Integrating-Factor Derivation for Single-State Reserves

Why the Integrating Factor Matters

The single-state Thiele equation describes how a reserve changes while a life remains in an active state. Its differential form is useful for describing the instantaneous balance, but a terminal reserve is often the most convenient condition for valuation: the reserve at a future time is known, and the reserve at an earlier time must be found. An integrating factor converts the differential equation into a formula for that earlier reserve as discounted future cash flows.

This tutorial uses a single active state with death as an exit. Cash flows while the life is active are continuous, and a death benefit is paid at the instant of death. The method also allows deterministic forces and cash-flow rates to vary with time. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let $V(t)$ be the reserve at time $t$, conditional on the insured being alive and in the active state. Let $\delta(t)$ be the force of interest, $\mu(t)$ the force of mortality, $b(t)$ the benefit payable on death at time $t$, and $c(t)$ the continuous net income while active (premium income less expenses). The single-state Thiele equation is $$ V'(t)=\delta(t)V(t)+c(t)-\mu(t)\bigl(b(t)-V(t)\bigr). $$ If the reserve at the terminal time $T$, conditional on being alive then, is $M$, the boundary condition is $V(T)=M$.

The equation assumes that the forces and cash-flow rates are deterministic functions of time, and that the stated death benefit is payable immediately on death. The reserve is conditional on being active at time $t$; it is not an average reserve across lives in different states. A lump-sum payment at a specified time must be handled as a jump or a boundary cash flow, not as a continuous rate $c(t)$.

Expanding the mortality term gives the linear differential equation $$ V'(t)-\bigl(\delta(t)+\mu(t)\bigr)V(t)=c(t)-\mu(t)b(t). $$ The combined coefficient $\delta(t)+\mu(t)$ is central to the integrating-factor solution. The formula below assumes a finite terminal time and integrable forces and cash-flow rates over the interval.

Deriving the Terminal-Value Formula

Set $a(t)=\delta(t)+\mu(t)$. Multiply the equation by an integrating factor chosen so that the left-hand side becomes the derivative of a product. Define $$ I(t)=\exp\left(-\int_0^t a(r)\,dr\right). $$ Then $I'(t)=-a(t)I(t)$, so the product rule gives $$ \frac{d}{dt}\bigl(I(t)V(t)\bigr) =I(t)\bigl(V'(t)-a(t)V(t)\bigr) =I(t)\bigl(c(t)-\mu(t)b(t)\bigr). $$ Integrate from $t$ to $T$: $$ I(T)V(T)-I(t)V(t) =\int_t^T I(s)\bigl(c(s)-\mu(s)b(s)\bigr)\,ds. $$ Rearranging and dividing by $I(t)$ yields the reserve at time $t$:

Theorem: Given the terminal condition $V(T)=M$, the single-state reserve at $t\leq T$ is $$ V(t)=\exp\left(-\int_t^T(\delta(r)+\mu(r))\,dr\right)M +\int_t^T \exp\left(-\int_t^s(\delta(r)+\mu(r))\,dr\right) \bigl(\mu(s)b(s)-c(s)\bigr)\,ds. $$

The sign of $c(s)$ is important: premium income reduces the reserve required, while expenses increase it because they reduce net income. Mortality enters through both the death-benefit amount and the chance that death occurs at each time.

Reading the Formula as Discounted Cash Flows

For $s\geq t$, define the conditional survival probability from $t$ to $s$ by $$ {}_{s-t}p_{x+t}=\exp\left(-\int_t^s\mu(r)\,dr\right), $$ where $x+t$ denotes the attained age at time $t$. Define the corresponding accumulation discount factor by $$ \exp\left(-\int_t^s\delta(r)\,dr\right). $$ Their product is exactly the kernel in the reserve formula: $$ \exp\left(-\int_t^s(\delta(r)+\mu(r))\,dr\right) =\exp\left(-\int_t^s\delta(r)\,dr\right){}_{s-t}p_{x+t}. $$ The formula therefore discounts each future amount for interest and weights it by the probability of remaining in the active state until the relevant time. The death-benefit part uses $\mu(s)b(s)\,ds$: conditional on being alive at $s$, the probability of death in a short interval of length $ds$ is approximately $\mu(s)\,ds$.

In this interpretation, the terminal amount $M$ is paid only if the insured survives to $T$. The continuous net income $c(s)$ is also received only while active, so it is subtracted inside the same survival-and-discount integral. If the terminal amount is not conditional on survival, or if benefits continue after death, this single-state formula does not describe the contract without modification.

Worked Example: A Terminal Reserve and a Level Death Benefit

A hypothetical policy is valued over two years. The annual force of interest is $\delta=0.03$. Future lifetime from the valuation date is exponential with constant force of mortality $\mu=0.02$. The death benefit is $b=80$, the continuous net income is $c=0.40$ per year, and the terminal reserve conditional on survival is $M=20$. Find $V(0)$.

Method. Apply the integrating-factor formula with $T=2$. Here $\delta+\mu=0.05$, and the net amount inside the integral is $\mu b-c=(0.02)(80)-0.40=1.20$.

The terminal component is $$ e^{-(0.05)(2)}(20)=20e^{-0.10}=18.096748. $$ The death-benefit less net-income component is $$ \int_0^2 e^{-0.05s}(1.20)\,ds =1.20\left(\frac{1-e^{-0.10}}{0.05}\right) =2.283902. $$ Thus $V(0)=18.096748+2.283902=20.380650$, or $\$20.38$ to the nearest cent.

Check. The death-benefit rate exceeds net income, so the integral adds to the terminal-value component. The resulting reserve is greater than the discounted terminal amount, as expected.

Common mistake avoided. Subtracting $c$ with the death benefit would incorrectly treat net premium income as an additional outgo. The correct integrand is $\mu b-c$.

Worked Example: A Change in Net Income During the Term

A hypothetical two-year contract has force of interest $\delta=0.02$ and exponential future lifetime with constant force of mortality $\mu=0.03$. The death benefit is $b=100$, the terminal reserve is $M=0$, and continuous net income is $c=1$ during year 1 and $c=2$ during year 2. Calculate $V(1)$ and $V(0)$.

Method. The combined force is $0.02+0.03=0.05$. At time 1, only the second-year cash-flow assumptions apply. The death-benefit rate is $\mu b=(0.03)(100)=3$, so the net outgo rate is $3-2=1$.

With one year remaining and no terminal reserve, $$ V(1)=\int_1^2 e^{-0.05(s-1)}(1)\,ds =\frac{1-e^{-0.05}}{0.05} =0.975412. $$ Therefore $V(1)=\mathbf{\$0.98}$ to the nearest cent.

At time 0, the net outgo rate is $3-1=2$ in year 1 and $3-2=1$ in year 2. Split the integral at the change in net income: $$ V(0)=2\int_0^1 e^{-0.05s}\,ds+\int_1^2 e^{-0.05s}\,ds. $$ The first part is $2(1-e^{-0.05})/0.05=1.950823$. The second part is $(e^{-0.05}-e^{-0.10})/0.05=0.927840$. Thus $V(0)=1.950823+0.927840=2.878663$, or $\$2.88$ to the nearest cent.

Check. The reserve is continuous at time 1 because there is no lump-sum cash flow at that time; only the continuous net-income rate changes. The value calculated at time 0 includes both years, while $V(1)$ includes only the remaining year.

Common mistake avoided. Using the first-year net income throughout the term ignores the stated change after year 1. The integral must be split wherever a cash-flow rate or force changes.

Worked Example: Piecewise Forces and a Nonzero Terminal Value

A hypothetical policy runs for two years and has a terminal reserve of $M=30$ if the insured is alive at year 2. During year 1, the force of interest is $\delta=0.02$, the mortality force is $\mu=0.01$, and the death benefit is $b=120$. During year 2, these are $\delta=0.04$, $\mu=0.02$, and $b=80$. The continuous net income is $c=0.50$ throughout. Mortality is piecewise exponential with the stated forces; interest is deterministic and piecewise constant. Find $V(0)$.

Method. Apply the terminal-value formula separately over each interval. The combined force is $0.03$ in year 1 and $0.06$ in year 2. The net outgo rates $\mu b-c$ are $0.01(120)-0.50=0.70$ and $0.02(80)-0.50=1.10$.

The terminal reserve is discounted through both years: $$ 30e^{-(0.03)(1)-(0.06)(1)}=30e^{-0.09}=27.417936. $$ The first-year death-benefit less net-income component is $$ 0.70\int_0^1 e^{-0.03s}\,ds =0.70\left(\frac{1-e^{-0.03}}{0.03}\right) =0.689604. $$ For the second-year component, the discount-and-survival factor already accumulated through year 1 is $e^{-0.03}$: $$ 1.10e^{-0.03}\int_1^2 e^{-0.06(s-1)}\,ds =1.10\left(\frac{e^{-0.03}-e^{-0.09}}{0.06}\right) =1.036096. $$ Adding the three terms gives $V(0)=27.417936+0.689604+1.036096=29.143636$, or $\$29.14$ to the nearest cent.

Check. Both interval outgo rates are positive, so both integral terms increase the reserve above the discounted terminal reserve. The year-2 term is discounted through year 1 as well as within year 2.

Common mistake avoided. Applying only the year-2 discount factor to the second-year term omits the survival and interest accumulation from time 0 to year 1. The factor $e^{-0.03}$ is required.

Common Errors and Scope of the Result

The integrating-factor formula is a prospective valuation identity obtained from the single-state reserve equation and the terminal condition. Its interpretation as a discounted expected value relies on the specified payment timing and assumptions. In particular, the death benefit is valued at the instant of death, and the terminal amount is conditional on being alive at $T$.

Reversing the sign of net income: use $\mu b-c$ in the prospective outgo integral, not $\mu b+c$.
Omitting mortality from the kernel: future cash flows are payable only if the life is still active, so discounting alone is not enough.
Using the wrong terminal condition: the term involving $M$ is present only when the stated terminal reserve is payable on survival to $T$.
Ignoring changes over time: when forces, benefits, or net income vary, use their appropriate values within each portion of the integral.
Applying the model to a different payment pattern: lump sums at fixed times or benefits payable in other states require appropriate jumps or additional state equations.

The main practical advantage is that one formula gives the reserve at any time $t$, not only at issue. It can be evaluated directly when the relevant integrals are manageable, or used as the basis for numerical integration when the forces or cash-flow rates are more complicated.

Key takeaway: Multiply the expanded Thiele equation by $\exp(-\int_0^t(\delta+\mu))$, integrate from the valuation time to the terminal time, and apply the terminal reserve. The resulting kernel discounts future cash flows for both interest and the probability of remaining in the active state.

Check Your Understanding

For each problem, use the single-state terminal-value formula and round monetary answers to the nearest cent.

A one-year hypothetical policy has $\delta=0.03$, constant exponential mortality force $\mu=0.02$, death benefit $b=50$, continuous net income $c=0.20$, and terminal reserve $M=0$. Find $V(0)$. (Answer: $\$0.78$)
A hypothetical policy is valued at time 1 and matures at time 3. From time 1 onward, $\delta=0.04$, the exponential mortality force is $\mu=0.01$, the death benefit is $100$, continuous net income is $0.50$, and $M=10$. Find $V(1)$. (Answer: $\$10.00$)
Over two years, $\delta=0.02$ and the exponential mortality force is $\mu=0.03$; the death benefit is $80$, $M=0$, and continuous net income is $0.40$ in year 1 and $0.80$ in year 2. Find $V(0)$. (Answer: $\$3.44$)
For each year of a two-year policy, the forces and benefit are constant: in year 1, $\delta=0.01,\ \mu=0.02,\ b=60$; in year 2, $\delta=0.03,\ \mu=0.01,\ b=60$. Mortality is piecewise exponential, $c=0$, and $M=5$. Find $V(0)$. (Answer: $\$6.41$)
A two-year hypothetical policy has zero force of interest, constant exponential mortality force $\mu=0.10$, death benefit $b=40$, continuous net income $c=0$, and terminal reserve $M=10$. Find $V(0)$. (Answer: $\$15.44$)

What You'll Learn