Disability Incidence, Recovery, and Mortality Intensities

What a Disability Intensity Measures

In a disability model, the same transition can have a different rate depending on the person’s current state. A person who is active may become disabled or die; a person who is disabled may recover or die. An intensity describes the instantaneous rate of one such transition, conditional on occupying its starting state. It is not, by itself, the probability that the transition occurs over a year or any other fixed period.

Use \(A\) for active, \(D\) for disabled, and \(\dagger\) for dead. For distinct states \(i\) and \(j\), the transition intensity from \(i\) to \(j\) at time \(t\) is defined by the short-interval conditional probability:

\mu_{ij}(t) = \lim_{h\downarrow 0} \frac{P(X(t+h)=j\mid X(t)=i)}{h}.

Here \(X(t)\) is the state occupied at time \(t\), and \(h\) is a positive interval tending to zero, measured in the same time unit as \(t\). In the usual continuous-time model, the probability of two or more transitions in an interval of length \(h\) is negligible compared with \(h\). Thus, for a small interval,

P(X(t+h)=j\mid X(t)=i) = \mu_{ij}(t)h+o(h).

The term \(o(h)\), divided by \(h\), tends to zero as \(h\) tends to zero. This approximation is local: an intensity of \(0.10\) per year does not mean there is exactly a 10% probability of transition during every year. To obtain a probability over a substantial interval, one must account for the possibility that the state changes during that interval.

Definition: In the active-disabled-dead model, \(\mu_{AD}(t)\) is the active-to-disabled intensity (disability incidence), \(\mu_{A\dagger}(t)\) is the active-to-dead intensity (active-state mortality), \(\mu_{DA}(t)\) is the disabled-to-active intensity (recovery), and \(\mu_{D\dagger}(t)\) is the disabled-to-dead intensity (disabled-state mortality). Each is conditional on being in its starting state at time \(t\).

Incidence, Recovery, and Mortality Are Different Events

Disability incidence concerns a transition from active to disabled. Its intensity, \(\mu_{AD}(t)\), describes the instantaneous rate of new disability among people who are active at time \(t\). It does not describe the fraction of all people who are disabled, because that fraction also depends on previous disability, recovery, and survival.

Recovery is the reverse state transition, from disabled to active. Its intensity, \(\mu_{DA}(t)\), applies only to people currently disabled. It is not a reduction to be subtracted from disability incidence: the two rates apply to different starting states and describe different events.

Mortality is state-specific in this model. The intensity \(\mu_{A\dagger}(t)\) applies to active people; \(\mu_{D\dagger}(t)\) applies to disabled people. They need not be equal. A person who is disabled is not counted as an active-to-dead transition, and a person who dies does not recover or return to an active state.

All four intensities have units of inverse time. If time is measured in years, a rate of \(0.08\) is expressed per year. If time is measured in months, a rate of \(0.08\) is expressed per month. A change of time unit requires a corresponding change of rate unit.

Worked Example: Reading Four State-Specific Rates

Consider a hypothetical model for a covered person. At a particular time, while active, the disability incidence intensity is \(0.08\) per year and the active-state mortality intensity is \(0.01\) per year. While disabled, the recovery intensity is \(0.30\) per year and the disabled-state mortality intensity is \(0.04\) per year. Interpret the rates and approximate the relevant probabilities over a short interval of \(0.01\) year.

Step 1: Identify the starting state for each rate. The \(0.08\) rate applies only while active and describes active-to-disabled transition. The \(0.01\) rate also applies only while active and describes active-to-dead transition. The \(0.30\) rate applies only while disabled and describes recovery. The \(0.04\) rate applies only while disabled and describes disabled-to-dead transition.

Step 2: Apply the short-interval approximation. For an intensity \(\mu\) that is effectively constant over this short interval, the corresponding transition probability is approximately \(\mu h\), where \(h=0.01\) year.

\begin{aligned} P(A\to D\text{ in }0.01\text{ year}\mid A\text{ now}) &\approx 0.08(0.01)=0.0008,\\ P(A\to\dagger\text{ in }0.01\text{ year}\mid A\text{ now}) &\approx 0.01(0.01)=0.0001,\\ P(D\to A\text{ in }0.01\text{ year}\mid D\text{ now}) &\approx 0.30(0.01)=0.0030,\\ P(D\to\dagger\text{ in }0.01\text{ year}\mid D\text{ now}) &\approx 0.04(0.01)=0.0004. \end{aligned}

The approximations are conditional on the current state. For example, \(0.0008\) is the approximate probability of becoming disabled over this short interval for someone active now; it is not a disability probability for someone already disabled.

Check. The rates have units per year and the interval is in years, so each product \(\mu h\) is dimensionless, as a probability must be. The short-interval probabilities are small, consistent with using a local-rate approximation.

Why Competing Intensities Determine the First Transition

When multiple transitions can occur from the current state, each intensity describes a different possible destination. For example, an active person may become disabled or die. The first event occurs at whichever transition happens first; the probability of a particular destination therefore depends both on its own intensity and on the competing intensity.

Suppose, as a simplifying assumption for a fixed interval, that the intensities out of a state are constant and that a transition ends the person’s time in that starting state. Let the possible destinations have intensities \(\mu_1,\ldots,\mu_r\), and write \(\lambda=\sum_{j=1}^r\mu_j\). If \(s(t)\) is the probability of remaining in the starting state for at least \(t\), then over a short additional interval \(h\), the probability of leaving is approximately \(\lambda h\). This gives the differential relation \(s'(t)=-\lambda s(t)\), with \(s(0)=1\). Solving yields \(s(t)=e^{-\lambda t}\).

The probability density that the first transition occurs at time \(t\) and is to destination \(j\) is the probability of still being in the starting state, multiplied by that destination’s instantaneous intensity. Integrating this density gives the probability of that first transition by time \(T\):

P(\text{first transition is to }j\text{ by }T) = \int_0^T e^{-\lambda t}\mu_j\,dt = \frac{\mu_j}{\lambda}\left(1-e^{-\lambda T}\right).

This result assumes constant competing intensities over the interval and no other exit routes from the starting state. It is not the general formula for time-varying rates; it is useful here to show why a single intensity is not enough to determine a finite-period probability when other transitions compete.

Worked Example: Disability or Death First While Active

A hypothetical active person is subject to constant disability incidence intensity \(0.20\) per year and active-state mortality intensity \(0.10\) per year. Calculate the probability that death is the first transition within two years, the probability that disability is the first transition within two years, and the probability of remaining active throughout the two years. Assume these are the only transitions out of the active state during the interval.

Method. Treat disability and death as competing first transitions. Their combined rate of exit from active is

\lambda=0.20+0.10=0.30\text{ per year}.

For \(T=2\) years, the exponent is \(\lambda T=0.30(2)=0.60\). The probability of no transition out of active over the full interval is

e^{-0.60}=0.5488116361.

The probability that death is first is its share of the combined intensity multiplied by the probability of any exit by two years:

\frac{0.10}{0.30}\left(1-e^{-0.60}\right) = \frac{1}{3}(0.4511883639) = 0.1503961213.

The probability that disability is first is

\frac{0.20}{0.30}\left(1-e^{-0.60}\right) = \frac{2}{3}(0.4511883639) = 0.3007922426.

Rounded to four decimal places, the probabilities are \(0.1504\) for death first, \(0.3008\) for disability first, and \(0.5488\) for remaining active. The method avoids the common mistake of treating \(0.10\) as the two-year probability of death: it is an intensity per year, and disability is a competing exit.

Check. The outcomes are mutually exclusive and exhaustive under the stated assumptions: \(0.1503961213+0.3007922426+0.5488116361=1.0000000000\), apart from displayed rounding.

Recovery and Disabled-State Mortality as Competing Exits

The same reasoning applies while a person is disabled. Recovery and death are separate possible exits from the disabled state. A higher recovery intensity increases the instantaneous rate of recovery among those still disabled; a higher disabled-state mortality intensity increases the instantaneous rate of death among those still disabled. Neither rate is a probability over a full year unless a probability calculation has also accounted for time and competing transitions.

Worked Example: Recovery Versus Death During Disability

A hypothetical disabled claimant has constant recovery intensity \(0.40\) per year and disabled-state mortality intensity \(0.10\) per year. Find the probability that recovery is the first transition within two years, and the probability that death is the first transition within two years. Assume these are the only exits from the disabled state during that period.

Step 1: Combine the exit intensities.

\lambda=0.40+0.10=0.50\text{ per year}.

Step 2: Calculate the probability of any exit by two years. The exponent is \(\lambda T=0.50(2)=1.00\), so the probability of still being disabled with no exit is

e^{-1.00}=0.3678794412, \qquad 1-e^{-1.00}=0.6321205588.

Step 3: Allocate exits by their intensity shares. Recovery accounts for \(0.40/0.50=0.8\) of the combined intensity. Thus

P(\text{recovery first by 2}) = 0.8(0.6321205588) = 0.5056964470.

Death accounts for \(0.10/0.50=0.2\) of the combined intensity:

P(\text{death first by 2}) = 0.2(0.6321205588) = 0.1264241118.

Rounded to four decimal places, the probabilities are \(0.5057\) and \(0.1264\), respectively. The remaining \(0.3679\) is the probability of no transition out of disabled during the two years. As a check, the three probabilities sum to \(1.0000\) at the displayed precision. The calculation avoids interpreting the recovery intensity \(0.40\) as a 40% two-year recovery probability.

Time Units and Common Interpretation Traps

The units attached to an intensity are part of its meaning. If an intensity is stated per month, multiply by an interval measured in months when using a short-interval approximation or a constant-rate formula. Converting a monthly intensity to an annual rate requires multiplying by 12, provided the model’s rate is constant under that time-unit conversion. Do not multiply by 12 and then also use a monthly interval.

For example, suppose the disability incidence intensity is \(0.015\) per month and active-state mortality intensity is \(0.002\) per month, both constant over one month. The combined exit intensity is \(0.017\) per month. The probability of any exit during the month is \(1-e^{-0.017}=0.0168563\), not \(1-e^{-0.204}\); the latter uses a one-year interval of 12 months. Of the monthly exits, disability is the first transition with probability

\frac{0.015}{0.017}(1-e^{-0.017}) \approx 0.0148732,

and death is first with probability

\frac{0.002}{0.017}(1-e^{-0.017}) \approx 0.0019831.

These probabilities sum to \(0.0168563\), the probability of any exit. They are conditional on being active at the start of the month and use the stated constant-intensity assumption.

Key takeaway: Identify both the transition and its starting state before interpreting an intensity. Active-to-disabled is incidence; disabled-to-active is recovery; active-to-dead and disabled-to-dead are distinct mortality intensities. Multiply by a time interval only for a short-interval approximation, and use a finite-interval calculation when required.

Check Your Understanding

For each problem, use the stated assumptions and keep the time units consistent.

At a particular time, an active person has disability incidence intensity \(0.06\) per year and active-state mortality intensity \(0.02\) per year, constant for \(0.1\) year. What is the probability that death is the first transition during that interval? Give four decimal places. (Answer: \(\frac{0.02}{0.08}(1-e^{-0.008})\approx 0.001992\), or \(0.0020\) to four decimal places.)
A model gives \(\mu_{DA}=0.25\) per year and \(\mu_{D\dagger}=0.05\) per year. State what each intensity measures and identify the starting state for both. (Answer: \(0.25\) per year is recovery from disabled to active; \(0.05\) per year is death from the disabled state; both are conditional on being disabled.)
A disabled person has constant recovery intensity \(0.50\) per year and disabled-state mortality intensity \(0.10\) per year. What is the probability that recovery is the first transition within one year? Give four decimals. (Answer: \(\frac{0.50}{0.60}(1-e^{-0.60})\approx 0.3760\).)
For an active person, disability incidence intensity is \(0.004\) per week and active-state mortality intensity is \(0.0005\) per week, both constant for four weeks. What is the probability that disability is the first transition during those four weeks? Give four decimals. (Answer: \(\frac{0.004}{0.0045}(1-e^{-0.018})\approx 0.0159\).)
For an active person, constant disability incidence and active-state mortality intensities are \(0.10\) and \(0.05\) per year, respectively. What is the probability that disability is the first transition within one year? Give four decimals, and state one assumption needed for the calculation. (Answer: \(\frac{0.10}{0.15}(1-e^{-0.15})\approx 0.0929\); the intensities are constant over the year and these are the only exits from active.)

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