One-Way Two-State Reserve Systems

Why One-Way Two-State Systems Are Useful

A one-way two-state model describes a policyholder who can move from an initial state into a second state, but cannot return during the term being valued. For example, a hypothetical contract might move from an active state into a long-term care state, with care benefits payable after entry. The key feature for reserving is that the reserve in the second state can be determined first: its future cash flows do not depend on the policyholder's earlier state.

This structure makes a multi-state Thiele calculation manageable without solving a fully coupled system. Starting with the terminal reserves, solve backward for the reserve in the absorbing state, then use that result to calculate the reserve in the transient state. The transition benefit must be included at the time of transition, together with the reserve immediately after transition.

Definition: Consider states 0 and 1 over $0\le t\le T$. A one-way two-state model has transition intensity $\lambda_{01}(t)=\lambda(t)$ from state 0 to state 1, and no transition out of state 1 during this term. State 0 is transient and state 1 is absorbing on $[0,T]$. Write $V_j(t)$ for the reserve at time $t$, conditional on being in state $j$, and $b_{01}(t)$ for a lump sum paid by the insurer on transition from 0 to 1. Let $c_j(t)$ be the continuous net cash-flow rate to the insurer in state $j$, with income positive and outgo negative. Let $\delta(t)$ be the force of interest.

The terminal conditions are $V_0(T)$ and $V_1(T)$, specified separately for the two states. They need not be equal: the contract may have different maturity benefits or other terminal values depending on the state at time $T$. In the equations below, $b_{01}$ is an outgo when positive, and $c_j$ follows the stated insurer-income convention.

The One-Way Thiele Equations

The multi-state Thiele equation, using continuous net cash flows and transition benefits, gives the following equations for this model:

\begin{aligned} V_1'(t)&=\delta(t)V_1(t)-c_1(t),\\ V_0'(t)&=\delta(t)V_0(t)-c_0(t) -\lambda(t)\bigl(b_{01}(t)+V_1(t)-V_0(t)\bigr)\\ &=\bigl(\delta(t)+\lambda(t)\bigr)V_0(t) -c_0(t)-\lambda(t)\bigl(b_{01}(t)+V_1(t)\bigr). \end{aligned}

There is no transition term in the equation for $V_1$, because state 1 has no exits in the model. In state 0, a transition causes an immediate payment $b_{01}(t)$ and leaves the policy in state 1 with reserve $V_1(t)$. The insurer's change in liability at transition is therefore $b_{01}(t)+V_1(t)-V_0(t)$. This is why the state-0 equation includes the transition intensity and the reserve difference.

Theorem: Given terminal values $V_0(T)$ and $V_1(T)$, define $D(t,s)=\exp\!\left(-\int_t^s\delta(r)\,dr\right)$. Then $$ \begin{aligned} V_1(t)&=D(t,T)V_1(T)+\int_t^T D(t,s)c_1(s)\,ds,\\ V_0(t)&=\exp\!\left(-\int_t^T[\delta(r)+\lambda(r)]\,dr\right)V_0(T)\\ &\quad+\int_t^T \exp\!\left(-\int_t^s[\delta(r)+\lambda(r)]\,dr\right) \bigl(c_0(s)+\lambda(s)[b_{01}(s)+V_1(s)]\bigr)\,ds. \end{aligned} $$

These formulas apply when the stated continuous-time model and its cash-flow conventions are appropriate, and the functions in the equations are integrable over the term. The formula for $V_0$ uses the already determined path $V_1(s)$; it does not assume that $V_1$ is constant.

Why the Backward Formulas Hold

The first equation can be rearranged as $V_1'(t)-\delta(t)V_1(t)=-c_1(t)$. Multiply by the integrating factor $\exp(-\int_0^t\delta(r)\,dr)$. The left side becomes the derivative of that factor times $V_1(t)$. Integrate from $t$ to $T$, use $V_1(T)$, and divide by the factor at $t$. This gives $V_1(t)=D(t,T)V_1(T)+\int_t^T D(t,s)c_1(s)\,ds$.

For state 0, write $a(t)=\delta(t)+\lambda(t)$ and $f(t)=c_0(t)+\lambda(t)[b_{01}(t)+V_1(t)]$, so $V_0'=aV_0-f$. Multiplication by $\exp(-\int_0^t a(r)\,dr)$ again turns the left side into a derivative. Integrating to $T$ and applying $V_0(T)$ gives the stated formula. The extra $\lambda$ in the discounting factor reflects both interest accumulation and the possibility of leaving state 0.

Record the state definitions and signs.
Identify the one-way transition, state-specific net cash-flow rates, and any lump sum payable on transition.

Apply the terminal conditions.
Write down $V_0(T)$ and $V_1(T)$ explicitly; do not assume either is zero unless the contract specifies it.

Solve state 1 backward.
Use its terminal reserve and its own interest and cash-flow terms to find $V_1(s)$ across the term.

Substitute into state 0.
Include $b_{01}(s)+V_1(s)$ in the transition term and integrate backward using $\delta+\lambda$.

Worked Calculations

Worked Example: Constant Cash Flows and a Transition Benefit

A hypothetical one-year contract has force of interest $\delta=0$, transition intensity $\lambda=1$ per year, and no transition out of state 1. The continuous net cash-flow rates are $c_0=\$20$ per year and $c_1=0$. A transition pays $\$10$, so $b_{01}=\$10$. The terminal reserves are $V_0(1)=\$0$ and $V_1(1)=\$100$. Find both reserves at time 0.

Method. Solve the state-1 equation first. Since its cash-flow rate and force of interest are both zero, $V_1(t)=\$100$ throughout the year.

For state 0, $\delta+\lambda=1$, and the forcing term is constant: $c_0+\lambda(b_{01}+V_1)=20+1(10+100)=\$130$ per year. The terminal state-0 reserve is zero, so

\begin{aligned} V_0(0) &=\int_0^1 e^{-s}(130)\,ds\\ &=130(1-e^{-1})\\ &=130(0.6321206\ldots)\\ &=\$82.18. \end{aligned}

Thus $V_1(0)=\$100.00$ and $V_0(0)=\$82.18$. The common mistake avoided here is treating the transition payment and the post-transition reserve as subtractions from the forcing term. Under the stated convention, a transition creates the insurer's outgo $b_{01}+V_1$, which enters the backward formula positively.

Worked Example: A State-1 Reserve That Changes Over Time

A two-year hypothetical contract has $\delta=0$ and $\lambda=1$ per year. The continuous net cash-flow rates are $c_0=\$10$ per year and $c_1=\$30$ per year. The transition benefit is $b_{01}=\$5$, and terminal reserves are $V_0(2)=\$0$ and $V_1(2)=\$20$. Calculate $V_1(s)$ and $V_0(0)$.

Method. In state 1, the reserve increases backward by the present value of its future continuous income. With zero interest, $V_1(s)=20+30(2-s)$. In particular, $V_1(0)=\$80$.

The state-0 discount factor is $e^{-s}$, since $\delta+\lambda=1$. Its forcing term at time $s$ is $10+1[5+20+30(2-s)]=95-30s$. Therefore,

\begin{aligned} V_0(0) &=\int_0^2 e^{-s}(95-30s)\,ds\\ &=95(1-e^{-2})-30\bigl(1-3e^{-2}\bigr)\\ &=65-5e^{-2}\\ &=65-5(0.1353353\ldots)\\ &=\$64.32. \end{aligned}

The integral $\int_0^2 se^{-s}\,ds=1-3e^{-2}$ follows by integration by parts. The answer is positive, consistent with the specified positive income rates, positive transition benefit, and positive terminal reserve in state 1. A common shortcut error is to replace $V_1(s)$ by $V_1(0)$ across the whole term; the transition may occur at any time, so the state-1 reserve must be evaluated at that transition time.

Worked Example: Transition Benefit With Interest and No State Income

A hypothetical five-year contract has effective annual interest rate $i=4\%$, so the force of interest is $\delta=\ln(1.04)=0.0392207$ per year. The transition intensity is constant at $\lambda=0.20$ per year. There are no continuous cash flows, both terminal reserves are zero, and a transition from state 0 pays $\$500$. Find $V_0(0)$.

Method. The state-1 reserve is zero at every time because it has neither future cash flows nor a nonzero terminal reserve. Thus the state-0 forcing term is $\lambda b_{01}=0.20(500)=\$100$ per year, and $\delta+\lambda=0.2392207$ per year. Applying the terminal formula,

\begin{aligned} V_0(0) &=\frac{100}{0.2392207} \left(1-e^{-5(0.2392207)}\right)\\ &=\frac{100}{0.2392207}(1-e^{-1.1961036})\\ &\approx \frac{100}{0.2392207}(1-0.302370)\\ &\approx \$291.63. \end{aligned}

The answer is rounded to the nearest cent. A frequent error is to use $i=0.04$ as the force of interest in a continuous-time equation. The correct force is $\ln(1.04)$; using the effective annual rate directly changes the reserve.

Common Sign and Boundary Checks

For a quick diagnostic, differentiate the proposed solution or substitute it into the differential equation and check the terminal values. In the state-0 equation, a positive transition benefit increases the value of expected future insurer outgo and therefore contributes positively to the backward reserve formula. A negative $c_j$, by contrast, represents a continuous outgo to the insurer and lowers the corresponding forcing term.

For example, in the first worked example, incorrectly using $c_0-\lambda(b_{01}+V_1)=20-110=-\$90$ as the forcing term would give $V_0(0)=-90(1-e^{-1})=-\$56.89$, not the correct $\$82.18$. That wrong calculation reverses the transition outgo's sign and produces a negative reserve despite the positive future outgoes. Also check that $\lambda$ and $\delta$ use the same time unit as $T$, and that the calculated functions satisfy $V_j(T)$ exactly.

Key takeaway: In a one-way system, the terminal reserve and future cash flows in state 1 determine $V_1(s)$ first. Then state 0 is valued using discounting at $\delta+\lambda$ and a forcing term of $c_0+\lambda(b_{01}+V_1)$. Preserve the cash-flow signs and the time units throughout.

Check Your Understanding

For each problem, use the one-way model and the stated insurer-income convention. Round monetary answers to the nearest cent where a decimal answer is requested.

A one-year contract has $\delta=0$, $\lambda=0.5$ per year, $c_0=c_1=0$, transition benefit $b_{01}=\$20$, $V_0(1)=0$, and $V_1(1)=\$80$. Find $V_0(0)$. (Answer: $\$39.35$.)
A two-year contract has $\delta=0$, $\lambda=0.5$ per year, $c_0=\$5$ per year, $c_1=\$10$ per year, $b_{01}=\$4$, and both terminal reserves equal zero. Find $V_1(s)$ and $V_0(0)$. (Answer: $V_1(s)=10(2-s)$, $V_0(0)=\$16.21$.)
A six-month contract has $\delta=0$, $\lambda=2$ per year, no continuous cash flows, $b_{01}=0$, $V_0(0.5)=0$, and $V_1(0.5)=\$50$. Find $V_0(0)$. (Answer: $\$31.61$.)
A one-year contract has $\delta=0$, $\lambda=1$ per year, $c_1=\$40$ per year, $c_0=0$, $b_{01}=0$, $V_1(1)=0$, and $V_0(1)=\$12$. Find $V_1(0)$ and $V_0(0)$. (Answer: $V_1(0)=\$40.00$, $V_0(0)=\$19.13$.)

What You'll Learn