Mortality Before and During Disability

Why the State at Death Matters

A disability model distinguishes a person who is active from one who is disabled. Death can occur from either state, and the mortality intensity need not be the same in both. For a disability income contract, the difference matters: death while active occurs before a disability claim has begun, whereas death while disabled occurs during a claim spell. These are distinct events, even when the question asks only for the probability of death by a given time.

This tutorial uses a continuous-time Markov model with active, disabled, and dead states. It focuses on death probabilities, not the value of benefits payable in those states. The formulas are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let \(A\), \(D\), and \(X\) denote the active, disabled, and dead states. In this model, \(A\) can transition to \(D\) or \(X\), and \(D\) can transition to \(A\) or \(X\). State \(X\) is absorbing: once dead, the individual remains dead. The death probability by time \(t\) is the probability of entering \(X\) by time \(t\). Death while active and death while disabled classify that event by the state occupied immediately before death.

Intensities and Death Probabilities

Write \(\alpha\) for the active-to-disabled intensity, \(\mu_A\) for the active-to-dead intensity, \(\rho\) for the disabled-to-active recovery intensity, and \(\mu_D\) for the disabled-to-dead intensity. In a continuous-time Markov model, an intensity is a transition rate conditional on occupying its originating state. For example, over a short interval of length \(h\), the probability of death in the next \(h\) units given that the person is active is \(\mu_A h+o(h)\).

For the constant-intensity model, using the state order \(A,D,X\) and the row-generator convention, the generator is

Q= \begin{pmatrix} -(\alpha+\mu_A)&\alpha&\mu_A\\ \rho&-(\rho+\mu_D)&\mu_D\\ 0&0&0 \end{pmatrix}.

The diagonal entries are the negatives of the total rates out of each transient state. Let \(p_A(t)\) and \(p_D(t)\) be the probabilities of occupying the active and disabled states at time \(t\), for a specified starting state. The probability of being alive at \(t\) is \(p_A(t)+p_D(t)\). Therefore,

P(\text{death by }t)=1-p_A(t)-p_D(t).

This total includes death after any number of disability onsets and recoveries. It is not, in general, the same as the probability of dying on the first transition out of the starting state.

Key result: The probability of death while active by time \(t\) is \(\int_0^t \mu_A p_A(s)\,ds\). The probability of death while disabled by time \(t\) is \(\int_0^t \mu_D p_D(s)\,ds\). These are probabilities of disjoint events, and their sum equals \(1-p_A(t)-p_D(t)\).

Why the State-Specific Integrals Work

The integral formula follows by considering a small interval from \(s\) to \(s+h\). The probability of being active at time \(s\) and dying in the next \(h\) units is \(p_A(s)\{\mu_A h+o(h)\}\). Summing these contributions over short intervals and taking the limit gives the active-state death probability. The same argument applies to the disabled state.

Track occupancy.
Find the probability of occupying each living state at time \(s\), accounting for all possible paths into that state.

Apply the state-specific mortality intensity.
Multiply the occupancy probability in each state by its mortality intensity to obtain the instantaneous probability flow into death.

Accumulate over the time interval.
Integrate each death flow from \(0\) to \(t\), then add them to obtain total death probability.

Equivalently, the living-state probabilities satisfy the forward equations

p_A'(t)=-(\alpha+\mu_A)p_A(t)+\rho p_D(t),\qquad p_D'(t)=\alpha p_A(t)-(\rho+\mu_D)p_D(t).

The initial conditions depend on the starting state. Starting active means \(p_A(0)=1,\ p_D(0)=0\); starting disabled means \(p_A(0)=0,\ p_D(0)=1\). These equations, or the transient part of \(e^{Qt}\), determine the occupancy probabilities used in the death calculations. With age-varying intensities, the same probability-flow reasoning applies, but the constant-intensity matrix-exponential formulas below do not.

Worked Calculations Under Constant Intensities

The following examples use a hypothetical disability model with constant intensities. They use the same transition structure but different questions: first, death from an active start; then, death from a disabled start. The examples also illustrate how the state at death is identified without mistaking claim incidence for claim prevalence.

Worked Example: Death by Two Years From an Active Start

A hypothetical model has \(\alpha=0.08\), \(\mu_A=0.12\), \(\rho=0.18\), and \(\mu_D=0.02\), all per year. Assume these intensities are constant for two years, the state process is Markov, and death is absorbing. An individual starts active. Calculate the probability of death by year 2, split by whether death occurs while active or disabled.

Method. Solve the transient occupancy equations and integrate the mortality flow from each state. This avoids treating the active mortality intensity as if it applied throughout the period, including after disability onset.

Here \(\alpha+\mu_A=0.20\) and \(\rho+\mu_D=0.20\). The two decay rates for the living-state probabilities are \(0.08\) and \(0.32\), since \(\sqrt{\alpha\rho}=\sqrt{0.08(0.18)}=0.12\). Thus, from an active start,

p_A(t)=\tfrac12(e^{-0.08t}+e^{-0.32t}),\qquad p_D(t)=\tfrac13(e^{-0.08t}-e^{-0.32t}).

At \(t=2\), \(e^{-0.16}=0.8521438\) and \(e^{-0.64}=0.5272924\). Therefore,

p_A(2)=\tfrac12(0.8521438+0.5272924)=0.6897181, \qquad p_D(2)=\tfrac13(0.8521438-0.5272924)=0.1082838.

The probability of being alive is \(0.6897181+0.1082838=0.7980019\), so the probability of death by year 2 is \(1-0.7980019=0.2019981\).

For the state-specific split, first integrate the occupancy probabilities:

\int_0^2p_A(s)\,ds =\tfrac12\left(\frac{1-e^{-0.16}}{0.08}+\frac{1-e^{-0.64}}{0.32}\right) =\tfrac12(1.8482026+1.4772112)=1.6627069, $$ $$ \int_0^2p_D(s)\,ds =\tfrac13\left(\frac{1-e^{-0.16}}{0.08}-\frac{1-e^{-0.64}}{0.32}\right) =\tfrac13(1.8482026-1.4772112)=0.1236638.

Consequently, death while active has probability \(0.12(1.6627069)=0.1995248\), and death while disabled has probability \(0.02(0.1236638)=0.0024733\). Their sum is \(0.2019981\), matching the total obtained from the probability of being alive. All probabilities are reported to four decimal places: \(0.1995\) for death while active, \(0.0025\) for death while disabled, and \(0.2020\) for total death.

Worked Example: Death by Two Years From a Disabled Start

Use the same hypothetical model and assumptions as in the previous example, but now suppose the individual starts disabled. Find the total death probability by year 2 and its split by state at death.

Method. Change the initial condition; do not reuse the active-start probabilities. The transition intensities are unchanged, but a disabled start changes the probability of occupying each state over time.

For a disabled start, the transient probabilities are

p_A(t)=\tfrac34(e^{-0.08t}-e^{-0.32t}),\qquad p_D(t)=\tfrac12(e^{-0.08t}+e^{-0.32t}).

Using \(e^{-0.16}=0.8521438\) and \(e^{-0.64}=0.5272924\),

p_A(2)=\tfrac34(0.8521438-0.5272924)=0.2436385, \qquad p_D(2)=\tfrac12(0.8521438+0.5272924)=0.6897181.

The alive probability is \(0.2436385+0.6897181=0.9333566\), giving total death probability \(1-0.9333566=0.0666434\).

The occupancy integrals are

\int_0^2p_A(s)\,ds=\tfrac34(1.8482026-1.4772112)=0.2782436, \qquad \int_0^2p_D(s)\,ds=\tfrac12(1.8482026+1.4772112)=1.6627069.

The probability of death while active is \(0.12(0.2782436)=0.0333892\). The probability of death while disabled is \(0.02(1.6627069)=0.0332541\). The sum is \(0.0666433\), which agrees with the total from the living-state probabilities apart from rounding. To four decimal places, total death is \(0.0666\), split approximately equally between death while active (\(0.0334\)) and death while disabled (\(0.0333\)).

Worked Example: No Recovery During the Observation Period

A separate hypothetical model has active-to-disability intensity \(\alpha=0.10\) per year, active mortality intensity \(\mu_A=0.10\), no recovery during the observation period (\(\rho=0\)), and disabled mortality intensity \(\mu_D=0.10\). Intensities are constant, and the individual starts active. Calculate the probability of death while disabled by year 1.

Method. With no recovery, the active state can only be left by disability onset or death. Its occupancy probability is \(p_A(t)=e^{-0.20t}\). Disabled occupancy is the accumulated flow of disability onsets, each multiplied by the probability of remaining disabled and alive until time \(t\):

p_D(t)=\int_0^t 0.10e^{-0.20s}e^{-0.10(t-s)}\,ds =e^{-0.10t}-e^{-0.20t}.

At year 1, \(p_A(1)=e^{-0.20}=0.8187308\) and \(p_D(1)=e^{-0.10}-e^{-0.20}=0.9048374-0.8187308=0.0861067\). Active-state death by year 1 is

\int_0^1 0.10p_A(s)\,ds =0.10\frac{1-e^{-0.20}}{0.20}=0.0906346.

The probability of death while disabled is the integral of \(0.10p_D(s)\). Since total death is \(1-p_A(1)-p_D(1)=0.0951626\), the disabled-state death probability is \(0.0951626-0.0906346=0.0045280\). To four decimal places it is \(0.0045\). The check is that \(0.0906346+0.0045280=0.0951626\), the total death probability. The no-recovery assumption is essential to the displayed occupancy formula; with recovery, paths from disabled back to active must also be included.

Common Mistake: Using the Wrong Mortality Intensity

A tempting shortcut is to calculate death from an active start as if the individual stayed active for the entire period. In the model of the first example, that wrong calculation gives \(1-e^{-0.12(2)}=0.2134\). It treats the active mortality intensity as the only way out of the active state and ignores disability onset and disabled-state mortality. It is not the total death probability under the three-state model.

The correction is to account for occupancy in both living states. The active intensity contributes only while the individual is active; the disabled intensity contributes only while disabled. For the first example, the correct state-specific probabilities sum to \(0.1995248+0.0024733=0.2019981\). A second related trap is to report \(\alpha/(\alpha+\mu_A)\) as the probability of death by a fixed time. That ratio is a first-exit event probability over an unlimited waiting time in a two-cause active-state competition; it is not a finite-horizon total death probability when recovery and later transitions are possible.

Key takeaway: To calculate death by time \(t\), first identify the starting state, then determine living-state occupancy over time. Multiply each occupancy probability by that state's mortality intensity and accumulate the resulting death flows. Total death is the sum of death while active and death while disabled, or equivalently one minus the probability of still occupying a living state at time \(t\).

Check Your Understanding

Use the stated assumptions in each problem. Probabilities are rounded to four decimal places.

A constant-intensity model has \(\alpha=0.08,\ \mu_A=0.12,\ \rho=0.18,\ \mu_D=0.02\) per year. Starting active, what are the probabilities of death while active and death while disabled by year 1? (Answer: \(0.1090\) while active; \(0.0007\) while disabled.)
In the same model, an individual starts disabled. What is the total probability of death by year 1? (Answer: \(0.0276\).)
A no-recovery model has \(\alpha=0.06,\ \mu_A=0.04,\ \mu_D=0.05\) per year, with constant intensities and an active start. What is the probability of death while disabled by year 2? (Answer: \(0.0054\).)
An individual starts active in a model where disability onset and active-state death have constant intensities \(0.03\) and \(0.07\) per year. What is the probability of death while active by year 3, before any disability onset? (Answer: \(0.1814\).)
In a model with recovery, can the probability of death by time \(t\) from a disabled start be calculated as \(1-e^{-\mu_Dt}\) if active-state mortality is positive? Explain the issue. (Answer: No. That expression assumes the individual remains disabled throughout; recovery can move the person to the active state, where active-state mortality also contributes.)

What You'll Learn