Why Total Intensity Is the Sum of Cause Intensities

Why Cause Intensities Add

In a competing decrement model, a member of a group may leave because of one of several causes. For example, an active policyholder might exit through death, surrender, or retirement. The causes are alternatives for the first exit from the group: for a given member, only one of them can be the cause of that exit. This structure makes it possible to describe the rate of all exits using the cause-specific rates.

The key point is that an intensity is defined through probabilities over a very short interval. For mutually exclusive exit causes, the probability of any exit over that interval is the sum of the probabilities of exiting by each cause. Dividing by the interval length and taking the limit gives the corresponding identity for intensities.

Definition: Suppose a member is in the group at time \(t\), and there are a finite number \(m\) of mutually exclusive causes of first exit. The cause-specific intensity for cause \(j\), denoted \(\mu^{(j)}(t)\), is the limit of the probability of exiting by cause \(j\) in the next interval of length \(h\), divided by \(h\), as \(h\) decreases to zero, provided that limit exists. The total intensity \(\mu^{(\cdot)}(t)\) is defined in the same way using the probability of exit by any cause.

The intensities are conditional on being in the group at time \(t\). They are instantaneous rates, not probabilities for a whole year or another fixed interval. Their units are the inverse of the time unit used for \(t\): if time is measured in years, an intensity is expressed per year.

Deriving the Sum Identity

Let \(C_j(h)\) be the event that the first exit occurs in the interval \((t,t+h]\) and is due to cause \(j\), conditional on being in the group at time \(t\). Let \(C(h)\) be the event that the first exit occurs in that interval, whatever its cause. Because every exit in this model has exactly one of the listed causes, the events \(C_1(h),\ldots,C_m(h)\) are mutually exclusive and their union is \(C(h)\). Therefore, for every \(h>0\),

P(C(h)\mid\text{in group at }t) = \sum_{j=1}^{m}P(C_j(h)\mid\text{in group at }t).

Divide both sides by \(h\). If each cause-specific intensity and the total intensity exist as finite limits when \(h\) decreases to zero, the limit of the finite sum is the sum of the limits. Hence

\mu^{(\cdot)}(t) = \sum_{j=1}^{m}\mu^{(j)}(t).

This result depends on the causes being a mutually exclusive and exhaustive classification of first exits from the group. It also uses a common time scale and a common group-at-risk condition for all the cause-specific intensities. If rates are measured using different time units or different populations at risk, they cannot be added until those definitions are made consistent.

Theorem: For a finite set of mutually exclusive and exhaustive causes of first exit, if the cause-specific intensities are defined conditional on the same group membership at time \(t\) and their limits exist, then the total intensity at \(t\) is their sum: \(\mu^{(\cdot)}(t)=\sum_{j=1}^{m}\mu^{(j)}(t)\).

Worked Applications

The identity is pointwise in time: it applies to the intensities at a specified time. When the cause-specific intensities vary with time, add their values at the time of interest. Do not substitute an interval probability for an intensity.

Worked Example: Aggregating Three Constant Cause Intensities

Method. Convert all cause rates to the same time unit and add them. A hypothetical income-protection contract has members in an active state. At a specified attained age, the annual cause-specific intensities of first exit are 0.018 for death, 0.042 for surrender, and 0.015 for retirement. Find the total intensity. Also give its first-order approximation to the probability of any exit during the next 0.01 year.

Step 1: Check the assumptions and units. The three causes are assumed to be mutually exclusive first exits, the intensities are all conditional on being active at the specified time, and each rate is measured per year.

Step 2: Add the cause-specific intensities.

\mu^{(\cdot)} =0.018+0.042+0.015 =0.075\text{ per year}.

Step 3: Use the intensity definition for a short interval. For small \(h\), the probability of any exit is \(\mu^{(\cdot)}h+o(h)\), where \(o(h)\) denotes terms whose ratio to \(h\) tends to zero as \(h\) tends to zero. With \(h=0.01\) year, the first-order term is

$$ 0.075(0.01)=0.00075. $$

Thus the total intensity is 0.075 per year, and the first-order approximation to the probability of exit in 0.01 year is 0.00075. This is only a short-interval approximation; the exact probability over a fixed interval cannot be obtained from an instantaneous intensity at one time without further assumptions about how the rates behave.

Check. The total is greater than each individual intensity and has units per year, as required. The common mistake avoided here is treating 0.075 as a one-year exit probability; it is an instantaneous rate.

Worked Example: Adding Time-Varying Cause Intensities

Method. Evaluate each cause-specific intensity at the requested duration, then add. In a hypothetical membership model, time \(t\) is measured in years since entry. The three cause intensities are

\mu^{(1)}(t)=0.008+0.001t,\qquad \mu^{(2)}(t)=0.012+0.0005t,\qquad \mu^{(3)}(t)=0.020-0.001t.

All rates are per year. Find the total intensity at duration \(t=6\), and state its first-order contribution to the probability of exit in the next 0.02 year.

Step 1: Evaluate each cause at \(t=6\).

\begin{aligned} \mu^{(1)}(6)&=0.008+0.001(6)=0.014,\\ \mu^{(2)}(6)&=0.012+0.0005(6)=0.015,\\ \mu^{(3)}(6)&=0.020-0.001(6)=0.014. \end{aligned}

Step 2: Add the instantaneous rates.

\mu^{(\cdot)}(6)=0.014+0.015+0.014=0.043\text{ per year}.

Step 3: Form the short-interval first-order term. For \(h=0.02\) year, that term is \(0.043(0.02)=0.00086\). Thus the first-order contribution to the exit probability is 0.00086; it is not a claim that this is the exact interval probability.

Check. Adding the functions first gives \(0.040+0.0005t\), which at \(t=6\) is also \(0.043\). The agreement confirms the arithmetic. The common mistake avoided is adding coefficients from the functions without evaluating the rates at the requested duration.

Worked Example: Recovering a Missing Cause Intensity

Method. Rearrange the total-intensity identity after confirming that all rates refer to the same group, time, and units. For a hypothetical group of active members at duration three years, the total intensity of first exit is 0.084 per year. The known cause-specific intensities are 0.019 per year for death and 0.027 per year for retirement. Find the intensity for the remaining cause, lapse.

Step 1: Write the identity for these three causes.

0.084=0.019+0.027+\mu^{(\mathrm{lapse})}.

Step 2: Isolate the unknown rate.

\begin{aligned} \mu^{(\mathrm{lapse})} &=0.084-0.019-0.027\\ &=0.084-0.046\\ &=0.038\text{ per year}. \end{aligned}

Step 3: Check the result. The inferred intensity is nonnegative, and \(0.019+0.027+0.038=0.084\). Therefore, the lapse intensity is 0.038 per year. A negative remainder would signal inconsistent data, mismatched definitions, or an omitted or duplicated cause; it would not represent a valid nonnegative intensity in this model.

The common mistake avoided is subtracting one known rate from the total and forgetting the second known cause. Writing the full sum before rearranging makes the accounting explicit.

Common Exam Trap: Mixing Units and Interpretations

A cause intensity is a rate conditional on being in the group at the instant under consideration. For example, if two cause intensities are reported as 12 and 8 exits per 1,000 members per year, their total is 20 exits per 1,000 members per year, or \(20/1000=0.020\) per year. The arithmetic sum is valid only after both values have the same denominator and time unit.

A frequent incorrect calculation is to add the numerical values 12 and 8 and report a total intensity of 20 per year. That calculation silently changes the scale: the given rates were per 1,000 members, not per member. The correction is to preserve the stated units while adding, then convert if required:

\frac{12}{1000}+\frac{8}{1000} =\frac{20}{1000} =0.020\text{ per year}.

Another error is to interpret this intensity as an exact probability over a year. The identity concerns instantaneous rates. A probability over a non-infinitesimal interval requires the behavior of the intensities over that interval and an appropriate probability calculation. Here the relevant conclusion is simply that, at the specified instant, the total intensity is the sum of the cause-specific intensities.

Why the Identity Matters

The total intensity provides a compact description of the overall rate of leaving a state, while the individual cause intensities retain the reason for exit. This separation is useful when a model needs both an overall exit rate and an allocation of exits among causes. It also gives an immediate consistency check: with nonnegative cause-specific intensities, the total must be at least as large as each individual cause intensity.

Before adding rates in an actuarial calculation, verify three items:

The causes are mutually exclusive categories of first exit and together cover the exits being modelled.
Each rate is conditional on the same group being at risk at the same time.
All rates use the same time unit and scale.

When these conditions hold and the limits exist, the result follows directly from the addition rule for mutually exclusive events. No independence assumption between the causes is needed: the model classifies a single first exit, so the cause events for that exit cannot occur together.

Check Your Understanding

For each problem, use the cause-specific rates only under the stated common risk-group and mutually exclusive first-exit assumptions.

Three cause intensities are 0.003, 0.007, and 0.002 per month. Find the total intensity per month and per year. (Answer: 0.012 per month; 0.144 per year.)
At duration \(t\) years, two cause intensities are \(\mu^{(1)}(t)=0.010+0.002t\) and \(\mu^{(2)}(t)=0.025-0.001t\), per year. A third is \(\mu^{(3)}(t)=0.006+0.0005t\), per year. Find the total intensity at \(t=3\). (Answer: 0.0455 per year.)
The total intensity is 0.073 per year. Three cause-specific intensities are 0.018, 0.024, and 0.011 per year; find the fourth. Verify the total. (Answer: 0.020 per year; \(0.018+0.024+0.011+0.020=0.073\).)
Two mutually exclusive causes have intensities of 5.4 and 2.1 exits per 1,000 members per year. Convert the total intensity to a per-member rate, then find its first-order contribution to the probability of exit over 0.01 year. (Answer: 0.0075 per year; 0.000075.)
At a specified time, three mutually exclusive cause-specific intensities are 0.016, 0.009, and 0.025 per year, all conditional on the same group at risk. State the total intensity and explain why adding them does not require an independence assumption. (Answer: 0.050 per year; the cause events for a single first exit are mutually exclusive, so their probabilities add.)

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