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Recurrent Disability with Recovery and Re-Entry

Build and solve state-value equations for disability benefits when recovery can be followed by another disability episode.

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What You'll Learn

  • Represent active, disabled, and dead statuses in a recurrent-disability model
  • Convert transition intensities and benefit rules into state-value equations
  • Include continuous disability payments and lump sums at transitions
  • Derive the value equations using a short-time recursion
  • Solve, check, and interpret values from each possible initial state

Why Recovery Does Not End the Valuation

In a recurrent-disability model, recovery returns a claimant to active status rather than ending coverage. The claimant may later become disabled again, so a valuation must allow for multiple disability episodes. A model with only one disability episode would omit those possible future payments and would value the benefit incorrectly.

The basic model has three states: active \(A\), disabled \(D\), and dead, which is absorbing. A transition from \(A\) to \(D\) represents the onset of disability; a transition from \(D\) to \(A\) represents recovery. Unless the contract specifies otherwise, each return to \(A\) restores the possibility of a later transition back to \(D\). The method below is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: A recurrent-disability model is a multiple-state model in which a claimant can move from active to disabled, recover to active, and become disabled again. The state-value \(V_A\) is the expected present value of future benefits when currently active; \(V_D\) is the corresponding value when currently disabled. In this tutorial, death has value zero unless a transition payment is explicitly stated.

Assume transition intensities are constant, the force of interest is constant at \(\delta\), and the future transition process depends on the current state, not on how many episodes have occurred or how long the claimant has occupied the state. Suppose a benefit is paid continuously at rate \(b_D\) while disabled. A lump sum may also be paid at a transition, such as an onset payment on \(A\to D\). These assumptions make \(A\) and \(D\) sufficient states for valuation. If benefits depend on episode number or duration, the state description must be expanded.

State-Value Equations

Write \(\lambda_{AD}\) for the intensity of disability onset, \(\mu_A\) for the intensity of death while active, \(\lambda_{DA}\) for the recovery intensity, and \(\mu_D\) for the intensity of death while disabled. Let \(c_{AD}\) and \(c_{DA}\) be lump sums paid immediately at onset and recovery, respectively; set either amount to zero if no such payment applies. The value of future benefits includes all possible later episodes.

Theorem: Under the constant-intensity, constant-force-of-interest model, with continuous benefit rate \(b_D\) in state \(D\), the state values satisfy
$$ (\delta+\lambda_{AD}+\mu_A)V_A =\lambda_{AD}(V_D+c_{AD}), \qquad (\delta+\lambda_{DA}+\mu_D)V_D =b_D+\lambda_{DA}(V_A+c_{DA}). $$

The terms on the left include discounting and every way to leave the current state. On the right, the active value receives an onset payment and continues with value \(V_D\) if disability begins. The disabled value includes the continuous benefit and the recovery payment and continuation value if recovery occurs. Death contributes no continuation value.

Why the Equations Hold

The equations follow by conditioning on what happens over a short interval of length \(h\). From state \(i\), the probability of a particular transition \(i\to j\) during that interval is \(\lambda_{ij}h+o(h)\); the probability of remaining in \(i\) is \(1-\sum_{j\ne i}\lambda_{ij}h+o(h)\). The discount factor over the interval is \(e^{-\delta h}=1-\delta h+o(h)\). Here, \(o(h)\) denotes a term whose ratio to \(h\) tends to zero as \(h\) tends to zero.

For a state with continuous benefit rate \(b_i\), a transition payment \(c_{ij}\), and future state value \(V_j\), the short-interval recursion is

$$ V_i=b_i h+e^{-\delta h} \left[ \left(1-\sum_{j\ne i}\lambda_{ij}h\right)V_i +\sum_{j\ne i}\lambda_{ij}h(V_j+c_{ij}) \right]+o(h). $$

Subtract \(V_i\), divide by \(h\), and let \(h\) tend to zero. The result is

$$ \left(\delta+\sum_{j\ne i}\lambda_{ij}\right)V_i =b_i+\sum_{j\ne i}\lambda_{ij}(V_j+c_{ij}). $$

Applying this expression to \(A\) and \(D\), with zero benefit rate while active and zero value after death, gives the stated equations. The self-consistency is essential: the value in state \(D\) includes benefits after any future recovery and subsequent disability onset.

Worked Valuations

Worked Example: Recurrent Coverage With an Onset Payment

A hypothetical policy pays $5,000 per year continuously while the claimant is disabled. It pays $1,000 immediately at each disability onset. The force of interest is \(\delta=0\). Per year, the transition intensities are \(\lambda_{AD}=0.20\), \(\mu_A=0.06\), \(\lambda_{DA}=0.30\), and \(\mu_D=0.09\). Find the values starting active and disabled. Assume each recovery restores eligibility, and there is no payment at recovery.

Method. Use the two state-value equations, including the onset payment in the \(A\to D\) transition term. Since \(\delta=0\), discounting does not reduce future values.

For an active claimant, the total exit intensity is \(0.20+0.06=0.26\). For a disabled claimant, it is \(0.30+0.09=0.39\). Thus

$$ 0.26V_A=0.20(V_D+1{,}000), \qquad 0.39V_D=5{,}000+0.30V_A. $$

Rearrange the disabled-state equation:

$$ V_D=\frac{5{,}000+0.30V_A}{0.39}. $$

Substitute this expression in the active-state equation:

$$ 0.26V_A =200+\frac{0.20}{0.39}(5{,}000+0.30V_A) =200+2{,}564.102564+0.153846154V_A. $$

Therefore \(0.106153846V_A=2{,}764.102564\), so \(V_A=26{,}038.6473\ldots\). Substituting back gives

$$ V_D=\frac{5{,}000+0.30(26{,}038.6473\ldots)}{0.39} =32{,}850.2415\ldots. $$

To the nearest cent, \(V_A=\$26{,}038.65\) and \(V_D=\$32{,}850.24\). Check the active equation: \(0.26V_A\approx \$6{,}770.05\), while \(0.20(V_D+\$1{,}000)\approx \$6{,}770.05\). Check the disabled equation: \(0.39V_D\approx \$12{,}811.59\), and \(\$5{,}000+0.30V_A\approx \$12{,}811.59\). The equations balance, subject to rounding.

Common mistake avoided. The active equation must use the disabled-state value that follows from the disabled equation. Using an inconsistent value for \(V_D\) can make one equation appear to balance while the other does not.

Worked Example: Positive Interest and Different Exit Rates

A hypothetical income contract pays $12,000 per year continuously while disabled and $500 immediately at each onset. The force of interest is \(\delta=0.04\). The annual transition intensities are \(\lambda_{AD}=0.15\), \(\mu_A=0.02\), \(\lambda_{DA}=0.40\), and \(\mu_D=0.10\). There is no recovery payment. Find \(V_A\) and \(V_D\).

Method. Add the force of interest to the relevant state’s total exit intensity, and include the onset lump sum only on the onset transition.

$$ 0.21V_A=0.15(V_D+500), \qquad 0.54V_D=12{,}000+0.40V_A. $$

The coefficients are \(0.04+0.15+0.02=0.21\) and \(0.04+0.40+0.10=0.54\). From the active equation,

$$ V_A=\frac{0.15V_D+75}{0.21} =\frac{5}{7}V_D+\frac{2{,}500}{7}. $$

Substitute into the disabled equation:

$$ 0.54V_D =12{,}000+0.40\left(\frac{5}{7}V_D+\frac{2{,}500}{7}\right) =12{,}142.857143+0.285714286V_D. $$

It follows that \(0.254285714V_D=12{,}142.857143\), giving \(V_D=47{,}752.80899\ldots\). Then

$$ V_A=\frac{5}{7}(47{,}752.80899\ldots)+\frac{2{,}500}{7} =34{,}466.29213\ldots. $$

To the nearest cent, \(V_D=\$47{,}752.81\) and \(V_A=\$34{,}466.29\). Checking, \(0.54V_D\approx \$25{,}786.52\), and \(12{,}000+0.40V_A\approx \$25{,}786.52\). Also, \(0.21V_A\approx \$7{,}237.92\), matching \(0.15(V_D+500)\approx \$7{,}237.92\).

Common mistake avoided. Omitting \(\delta\) from the left-side coefficient treats future payments as if they were undiscounted. The interest force belongs in both state equations, not only in the equation for the disabled state.

Worked Example: Payments at Onset and Recovery

A hypothetical plan pays $6,000 per year continuously while disabled. It pays $800 at each onset and $200 at each recovery. The force of interest is \(\delta=0.05\). The annual intensities are \(\lambda_{AD}=0.10\), \(\mu_A=0.05\), \(\lambda_{DA}=0.25\), and \(\mu_D=0.10\). Find the value from each state and the expected value for a group that is initially 70% active and 30% disabled.

Method. Attach each lump sum to the transition on which it is paid. The state-value equations are

$$ 0.20V_A=0.10(V_D+800), \qquad 0.40V_D=6{,}000+0.25(V_A+200). $$

The coefficients are \(0.05+0.10+0.05=0.20\) and \(0.05+0.25+0.10=0.40\). Rearranging the first equation gives \(V_A=0.5V_D+400\). Substitute this in the second:

$$ 0.40V_D =6{,}000+0.25(0.5V_D+400+200) =6{,}150+0.125V_D. $$

Thus \(0.275V_D=6{,}150\), so \(V_D=\$22{,}363.6364\ldots\). Then \(V_A=0.5(22{,}363.6364\ldots)+400=\$11{,}581.8182\ldots\). To the nearest cent, the values are \(V_D=\$22{,}363.64\) and \(V_A=\$11{,}581.82\).

For the stated initial mixture, the expected value is

$$ 0.70V_A+0.30V_D =0.70(11{,}581.8182\ldots)+0.30(22{,}363.6364\ldots) =\$14{,}816.36. $$

As a check, \(0.40V_D\approx \$8{,}945.45\), which equals \(6{,}000+0.25(V_A+200)\). Also, \(0.20V_A\approx \$2{,}316.36\), which equals \(0.10(V_D+800)\). The mixture value lies between the two state values, as it must.

Common mistake avoided. The $200 recovery payment is included in the disabled-state equation because recovery occurs from \(D\). It is not included in the active-state equation, which represents a claimant currently in \(A\).

Interpretation and Exam Checks

A recurrent model counts possible future episodes through the continuation values, not by adding a separate value for each episode in advance. After recovery, the value \(V_A\) includes the possibility of another onset; after that onset, \(V_D\) includes the possibility of another recovery. The coupled equations account for this repetition.

Before solving, check that the state definitions match the benefit design. In particular, verify whether recovery restores eligibility, whether a lump sum is payable on every onset or only the first, and whether continuous benefits depend on time disabled. If a maximum number of episodes, a cumulative benefit limit, or a duration-dependent payment applies, the two-state equations above are not sufficient: additional states or duration-dependent methods are needed.

  • Use annual transition intensities with an annual force of interest, unless the problem explicitly specifies another time unit.
  • Include every transition out of a state in its total exit intensity, including death.
  • Include a transition payment on the right-hand side with that transition’s intensity.
  • Do not add a continuation value after death when death is absorbing and has no benefit.
  • Substitute the solved values into both original equations as a numerical check.

Check Your Understanding

For each problem, assume constant transition intensities and force of interest, an absorbing death state with no death payment, and recurrent eligibility after recovery. Round money to the nearest cent.

  1. A hypothetical benefit pays $9,000 per year continuously while disabled. Let \(\delta=0.03\), \(\lambda_{AD}=0.12\), \(\mu_A=0.03\), \(\lambda_{DA}=0.30\), and \(\mu_D=0.07\). There are no transition payments. Find \(V_A\) and \(V_D\). (Answer: \(V_A=\$30{,}000.00,\ V_D=\$45{,}000.00\))
  2. A plan pays $8,000 per year continuously while disabled and $600 at each onset, but nothing at recovery. Let \(\delta=0.04\), \(\lambda_{AD}=0.10\), \(\mu_A=0.02\), \(\lambda_{DA}=0.20\), and \(\mu_D=0.08\). Find both state values. (Answer: \(V_A=\$26{,}256.41,\ V_D=\$41{,}410.26\))
  3. A contract pays $10,000 per year continuously while disabled and $300 at each recovery, with no onset payment. Let \(\delta=0.02\), \(\lambda_{AD}=0.08\), \(\mu_A=0.04\), \(\lambda_{DA}=0.50\), and \(\mu_D=0.10\). Find both state values. (Answer: \(V_A=\$48{,}333.33,\ V_D=\$84{,}583.33\))
  4. A hypothetical plan pays $5,000 per year continuously while disabled, $400 at each onset, and $100 at each recovery. Let \(\delta=0.03\), \(\lambda_{AD}=0.10\), \(\mu_A=0.02\), \(\lambda_{DA}=0.20\), and \(\mu_D=0.05\). Find both state values and the expected value for an initial group that is 40% active and 60% disabled. (Answer: \(V_A=\$23{,}084.85,\ V_D=\$34{,}227.27\); mixture value \(=\$29{,}770.30\))
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