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Exam ALTAM · Long-Term Care Models · Tutorial 195 of 999

Long-Term Care Model Validation and Common Errors

Use a structured validation process to check that an LTC model’s states, transitions, probabilities, and benefit payments are internally consistent.

Intermediate 12 min read

What You'll Learn

  • Check transition directions against the stated state definitions and matrix convention
  • Validate intensity-matrix diagonals and row sums
  • Test short-interval state probabilities for nonnegativity and conservation
  • Match benefit eligibility to the contract’s trigger and payment date
  • Distinguish recovery, relapse, and death in state and payment calculations
  • Diagnose common timing and rate-unit errors in LTC valuations

Why LTC Model Validation Matters

A long-term care (LTC) valuation can produce a plausible-looking answer even when its state transitions or payment rules are wrong. A recovery rate entered as a death rate, a monthly probability confused with an annual intensity, or a benefit paid after death can each distort a result without making the arithmetic look obviously unreasonable. Validation therefore starts with the model structure, not with the final present value.

This tutorial gives a systematic way to check transition directions, intensity matrices, state probabilities, eligibility conditions, payment timing, recovery, and death. The methods are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: In a continuous-time Markov model, the insured occupies one state at time \(t\). The intensity \(q_{ij}(t)\), for \(i\ne j\), is the instantaneous rate of transition from state \(i\) to state \(j\), conditional on being in state \(i\) at time \(t\). Under a constant-intensity model, the rates do not vary with time. The intensity matrix \(Q(t)\) has off-diagonal entries \(q_{ij}(t)\) and diagonal entries chosen so each row sums to zero.

The state labels and matrix convention must be explicit. Here, rows denote the state currently occupied and columns denote the destination state. For example, \(q_{CR}\) means a transition from care to recovery; it does not mean a transition from recovery to care. Death is usually represented as an absorbing state: once entered, no further transitions occur.

A Validation Sequence

A useful audit follows the model from its assumptions through to its cash flows. Each check targets a different kind of error; passing one check does not guarantee that the others pass.

1
Name the states and allowed transitions.
State what each label means and whether each transition is possible. In particular, distinguish care-to-recovery from recovery-to-care.
2
Check the intensity matrix.
Off-diagonal entries must be nonnegative transition intensities. Each diagonal entry must be the negative sum of the other entries in its row.
3
Check probabilities and units.
State probabilities must be between zero and one and sum to one. Intensities must use the same time unit as the calculation interval.
4
Apply the contract definition of eligibility.
Identify which states qualify for payment, and whether eligibility is tested at entry, throughout a period, or on the payment date.
5
Place each cash flow on its actual payment date.
Account for recovery and death before that date if the contract requires the insured to be eligible then.

For a row-vector of state probabilities \(p(t)\) and a constant intensity matrix \(Q\), the forward equation is \(p'(t)=p(t)Q\). For a time-varying intensity matrix, it becomes \(p'(t)=p(t)Q(t)\). This equation tracks the probability mass entering and leaving each state. It does not, by itself, determine which states trigger a benefit; that comes from the contract.

Key check: For any valid intensity matrix, \(q_{ij}\ge 0\) for \(i\ne j\), \(q_{ii}=-\sum_{j\ne i}q_{ij}\), and every row sums to zero. For a valid state-probability vector, every component is nonnegative and the components sum to one.

Why the Row Sums Must Be Zero

Over a short interval of length \(h\), conditional on starting in state \(i\), the probability of moving to a different state \(j\) is approximately \(q_{ij}h\). The probability of remaining in state \(i\) is therefore approximately one minus the sum of those probabilities. This gives the diagonal entry.

$$ P_{ii}(h)\approx 1-h\sum_{j\ne i}q_{ij}, \qquad q_{ii}=-\sum_{j\ne i}q_{ij}. $$

The same requirement ensures probability conservation. Let \(\mathbf{1}\) be a column vector of ones. Because the row sums of \(Q\) are zero, \(Q\mathbf{1}=0\). Thus

$$ \frac{d}{dt}\{p(t)\mathbf{1}\} =p'(t)\mathbf{1} =p(t)Q\mathbf{1} =0. $$

So if the initial state probabilities sum to one, their total remains one. This is a structural check, not a substitute for checking each probability or the benefit rules.

Worked Example: Repairing an LTC Intensity Matrix

Consider a hypothetical model with states \(A\) (active and not receiving care), \(C\) (receiving eligible care), \(R\) (recovered and not receiving care), and \(D\) (dead). Assume constant intensities per year. The allowed transitions are \(A\to C\) at \(0.12\), \(A\to D\) at \(0.01\), \(C\to R\) at \(0.25\), \(C\to D\) at \(0.10\), \(R\to C\) at \(0.04\), and \(R\to D\) at \(0.02\). Death is absorbing. An analyst records the \(C\)-row diagonal as \(-0.30\). Audit and correct the matrix, then approximate the state probabilities after \(0.1\) year for a person starting in care.

Method. Use rows as origin states and columns as destination states, in the order \(A,C,R,D\). The \(C\)-row off-diagonal entries are \(0\), \(0.25\), and \(0.10\). Their sum is \(0.35\), so the diagonal must be \(-0.35\), not \(-0.30\). The corrected matrix is

$$ Q= \begin{pmatrix} -0.13&0.12&0&0.01\\ 0&-0.35&0.25&0.10\\ 0&0.04&-0.06&0.02\\ 0&0&0&0 \end{pmatrix}. $$

The row sums are \(-0.13+0.12+0.01=0\), \(0-0.35+0.25+0.10=0\), \(0+0.04-0.06+0.02=0\), and \(0\). The off-diagonal entries have the stated transition directions, and the death row is zero because death is absorbing.

For a short interval \(h\), use the first-order approximation \(p(h)\approx p(0)+h\,p(0)Q\). Starting in care gives \(p(0)=(0,1,0,0)\). Therefore \(p(0)Q=(0,-0.35,0.25,0.10)\), and

$$ p(0.1)\approx(0,1,0,0)+0.1(0,-0.35,0.25,0.10) =(0,0.965,0.025,0.010). $$

The approximate probabilities are nonnegative and sum to \(0.965+0.025+0.010=1.000\). If the analyst had used \(-0.30\), the same calculation would give a total of \(1.005\), which is impossible. This is a short-interval approximation, not an exact calculation for a long interval. The common mistake is to set a diagonal to an independently chosen value rather than to the negative sum of the outgoing intensities.

Check Eligibility and Payment Timing Separately

A correct transition model can still be paired with an incorrect benefit calculation. Write the eligibility rule in words before calculating: for example, “a payment is made only if the insured is alive and in an eligible care state on the payment date.” That rule differs from paying for any time spent in care during the period, or paying immediately upon care entry.

Payment timing also determines which probability is relevant. A benefit paid continuously while care continues depends on the time spent in care. A benefit paid at the end of a month, conditional on being in care then, depends on the probability of occupying the eligible state on that date. If recovery or death ends eligibility, both are exits from the eligible state for this purpose.

Worked Example: A Quarterly Payment Conditional on Remaining in Care

A hypothetical LTC contract pays \( \$900\) at the end of a three-month period if the insured is alive and still receiving eligible care on that date. At the start of the period, the insured is in care. Assume constant annual intensities of \(0.24\) from care to recovery and \(0.12\) from care to death. There are no other exits from care. Find the expected payment for the period, before interest.

Method. The total rate of leaving care is \(0.24+0.12=0.36\) per year. Under constant intensities, the probability of remaining in care for \(t\) years is \(e^{-0.36t}\). Three months is \(t=3/12=0.25\) year, so the probability of being in care at the payment date is

$$ P(\text{eligible at }0.25) =e^{-0.36(0.25)} =e^{-0.09} =0.913931. $$

The payment is made only when that eligibility condition is met. Thus the expected payment is \( \$900(0.913931)=\$822.538\), or \(\$822.54\) to the nearest cent.

The result is a payment-date probability, not the probability of ever entering care: the insured started in care already. It also counts both recovery and death as exits before payment. A common error is to treat the annual intensity \(0.36\) as a three-month exit probability. The probability of remaining in care is instead \(e^{-0.36(0.25)}\), which reflects the duration and the continuous-time assumption.

Recovery, Relapse, and Death Are Distinct

Recovery is not the same event as death. A recovered person may be alive but ineligible for a care benefit; if relapse is allowed, that person may later return to care. A model that removes recovered lives permanently, or that counts them as continuing care claimants, can misstate both benefit probabilities and expected payments. Likewise, death ends future eligibility even if the insured was receiving care immediately before death.

When checking a model with recovery and relapse, verify that the care-to-recovery and recovery-to-care intensities occupy the correct cells. Then check whether death intensities apply in every living state and whether the contract pays at the time of care, at the end of a period, or only after a qualifying assessment. These are separate model and contract questions.

Worked Example: Relapse, Death, and an End-of-Year Benefit

A hypothetical policy pays \( \$5{,}000\) at the end of one year if the insured is alive and in care on that date. The insured starts in care. Assume constant annual intensities: care-to-recovery \(0.40\), recovery-to-care \(0.20\), and death \(0.10\) from either living state. Assume the death intensity is independent of the care/recovery transitions, as indicated by its equal value in both living states. The effective annual interest rate is \(i=5\%\). Find the expected present value of the year-end payment.

Method. First ignore death and calculate the probability of being in care after one year in the two-state care/recovery process. Its intensity matrix is

$$ Q_{CR}= \begin{pmatrix} -0.40&0.40\\ 0.20&-0.20 \end{pmatrix}. $$

The long-run care probability for this two-state process is \(0.20/(0.40+0.20)=1/3\). Starting in care, the care probability at time \(t\), conditional on following this process, is

$$ \frac{1}{3}+\left(1-\frac{1}{3}\right)e^{-(0.40+0.20)t}. $$

At \(t=1\), this is \(1/3+(2/3)e^{-0.6}=0.699208\). The probability of surviving the independent death intensity for one year is \(e^{-0.10}=0.904837\). Therefore the unconditional probability of being alive and in care at year-end is

$$ 0.699208(0.904837)=0.632669. $$

The expected year-end payment is \( \$5{,}000(0.632669)=\$3{,}163.35\) to the nearest cent. Since the payment is made at the end of year one, its present value is divided by \(1+i=1.05\):

$$ \operatorname{EPV} =\frac{\$3{,}163.35}{1.05} =\$3{,}012.71. $$

As a probability check, the probability of being alive in either care or recovery at year-end is \(e^{-0.10}=0.904837\). The corresponding recovery probability is \(0.904837-0.632669=0.272168\), and the death probability is \(1-0.904837=0.095163\). These three probabilities sum to one, up to rounding. The common mistake is to treat recovery as death or as continued eligibility; recovery is a separate living state, and only the care state triggers this particular payment.

Common Errors to Catch Before Valuation

Potential errorWhy it failsValidation question
Reversing a transition directionA row-to-column convention assigns a distinct meaning to each matrix entry.Does each nonzero entry agree with its origin and destination states?
Using an incorrect diagonalThe resulting row may not sum to zero, violating probability conservation.Is each diagonal the negative sum of the other entries in its row?
Using an intensity as a period probabilityAn intensity is a rate, not a probability for an arbitrary interval.Have the rate’s time unit and the interval length been used consistently?
Paying a benefit after death or recoveryThe cash flow may require eligibility on a specific date or throughout a period.Does the assumed state at the actual payment date satisfy the contract trigger?
Omitting relapse or counting recovery as permanent exitEligible care can recur if the model allows a transition back from recovery.Are recovery and relapse represented as separate, correctly directed transitions?

For a final reasonableness check, a benefit probability should not exceed one, and expected payments should not exceed the maximum payment multiplied by the probability of eligibility when a single payment is possible. A present value should also be no greater than its undiscounted expected payment when the only discounting is at a positive interest rate and the payment is made in the future.

Check Your Understanding

For each problem, state the relevant model check or calculation and compare your result with the answer.

  1. A care state has annual intensities \(0.18\) to recovery and \(0.07\) to death, with no other exits. What must the care-state diagonal be in the intensity matrix? (Answer: \(-0.25\) per year.)
  2. A person starts in care. The annual intensities are \(0.30\) to recovery and \(0.10\) to death. Use the first-order short-interval approximation over \(0.2\) year to give the probabilities of care, recovery, and death, in that order. (Answer: \((0.920,\,0.060,\,0.020)\).)
  3. A contract pays \(\$800\) at the end of one month if the insured, who starts in care, is still in care. The constant annual exit intensities are \(0.30\) to recovery and \(0.20\) to death. Find the expected payment before interest, to the nearest cent. (Answer: \(\$767.35\).)
  4. A two-state care/recovery model has care-to-recovery intensity \(0.30\) and recovery-to-care intensity \(0.10\) per year, with no death. Starting in care, what is the probability of being in care after one year, to four decimals? (Answer: \(0.7527\).)
  5. A single payment of \(\$1{,}200\) is made at year-end only if the insured is alive and eligible then. The probability of being eligible at year-end is \(0.62\), and the effective annual interest rate is \(4\%\). Find the expected present value to the nearest cent. (Answer: \(\$715.38\).)
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