Associated Probabilities and Decrement Forces

Why Probabilities and Forces Must Be Kept Separate

In an associated single-decrement table, a cause is considered as if it were the only decrement operating. Its annual probability describes whether that decrement occurs during a year in the associated model. A decrement force describes the instantaneous rate of exit at a particular time, conditional on remaining in that model until then. They are related, but they are not interchangeable: the relationship depends on how exits are distributed within the year.

This distinction matters when converting an associated table into probabilities for a population subject to competing causes. The annual associated probabilities alone do not specify when exits occur during the year. UDD and constant force make different within-year assumptions, so they generally produce different observed probabilities even when they start with the same annual associated probabilities.

Definition: For cause \(j\), let \(q'_j\) be its associated probability of decrement during one year when all other causes are removed. Let \(\mu'_j(t)\) be its associated decrement force at time \(t\) within that year, conditional on no decrement of cause \(j\) before time \(t\). The annual probability \(q'_j\) is dimensionless; the force \(\mu'_j(t)\) is a rate per year and may vary with \(t\).

The prime marks an associated, single-decrement quantity. By contrast, \(q_j\) denotes the probability of an observed exit from cause \(j\) when all competing causes are present. In general, \(q_j\ne q'_j\): a person may leave from another cause before cause \(j\) can occur. The force also depends on the model. Under UDD it varies over the year, while under constant force it is constant.

Associated Force Under UDD

Under the uniform distribution of decrements (UDD) assumption for cause \(j\), exits from that cause are uniformly distributed over the year in its associated model. If the associated annual probability is \(q'_j\), the associated survival probability to time \(t\), for \(0\le t\le1\), is \(1-tq'_j\). The associated force follows from the definition of a force as the instantaneous decrement rate among those still exposed.

\mu'_j(t) = -\frac{d}{dt}\log(1-tq'_j) = \frac{q'_j}{1-tq'_j}, \qquad 0\le t<1.

Thus, under UDD, the annual probability is not itself the force. The force starts at \(q'_j\) at the beginning of the year and increases as the associated survivors are depleted. For \(q'_j>0\), its value at the end of the year is interpreted as a left-hand limit; the formula is not evaluated at \(t=1\) when the survival probability is zero.

Write associated survival.
Under UDD, the probability of surviving cause \(j\) alone to time \(t\) is \(1-tq'_j\).

Differentiate survival.
The associated decrement density is \(-d(1-tq'_j)/dt=q'_j\).

Divide density by survival.
The force is \(q'_j/(1-tq'_j)\), not a constant equal to \(q'_j\).

The derivation uses the standard relation between a survival function \(S(t)\) and its force: \(\mu(t)=-S'(t)/S(t)\) whenever \(S(t)>0\). Here, \(S(t)=1-tq'_j\), so \(-S'(t)=q'_j\), giving the stated force. Integrating that force over the year recovers the annual probability:

1-\exp\left(-\int_0^1\mu'_j(t)\,dt\right) = 1-\exp\left(-\int_0^1\frac{q'_j}{1-tq'_j}\,dt\right) = 1-\exp\bigl(\log(1-q'_j)\bigr) = q'_j.

Worked Example: Reading the UDD Force Within a Year

A hypothetical disability decrement has annual associated probability \(q'_1=0.20\). Assume UDD for the associated decrement throughout a one-year interval, with no other decrement in this single-decrement model. There is no interest assumption involved. Find the associated force at time \(t=0.5\), and compare it with the annual associated probability.

Method. Use the UDD force formula at the specified time. The common mistake to avoid is treating the annual probability as a constant force.

At \(t=0.5\), associated survival is \(1-(0.5)(0.20)=0.90\). The associated force is therefore

\mu'_1(0.5) = \frac{0.20}{1-(0.5)(0.20)} = \frac{0.20}{0.90} = 0.2222\text{ per year}.

The annual associated probability is \(0.20\), not \(0.2222\). The latter is an instantaneous rate at the midpoint. As a check, the force at the start of the year is \(0.20/(1-0)=0.20\) per year, which is lower than the midpoint force, as expected because fewer associated lives remain at the midpoint.

Observed Probabilities Under UDD Competing Decrements

Suppose there are \(m\) causes, each with a UDD associated decrement model, and the associated decrements operate independently in the construction of the multiple-decrement model. For cause \(j\), the associated decrement density at time \(t\) is \(q'_j\). Each other cause \(k\) has associated survival \(1-tq'_k\) to time \(t\). An observed exit from cause \(j\) at time \(t\) requires survival from every other cause up to that time.

Theorem: Under independent UDD associated decrements, the observed annual probability of exit from cause \(j\) is

q_j = q'_j\int_0^1\prod_{k\ne j}(1-tq'_k)\,dt.

The product inside the integral accounts for survival from all competing causes. When there are exactly two causes, the product contains one factor, so the integral gives a simple exact expression:

q_1=q'_1\left(1-\frac{q'_2}{2}\right), \qquad q_2=q'_2\left(1-\frac{q'_1}{2}\right).

For three or more causes, the product has higher-order terms. The two-cause expression must not be applied as an exact general formula. For example, with three causes, the integral for cause 1 includes the product of the two competing survival terms, not merely a first-order adjustment.

Worked Example: Three UDD Causes and the Full Product

A hypothetical policy population has three competing exit causes with annual associated probabilities \(q'_1=q'_2=q'_3=0.20\). Assume independent UDD associated decrements over a one-year interval. There is no interest assumption, and there is no decrement outside these three causes. Find the observed annual probability for cause 1 and check the total probability of exit.

Method. Use the exact product-integral formula. The common mistake to avoid is using the exact two-cause shortcut when there are three causes.

For cause 1, the other two associated survival terms are both \(1-0.20t\). Thus,

\begin{aligned} q_1 &=0.20\int_0^1(1-0.20t)^2\,dt\\ &=0.20\int_0^1(1-0.40t+0.04t^2)\,dt\\ &=0.20\left(1-0.20+\frac{0.04}{3}\right)\\ &=0.20(0.8133333\ldots)\\ &=0.1626667\ldots. \end{aligned}

By symmetry, \(q_2=q_3=0.1626667\ldots\). The probability of no decrement is the product of the three associated survivals to the end of the year: \(0.80^3=0.512\). The total exit probability is therefore \(1-0.512=0.488\). The cause probabilities sum to \(3(0.1626667\ldots)=0.488\), confirming the calculation.

A first-order shortcut would give \(0.20(1-0.20)=0.16\), which is not exact. The missing term comes from multiplying the two competing survival factors before integrating.

Associated Force Under Constant Force

Under the constant force assumption for cause \(j\), its associated force is a constant \(\mu'_j\) throughout the year. Its associated survival to time \(t\) is \(\exp(-\mu'_jt)\). At the end of the year, the associated probability of decrement is one minus survival. Hence the conversion in either direction is:

q'_j=1-e^{-\mu'_j}, \qquad \mu'_j=-\log(1-q'_j).

This conversion applies when the associated decrement has constant force over the year. It is not the UDD force formula: under UDD, the force is \(q'_j/(1-tq'_j)\), which varies with time. The two models can share the same annual associated probability while assigning different timing to exits within the year.

For multiple causes under constant forces, the total force is \(\sum_k\mu'_k\). The probability of no exit from any cause during the year is \(\exp(-\sum_k\mu'_k)\). Cause \(j\)'s observed probability is its share of the total force multiplied by the probability of any exit:

q_j = \frac{\mu'_j}{\sum_k\mu'_k} \left(1-e^{-\sum_k\mu'_k}\right), \qquad \sum_k\mu'_k>0.

This result follows by integrating the cause-\(j\) exit density. At time \(t\), survival from every cause is \(\exp(-t\sum_k\mu'_k)\), and the instantaneous cause-\(j\) rate is \(\mu'_j\). Therefore, \(q_j=\int_0^1\mu'_j\exp(-t\sum_k\mu'_k)\,dt\), which evaluates to the displayed expression. If all forces are zero, all observed exit probabilities are zero; the ratio form is not needed.

Worked Example: Converting Associated Probabilities Under Constant Force

A hypothetical contract has two competing exit causes, disability and surrender, with annual associated probabilities \(q'_1=0.10\) and \(q'_2=0.20\). Assume constant force for each cause over a one-year interval, and no other cause of exit. There is no interest assumption. Find both associated forces and the observed annual cause probabilities.

Method. First convert each associated probability to its constant associated force; then use the force-share formula. The common mistake to avoid is using the associated probabilities directly as force shares.

The associated forces are

\begin{aligned} \mu'_1&=-\log(1-0.10)=-\log(0.90)=0.1053605,\\ \mu'_2&=-\log(1-0.20)=-\log(0.80)=0.2231436. \end{aligned}

The total force is \(0.1053605+0.2231436=0.3285041\) per year. The probability of any exit is \(1-e^{-0.3285041}=1-(0.90)(0.80)=0.28\). The force share for cause 1 is \(0.1053605/0.3285041=0.3207\), and the share for cause 2 is \(0.2231436/0.3285041=0.6793\). Therefore,

q_1=(0.3207)(0.28)=0.0898, \qquad q_2=(0.6793)(0.28)=0.1902.

As a check, \(q_1+q_2=0.2800\), equal to the probability of any exit. Also, the observed cause probabilities are below their associated probabilities, since each cause competes with the other.

Worked Example: Recovering Associated Forces from Observed Exits

In a hypothetical one-year study with two competing causes, the observed probabilities are \(q_1=0.12\) and \(q_2=0.08\). Assume constant forces for both causes and no other exits. There is no interest assumption. Recover the associated forces and associated annual probabilities.

Method. Under constant force, first find the total force from the probability of no exit; then allocate it in proportion to the observed cause probabilities. The common mistake to avoid is equating an observed probability directly with an associated probability.

The probability of no exit is \(1-0.12-0.08=0.80\), so the total force is

\mu'_1+\mu'_2=-\log(0.80)=0.2231436.

The observed cause shares are \(0.12/0.20=0.60\) and \(0.08/0.20=0.40\). Thus,

\begin{aligned} \mu'_1&=0.60(0.2231436)=0.1338861,\\ \mu'_2&=0.40(0.2231436)=0.0892574. \end{aligned}

Convert each force to its associated annual probability:

\begin{aligned} q'_1&=1-e^{-0.1338861}=0.1253,\\ q'_2&=1-e^{-0.0892574}=0.0854. \end{aligned}

To check, the cause shares of the total force are \(0.60\) and \(0.40\), and the probability of any exit is \(0.20\). The resulting observed probabilities are \(0.60(0.20)=0.12\) and \(0.40(0.20)=0.08\), as given.

Exam Checks and Common Traps

Before using a conversion, identify which quantities are associated and which are observed, and state the within-year assumption. The annual associated probability is an endpoint probability for a single-decrement model; the associated force is an instantaneous rate. In particular:

Under UDD, use \(\mu'_j(t)=q'_j/(1-tq'_j)\); do not treat \(q'_j\) as a constant force.
Under constant force, use \(\mu'_j=-\log(1-q'_j)\); do not use this as the UDD force at every time.
For UDD competing decrements, retain the full product over the other causes. The two-cause simplification is not exact for three or more causes.
Check the sum of observed cause probabilities against the probability of at least one exit.

These distinctions are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. They are especially useful when a problem gives associated probabilities but asks for observed competing-cause probabilities, or gives observed exits and asks for an associated constant-force basis.

Check Your Understanding

Use the stated within-year assumption in each problem. Probabilities in the answers are rounded to four decimal places unless an exact expression is given.

A single associated decrement has \(q'_1=0.25\) under UDD. Find its force at \(t=0.4\). (Answer: \(0.2778\) per year.)
Two causes have UDD associated probabilities \(q'_1=0.10\) and \(q'_2=0.40\). Find both observed annual probabilities and the probability of no exit. (Answer: \(q_1=0.0800,\ q_2=0.3800\), and no exit \(=0.5400\).)
Two causes have constant associated forces and annual associated probabilities \(q'_1=0.20\) and \(q'_2=0.10\). Find the observed annual probability for cause 1. (Answer: \(0.1902\).)
Three causes have UDD associated probabilities \(0.10,\ 0.20,\ 0.30\). Find the observed annual probability for cause 1 using the full product integral. (Answer: \(0.0770\).)
Under constant force, observed probabilities are \(q_1=0.06\) and \(q_2=0.09\), with no other exits. Find the two associated annual probabilities. (Answer: \(q'_1=0.0629,\ q'_2=0.0929\).)

What You'll Learn