Why Intensities Are Not Interval Probabilities

Introduction

In a decrement model, an intensity describes how quickly exits are occurring at an instant, among members who have not yet exited. A probability describes the chance of an event occurring over a specified interval. The two quantities are connected, but they are not interchangeable: an intensity is not itself the probability of exit during a year, a month, or any other finite period.

That distinction matters whenever a calculation converts rates into cohort probabilities. Using a rate as if it were an interval probability can produce an incorrect answer, or even a number greater than one. The key is to account for the time exposed to the risk and for the possibility that an exit happens before the interval ends.

Definition: For an individual currently in the active group at time \(t\), the cause-specific intensity \(\lambda_j(t)\) for cause \(j\) is the instantaneous rate of exit from that group due to cause \(j\). In a short interval of length \(\Delta t\), its defining approximation is \(P(\text{exit from cause }j\text{ in }[t,t+\Delta t)\mid\text{active at }t)=\lambda_j(t)\Delta t+o(\Delta t)\), where \(o(\Delta t)/\Delta t\) tends to zero as \(\Delta t\) tends to zero. An intensity is measured per unit of time; a finite-interval exit probability is dimensionless.

Intensity and Probability Answer Different Questions

A cause-specific intensity is conditional on still being in the group at the instant when the rate is considered. It describes the local rate at which exits of that cause occur. A probability over an interval instead starts with the group at the beginning of the interval and asks what fraction exits from a cause before the interval ends.

For example, an intensity of \(0.8\) per year does not mean that 80% of the group exits from that cause during a year. Some members who are exposed at the start may exit earlier, and each member can exit at most once. The intensity must be combined with the probability of remaining exposed up to each point in the interval.

For multiple causes, write \(\lambda_j(t)\) for the intensity of cause \(j\), and let the total intensity be \(\lambda_\bullet(t)=\sum_j\lambda_j(t)\). Under the competing-decrement framework, the probability of remaining in the group from time \(0\) to time \(t\) is

{}_tp=\exp\left(-\int_0^t\lambda_\bullet(s)\,ds\right).

Here \({}_tp\) denotes the probability of remaining in the active group for \(t\) time units, conditional on being in it at time 0. The probability of exiting from cause \(j\) during \([0,h]\) is

q_j(0,h)=\int_0^h {}_tp\,\lambda_j(t)\,dt.

The integrand has a direct interpretation: \({}_tp\) is the probability of still being exposed at time \(t\), and \(\lambda_j(t)\,dt\) is the approximate conditional probability of an exit from cause \(j\) in the next small interval, given exposure at time \(t\). The integral accumulates these probabilities across the interval.

Why the Conversion Formula Holds

The formula follows by dividing the interval into short pieces. At the start of a piece near time \(t\), the probability of still being in the group is approximately \({}_tp\). Conditional on being there, the probability of a cause-\(j\) exit in a piece of width \(\Delta t\) is approximately \(\lambda_j(t)\Delta t\). Therefore the unconditional probability of a cause-\(j\) exit in that piece is approximately \({}_tp\,\lambda_j(t)\Delta t\). Adding these contributions and taking the limit as the pieces become arbitrarily short gives the integral.

If all cause-specific intensities are constant over an interval of length \(h\), let their sum be the constant \(\lambda_\bullet\). Then survival to time \(t\) is \(e^{-\lambda_\bullet t}\), and integration gives

q_j(0,h)=\int_0^h e^{-\lambda_\bullet t}\lambda_j\,dt =\frac{\lambda_j}{\lambda_\bullet}\left(1-e^{-\lambda_\bullet h}\right).

This constant-intensity result requires constant rates over the interval and a fixed set of competing causes. It does not say that the intensity equals the exit probability. Instead, the cause-specific probability is the total probability of any exit multiplied by the cause's share of the total intensity. For time-varying rates, the integral form applies; the constant-rate expression cannot be used without its constant-rate assumption.

Worked Example: A Single Cause Over One Year

A hypothetical group is subject to one decrement cause with a constant intensity of \(1.4\) exits per person-year. Find the probability that a member of the group exits from this cause during one year, assuming the member is in the group at the start.

Method. With one cause, the total intensity is also \(1.4\) per year. Apply the constant-intensity formula with \(\lambda_j=\lambda_\bullet=1.4\) and \(h=1\):

q(0,1)=\frac{1.4}{1.4}\left(1-e^{-1.4(1)}\right) =1-e^{-1.4}.

Using \(e^{-1.4}=0.24659696\), the one-year exit probability is \(1-0.24659696=0.75340304\), or 0.7534 to four decimal places.

The probability of remaining in the group is \(e^{-1.4}=0.2466\). The check is \(0.7534+0.2466=1.0000\), allowing for rounding. The intensity is \(1.4\) per year, while the probability of exit is \(0.7534\). The intensity exceeds one because it is a rate per unit time, not a probability. The common error of reporting 1.4 as the one-year exit probability would be invalid because it exceeds one.

Competing Causes Over a Finite Interval

With competing causes, an exit from any cause prevents a later exit from another cause during the same spell in the active group. Consequently, each cause-specific probability depends not only on its own intensity but also on the total intensity, since the total rate determines how many members remain exposed as time passes.

The probabilities must obey conservation: the probability of remaining plus the probabilities of exiting from all causes must equal one. For constant intensities over an interval of length \(h\), the total probability of any exit is \(1-e^{-\lambda_\bullet h}\), and the cause-specific probabilities divide that total according to the intensity shares. An intensity share is not generally the probability of that cause's exit over the interval; it is the cause's share of exits when the intensities are constant.

Worked Example: Mortality and Withdrawal Competing for One Year

A hypothetical long-term contract has two constant decrement intensities over the next year: mortality at \(0.3\) per year and withdrawal at \(0.2\) per year. For a contract in force at the start of the year, calculate the probability of remaining in force, the probability of mortality, and the probability of withdrawal.

Step 1: Find the total intensity.

\lambda_\bullet=0.3+0.2=0.5\text{ per year}.

Step 2: Find the probability of remaining in force.

{}_1p=e^{-0.5(1)}=e^{-0.5}=0.60653066\approx 0.6065.

Step 3: Find the total probability of exit.

1-{}_1p=1-0.60653066=0.39346934.

Step 4: Allocate exits by cause. Mortality's share of the total intensity is \(0.3/0.5=0.6\), and withdrawal's share is \(0.2/0.5=0.4\). Thus

q_{\text{mortality}}(0,1)=0.6(0.39346934)=0.23608160\approx 0.2361, $$ $$ q_{\text{withdrawal}}(0,1)=0.4(0.39346934)=0.15738774\approx 0.1574.

Check. The probabilities sum to \(0.6065+0.2361+0.1574=1.0000\). The rates \(0.3\) and \(0.2\) are not the one-year probabilities: the correct probabilities are \(0.2361\) and \(0.1574\). The calculation uses the cause shares only after finding the total probability of exit.

Key result: Under constant competing intensities, the probability of exit from cause \(j\) over \(h\) time units is \(\frac{\lambda_j}{\lambda_\bullet}(1-e^{-\lambda_\bullet h})\), not \(\lambda_j h\). The latter can be a useful short-interval approximation when the total probability of exit is small, but it is not the exact finite-interval probability.

Worked Example: Three Years With Two Competing Causes

A hypothetical benefit remains payable while a member is in an active status. For three years, the constant intensity of transition to a care status is \(0.12\) per year and the constant intensity of death is \(0.08\) per year. Starting in the active status, find the probability of each exit over three years and the probability of remaining active.

Step 1: Combine the intensities and calculate survival. The total intensity is \(0.12+0.08=0.20\) per year, so

{}_3p=e^{-0.20(3)}=e^{-0.6}=0.54881164\approx 0.5488.

Step 2: Calculate the total probability of exit.

1-{}_3p=1-0.54881164=0.45118836.

Step 3: Allocate exits by cause. The care-status intensity share is \(0.12/0.20=0.6\), and the death intensity share is \(0.08/0.20=0.4\). Therefore

q_{\text{care}}(0,3)=0.6(0.45118836)=0.27071302\approx 0.2707, $$ $$ q_{\text{death}}(0,3)=0.4(0.45118836)=0.18047534\approx 0.1805.

Check. The rounded probabilities satisfy \(0.5488+0.2707+0.1805=1.0000\). Among members who exit during the three years, the care-status share is \(0.27071302/0.45118836=0.6000\). The exit probability of \(0.2707\) is not \(0.12\), nor is it the three-year product \(0.12(3)\); exposure to the care-status risk ends when either cause removes a member from the active group.

Common Mistakes and Practical Checks

The most important exam trap is to read a cause-specific intensity as an interval probability. For instance, with constant intensities \(0.3\) and \(0.2\) per year over one year, the incorrect calculation might assign mortality probability \(0.3\) and withdrawal probability \(0.2\). It would then assign a total exit probability of \(0.5\) and remaining probability of \(0.5\). The correct remaining probability is \(e^{-0.5}=0.6065\), and the correct cause probabilities are \(0.2361\) and \(0.1574\). The wrong calculation treats each member as if exposed to both full-year rates, regardless of whether an exit has already occurred.

A related error is to multiply an intensity by the interval length and call the result exact. For a single constant intensity of \(0.8\) per year over one year, this gives \(0.8\), while the exact probability is \(1-e^{-0.8}=0.5507\). The product \(\lambda h\) approximates the probability only when the interval is short enough that the chance of exit is small. It is not a general conversion rule.

Before accepting a calculation, check the following:

Units: an intensity is expressed per unit of time, so the interval length must use the same time unit.
Probability bounds: every finite-interval probability must lie between zero and one, even when an intensity is greater than one per year.
Competing causes: use the total intensity to find remaining probability, then account for each cause's intensity.
Conservation: add the probability of remaining to all cause-specific exit probabilities; the total should equal one.

These checks are useful in valuation and cohort calculations because a rate error can affect every later quantity built from the exit probabilities. This distinction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume members start in the active group and use the stated constant intensities throughout the interval. Give probabilities to four decimal places.

A hypothetical service contract has one decrement cause with intensity \(0.45\) per year. Find the probability of exit over two years, and state why multiplying the intensity by two is not the exact answer. (Answer: \(1-e^{-0.9}=0.5934\); \(0.45(2)=0.9\) is not the exact probability.)
A hypothetical benefit has two competing causes with intensities \(0.18\) and \(0.12\) per year over two years. Find the probability of remaining and the probability of exit from each cause. (Answer: remaining \(e^{-0.6}=0.5488\); cause 1 \(0.2707\); cause 2 \(0.1805\).)
A hypothetical membership group has one exit cause with intensity \(1.1\) per year. Find the exit probability over six months, using years as the time unit. (Answer: \(1-e^{-1.1(0.5)}=0.4231\).)
A hypothetical policy has competing intensities of \(0.24\) per year for one exit cause and \(0.16\) per year for another, over one year. Find the probability of remaining, both cause-specific exit probabilities, and the first cause's share among exits. (Answer: remaining \(0.6703\); first cause \(0.1978\); second cause \(0.1319\); first cause's share among exits \(0.6000\).)
A hypothetical one-cause model has constant intensity \(0.7\) per year. Calculate the exit probability over one year and over one quarter of a year, then explain why multiplying the quarterly probability by four does not give the one-year probability. (Answer: one year \(1-e^{-0.7}=0.5034\); one quarter \(1-e^{-0.175}=0.1605\); four times the quarterly probability is approximately \(0.6422\), not \(0.5034\), because the quarterly probability is conditional on remaining exposed at that quarter's start.)

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