Why Interpolation Assumptions Matter
A decrement table commonly gives the probability that a member of a group exits during a full year. It does not, by itself, specify how exits are distributed within that year. When a calculation needs survival or exits after only part of a year, an interpolation assumption is required. Two standard assumptions are uniform distribution of decrements (UDD) and constant force of decrement.
Both methods match the same whole-year probability, but they describe different patterns of exits during the year. Consequently, a fractional-year survival estimate under one method will generally differ from the estimate under the other. This tutorial compares the assumptions for a single decrement probability, or for the total exit probability in a decrement group. It does not allocate total exits among separate causes.
The Two Fractional-Year Models
Under UDD, exits are spread uniformly over the year among the members present at its start. Equivalently, conditional on exiting during that year, the time of exit is uniformly distributed over the year. The proportion of the original group that has exited by time \(t\) is therefore \(tq_x\).
Under the constant-force assumption, the force of decrement is constant throughout the year. For \(q_x<1\), its value is chosen so that the one-year survival probability is \(p_x\). Thus the fractional-year survival probability is exponential in duration.
The superscript CF identifies the constant-force interpolation. The \(q_x\) in these formulas is a one-year probability, not a force. Under constant force, the corresponding force is \(-\ln(1-q_x)\); using \(q_x\) itself as the force would not, in general, reproduce the given annual probability.
| Feature | UDD | Constant force |
|---|---|---|
| Fractional survival | \(1-tq_x\) | \((1-q_x)^t\) |
| Fractional exit probability | \(tq_x\) | \(1-(1-q_x)^t\) |
| Within-year description | Exits occur uniformly over the year | The force of decrement is constant |
| At the year end | Survival is \(1-q_x\) | Survival is \(1-q_x\) |
Why the Estimates Differ
The formulas agree at \(t=0\) and \(t=1\), and they also agree throughout the year if \(q_x=0\). For an interior time \(0<t<1\) and a nonzero annual exit probability less than one, the estimates are generally not equal. In that case, UDD gives higher survival and lower fractional exit probability than constant force.
This ordering follows from the weighted arithmetic-geometric mean inequality. Apply it to the positive numbers \(1-q_x\) and \(1\), with weights \(t\) and \(1-t\):
When \(0<t<1\) and \(q_x>0\), the two numbers being averaged are different, so the inequality is strict. Subtracting each survival probability from one reverses the ordering for exit probabilities. This explains the difference, rather than merely giving a numerical comparison.
The assumptions also imply different patterns in time. Under UDD, the fractional exit probability from the start of the year increases linearly with \(t\). Its associated instantaneous force is \(q_x/(1-tq_x)\), which increases during the year when \(q_x>0\). Under constant force, the force remains fixed at \(-\ln(1-q_x)\), so survival follows an exponential curve. The two curves share their endpoints, but generally do not trace the same path between them.
Worked Calculations
Worked Example: Comparing Fractional Survival for a Hypothetical Group
A hypothetical decrement table gives an annual exit probability \(q_x=0.25\). Estimate the probability of remaining in the group for \(0.4\) years under both UDD and constant force, and compare the expected numbers remaining from an initial group of 20,000.
Method. Apply each interpolation formula to the same annual probability and duration. The annual survival probability is \(p_x=1-0.25=0.75\).
UDD survival.
The corresponding exit probability is \(1-0.90=0.10\). From 20,000 members, the expected number remaining is \(20{,}000(0.90)=18{,}000\), and the expected number exited is \(20{,}000(0.10)=2{,}000\).
Constant-force survival.
The constant force is \(-\ln(0.75)=0.287682\ldots\) per year, consistent with the same calculation \(e^{-0.4(0.287682\ldots)}=0.891301\ldots\). The exit probability is \(1-0.891301\ldots=0.108699\ldots\). The expected number remaining is \(20{,}000(0.891301\ldots)=17{,}826.02\), and the expected number exited is \(2{,}173.98\).
Check. Under each method, expected remaining plus expected exited equals 20,000. Constant force estimates about 173.98 fewer members remaining than UDD. Probabilities and expected counts are reported to four decimal places and two decimal places, respectively. The common mistake to avoid is treating the annual exit probability \(0.25\) as a constant force; the force used here is \(-\ln(0.75)\), not \(0.25\).
Worked Example: Exit Probability Between Two Fractional Times
For a hypothetical group with annual exit probability \(q_x=0.20\), find the probability of exiting between \(0.2\) and \(0.7\) years, conditional on being in the group at time \(0.2\). Calculate it under both assumptions.
Method. For either model, conditional exit probability over the interval is one minus the ratio of survival to time \(0.7\) to survival to time \(0.2\). Under UDD, first calculate both survival probabilities from the start of the year:
Thus, conditional on surviving to \(0.2\), the probability of surviving to \(0.7\) is \(0.86/0.96\). The conditional probability of exit is
Under constant force, the survival ratio over the interval is \((1-q_x)^{0.7}/(1-q_x)^{0.2}=(0.8)^{0.5}\). Hence the conditional exit probability is
The answers are 0.1042 under UDD and 0.1056 under constant force, to four decimal places. The constant-force value is larger, as expected from the survival ordering. A common error is to use the exit probability from time zero to \(0.7\) as the answer; that probability does not condition on survival to \(0.2\).
Worked Example: Inferring the Annual Probability From a Midyear Estimate
A hypothetical study reports a fractional-year survival ratio of \(0.90\) at \(t=0.5\). Find the annual exit probability implied by this observation under UDD and under constant force. Then compare the resulting year-end survival probabilities.
UDD method. Set the UDD survival formula equal to the observed value and solve for \(q_x\):
The implied annual survival probability is \(1-0.20=0.80\), so the annual exit probability is 0.20.
Constant-force method. Solve \((1-q_x)^{0.5}=0.90\) by squaring both sides:
The implied annual survival probability is 0.81, and the implied annual exit probability is 0.19. Both methods reproduce the observed midyear survival ratio of 0.90, but they imply different year-end values. This illustrates why an observation within the year does not determine the annual probability without an interpolation assumption. The calculation avoids the trap of applying the UDD linear formula to a constant-force model, or vice versa.
Applying the Assumptions Carefully
Before interpolating, identify the probability represented in the table and the duration required by the problem. In particular, \(q_x\) is the probability of exit over the full year, while \({}_tq_x\) is the probability of exit over only the first \(t\) years. They are not interchangeable. The relevant model should be stated before substituting values.
Confirm which group and decrement the table describes, and record its one-year exit probability \(q_x\).
Use UDD if exits are assumed uniformly distributed through the year; use constant force if the force is assumed constant.
Use \(1-tq_x\) under UDD or \((1-q_x)^t\) under constant force.
For an interval starting after time zero, divide the later survival probability by the survival probability at the interval's start, then subtract from one to obtain conditional exit.
At \(t=0\), survival must be one; at \(t=1\), it must be \(1-q_x\). For \(0<t<1\) and \(0<q_x<1\), UDD survival should exceed constant-force survival.
When \(q_x=1\), UDD still gives \({}_tp_x=1-t\), but constant force would require an infinite force to produce zero survival at the year end. The usual finite-force formula \((1-q_x)^t\) is therefore stated for \(q_x<1\); the endpoint case must be handled according to the model and the problem's assumptions. For ordinary table probabilities with \(0\leq q_x<1\), both formulas are defined and agree at the full-year endpoint.
Check Your Understanding
For each question, state or use the specified interpolation assumption. Report probabilities to four decimal places where needed.
- A hypothetical group has annual exit probability \(q_x=0.10\). Find the survival probability at \(t=0.6\) under UDD and constant force. (Answer: UDD \(0.9400\); constant force \(0.9387\).)
- A hypothetical table has \(q_x=0.36\). Calculate fractional survival at \(t=0.5\) under each assumption and identify which gives the higher exit probability. (Answer: UDD survival \(0.8200\); constant-force survival \(0.8000\); constant force gives the higher exit probability.)
- For a hypothetical group with \(q_x=0.25\), calculate the probability of exit between \(t=0.4\) and \(t=0.8\), conditional on survival to \(t=0.4\), under each assumption. (Answer: UDD \(0.1111\); constant force \(0.1087\).)
- A hypothetical UDD model gives survival \(0.90\) after \(t=0.25\). Find its annual exit probability and its year-end survival probability. (Answer: annual exit probability \(0.4000\); year-end survival probability \(0.6000\).)
- A hypothetical constant-force model gives survival \(0.88\) after \(t=0.5\). Find its annual survival and exit probabilities. (Answer: annual survival probability \(0.7744\); annual exit probability \(0.2256\).)