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Exam ALTAM · Decrement Table Calculations · Tutorial 294 of 999

Common Errors in Decrement Table Calculations

Use the stated decrement model to identify the right probability, conversion, and fractional-age calculation—and to spot results that cannot be correct.

Intermediate 10 min read

What You'll Learn

  • Distinguish cause-specific force rates from annual decrement probabilities.
  • Check whether a decrement conversion uses constant forces or another stated assumption.
  • Calculate conditional exit probabilities using the correct starting population.
  • Apply UDD and constant-force fractional-age formulas consistently.
  • Reconcile cause-specific decrements with total decrements and survival.
  • Diagnose common calculation errors with explicit checks.

Why Decrement Calculations Go Wrong

A multiple-decrement table describes a population exposed to several competing causes of exit. A member who exits from one cause is no longer exposed to the other causes in that period. Many errors arise from overlooking that competition: treating a rate as a probability, converting a table without specifying the assumption, or using the wrong group as the denominator in a conditional probability.

Fractional-age calculations require the same care. A table may give annual probabilities, but probabilities within the year depend on an assumption about how decrements are distributed over time. The uniform distribution of decrements (UDD) and constant-force assumptions generally produce different answers. Neither should be inserted silently into a calculation.

Definition: For a group of \(l_x\) members at the start of a year, let \(d_x^{(j)}\) be the number exiting during the year from cause \(j\), and define \(q_x^{(j)}=d_x^{(j)}/l_x\). The total decrement probability is \(q_x=\sum_j q_x^{(j)}\), and the probability of remaining in the group to year-end is \(p_x=1-q_x\). These are probabilities over the year, not instantaneous rates.

A cause-specific force of decrement, written here as \(\mu^{(j)}(t)\), is an instantaneous rate among members still in the group at time \(t\). A force has units of inverse time; it is not itself the probability of exiting from that cause during a year. In general, the annual probability depends on all competing forces operating while a member remains exposed.

Rates, Probabilities, and Conversion Assumptions

When the cause-specific forces are constant over a year, write them as \(\mu_1,\ldots,\mu_m\), and let \(\mu=\sum_j\mu_j\). The probability of surviving all causes for one year is \(e^{-\mu}\), so \(q=1-e^{-\mu}\). The probability of exit from cause \(j\) is not \(1-e^{-\mu_j}\), because that expression acts as if cause \(j\) were the only cause operating.

$$ q^{(j)}=\int_0^1 e^{-\mu t}\mu_j\,dt =\frac{\mu_j}{\mu}(1-e^{-\mu}), \qquad q=\sum_j q^{(j)}=1-e^{-\mu}. $$

The integral counts cause-\(j\) exits at time \(t\): the member must first survive all causes to \(t\), with probability \(e^{-\mu t}\), and then exit from cause \(j\) in an infinitesimal interval, at rate \(\mu_j\). This also explains why adding the cause-specific forces and then treating each force as an annual probability is generally wrong.

An associated single-decrement probability asks what the probability would be if one cause operated alone, under a specified model. Under constant forces, removing all causes other than \(j\) gives \(q'^{(j)}=1-e^{-\mu_j}\). That is a different quantity from the multiple-decrement probability \(q^{(j)}\). A conversion from observed multiple-decrement probabilities to associated single-decrement probabilities is not unique unless its assumptions are stated.

Worked Example: Converting Constant Forces Without Calling Them Probabilities

Method. Use the competing constant-force model for one year. A hypothetical group has constant cause-specific forces \(\mu_1=0.10\) and \(\mu_2=0.20\) per year. Find the annual probability of each cause, the total exit probability, and the associated single-decrement probabilities. Identify the error in using the forces directly as annual probabilities.

The total force is \(\mu=0.10+0.20=0.30\). The one-year survival probability is \(e^{-0.30}=0.740818\), so the total exit probability is \(1-0.740818=0.259182\).

The cause-1 share of the total force is \(0.10/0.30=1/3\). Therefore, \(q^{(1)}=(1/3)(0.259182)=0.086394\). The cause-2 share is \(0.20/0.30=2/3\), so \(q^{(2)}=(2/3)(0.259182)=0.172788\). As a check, \(0.086394+0.172788=0.259182\).

For the associated single-decrement probabilities, remove the competing cause in each calculation: \(q'^{(1)}=1-e^{-0.10}=0.095163\) and \(q'^{(2)}=1-e^{-0.20}=0.181269\). They are not the multiple-decrement probabilities just calculated.

Common mistake and correction. Setting the annual cause probabilities equal to \(0.10\) and \(0.20\) gives a total of \(0.30\), which contradicts the constant-force total \(1-e^{-0.30}=0.259182\). The correction is to account for survival under all causes before an exit can occur. Rounded to four decimal places, the multiple-decrement probabilities are \(0.0864\) and \(0.1728\), summing to \(0.2592\).

A useful diagnostic is to check that the cause-specific probabilities sum to the total probability and that total probability plus survival equals one. These checks detect arithmetic and closure errors, but they do not establish that the conversion assumption was appropriate. The assumed model must also match the question.

Conditional Probabilities Need the Right Denominator

For a decrement table, \(q_x^{(j)}\) is conditional on being in the group at the start of the year. Among those who exit during the year, the conditional probability that cause \(j\) is responsible is instead \(q_x^{(j)}/q_x\), provided \(q_x>0\). Among members who survive to a later year, the later-year decrement probability is conditional on that survival; it is not an unconditional probability for the original cohort.

Worked Example: Distinguishing Cohort and Exit-Conditioned Probabilities

Method. State the event and denominator before dividing. In a hypothetical table, 1,000 members are exposed at the start of year 1. During that year 120 exit from cause A, 80 from cause B, and 800 remain. In year 2, conditional on being alive and in the group at the start of that year, the probabilities are \(q_2^{(A)}=0.10\) and \(q_2^{(B)}=0.05\). Find the probability that a year-1 member exits from A during year 2, and the probability of A among all year-2 exits.

At the start of year 2, the 800 survivors are \(800/1{,}000=0.80\) of the original cohort. The probability for an original member of both surviving year 1 and exiting from A in year 2 is \(0.80(0.10)=0.08\). Equivalently, \(800(0.10)=80\) members exit from A in year 2, and \(80/1{,}000=0.08\).

The total year-2 exit probability conditional on being present at the start of year 2 is \(0.10+0.05=0.15\). Thus the probability of A conditional on exiting in year 2 is \(0.10/0.15=2/3\). The counts give the same result: 80 of the \(800(0.15)=120\) exits are from A, and \(80/120=2/3\).

Common mistake and correction. The value \(0.10\) answers the question conditional on being in the year-2 group, not for a member of the original cohort. The original-cohort probability is \(0.08\); the exit-conditioned probability is \(2/3\). They differ because their denominators describe different populations.

For multiple years, multiply conditional survival probabilities along the path before applying a later conditional decrement probability. For example, if a member must survive a first year with survival probability \(p_1\) before facing a second-year cause probability \(q_2^{(j)}\), the probability of that sequence is \(p_1q_2^{(j)}\). This does not mean that the second-year probability itself has changed; it means the original cohort includes members who did not reach the second year.

Fractional Ages: Do Not Mix Models

Annual probabilities do not determine fractional-year probabilities without an assumption. Under UDD, decrement counts are assumed to be uniformly distributed through the year: for each cause, the number of decrements accumulated by time \(t\) is \(t\) times its annual count. If annual cause probabilities are \(q^{(j)}\) and total \(q=\sum_jq^{(j)}\), then the probability of surviving to time \(t\) is \(1-tq\), and the probability of cause \(j\) during the first \(t\) years is \(tq^{(j)}\).

$$ {}_tp=1-tq,\qquad {}_tq^{(j)}=tq^{(j)} \quad\text{(UDD, for }0\le t\le1\text{).} $$

The conditional probability of cause \(j\) in the interval from \(s\) to \(t\), given survival to \(s\), is the cause-\(j\) probability over that interval divided by the probability of survival to \(s\): \((t-s)q^{(j)}/(1-sq)\), for \(0\le s<t\le1\). The denominator matters: an interval probability measured against the original cohort is not yet conditional on reaching the interval.

Under constant total force, annual survival \(p=1-q\) instead implies survival to time \(t\) of \(p^t\). If cause-specific forces are constant, their shares of the total force allocate the fractional-period exits. Applying UDD to one part of a calculation and constant force to another generally fails to preserve the stated annual table.

Worked Example: Applying UDD Consistently at Midyear

Method. Use UDD for both causes and for both half-years. A hypothetical annual table has \(q^{(A)}=0.12\) and \(q^{(B)}=0.08\), so \(q=0.20\). Find the probability of surviving to midyear, the probability of cause A in the first half-year, and the conditional probability of cause A in the second half-year given survival to midyear.

Under UDD, survival to \(t=0.5\) is \(1-(0.5)(0.20)=0.90\). The first-half probability of cause A, measured against the initial group, is \((0.5)(0.12)=0.06\). Cause B in the first half is \((0.5)(0.08)=0.04\), so total first-half exits are \(0.10\). The survival and exit probabilities sum to \(0.90+0.10=1.00\).

Under UDD, cause-A decrements in the second half are also \(0.5(0.12)=0.06\) of the initial group. Conditional on being among the \(0.90\) who survive to midyear, the cause-A probability in the second half is \(0.06/0.90=0.066666\ldots\), or \(0.0667\) to four decimal places.

Common mistake and correction. Using constant-force survival at midyear would give \(\sqrt{0.80}=0.8944\), not the UDD value \(0.90\). That is not an alternative intermediate step within this UDD calculation; it changes the model. The stated annual probabilities are recovered under UDD because the two half-year decrements for each cause add to that cause’s annual decrement.

A Practical Error-Checking Sequence

When a calculation appears plausible but its assumptions or denominator are unclear, use a short sequence of checks before accepting it. The aim is to identify the model first, then test whether the resulting quantities agree with that model.

1
Label the quantity.
Write whether each input is a force, an annual probability, a count, or a probability conditional on a particular starting group.
2
State the assumption.
For a conversion or fractional-age calculation, identify the assumed force pattern or interpolation rule. If the information does not determine a unique conversion, say so.
3
Write the denominator.
For each conditional probability, identify the population on which it is based: the original cohort, those surviving to a year, or those exiting during a year.
4
Check closure.
Confirm that cause probabilities sum to the total decrement probability and that total decrement plus survival equals one on the same basis.

These checks are particularly useful when a problem presents a table and asks for a conditional or fractional-year result. A number can be arithmetically correct under one model but wrong for the model specified in the problem. State the assumption beside the calculation, retain the correct conditioning population, and round only after the checks are complete. This error-diagnosis approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Key takeaway: A decrement force, an annual cause probability, an associated single-decrement probability, and a probability conditional on exiting are different quantities. Label the quantity, name the model, and verify the denominator before calculating.

Check Your Understanding

For each problem, specify the assumption used before calculating. Answers are rounded as indicated.

  1. A hypothetical population has two constant cause-specific forces per year, \(0.04\) and \(0.06\). Find the total one-year exit probability and each cause-specific one-year probability. (Answer: total \(0.0952\); cause 1 \(0.0381\); cause 2 \(0.0571\).)
  2. In a one-year table, \(q^{(A)}=0.18\) and \(q^{(B)}=0.12\). Find the probability of A among exits, and state the probability of A for a member present at the start of the year. (Answer: among exits \(0.6000\); start-of-year probability \(0.1800\).)
  3. Under constant cause-specific forces, observed annual probabilities are \(q^{(A)}=0.10\) and \(q^{(B)}=0.15\). Find the associated single-decrement probabilities. (Answer: \(q'^{(A)}=0.1087\), \(q'^{(B)}=0.1585\), to four decimal places.)
  4. Under UDD, annual probabilities are \(q^{(A)}=0.09\) and \(q^{(B)}=0.06\). Find the conditional probability of cause A during the second half-year given survival to midyear. (Answer: \(0.045/0.925=0.0486\), to four decimal places.)
  5. Under constant cause-specific forces, annual probabilities are \(q^{(A)}=0.08\) and \(q^{(B)}=0.12\). Find the first-half-year probability of cause A, measured against the original group. (Answer: \(0.0422\), to four decimal places.)
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