Recovery, Recurrence, and Multiple Claim Spells

Why Recovery Changes the Claim Model

A disability contract may terminate a claim when an insured recovers, but recovery does not necessarily end coverage. If the insured later becomes disabled again, a new claim spell may begin. The expected cost therefore depends not only on the chance of a first onset, but also on how often the insured moves between active and disabled states during the coverage period.

A multiple-state model handles this recurrence directly. The insured can move from active to disabled, recover from disabled to active, and then experience another onset. This tutorial focuses on how to represent those repeated spells and calculate expected payments associated with them. Any products and data below are hypothetical.

Definition: A disability spell is a continuous period spent in the disabled state, starting with a transition from active to disabled and ending with recovery or another specified claim-terminating event. A recurrent spell is a later disability spell following a recovery and a return to the active state. In a model with active state $A$ and disabled state $D$, each transition $A\to D$ starts a spell, and each transition $D\to A$ ends that spell by recovery.

A Two-State Model with Re-entry

First consider a continuous-time Markov model with two states: active $A$ and disabled $D$. Let $\alpha$ be the constant intensity of transition from $A$ to $D$, and let $\rho$ be the constant intensity of recovery from $D$ to $A$. Intensities are measured per year. Assume they are positive and constant, and that the future transition probabilities depend on the current state rather than on the duration already spent in that state. This last condition is the Markov assumption.

For a short interval of length $h$, the conditional transition probabilities are $\alpha h+o(h)$ from active to disabled and $\rho h+o(h)$ from disabled to active. Here $o(h)$ denotes a term that becomes negligible relative to $h$ as $h$ tends to zero. Remaining in the current state has the complementary probability to first order. In this two-state model, recovery returns the insured to the same active state from which another onset can occur.

Q= \begin{pmatrix} -\alpha & \alpha\\ \rho & -\rho \end{pmatrix}, \qquad P'(t)=P(t)Q,\qquad P(0)=I.

The rows and columns of the generator $Q$ are ordered $A,D$, and $P(t)$ is the transition-probability matrix over $t$ years. These equations assume constant intensities and no additional states such as death. If death or another absorbing state is included, the generator and probability equations must include that state as well; the two-state formulas below then do not apply unchanged.

Key result: If the insured starts active, the probability of being active at time $t$ is $p_A(t)=\dfrac{\rho}{\alpha+\rho}+\dfrac{\alpha}{\alpha+\rho}e^{-(\alpha+\rho)t}$. The probability of being disabled is $p_D(t)=1-p_A(t)$. These are state probabilities at time $t$, not probabilities of having experienced a disability by that time.

Why the State-Probability Formula Holds

Write $p_A(t)=P(\text{active at }t)$. Over a short interval, the insured can leave the active state through onset, or enter it through recovery. The forward equation for active-state occupancy is therefore $p_A'(t)=-\alpha p_A(t)+\rho p_D(t)$. Since the model has only two states, $p_D(t)=1-p_A(t)$, giving a single differential equation.

p_A'(t)=\rho-(\alpha+\rho)p_A(t).

The equilibrium solution of this equation is $\rho/(\alpha+\rho)$. Subtracting that constant from both sides gives a homogeneous equation for the difference between $p_A(t)$ and its equilibrium value. Its solution is a constant times $e^{-(\alpha+\rho)t}$. Since the insured starts active, $p_A(0)=1$, which determines that constant as $\alpha/(\alpha+\rho)$. Thus

p_A(t)=\frac{\rho}{\alpha+\rho} +\left(1-\frac{\rho}{\alpha+\rho}\right)e^{-(\alpha+\rho)t} =\frac{\rho}{\alpha+\rho} +\frac{\alpha}{\alpha+\rho}e^{-(\alpha+\rho)t}.

The same reasoning works for a different starting state, but the initial condition changes. For example, if the insured starts disabled, then $p_A(0)=0$ and $p_A(t)=\frac{\rho}{\alpha+\rho}(1-e^{-(\alpha+\rho)t})$. In either case, state occupancy can be used to value payments made in a state or at transitions out of a state.

Expected Number of Recurrent Onsets

Let $N_{AD}(T)$ count every transition from active to disabled during the interval from time 0 to $T$. A return to active after recovery makes a later transition $A\to D$ another count. Under the constant-intensity model, the expected number of onsets is

E[N_{AD}(T)]=\int_0^T \alpha\,p_A(t)\,dt.

To justify this result, conditional on being active at time $t$, the probability of an onset during the next short interval of length $h$ is $\alpha h+o(h)$. The unconditional probability of an onset in that interval is therefore $p_A(t)\alpha h+o(h)$. Summing these expected increments over short intervals and taking the limit gives the integral. This is an expected count: it can exceed one. It is not the probability of at least one onset.

When the insured starts active, substituting the state-probability formula gives

E[N_{AD}(T)] =\alpha\left[ \frac{\rho}{\alpha+\rho}T +\frac{\alpha}{(\alpha+\rho)^2} \left(1-e^{-(\alpha+\rho)T}\right) \right].

This expression counts all onsets, including those after recovery. It assumes there is no contract limit on the number of claims and that the insured remains eligible to make a claim whenever active. A maximum benefit period or another termination rule changes the claim model and may require additional states or duration tracking.

Worked Example: Expected Payments at Each New Onset

A hypothetical policyholder starts active. The annual onset and recovery intensities are $\alpha=0.20$ and $\rho=0.30$, respectively. The policy pays $500 immediately at each new onset during a two-year term. Assume no mortality, no discounting, no limit on the number of spells, and no payment at recovery. Find the expected total onset payments.

Method. Calculate the expected number of $A\to D$ transitions from active-state occupancy, then multiply by the payment per transition. The payment is made at onset, so each transition contributes one $500 payment.

The total transition intensity between the two states is $\alpha+\rho=0.50$. Starting active,

p_A(t)=\frac{0.30}{0.50}+\frac{0.20}{0.50}e^{-0.50t} =0.60+0.40e^{-0.50t}.

The expected active time over the two years is

\int_0^2 p_A(t)\,dt =0.60(2)+0.40\frac{1-e^{-1}}{0.50} =1.20+0.40(1.264241) =1.705696\text{ years}.

Thus the expected number of onsets is

E[N_{AD}(2)]=0.20(1.705696)=0.341139.

The expected total payment is $500(0.341139289)=\$170.569645$, or $170.57 to the nearest cent. The result is less than one expected payment because many policyholders would have no onset during the term, while some could have more than one. The calculation avoids the mistake of counting only first onsets: after recovery, the insured is again exposed to the onset intensity.

Benefits Paid While Disabled

A different benefit design pays income continuously while the insured is disabled. Such a benefit is driven by disabled-state occupancy, not by the number of claims. If the payment rate is $b$ dollars per year while disabled and the force of interest is $\delta$, the present value of payments over a term of $T$ years is the integral of the discounted payment rate times the probability of being disabled.

E[\text{present value}] =\int_0^T e^{-\delta t}b\,p_D(t)\,dt.

This formula assumes the income is paid continuously at rate $b$, discounting uses a constant force of interest, and the state probabilities follow the specified model. It does not describe a benefit paid only at the start or end of each spell. For a discrete payment schedule, the expected payment at each payment time must instead be weighted by the probability of being eligible at that time.

Worked Example: Discounted Income Across Repeated Spells

A hypothetical contract starts with an active insured and pays disability income continuously at $500 per year whenever the insured is disabled. The term is two years. Assume constant intensities $\alpha=0.20$ and $\rho=0.30$ per year, no mortality, and a constant force of interest $\delta=0.05$. There is no waiting period, and all disability spells are covered. Find the expected present value.

Method. Use disabled-state occupancy at each time and discount the income paid at that time. From the first example’s state model,

p_D(t)=1-p_A(t)=0.40(1-e^{-0.50t}).

Therefore the expected present value is

500\int_0^2 e^{-0.05t}0.40(1-e^{-0.50t})\,dt =200\left[ \frac{1-e^{-0.10}}{0.05} -\frac{1-e^{-1.10}}{0.55} \right].

The first integral term is $(1-e^{-0.10})/0.05=1.903251639$. The second is $(1-e^{-1.10})/0.55=1.212961666$. Their difference is $0.690289973$, and multiplying by 200 gives $\$138.057995$. The expected present value is $138.06 to the nearest cent.

As a check on the occupancy calculation, at time 2 the disabled-state probability is $0.40(1-e^{-1})=0.252848$, which is between zero and one. The present value uses the entire path of disabled occupancy, not just the probability at the end of the term. Treating each spell as a separate fixed payment would not match this continuous-income benefit design.

Discounted Payments at Onset

If a fixed amount is paid at each transition into disability, discounting applies at the transition time. Under constant $\alpha$, the expected present value of an onset payment $C$ over $T$ years is

E[\text{present value}] =C\int_0^T e^{-\delta t}\alpha p_A(t)\,dt.

The factor $\alpha p_A(t)\,dt$ represents the expected number of onset transitions in a short interval around $t$; the factor $e^{-\delta t}$ discounts a payment made then. The formula values each onset payment at its actual transition time in expectation. It assumes the same payment is made at every onset and that there is no claim cap or other payment condition.

Worked Example: Discounting a Payment for Every Spell

A hypothetical policy starts with the insured active and pays $1,000 immediately at every onset during a one-year term. The annual onset intensity is $\alpha=0.20$, the recovery intensity is $\rho=0.30$, and the force of interest is $\delta=0.10$. Assume no mortality and no claim limit. Find the expected present value of onset payments.

Method. Discount each onset by its time, weight it by the active-state probability, and integrate. Here $\alpha+\rho=0.50$, so $p_A(t)=0.60+0.40e^{-0.50t}$. Then

\begin{aligned} E[\text{present value}] &=1000(0.20)\int_0^1 e^{-0.10t} \left(0.60+0.40e^{-0.50t}\right)\,dt\\ &=200\left[ 0.60\frac{1-e^{-0.10}}{0.10} +0.40\frac{1-e^{-0.60}}{0.60} \right]. \end{aligned}

The first integral contribution is $0.60(0.951625820)=0.570975492$. The second is $0.40(0.751980607)=0.300792243$. Their sum is $0.871767735$. Multiplication by 200 gives $\$174.353547$, so the expected present value is $174.35 to the nearest cent.

The calculation discounts the onset payment at its transition time; it does not discount every payment as if it were made at the end of the year. It also includes transitions after a recovery, since $p_A(t)$ includes both insureds who have never been disabled and those who have recovered.

Common Exam Traps and Practical Checks

Repeated-spell questions often turn on what is being counted or paid. The expected number of transitions into disability, the probability of a first onset, the probability of being disabled at a particular time, and the expected time spent disabled are different quantities. Match the quantity to the contract benefit before setting up a calculation.

Counting only the first onset. A formula for first-onset probability does not count later claims after recovery. For recurrent onset payments, use the expected onset intensity $\alpha p_A(t)$.
Using disabled occupancy for an onset benefit. Continuous income while disabled is weighted by $p_D(t)$. A payment at onset is weighted by $\alpha p_A(t)$.
Confusing a state probability with an event probability. $p_D(T)$ is the chance of being disabled at time $T$, not the chance of having had any spell by time $T$. Someone may have had a spell and recovered before $T$.
Ignoring payment timing. An immediate onset payment is discounted at the onset time. Continuous income is integrated over time, with discounting at each payment instant.
Applying a model beyond its assumptions. The two-state formulas presume constant intensities, no mortality, and no claim-duration dependence. Add states or use a more suitable model if death, a spell limit, or duration-based termination affects eligibility.

For example, in the first worked example, replacing recurrent onset payments with a calculation for first-onset probability would miss the post-recovery exposure. The correction is to integrate the onset intensity over all times when the insured is active: $\int_0^T \alpha p_A(t)\,dt$. That expected count is then multiplied by the amount paid per onset.

Key takeaway: Re-entry after recovery means that disability claims are recurrent transitions, not merely a one-time onset event. Use active-state occupancy to count onset payments and disabled-state occupancy to value income paid while disabled, applying discounting at the relevant payment times.

Check Your Understanding

For each problem, state the relevant model quantity before calculating. Unless stated otherwise, assume constant intensities, no mortality, and no claim limit.

An insured starts active with onset intensity $0.10$ and recovery intensity $0.40$ per year. Over one year, find the expected number of onsets. (Answer: 0.0957 onsets)
An insured starts active. The annual onset and recovery intensities are $0.30$ and $0.20$. Find the probability of being disabled at time 2. (Answer: 0.3793)
A contract pays $800 per year continuously while disabled for one year. The insured starts active, with onset intensity $0.20$ and recovery intensity $0.30$. There is no interest. Find the expected total benefit. (Answer: $68.18)
A contract pays $400 immediately at every onset over two years. The insured starts active, with onset intensity $0.10$, recovery intensity $0.40$, and force of interest $0.05$. Find the expected present value. (Answer: $70.61)
An insured starts disabled, with onset and recovery intensities both $0.25$ per year. Find the expected number of new onsets during the next two years. (Answer: 0.0920 onsets)

What You'll Learn