Tutorials › Exam ALTAM - Advanced Long-Term Actuarial Mathematics › Remaining Probability Under Time-Varying Intensities

Home  /  Tutorials  /  Exam ALTAM / Competing Decrement Models

Exam ALTAM · Competing Decrement Models · Tutorial 203 of 999

Remaining Probability Under Time-Varying Intensities

Learn to combine time-varying cause-specific intensities, integrate the total intensity over time, and calculate the probability of remaining in the group.

Intermediate 10 min read

What You'll Learn

  • State the assumptions behind a time-varying competing-decrement model.
  • Derive remaining probability from the survival differential equation.
  • Integrate cause-specific intensities to obtain cumulative total intensity.
  • Evaluate the remaining probability for linear, polynomial, and exponential rates.
  • Recognize time-unit and rate-versus-probability errors in calculations.

Why Time-Varying Intensities Need an Integral

In a competing-decrement model, a person can leave an initial group through one of several mutually exclusive causes. The chance of remaining in the group depends on the combined effect of all causes over the time interval. When the cause-specific intensities change with time, a single constant rate does not describe that effect: the intensity at each instant contributes over the time it applies.

This tutorial derives the remaining-probability formula for time-varying intensities and applies it to several functional forms. The central calculation is to add the cause-specific intensities at each time, integrate that total over the interval, and exponentiate the negative integral. The result is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let \(T\) be the time of the first exit from an initial group, and let there be \(J\) competing exit causes. For cause \(j\), the cause-specific intensity at time \(t\), conditional on still being in the group immediately before \(t\), is \(\mu_j(t)\). The total intensity is \(\mu(t)=\sum_{j=1}^{J}\mu_j(t)\). Define the remaining probability by \(R(t)=P(T>t)\).

Assume the intensity functions are nonnegative and integrable over the interval of interest. They describe the instantaneous exit rates for a person who is still in the group; they are not probabilities of exit over a whole year or other interval. The model concerns the first exit from the initial group. If a person can exit and later re-enter, this formula alone does not give the probability of occupying the group at a later time.

Deriving the Remaining-Probability Formula

Over a short interval from \(t\) to \(t+h\), the conditional probability of exiting through any cause is approximately the total intensity at \(t\) multiplied by the interval length. More precisely, under the continuous-time intensity model, \(P(t<T\leq t+h\mid T>t)=\mu(t)h+o(h)\), where \(o(h)/h\) tends to zero as \(h\) tends to zero. Therefore the conditional probability of remaining in the group throughout that short interval is \(1-\mu(t)h+o(h)\).

The probability of remaining until \(t+h\) is the probability of having remained until \(t\), multiplied by the conditional probability of surviving the next short interval. Thus,

$$ R(t+h)=R(t)\{1-\mu(t)h+o(h)\}. $$

Subtract \(R(t)\), divide by \(h\), and take the limit as \(h\) decreases to zero. This gives the survival differential equation. Since \(R(0)=1\), separating variables and integrating from 0 to \(t\) yields the result.

$$ \frac{dR(t)}{dt}=-\mu(t)R(t),\qquad \frac{d}{dt}\log R(t)=-\mu(t), $$
$$ \log R(t)=-\int_0^t \mu(s)\,ds =-\int_0^t\sum_{j=1}^{J}\mu_j(s)\,ds, \qquad R(t)=\exp\left(-\int_0^t\sum_{j=1}^{J}\mu_j(s)\,ds\right). $$
Theorem: If a person is in the group at time \(s\), the probability of remaining in it through time \(t\), where \(t\geq s\), is \( \exp\left(-\int_s^t\mu(u)\,du\right) \), with \(\mu(u)\) the total intensity at time \(u\). If the intensities are functions of time since entry, use that duration as the integration variable. For constant total intensity \(\mu\), this reduces to \(e^{-\mu(t-s)}\).

The formula depends on the cumulative total intensity over the interval, not just the endpoint intensity. Adding the cause-specific intensities before integrating is valid because the instantaneous exits through the causes are mutually exclusive. Equivalently, one can integrate each cause-specific intensity and add the integrals; both methods give the same cumulative total intensity.

Applying the Formula

A reliable calculation follows the same sequence each time: identify the time unit and interval, write the total intensity, integrate it over that interval, and then exponentiate its negative. Keep the intensity functions in rate-per-time units throughout. The integral is dimensionless, as an exponent must be.

1
Set the time scale.
Confirm whether \(t\) is measured in years, months, or another unit, and make sure each intensity uses that same unit.
2
Form the total intensity.
Add the cause-specific intensities as functions of time, rather than adding interval probabilities.
3
Integrate over the full interval.
Evaluate the cumulative total intensity from the starting time to the ending time.
4
Exponentiate and check.
Calculate the exponential of the negative cumulative intensity. The answer must lie between 0 and 1 and cannot exceed 1.

Worked Example: Three Causes With Linear Intensities

A hypothetical insured is in an active-care group at time 0. Time \(t\) is measured in years. While the insured remains in the group, the cause-specific intensities are \(\mu_1(t)=0.01\), \(\mu_2(t)=0.02+0.01t\), and \(\mu_3(t)=0.005t\), all per year. Find the probability of remaining in the group for two years.

Method. Add the intensities at each time, integrate the resulting total intensity over years 0 to 2, and take the exponential of its negative.

The total intensity is \(\mu(t)=0.01+(0.02+0.01t)+0.005t=0.03+0.015t\). Integrating by cause makes each contribution explicit:

$$ \int_0^2\mu_1(t)\,dt=0.01(2)=0.02, \qquad \int_0^2\mu_2(t)\,dt=0.02(2)+0.01\frac{2^2}{2}=0.04+0.02=0.06, $$
$$ \int_0^2\mu_3(t)\,dt=0.005\frac{2^2}{2}=0.01, \qquad \int_0^2\mu(t)\,dt=0.02+0.06+0.01=0.09. $$

Therefore the remaining probability is \(R(2)=e^{-0.09}=0.913931\ldots\), or 0.9139 to four decimal places. The cumulative intensity is positive, so the probability is less than 1, as required. The common mistake is to misintegrate the linear term: \(\int_0^2 0.015t\,dt=0.015(2^2/2)=0.03\), not 0.06.

Worked Example: A Decreasing Exponential Intensity

In a hypothetical post-treatment group, time \(t\) is measured in years and the two cause-specific intensities are \(\mu_1(t)=0.02+0.01t\) and \(\mu_2(t)=0.01e^{-t}\), per year. Find the probability of remaining in the group for two years. Give the probability to four decimal places.

Method. Integrate the two cause-specific intensities separately. The linear part has an elementary polynomial integral; the exponential part uses \(\int e^{-t}\,dt=-e^{-t}\).

$$ \int_0^2\mu_1(t)\,dt =0.02(2)+0.01\frac{2^2}{2} =0.04+0.02=0.06. $$

For the second cause,

$$ \int_0^2\mu_2(t)\,dt =0.01[-e^{-t}]_0^2 =0.01(1-e^{-2}) =0.00864665\ldots. $$

The cumulative total intensity is \(0.06+0.00864665\ldots=0.06864665\ldots\). Hence \(R(2)=e^{-0.06864665\ldots}=0.9336565\ldots\), which is 0.9337 to four decimal places. This value is below 1 and above the probability \(e^{-0.06}\) that would result from cause 1 alone, which is a useful check because cause 2 adds a nonnegative exit intensity. A common error is to treat the decreasing intensity \(0.01e^{-t}\) as a constant \(0.01\) for both years; that would incorrectly assign it an integral of 0.02 instead of \(0.00864665\ldots\).

Worked Example: Combining Linear and Quadratic Intensities

A hypothetical disability model tracks a group for up to three years. Let time \(t\) be measured in years, with cause-specific intensities \(\mu_1(t)=0.03+0.004t\) and \(\mu_2(t)=0.002t^2\), per year. Calculate the probability of remaining in the group at three years.

Method. Integrate the constant and linear terms for the first cause, then integrate the quadratic term for the second cause. Add the integrals before applying the exponential.

$$ \int_0^3\mu_1(t)\,dt =0.03(3)+0.004\frac{3^2}{2} =0.09+0.018=0.108, $$
$$ \int_0^3\mu_2(t)\,dt =0.002\frac{3^3}{3} =0.002(9)=0.018. $$

Thus the cumulative total intensity is \(0.108+0.018=0.126\), and the remaining probability is \(R(3)=e^{-0.126}=0.8816186\ldots\), or 0.8816 to four decimal places. The result is between 0 and 1, and it is less than the remaining probability under cause 1 alone because cause 2 adds an additional nonnegative intensity. The common mistake is to integrate \(0.002t^2\) as though its integral were \(0.002t^2\); the antiderivative is \(0.002t^3/3\).

Interpretation and Common Exam Traps

The exponent contains the total intensity accumulated throughout the interval. If a total intensity increases with time, later portions of the interval contribute more per unit time than earlier portions; if it decreases, they contribute less. The formula accounts for this variation without requiring a constant-rate approximation. For a model starting at time \(s\) rather than 0, use the same calculation with limits \(s\) and \(t\).

A frequent trap is confusing an intensity with an interval probability. An intensity is an instantaneous rate conditional on still being in the group. It can vary with time and is not itself the probability of exiting during a year. The exponential-integral formula converts the accumulated rate into the probability of remaining. Another trap is mixing units: if \(t\) is in months, the intensities must be per month, or the time must be converted consistently before integration.

The formula also makes a useful comparison possible. If one cause-specific intensity is increased at any times while the others are unchanged, the cumulative total intensity cannot decrease. Since \(e^{-x}\) decreases as \(x\) increases, the remaining probability cannot increase. This is a mathematical check on sensitivity calculations and on the direction of a numerical answer.

Key takeaway: For time-varying competing causes, remaining probability is the exponential of the negative integrated total intensity. Add the rates as functions of time, integrate over the correct interval and time scale, and only then exponentiate.

Check Your Understanding

For each problem, time is measured in the stated unit and all intensities are conditional on remaining in the group. Round probabilities to four decimal places unless another rounding instruction is given.

  1. Two causes have intensities \(\mu_1(t)=0.015\) and \(\mu_2(t)=0.01+0.005t\) per year. Find the probability of remaining for four years. (Answer: 0.8694)
  2. For \(0\leq t\leq2\) years, let \(\mu_1(t)=0.02t\) and \(\mu_2(t)=0.01t^2\), per year. Find the probability of remaining through year 2. (Answer: 0.9355)
  3. For a hypothetical group, \(\mu_1(t)=0.01+0.002t\) and \(\mu_2(t)=0.004e^{-t}\), per year, with \(t\) in years. Find the probability of remaining for three years. (Answer: 0.9581)
  4. Time is measured in months. Two cause-specific intensities are \(\mu_1(t)=0.002\) and \(\mu_2(t)=0.001+0.0005t\), per month. Find the probability of remaining for 12 months. (Answer: 0.9305)
  5. A single total intensity is \(\mu(t)=0.02+0.01t\), per year, where \(t\) is in years. Find the time \(t\) at which the remaining probability first reaches 0.9. Give the answer to three decimal places. (Answer: 3.007 years)

Previous

Next

Back to Tutorials
Clinical Biostats