What the Associated Table Measures
In a multiple-decrement model, an individual may leave through more than one cause. An associated single-decrement table isolates one specified cause, say cause \(j\), and describes the hypothetical experience when that is the only cause of exit operating. The table is therefore not the observed experience of a group exposed to all causes at once.
Once an associated table is supplied, its one-year survival and exit probabilities are obtained directly from the lives at the start of the interval and the exits during it. These calculations do not require choosing an assumption about how competing causes operate within the year. That kind of assumption is needed to construct or convert tables in the first place, but it is not needed to calculate probabilities from a completed associated table.
The prime marks an associated single-decrement quantity, while the superscript \(j\) identifies the cause retained in the table. It is useful to keep both marks: \(q_x^{\prime(j)}\) is not generally the same as the observed probability \(q_x^{(j)}\) of exiting from cause \(j\) in a multiple-decrement model. The associated probability is conditional on being in the associated table at age \(x\); the observed probability is measured among lives exposed to all active causes.
The lives in an actuarial table may be expected counts rather than a literal cohort of people. The ratios still have the same interpretation: they give the one-interval probabilities represented by that table. If the table is stated using a radix other than one, use the lives and exits on that same scale. The scale cancels in the ratio.
Calculating Both Probabilities from One Interval
There are two common ways a table presents the necessary information. It may give the starting lives and exits, or it may give the numbers of lives at the start and end of the interval. In the first case calculate \(q_x^{\prime(j)}\) directly and use the remaining fraction for \(p_x^{\prime(j)}\). In the second case use the successive survivor values for \(p_x^{\prime(j)}\), and obtain \(q_x^{\prime(j)}\) either from the decrement count or as its complement.
Worked Example: Probabilities from an Exit Count
A hypothetical associated table retains only cause \(j\). At age 55, it has 4,800 lives and records 84 exits from cause \(j\) before age 56. Calculate the associated exit and survival probabilities. Also calculate the expected numbers surviving and exiting from a group of 1,600 lives governed by the same table. There is no interest-rate assumption or separate mortality distribution: the table values specify the one-cause decrement experience.
Method. Divide the exits by the lives at risk to obtain the exit probability. Subtract the exits from the starting lives to obtain the next-age lives, then divide by the starting lives to obtain the survival probability.
The number remaining at age 56 in the table is $$ l_{56}^{\prime(j)}=4{,}800-84=4{,}716. $$ Therefore $$ q_{55}^{\prime(j)}=\frac{84}{4{,}800}=0.0175, \qquad p_{55}^{\prime(j)}=\frac{4{,}716}{4{,}800}=0.9825. $$ For 1,600 lives, the expected exits are \(1{,}600(0.0175)=28\), and the expected survivors are \(1{,}600(0.9825)=1{,}572\). These add to the original 1,600.
Check and trap avoided. The probabilities add to \(0.0175+0.9825=1\). A common mistake is to divide 84 by the next-age lives, 4,716. The denominator must be the lives exposed at the start of the interval, 4,800.
Worked Example: Two Consecutive Intervals in an Associated Table
A hypothetical associated table for a long-term care exit cause lists 12,500 lives at age 70, 12,125 at age 71, and 11,640 at age 72. Calculate the exit and survival probabilities for each year and the probability of surviving both years in this associated table. No interest rate is relevant to these probability calculations, and no separate distributional assumption is used; the successive table values are given.
Method. For each interval, take the decrease in lives as the exits. Divide that decrease by the interval's starting lives to get the exit probability. The survival probability is the next-age lives divided by the starting lives.
From age 70 to 71, the number of exits is \(12{,}500-12{,}125=375\). Thus $$ q_{70}^{\prime(j)}=\frac{375}{12{,}500}=0.03, \qquad p_{70}^{\prime(j)}=\frac{12{,}125}{12{,}500}=0.97. $$ From age 71 to 72, the number of exits is \(12{,}125-11{,}640=485\). Thus $$ q_{71}^{\prime(j)}=\frac{485}{12{,}125}=0.04, \qquad p_{71}^{\prime(j)}=\frac{11{,}640}{12{,}125}=0.96. $$ To survive both years, an individual must survive the first interval and then the second. The two-year probability is $$ p_{70}^{\prime(j)}p_{71}^{\prime(j)} =0.97(0.96)=0.9312. $$ The table gives the same result directly: \(11{,}640/12{,}500=0.9312\).
Check and trap avoided. In each interval, survival plus exit is one: \(0.97+0.03=1\) and \(0.96+0.04=1\). Do not add the two annual exit probabilities to calculate two-year survival or exit. The interval probabilities have different starting populations; survival over both intervals is the product of the two survival probabilities.
Why the Probabilities Must Sum to One
The complement relationship follows from how an associated single-decrement interval is defined. Consider the lives present at age \(x\). Under the table's conditions, each either exits from cause \(j\) before age \(x+1\), or remains in the table at age \(x+1\). These outcomes cannot both occur, and together they include every life in the starting group. Thus the starting lives split into exits and survivors.
Divide every term by the positive starting number \(l_x^{\prime(j)}\). By the definitions of the two probabilities, the result is
Equivalently, either probability can be found from the other:
This identity is also a practical check on calculations. If the two probabilities do not sum to one, check the interval boundaries, the starting-life denominator, and whether the supplied table really has only the one decrement and no additions or other exits during that interval.
Worked Example: Using a Scaled Table and Checking the Cause
A hypothetical associated table for a disability exit cause uses a radix of 100,000. It lists 100,000 lives at age 45 and 96,250 at age 46. Calculate the associated probabilities, then find the expected exits and survivors among 2,400 lives at age 45. The interval is exactly one year; no interest rate, mortality distribution, or additional decrement assumption is used beyond the stated associated table.
Method. Recover the decrement count from the decrease in table lives, calculate both probabilities using the starting lives, and then apply those probabilities to the group of 2,400.
The table decrement is $$ d_{45}^{\prime(j)}=100{,}000-96{,}250=3{,}750. $$ Therefore $$ q_{45}^{\prime(j)}=\frac{3{,}750}{100{,}000}=0.0375, \qquad p_{45}^{\prime(j)}=\frac{96{,}250}{100{,}000}=0.9625. $$ For 2,400 starting lives, expected exits are \(2{,}400(0.0375)=90\). Expected survivors are \(2{,}400(0.9625)=2{,}310\). The counts reconcile: \(90+2{,}310=2{,}400\).
Check and trap avoided. The probabilities sum to \(0.0375+0.9625=1\). The radix is not the number of lives in the smaller group; it is simply the scale used to report the table. The probabilities are unchanged if the table is rescaled, because both numerator and denominator change by the same factor.
Associated Probabilities Are Not Observed Cause Probabilities
The distinction between associated and observed quantities is important in multiple-decrement work. An observed cause-specific probability describes the chance of exiting from cause \(j\) during an interval while the other causes are also operating. An associated probability describes the corresponding one-cause table. Consequently, an observed probability cannot automatically be substituted for \(q_x^{\prime(j)}\), nor can it automatically be used to calculate \(p_x^{\prime(j)}\).
For example, if an observed multiple-decrement table reports \(q_x^{(j)}=0.03\), that number alone does not establish that \(q_x^{\prime(j)}=0.03\). To construct an associated probability from observed competing-decrement data requires additional information or a specified conversion assumption. If the associated table itself is given, however, its probabilities follow from its own exits and lives as shown here. This is the boundary between reading a completed associated table and converting a multiple-decrement table.
| Quantity | What the interval describes | How it is obtained here |
|---|---|---|
| \(q_x^{\prime(j)}\) | Exit through cause \(j\) in the associated one-cause table | \(d_x^{\prime(j)}/l_x^{\prime(j)}\) |
| \(p_x^{\prime(j)}\) | Remain in the associated one-cause table to age \(x+1\) | \(l_{x+1}^{\prime(j)}/l_x^{\prime(j)}\) |
| \(q_x^{(j)}\) | Observed exit through cause \(j\) with competing causes operating | Not interchangeable with the associated probability |
Calculation Checklist
When asked to calculate probabilities from an associated table, use a consistent sequence:
Confirm that the data refer to the associated table for the specified cause \(j\), and identify the starting age \(x\) and next age \(x+1\).
Use \(l_x^{\prime(j)}\) as the denominator. If the decrement count is not listed, calculate it as \(l_x^{\prime(j)}-l_{x+1}^{\prime(j)}\).
Divide exits by starting lives for \(q_x^{\prime(j)}\), and divide next-age lives by starting lives for \(p_x^{\prime(j)}\).
Verify that the probabilities sum to one and that the implied exit and survivor counts add to the starting lives.
A useful exam habit is to write the count identity before dividing: starting lives equal exits plus survivors. It makes the denominator clear, exposes an inconsistent table row, and supplies an immediate check on the final probabilities. This material is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use the specified associated table rather than assuming that observed competing-decrement probabilities are interchangeable with associated probabilities.
- An associated table for cause \(j\) has 9,000 lives at age 62 and 270 exits during the year. Find \(q_{62}^{\prime(j)}\) and \(p_{62}^{\prime(j)}\). (Answer: \(q_{62}^{\prime(j)}=0.0300,\ p_{62}^{\prime(j)}=0.9700\).)
- An associated table lists 14,000 lives at age 50 and 13,580 at age 51. Find the exit count, exit probability, and survival probability for the year. (Answer: 420 exits; \(q_{50}^{\prime(j)}=0.0300,\ p_{50}^{\prime(j)}=0.9700\).)
- For two consecutive years in an associated table, the survival probabilities are 0.98 and 0.95. Find the probability of surviving both years and the probability of exiting at least once during those years. (Answer: two-year survival \(=0.9310\); probability of at least one exit \(=0.0690\).)
- An associated table has 6,000 lives at age 40 and an exit probability of 0.042 for the year. Find the number of exits and the number surviving to age 41. (Answer: 252 exits and 5,748 survivors.)
- An observed multiple-decrement table reports an exit probability of 0.027 from cause \(j\). Separately, the associated table for cause \(j\) has 10,000 lives at age \(x\) and 310 exits before age \(x+1\). Calculate the associated probabilities and state which exit probability applies to the one-cause table. (Answer: \(q_x^{\prime(j)}=0.0310,\ p_x^{\prime(j)}=0.9690\); the associated exit probability is 0.0310, not the observed 0.027.)