Why Care-Level Reserves Need a System
A long-term care contract may pay different continuous benefit rates at different levels of care. A policyholder may also recover from care, move to a more intensive care level, or die. Because the future cash flows depend on the policyholder’s current state, one reserve value is not enough: the contract has a reserve for each living state.
This tutorial formulates those state-specific reserves and shows how transition intensities enter each equation. The model is a continuous-time Markov model: conditional on the current state and time, transition probabilities over a short interval are determined by the specified transition intensities. Rates and benefit amounts below are hypothetical. The methods are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Set Up the State-Specific Cash Flows
For a living state \(j\), let \(g_j(t)\) be the continuous net outgo rate: benefits and expenses paid continuously, minus premiums received continuously. Thus a care benefit of $12,000 per year with no other cash flow gives \(g_j=12{,}000\), while a premium of $1,000 per year and no benefit gives \(g_j=-1{,}000\). This sign convention keeps premiums and benefits in the same equation.
Let \(\lambda_{jk}(t)\) be the transition intensity from living state \(j\) to state \(k\). In the equations below, living-to-living transitions have no immediate lump-sum payment. Let \(\mu_j(t)\) be the intensity of death from state \(j\), and let \(B_j(t)\) be the benefit paid immediately if death occurs from that state. Death is represented by the absorbing state \(D\), with \(V_D=0\). The general equation for this model is:
The sum includes each living state directly reachable from \(j\). For example, if recovery from \(C_1\) leads to \(H\), and worsening leads from \(C_1\) to \(C_2\), both transitions appear in the \(C_1\) equation. A state with no direct transition to another state has no corresponding term; do not insert a transition merely because it is possible later by way of another state.
Why the Thiele Equation Has These Signs
A short-interval valuation explains the equation and its signs. Consider a time interval of length \(h\) while the policyholder is in living state \(j\). To first order in \(h\), the probability of a transition from \(j\) to \(k\) is \(\lambda_{jk}(t)h\), the probability of death is \(\mu_j(t)h\), and the probability of remaining in \(j\) is one minus the sum of these probabilities. The continuous net outgo over the interval is approximately \(g_j(t)h\).
Discount future reserves back across the interval at force of interest \(\delta(t)\), equate the resulting expected value plus the interval’s outgo to \(V_j(t)\), and retain terms of first order in \(h\). A transition to \(k\) replaces the future reserve \(V_j\) by \(V_k\); a death replaces it by zero and pays \(B_j\). Rearranging gives:
The signs also have a direct interpretation. Continuous benefits increase net outgo and therefore enter with a negative sign in the differential equation; premiums reduce net outgo and enter positively. A transition toward a state with a lower reserve contributes positively through \(V_j-V_k\), while a death benefit larger than the current reserve contributes negatively through \(V_j-B_j\). A derivative describes how the reserve changes as calendar time moves forward; it is not itself the reserve or the expected benefit amount.
Writing the Long-Term Care System
For a common care-level structure, allow transitions \(H\to C_1\), \(C_1\to H\), \(C_1\to C_2\), and \(C_2\to C_1\), as well as death from each living state. Denote the corresponding living-state intensities by \(\alpha,\rho_1,\gamma,\rho_2\), respectively. Denote the state-specific death intensities by \(\mu_H,\mu_1,\mu_2\). Then the equations are:
Here \(V_1\) and \(V_2\) abbreviate \(V_{C_1}\) and \(V_{C_2}\). The healthy state has no direct recovery term in this structure, and the highest care state has no direct worsening term. Those omissions reflect the stated model, not a general rule about care contracts. If a product permits another transition, add its intensity and reserve difference to the equation for the state it leaves.
For a contract ending at time \(T\), specify a terminal reserve for each living state. If no benefit or expense remains after \(T\), the usual terminal conditions are \(V_H(T)=V_1(T)=V_2(T)=0\). If a maturity payment or other cash flow is due at \(T\), the terminal values must reflect it. These are backward differential equations: the terminal conditions are at the end of the coverage period, and the reserves are found at earlier times.
Worked Example: Forming and Evaluating Three Reserve Derivatives
A hypothetical contract has states \(H,C_1,C_2,D\). Use thousands of dollars and years. The force of interest is \(\delta=0.04\). Healthy policyholders pay a continuous premium of $1.2 thousand per year, and care benefits are $12 thousand and $24 thousand per year in \(C_1\) and \(C_2\). The living-state intensities are \(H\to C_1:0.08\), \(C_1\to H:0.20\), \(C_1\to C_2:0.10\), and \(C_2\to C_1:0.15\), all per year. Death intensities are \(0.01,0.03,0.05\) per year in \(H,C_1,C_2\), respectively, with a $100 thousand death benefit in each state. At a particular time the reserves are \(V_H=20,V_1=50,V_2=80\). Find the three reserve derivatives.
Method. Use the state-specific Thiele equations, with all money in thousands. Net outgo is \(g_H=-1.2\), \(g_1=12\), and \(g_2=24\). The negative healthy-state net outgo reflects premium income.
For the healthy state:
For the first care level, recovery leads to \(H\), worsening leads to \(C_2\), and death triggers the death benefit:
For the second care level:
The derivatives are \(-$1.2\), \(-$8.5\), and \(-$17.3\) thousand per year, respectively. Each negative derivative means the reserve is decreasing locally as time advances at the stated reserve values and rates. This calculation is a local derivative evaluation; the three numbers alone do not determine the reserve path. A common error is to use \(B_j-V_j\) for the death term while retaining the equation above. That reverses the death contribution: the correct term is \(\mu_j(V_j-B_j)\), which here is negative whenever the death benefit exceeds the reserve.
Constant-Force Reserves for Perpetual Care Benefits
A useful special case has constant forces, constant continuous benefits, no premiums, and no terminal date: care benefits continue for as long as the policyholder remains alive. If the reserves are constant in time, each derivative is zero. Setting the derivative equations equal to zero produces linear equations for the reserves. This is a stationary special case, not the general solution for a finite-term contract.
Worked Example: Solving a Perpetual Three-State Care Reserve
A hypothetical contract pays continuous benefits of $10 thousand per year in \(C_1\) and $20 thousand per year in \(C_2\), with no benefit in \(H\). There are no premiums or death benefits. The force of interest is \(0.05\). The intensities per year are \(H\to C_1:0.10\), death from \(H:0.05\), \(C_1\to H:0.20\), \(C_1\to C_2:0.10\), death from \(C_1:0.05\), \(C_2\to C_1:0.25\), and death from \(C_2:0.05\). Find the stationary reserves.
Method. Use \(g_H=0\), \(g_1=10\), \(g_2=20\), set all derivatives to zero, and use \(V_D=0\) with zero death benefits. In state \(H\):
In state \(C_1\), the total coefficient on \(V_1\) is the interest force plus all three outgoing intensities, \(0.05+0.20+0.10+0.05=0.40\). Therefore:
In state \(C_2\), the total coefficient on \(V_2\) is \(0.05+0.25+0.05=0.35\), so:
Multiply the \(C_1\) equation by 10 to get \(3V_1-V_2=100\). Multiply the \(C_2\) equation by 20 to get \(7V_2-5V_1=400\). Substitute \(V_2=3V_1-100\) into the second equation:
Thus \(V_2=3(68.75)-100=106.25\), and \(V_H=0.5(68.75)=34.375\), all in thousands of dollars. Check the \(C_1\) equation: \(0.30(68.75)-0.10(106.25)=20.625-10.625=10\). Check the \(C_2\) equation: \(0.35(106.25)-0.25(68.75)=37.1875-17.1875=20\). Both balances hold. The lower healthy-state reserve is consistent with the possibility of entering care, but reserves should be determined from the equations rather than guessed from benefit amounts.
Time Units, Boundaries, and Exam Checks
Intensities, the force of interest, and continuous benefit rates must use the same time unit. If the model uses years, an intensity stated per month must be converted to a per-year intensity before combining it with annual rates. Likewise, a monthly continuous benefit rate of $900 per month corresponds to $10,800 per year, not $900 per year. Do not confuse this continuous-rate calculation with a contract that actually pays a lump sum at the end of each month.
A useful consistency check is to inspect each transition term separately. A transition from \(j\) to \(k\) contributes \(\lambda_{jk}(V_j-V_k)\); if the destination reserve is greater than the origin reserve, the term is negative. Death contributes \(\mu_j(V_j-B_j)\). The outgoing transition intensities, including death, must each appear once in the equation for the state being left. Also check that a premium is entered as negative net outgo and that the terminal condition matches the contract’s actual final cash flows.
Worked Example: Converting a Monthly Care Rate Before Evaluating the Equation
A hypothetical \(C_2\) benefit is paid continuously at an equivalent rate of $1,250 per month. Use years and thousands of dollars. At the time of valuation, \(V_2=60\), \(V_1=30\), and \(V_D=0\). The force of interest is \(0.05\), recovery from \(C_2\) to \(C_1\) has intensity \(0.10\) per year, and death has intensity \(0.04\) per year with a $40 thousand death benefit. There are no other cash flows or transitions from \(C_2\). Find \(V_2'\).
Method. Convert the continuous benefit rate first: \(12(1.25)=15\) thousand dollars per year. Therefore \(g_2=15\). Apply the \(C_2\) equation:
The reserve derivative is \(-$8.2\) thousand per year. The check is dimensionally consistent: each term is a reserve in thousands multiplied by a rate per year, or a benefit flow in thousands per year. A common wrong calculation uses \(g_2=1.25\), as though the stated monthly amount were already an annual benefit rate. That gives \(3-1.25+3+0.8=5.55\) thousand per year, which is wrong because it understates the annualized continuous benefit by a factor of 12.
Check Your Understanding
Use the stated reserve convention and report monetary calculations in thousands of dollars unless otherwise specified.
- In \(H\), \(V_H=15\) and \(V_1=35\). The force of interest is \(0.04\), the continuous premium is $0.5 thousand per year, \(H\to C_1\) has intensity \(0.10\), and death has intensity \(0.02\) with a $50 thousand death benefit. Find \(V_H'\). (Answer: \(-1.600\) thousand per year)
- In \(C_1\), let \(V_1=42\), \(V_H=18\), and \(V_2=68\). The force of interest is \(0.03\), the continuous care benefit is $9.6 thousand per year, recovery intensity is \(0.20\), worsening intensity is \(0.10\), and death intensity is \(0.04\) with a $30 thousand death benefit. Find \(V_1'\). (Answer: \(-5.660\) thousand per year)
- A perpetual contract has no death benefit, a continuous benefit of $12 thousand per year in \(C_1\), and no benefits in \(H\). The force of interest is \(0.04\); intensities are \(H\to C_1:0.08\), death from \(H:0.02\), \(C_1\to H:0.16\), and death from \(C_1:0.04\). There is no \(C_2\) state. Find the stationary reserves in \(C_1\) and \(H\). (Answer: \(V_1=80.769\) thousand and \(V_H=46.154\) thousand)
- A \(C_2\) benefit is paid continuously at an equivalent rate of $1,250 per month. The force of interest is \(0.05\), \(V_2=60\), \(V_1=30\), recovery intensity is \(0.10\) per year, and death intensity is \(0.04\) per year with a $40 thousand death benefit. Find \(V_2'\), with no other cash flows or transitions from \(C_2\). (Answer: \(-8.200\) thousand per year)