The Constant Force Assumption for Decrements

Why Constant Forces Matter

In a multiple-decrement model, a person or policy can leave the active state for one of several competing causes. The forces assigned to those causes describe how quickly each cause removes an individual who is still active. If each force is constant during the year, the separate forces combine into a particularly simple expression for the probability of remaining active.

The key is to combine forces, not annual decrement probabilities. A force is an instantaneous rate conditional on still being active; an annual probability accumulates the effect of that rate over time. Confusing the two can give an incorrect aggregate survival probability. The constant-force assumption is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Suppose there are $m$ competing causes of decrement during the year of age $x$. For cause $j$, let $\mu_x^{(j)}$ be its force of decrement at duration $t$, conditional on being active immediately before $t$. The constant-force assumption states that each $\mu_x^{(j)}$ is constant for $0\leq t\leq 1$. The aggregate force is the sum of the cause-specific forces: $$ \mu_x^{(\tau)}=\sum_{j=1}^{m}\mu_x^{(j)}. $$ Here $\tau$ denotes termination from the active state by any cause.

The forces are measured per unit of time. If the time variable is in years, a force such as $0.04$ is an annual force. It is not automatically a 4% annual probability. Under this assumption, the probability of remaining active for $t$ years, for $0\leq t\leq 1$, is

{}_tp_x^{(\tau)} =\exp\left(-t\sum_{j=1}^{m}\mu_x^{(j)}\right).

The corresponding probability of at least one decrement by time $t$ is $1-{}_tp_x^{(\tau)}$. This is an aggregate probability: it does not identify which cause produces the decrement. The calculation applies over a period for which the stated forces remain constant; if the forces change over time, the constant-force expression over the whole period does not apply without adjusting for that variation.

Associated Probabilities and Forces

An associated single-decrement table considers one cause on its own, with the other causes removed. In that table, ${}_1q_x^{(j)}$ is the probability of decrement from cause $j$ during the year if cause $j$ were the only decrement operating. Under a constant force for that cause, its associated one-year survival probability is $e^{-\mu_x^{(j)}}$, so

{}_1q_x^{(j)}=1-e^{-\mu_x^{(j)}}, \qquad \mu_x^{(j)}=-\ln\left(1-{}_1q_x^{(j)}\right).

These formulas concern the associated single-decrement table. An associated probability is not the same as the observed probability of that cause when the causes compete. In the multiple-decrement model, every cause acts on the same population that remains active, and a decrement from any cause prevents subsequent decrements from the other causes for that individual. The aggregate survival relation combines the forces directly; determining the probability assigned to each cause requires additional calculations.

Worked Example: Combining Three Constant Forces

Hypothetical scenario. In a model for a one-year period, an active policy can terminate through death, disability, or surrender. Assume the cause-specific forces are constant and equal to $0.040$, $0.025$, and $0.015$ per year, respectively. Find the aggregate force, the probability the policy remains active for the year, and the probability of any termination.

Method. Add the cause-specific forces to obtain the aggregate force, then apply the constant-force survival formula. This avoids treating a force as if it were an annual probability.

Step 1: Add the forces.

\mu_x^{(\tau)} =0.040+0.025+0.015 =0.080\text{ per year}.

Step 2: Calculate one-year survival. With $t=1$ year,

{}_1p_x^{(\tau)} =e^{-1(0.080)} =e^{-0.080} =0.9231163.

Step 3: Calculate the probability of any termination.

1-{}_1p_x^{(\tau)} =1-0.9231163 =0.0768837.

Rounded to four decimal places, the one-year survival probability is $0.9231$, and the probability of any termination is $0.0769$. The check $0.9231163+0.0768837=1.0000000$ confirms that the two aggregate outcomes are complementary.

Common mistake avoided: Adding $0.040+0.025+0.015$ and calling $0.080$ an 8% annual termination probability is incorrect. The sum is a force; the probability of termination is $1-e^{-0.080}$, not $0.080$.

Why the Aggregate Force Is a Sum

The addition rule follows from the conditional rates of exit. Over a short interval of length $h$, the probability of exit from cause $j$, conditional on being active at the start of the interval, is approximately $\mu_x^{(j)}h$. The probability of exit from any cause over that interval is therefore approximately the sum of these terms. The chance of two or more exits within the same very short interval is of smaller order than $h$, so it does not affect the instantaneous force.

Let $S(t)={}_tp_x^{(\tau)}$. Over a short interval, the probability of staying active falls at the aggregate rate. Thus

S(t+h)-S(t) \approx -S(t)\left(\sum_{j=1}^{m}\mu_x^{(j)}\right)h.

Divide by $h$ and take the limit as $h$ approaches zero. Since the forces are constant, this gives the differential equation

\frac{dS(t)}{dt} =-\mu_x^{(\tau)}S(t), \qquad S(0)=1.

Separate variables and integrate from $0$ to $t$:

\int_1^{S(t)}\frac{1}{s}\,ds =-\mu_x^{(\tau)}\int_0^t du, \qquad \ln S(t)=-\mu_x^{(\tau)}t.

Exponentiating yields $S(t)=e^{-t\mu_x^{(\tau)}}$, which is the stated aggregate survival formula. Since $\mu_x^{(\tau)}=\sum_j\mu_x^{(j)}$, the cause-specific forces add inside the exponent. This argument assumes the causes are mutually exclusive routes out of the active state and that the specified forces apply to individuals who are still active.

Worked Example: From Associated Probabilities to Aggregate Survival

Hypothetical scenario. A model has two competing causes. In the associated single-decrement tables, the one-year decrement probabilities are $0.10$ for cause 1 and $0.20$ for cause 2. Assume each associated cause has a constant force during the year. Find the two forces and the aggregate one-year survival probability when both causes operate.

Method. Convert each associated probability to its force using $\mu=-\ln(1-q)$. Add the forces and use the aggregate survival formula. This distinguishes associated probabilities from a competing-model probability.

Step 1: Convert the associated probabilities. For cause 1,

\mu_x^{(1)} =-\ln(1-0.10) =-\ln(0.90) =0.1053605\text{ per year}.

For cause 2,

\mu_x^{(2)} =-\ln(1-0.20) =-\ln(0.80) =0.2231436\text{ per year}.

Step 2: Add the forces and calculate aggregate survival.

\mu_x^{(\tau)} =0.1053605+0.2231436 =0.3285041,

{}_1p_x^{(\tau)} =e^{-0.3285041} \approx 0.7200.

A direct check uses the associated survival probabilities: $e^{-\mu_x^{(1)}}=0.90$ and $e^{-\mu_x^{(2)}}=0.80$. Their product is $0.90(0.80)=0.7200$, agreeing with the calculation from the summed forces. To four decimal places, aggregate one-year survival is $0.7200$, and the probability of any decrement is $1-0.7200=0.2800$.

Common mistake avoided: The sum $0.10+0.20=0.30$ is not the aggregate decrement probability. Those are probabilities from separate associated tables. With both causes operating under the stated constant-force model, aggregate survival is $0.90(0.80)=0.72$, giving aggregate decrement probability $0.28$.

Time Units and the Limits of the Assumption

A force and the time multiplied by it must use reciprocal units. If a force is stated per month, use time in months; if the calculation is in years, first express the force per year. The model also requires the forces to be constant over the period being calculated. A constant force for the first year does not, by itself, establish the force for a later year.

Worked Example: Monthly Forces and a Quarter-Year Check

Hypothetical scenario. During a policy year, the constant monthly forces for two competing causes are $0.006$ and $0.004$ per month. Assume both forces apply throughout the year and the policy begins active. Find the probability of remaining active for one year and for three months. Check that the quarterly result is consistent with the annual result.

Method. Add the monthly forces and use months as the time unit. For the annual calculation, either use 12 months directly or convert the aggregate force to a per-year rate.

Step 1: Combine the forces.

\mu^{(\tau)}=0.006+0.004=0.010\text{ per month}.

Step 2: Calculate one-year survival. One year is 12 months, so

{}_{12}p^{(\tau)} =e^{-12(0.010)} =e^{-0.12} =0.8869204.

The equivalent annual aggregate force is $12(0.010)=0.120$ per year, giving the same expression $e^{-0.120}$. The one-year probability of any decrement is $1-0.8869204=0.1130796$.

Step 3: Calculate three-month survival and check consistency.

{}_{3}p^{(\tau)} =e^{-3(0.010)} =e^{-0.030} =0.9704455.

Under the same constant forces, three successive three-month periods have survival $(0.9704455)^4$, which is approximately $0.8869204$, matching the one-year result. Rounded to four decimal places, survival is $0.8869$ for one year and $0.9704$ for three months; the corresponding decrement probabilities are $0.1131$ and $0.0296$.

Common mistake avoided: Using $t=1$ with the monthly force would calculate one month of exposure, not one year. The annual exponent is $12(0.010)$, because the force is stated per month.

Common Checks and Interpretation

Before applying a constant-force calculation, check that the stated rates are forces rather than annual probabilities, that every force uses the same time unit, and that the assumption applies over the duration being evaluated. The result is aggregate survival only. It does not say how many exits arise from each cause, and an observed multiple-decrement probability for one cause should not be substituted directly as though it were an associated single-decrement probability.

Check the unit: the product of force and duration in the exponent must be dimensionless.
Check the range: $e^{-t\mu_x^{(\tau)}}$ lies between 0 and 1 for $t\geq 0$ and nonnegative forces.
Check complements: aggregate survival plus the probability of any decrement must equal 1.
Check the model: if a cause-specific force varies within the period, the constant-force formula for that cause over the entire period is not justified.

Key takeaway: Under constant cause-specific forces, the aggregate force is their sum and aggregate survival over duration $t$ is $e^{-t\sum_j\mu_x^{(j)}}$. Keep forces distinct from annual probabilities, and keep time units consistent.

Check Your Understanding

For each hypothetical model, apply the constant-force assumption and keep the stated time units consistent.

Two constant forces during a year are $0.020$ and $0.050$ per year. Find one-year aggregate survival and the probability of any decrement. (Answer: survival $0.9324$; any decrement $0.0676$.)
Two associated single-decrement one-year probabilities are $0.04$ and $0.06$. Under constant forces, find aggregate survival when both causes operate. (Answer: $0.96(0.94)=0.9024$, so aggregate decrement probability is $0.0976$.)
Two competing causes have constant monthly forces $0.002$ and $0.003$. Find one-year aggregate survival and one-year probability of any decrement. (Answer: $e^{-12(0.005)}=0.9418$; any decrement $0.0582$.)
Three constant forces are $0.03$, $0.02$, and $0.05$ per year. Find the probability of remaining active for two years, assuming the same forces apply throughout both years. (Answer: $e^{-2(0.10)}=0.8187$.)
A model has aggregate one-year survival $0.85$. Its first cause has constant force $0.070$ per year, and there are two causes in total. Find the constant force for the second cause. (Answer: $-\ln(0.85)-0.070=0.0925$ per year, to four decimal places.)

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