From Cause-Specific Rates to a One-Year Probability
A multiple-decrement model follows members of a group who can leave through several mutually exclusive causes. A rate describes the instantaneous pace of exit among those still in the group; a probability describes the chance of an exit over a specified period. Those quantities are related, but they are not interchangeable. In particular, adding annual cause-specific forces gives a total force, not the total annual exit probability.
This tutorial considers a single year and converts cause-specific annual forces into the probability of any exit and the probability of remaining in the group for that year. The calculation is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. Unless stated otherwise, the forces are constant over the year.
The Constant-Force Conversion
The constant-force assumption means that, throughout the year, each cause-specific force stays at its stated value. The member leaves the group upon the first decrement, so the relevant event for remaining is avoiding every cause throughout the year. Cause-specific forces are conditional rates among those still in the group; they are not probabilities for separate populations that can simply be added as annual probabilities.
For a short interval of length \(dt\), the chance of an exit from cause \(j\), conditional on still being exposed at the start of the interval, is approximately \(\mu_x^{(j)}dt\). The chance of an exit from any cause is therefore approximately the sum of those terms. This gives the total force \(\mu_x^{(\tau)}\). Under constant total force, the remaining probability declines exponentially:
The rates in the exponent are annual forces, so their sum is dimensionless over the one-year period. If the period has length \(t\) years and the forces remain constant, the corresponding remaining probability is \(\exp(-t\sum_j\mu_x^{(j)})\). Do not use the one-year formula unchanged for a different exposure period.
Why the Exponential Formula Holds
Let \(S(t)\) be the probability of remaining in the group for \(t\) years from the start of the year, with \(S(0)=1\). During a short interval \(dt\), conditional on still being in the group, the probability of an exit is \(\mu_x^{(\tau)}dt+o(dt)\), where \(o(dt)\) denotes terms that become negligible relative to \(dt\) as \(dt\) approaches zero. Thus the probability of surviving the short interval is \(1-\mu_x^{(\tau)}dt+o(dt)\). It follows that:
Solving this differential equation with \(S(0)=1\) gives \(S(t)=\exp(-t\mu_x^{(\tau)})\). Setting \(t=1\) yields the one-year remaining probability. The total exit probability is its complement because, over the year, exactly one of two outcomes occurs: the member remains in the group or exits from at least one cause.
Worked Example: Two Constant Cause-Specific Forces
A hypothetical membership group has two possible exits during the next year. The constant annual force of exit from cause A is \(0.020\), and from cause B is \(0.030\). Find the probability of remaining in the group for the year and the probability of exiting from any cause. State the method and avoid treating forces as annual probabilities.
Step 1: Add the forces. The total annual force is \(\mu_x^{(\tau)}=0.020+0.030=0.050\).
Step 2: Find the probability of remaining. Apply the constant-force formula: \({}_1p_x^{(\tau)}=e^{-0.050}=0.9512294\ldots\).
Step 3: Take the complement. \({}_1q_x^{(\tau)}=1-0.9512294\ldots=0.0487706\ldots\). To four decimal places, the probability of remaining is \(0.9512\), and the probability of any exit is \(0.0488\).
Check and common mistake avoided. The probabilities sum to \(0.9512+0.0488=1.0000\). Adding the forces is correct; treating the sum \(0.050\) as the exit probability would give \(0.0500\), which is not the constant-force result.
Rates Reported in Different Units
Before adding cause-specific rates, express every rate in the same time unit and on the same scale. For example, a rate stated as a number of exits per 1,000 exposure-years must be divided by 1,000 to express it per exposure-year. A percentage must be written as a decimal. A unit mismatch can materially change the exponent and therefore the probability.
The calculation also depends on what the word “rate” means. The formula above applies when the supplied values are constant forces of decrement. If a table instead supplies annual decrement probabilities, those are probabilities rather than forces; the appropriate calculation depends on the model and assumptions behind that table. In particular, do not exponentiate a probability as though it were a force, or infer that a central rate is a constant force without an explicit assumption.
Worked Example: Converting Rates Per 1,000 Exposure-Years
For a hypothetical group, the constant annual forces of exit are reported as 14 per 1,000 exposure-years for cause A and 9 per 1,000 exposure-years for cause B. Find the one-year probability of any exit and the probability of remaining. Identify the unit conversion that is essential.
Step 1: Convert to forces per year. Cause A has force \(14/1000=0.014\) per year, and cause B has force \(9/1000=0.009\) per year.
Step 2: Add the converted forces. The total annual force is \(0.014+0.009=0.023\).
Step 3: Convert the force to a remaining probability. \({}_1p_x^{(\tau)}=e^{-0.023}=0.977262\ldots\), which rounds to \(0.9773\) to four decimal places.
Step 4: Find the exit probability. \({}_1q_x^{(\tau)}=1-0.977262\ldots=0.022738\ldots\), which rounds to \(0.0227\) to four decimal places.
Check and common mistake avoided. The rounded probabilities total \(0.9773+0.0227=1.0000\). Using \(14+9=23\) directly in the exponent would confuse a rate per 1,000 with a rate per year and produce an invalid result.
Solving for an Unknown Cause-Specific Force
The same relationship can be rearranged when the total exit probability is known and one cause-specific force is unknown. First convert the total exit probability to a remaining probability. Then take the natural logarithm to recover the total force, and subtract the known cause-specific forces. The resulting unknown force must be nonnegative and cannot exceed the total force.
Worked Example: Recovering a Missing Cause-Specific Force
A hypothetical group has two constant causes of exit. The one-year probability of any exit is \(0.1000\), and the force for cause A is \(0.040\) per year. Find the force for cause B. Report the force to four significant figures and check it by reconstructing the exit probability.
Step 1: Find the remaining probability. \({}_1p_x^{(\tau)}=1-0.1000=0.9000\).
Step 2: Recover the total force. \(\mu_x^{(\tau)}=-\ln(0.9000)=0.1053605\ldots\) per year.
Step 3: Subtract the known force. \(\mu_x^{(B)}=0.1053605\ldots-0.040=0.0653605\ldots\) per year. To four significant figures, this is \(0.06536\) per year.
Step 4: Check by reconstruction. The unrounded forces sum to \(0.040+0.0653605\ldots=0.1053605\ldots\), so the remaining probability is \(e^{-0.1053605\ldots}=0.9000\) and the total exit probability is \(1-0.9000=0.1000\).
Method and common mistake avoided. The method is to take the negative logarithm of the remaining probability, not of the exit probability. Subtracting the known force from the resulting total force gives the unknown force. The answer is positive, as required.
When the Constant-Force Formula Does Not Apply
If cause-specific forces vary during the year, their values at a single point do not determine the one-year probability. The total force at each time is still the sum of the cause-specific forces at that time, but the accumulated force over the year is an integral. In that case, remaining probability is the exponential of the negative integrated total force:
Thus, one may use the constant-force formula only when each stated cause-specific force is constant over the period, or when the data justify an equivalent constant force. If annual probabilities are supplied instead, use the probability framework indicated by the problem rather than silently applying this force-based formula.
Common Exam Traps
- Adding forces and calling the sum a probability. For total force \(0.050\), the one-year exit probability is \(1-e^{-0.050}=0.0488\), not \(0.0500\).
- Using the exit probability in the logarithm. Recover total force as \(-\ln(1-{}_1q_x^{(\tau)})\), not \(-\ln({}_1q_x^{(\tau)})\).
- Mixing units. Convert each value to a common per-year force before summing or exponentiating.
- Confusing rates with decrement probabilities. The exponential conversion is for forces under the stated model; do not apply it automatically to annual probabilities or other rate measures.
- Rounding too early. Keep the total force and exponential at full calculator precision, then round the final probabilities or rates as requested.
The main practical value of the conversion is that it turns a set of cause-specific forces into a consistent all-cause probability for a specified period. That total probability is also a useful check: the probability of remaining and the probability of any exit must add to one. Keep the model, units, and time period explicit, and the calculation is usually short and reliable.
Check Your Understanding
For each problem, assume the stated cause-specific rates are constant forces unless the problem says otherwise. Probabilities are rounded to four decimal places.
- A hypothetical group has annual cause-specific forces \(0.010\) and \(0.040\). Find the total force, the one-year probability of remaining, and the probability of any exit. (Answer: total force \(0.050\); remaining \(0.9512\); exit \(0.0488\).)
- Two causes have forces of 14 and 9 per 1,000 exposure-years. Convert the rates and find the one-year exit probability. (Answer: forces \(0.014\) and \(0.009\) per year; exit probability \(0.0227\).)
- A group has three constant cause-specific forces, \(0.025\), \(0.015\), and \(0.010\) per year. Find the probability of remaining for one year and the probability of any exit. (Answer: remaining \(0.9512\); exit \(0.0488\).)
- The one-year probability of any exit is \(0.0800\). One of two constant cause-specific forces is \(0.025\) per year. Find the other force to four significant figures. (Answer: \(-\ln(0.9200)-0.025=0.05838\) per year.)
- During the first half-year, two cause-specific forces are \(0.040\) and \(0.020\) per year. During the second half-year, they are \(0.020\) and \(0.040\) per year. Find the probability of remaining for the full year and the probability of any exit. (Answer: integrated total force \(0.060\); remaining \(e^{-0.060}=0.9418\); exit \(0.0582\).)