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Exam ALTAM · Decrement Table Calculations · Tutorial 267 of 999

Survival Across Several Ages and Durations

Build multi-year survival and survivor counts by chaining the correct annual probabilities, then use cohort counts to check or recover decrement rates.

Intermediate 9 min read

What You'll Learn

  • Define annual total exit and survival probabilities from cause-specific decrement probabilities.
  • Chain probabilities across ages to calculate survival over several years.
  • Convert multi-year survival probabilities into survivor counts.
  • Calculate annual probabilities from decrement counts using the correct starting population.
  • Recover a missing annual or cause-specific probability from cohort survivor counts.
  • Avoid adding annual exit probabilities when calculating multi-year survival.

Introduction

A probability of remaining in a group for one year is conditional on being in that group at the start of the year. To calculate survival across several years, the annual probabilities must therefore be applied successively: the population entering each year is the population that survived all earlier years. This principle connects annual multiple-decrement table entries to survival over longer durations and to the number of survivors at a later age.

The calculation is straightforward once the denominator and time interval are kept clear. The main trap is to add annual exit probabilities across different years, as though they were competing outcomes in one interval. They are not: each year has a new conditional probability based on the lives still in the group.

Annual Survival in a Multiple-Decrement Table

Suppose a group is subject to \(J\) mutually exclusive exit causes during each year. For an individual who is in the group at exact age \(x\), let \(q_x^{(j)}\) be the probability of exiting from cause \(j\) between ages \(x\) and \(x+1\). Assume that the listed causes are exhaustive for the group: an individual either remains in the group through the year or exits by exactly one of those causes. Re-entry is not allowed.

Definition: The total annual exit probability at age \(x\) is \(q_x=\sum_{j=1}^{J}q_x^{(j)}\). The annual probability of remaining in the group through the interval is \(p_x=1-q_x\). Both probabilities are conditional on being in the group at exact age \(x\). If \(l_x\) denotes the number of lives in the group at that age, then the table count relation is \(l_{x+1}=l_xp_x=l_x-\sum_{j=1}^{J}d_x^{(j)}\), where \(d_x^{(j)}\) is the number exiting from cause \(j\) during that interval.

The time unit here is one year. The age subscript identifies the interval, not a probability that applies to every future year. In general, both the total exit probability and the cause mix can vary with age. If probabilities are applied to a hypothetical cohort, the resulting survivor count can be an expected count and need not be an integer. Counts taken directly from an observed decrement table, by contrast, are actual tabulated counts.

For a single individual in one year, the outcomes form a categorical distribution: remain in the group with probability \(p_x\), or exit from cause \(j\) with probability \(q_x^{(j)}\). These probabilities sum to one. No assumption that outcomes in different years are independent is required to chain the conditional survival probabilities.

Chaining Survival Across Years

Write \(n\) for the number of complete annual intervals. To remain in the group for \(n\) years, an individual must survive the first interval, then the second conditional on reaching it, and so on. Thus the probability of surviving from age \(x\) to age \(x+n\) is the product of the annual survival probabilities at the ages encountered.

$$ {}_np_x^{\text{group}}=\prod_{r=0}^{n-1}p_{x+r} =\prod_{r=0}^{n-1}\left(1-\sum_{j=1}^{J}q_{x+r}^{(j)}\right). $$

The superscript “group” emphasizes that this is survival in the specified multiple-decrement group, meaning no exit by any listed cause. It is not survival from just one selected cause. If the starting count is \(l_x\), the corresponding number remaining at age \(x+n\) is

$$ l_{x+n}=l_x\,{}_np_x^{\text{group}} =l_x\prod_{r=0}^{n-1}p_{x+r}. $$
1
Find each annual total exit probability.
Add the mutually exclusive cause-specific probabilities for that age, or divide the total exits by the lives entering the interval.
2
Convert to annual survival.
For each interval, subtract the total exit probability from one.
3
Chain the intervals in order.
Multiply the annual survival probabilities from the starting age through the final interval.
4
Apply the starting count and check.
Multiply the chained probability by \(l_x\). Check that the result does not exceed the starting count and agrees with sequentially calculating the survivors year by year.

Why the Product Formula Holds

Let \(A_r\) be the event that an individual remains in the group through the interval from age \(x+r\) to \(x+r+1\). Remaining for \(n\) years requires all of \(A_0,A_1,\ldots,A_{n-1}\) to occur. The probability chain rule gives

$$ P(A_0\cap A_1\cap\cdots\cap A_{n-1}) =P(A_0)\,P(A_1\mid A_0)\cdots P(A_{n-1}\mid A_0\cap\cdots\cap A_{n-2}). $$

On the condition that the individual has remained in the group to age \(x+r\), the probability of remaining through the next interval is \(p_{x+r}\). Substituting these conditional probabilities into the chain rule yields the product formula. This argument uses conditional probabilities; it does not require independent annual outcomes.

Worked Example: Projecting Survivors Through Three Ages

Method and assumptions. Consider a hypothetical group of 10,000 lives at exact age 60. The model has annual, age-specific categorical outcomes: remain in the group, or exit from one of two exhaustive causes. For ages 60, 61, and 62, the cause-specific probabilities are respectively \((0.04,0.06)\), \((0.08,0.12)\), and \((0.10,0.15)\). There is no interest calculation, and no separate mortality assumption is needed beyond these two specified exit causes; any exit included in the model is represented by one of them.

At age 60. The total exit probability is \(0.04+0.06=0.10\), so \(p_{60}=1-0.10=0.90\). The number surviving to age 61 is \(10{,}000(0.90)=9{,}000\).

At age 61. The total exit probability is \(0.08+0.12=0.20\), so \(p_{61}=1-0.20=0.80\). The number surviving to age 62 is \(9{,}000(0.80)=7{,}200\).

At age 62. The total exit probability is \(0.10+0.15=0.25\), so \(p_{62}=1-0.25=0.75\). The number surviving to age 63 is \(7{,}200(0.75)=5{,}400\).

The chained survival probability is \(0.90(0.80)(0.75)=0.54\), and \(10{,}000(0.54)=5{,}400\), matching the sequential count. The result is checked both ways. The common mistake avoided is adding the three annual total exit probabilities and treating their sum as the three-year exit probability.

Using Decrement Counts as Annual Probabilities

A multiple-decrement table may give lives entering an interval and exits by cause instead of probabilities. For any one age interval, each cause-specific probability uses the number entering that interval as its denominator. The number entering the next interval is reduced by all exits from the current one. This changing denominator is essential when calculating annual probabilities from counts.

Worked Example: Chaining Probabilities From Table Counts

Method and assumptions. A hypothetical table records 1,200 lives entering the year from age 60 to 61. During that year, 60 exit from cause A and 120 from cause B. Of those remaining at age 61, 51 exit from A and 102 from B during the year from age 61 to 62. The two causes are exhaustive, the year is the time unit, and each individual has categorical annual outcomes with the probabilities implied by the table counts. There is no interest or additional mortality assumption.

First interval. The total exits are \(60+120=180\). The total exit probability is \(180/1{,}200=0.15\), and the survival probability is \(1-0.15=0.85\). The count reaching age 61 is \(1{,}200-180=1{,}020\). Separately, the cause-specific probabilities are \(60/1{,}200=0.05\) and \(120/1{,}200=0.10\).

Second interval. The denominator is now 1,020, not 1,200. The total exits are \(51+102=153\), giving total exit probability \(153/1{,}020=0.15\) and survival probability \(0.85\). The count reaching age 62 is \(1{,}020-153=867\). The cause-specific probabilities are \(51/1{,}020=0.05\) and \(102/1{,}020=0.10\).

The two-year survival probability is \(0.85(0.85)=0.7225\). Applying it to the starting count gives \(1{,}200(0.7225)=867\), as in the table. The common mistake avoided is using the original 1,200 as the second-year denominator. Doing so would give an incorrect second-year exit probability of \(153/1{,}200=0.1275\).

Recovering a Missing Probability

The same product relation can be rearranged when survivor counts or some annual probabilities are known. First calculate the survival ratio from the start of the period to its end. Then divide out the known annual survival factors to obtain the missing factor. If one cause-specific probability is also known, subtract it from the inferred total exit probability to find the other cause probability.

Worked Example: Finding a Missing Cause-Specific Probability

Method and assumptions. In a hypothetical annual table, 16,000 lives enter age 70 and 10,080 reach age 73. The total exit probabilities at ages 70 and 71 are 0.10 and 0.20. At age 72 there are two exhaustive categorical exit causes; the probability for cause A is 0.05 and the probability for cause B is unknown. All probabilities are conditional on being in the group at the start of the corresponding year. No interest or other mortality assumption is used.

The observed three-year survival ratio is \(10{,}080/16{,}000=0.63\). The known annual survival probabilities are \(p_{70}=1-0.10=0.90\) and \(p_{71}=1-0.20=0.80\). Let \(p_{72}\) be the missing annual survival probability. Chaining the three intervals gives

$$ 0.90(0.80)p_{72}=0.63,\qquad p_{72}=\frac{0.63}{0.72}=0.875. $$

Therefore the total exit probability at age 72 is \(q_{72}=1-0.875=0.125\). Since the two causes are exhaustive, \(q_{72}^{(A)}+q_{72}^{(B)}=q_{72}\), so \(q_{72}^{(B)}=0.125-0.05=0.075\).

For a check, 16,000 lives become \(16{,}000(0.90)=14{,}400\), then \(14{,}400(0.80)=11{,}520\), and finally \(11{,}520(0.875)=10{,}080\). The recovered cause B probability corresponds to \(11{,}520(0.075)=864\) exits; cause A corresponds to \(11{,}520(0.05)=576\). Together, 1,440 exits leave \(11{,}520-1{,}440=10{,}080\) survivors. The common mistake avoided is subtracting the known cause A probability directly from the three-year exit probability; the 0.125 total exit probability applies only to the age-72 interval.

Common Calculation Traps

Annual exit probabilities refer to different conditional populations and different intervals. The following errors can produce plausible-looking but incorrect results:

  • Adding annual exits across years. If annual exit probabilities are 0.10 and 0.20, two-year survival is \((1-0.10)(1-0.20)=0.72\), not \(1-(0.10+0.20)=0.70\). The latter incorrectly treats exits in separate years as outcomes in one interval.
  • Using the wrong denominator. When converting counts to probabilities, use the lives entering that specific interval, after earlier exits have been removed.
  • Summing cause-specific probabilities across different years before converting to survival. Sum the causes within each year to obtain that year’s total exit probability; then subtract from one and multiply the resulting annual survival probabilities.
  • Confusing the starting count with the survival probability. The product gives a probability. Multiply by the starting count to obtain a survivor count.

The method matters whenever a later calculation depends on the number still exposed to decrements at a future age. A consistent sequence of conditional annual probabilities gives both the multi-year probability of remaining in the group and the corresponding survivor count.

Check Your Understanding

For each problem, treat the listed causes as exhaustive annual exits and use the specified interval probabilities conditional on entering that interval.

  1. A hypothetical cohort has 5,000 lives at age 40. Total exit probabilities at ages 40, 41, and 42 are 0.08, 0.15, and 0.10. Find the probability of reaching age 43 and the expected number who do. (Answer: survival probability 0.7038; 3,519 survivors)
  2. There are 8,000 lives at age 55. The cause-specific exit probabilities from age 55 to 56 are 0.03 and 0.07; from 56 to 57 they are 0.06 and 0.14. Find the number expected at age 57. (Answer: 5,760)
  3. A cohort of 9,000 has annual survival probability 0.90 in the first year. At the end of two years, 6,480 remain. Find the second-year survival probability and total exit probability. (Answer: survival probability 0.80; total exit probability 0.20)
  4. A table has 10,000 lives at age 30 and 6,840 at age 32. The age-30 total exit probability is 0.10. Find the age-31 total exit probability. (Answer: 0.24)
  5. A hypothetical group of 25,000 has two cause-specific exit probabilities at each of ages 50, 51, and 52: respectively \((0.02,0.03)\), \((0.04,0.06)\), and \((0.08,0.12)\). Find the expected number remaining at age 53. (Answer: 17,100)

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