Converting Observed Probabilities Under Constant Force

Why the Conversion Is Needed

In a multiple-decrement model, each observed cause probability reflects the presence of all competing causes: an individual can exit for a particular cause only if they have not already exited for another. An associated single-decrement probability answers a different question: what would the probability for that cause be if it acted alone, under the specified force assumption? Converting between them requires a model for exits within the interval.

This tutorial uses constant forces over a one-year interval. The method is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. Its central calculation has two stages: find the total force implied by the observed exits, then divide that force among causes according to their observed shares.

Definition: Let \(q_j\) be the observed probability of exit from cause \(j\) during a one-year interval, among individuals active at its start. Write \(Q=\sum_j q_j\) for the total observed exit probability. Under constant cause forces \(\mu_j\), the total force is \(\mu=\sum_j\mu_j\). The associated single-decrement probability for cause \(j\), denoted \(q_j'\), is the probability of exit from cause \(j\) over the year if that cause acts alone at force \(\mu_j\).

The Constant-Force Conversion

Assume the cause forces are constant throughout the year, the causes are mutually exclusive exit routes, and there are no simultaneous exits. The probability of remaining active to time \(t\), for \(0\le t\le1\), is \(e^{-\mu t}\). The observed cause probability is therefore

q_j=\int_0^1 e^{-\mu t}\mu_j\,dt =\frac{\mu_j}{\mu}\left(1-e^{-\mu}\right),\qquad \mu>0.

Summing this equation over causes gives \(Q=1-e^{-\mu}\). If \(0<Q<1\), the conversion is

\mu=-\ln(1-Q),\qquad \mu_j=\mu\frac{q_j}{Q},\qquad q_j'=1-e^{-\mu_j}.

The first equation determines the total force. The second assigns each cause the same share of the total force as its share of observed exits. The last equation converts that cause force to the probability it would produce acting alone. The result \(q_j'\) is not generally equal to the observed \(q_j\): competing causes reduce the time during which cause \(j\) can produce an observed exit.

Check the observed probabilities.
Compute \(Q=\sum_j q_j\). The data must have \(0\le Q\le1\), with each \(q_j\ge0\).

Find the total force.
For \(0<Q<1\), calculate \(\mu=-\ln(1-Q)\).

Allocate force by cause.
Calculate each share \(q_j/Q\), then \(\mu_j=\mu(q_j/Q)\).

Convert to associated probabilities.
Calculate \(q_j'=1-e^{-\mu_j}\), retaining full precision until the final rounding.

Why the Formula Holds

The formula follows directly from the probability of surviving all causes up to time \(t\). Over a short interval of length \(dt\), an individual still active at time \(t\) exits for cause \(j\) with probability approximately \(\mu_jdt\). Thus the probability of an observed exit for that cause in \([t,t+dt]\), measured from the start of the year, is \(e^{-\mu t}\mu_jdt\). Integrating over the year gives the observed \(q_j\).

\sum_j q_j =\int_0^1 e^{-\mu t}\sum_j\mu_j\,dt =\int_0^1 e^{-\mu t}\mu\,dt =1-e^{-\mu}.

For \(Q>0\), dividing the cause-specific integral by the total integral gives \(q_j/Q=\mu_j/\mu\). This is why the observed cause share is used to allocate the total force. In the associated single-decrement setting, no competing cause removes an individual first, so survival under cause \(j\) alone for one year is \(e^{-\mu_j}\); its exit probability is \(1-e^{-\mu_j}\).

Worked Example: Two Observed Causes

In a hypothetical one-year decrement study, the observed exit probabilities are \(q_1=0.10\) and \(q_2=0.20\). Assume constant cause forces over the year. Find the total force, each cause force, and the associated single-decrement probabilities. Give probabilities to four decimal places and forces to four significant figures.

Method. First convert the total observed probability to a total force, then allocate by observed cause shares.

The total exit probability is \(Q=0.10+0.20=0.30\). Therefore

\mu=-\ln(1-0.30)=-\ln(0.70)=0.3566749.

The observed shares are \(q_1/Q=0.10/0.30=1/3\) and \(q_2/Q=0.20/0.30=2/3\). Hence

\mu_1=0.3566749\left(\frac13\right)=0.1188916,\qquad \mu_2=0.3566749\left(\frac23\right)=0.2377833.

Convert each force to its single-decrement probability:

q_1'=1-e^{-0.1188916}=1-0.70^{1/3}=0.1120960, $$ $$ q_2'=1-e^{-0.2377833}=1-0.70^{2/3}=0.2116265.

Thus \(\mu=0.3567\), \(\mu_1=0.1189\), \(\mu_2=0.2378\), \(q_1'=0.1121\), and \(q_2'=0.2116\). As a check, the forces sum to \(0.3566749\), and \(1-e^{-0.3566749}=0.30\), the observed total. The associated probabilities sum to \(0.3237\), which need not equal the observed total: each associated probability is calculated with its cause acting alone.

Worked Example: Three Causes and Unequal Shares

A hypothetical portfolio has observed one-year exit probabilities \(q_1=0.12\), \(q_2=0.08\), and \(q_3=0.05\). Assume constant forces. Calculate each associated probability, giving probabilities to four decimal places and forces to four significant figures.

Method. Use the observed total to determine the common total force, then allocate it according to the three observed shares.

First, \(Q=0.12+0.08+0.05=0.25\), so

\mu=-\ln(1-0.25)=-\ln(0.75)=0.2876821.

The cause shares are \(0.12/0.25=0.48\), \(0.08/0.25=0.32\), and \(0.05/0.25=0.20\). The corresponding cause forces are

\mu_1=0.2876821(0.48)=0.1380874,\quad \mu_2=0.2876821(0.32)=0.0920583,\quad \mu_3=0.2876821(0.20)=0.0575364.

Therefore,

q_1'=1-e^{-0.1380874}=0.1290,\quad q_2'=1-e^{-0.0920583}=0.0879,\quad q_3'=1-e^{-0.0575364}=0.0559.

The forces add to \(0.2876821\), and their associated shares are \(0.48\), \(0.32\), and \(0.20\). The associated probabilities sum to \(0.2728\), not \(0.25\); that difference is consistent with removing the competition in each single-decrement calculation.

Worked Example: A Common Error in Treating Observed Probabilities

Suppose observed probabilities are \(q_1=0.15\) and \(q_2=0.10\). A calculation mistakenly treats each observed probability as if it were already an associated single-decrement probability, setting the forces to \(-\ln(0.85)\) and \(-\ln(0.90)\). Show the consequence and calculate the correct associated probabilities.

Incorrect calculation. The mistaken forces are \(\mu_1=-\ln(0.85)=0.1625189\) and \(\mu_2=-\ln(0.90)=0.1053605\). Their sum is \(0.2678794\). The total probability implied by these competing forces is

1-e^{-(\mu_1+\mu_2)} =1-(0.85)(0.90)=0.235.

This is not the observed total \(0.15+0.10=0.25\). The error is applying the single-decrement conversion separately to observed competing-decrement probabilities.

Correct calculation. Here \(Q=0.25\), so \(\mu=-\ln(0.75)=0.2876821\). The observed shares are \(0.15/0.25=0.60\) and \(0.10/0.25=0.40\). Thus

\mu_1=0.2876821(0.60)=0.1726092,\qquad \mu_2=0.2876821(0.40)=0.1150728.

The associated probabilities are

q_1'=1-e^{-0.1726092}=1-0.75^{0.60}=0.1585, $$ $$ q_2'=1-e^{-0.1150728}=1-0.75^{0.40}=0.1087.

The corrected forces sum to \(0.2876821\), giving total observed probability \(1-e^{-0.2876821}=0.25\). This check confirms that the conversion matches the supplied competing-decrement data.

Zero Total Exits and Boundary Checks

The formula for allocating force uses \(q_j/Q\), so it cannot be used by direct division when \(Q=0\). If all observed probabilities are nonnegative and sum to zero, each \(q_j\) is zero. Under finite nonnegative constant forces, the total force must then be zero, and every cause force and associated probability is zero. The cause shares are undefined in this boundary case, but no allocation is needed.

At the other boundary, \(Q=1\) would require \(\mu=-\ln(0)\), which is infinite. No finite constant total force produces an exact one-year exit probability of one. Thus the finite-force conversion applies for \(0<Q<1\). A value \(Q>1\), or any negative cause probability, is invalid as a set of probabilities and must not be repaired by applying the conversion formula.

Worked Example: Zero Observed Exits

A hypothetical table records \(q_1=0\), \(q_2=0\), and \(q_3=0\) over one year. Under finite, nonnegative constant cause forces, determine the total force, cause forces, and associated probabilities.

The total is \(Q=0+0+0=0\). The total exit probability under constant forces is \(1-e^{-\mu}\). Setting this equal to zero gives \(e^{-\mu}=1\), so \(\mu=0\). Since \(\mu=\mu_1+\mu_2+\mu_3\) and all forces are nonnegative, each force must be zero:

\mu_1=\mu_2=\mu_3=0.

Consequently,

q_1'=1-e^0=0,\qquad q_2'=1-e^0=0,\qquad q_3'=1-e^0=0.

The result is internally consistent: all cause forces sum to zero, and all observed and associated exit probabilities are zero. Do not attempt to calculate \(q_j/Q\); division by zero is not part of the boundary-case method.

Common Exam Traps

Confusing observed and associated probabilities. The observed \(q_j\) includes competition. The associated \(q_j'\) is calculated from the force \(\mu_j\) with other causes removed.
Using \(-\ln(1-q_j)\) for each observed cause. This treats each observed probability as an associated probability. Instead, first calculate \(\mu\) from the total \(Q\), then allocate it by \(q_j/Q\).
Dividing by zero when \(Q=0\). Set all forces and associated probabilities to zero under the stated finite, nonnegative-force model.
Accepting a total above one or an exact total of one without comment. \(Q>1\) is invalid; \(Q=1\) has no finite total force under this model.
Rounding too early. Keep the logarithm and cause forces unrounded while calculating \(q_j'\), then round the final values as requested.

Key takeaway: For \(0<Q<1\), calculate \(\mu=-\ln(1-Q)\), allocate \(\mu_j=\mu q_j/Q\), and obtain \(q_j'=1-e^{-\mu_j}\). When \(Q=0\), all forces and associated probabilities are zero; when \(Q=1\), no finite-force conversion exists.

Check Your Understanding

For each problem, assume constant nonnegative cause forces over a one-year interval. Give probabilities to four decimal places where applicable.

Observed probabilities are \(0.04\) and \(0.06\). Find the total force, both cause forces, and both associated probabilities. (Answer: \(\mu=0.1054\); \(\mu_1=0.04214\), \(\mu_2=0.06322\); \(q_1'=0.0413\), \(q_2'=0.0613\).)
Three observed probabilities are \(0.05\), \(0.10\), and \(0.15\). Find the associated probabilities. (Answer: \(q_1'=0.0577\), \(q_2'=0.1121\), \(q_3'=0.1633\).)
Observed probabilities for two causes are both zero. What are the cause forces and associated probabilities under the finite-force model? (Answer: \(\mu_1=\mu_2=0\) and \(q_1'=q_2'=0\).)
Two observed cause probabilities are \(0.12\) and \(0.18\). Find their associated probabilities. (Answer: \(q_1'=1-0.70^{0.4}=0.1330\), \(q_2'=1-0.70^{0.6}=0.1927\).)
A calculation treats observed probabilities \(0.08\) and \(0.12\) as single-decrement probabilities and assigns forces \(-\ln(0.92)\) and \(-\ln(0.88)\). What total observed probability do those forces imply, and what are the correct associated probabilities? (Answer: the mistaken forces imply \(1-(0.92)(0.88)=0.1904\); the correct associated probabilities are \(1-0.80^{0.4}=0.0854\) and \(1-0.80^{0.6}=0.1253\).)

What You'll Learn