Recovering Observed Exits From Constant Forces

Why Constant Forces Require Two Different Exit Calculations

When several causes can remove a person from a population, an exit from one cause may occur only if no competing cause has already produced an exit. A constant force for a cause therefore does not, by itself, equal the observed probability of exit from that cause. It is a rate governing the decrement while the person remains exposed to all causes.

This distinction matters when constructing or interpreting associated single-decrement tables. The observed multiple-decrement probabilities describe what happens with all causes operating together. An associated probability describes a hypothetical table in which the cause of interest operates alone. Both can be calculated from the same supplied constant forces, but they answer different questions.

Definition: Suppose there are \(r\) mutually exclusive causes of decrement, with constant cause-specific forces \(\mu_1,\ldots,\mu_r\) over an interval of \(t\) years. The aggregate force is \(M=\sum_{j=1}^r\mu_j\). The aggregate survival probability is the probability of remaining in the initial state for the full interval with no decrement from any cause. The observed exit probability for cause \(j\) is the probability of leaving by cause \(j\) while all causes operate. The associated single-decrement probability for cause \(j\) is the probability of leaving by cause \(j\) over the interval in the hypothetical model where all other causes are removed.

The model assumes that the cause-specific forces are constant throughout the interval, measured on the same time scale, and apply while the individual remains in the initial state. Causes are competing exits: after any exit, the individual is no longer exposed to the other decrements in that interval. There are no entries back into the state and no additional decrement causes beyond those specified. The formulas below do not apply unchanged if the forces vary with time; in that setting, the cumulative forces must be integrated over time.

Aggregate Survival and Observed Exits

With all causes operating, the instantaneous force of leaving the initial state is the sum \(M\). Under a constant aggregate force, the probability of surviving \(t\) years is exponential. An observed exit from cause \(j\) can happen at any time \(s\) in the interval: the individual must survive all causes to time \(s\), then leave through cause \(j\) in the next small time interval.

{}_tp_{\text{all}}=e^{-Mt}, \qquad {}_tq_{\text{all}}=1-e^{-Mt}.

Here \({}_tp_{\text{all}}\) denotes survival with no decrement from any cause, and \({}_tq_{\text{all}}\) denotes an exit from at least one cause by time \(t\). If \(\mu_j\) is the constant force for cause \(j\), its observed exit probability is

{}_tq_j^{\,\text{obs}} =\int_0^t e^{-Ms}\mu_j\,ds =\frac{\mu_j}{M}\left(1-e^{-Mt}\right), \qquad M>0.

The integral includes both requirements: \(e^{-Ms}\) is the probability of surviving every cause until time \(s\), while \(\mu_j\,ds\) is the approximate probability of exiting by cause \(j\) during the next small interval. If \(M=0\), all the cause forces are zero, survival is 1, and every exit probability is 0; the displayed fraction is not used.

Associated Probabilities: Remove the Competing Causes

For the associated single-decrement probability for cause \(j\), retain \(\mu_j\) and remove every other cause. The force in this hypothetical model is therefore \(\mu_j\), not the aggregate force \(M\). Its survival probability for \(t\) years is \(e^{-\mu_jt}\), so its probability of decrement is

{}_tq_j^{\,\text{assoc}}=1-e^{-\mu_jt}.

This is not generally the same as the observed probability \({}_tq_j^{\,\text{obs}}\). In the observed model, a person who exits from another cause first cannot later exit from cause \(j\) during the interval. In the associated model, those other causes are absent, so the individual remains exposed to cause \(j\) for longer on average.

Key distinction: Use the aggregate force \(M\) in the survival term when calculating observed exits under competition. Use only the force \(\mu_j\) when calculating the associated probability for cause \(j\) with competitors removed.

Why the Observed-Exit Formula Holds

The observed-exit formula follows by dividing the interval into short pieces and summing the probability of an exit from cause \(j\) in each piece, conditional on having survived to its start. In the limit, this sum becomes an integral.

Find survival to time \(s\).
Under constant forces, the aggregate force is \(M\), so the probability of no exit before time \(s\) is \(e^{-Ms}\).

Represent an exit from cause \(j\).
Conditional on still being present at time \(s\), the probability of an exit from cause \(j\) in a short interval of width \(ds\) is approximately \(\mu_j\,ds\).

Accumulate over the interval.
Multiply survival to \(s\) by the cause-\(j\) exit contribution and integrate from 0 to \(t\): \(\int_0^t e^{-Ms}\mu_j\,ds\).

Evaluate the integral.
For \(M>0\), the integral is \(\mu_j(1-e^{-Mt})/M\), giving the observed probability.

The calculation also provides an essential check. Summing over all causes gives

\sum_{j=1}^r {}_tq_j^{\,\text{obs}} =\int_0^t e^{-Ms}\left(\sum_{j=1}^r\mu_j\right)ds =\int_0^t e^{-Ms}M\,ds =1-e^{-Mt} ={}_tq_{\text{all}}.

Thus the observed exit probabilities must add to the probability of any exit. The associated probabilities do not generally add to this total: each is calculated in a separate hypothetical model with only one cause present.

Worked Example: Two Causes Over One Year

A hypothetical insured population is subject to two competing decrements. The constant annual force for cause 1 is \(\mu_1=0.10\), and for cause 2 it is \(\mu_2=0.20\). Calculate one-year survival, each observed exit probability, and each associated single-decrement probability. Give probabilities to four decimal places.

Method. First combine the forces for the observed multiple-decrement model. Then calculate observed exits using aggregate survival and calculate associated probabilities using each cause force alone.

The aggregate force is \(M=0.10+0.20=0.30\). One-year survival from all causes is

{}_1p_{\text{all}}=e^{-0.30}=0.7408182\ldots \approx 0.7408.

The probability of any exit is \(1-0.7408182\ldots=0.2591818\ldots\). For cause 1, the observed probability is

{}_1q_1^{\,\text{obs}} =\frac{0.10}{0.30}(1-e^{-0.30}) =\frac{1}{3}(0.2591818\ldots) =0.0863939\ldots \approx 0.0864.

For cause 2,

{}_1q_2^{\,\text{obs}} =\frac{0.20}{0.30}(1-e^{-0.30}) =\frac{2}{3}(0.2591818\ldots) =0.1727879\ldots \approx 0.1728.

With the competing cause removed, the associated probabilities are

{}_1q_1^{\,\text{assoc}}=1-e^{-0.10}=0.0951626\ldots \approx 0.0952, \qquad {}_1q_2^{\,\text{assoc}}=1-e^{-0.20}=0.1812692\ldots \approx 0.1813.

Check. The observed exits sum to \(0.0863939\ldots+0.1727879\ldots=0.2591818\ldots\), equal to \(1-{}_1p_{\text{all}}\). The associated probabilities are larger than their corresponding observed probabilities here because the competing decrement has been removed. The common mistake avoided is using \(1-e^{-\mu_j}\) as the observed exit probability; that expression is associated, not observed.

Worked Example: Three Causes Over Two Years

For a hypothetical cohort, three constant annual forces operate for two years: \(\mu_1=0.05\), \(\mu_2=0.10\), and \(\mu_3=0.15\). Find survival, all three observed exit probabilities, and all three associated probabilities. Give probabilities to four decimal places.

The aggregate force is \(M=0.05+0.10+0.15=0.30\) per year. The interval is two years, so the exponent uses \(Mt=0.30(2)=0.60\), not \(0.30\). Therefore,

{}_2p_{\text{all}}=e^{-0.60}=0.5488116\ldots \approx 0.5488, \qquad {}_2q_{\text{all}}=1-e^{-0.60}=0.4511884\ldots.

The observed exit probabilities are calculated using the same aggregate survival term:

{}_2q_1^{\,\text{obs}} =\frac{0.05}{0.30}(0.4511884\ldots) =0.0751981\ldots \approx 0.0752,

{}_2q_2^{\,\text{obs}} =\frac{0.10}{0.30}(0.4511884\ldots) =0.1503961\ldots \approx 0.1504,

{}_2q_3^{\,\text{obs}} =\frac{0.15}{0.30}(0.4511884\ldots) =0.2255942\ldots \approx 0.2256.

For the associated probabilities, each cause operates alone for two years. The cumulative forces are \(0.05(2)=0.10\), \(0.10(2)=0.20\), and \(0.15(2)=0.30\):

{}_2q_1^{\,\text{assoc}}=1-e^{-0.10}=0.0952,\quad {}_2q_2^{\,\text{assoc}}=1-e^{-0.20}=0.1813,\quad {}_2q_3^{\,\text{assoc}}=1-e^{-0.30}=0.2592.

Check. The observed probabilities sum to \(0.0751981\ldots+0.1503961\ldots+0.2255942\ldots=0.4511884\ldots\), exactly the total exit probability before rounding. A common time-unit error is to use a one-year exponent for a two-year interval. The correct exponent is always force multiplied by the duration in years.

Worked Example: Annual Forces Over Six Months

A hypothetical long-term care model has two constant forces, measured per year: \(\mu_A=0.24\) for transition to care and \(\mu_B=0.12\) for another exit. Calculate survival, the observed exit probabilities, and the associated probabilities over six months. Give probabilities to four decimal places.

Method. Convert six months to \(t=0.5\) years before applying the annual forces. The aggregate force is \(M=0.24+0.12=0.36\) per year, and the cumulative aggregate force for the interval is \(0.36(0.5)=0.18\).

{}_tp_{\text{all}}=e^{-0.18}=0.8352702\ldots \approx 0.8353, \qquad {}_tq_{\text{all}}=1-e^{-0.18}=0.1647298\ldots.

For cause A, the observed exit probability is

{}_tq_A^{\,\text{obs}} =\frac{0.24}{0.36}(0.1647298\ldots) =\frac{2}{3}(0.1647298\ldots) =0.1098199\ldots \approx 0.1098.

For cause B,

{}_tq_B^{\,\text{obs}} =\frac{0.12}{0.36}(0.1647298\ldots) =\frac{1}{3}(0.1647298\ldots) =0.0549099\ldots \approx 0.0549.

For the associated probabilities, use each annual force alone over \(0.5\) years:

{}_tq_A^{\,\text{assoc}}=1-e^{-0.24(0.5)} =1-e^{-0.12}=0.1131, \qquad {}_tq_B^{\,\text{assoc}}=1-e^{-0.12(0.5)} =1-e^{-0.06}=0.0582.

Check. The observed probabilities add to \(0.1098199\ldots+0.0549099\ldots=0.1647298\ldots\), equal to the total exit probability. Using \(t=1\) because the forces are annual would incorrectly calculate a full-year probability. Annual units specify the force scale; they do not change the six-month duration.

Common Traps and Practical Checks

The main challenge is not evaluating an exponential. It is selecting the correct force and interval for the quantity being requested. Before substituting numbers, identify whether the question asks about the actual model with competing causes or a hypothetical associated table.

Using the cause force alone for an observed exit. This ignores the possibility that another decrement removes the individual first. Use \(M\) in the survival term for observed exits.
Using the aggregate force in an associated probability. An associated table removes competitors, so the force is \(\mu_j\), not \(M\).
Mixing time units. If forces are annual, express the interval in years. For a duration of \(t\) years, the exponent is force multiplied by \(t\).
Forgetting the total-exit check. The sum of observed exits must equal \(1-e^{-Mt}\). Small differences may occur only from rounding.
Adding associated probabilities as if they were observed exits. Each associated probability comes from a different hypothetical model. Their sum is not generally the probability of any exit in the competing model.

These calculations provide a direct way to translate constant cause-specific forces into probabilities without confusing the actual multiple-decrement experience with a single-decrement comparison. This distinction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, calculate the requested probabilities and round to four decimal places.

Two causes have constant annual forces \(0.08\) and \(0.12\). Over one year, find survival, each observed exit probability, and each associated probability. (Answer: survival \(0.8187\); observed exits \(0.0725,\ 0.1088\); associated probabilities \(0.0769,\ 0.1131\).)
Three causes have constant annual forces \(0.04,\ 0.06,\ 0.10\). Over one year, find aggregate survival, the three observed exits, and the three associated probabilities. (Answer: survival \(0.8187\); observed exits \(0.0363,\ 0.0544,\ 0.0906\); associated probabilities \(0.0392,\ 0.0582,\ 0.0952\).)
Two causes have constant annual forces \(0.30\) and \(0.10\). Over two years, calculate the observed exit probabilities and the associated probabilities. (Answer: observed exits \(0.4130,\ 0.1377\); associated probabilities \(0.4512,\ 0.1813\).)
Two constant annual forces are \(0.09\) and \(0.03\). For a three-month interval, calculate survival, the two observed exits, and the two associated probabilities. (Answer: survival \(0.9704\); observed exits \(0.0222,\ 0.0074\); associated probabilities \(0.0222,\ 0.0075\).)
Three causes have constant annual forces \(0.02,\ 0.05,\ 0.03\). Over 18 months, calculate survival, all observed exits, and all associated probabilities. (Answer: survival \(0.8607\); observed exits \(0.0279,\ 0.0696,\ 0.0418\); associated probabilities \(0.0296,\ 0.0723,\ 0.0440\).)

What You'll Learn