero, as required, and is nonnegative for $t\geq 0$. Discounting and integrating over the infinite term gives the discounted expected time spent in $B$:

\begin{aligned} \int_0^\infty e^{-0.10t}p_B(t)\,dt &=2\int_0^\infty\left(e^{-0.20t}-e^{-0.30t}\right)\,dt\\ &=2\left(\frac{1}{0.20}-\frac{1}{0.30}\right)\\ &=2\left(5-\frac{10}{3}\right) =\frac{10}{3}\text{ years}. \end{aligned}

Therefore the EPV is $9{,}000(10/3)=\$30{,}000.00$. As a check, the discounted duration is positive and finite. A common mistake is to use the discounted entry flow $0.20p_A(t)$ in place of occupancy. That would give $\int_0^\infty e^{-0.10t}0.20e^{-0.20t}\,dt=0.20/0.30=2/3$, and an incorrect value of $6,000.00. It counts discounted entries, not the duration of benefit payments after entry.

Worked Example: A Finite Payment Term from the Benefit State

A hypothetical policyholder starts in benefit state $B$. The only transition is from $B$ to absorbing state $C$, at intensity $0.10$ per year; no benefit is paid in $C$. The benefit rate is $4,000 per year and payments stop after $T=5\ln 2$ years, whether or not the policyholder remains in $B$. The effective annual interest rate is $i=e^{0.10}-1=0.1052$ to four significant figures. Find the EPV, rounding the final amount to cents.

The force of interest is $\delta=0.10$. Since the policyholder starts in $B$ and leaves it at constant intensity $0.10$, the occupancy probability is the probability of surviving in $B$ to time $t$:

p_B(t)=e^{-0.10t}.

The benefit can be paid only before both the transition out of $B$ and the end of the specified term. Thus the discounted occupancy integral is

\begin{aligned} \int_0^T e^{-0.10t}p_B(t)\,dt &=\int_0^{5\ln 2}e^{-0.20t}\,dt\\ &=\frac{1-e^{-0.20(5\ln 2)}}{0.20}\\ &=\frac{1-e^{-\ln 2}}{0.20} =\frac{1-0.5}{0.20}=2.5\text{ years}. \end{aligned}

Multiplying by the annual rate gives $4{,}000(2.5)=\$10{,}000.00$. The term is $5\ln 2\approx3.4657$ years, but the calculation uses the exponential survival and discount factors over that exact term. A common mistake is to ignore the finite payment limit and integrate to infinity; that would produce $4{,}000/0.20=\$20{,}000.00$, twice the correct value.

Worked Example: Uncertain Initial State

At time zero, a hypothetical policyholder is in benefit state $B$ with probability $0.40$ and in state $A$ with probability $0.60$. From $A$, the policyholder moves to $B$ at intensity $0.20$ per year. From $B$, the policyholder moves to absorbing nonpayment state $C$ at intensity $0.10$ per year. There are no other transitions. A continuous benefit of $6,000 per year is paid in $B$ for an unlimited term. The force of interest is $\delta=0.10$, corresponding to effective annual rate $i=e^{0.10}-1=0.1052$ to four significant figures. Find the EPV.

First find the occupancy probability conditional on each possible initial state. Starting in $B$, the probability of still being there at time $t$ is $e^{-0.10t}$. Starting in $A$, the probability of being in $B$ is $2(e^{-0.10t}-e^{-0.20t})$, as derived in the sequential-transition example. Weighting these conditional probabilities by their initial probabilities gives

\begin{aligned} p_B(t) &=0.40e^{-0.10t} +0.60\left[2(e^{-0.10t}-e^{-0.20t})\right]\\ &=1.60e^{-0.10t}-1.20e^{-0.20t}. \end{aligned}

At $t=0$, this expression gives $1.60-1.20=0.40$, matching the initial probability of being in $B$. The discounted expected occupancy is

\begin{aligned} \int_0^\infty e^{-0.10t}p_B(t)\,dt &=\frac{1.60}{0.20}-\frac{1.20}{0.30}\\ &=8-4=4\text{ years}. \end{aligned}

Hence the EPV is $6{,}000(4)=\$24{,}000.00$. A common mistake is to value the benefit as though every policyholder starts in $B$. That would give $6{,}000/(0.10+0.10)=\$30{,}000.00$. The correct calculation respects the initial mixture and includes the possibility of later entry from $A$.

Applying the Method Reliably

For a more general model, the occupancy probability may have to be obtained from a system of Kolmogorov equations or from the transition-probability matrix. Once $p_B(t)$ is known, the valuation step is unchanged. If payments are made in several states, add the expected payment rates state by state. For example, if state $j$ has rate $b_j(t)$, the EPV is the sum of the integrals of $e^{-\delta t}b_j(t)p_j(t)$ over the relevant states and term.

The integral can be evaluated analytically when the occupancy probabilities are sums of exponential terms. For $\lambda>0$,

\int_0^T e^{-\lambda t}\,dt=\frac{1-e^{-\lambda T}}{\lambda}, \qquad \int_0^\infty e^{-\lambda t}\,dt=\frac{1}{\lambda}.

These identities explain why a constant-force Markov model often leads to a sum of simple discounted exponential integrals. For a finite term, retain the upper-limit terms; for an unlimited term, each exponential in the discounted integrand must decay sufficiently fast for the integral to converge. If interest or transition intensities vary with time, the same occupancy principle applies, but the relevant state probabilities and discount factors may no longer reduce to constant-rate exponential expressions.

Identify the payment state and term.
Specify which states generate continuous payments, the annual rate in each state, and the time interval during which payments can be made.

Set the time and interest conventions.
Express transition intensities per year and use $\delta=\ln(1+i)$ when the input is an effective annual rate.

Find the occupancy probability.
Use the state-transition model and initial distribution to obtain $p_j(t)$ for every payment state $j$.

Discount and integrate occupancy.
Evaluate $\int e^{-\delta t}b_j(t)p_j(t)\,dt$ over the payment term and sum across payment states if needed.

Check the interpretation.
Verify initial occupancy, signs, units, term limits, and that the result values time in payment states rather than transition counts.

Common Traps and Interpretation

The value of a continuous state benefit is a rate multiplied by discounted expected time in the payment state. That interpretation provides a useful unit check: the integral of discounted occupancy has units of years, and multiplying it by dollars per year gives dollars. It also clarifies several common errors.

Using an entry probability or transition intensity instead of occupancy. Entry into a state is a transition event. Payments that continue after entry depend on how long the process remains in the payment state.
Forgetting discounting in the time integral. The factor $e^{-\delta t}$ applies at each payment time, not once to the total undiscounted duration.
Using the wrong interest conversion. Under an effective annual rate $i$, continuous discounting uses $\delta=\ln(1+i)$, not $i$ itself.
Ignoring a payment-term limit. A finite contract term requires integration only to $T$, even when the state may be occupied beyond $T$.
Overlooking initial-state uncertainty. The unconditional occupancy probability must reflect the initial distribution, including both initial benefit-state occupancy and later entries.

The method is useful whenever cash flows accrue while a contractholder occupies a specified state. It separates the state-model task—finding occupancy probabilities—from the valuation task—discounting and integrating the resulting expected payment rate. This makes it possible to check each part independently before combining them.

Key takeaway: For a continuous benefit payable only while the policyholder occupies state $B$, value the benefit by integrating its discounted rate against $P(X_t=B)$ over the payment term. Occupancy, not entry flow, determines the amount accrued.

Check Your Understanding

For each problem, identify the discounted occupancy integral before multiplying by the benefit rate.

A policyholder starts in benefit state $B$ and exits to a nonpayment absorbing state at intensity $0.15$ per year. The benefit is $3,000 per year for an unlimited term. The force of interest is $0.05$. Find the EPV. (Answer: $15,000.00)
A process starts in $A$, moves to benefit state $B$ at intensity $0.25$, and exits from $B$ to an absorbing nonpayment state at intensity $0.15$. A continuous benefit of $5,000 per year is payable indefinitely. The force of interest is $0.10$. Find the EPV. (Answer: $14,285.71)
At time zero, a policyholder is in $B$ with probability $0.30$ and in absorbing nonpayment state $C$ with probability $0.70$. The exit intensity from $B$ is $0.05$. A benefit of $2,000 per year is payable for 10 years. The force of interest is $0.05$. Find the EPV to the nearest cent. (Answer: $3,792.72)
A process starts in $A$, enters benefit state $B$ at intensity $0.30$, and exits $B$ to a nonpayment absorbing state at intensity $0.20$. The benefit is $8,000 per year for an unlimited term, and the force of interest is $0.10$. Find the EPV. (Answer: $20,000.00)
At time zero, a policyholder is in benefit state $B$ with probability $0.50$, or benefit state $D$ with probability $0.50$. State $B$ exits to a nonpayment absorbing state at intensity $0.10$, and state $D$ exits to a nonpayment absorbing state at intensity $0.20$. The benefit rates are $4,000 per year in $B$ and $6,000 per year in $D$, payable indefinitely. The force of interest is $0.10$. Find the combined EPV. (Answer: $20,000.00)

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