Common Errors in Disability Income Valuations

Why Disability Valuations Need an Error Check

A disability income valuation depends on more than the benefit amount and interest rate. The result also depends on which state the insured occupies, when eligibility begins and ends, and whether a payment is continuous or made at specified dates. A formula can look plausible while answering the wrong question—for example, counting first disability claims when the question asks for all disability entries, or valuing payments after recovery.

This tutorial focuses on four checks: distinguish incidence from prevalence, allow for recovery and possible re-entry, account for recovery and death during a waiting period, and use the stated payment convention. The calculations use hypothetical contracts and assumptions.

Definition: Incidence describes transitions into disability during a period. Prevalence describes the probability, or proportion of lives, disabled at a specified time. A life can contribute to incidence more than once if recovery returns the life to an eligible active state and another disability entry is possible.

Incidence Is Not the Same as Prevalence

Suppose a model has two living states: active $A$ and disabled $D$. Disability incidence occurs at constant intensity $\alpha$ from $A$ to $D$, and recovery occurs at constant intensity $\beta$ from $D$ to $A$. There is no death in this example. The transition intensities are measured per year, and the process starts active.

Let $p_A(t)$ and $p_D(t)$ be the probabilities of occupying states $A$ and $D$ at time $t$. The expected number of new disability entries per life over $[0,T]$ is the incidence intensity multiplied by active occupancy and integrated over time:

E[N(T)] = \int_0^T \alpha p_A(t)\,dt.

This is an expected count of entries, not a probability of being disabled. By contrast, $p_D(T)$ is the probability of being disabled at the endpoint. The distinction matters especially when recovery permits another entry.

Derivation: Over a short interval of length $h$, conditional on being active at time $t$, the probability of an entry is $\alpha h+o(h)$. Thus the unconditional expected entries during that interval are $p_A(t)\alpha h+o(h)$. Summing over short intervals and taking the limit gives $\int_0^T\alpha p_A(t)\,dt$. If recovery is possible, $p_A(t)$ must include lives that recovered; it is not just the probability of having remained continuously active since time zero.

Worked Example: Counting Entries When Recovery Allows Re-Entry

A hypothetical group of 1,000 people begins active. Disability incidence has constant intensity $\alpha=0.20$ per year, recovery has constant intensity $\beta=0.30$ per year, and there is no mortality. Find (a) the expected number of disability entries during the first year and (b) the expected number disabled at the end of the year.

Method. Use the two-state transition model, so recovered lives return to the active population at risk of a new disability entry. The state probabilities satisfy

p_A'(t)=-0.20p_A(t)+0.30p_D(t),\qquad p_D'(t)=0.20p_A(t)-0.30p_D(t),

with $p_A(0)=1$ and $p_D(0)=0$. Since $p_A(t)+p_D(t)=1$, the second equation becomes $p_D'(t)=0.20-0.50p_D(t)$. Solving with the initial condition gives

p_D(t)=\frac{0.20}{0.50}\left(1-e^{-0.50t}\right) =0.40\left(1-e^{-0.50t}\right), \qquad p_A(t)=0.60+0.40e^{-0.50t}.

(a) The expected entries per person are

\begin{aligned} E[N(1)] &=\int_0^1 0.20p_A(t)\,dt\\ &=0.20\left[0.60+0.40\frac{1-e^{-0.50}}{0.50}\right]\\ &=0.20\left[0.60+0.40(0.7869387)\right]\\ &=0.20(0.9147755)=0.1829551. \end{aligned}

The group therefore has $1{,}000(0.1829551)=182.9551$ expected entries, or 182.96 entries to two decimal places. (b) At time 1,

p_D(1)=0.40(1-e^{-0.50}) =0.40(0.3934693)=0.1573877.

The expected number disabled at the endpoint is $1{,}000(0.1573877)=$ 157.39 people, to two decimal places. As a check, the expected entries exceed the expected endpoint disabled count because some entrants recover before year-end, while some people can enter more than once.

Common mistake avoided. Treating the number of first entries as the expected number of all entries would give $1{,}000(1-e^{-0.20})=181.27$. That is the expected count of people with a first entry in the year, not the expected total entry count when recovery permits re-entry.

Waiting Periods: Recovery and Death Both Matter

A waiting period is a test of eligibility, not merely a delay applied to an otherwise certain payment. If a benefit is payable only when the insured remains disabled and alive through the waiting period, both recovery and disabled-state mortality can terminate eligibility before the first payment. Under constant, competing transition intensities, the probability of remaining disabled and alive for $t$ years is the exponential of minus the sum of those intensities times $t$.

For example, if recovery intensity is $\rho$ and disabled-state mortality intensity is $\mu_D$, then, conditional on being disabled and alive at the start of the period, the probability of still being disabled and alive after $w$ years is $e^{-(\rho+\mu_D)w}$. This formula assumes constant intensities during the period and no transition back into the state that qualifies for payment. It is not the probability of recovery alone.

Worked Example: Valuing a Benefit After a Waiting Period

A hypothetical insured is disabled and alive at time 0. Recovery has constant intensity $0.30$ per year and disabled-state mortality has constant intensity $0.10$ per year. A benefit of $12,000 per year is payable continuously, beginning after a waiting period of 0.25 years, and stops immediately upon recovery or death. The force of interest is $\delta=0.04$ per year. Find the expected present value at time 0.

Method. First calculate eligibility at the end of the waiting period. The combined intensity of leaving the disabled-alive state is $0.30+0.10=0.40$. The probability of remaining eligible to the end of the wait is

e^{-0.40(0.25)}=e^{-0.10}=0.9048374.

The probability of recovery during the wait is the recovery share of the total exit intensity multiplied by the probability of any exit:

\frac{0.30}{0.40}(1-e^{-0.10}) =0.75(0.0951626)=0.0713719.

Similarly, the probability of death during the wait is

\frac{0.10}{0.40}(1-e^{-0.10}) =0.25(0.0951626)=0.0237906.

The three probabilities check: $0.9048374+0.0713719+0.0237906=1.0000000$, subject to rounding.

To value the continuous benefit, at time $t\geq0.25$ the probability of still being disabled and alive is $e^{-0.40t}$, and the discount factor is $e^{-0.04t}$. Therefore

\begin{aligned} \operatorname{EPV} &=12{,}000\int_{0.25}^{\infty}e^{-0.04t}e^{-0.40t}\,dt\\ &=12{,}000\frac{e^{-0.44(0.25)}}{0.44}\\ &=12{,}000\frac{e^{-0.11}}{0.44}\\ &=12{,}000\frac{0.8958341}{0.44}\\ &=\$24{,}431.84. \end{aligned}

The present value is $24,431.84, rounded to the nearest cent. The integral starts at the end of the waiting period and continues only while the insured remains disabled and alive.

Common mistake avoided. Omitting recovery and mortality from the termination condition gives $12{,}000e^{-0.04(0.25)}/0.04=\$297{,}014.95$. This treats the benefit as continuing indefinitely after eligibility has ended. The correct valuation includes both termination intensities in the survival probability.

Payment Convention: Value Each Scheduled Payment at Its Own Date

A continuous benefit rate and a sequence of installments are different payment models. For a continuous benefit, value the payment rate over eligible time. For installments, consider each stated payment date separately and include the probability that the insured satisfies the payment condition at that date. Do not replace one convention with another merely because the annualized amounts look similar.

Worked Example: Scheduled Installments Conditional on Continued Disability

A hypothetical policy pays $800 at times 0.50, 0.75, and 1.00 years after disability begins. Each installment is payable only if the insured is disabled and alive immediately before its payment. Recovery and disabled-state mortality intensities are constant at $0.12$ and $0.08$ per year, respectively. The force of interest is $\delta=0.04$ per year. Find the expected present value at claim inception.

Method. Starting disabled and alive, the probability of still being disabled and alive at time $t$ is $e^{-(0.12+0.08)t}=e^{-0.20t}$. The discount factor is $e^{-0.04t}$. Thus the expected present value of an installment at time $t$ is $800e^{-0.24t}$. The three contributions are

\begin{aligned} t=0.50:\quad &800e^{-0.24(0.50)} =800e^{-0.12}=800(0.8869204)=\$709.54,\\ t=0.75:\quad &800e^{-0.24(0.75)} =800e^{-0.18}=800(0.8352702)=\$668.22,\\ t=1.00:\quad &800e^{-0.24(1.00)} =800e^{-0.24}=800(0.7866279)=\$629.30. \end{aligned}

Adding the unrounded contributions gives

800(0.8869204+0.8352702+0.7866279) =800(2.5088185)=\$2{,}007.05.

The expected present value is $2,007.05, rounded to the nearest cent. As a check, each payment contribution is less than $800 because both the chance of remaining eligible and the discount factor are less than 1.

Common mistake avoided. Discounting but ignoring recovery and death would give $800(e^{-0.02}+e^{-0.03}+e^{-0.04})=\$2{,}329.15$. That calculation assumes every installment is paid, even when the insured has recovered or died. Each payment must be weighted by the probability of satisfying its condition on its own payment date.

A Practical Audit Before Accepting a Result

Before finalizing a disability income valuation, identify the modeled state at the start, each transition that changes eligibility, and the payment convention. Then match the probability calculation to the question being asked. In particular, an expected number of claim entries, a probability of being disabled at a date, and a present value of payments are distinct quantities.

Question being answered	Calculation to check	Common error
How many new disability entries are expected?	Integrate incidence intensity times active-state occupancy.	Use only uninterrupted active survival when recovery can return people to active status.
How many are disabled at a specified time?	Calculate the disabled-state occupancy probability at that time.	Substitute the number of entries for the disabled-state probability.
Does the insured qualify after a waiting period?	Include every transition that ends eligibility during the wait.	Ignore death or recovery before the first payable date.
What is the value of scheduled installments?	For each date, multiply the amount by eligibility probability and discount.	Use a continuous payment formula or omit disabled-state survival.

A useful final check is dimensional and logical. Intensities must use the same time unit as the waiting period and payment dates. Probabilities must lie between 0 and 1. A payment made only while disabled and alive cannot have an expected present value greater than the value obtained by assuming it is certainly paid at every scheduled date.

Key takeaway: State occupancy determines who is eligible; transition intensities determine how eligibility changes; and payment timing determines when eligible benefits are valued. Keep these three pieces separate, then combine them explicitly.

Check Your Understanding

For each hypothetical model, identify the relevant state or payment condition before calculating.

A life starts active, with disability incidence intensity $0.10$ per year and recovery intensity $0.20$ per year; there is no mortality. What is the probability of being disabled at time 2? (Answer: 0.1504)
A disabled claim has recovery intensity $0.20$ and mortality intensity $0.10$, both constant per year. What is the probability that the insured remains disabled and alive through a 0.50-year waiting period? (Answer: 0.8607)
An insured is disabled and alive now. Recovery intensity is $0.20$ and disabled-state mortality intensity is $0.05$, both constant. What is the probability of still being disabled and alive after 0.75 years? (Answer: 0.8290)
A policy pays $500 at times 0.50 and 1.00 years, only if the insured is disabled and alive immediately before each payment. Recovery and disabled-state mortality intensities are each $0.10$ per year, and $\delta=0.05$. Find the expected present value at claim inception. (Answer: $830.65)
A hypothetical contract pays continuously at $9,000 per year beginning at time 0.25, provided the insured remains disabled and alive. Recovery intensity is $0.15$, disabled-state mortality intensity is $0.05$, and $\delta=0.03$, all per year. Find the expected present value at time 0. (Answer: $36,943.90)

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