Continuous Payments During State Occupancy

Introduction

A long-term benefit may accrue continuously while a covered person remains in a specified condition. For example, a hypothetical disability income contract might pay at a stated annual rate only while the insured is disabled. The payment is not triggered by the instant of becoming disabled; it accumulates over the time spent in the disabled state. Valuing it therefore requires the probability of occupying that state at each time, not just the probability of a particular transition.

This tutorial develops the expected present value (EPV) of such continuous state-occupancy payments. The method uses the state probabilities of a continuous-time Markov chain and discounts each payment made at time $t$ by the force of interest. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: A continuous state-occupancy payment at rate $b_j(t)$ is a payment accruing at $b_j(t)$ dollars per year at time $t$ if the process occupies state $j$ at that time. If $I_j(t)$ equals 1 when the process is in state $j$, and 0 otherwise, the accumulated payment over $[0,n]$ is $\int_0^n b_j(t)I_j(t)\,dt$. The rate $b_j(t)$ is a payment rate, not a lump sum paid on entering or leaving the state.

The Occupancy-Payment Formula

Suppose a process has a finite set of states and follows a continuous-time Markov chain. Let $p_j(t)$ be the probability that it occupies state $j$ at time $t$, given the specified initial state or initial distribution. State probabilities can be obtained from the transition probabilities or Kolmogorov equations. Assume that the payment rates are deterministic functions of time and state, that they are integrable over the term, and that the force of interest $\delta$ is constant. For payments in a set of eligible states $J$, the EPV over term $n$ is

\operatorname{EPV} = \int_0^n e^{-\delta t} \sum_{j\in J} b_j(t)p_j(t)\,dt.

Here $b_j(t)$ is the rate paid if the process is in state $j$, and $p_j(t)$ is the probability of being in that state. If the payment rate is constant within each eligible state, write $b_j(t)=b_j$. The formula permits the payment rate to differ between states. It does not include a separate amount payable at the instant of a transition.

Why the Formula Holds

The formula follows by writing the present value as a random integral, then taking its expectation. For each state $j$, the discounted payment made at time $t$ is $e^{-\delta t}b_j(t)I_j(t)\,dt$. Since the time horizon and rates are integrable, expectation can be interchanged with integration. Also, $E[I_j(t)]=P(\text{state }j\text{ at }t)=p_j(t)$. Therefore,

Write the present value.
The total present value from eligible states is $\int_0^n e^{-\delta t}\sum_{j\in J} b_j(t)I_j(t)\,dt$.

Take expectations.
Interchanging expectation and integration gives $\int_0^n e^{-\delta t}\sum_{j\in J} b_j(t)E[I_j(t)]\,dt$.

Use the occupancy probability.
Because $E[I_j(t)]=p_j(t)$, the expression becomes $\int_0^n e^{-\delta t}\sum_{j\in J}b_j(t)p_j(t)\,dt$, as stated.

This argument explains why the payment rate is multiplied by an occupancy probability. It is the expected payment rate at time $t$, before discounting. The intensity of a transition is not a substitute for that probability: an intensity describes an instantaneous transition rate conditional on the current state.

Worked Example: Reversible Movement Between Two States

A hypothetical contract has states $A$ and $B$. The process starts in $A$, moves from $A$ to $B$ at intensity $0.3$ per year, and moves from $B$ to $A$ at intensity $0.3$ per year. A benefit of \$6,000 per year is paid continuously while the process is in $B$, for two years. The effective annual interest rate is $i=e^{0.2}-1$, so the force of interest is $\delta=0.2$. Find the EPV.

Method. First find the probability of occupying $B$, then integrate the discounted expected payment rate. For this homogeneous two-state chain, the occupancy probability starting from $A$ is

p_B(t)=\frac{1}{2}\left(1-e^{-0.6t}\right).

The payment rate is \$6,000 in $B$ and zero in $A$, so

\begin{aligned} \operatorname{EPV} &=\int_0^2 e^{-0.2t}(6000)p_B(t)\,dt\\ &=3000\int_0^2\left(e^{-0.2t}-e^{-0.8t}\right)\,dt\\ &=3000\left[ \frac{1-e^{-0.4}}{0.2} -\frac{1-e^{-1.6}}{0.8} \right]. \end{aligned}

The first integral is $(1-e^{-0.4})/0.2=1.6483998$. The second is $(1-e^{-1.6})/0.8=0.9976294$. Their difference is $0.6507704$, and multiplying by \$3,000 gives \$1,952.3112. Rounded to the nearest cent, the EPV is \$1,952.31.

Check. The probability of $B$ starts at zero and approaches one-half, as expected for equal rates in both directions. The computed EPV is positive and less than the maximum possible present value of two years of certain \$6,000 annual payments, which is consistent with payments being made only while in $B$. This method avoids treating the process as if it could enter $B$ only once.

Different Payment Rates in Absorbing States

When a process leaves an initial state for one of several absorbing states, the occupancy probabilities in the absorbing states can be used directly. An absorbing state, once entered, is never left. In particular, if the process starts in $A$ and exits at total intensity $\lambda$, the probability of having exited $A$ by time $t$ is $1-e^{-\lambda t}$. If exit destinations have intensities $\lambda_j$, then the probability of occupying absorbing state $j$ at time $t$ is $(\lambda_j/\lambda)(1-e^{-\lambda t})$, where $\lambda=\sum_j\lambda_j$.

Worked Example: Payments in Two Absorbing Conditions

A hypothetical process starts in state $A$ and can move to either absorbing state $D$ or absorbing state $C$. The transition intensities are $0.1$ per year from $A$ to $D$, and $0.4$ per year from $A$ to $C$. A continuous benefit is paid for one year at \$3,000 per year in $D$, and \$7,000 per year in $C$. No benefit is paid in $A$. The force of interest is $\delta=0.1$. Find the EPV.

The total exit intensity is $\lambda=0.1+0.4=0.5$. The destination proportions are $0.1/0.5=0.2$ for $D$ and $0.4/0.5=0.8$ for $C$. Thus

p_D(t)=0.2(1-e^{-0.5t}), \qquad p_C(t)=0.8(1-e^{-0.5t}).

The expected payment rate at time $t$ is

3000p_D(t)+7000p_C(t) = \bigl(3000(0.2)+7000(0.8)\bigr)(1-e^{-0.5t}) = 6200(1-e^{-0.5t}).

Discount and integrate over one year:

\begin{aligned} \operatorname{EPV} &=6200\int_0^1 e^{-0.1t}(1-e^{-0.5t})\,dt\\ &=6200\left[ \frac{1-e^{-0.1}}{0.1} -\frac{1-e^{-0.6}}{0.6} \right]. \end{aligned}

The first term in brackets is $0.9516258$, and the second is $0.7519806$. The difference is $0.1996452$; multiplying by \$6,200 gives \$1,237.8002. Therefore the EPV, to the nearest cent, is \$1,237.80.

Check. At any time, $p_D(t)+p_C(t)=1-e^{-0.5t}$, the probability of having left $A$. The expected payment rate conditional on having left is \$6,200, the intensity-weighted destination average. This method avoids applying the full \$7,000 rate to every exit or using the total exit intensity as though it were an occupancy probability.

Payments at Different Rates in Different States

The formula also applies when payments are made in more than one state, including the initial state. Consider a chain with one transition from $A$ to absorbing state $B$. If its intensity is $\lambda$, the occupancy probabilities are $p_A(t)=e^{-\lambda t}$ and $p_B(t)=1-e^{-\lambda t}$. The expected payment rate is the sum of each state’s rate multiplied by its occupancy probability. This makes the calculation a weighted sum, not a choice of just one benefit rate.

Worked Example: A Payment Rate That Changes with State

A hypothetical contract’s process starts in $A$, moves to absorbing state $B$ at intensity $0.25$ per year, and remains in $B$ after entry. It pays continuously at \$1,000 per year while in $A$ and \$4,000 per year while in $B$, for two years. The effective annual interest rate is $i=e^{0.05}-1$, giving $\delta=0.05$. Find the EPV.

Here $p_A(t)=e^{-0.25t}$ and $p_B(t)=1-e^{-0.25t}$. The expected payment rate is

1000p_A(t)+4000p_B(t) =1000e^{-0.25t}+4000(1-e^{-0.25t}) =4000-3000e^{-0.25t}.

Therefore,

\begin{aligned} \operatorname{EPV} &=\int_0^2 e^{-0.05t}(4000-3000e^{-0.25t})\,dt\\ &=4000\frac{1-e^{-0.1}}{0.05} -3000\frac{1-e^{-0.6}}{0.3}. \end{aligned}

The first term is $4000(1.9032516)=\$7,613.0066$. The second is $3000(1.5039612)=\$4,511.8836$. Subtracting gives \$3,101.1230, so the EPV to the nearest cent is \$3,101.12.

Check. The payment rate is initially \$1,000 per year and increases in expectation as probability moves to $B$, where the rate is \$4,000. The EPV is between the values obtained by paying either rate with certainty for the full two years. This method avoids discounting the two state rates as if each applied for the entire term.

Common Exam Trap: Intensity Is Not Occupancy Probability

A transition intensity has units of transitions per year, conditional on occupying the origin state. It is not itself a probability. In the reversible two-state example, the incorrect substitution $p_B(t)=0.3t$ treats the rate of transition into $B$ as if it were the probability of being in $B$, while ignoring both the possibility of returning to $A$ and the restriction that a probability cannot exceed 1.

That incorrect substitution would give

6000\int_0^2 e^{-0.2t}(0.3t)\,dt = 1800\frac{1-(1+0.4)e^{-0.4}}{0.2^2} \approx \$2,769.84,

which is not the EPV. The correct occupancy probability is $p_B(t)=\tfrac12(1-e^{-0.6t})$, yielding \$1,952.31. The correction is to find the probability of occupying each eligible state from the multi-state model first, and only then apply the continuous-payment formula.

Key takeaway: For continuous payments made only while states are occupied, multiply each state’s payment rate by its occupancy probability, discount at the force of interest, and integrate over the payment term. A transition-triggered lump sum is a different cash-flow pattern and is not included in this occupancy formula.

Check Your Understanding

For each problem, use state-occupation probabilities and continuous discounting. Dollar answers are rounded to the nearest cent.

A process starts in $A$, moves to absorbing state $B$ at intensity $0.2$ per year, and pays \$2,500 per year continuously in $B$ for one year. The force of interest is $0.1$. Find the EPV. (Answer: \$219.22)
A two-state process starts in $A$; the intensity from $A$ to $B$ and from $B$ to $A$ is $0.4$ per year. A \$4,000 annual rate is paid continuously in $B$ for one year. The force of interest is $0.2$. Find the EPV. (Answer: \$548.45)
A process starts in $A$, moves to absorbing state $B$ at intensity $0.3$ per year, and pays \$1,500 per year in $A$ and \$4,500 per year in $B$ for two years. The force of interest is $0.1$. Find the EPV. (Answer: \$4,027.08)
A process starts in $A$ and moves to either absorbing state $D$ or absorbing state $C$ at respective intensities $0.2$ and $0.3$ per year. Rates of \$2,000 per year in $D$ and \$5,000 per year in $C$ are paid for one year. The force of interest is $0.1$. Find the EPV. (Answer: \$758.65)
A process starts in $A$, moves to absorbing state $B$ at intensity $0.5$ per year, and pays in $B$ only. The payment rate is \$1,000 per year from time 0 to 1, then \$2,000 per year from time 1 to 2. There is no interest. Find the EPV. (Answer: \$1,258.46)

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