Why Modified Reserves Must Be State-Specific
A policyholder's current state can materially change the value of future benefits and premiums. A hypothetical long-term care contract, for example, may have one set of future cash flows while the insured is active and another while the insured is receiving care. A modified valuation basis may also use different premiums, expenses, benefits, interest, or transition intensities from the contractual basis. The reserve must therefore be calculated for each state using the cash flows and assumptions specified for that valuation basis.
This tutorial develops the state-by-state prospective reserve and its differential equation. The essential discipline is to identify, for each state, the modified continuous cash flow, the modified payment at each transition, the intensity of that transition, and the reserve in the destination state.
Assume a continuous-time Markov model with a finite set of states. For distinct states \(i\) and \(j\), let \(\mu_{ij}^M(t)\) be the modified-basis transition intensity from \(i\) to \(j\). Conditional on being in state \(i\) at time \(t\), the probability of transition to \(j\) during the next interval of length \(h\) is \(\mu_{ij}^M(t)h+o(h)\). Assume the intensities and valuation cash flows are suitably integrable, and use a specified force of interest \(\delta^M(t)\). The equations below apply between payment boundaries; a lump sum at a boundary must be included in the boundary condition.
The Multiple-State Thiele Equation
For a state \(i\), the reserve changes both because time passes and because the policy may leave that state. The modified-basis Thiele equation is
The term in square brackets is the net change in the insurer's liability on a transition: pay \(b_{ij}^M(t)\), then replace the reserve for state \(i\) by the reserve for state \(j\). A transition to an absorbing state with no future liability has \(V_j^M(t)=0\), but its transition payment must still be included if one is specified.
A short derivation confirms the signs. Over an interval of length \(h\), conditional on state \(i\) at \(t\), either there is no transition or one transition to a state \(j\). Discounting over that interval, the reserve satisfies, to first order,
Use \(e^{-\delta h}=1-\delta h+o(h)\) and \(V_i^M(t+h)=V_i^M(t)+h\,dV_i^M(t)/dt+o(h)\). Subtract \(V_i^M(t)\), divide by \(h\), and let \(h\) approach zero. The resulting equation is the multiple-state Thiele equation above. In particular, positive net outgo \(g_i^M\) lowers the reserve derivative, while a premium, which makes \(g_i^M\) smaller, raises it relative to the otherwise identical cash-flow pattern.
Worked Calculations
Worked Example: One-Year Reserves with Active, Disabled, and Death States
A hypothetical one-year contract has active state \(A\), disabled state \(D\), and absorbing death state \(X\). On the modified basis, the constant intensities per year are \(\mu_{AD}^M=0.20\), \(\mu_{AX}^M=0.10\), and \(\mu_{DX}^M=0.30\). There are no continuous cash flows or transition payments. The modified force of interest is \(\delta^M=0.05\). At year 1, the reserve is $1,000 in \(A\), $1,600 in \(D\), and zero in \(X\). Find the reserves at time 0 in states \(A\) and \(D\).
Method. Calculate the one-year state probabilities under the modified intensities, value the terminal reserve conditional on the initial state, and discount it. Starting in \(A\), the total rate of leaving \(A\) is \(0.20+0.10=0.30\), so
To be in \(D\) at year 1, the transition from \(A\) to \(D\) must occur at time \(s\), and the policy must then remain in \(D\) to year 1. Since the total exit rate from \(D\) is also \(0.30\),
The remaining probability is death: \(1-0.740818-0.148164=0.111018\). It has no terminal reserve. The undiscounted expected terminal reserve starting in \(A\) is
Thus \(V_A^M(0)=e^{-0.05}(977.880)=\$930.19\), rounded to the nearest cent. Starting in \(D\), the probability of remaining there for one year is \(e^{-0.30}=0.740818\), so
The calculation is checked by conditioning only on states with nonzero terminal reserves; the death-state probability contributes zero. A common mistake is to use the active-state terminal amount for every policy alive at year 1. The disabled-state amount must instead be weighted by the probability of being in \(D\).
Worked Example: Including a Payment at Disability Onset
Use states \(A,D,X\) for active, disabled, and death. The modified intensities per year are \(\mu_{AD}^M=0.20\), \(\mu_{AX}^M=0.10\), and \(\mu_{DX}^M=0.10\). The modified force of interest is \(0.05\). There are no continuous cash flows. At year 1 the terminal reserves are $900 in \(A\), $1,400 in \(D\), and zero in \(X\). A modified-basis outgo of $100 is paid immediately upon transition \(A\to D\). Calculate \(V_A^M(0)\) and \(V_D^M(0)\).
Method. For a policy starting in \(A\), the probability of remaining active to year 1 is \(e^{-0.30}=0.740818\). The probability of being disabled at year 1 is obtained by integrating over the onset time \(s\). The active exit rate is \(0.30\), and the disabled death intensity is \(0.10\):
The expected terminal reserve before discounting is therefore \((0.740818)(900)+(0.164019)(1400)=666.736+229.627=896.363\). Discounting gives \(e^{-0.05}(896.363)=\$852.65\).
The transition payment is valued separately. The probability density of a transition from \(A\) to \(D\) at time \(s\), before any other exit from \(A\), is \(0.20e^{-0.30s}\). Discounting the $100 payment to time 0 gives
Consequently \(V_A^M(0)=852.65+16.87=\$869.52\), rounded to the nearest cent. Starting in \(D\), the probability of surviving in that state to year 1 is \(e^{-0.10}\). Hence \(V_D^M(0)=e^{-0.05}e^{-0.10}(1400)=1400e^{-0.15}=\$1{,}204.99\). The onset payment is included only for a policy starting in \(A\) that experiences the transition; it is not an extra terminal benefit.
Worked Example: State-Specific Reserve Derivatives at a Boundary
At a particular time, a hypothetical modified basis has states \(A,D,X\), with \(X\) absorbing and \(V_X^M=0\). The force of interest is \(0.04\). The current reserves are \(V_A^M=500\) and \(V_D^M=800\). In state \(A\), benefits and expenses total $50 per year and premiums are $100 per year. In state \(D\), benefits and expenses total $220 per year and there is no premium. The intensities are \(\mu_{AD}^M=0.15\), \(\mu_{AX}^M=0.05\), and \(\mu_{DX}^M=0.25\). A $300 payment is made on transition \(A\to D\); other transition payments are zero. Find the two reserve derivatives.
Method. First form net continuous outgo: \(g_A^M=50-100=-50\) and \(g_D^M=220\). For state \(A\), the transition terms are \(0.15(300+800-500)=90\) and \(0.05(0+0-500)=-25\). Their sum is \(65\). Therefore
For state \(D\), the only transition term is \(0.25(0+0-800)=-200\). Thus
Both transition payments and destination-state reserves are included. A frequent sign error is to subtract \(V_j^M-V_i^M\) without also treating the transition payment consistently. Using the bracketed net change \(b_{ij}^M+V_j^M-V_i^M\) avoids that error.
Applying the Equations Reliably
A multiple-state modified reserve calculation is best organized state by state. The assumptions need not be the same in every state: for example, a modified basis may specify a premium only while active, a different expense rate while disabled, or distinct transition intensities by state. Each state equation uses its own \(g_i^M\) and all applicable outgoing transitions. The reserve in a destination state enters even when the transition itself has no payment.
For each state, identify every possible next state, including absorbing states, and record the modified transition intensity.
In each state calculate benefits and expenses less premiums as \(g_i^M\). Separately record transition outgo \(b_{ij}^M\).
Specify the terminal reserve for each state under the modified basis before solving backward.
Use the same force of interest, time units, and sign convention throughout; verify that every outgoing transition appears once in the originating state's equation.
The state-specific structure matters because a reserve conditional on being active is not interchangeable with one conditional on being disabled. Even if the valuation interest rate is common, different benefit streams, premiums, transition payments, or intensities produce different prospective values. When a model is time-dependent, the same equation applies with time-varying quantities, but the reserve generally requires solving a system of differential equations rather than using a constant-rate shortcut.
The formulas require a coherent Markov transition model and valuation assumptions that define the future cash flows conditional on the current state. They do not permit an intensity to be treated as a one-year probability without conversion. For a constant total exit intensity \(\mu\), the probability of no exit over one year is \(e^{-\mu}\), not \(1-\mu\), except as a small-intensity approximation.
Check Your Understanding
Use the stated modified basis in each problem. Answers are rounded to the nearest cent where monetary amounts are requested.
- A one-year model has active and disabled states, with death absorbing. The intensities are \(\mu_{AD}^M=0.10\), \(\mu_{AX}^M=0.10\), and \(\mu_{DX}^M=0.20\). There are no interim cash flows, \(\delta^M=0.03\), and terminal reserves are $500 in active and $900 in disabled. Find both time-0 reserves. (Answer: \(V_A^M(0)=\$468.77;\ V_D^M(0)=\$715.08\).)
- At a given time, \(V_A^M=300\), \(V_B^M=500\), and \(V_X^M=0\). The force is \(0.05\), \(g_A^M=-20\), \(\mu_{AB}^M=0.20\), \(\mu_{AX}^M=0.10\), and the transition payment \(A\to B\) is $50. Find \(dV_A^M/dt\). (Answer: \(15\) dollars per year.)
- Over one year, state \(A\) can transition to \(B\) at intensity \(0.10\) or death at intensity \(0.10\); \(B\) has death intensity \(0.10\). The force is \(0.02\), terminal reserves are $1,000 in \(A\), $1,200 in \(B\), and zero in death, and a $40 payment is made at \(A\to B\). There are no continuous cash flows. Find the reserve starting in \(A\). (Answer: \(\$907.39\).)
- At a given time, \(V_C^M=700\) and the only transition from \(C\) is to an absorbing state \(X\), with \(\mu_{CX}^M=0.12\). Let \(\delta^M=0.03\), \(g_C^M=40\), and \(b_{CX}^M=100\). Calculate \(dV_C^M/dt\). (Answer: \(-\$28\) per year.)
- Explain why a transition payment at \(A\to D\) belongs in the equation for \(V_A^M\), and state the term in that equation that accounts for the reserve after transition. (Answer: It is an outgo incurred when leaving \(A\), so it appears as \(b_{AD}^M\) in \(\mu_{AD}^M[b_{AD}^M+V_D^M-V_A^M]\); \(V_D^M\) is the destination-state reserve.)