Tutorials › Exam ALTAM - Advanced Long-Term Actuarial Mathematics › Continuous Disability Income Annuities

Exam ALTAM · Disability Income Models · Tutorial 140 of 999

Continuous Disability Income Annuities

Learn to value a continuous income stream by integrating its discounted payment rate against the probability of being disabled at each time.

Intermediate 10 min read

What You'll Learn

  • Express the present value of continuous income as an integral over time.
  • Use the probability of occupying the disabled state, rather than disability incidence alone.
  • Derive occupancy probabilities for constant disability and recovery intensities.
  • Include mortality and force of interest consistently in an EPV calculation.
  • Identify when an occupancy probability is insufficient to describe a payment rule.

Introduction

A continuous disability income benefit is paid in proportion to the time a policyholder occupies the disabled state. Its expected present value therefore depends not just on whether disability has ever occurred, but on the probability of being disabled at each time. A policyholder may recover after becoming disabled, and a recovered policyholder may later become disabled again. The relevant probability is the probability of occupying the disabled state at time \(t\), including every path through the model that leads to that state.

This tutorial derives the valuation formula for a continuous benefit paid while disabled and applies it to models with recovery and mortality. The methods are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Continuous Income and State Occupancy

Let \(D\) denote the disabled state and let \(T\) be the end of the benefit valuation period. Suppose the benefit is paid continuously at a deterministic rate \(b(t)\) dollars per year whenever the policyholder is disabled at time \(t\). For example, a level benefit of \(\$7{,}200\) per year has payment rate \(b(t)=7{,}200\). If the policyholder is active or dead, this benefit rate is zero.

Definition: Let \(p_D(t)\) be the probability that the policyholder is in the disabled state at time \(t\), given the specified initial state or initial distribution. The expected present value (EPV) of a continuous disability income benefit over \([0,T]\), under constant force of interest \(\delta\), is $$ \operatorname{EPV}=\int_0^T e^{-\delta t}b(t)p_D(t)\,dt. $$

Here, \(\delta\) is the annual force of interest, so the discount factor for a payment at time \(t\) is \(e^{-\delta t}\). The formula assumes the payment rate is determined by the state occupied at each instant and that payments stop when the policyholder leaves the disabled state. The probabilities \(p_D(t)\) must be calculated from the transition model and the stated initial condition.

Why the Occupancy Integral Gives the EPV

Let \(Z_D(t)\) equal 1 when the policyholder is disabled at time \(t\), and 0 otherwise. For a particular path through the model, the present value of the income is $$ \int_0^T e^{-\delta t}b(t)Z_D(t)\,dt. $$ Taking expectations gives the EPV. Since the discounted benefit rate is nonnegative, expectation and integration can be interchanged. Also, \(E[Z_D(t)]=P(\text{disabled at }t)=p_D(t)\). Therefore,

$$ E\left[\int_0^T e^{-\delta t}b(t)Z_D(t)\,dt\right] = \int_0^T e^{-\delta t}b(t)E[Z_D(t)]\,dt = \int_0^T e^{-\delta t}b(t)p_D(t)\,dt. $$

This argument explains why an expected payment rate at time \(t\) is \(b(t)p_D(t)\): the full rate is paid if disabled, and zero is paid otherwise. It also explains a key distinction: the probability of first becoming disabled by time \(t\) is not generally \(p_D(t)\). A first-onset probability does not account for whether the person recovered before time \(t\), or whether a later disability spell began.

Constant Disability and Recovery Intensities

Consider first a two-state model with active state \(A\) and disabled state \(D\), no mortality, and constant transition intensities \(\alpha\) from \(A\) to \(D\) and \(\rho\) from \(D\) to \(A\). Starting active, let \(p_D(t)=P(X(t)=D\mid X(0)=A)\). Over a short interval, the probability of entering \(D\) from \(A\) is approximately \(\alpha\) times the interval length, while the probability of leaving \(D\) is approximately \(\rho\) times that length. Thus the forward equation is

$$ p_D'(t)=\alpha(1-p_D(t))-\rho p_D(t) =\alpha-(\alpha+\rho)p_D(t),\qquad p_D(0)=0. $$

Solving this differential equation gives \(p_D(t)=\frac{\alpha}{\alpha+\rho}(1-e^{-(\alpha+\rho)t})\). Substituting this occupancy probability into the EPV integral gives a convenient closed form for a level continuous benefit.

Result: With constant \(\alpha\) and \(\rho\), no mortality, a policyholder initially active, a level annual benefit rate \(b\), and constant force of interest \(\delta\), $$ \operatorname{EPV} = b\frac{\alpha}{\alpha+\rho} \left[ \frac{1-e^{-\delta T}}{\delta} - \frac{1-e^{-(\delta+\alpha+\rho)T}}{\delta+\alpha+\rho} \right]. $$

The expression follows by writing the integrand as \(b\frac{\alpha}{\alpha+\rho}(e^{-\delta t}-e^{-(\delta+\alpha+\rho)t})\) and integrating each exponential from 0 to \(T\). If \(\delta=0\), the term \((1-e^{-\delta T})/\delta\) is interpreted by its limit \(T\). This formula does not apply unchanged if the intensities vary with time, mortality is present, or the payment rule depends on more than the current state.

Worked Example: Level Income With Recovery

A hypothetical contract starts with the insured active. Disability occurs at constant intensity \(\alpha=0.08\) per year, and recovery occurs at constant intensity \(\rho=0.12\) per year. There is no mortality. A continuous benefit of \(\$7{,}200\) per year is paid while disabled for 3 years. The force of interest is \(\delta=0.05\). Find the EPV to the nearest cent.

Method. Use the active-disabled model and integrate the discounted benefit rate against the disabled-state occupancy probability. The total rate of leaving either state is \(\alpha+\rho=0.20\), and the long-run disabled-state proportion is \(\alpha/(\alpha+\rho)=0.08/0.20=0.40\). Hence

$$ p_D(t)=0.40(1-e^{-0.20t}). $$

The EPV is therefore

$$ \begin{aligned} \operatorname{EPV} &=7{,}200\int_0^3 e^{-0.05t}\,0.40(1-e^{-0.20t})\,dt\\ &=2{,}880\left[\frac{1-e^{-0.15}}{0.05}-\frac{1-e^{-0.75}}{0.25}\right]. \end{aligned} $$

Using \(e^{-0.15}=0.8607079764\) and \(e^{-0.75}=0.4723665527\), the two terms in brackets are

$$ \frac{1-0.8607079764}{0.05}=2.7858404715,\qquad \frac{1-0.4723665527}{0.25}=2.1105337890. $$

Their difference is \(0.6753066825\), so the EPV is \(2{,}880(0.6753066825)=1{,}944.8833\). The answer, rounded only at the final step to the nearest cent, is \(\$1{,}944.88\). The calculation allows for repeated movement between active and disabled states through the occupancy probability; it does not assume that the first disability spell lasts to the end of the term.

Including Mortality

If death is included as an absorbing state, the same EPV integral applies, but \(p_D(t)\) must reflect the probability of still being alive and disabled. For a constant-intensity model with active-to-disabled intensity \(\alpha\), disabled-to-active intensity \(\rho\), and death intensities, the occupancy probability is obtained from the multi-state transition model. It is not generally correct to use the no-mortality formula and then discount for mortality separately: death can occur at different rates in the active and disabled states, and the transition probabilities already describe those competing exits.

Worked Example: Income for an Insured Already Disabled

A hypothetical insured is disabled at time 0. While disabled, recovery occurs at constant intensity \(0.25\) per year and death at constant intensity \(0.05\) per year. Once recovered, the insured does not return to disability under this model. A benefit of \(\$18{,}000\) per year is paid continuously while disabled for 5 years. The force of interest is \(\delta=0.04\). Find the EPV to the nearest cent.

Method. Starting disabled, the insured remains disabled until either recovery or death. The total intensity of leaving the disabled state is \(0.25+0.05=0.30\), so

$$ p_D(t)=e^{-0.30t}. $$

The EPV is

$$ \begin{aligned} \operatorname{EPV} &=18{,}000\int_0^5 e^{-0.04t}e^{-0.30t}\,dt\\ &=18{,}000\int_0^5 e^{-0.34t}\,dt\\ &=18{,}000\frac{1-e^{-1.70}}{0.34}. \end{aligned} $$

Since \(e^{-1.70}=0.1826835241\), the integral is \((1-0.1826835241)/0.34=0.8173164759/0.34=2.4038722821\). Thus the unrounded EPV is \(18{,}000(2.4038722821)=43{,}269.7011\), giving \(\$43{,}269.70\) to the nearest cent. Recovery and death both terminate this benefit, so both intensities belong in the disabled-state exit rate.

Worked Example: Equal Mortality Intensities in Both Living States

A hypothetical policyholder starts active. The disability intensity is \(0.10\) per year and the recovery intensity is \(0.20\) per year. Death occurs at intensity \(0.10\) per year from either active or disabled state. A continuous benefit of \(\$9{,}000\) per year is payable while disabled for 3 years. The force of interest is \(\delta=0.05\). Find the EPV to the nearest cent.

Method. Conditional on being alive, the active-disabled transitions have total rate \(0.10+0.20=0.30\), so the probability of being disabled conditional on survival is \(\frac{0.10}{0.30}(1-e^{-0.30t})\). Because the death intensity is the same in both living states, the probability of being alive to time \(t\) is \(e^{-0.10t}\). Therefore the unconditional disabled-state probability is

$$ p_D(t)=e^{-0.10t}\frac{1}{3}(1-e^{-0.30t}). $$

Multiplying by the benefit and discount factors,

$$ \begin{aligned} \operatorname{EPV} &=9{,}000\int_0^3 e^{-0.05t}e^{-0.10t}\frac{1}{3}(1-e^{-0.30t})\,dt\\ &=3{,}000\left[\frac{1-e^{-0.45}}{0.15}-\frac{1-e^{-1.35}}{0.45}\right]. \end{aligned} $$

Using \(e^{-0.45}=0.6376281516\) and \(e^{-1.35}=0.2592402606\), the terms are \(0.3623718484/0.15=2.4158123227\) and \(0.7407597394/0.45=1.6461327542\). Their difference is \(0.7696795685\). Thus the EPV is \(3{,}000(0.7696795685)=2{,}309.0387\), or \(\$2{,}309.04\) to the nearest cent. The common mortality intensity factors out as a survival probability; this simplification depends on its being the same in both living states.

Changing Benefit Rates and Limits of the Formula

The occupancy method does not require a level benefit. If a deterministic benefit rate changes at a known time, use its appropriate value on each time interval and integrate each part. For example, if \(b(t)=b_1\) up to time \(s\) and \(b(t)=b_2\) afterward, then the EPV is the sum of the integral over \([0,s]\) and the integral over \([s,T]\), each using the same \(p_D(t)\) but its applicable benefit rate.

The basic formula assumes that the rate at time \(t\) can be determined from the current state and time. If a contract pays a different amount according to how long the current disability spell has lasted, then knowing only the probability of being disabled at time \(t\) may not identify the expected payment rate. The model must then distinguish the relevant disability-duration conditions, or otherwise calculate the corresponding conditional payment amounts.

Common Exam Trap: Using First Onset or Endpoint Occupancy

A frequent error is to value a continuous benefit using the probability of being disabled at the end of the term, as if that probability applied throughout the term. In the first worked example, the disabled-state probability at year 3 is

$$ p_D(3)=0.40(1-e^{-0.60})=0.1804753456. $$

The incorrect calculation \(7{,}200(3)p_D(3)\) gives approximately \(\$3{,}898.27\). It treats the year-3 occupancy probability as the occupancy probability at every earlier time and ignores discounting. The correct calculation integrates \(7{,}200e^{-0.05t}p_D(t)\) from 0 to 3 and gives \(\$1{,}944.88\). Endpoint occupancy is only one value of the probability function; it is not a substitute for the full time path.

  • Use occupancy, not incidence. First-onset probability does not account for recovery or subsequent disability spells.
  • Apply interest continuously. Under force \(\delta\), discount at time \(t\) by \(e^{-\delta t}\), not by an annual discount factor raised to an inconsistent time unit.
  • Include all exits from disability. Recovery and disabled-state mortality both reduce the probability of continuing to receive the benefit.
  • Keep the units consistent. If \(b(t)\) is dollars per year, \(dt\) is measured in years and the integral is in dollars.
Key takeaway: For a continuous income benefit paid only while disabled, value the instantaneous benefit rate at each time by multiplying it by the probability of occupying the disabled state, discounting it, and integrating over the payment period.

Check Your Understanding

For each problem, use the occupancy probability appropriate to the stated initial state and transition assumptions. Round monetary answers to the nearest cent.

  1. A policyholder starts active in a two-state model with no mortality. Disability and recovery intensities are \(0.06\) and \(0.14\) per year. A benefit of \(\$10{,}000\) per year is paid continuously while disabled for 2 years, and \(\delta=0.04\). Find the EPV. (Answer: \(\$1{,}001.07\))
  2. An insured starts disabled. Recovery and death intensities are \(0.18\) and \(0.02\) per year, respectively. A continuous benefit of \(\$15{,}000\) per year is paid while disabled for 4 years, with \(\delta=0.05\). Find the EPV. (Answer: \(\$37{,}927.23\))
  3. A policyholder starts active in a no-mortality model with disability intensity \(0.12\) and recovery intensity \(0.18\) per year. A level continuous benefit of \(\$8{,}000\) per year is payable while disabled for 3 years, and \(\delta=0.06\). Find the EPV. (Answer: \(\$2{,}915.33\))
  4. A policyholder starts active. Disability and recovery intensities are \(0.09\) and \(0.21\) per year, and the death intensity is \(0.03\) per year from either living state. A continuous benefit of \(\$12{,}000\) per year is paid while disabled for 2 years, with \(\delta=0.04\). Find the EPV. (Answer: \(\$1{,}631.18\))
  5. A policyholder starts active in a no-mortality model with disability and recovery intensities both equal to \(0.10\) per year. The continuous benefit rate is \(\$6{,}000\) per year from time 0 to 1, then \(\$9{,}000\) per year from time 1 to 2, provided the policyholder is disabled. If \(\delta=0.05\), find the EPV. (Answer: \(\$1{,}346.26\))
Clinical Biostats