Multi-Year Decrement Probabilities With Changing Rates

Why Changing Annual Probabilities Need Careful Weighting

In a multiple-decrement model, an individual can leave an active group for one of several mutually exclusive causes. If the annual probabilities change with age or duration, the probability of surviving several years is not found by repeating one fixed annual survival probability. Further, a cause-specific probability over the whole period must account for how many individuals are still exposed to that cause in each year.

The central calculation has two parts: multiply the appropriate annual survival probabilities to find the probability of reaching later years, then weight each year's cause-specific exit probability by the probability of reaching that year. This applies when annual rates are specified for successive years of age or duration and are conditional on being active at the beginning of each year.

Definition: Suppose there are $m$ mutually exclusive exit causes. Let $q_{x+r}^{(j)}$ be the probability that an active individual exits from cause $j$ during the year beginning at age or duration $x+r$, conditional on being active at the start of that year. The probability of remaining active through that year is $$ p_{x+r}^{(\tau)}=1-\sum_{j=1}^{m}q_{x+r}^{(j)}. $$ Here $\tau$ denotes exit from any cause. The annual probabilities must satisfy $q_{x+r}^{(j)}\geq 0$ and $\sum_j q_{x+r}^{(j)}\leq 1$.

The annual probabilities are conditional on reaching the start of the year. Thus, a cause probability in year three is not itself the probability, from the original starting point, of exiting in year three. The latter also depends on surviving the first two years.

Multi-Year Formulas

For $k$ successive years, the probability of remaining active to the start of year $k+1$ is the product of the annual all-cause survival probabilities. The cumulative probability of exit from cause $j$ during those $k$ years is the sum of the year-specific cause probabilities, each weighted by the probability of reaching that year.

{}_kp_x^{(\tau)} =\prod_{r=0}^{k-1}p_{x+r}^{(\tau)}

{}_kq_x^{(j)} =\sum_{r=0}^{k-1}{}_rp_x^{(\tau)}q_{x+r}^{(j)}, \qquad {}_0p_x^{(\tau)}=1.

In the second formula, the term with $r=0$ is the probability of exiting from cause $j$ in the first year. The term with $r=1$ is the probability of surviving the first year and then exiting from cause $j$ in the second year, and so on. These are probabilities measured from the original starting point, so they can be added.

Theorem: If the causes are mutually exclusive and exhaustive for exits from the active group, then $$ {}_kp_x^{(\tau)}+\sum_{j=1}^{m}{}_kq_x^{(j)}=1. $$ This closure identity is a useful check: by the end of the period, the individual either remains active or has exited through exactly one of the causes.

Why the Product and Weighted Sum Work

The formulas follow by splitting outcomes year by year. To reach the start of year $r+1$, the individual must remain active in every one of the first $r$ years. Conditional on reaching each year's start, the probability of remaining active that year is $p_{x+r}^{(\tau)}$. Applying the multiplication rule for successive conditional probabilities gives the product for survival.

For cause $j$, exit during year $r+1$ requires both reaching that year and then exiting from cause $j$ within it. Its probability from the original starting point is therefore ${}_rp_x^{(\tau)}q_{x+r}^{(j)}$. The events of exiting from cause $j$ in different years are disjoint, so their probabilities add. The same reasoning across all causes and the final survival event gives the closure identity.

Check the annual inputs.
For each year, add the cause probabilities. Their sum cannot exceed one.

Find annual survival.
Subtract the sum of all cause probabilities from one.

Track the probability or number reaching each year.
Multiply the prior year-start value by that year's all-cause survival probability.

Accumulate cause-specific exits.
In each year, multiply the year-start value by that year's cause probability, then add across years.

Check closure.
Final survival plus all cumulative cause exits must equal the original probability or cohort size.

Worked Calculations

Worked Example: Three Years With Changing Cause Probabilities

A hypothetical group begins with 1,000 active members. For the next three years, the probabilities of exit from causes 1 and 2, conditional on being active at the start of the year, are respectively $(0.04,0.02)$, $(0.03,0.05)$, and $(0.06,0.02)$. Find the probability of reaching each later year, final survival, and cumulative exits by cause.

Method. Calculate each year's all-cause survival, propagate the year-start cohort, and weight each cause probability by that year's starting cohort.

Year 1. All-cause survival is $1-0.04-0.02=0.94$. Of the 1,000 members, $1{,}000(0.04)=40$ exit from cause 1 and $1{,}000(0.02)=20$ from cause 2. The year-2 starting cohort is $1{,}000(0.94)=940$.

Year 2. All-cause survival is $1-0.03-0.05=0.92$. Cause 1 exits are $940(0.03)=28.2$; cause 2 exits are $940(0.05)=47$. The year-3 starting cohort is $940(0.92)=864.8$.

Year 3. All-cause survival is $1-0.06-0.02=0.92$. Cause 1 exits are $864.8(0.06)=51.888$, and cause 2 exits are $864.8(0.02)=17.296$. Final survivors are $864.8(0.92)=795.616$.

The cumulative cause 1 exits are $40+28.2+51.888=120.088$, or $0.120088$ of the original group. The cumulative cause 2 exits are $20+47+17.296=84.296$, or $0.084296$. The probabilities of reaching years 2 and 3 are $0.94$ and $0.94(0.92)=0.8648$; final survival is $0.94(0.92)(0.92)=0.795616$.

Check. The counts sum to $120.088+84.296+795.616=1{,}000$, and the probabilities sum to $0.120088+0.084296+0.795616=1$. The common mistake avoided here is adding annual cause probabilities without weighting later years by the cohort still active.

Worked Example: Cumulative Probabilities From Duration-Based Rates

A hypothetical contract group has two possible exits, A and B. Over three successive years, the conditional annual probabilities for A are $0.02, 0.04, 0.05$, and for B they are $0.03, 0.02, 0.04$. Calculate the three-year survival probability and cumulative exit probability for each cause.

Method. Use the annual survival factors first, because each later year's cause probabilities apply only to those who reach that year.

Annual survival probabilities are $1-0.02-0.03=0.95$, $1-0.04-0.02=0.94$, and $1-0.05-0.04=0.91$. The probabilities of reaching the starts of years 1, 2, and 3 are $1$, $0.95$, and $0.95(0.94)=0.893$. The probability of surviving all three years is $0.893(0.91)=0.81263$.

For cause A, the cumulative probability is $1(0.02)+0.95(0.04)+0.893(0.05)=0.02+0.038+0.04465=0.10265$. For cause B, it is $1(0.03)+0.95(0.02)+0.893(0.04)=0.03+0.019+0.03572=0.08472$.

Check. $0.81263+0.10265+0.08472=1$. Thus the three-year probabilities are $0.81263$ for survival, $0.10265$ for cause A, and $0.08472$ for cause B. The common mistake avoided is applying the year-3 probabilities to the original group rather than to the $0.893$ probability that reaches year 3.

Worked Example: Four-Year Cohort With Age-Varying Rates

A hypothetical cohort of 2,500 members is followed for four years. For causes 1 and 2, the conditional annual exit probabilities are $(0.02,0.01)$, $(0.03,0.02)$, $(0.04,0.03)$, and $(0.05,0.01)$. Calculate final survivors and cumulative exits by cause, then express each as a probability from the original cohort.

Method. Track the number active at each year start. This is equivalent to tracking survival probabilities, but makes the cause-specific exit counts explicit.

The annual survival probabilities are $0.97, 0.95, 0.93,$ and $0.94$. The year-start cohort sizes are $2{,}500$, $2{,}500(0.97)=2{,}425$, $2{,}425(0.95)=2{,}303.75$, and $2{,}303.75(0.93)=2{,}142.4875$. Final survivors are $2{,}142.4875(0.94)=2{,}013.93825$.

Cause 1 exits by year are $2{,}500(0.02)=50$, $2{,}425(0.03)=72.75$, $2{,}303.75(0.04)=92.15$, and $2{,}142.4875(0.05)=107.124375$. Their total is $322.024375$.

Cause 2 exits by year are $2{,}500(0.01)=25$, $2{,}425(0.02)=48.5$, $2{,}303.75(0.03)=69.1125$, and $2{,}142.4875(0.01)=21.424875$. Their total is $164.037375$.

Dividing by the original 2,500 gives cause 1 probability $322.024375/2{,}500=0.12880975$, cause 2 probability $164.037375/2{,}500=0.06561495$, and survival probability $2{,}013.93825/2{,}500=0.8055753$. To four decimal places, these are $0.1288$, $0.0656$, and $0.8056$.

Check. The counts add to $322.024375+164.037375+2{,}013.93825=2{,}500$. The unrounded probabilities add to one. The common mistake avoided is rounding intermediate cohort sizes before using them in later years.

Common Exam Trap: Missing a Year-Start Weight

Suppose cause A probabilities over three years are $0.04, 0.06, 0.03$, and cause B probabilities are $0.03, 0.02, 0.07$. The annual all-cause survival probabilities are $0.93, 0.92,$ and $0.90$. The probabilities reaching the starts of years 2 and 3 are $0.93$ and $0.93(0.92)=0.8556$, respectively.

A tempting but incorrect calculation for cause B is $0.03+0.93(0.02)+0.8556(0.07)=0.107892$. The arithmetic in this expression is itself wrong: $0.93(0.02)=0.0186$, so the correct calculation is $0.03+0.0186+0.059892=0.108492$. The cause A probability is $0.04+0.93(0.06)+0.8556(0.03)=0.121468$, and survival is $0.93(0.92)(0.90)=0.77004$. The check is $0.108492+0.121468+0.77004=1$.

This illustrates two checks worth making separately: first, use the correct probability of reaching each year; second, recompute the weighted terms before accepting a rounded total. A missing factor or a small multiplication error can break closure.

Practical Checks and Applicability

These formulas apply when each year's cause probabilities describe exits during that year, conditional on being active at its beginning, and the causes are mutually exclusive. Annual probabilities may vary freely by age or duration; they need not be constant. If the cause probabilities are supplied as unconditional probabilities from the original cohort instead, they should not be multiplied by year-start survival again.

For an exam calculation, preserve sufficient precision in the products and sums, then round only the requested final results. Useful checks include:

Each year's cause probabilities sum to no more than one.
Year-start survival cannot increase from one year to the next.
Every cumulative cause probability is nonnegative.
Final survival plus all cumulative cause probabilities equals one, apart from rounding.

Multi-year decrement calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. Their main practical role is to translate changing annual decrement assumptions into coherent probabilities for a period longer than one year.

Check Your Understanding

For each problem, use the annual probabilities as conditional on being active at the start of the stated year. Check that survival and cause-specific exits close to one.

Over two years, cause 1 probabilities are $0.08$ and $0.03$, and cause 2 probabilities are $0.02$ and $0.04$. Find final survival and both cumulative cause probabilities. (Answer: survival $0.8370$; cause 1 $0.1070$; cause 2 $0.0560$.)
Over three years, cause A probabilities are $0.01, 0.05, 0.04$, and cause B probabilities are $0.04, 0.01, 0.03$. Find final survival and the cumulative probability for each cause. (Answer: survival $0.83049$; A $0.09322$; B $0.07629$.)
Over four years, cause 1 probabilities are $0.03, 0.02, 0.05, 0.04$, and cause 2 probabilities are $0.02, 0.03, 0.01, 0.02$. Find final survival and cumulative exits by cause. (Answer: survival $0.797449$; cause 1 $0.128059$; cause 2 $0.074492$.)
Over three years, causes A, B, and C have annual probabilities $(0.02,0.01,0.02)$, $(0.03,0.04,0.01)$, and $(0.04,0.02,0.03)$, respectively. Find the final survival and each cumulative cause probability. (Answer: survival $0.79534$; A $0.08346$; B $0.06548$; C $0.05572$.)
A cohort of 800 members has cause A probabilities $0.05, 0.02, 0.04$ and cause B probabilities $0.03, 0.06, 0.01$ over three years. Find final survivors and cumulative exit counts by cause. (Answer: survivors $643.264$; cause A $81.8048$; cause B $74.9312$; total $800$.)

What You'll Learn