Two-Cause Calculations With Mortality and Withdrawal

Why Two-Cause Calculations Matter

A person covered by a long-term contract may leave the active group for more than one reason. In a hypothetical disability-income contract, for example, a covered person may die or withdraw from coverage. Each cause removes the person from the group at risk of either cause. The probability of leaving by one cause must therefore account for the competing possibility of leaving by the other.

This tutorial focuses on calculations with two causes, mortality and withdrawal, when both cause-specific intensities are constant over the period considered. The key steps are to add the intensities to obtain the total intensity, calculate the probability of remaining, and then allocate the probability of exit between the two causes. All rates in the examples are annual forces, and the time period is measured in years.

Definition: Suppose a member of a group can leave by mortality or withdrawal. Let \(\mu\) be the constant mortality intensity and \(\omega\) the constant withdrawal intensity. The total exit intensity is \(\lambda=\mu+\omega\). For a period of \(t\) years, the probability of remaining in the group is \(e^{-\lambda t}\). The probability of exit by mortality is \(\frac{\mu}{\lambda}(1-e^{-\lambda t})\), and the probability of exit by withdrawal is \(\frac{\omega}{\lambda}(1-e^{-\lambda t})\), provided \(\lambda>0\).

These formulas assume that the two causes are the only ways to leave the group, and that each cause-specific intensity stays constant throughout the period. An intensity is a rate per unit of time, not a probability of exit over the whole period. The formulas account for the fact that a person who has already exited by one cause cannot later exit by the other within the model.

Why the Cause-Specific Probabilities Have This Form

The total intensity \(\lambda\) governs the probability of remaining. Over \(t\) years, that probability is \(e^{-\lambda t}\). At time \(s\), the probability of still being in the group is \(e^{-\lambda s}\). Among those still in the group, mortality exits occur at intensity \(\mu\), so the unconditional density of mortality exit at time \(s\) is \(\mu e^{-\lambda s}\). Integrating this density over the period gives the mortality-exit probability.

P(\text{mortality exit by }t) =\int_0^t \mu e^{-\lambda s}\,ds =\frac{\mu}{\lambda}(1-e^{-\lambda t}).

The same argument applies to withdrawal, replacing \(\mu\) with \(\omega\). This also explains why the cause-specific probabilities are allocated in proportion to the intensities: among all exits over the period, the mortality share is \(\mu/\lambda\), and the withdrawal share is \(\omega/\lambda\). The calculation uses the intensity at each time that a person remains exposed; it does not assume that every person experiences both causes.

P(\text{withdrawal exit by }t) =\int_0^t \omega e^{-\lambda s}\,ds =\frac{\omega}{\lambda}(1-e^{-\lambda t}).

As a check, the three possible outcomes—remaining, mortality exit, and withdrawal exit—have probabilities that sum to one:

e^{-\lambda t} +\frac{\mu}{\lambda}(1-e^{-\lambda t}) +\frac{\omega}{\lambda}(1-e^{-\lambda t}) =1.

Applying the Formulas to Mortality and Withdrawal

For a given period, first calculate \(\lambda=\mu+\omega\). Then calculate the remaining probability \(e^{-\lambda t}\) and the total exit probability \(1-e^{-\lambda t}\). Finally, multiply that total exit probability by each cause’s share of the total intensity. Keeping the remaining probability separate from the cause-specific exits helps prevent double counting.

State the assumptions and time unit.
Confirm that mortality and withdrawal intensities are constant, and express the period in the same time unit as the intensities.

Calculate the total intensity.
Add the mortality and withdrawal intensities: \(\lambda=\mu+\omega\).

Calculate remaining and total exit probabilities.
Use \(e^{-\lambda t}\) for remaining and \(1-e^{-\lambda t}\) for exit by either cause.

Split exits by cause and check.
Multiply the total exit probability by \(\mu/\lambda\) and \(\omega/\lambda\); verify that all three outcomes sum to one.

Worked Example: Eight Years of Coverage

A hypothetical closed group has a constant annual mortality intensity of \(0.025\) and a constant annual withdrawal intensity of \(0.015\). Calculate the probabilities of remaining in the group, exiting by mortality, and exiting by withdrawal over eight years. Then calculate the expected numbers in each outcome from an initial cohort of 500. Treat members’ outcomes as governed by the stated model; report probabilities to four decimals and expected counts to two decimals.

Method and assumptions. The only exits are mortality and withdrawal. Both annual intensities are constant for eight years. First add the intensities:

\lambda=0.025+0.015=0.040.

The total exit probability over eight years is:

1-e^{-\lambda t} =1-e^{-(0.040)(8)} =1-e^{-0.32} =1-0.7261490 =0.2738510.

The mortality share of the total intensity is \(0.025/0.040=0.625\), and the withdrawal share is \(0.015/0.040=0.375\). Therefore:

P(\text{mortality exit}) =0.625(0.2738510) =0.1711569 \approx 0.1712, $$ $$ P(\text{withdrawal exit}) =0.375(0.2738510) =0.1026941 \approx 0.1027.

The remaining probability is \(e^{-0.32}=0.7261490\), or \(0.7261\). Multiplying each unrounded probability by 500 gives expected counts:

500(0.7261490)=363.0745\approx363.07,\qquad 500(0.1711569)=85.5784\approx85.58, $$ $$ 500(0.1026941)=51.3471\approx51.35.

Check. The rounded probabilities sum to \(0.7261+0.1712+0.1027=1.0000\). The expected counts sum to \(363.07+85.58+51.35=500.00\). This method avoids calculating mortality as if withdrawal were absent.

Worked Example: Unequal Cause Shares in a Three-Year Period

A hypothetical membership group has constant annual mortality intensity \(0.040\) and constant annual withdrawal intensity \(0.060\). Over three years, calculate the probability of each outcome and the expected number of each outcome from 1,200 members. Give probabilities to four decimals and counts to two decimals.

Method and assumptions. Mortality and withdrawal are the only exit causes, and their intensities remain constant for all three years. The total intensity is:

\lambda=0.040+0.060=0.100.

The remaining probability is:

e^{-\lambda t}=e^{-(0.100)(3)}=e^{-0.3}=0.7408182\approx0.7408.

Thus the total probability of exiting by either cause is \(1-0.7408182=0.2591818\). The mortality share is \(0.040/0.100=0.4\), and the withdrawal share is \(0.060/0.100=0.6\). The cause-specific probabilities are:

P(\text{mortality exit})=0.4(0.2591818)=0.1036727\approx0.1037, $$ $$ P(\text{withdrawal exit})=0.6(0.2591818)=0.1555091\approx0.1555.

For the cohort, multiply the unrounded probabilities by 1,200:

1{,}200(0.7408182)=888.9819\approx888.98\text{ remaining}, $$ $$ 1{,}200(0.1036727)=124.4073\approx124.41\text{ mortality exits}, $$ $$ 1{,}200(0.1555091)=186.6109\approx186.61\text{ withdrawal exits}.

Check. The probabilities sum to \(0.7408+0.1037+0.1555=1.0000\), and the rounded expected counts sum to \(888.98+124.41+186.61=1{,}200.00\). The larger withdrawal intensity correctly produces the larger share of exits.

Worked Example: Cause Probabilities for a Longer Horizon

A hypothetical group of 800 policyholders is subject to constant annual mortality intensity \(0.015\) and withdrawal intensity \(0.025\) for 12 years. Calculate the probability of each outcome and the corresponding expected cohort counts. Report probabilities to four decimals and counts to two decimals.

Method and assumptions. Both intensities remain constant for the full 12-year period, and these are the only causes of exit. The total intensity is \(0.015+0.025=0.040\), so the exponent is \((0.040)(12)=0.48\). The remaining probability is:

e^{-0.48}=0.6187834\approx0.6188.

The total exit probability is \(1-0.6187834=0.3812166\). The mortality and withdrawal shares are respectively \(0.015/0.040=0.375\) and \(0.025/0.040=0.625\). Hence:

P(\text{mortality exit})=0.375(0.3812166)=0.1429562\approx0.1430, $$ $$ P(\text{withdrawal exit})=0.625(0.3812166)=0.2382604\approx0.2383.

For 800 policyholders, the expected counts are:

800(0.6187834)=495.0267\approx495.03\text{ remaining}, $$ $$ 800(0.1429562)=114.3650\approx114.37\text{ mortality exits}, $$ $$ 800(0.2382604)=190.6083\approx190.61\text{ withdrawal exits}.

Check. The unrounded outcome probabilities sum to \(0.6187834+0.1429562+0.2382604=1.0000000\). The rounded expected counts sum to \(495.03+114.37+190.61=800.01\); the one-hundredth difference from 800 is due to rounding each count separately. The unrounded counts sum to 800.

Common Exam Trap: Ignoring the Competing Cause

A tempting but incorrect calculation is to treat the mortality intensity as though mortality were the only exit. That would assign mortality probability \(1-e^{-\mu t}\). In the first example, this incorrect calculation would give:

1-e^{-(0.025)(8)} =1-e^{-0.20} =1-0.8187308 =0.1812692.

This is not the mortality-exit probability in the two-cause model. It is the probability of mortality exit in a model with no withdrawal. When withdrawal is present, some people leave before they can experience mortality exit during the period. The correct mortality probability is:

\frac{0.025}{0.025+0.015} \left(1-e^{-(0.025+0.015)(8)}\right) =0.1711569.

Another error is to treat \(0.025\) and \(0.015\) as eight-year probabilities and add them. They are annual intensities, so the model uses their sum in the exponential and then allocates total exits by their shares. Always state the time unit and use the given period in that same unit.

Key takeaway: With two constant competing causes, calculate remaining probability using the sum of the intensities. Allocate the resulting total exit probability in proportion to each cause’s intensity. Check that remaining, mortality exit, and withdrawal exit add to one.

Check Your Understanding

In each problem, assume the stated mortality and withdrawal intensities are constant, are measured per year, and describe the only two ways to leave the group. Round probabilities to four decimals and counts to two decimals where requested.

A group has annual mortality intensity \(0.030\) and withdrawal intensity \(0.020\). Find the remaining, mortality-exit, and withdrawal-exit probabilities over five years. (Answer: \(0.7788,\ 0.1327,\ 0.0885\), respectively.)
A hypothetical contract group has mortality intensity \(0.010\) and withdrawal intensity \(0.040\). Find the three outcome probabilities over ten years. (Answer: remaining \(0.6065\), mortality exit \(0.0787\), withdrawal exit \(0.3148\).)
For 600 members, the annual mortality and withdrawal intensities are \(0.020\) and \(0.030\). Find expected remaining, mortality-exit, and withdrawal-exit counts after four years. (Answer: \(491.24,\ 43.50,\ 65.26\), respectively.)
A group has annual mortality intensity \(0.045\) and withdrawal intensity \(0.015\). Find the three outcome probabilities over six years. (Answer: remaining \(0.6977\), mortality exit \(0.2267\), withdrawal exit \(0.0756\).)
Constant annual mortality and withdrawal intensities are \(0.012\) and \(0.008\). Find the period at which the probability of remaining is \(0.8000\), then find the two cause-specific exit probabilities over that period. (Answer: \(11.1572\) years; mortality exit \(0.1200\), withdrawal exit \(0.0800\).)

What You'll Learn