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A Worked Constant-Force Conversion

Learn to convert a multi-cause annual decrement table into associated probabilities under constant force and verify that the implied forces reproduce its aggregate survival and cause shares.

Intermediate 10 min read

What You'll Learn

  • Distinguish observed competing-decrement probabilities from associated single-decrement probabilities.
  • Check that observed probabilities form a valid annual decrement table before converting them.
  • Allocate the total constant force across causes using their shares of observed exits.
  • Calculate each associated probability from its cause-specific force.
  • Verify the reconstructed aggregate survival and relative cause shares.

Purpose of a Constant-Force Conversion

A multiple-decrement table records exits from a group exposed to several causes during the same period. Its observed probability for a cause reflects both that cause and the competing causes: a person can exit for only one cause, and an earlier exit prevents a later one. An associated single-decrement probability answers a different question. It is the probability of exit from one cause over the period in a hypothetical model where the other causes have been removed.

Under the constant-force assumption, the observed annual table can be converted into associated probabilities by first finding the total force consistent with aggregate survival, then allocating that force to causes in proportion to their observed exit probabilities. The resulting associated probabilities generally do not add to the observed total exit probability. The conversion is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: For one year of exposure, let \(q_j\) be the observed probability of exit from cause \(j\), for \(j=1,\ldots,m\). Let \(p=1-\sum_{j=1}^m q_j\) be the observed probability of surviving the year without an exit from any cause. The associated single-decrement probability for cause \(j\), under constant forces, is the one-year exit probability in the model that retains cause \(j\)'s force and removes the other causes.

The conversion requires a valid table: each \(q_j\) must be nonnegative and \(\sum_j q_j<1\), so that \(p>0\). The model assumes that each cause has a constant force throughout the year, that the forces operate simultaneously while the individual remains exposed, and that the same cause-specific force is retained when constructing its associated single-decrement table. If the table covers a period other than one year, the duration must be accounted for; the formulas below use a one-year period.

The Conversion Formula

Write \(\lambda\) for the total force while all causes operate, and \(\mu_j\) for the force of cause \(j\). The total force is the sum of the cause-specific forces. Under constant forces, the probability of no exit by time \(t\) within the year is \(e^{-\lambda t}\).

$$ \lambda=\sum_{j=1}^m \mu_j=-\ln p,\qquad \mu_j=\lambda\frac{q_j}{\sum_{k=1}^m q_k},\qquad q_j'=1-e^{-\mu_j} =1-p^{\,q_j/\sum_{k=1}^m q_k}. $$

Here \(q_j'\) denotes the associated single-decrement probability. The cause share \(q_j/\sum_k q_k\) is used only when there is at least one observed exit; if all observed exit probabilities are zero, every cause force is zero and every associated probability is zero.

Why the formula holds: With all causes operating, the probability of exit from cause \(j\) during the year is the integral of the probability of still being exposed at time \(t\), multiplied by the instantaneous cause-\(j\) exit rate:
$$ q_j=\int_0^1 e^{-\lambda t}\mu_j\,dt =\frac{\mu_j}{\lambda}(1-e^{-\lambda}),\qquad \lambda>0. $$

The corresponding probability of no exit is \(p=e^{-\lambda}\), so \(1-e^{-\lambda}=1-p=\sum_k q_k\). Dividing the expression for \(q_j\) by the sum of all observed exit probabilities gives \(\mu_j/\lambda=q_j/\sum_k q_k\). Also, solving \(p=e^{-\lambda}\) gives \(\lambda=-\ln p\). With cause \(j\) alone, survival for one year is \(e^{-\mu_j}\), so its associated exit probability is \(1-e^{-\mu_j}\). Substitution gives the stated conversion.

Worked Conversions

Worked Example: Two Causes With Equal Aggregate Survival

A hypothetical one-year table has observed exit probabilities \(q_1=0.30\) and \(q_2=0.20\). Assume constant cause-specific forces over the year. Find both associated probabilities, then check the implied aggregate survival and cause shares.

Method and validity check. The aggregate exit probability is \(0.30+0.20=0.50\), so the probability of no exit is \(p=1-0.50=0.50\). Both cause probabilities are nonnegative and their sum is less than one, so the table is valid.

Total force. The force that produces one-year survival \(0.50\) is

$$ \lambda=-\ln(0.50)=0.693147. $$

Cause shares and forces. Cause 1 has share \(0.30/0.50=0.60\), and cause 2 has share \(0.20/0.50=0.40\). Therefore,

$$ \mu_1=0.693147(0.60)=0.415888,\qquad \mu_2=0.693147(0.40)=0.277259. $$

Associated probabilities. Retain each force alone in turn:

$$ q_1'=1-e^{-0.415888}=1-0.659754=0.340246, $$
$$ q_2'=1-e^{-0.277259}=1-0.757858=0.242142. $$

Rounded to four decimal places, the associated probabilities are \(0.3402\) and \(0.2421\).

Check. The forces sum to \(0.415888+0.277259=0.693147\), so the reconstructed aggregate survival is \(e^{-0.693147}=0.5000\). Their shares of total force are \(0.60\) and \(0.40\), matching the observed cause shares \(0.30/0.50\) and \(0.20/0.50\). The wrong shortcut of treating the observed cause probabilities as the associated probabilities would give \(0.30\) and \(0.20\); that ignores the competing-cause effect in the original table.

Worked Example: Three Causes With Repeated Probabilities

A hypothetical annual decrement table reports \(q_1=0.10\), \(q_2=0.20\), and \(q_3=0.10\). Under constant cause forces, calculate the three associated probabilities and verify that the original aggregate survival and cause shares are reproduced.

Method and validity check. The aggregate exit probability is \(0.10+0.20+0.10=0.40\), leaving \(p=0.60\). The probabilities are nonnegative and sum to less than one, so the conversion is defined.

Total force and shares. The aggregate force is

$$ \lambda=-\ln(0.60)=0.510826. $$

The cause shares are \(0.10/0.40=0.25\), \(0.20/0.40=0.50\), and \(0.10/0.40=0.25\). Thus,

$$ \mu_1=0.510826(0.25)=0.127706,\quad \mu_2=0.510826(0.50)=0.255413,\quad \mu_3=0.510826(0.25)=0.127706. $$

Associated probabilities. Apply \(q_j'=1-e^{-\mu_j}\) to each force:

$$ q_1'=1-e^{-0.127706}=0.1199,\qquad q_2'=1-e^{-0.255413}=0.2254,\qquad q_3'=1-e^{-0.127706}=0.1199. $$

The results are rounded to four decimal places. In particular, the repeated forces for causes 1 and 3 produce equal associated probabilities.

Check. Adding the forces gives \(0.127706+0.255413+0.127706=0.510825\), equal to \(0.510826\) to the displayed precision; the unrounded forces sum to \(-\ln(0.60)\). Hence aggregate survival is \(e^{-\lambda}=0.60\). Force shares are \(0.25,0.50,0.25\), which match the observed exit shares \(0.10/0.40,0.20/0.40,0.10/0.40\). Do not add the associated probabilities and interpret their sum as the original aggregate exit probability: associated probabilities come from separate single-cause models, not one simultaneous competing-cause model.

Worked Example: Unequal Shares Across Three Causes

For a hypothetical one-year table, the observed probabilities are \(q_1=0.08\), \(q_2=0.12\), and \(q_3=0.20\). Assume constant forces during the year. Find the associated probabilities and check the total survival and the relative cause shares.

Method and validity check. The total observed exit probability is \(0.08+0.12+0.20=0.40\), and the no-exit probability is \(p=0.60\). Thus the table is valid, and the total force is

$$ \lambda=-\ln(0.60)=0.510826. $$

Allocate the force. The observed cause shares are \(0.08/0.40=0.20\), \(0.12/0.40=0.30\), and \(0.20/0.40=0.50\). The cause-specific forces are therefore

$$ \mu_1=0.510826(0.20)=0.102165,\quad \mu_2=0.510826(0.30)=0.153248,\quad \mu_3=0.510826(0.50)=0.255413. $$

Calculate the associated probabilities.

$$ q_1'=1-e^{-0.102165}=0.0971,\qquad q_2'=1-e^{-0.153248}=0.1421,\qquad q_3'=1-e^{-0.255413}=0.2254. $$

Each probability is rounded to four decimal places.

Check. The cause forces sum to \(0.510826\), giving aggregate survival \(e^{-0.510826}=0.6000\). Their shares are \(0.20,0.30,0.50\), equal to the observed exit shares. A common error is to allocate the total force equally just because there are three causes. That would give each cause a force of \(\lambda/3\) and would fail to preserve the observed relative cause shares.

Checks and Exam Traps

A reliable conversion keeps three quantities distinct: observed cause probabilities \(q_j\), cause-specific forces \(\mu_j\), and associated probabilities \(q_j'\). Observed probabilities describe exits in the presence of all causes. Forces are the constant rates used to model those competing exits. Associated probabilities are calculated after retaining only one force at a time.

1
Check the table.
Confirm \(q_j\geq 0\) for every cause and calculate \(p=1-\sum_j q_j\). The conversion requires \(p>0\).
2
Find total force.
Use \(\lambda=-\ln p\), with the year as the exposure period.
3
Allocate by observed shares.
For each cause, calculate \(\mu_j=\lambda q_j/\sum_k q_k\), provided the total exit probability is positive.
4
Calculate and verify.
Find \(q_j'=1-e^{-\mu_j}\). Check that the forces sum to \(\lambda\), that \(e^{-\lambda}=p\), and that the force shares equal the observed cause shares.

The checks are useful because a plausible-looking associated probability can still come from a conversion error. Using \(q_j\) directly as \(q_j'\) fails to account for competing causes. Using \(\sum_j q_j\) in place of \(p\) when taking the logarithm confuses aggregate exit probability with aggregate survival. Allocating the total force equally across causes also fails unless their observed exit probabilities are equal.

If \(\sum_jq_j=0\), there are no observed exits and \(p=1\). Then \(\lambda=0\), all cause forces are zero, and all associated probabilities are zero. If \(p=0\), the table has no observed survivors; a finite total constant force cannot produce zero survival over a finite one-year period, so this conversion does not yield finite cause forces.

Key takeaway: Under constant forces, convert observed probabilities by finding the total force from the observed survival probability, allocating that force in proportion to observed cause exits, and converting each cause force to a single-decrement probability. Verify aggregate survival and cause shares, and do not mistake associated probabilities for observed competing-decrement probabilities.

Check Your Understanding

For each hypothetical annual table, apply the constant-force conversion. Round associated probabilities to four decimal places.

  1. A table has two observed exit probabilities, \(0.15\) and \(0.25\). Find the two associated probabilities. (Answer: \(0.1744,\ 0.2733\))
  2. Three causes have observed probabilities \(0.06,\ 0.09,\ 0.15\). Find the cause shares and the three associated probabilities. (Answer: Shares \(0.20,\ 0.30,\ 0.50\); associated probabilities \(0.0689,\ 0.1015,\ 0.1633\))
  3. Two causes have observed probabilities \(0.20\) and \(0.30\). Find the total force, the cause-specific forces, and the associated probabilities. (Answer: Total force \(0.6931\); cause forces \(0.4159,\ 0.2773\); associated probabilities \(0.3402,\ 0.2421\))
  4. Three causes have observed probabilities \(0.04,\ 0.16,\ 0.20\). Find the associated probabilities and state the reconstructed aggregate survival. (Answer: Associated probabilities \(0.0498,\ 0.1848,\ 0.2254\); aggregate survival \(0.6000\))
  5. A two-cause table has observed probabilities \(0.12\) and \(0.08\). Find the total force and associated probabilities. (Answer: Total force \(0.2231\); associated probabilities \(0.1253,\ 0.0854\))

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