Cause-Specific Exit Probabilities from Competing Intensities

Why Competing Intensities Need to Be Combined

When several events can end a policy’s in-force status, the probability of exit by one particular cause depends on whether the policy remains in force long enough for that cause to occur. For example, a hypothetical contract might terminate because of death or lapse. A death intensity describes the instantaneous rate of death among contracts still in force; it is not itself the probability of death over a year. Nor can the death probability be calculated without accounting for lapse, which can remove a contract before death occurs.

This tutorial derives the probability of remaining in force and the probability of exit by each cause over a specified interval. The model is for a single initial in-force state with mutually exclusive exit causes and no re-entry during the interval. If a model includes transitions into other states or permits re-entry, those transitions must also be represented; the formulas here cannot simply be used as if the process had only absorbing exits.

Definition: Let \(\mu_j(t)\) be the cause-specific intensity of exit by cause \(j\) at time \(t\), conditional on being in force immediately before \(t\). For \(m\) possible exit causes, the total exit intensity is \(\mu_{\text{tot}}(t)=\sum_{j=1}^{m}\mu_j(t)\). The probability of remaining in force from time \(s\) to time \(t\), conditional on being in force at \(s\), is denoted here by \(S(s,t)\).

The cause-specific intensities are nonnegative and locally integrable over the interval. Under the continuous-time model, the probability of two or more exits in an infinitesimally short interval is negligible relative to the interval’s length. The causes describe alternative first exits from force, not independent events that can each occur to the same contract as its first exit.

Survival and Cause-Specific Exit Probabilities

Over a short interval of length \(h\), conditional on being in force at time \(t\), the probability of exit by cause \(j\) is approximately \(\mu_j(t)h\). The probability of any exit is therefore approximately \(\mu_{\text{tot}}(t)h\), so the probability of remaining in force is approximately \(1-\mu_{\text{tot}}(t)h\). Accumulating this survival over the interval gives the exponential of the integrated total intensity.

S(s,t)=\exp\left(-\int_s^t \mu_{\text{tot}}(u)\,du\right) =\exp\left(-\int_s^t\sum_{j=1}^{m}\mu_j(u)\,du\right).

For an exit by cause \(j\) at time \(u\), the contract must first have remained in force from \(s\) to \(u\), and then exit by cause \(j\) at that time. The probability density of that outcome is \(S(s,u)\mu_j(u)\). Integrating this density over the interval gives the cause-specific probability. In particular, these are probabilities conditional on being in force at the start time \(s\).

P(\text{exit by cause }j\text{ in }(s,t]\mid\text{in force at }s) =\int_s^t S(s,u)\mu_j(u)\,du.

The outcomes are mutually exclusive and exhaustive: the contract either remains in force through \(t\), or exits by exactly one of the modeled causes during \((s,t]\). Consequently, the cause-specific exit probabilities plus the in-force probability must sum to one. This identity is also a useful calculation check.

S(s,t)+\sum_{j=1}^{m}\int_s^t S(s,u)\mu_j(u)\,du=1.

Why the formula holds: The survival function satisfies the differential equation \(\frac{\partial}{\partial t}S(s,t)=-S(s,t)\mu_{\text{tot}}(t)\), with \(S(s,s)=1\). Its solution is the exponential of the negative integrated total intensity. The probability density of exit by cause \(j\) at \(u\) is survival to \(u\), multiplied by the cause-\(j\) intensity at \(u\). Integrating that density gives the stated cause-specific probability.

Worked Calculations

Worked Example: Constant Competing Intensities

A hypothetical policy is in force at time 0. Over the next two years, its constant death intensity is \(0.03\) per year and its constant lapse intensity is \(0.05\) per year. Find the probability of remaining in force for two years and the probabilities of exit by death and by lapse during that period.

Method. Add the cause-specific intensities to obtain the total intensity, then integrate survival times each cause-specific intensity. The intensities are constant and measured per year; the interval length is two years.

The total intensity is \(0.03+0.05=0.08\) per year. Thus the probability of remaining in force is

S(0,2)=e^{-0.08(2)}=e^{-0.16}=0.8521438.

For constant intensities, each cause’s probability is its share of the total intensity multiplied by the probability of any exit. Here the probability of any exit is \(1-0.8521438=0.1478562\). The death probability is

\frac{0.03}{0.08}(0.1478562)=0.375(0.1478562)=0.0554461.

The lapse probability is

\frac{0.05}{0.08}(0.1478562)=0.625(0.1478562)=0.0924101.

Check. The probabilities sum to \(0.8521438+0.0554461+0.0924101=1.0000000\). To four decimal places, the in-force, death-exit, and lapse-exit probabilities are \(0.8521\), \(0.0554\), and \(0.0924\), respectively. The calculation avoids treating the two exit causes as separate events that can both terminate the same policy.

Worked Example: Intensities That Change at the End of Year One

A hypothetical contract starts in force at time 0. During year 1, the death intensity is \(0.02\) and the lapse intensity is \(0.06\) per year. During year 2, the death intensity is \(0.04\) and the lapse intensity is \(0.02\) per year. Calculate the probability of remaining in force through year 2 and the probability of exit by each cause during the two-year period.

Method. Calculate survival and cause-specific exits separately on each constant-intensity segment. For the second segment, multiply its conditional probabilities by the probability of surviving the first year.

The total intensity is \(0.08\) in year 1 and \(0.06\) in year 2. Survival through both years is

S(0,2)=e^{-0.08}e^{-0.06}=e^{-0.14}=0.8693582.

In year 1, the probability of any exit is \(1-e^{-0.08}=0.0768837\). The death and lapse probabilities over that year, conditional on being in force at time 0, are

P(\text{death in year 1})=\frac{0.02}{0.08}(1-e^{-0.08}) =0.25(0.0768837)=0.0192209, $$ $$ P(\text{lapse in year 1})=\frac{0.06}{0.08}(1-e^{-0.08}) =0.75(0.0768837)=0.0576627.

The probability of reaching year 2 in force is \(e^{-0.08}=0.9231163\). Conditional on reaching year 2, the probability of any exit during that year is \(1-e^{-0.06}=0.0582355\). The year-2 death and lapse probabilities, conditional on being in force at time 0, are therefore

P(\text{death in year 2}) =e^{-0.08}\frac{0.04}{0.06}(1-e^{-0.06}) =0.9231163\left(\frac{2}{3}\right)(0.0582355) =0.0358387, $$ $$ P(\text{lapse in year 2}) =e^{-0.08}\frac{0.02}{0.06}(1-e^{-0.06}) =0.9231163\left(\frac{1}{3}\right)(0.0582355) =0.0179194.

Adding the segment probabilities gives total death probability \(0.0192209+0.0358387=0.0550596\) and total lapse probability \(0.0576627+0.0179194=0.0755821\). Check. \(0.8693582+0.0550596+0.0755821=1.0000000\), up to the displayed precision. Rounded to four decimals, the probabilities are \(0.8694\) in force, \(0.0551\) death exit, and \(0.0756\) lapse exit. A common error is to apply the year-2 exit probabilities to the original population rather than only to the fraction that survived year 1.

Worked Example: Proportional Time-Varying Intensities

For a hypothetical one-year coverage period, let the death intensity at duration \(t\) years be \(0.02(1+t)\) and the withdrawal intensity be \(0.04(1+t)\), for \(0\leq t\leq1\). Find the one-year in-force probability and the probability of exit by each cause.

Method. Integrate the total intensity for survival. Since the two cause-specific intensities remain in a fixed ratio at every time, each cause receives the same fraction of the total exit probability as its fraction of the total intensity.

The total intensity is \(0.06(1+t)\). Its integral over the year is

\int_0^1 0.06(1+t)\,dt =0.06\left[t+\frac{t^2}{2}\right]_0^1 =0.06(1.5)=0.09.

Therefore, survival is \(S(0,1)=e^{-0.09}=0.9139312\), and the probability of any exit is \(1-e^{-0.09}=0.0860688\). The death intensity is one-third of the total intensity at every duration, while the withdrawal intensity is two-thirds. Hence

P(\text{death exit})=\frac{1}{3}(0.0860688)=0.0286896, \qquad P(\text{withdrawal exit})=\frac{2}{3}(0.0860688)=0.0573792.

Check. \(0.9139312+0.0286896+0.0573792=1.0000000\). To four decimals, the probabilities are \(0.9139\) in force, \(0.0287\) death exit, and \(0.0574\) withdrawal exit. This proportional-share shortcut works here because the intensity ratio is constant over time; it is not generally valid when the relative sizes of the cause-specific intensities change.

Checks, Conditions, and an Exam Trap

The main operational distinction is between an intensity and an interval probability. An intensity is a rate conditional on being in force at a given instant. An interval exit probability accounts both for that rate and for the chance of surviving in force up to the potential exit time. If intensities vary over time, the survival weighting inside the integral matters.

For constant cause-specific intensities \(\mu_1,\ldots,\mu_m\) over an interval of length \(h\), write \(\mu_{\text{tot}}=\sum_j\mu_j\). The probability of exit by cause \(j\) is \(\frac{\mu_j}{\mu_{\text{tot}}}(1-e^{-\mu_{\text{tot}}h})\), provided \(\mu_{\text{tot}}>0\). If every cause intensity is zero, survival is one and all exit probabilities are zero. The constant-intensity expression should not be applied to a period over which the intensities change unless the period is split into segments or the appropriate integrals are evaluated.

Common exam trap: Do not calculate each cause’s probability as though it were the only exit cause and then add those probabilities. For example, with death intensity \(0.03\), lapse intensity \(0.05\), and a two-year term, using \(1-e^{-0.03(2)}\) for death and \(1-e^{-0.05(2)}\) for lapse ignores the competing cause in each survival calculation. The correction is to use the total intensity \(0.08\) for survival and weight the total exit probability by each cause’s share, giving \(0.0554\) for death and \(0.0924\) for lapse, not the isolated-cause values.

A final check is to add the in-force probability to all cause-specific exit probabilities. If they do not sum to one, check whether the total intensity includes every modeled exit, whether later segments were weighted by survival to their start, and whether the interval units match the intensity units. These checks are especially useful in integrated calculations aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, calculate the requested probabilities conditional on being in force at the stated interval’s start, unless otherwise specified. Probabilities in the answers are rounded to four decimals.

A hypothetical policy has constant death and lapse intensities of \(0.01\) and \(0.03\) per year, respectively. Find the one-year survival probability and the probability of exit by each cause. (Answer: survival \(0.9608\); death \(0.0098\); lapse \(0.0294\).)
Over a three-year interval, a contract has constant competing exit intensities of \(0.05\) per year for death and \(0.05\) per year for surrender. Find the probability of remaining in force and the probability of exit by each cause. (Answer: survival \(0.7408\); death \(0.1296\); surrender \(0.1296\).)
During year 1, a hypothetical membership contract has death and withdrawal intensities \(0.03\) and \(0.01\) per year. During year 2, the intensities are \(0.02\) and \(0.04\) per year. Find the two-year survival probability and the total probabilities of death and withdrawal. (Answer: survival \(0.9048\); death \(0.0481\); withdrawal \(0.0471\).)
Over one year, let two cause-specific intensities be \(0.01(1+t)\) and \(0.03(1+t)\), where \(t\) is measured in years from the start of coverage. Find survival and each cause-specific exit probability. (Answer: survival \(0.9418\); first cause \(0.0146\); second cause \(0.0437\).)
A contract is known to be in force at time 1. From time 1 to time 2, its constant death and lapse intensities are \(0.02\) and \(0.04\) per year. Find the conditional probability of surviving the interval and the conditional probability of exit by each cause. (Answer: survival \(0.9418\); death \(0.0194\); lapse \(0.0388\).)

What You'll Learn