Why the Endpoint Determines the Calculation
A Thiele reserve equation describes how a reserve changes through time, but it does not by itself specify which solution applies to a particular contract. The contract’s endpoint supplies the necessary boundary condition: once the policy has ended and all final cash flows have been made, no future policy liability remains. The reserve immediately before those cash flows, however, may be positive, negative, or state-dependent.
This distinction matters in practical calculations. A maturity benefit payable at time \(T\) is still an obligation immediately before payment, even though the contract has no reserve after the payment. Likewise, a premium due at maturity affects the reserve before the endpoint transactions. The reserve equation is therefore solved from the endpoint back toward issue, using the relevant terminal value as its starting value. This is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Setting the Terminal Reserve Correctly
The notation \(T-\) means immediately before the transactions at time \(T\); \(T+\) means immediately after them. A contract that ends at \(T\) has no future policy cash flows after the endpoint, so its post-termination reserve is zero. Its pre-payment reserve is determined by what is about to be paid or received. If there are no endpoint payments, the pre-payment reserve is also zero.
For a contract with several possible in-force states, set the boundary condition separately for each state. A maturity benefit may differ by state, and a premium due at maturity may be payable only in certain states. The calculation must follow the stated payment timing: do not subtract a premium that is not due, or include a benefit payable only after the policy has already terminated.
| Endpoint situation | Reserve just before endpoint transactions | Reserve after termination |
|---|---|---|
| No endpoint payment or receipt | Zero | Zero |
| Benefit payable in state \(j\) | Benefit due in state \(j\) | Zero |
| Benefit and premium both due in state \(j\) | Benefit less premium in state \(j\) | Zero |
| Different benefits by in-force state | Set a separate value for each state | Zero in each terminated state |
These conditions presume that the contract ends at \(T\) and that the reserve is defined immediately before all endpoint transactions. If the problem instead defines the reserve after a particular transaction, or coverage continues after \(T\), the boundary must reflect that convention and any remaining obligations.
Worked Example: Netting a Final Premium Against a Maturity Benefit
A hypothetical one-year policy pays a maturity benefit of $900 at time \(T=1\) if the insured is active. A final premium of $100 is also due at \(T\), immediately before the benefit is paid, but only if the insured is active. If disabled at \(T\), the policy instead pays a $1,200 termination benefit and collects no final premium. Find the terminal reserves immediately before the endpoint transactions and immediately after termination.
Method. Set the reserve in each state equal to the net endpoint liability immediately before the transactions. There are no future obligations once the policy has terminated.
In the active state, the insurer will pay $900 and collect $100. Thus \(V_A(1-)=900-100=\$800\).
In the disabled state, the insurer will pay $1,200 and collect no premium. Thus \(V_D(1-)=1,200-0=\$1,200\).
After the endpoint transactions, the contract has ended in either state, so \(V_A(1+)=V_D(1+)=\$0\). The calculation avoids treating the active-state boundary as $900: that would omit a premium explicitly due at the endpoint.
Why Reserve Equations Are Solved Backward
For a single state, write the reserve equation in the form \(V'(t)=f(t,V(t))\), where \(f\) is the reserve derivative specified by the contract’s interest and cash-flow assumptions. The endpoint value \(V(T-)\) is known from the benefits and premiums due at \(T\). To find an earlier reserve, move from \(T\) toward time zero. In general, starting at time zero would require an initial reserve value that is precisely what the calculation is intended to determine.
A short Taylor expansion explains the basic numerical step. For a small time interval \(h>0\), expand the reserve at an earlier time around the known later time \(t\):
Substituting the differential equation \(V'(t)=f(t,V(t))\) and ignoring the second-order error gives
This is explicit Euler applied backward in time: the derivative is evaluated using the reserve at the known later endpoint of the step. It is not the implicit backward Euler method. The implicit method instead evaluates the derivative at the unknown earlier value, using \(V(t-h)=V(t)-h f(t-h,V(t-h))\), and generally requires solving an equation for \(V(t-h)\).
Identify whether the boundary reserve is measured before or after endpoint transactions, and list which benefits and premiums are due.
Calculate a separate \(V_j(T-)\) for each in-force state \(j\). After the contract’s final transactions, set the terminated contract’s reserve to zero.
For explicit Euler, calculate \(f(t,V(t))\) using the reserve at the later time in the step.
Use \(V(t-h)\approx V(t)-h f(t,V(t))\), then repeat with the newly calculated reserve.
Worked Backward Calculations
Worked Example: A One-Year Maturity Benefit With Continuous Premiums
A hypothetical one-year contract has no mortality or other decrements. It pays a maturity benefit of $1,000 at \(T=1\) and receives continuous premiums at $100 per year over \([0,1)\). The force of interest is constant at \(\delta=0.05\). Under the stated cash-flow convention, the reserve equation is \(V'=0.05V+100\). Use two explicit Euler steps of length \(h=0.5\) to estimate \(V(0.5)\) and \(V(0)\).
Method. The maturity benefit gives \(V(1-)=\$1,000\), because there is no premium due at the endpoint. Step backward from that known value, evaluating the derivative at the later time in each step.
At \(t=1\), the derivative is \(V'(1-)=0.05(1,000)+100=50+100=150\). Therefore, \(V(0.5)\approx1,000-0.5(150)=1,000-75=\$925.00\).
At \(t=0.5\), the derivative using the estimated reserve is \(V'(0.5)\approx0.05(925)+100=46.25+100=146.25\). The next backward step gives \(V(0)\approx925-0.5(146.25)=925-73.125=\$851.875\), or $851.88 to the nearest cent.
The result is directionally consistent: the derivative is positive at both known step values, so moving backward subtracts a positive increment and the reserve falls. The method avoids reversing the sign of the derivative while stepping backward. The approximation is numerical; reducing the step size is one way to assess the effect of discretization.
Worked Example: Backward Steps for Two In-Force States
A hypothetical contract ends at \(T=1\). At the endpoint, the maturity benefit is $500 in the active state \(A\) and $800 in the disabled state \(D\). There are no transition payments or premiums. The force of interest is \(0.04\), and the transition intensities are constant: \(q_{AD}=0.10\) per year and \(q_{DA}=0.20\) per year. For this contract, the reserve equations are \(V_A'=0.04V_A+0.10(V_A-V_D)\) and \(V_D'=0.04V_D+0.20(V_D-V_A)\). Use two explicit Euler steps of length \(0.25\).
Method. Start with the state-specific terminal values \(V_A(1-)=500\) and \(V_D(1-)=800\). At each step, evaluate both derivatives using the pair of reserves at the known later time.
At \(t=1\), \(V_A'=0.04(500)+0.10(500-800)=20-30=-10\), and \(V_D'=0.04(800)+0.20(800-500)=32+60=92\). Thus the first step gives \(V_A(0.75)\approx500-0.25(-10)=\$502.50\) and \(V_D(0.75)\approx800-0.25(92)=\$777.00\).
At \(t=0.75\), use those newly calculated values: \(V_A'=0.04(502.50)+0.10(502.50-777.00)=20.10-27.45=-7.35\), and \(V_D'=0.04(777.00)+0.20(777.00-502.50)=31.08+54.90=85.98\). The second step gives \(V_A(0.50)\approx502.50-0.25(-7.35)=502.50+1.8375=\$504.34\) and \(V_D(0.50)\approx777.00-0.25(85.98)=777.00-21.495=\$755.51\), rounded to the nearest cent.
As a check, the active-state derivative is negative in both steps, so stepping backward increases its reserve. The disabled-state derivative is positive in both steps, so stepping backward decreases its reserve. These directions match the computed changes. The example also illustrates why one terminal value cannot stand in for all states.
Worked Example: Separating a Terminal Payment From a Backward Step
A hypothetical one-year contract pays a $900 maturity benefit and collects a final premium of $100 at \(T=1\), with the premium due immediately before the benefit. Continuous premiums of $40 per year are collected before \(T\). There are no decrements, the force of interest is \(0.02\), and the reserve equation before maturity is \(V'=0.02V+40\). Estimate the reserve at \(t=0.5\) using one explicit Euler step.
Method. First form the pre-transaction terminal reserve by netting the simultaneous endpoint cash flows. Then use that value as the later endpoint in the backward step.
Immediately before the transactions, \(V(1-)=900-100=\$800\). After the premium and benefit have been paid and the contract has ended, the reserve is zero; that post-termination value is not the starting value for the backward integration over the in-force contract.
The derivative at \(t=1-\) is \(V'(1-)=0.02(800)+40=16+40=56\). With \(h=0.5\), \(V(0.5)\approx800-0.5(56)=800-28=\$772.00\).
The calculation avoids substituting the post-termination reserve of zero for the pre-payment boundary. Doing so would incorrectly erase the $800 net amount due at maturity and would not represent the value of obligations just before the endpoint cash flows.
Common Errors and Practical Checks
Terminal conditions are easy to misread when a problem mentions both a final payment and termination. Before calculating, write down the reserve timing and the endpoint transactions explicitly. Then keep that convention unchanged through every numerical step.
- Starting from the wrong side of the endpoint. A maturity benefit belongs in the reserve just before payment, not in the post-termination reserve.
- Ignoring the final premium. If a premium is due at the endpoint before the reserve is measured, net it against the benefit. Do not include it if the contract says it is not due.
- Using one boundary for every state. State-contingent endpoint payments require state-contingent terminal reserves.
- Using the wrong Euler label or update. The update \(V(t-h)\approx V(t)-h f(t,V(t))\) evaluates the derivative at the known later endpoint. It is explicit Euler applied backward in time, not implicit backward Euler.
- Using an unsuitable step size without checking. A numerical step is an approximation. Check signs, units, cash-flow timing, and arithmetic; if accuracy is important, compare calculations using smaller steps.
The terminal condition makes the reserve equation a boundary-value calculation: the contract endpoint is known, while earlier values are required. Moving backward preserves that known endpoint as the anchor and reveals the reserve at earlier durations. Getting the endpoint timing right is essential before any numerical method can be trusted.
Check Your Understanding
For each question, state the endpoint convention before calculating.
- A contract ends at \(T=2\) and pays a $750 benefit at \(T\). No premium or other payment is due then. What are the reserves immediately before the benefit and after termination? (Answer: \(V(2-)=\$750\); \(V(2+)=\$0\))
- A one-year contract has \(V(1-)=\$500\) and \(V'=0.04V+30\). Use one explicit Euler step of length \(0.25\) backward from \(t=1\). (Answer: \(V(0.75)\approx\$487.50\))
- At \(T=1\), two states have terminal reserves \(V_A(1-)=\$300\) and \(V_D(1-)=\$500\). At the endpoint the derivatives are given by \(V_A'=0.02V_A+0.05(V_A-V_D)\) and \(V_D'=0.02V_D+0.10(V_D-V_A)\). Use one explicit Euler step of length \(0.1\) backward. (Answer: \(V_A(0.9)\approx\$300.40\), \(V_D(0.9)\approx\$497.00\))
- A maturity benefit of $1,050 and a final premium of $150 are both due at \(T\), with the premium collected immediately before the benefit. Find the reserve just before the transactions and after termination. (Answer: \(V(T-)=\$900\); \(V(T+)=\$0\))
- A reserve satisfies \(V'=0.10V\), with \(V(1)=\$100\). Using explicit Euler applied backward with \(h=0.5\), calculate \(V(0.5)\). Why is this not the implicit backward Euler step? (Answer: \(V(0.5)\approx\$95.00\); the derivative is evaluated at the known later value \(V(1)\), rather than at the unknown \(V(0.5)\))