Recovery and Return-to-Active Reserve Transitions

Why Recovery Changes More Than the Disabled-State Reserve

When a disabled policyholder recovers, the policy may return to its active state. That change can reinstate future premiums, restore coverage for future insured events, or do both. A reserve calculation must therefore follow the recovery transition into the active-state reserve, rather than treating recovery as the end of all future obligations.

The central calculation is conditional: from the disabled state, weight the future value following recovery by the recovery probability and the future value following continued disability by the persistence probability. The recovery branch uses the active-state reserve at the next valuation time. Whether this makes the disabled-state reserve larger or smaller depends on the values of the two branches.

Definition: $V_A(t)$ and $V_D(t)$ are prospective reserves at time $t$, conditional on the policy being active or disabled, respectively. They are measured immediately before any premium due at time $t$. A reserve is the expected present value of future benefits and expenses less the expected present value of future premiums, conditional on the state at time $t$.

Use an annual model with active state $A$ and disabled state $D$. During each year, the policy moves from its state at the start of the year to a state at the end. Let $p_{ij}(t)$ be the probability of moving from state $i$ to state $j$ during the year from $t$ to $t+1$. The probabilities out of each state sum to 1. Let $b_{ij}(t+1)$ be a benefit paid at the end of the year for the transition from $i$ to $j$. Let $P_A(t)$ be the premium paid immediately at time $t$ when active. In the examples, premiums are waived while disabled.

The annual effective interest rate is $i$, so $v=1/(1+i)$. This model assumes the stated transition probabilities and cash flows apply for the year, and that the next reserve is measured immediately before any premium at $t+1$. Transition benefits are paid at the year-end before that next reserve valuation. If a contract uses different payment dates or state-dependent premiums, the cash-flow timing and recurrence must be adjusted accordingly.

The Conditional Reserve Recursion

Conditioning on the state reached at $t+1$ gives the one-year reserve equations. An active policy pays its current premium, then has either an active or disabled future reserve. A disabled policy pays no premium under the waiver assumption, and recovery leads to the active future reserve.

V_A(t)=-P_A(t)+v\left[p_{AA}(t)\{b_{AA}(t+1)+V_A(t+1)\} +p_{AD}(t)\{b_{AD}(t+1)+V_D(t+1)\}\right]

V_D(t)=v\left[p_{DA}(t)\{b_{DA}(t+1)+V_A(t+1)\} +p_{DD}(t)\{b_{DD}(t+1)+V_D(t+1)\}\right]

Here $p_{DA}(t)$ is the recovery probability and $p_{DD}(t)$ is the probability of remaining disabled. A recovery benefit, if payable, is included in $b_{DA}$. The active reserve $V_A(t+1)$ already reflects future active coverage and any premiums payable at $t+1$; do not add those future cash flows a second time in the recovery branch.

Why the disabled-state equation holds: Conditional on being disabled at time $t$, there are two possible next states in this two-state model. With probability $p_{DA}(t)$, the policy recovers, so the year-end value is the transition benefit $b_{DA}(t+1)$ plus the reserve $V_A(t+1)$. With probability $p_{DD}(t)$, disability continues, so the corresponding value is $b_{DD}(t+1)+V_D(t+1)$. The expected year-end amount is discounted by $v$. There is no premium term because premiums are waived while disabled. This gives the stated recursion by conditioning on the next state.

Worked Calculations

Worked Example: A One-Year Recovery Branch

A hypothetical policy is valued at 5% effective annual interest. At time $t+1$, its reserves are $V_A(t+1)=\$20$ and $V_D(t+1)=\$50$. A disabled policyholder recovers with probability $0.30$; otherwise, disability continues. A \$10 benefit is paid at year-end if disability continues, and no benefit is paid on recovery. Calculate the disabled reserve at $t$. Also calculate the active reserve if the active-state premium is \$12, the active-to-active probability is $0.80$, the active-to-disabled probability is $0.20$, and a \$5 benefit is paid on the active-to-disabled transition.

Method. Apply the conditional recursion separately to each starting state. The annual discount factor is $v=1/1.05=0.95238095$.

For a policy disabled at $t$, the recovery branch has probability $0.30$ and value $0+20=\$20$. The persistence branch has probability $0.70$ and value $10+50=\$60$. Thus the expected year-end value is $0.30(20)+0.70(60)=6+42=\$48$, and

V_D(t)=\frac{48}{1.05}=\$45.714285\ldots\approx \$45.71.

For an active policy, the current premium is subtracted at time $t$. The active branch has value $0+20=\$20$; the disability branch has value $5+50=\$55$. Therefore,

V_A(t)=-12+\frac{0.80(20)+0.20(55)}{1.05} =-12+\frac{16+11}{1.05} =-12+25.714285\ldots =\$13.71.

Check. The transition probabilities in each starting state sum to 1. The disabled calculation includes no current premium, while the active calculation subtracts the \$12 premium. The results are rounded to cents only at the final step.

Common mistake avoided: Subtracting the \$12 premium from $V_D(t)$ would incorrectly charge a premium during disability. It would give \$33.71 instead of \$45.71, even though the contract waives that premium.

Worked Example: Recovery Leads to a Premium-Paying Active Reserve

A hypothetical two-year policy is valued at 4% effective annual interest. At time $t+1$, there is one year remaining. An active policy then pays a \$5 premium. During that final year, it remains active with probability $0.90$, or becomes disabled with probability $0.10$. A \$100 benefit is paid at year-end on the active-to-disabled transition. There are no other benefits in that year, and both reserves at maturity are zero. Calculate $V_A(t+1)$. At $t+1$, a disabled policy recovers with probability $0.40$; otherwise, it remains disabled. A \$30 benefit is paid at year-end if disability continues. Calculate $V_D(t+1)$. Finally, at time $t$, a disabled policy recovers with probability $0.25$, and otherwise remains disabled; a \$25 benefit is paid at year-end on continued disability. Calculate $V_D(t)$.

Method. Work backward from the zero maturity reserves. Use $v=1/1.04=25/26$.

At $t+1$, the active reserve is the current premium outflow plus the discounted expected final-year benefit:

V_A(t+1)=-5+\frac{0.90(0)+0.10(100)}{1.04} =-5+\frac{10}{1.04} =-5+\frac{250}{26} =\frac{120}{26} \approx \$4.6154.

For a policy disabled at $t+1$, recovery leads to a zero terminal reserve, while continued disability produces the \$30 benefit. Hence

V_D(t+1)=\frac{0.40(0)+0.60(30)}{1.04} =\frac{18}{1.04} =\frac{450}{26} \approx \$17.3077.

At time $t$, the recovery branch leads to $V_A(t+1)=120/26$, not to the disabled reserve. The continued-disability branch has value $25+450/26=1100/26$. Therefore,

\begin{aligned} V_D(t) &=\frac{0.25(120/26)+0.75(1100/26)}{1.04}\\ &=\frac{30/26+825/26}{1.04} =\frac{855}{26}\cdot\frac{25}{26} =\frac{21375}{676}\\ &\approx \$31.62. \end{aligned}

Check. At $t$, the recovery and persistence probabilities sum to 1. The recovery branch carries forward the active reserve, which includes the premium payable at $t+1$ and the active policy’s remaining coverage. The final reserve is rounded to cents.

Common mistake avoided: Replacing $V_A(t+1)$ with zero would treat recovery as policy termination. Here recovery restores active status, so its future premiums and benefits are represented by the active reserve.

Worked Example: Testing Whether More Recovery Raises the Reserve

At 3% effective annual interest, suppose the next active reserve is \$80 and the next disabled reserve is \$40. From disability, recovery has probability $q$; continued disability has probability $1-q$. There is no benefit on recovery and a \$12 benefit on continued disability. Calculate the disabled reserve for $q=0.25$ and $q=0.50$, and explain the direction of change.

Method. Keep the two branch values explicit before changing the recovery probability. The recovery branch is $0+80=\$80$. The continued-disability branch is $12+40=\$52$. Thus, for any $q$ between 0 and 1,

V_D(t;q)=\frac{q(80)+(1-q)(52)}{1.03} =\frac{52+28q}{1.03}.

For $q=0.25$, the expected year-end value is $52+28(0.25)=59$, so $V_D(t;0.25)=59/1.03=\$57.28$. For $q=0.50$, it is $52+28(0.50)=66$, so $V_D(t;0.50)=66/1.03=\$64.08$. Both answers are rounded to cents.

Check and interpretation. The recovery branch is \$28 more costly than the continued-disability branch. Raising $q$ by 0.25 therefore raises the reserve by $0.25(28)/1.03=\$6.80$, consistent with the difference between the rounded reserves. Greater recovery probability does not necessarily lower a reserve: its direction depends on the relative values of the recovery and persistence branches.

Common mistake avoided: Assuming recovery must reduce the reserve because disability benefits may stop. Here the recovered policy returns to an active reserve of \$80, which exceeds the \$52 value of continued disability.

When Recovery Raises or Lowers the Reserve

For fixed future reserves and transition benefits, write the disabled reserve as a function of the recovery probability $q=p_{DA}(t)$. Since $p_{DD}(t)=1-q$, the recurrence becomes

V_D(t;q)=v\left[q\{b_{DA}+V_A(t+1)\} +(1-q)\{b_{DD}+V_D(t+1)\}\right].

The change in reserve for a small change in $q$ is determined by the difference between the two branch values:

\frac{\partial V_D(t;q)}{\partial q} =v\left[\{b_{DA}+V_A(t+1)\}-\{b_{DD}+V_D(t+1)\}\right].

This follows by expanding the expression: its terms involving $q$ are $vq(b_{DA}+V_A(t+1))-vq(b_{DD}+V_D(t+1))$; the remaining terms do not depend on $q$. The derivative is positive if the recovery branch has the greater value, negative if the persistence branch has the greater value, and zero if the branch values are equal. This comparison is valid while the other inputs remain fixed; if changing recovery assumptions also changes other transition probabilities or benefit amounts, those changes must be included separately.

Practical Checks for State-Contingent Reserves

A reliable calculation keeps the state, cash-flow timing, and transition branch visible at every step. In particular, distinguish a premium due at the current valuation time from premiums included in a future active-state reserve. A premium due only when active belongs in the active-state cash flows; it is not charged in the disabled equation merely because recovery may happen later.

Check the state probabilities. For the two-state model, $p_{DA}+p_{DD}=1$ and $p_{AA}+p_{AD}=1$.
Use the correct next-state reserve. Recovery leads to $V_A(t+1)$, while continued disability leads to $V_D(t+1)$.
Place benefits on the stated transition. A benefit on continued disability belongs in the $D$-to-$D$ branch; a recovery benefit belongs in the $D$-to-$A$ branch.
Match the premium rule. If premiums are waived in disability, there is no current premium term in $V_D(t)$. If the contract instead charges a premium after recovery at a specified date, use that contract timing.
Compare branch values before judging direction. Recovery increases the reserve only when its benefit-plus-reserve value exceeds the corresponding value for continued disability.

These checks are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus: the essential skill is not merely inserting a recovery probability, but matching each possible transition to its appropriate benefit, premium rule, and future state reserve.

Check Your Understanding

For each problem, use the stated annual effective interest rate and follow the timing conventions in the problem.

A disabled policyholder recovers with probability 0.40 and otherwise remains disabled. At the next valuation time, $V_A=\$30$ and $V_D=\$20$. A \$8 benefit is paid on continued disability, with no recovery benefit. At 5% interest, find the current disabled reserve. (Answer: \$27.43)
An active policy pays a \$7 premium now. At year-end it remains active with probability 0.75 or becomes disabled with probability 0.25. The transition benefits are \$0 and \$12, respectively, and the next reserves are $V_A=\$24$ and $V_D=\$40$. At 4% interest, find the current active reserve. (Answer: \$22.81)
At 2% interest, a disabled policy recovers with probability 0.30. The next active reserve is \$50. Continued disability has a \$10 benefit and a next disabled reserve of \$35. Find the current disabled reserve. (Answer: \$45.59)
At 4% interest, the recovery branch has a year-end benefit-plus-reserve value of \$72, and the continued-disability branch has a value of \$48. What recovery probability gives a disabled reserve of \$57.69? (Answer: 0.5000)

What You'll Learn