The Uniform Distribution of Decrements Assumption

Why the Within-Year Assumption Matters

An associated single-decrement table describes a hypothetical decrement operating alone, even when its annual probability has been obtained in a multiple-decrement setting. The annual probability specifies how many exits occur over a whole year, but not when those exits occur within that year. To calculate at a fractional duration, a further assumption is needed. The uniform distribution of decrements (UDD) assumption supplies one: exits are spread uniformly through the year in the associated single-decrement table.

This assumption concerns the distribution of a particular cause’s exits in its associated table. It does not assert that observed exits from several competing causes are uniformly distributed as a combined group, and it does not by itself convert associated probabilities into observed multiple-decrement probabilities.

Definition: Consider an associated single-decrement table for cause \(j\) over the year from age \(x\) to age \(x+1\). Let \(q_x^{(j)}\) be the probability of decrement from cause \(j\) over the complete year, conditional on being alive and exposed to that decrement at age \(x\). Under UDD, the fraction of the year’s decrement exits that have occurred by time \(t\), for \(0\le t\le 1\), is \(t\). Equivalently, the fractional-year decrement probability is \({}_tq_x^{(j)}=tq_x^{(j)}\), and the associated survival probability is \({}_tp_x^{(j)}=1-tq_x^{(j)}\).

Here \(t\) is measured in years. Thus \(t=0.4\) means four-tenths of the way through the year, not four months unless the time is first converted to years. The annual probability \(q_x^{(j)}\) is fixed for the interval; UDD specifies how that probability accumulates between the integer-age endpoints.

Deriving the Linear Exit and Survival Pattern

The UDD formulas follow directly by considering a cohort of \(l_x^{(j)}\) lives at the start of the year in the associated table. The expected number of exits over the full year is \(l_x^{(j)}q_x^{(j)}\). If exits are uniformly distributed over the year, the fraction of those exits occurring by time \(t\) is \(t\). Subtracting those exits from the initial cohort gives the survivors at time \(t\).

d_x^{(j)}=l_x^{(j)}q_x^{(j)},\qquad \text{exits by time }t=t\,d_x^{(j)},\qquad l_{x+t}^{(j)}=l_x^{(j)}(1-tq_x^{(j)}).

Dividing the number alive at time \(t\) by the number alive at the start gives the fractional-year survival probability. This also shows why the survival curve is linear under UDD: its value falls from \(1\) at \(t=0\) to \(1-q_x^{(j)}\) at \(t=1\), at a constant rate in \(t\).

{}_tp_x^{(j)} =\frac{l_{x+t}^{(j)}}{l_x^{(j)}} =1-tq_x^{(j)}, \qquad {}_tq_x^{(j)}=tq_x^{(j)}.

A further implication helps distinguish UDD from a constant force assumption. The unconditional fraction of the starting cohort exiting in an interval of length \(h\) is \(h q_x^{(j)}\), provided the interval lies within this year. But the conditional probability of exiting during that interval depends on how many lives have survived to its beginning. For \(0\le s\le t\le1\), that conditional probability is

P(\text{decrement in }(s,t]\mid\text{survives to }s) =\frac{(t-s)q_x^{(j)}}{1-sq_x^{(j)}}.

The numerator is the share of the starting cohort’s exits assigned to that interval. The denominator is the share still alive at time \(s\). Consequently, equal-length intervals have equal unconditional exits under UDD, but generally do not have equal conditional exit probabilities given survival to the start of each interval.

Worked Applications

Worked Example: Fractional-Year Survival in an Associated Table

A hypothetical associated single-decrement table has \(5{,}000\) lives at age \(x\), and its annual probability for cause \(j\) is \(q_x^{(j)}=0.18\). Assume UDD. Find the expected exits by time \(0.35\), the number alive then, and the conditional probability of decrement between time \(0.35\) and the end of the year.

Method. Apply the linear exit rule to the full-year exits, then divide the exits remaining after time \(0.35\) by the lives still exposed at that time.

The expected number of exits over the full year is

$$ 5{,}000(0.18)=900. $$

By time \(0.35\), the fraction of those exits is \(0.35\), so the expected exits are

$$ 0.35(900)=315. $$

The expected number alive at time \(0.35\) is therefore

$$ 5{,}000-315=4{,}685. $$

The exits remaining for the rest of the year are \(900-315=585\). Hence the conditional probability of decrement between \(0.35\) and \(1\), given survival to \(0.35\), is

\frac{585}{4{,}685}=0.124866\ldots \approx 0.1249.

Thus there are 315 expected exits by time \(0.35\), 4,685 expected survivors then, and a conditional remaining-year decrement probability of 0.1249. The rounding is to four decimal places. The calculation avoids confusing the remaining unconditional probability \(0.65(0.18)=0.117\) with the probability conditional on still being alive at time \(0.35\).

Worked Example: Conditional Exit Probability Over a Fractional Interval

In a hypothetical associated table, \(2{,}500\) lives are exposed at age \(x\), and \(q_x^{(j)}=0.24\). Under UDD, calculate the probability that a life survives to time \(0.20\) and then exits from cause \(j\) before time \(0.65\), conditional on survival to time \(0.20\). Also find the probability of surviving from time \(0.20\) through time \(0.65\), given survival to time \(0.20\).

Method. First find the number still alive at time \(0.20\). Then use the linear exit pattern to count exits in the interval and condition on the lives alive at its start.

At time \(0.20\), the fraction of the initial cohort that has exited is

$$ 0.20(0.24)=0.048. $$

So the expected number of exits by that time is \(2{,}500(0.048)=120\), leaving \(2{,}500-120=2{,}380\) lives. By time \(0.65\), the fraction of the initial cohort exited is

$$ 0.65(0.24)=0.156, $$

corresponding to \(2{,}500(0.156)=390\) exits. Therefore, the number exiting between \(0.20\) and \(0.65\) is \(390-120=270\). Conditional on survival to \(0.20\), the probability of exit in that interval is

\frac{270}{2{,}380}=0.113445\ldots \approx 0.1134.

The conditional probability of surviving that interval is the complement:

1-0.113445\ldots=0.886554\ldots\approx0.8866.

The results are 0.1134 for exit and 0.8866 for survival, rounded to four decimal places. As a check, the probabilities add to \(1.0000\) after rounding.

Worked Example: Keeping Associated Tables Separate

Two hypothetical associated single-decrement tables are supplied for different causes. Table A starts with \(1{,}200\) lives and has annual probability \(q_x^{(A)}=0.10\); Table B starts with \(800\) lives and has annual probability \(q_x^{(B)}=0.25\). Assume UDD in each table. For each table, find the expected exits from time \(0.40\) to time \(0.80\), and the conditional probability of exit in that interval given survival to time \(0.40\).

Method. Work within each associated table separately. Under UDD, the interval contains \(0.80-0.40=0.40\) of that table’s annual exits. Divide the interval exits by the number alive at time \(0.40\) to obtain the conditional probability.

Table A. Full-year expected exits are

$$ 1{,}200(0.10)=120. $$

Expected exits from time \(0.40\) to \(0.80\) are \(0.40(120)=48\). By time \(0.40\), expected exits are \(0.40(120)=48\), leaving \(1{,}200-48=1{,}152\) lives. The conditional exit probability in the requested interval is

\frac{48}{1{,}152}=0.041666\ldots\approx0.0417.

Table B. Full-year expected exits are \(800(0.25)=200\). Expected exits from time \(0.40\) to \(0.80\) are \(0.40(200)=80\). By time \(0.40\), expected exits are also \(80\), leaving \(800-80=720\) lives. The conditional exit probability in the interval is

\frac{80}{720}=0.111111\ldots\approx0.1111.

The expected interval exits are 48 in Table A and 80 in Table B; the conditional probabilities are 0.0417 and 0.1111, respectively, rounded to four decimal places. These are calculations within separate associated tables. Adding their lives or exits would not, without additional information and a multiple-decrement model, produce a combined observed decrement probability.

UDD Compared With Other Within-Year Patterns

UDD is a distributional assumption about exits, not a claim that the conditional instantaneous rate of exit is constant. For \(q_x^{(j)}<1\), differentiating the UDD survival function gives the associated force at fractional duration \(t\):

\mu_{x+t}^{(j)} =-\frac{d}{dt}\ln({}_tp_x^{(j)}) =\frac{q_x^{(j)}}{1-tq_x^{(j)}}.

The numerator is constant, but the denominator decreases as the year progresses. Thus the force increases with \(t\) when \(q_x^{(j)}>0\), even though the unconditional exits per unit of the original cohort are uniform. A constant force assumption instead specifies a constant conditional instantaneous rate; it does not produce the same linear survival curve.

If \(q_x^{(j)}=0\), there are no exits and survival remains \(1\) throughout the year. If \(q_x^{(j)}=1\), the UDD survival formula gives \({}_tp_x^{(j)}=1-t\), which is valid through the interval and reaches zero at its endpoint. The force expression is only finite for \(t<1\) in this boundary case; the survival and exit formulas remain the appropriate way to handle the interval endpoint.

Common Exam Trap

A frequent error is to apply the annual probability \(q_x^{(j)}\) unchanged to every fraction of the year. Another is to confuse the unconditional interval probability with the conditional probability for someone known to have survived to the interval’s start.

Common mistake and correction: Suppose \(q_x^{(j)}=0.20\), and the question asks for the probability of exit between \(0.25\) and \(0.75\), given survival to \(0.25\). A wrong calculation is to use \(0.20\) as the interval probability, or to report \(0.50(0.20)=0.10\) as the conditional answer. Under UDD, \(0.10\) is the unconditional fraction of the starting cohort exiting during that interval. The fraction surviving to \(0.25\) is \(1-0.25(0.20)=0.95\), so the required conditional probability is \(0.10/0.95=0.1053\), rounded to four decimal places.

A reliable procedure is to identify the associated table and cause first, express every time point as a fraction of the one-year interval, calculate the unconditional exits or survivors at those points, and only then condition if the question specifies survival to an intermediate time. This keeps the UDD calculation separate from the later task of relating associated tables to observed competing decrements. The distinction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

Assume UDD in each stated associated single-decrement table. Give probabilities to four decimal places where needed.

An associated table has \(q_x^{(j)}=0.16\). What is the probability of decrement by time \(0.30\)? (Answer: \(0.0480\))
A table starts with \(1{,}800\) lives and has \(q_x^{(j)}=0.25\). How many are expected to have exited by time \(0.40\), and how many are expected to remain? (Answer: 180 exits; 1,620 survivors)
For \(q_x^{(j)}=0.10\), find the conditional probability of exit between times \(0.30\) and \(0.80\), given survival to \(0.30\). (Answer: \(0.0515\))
An associated table has \(q_x^{(j)}=0.20\). At what fractional time \(t\) is the survival probability \(0.91\)? (Answer: \(t=0.45\) years)
For \(q_x^{(j)}=0.12\), calculate the conditional probability of exit between times \(0.25\) and \(0.75\), given survival to \(0.25\), and state the unconditional probability of exit in that interval from the start of the year. (Answer: Conditional \(0.0619\); unconditional \(0.0600\))

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