UDD Calculations from lx, dx, and qx

Using Life-Table Inputs at Fractional Ages

A life table usually reports values at integer ages, but a calculation may involve someone aged \(x+0.3\) or a duration of \(1.6\) years. Under the uniform distribution of deaths assumption (UDD), the integer-age entries are enough to calculate the corresponding fractional-age survivor count and probability. The essential step is to identify which integer-age year contains each fractional part.

The same calculation may be presented using \(l_x\), \(d_x\), or \(q_x\). These are linked quantities, but they are not interchangeable without conversion: \(l_x\) is the number alive at exact integer age \(x\), \(d_x\) is the number who die between ages \(x\) and \(x+1\), and \(q_x\) is the probability that a life aged \(x\) dies before age \(x+1\). This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: In a life table, \(l_x\) denotes the number alive at exact integer age \(x\), and \(d_x=l_x-l_{x+1}\) denotes the number who die between ages \(x\) and \(x+1\). The one-year death probability is \(q_x=d_x/l_x\), provided \(l_x>0\). Under UDD, deaths are distributed uniformly through each one-year age interval.

UDD Formulas from the Three Life-Table Inputs

Let \(0\le t\le 1\). Under UDD, the interpolated number alive at age \(x+t\) is the straight-line interpolation between \(l_x\) and \(l_{x+1}\). Since \(l_{x+1}=l_x-d_x\), this can be written in three equivalent ways:

l_{x+t}=l_x-t d_x=l_x(1-tq_x), \qquad d_x=l_x-l_{x+1}, \qquad q_x=\frac{d_x}{l_x}.

The probability of surviving from exact age \(x\) to age \(x+t\) is the fraction of the age-\(x\) group still alive at \(x+t\). The probability of dying before reaching \(x+t\) is the complementary fraction:

{}_t p_x=\frac{l_{x+t}}{l_x}=1-tq_x, \qquad {}_t q_x=1-{}_t p_x=tq_x.

These fractional-year formulas apply to a segment within one integer-age year under UDD. If a calculation spans more than one year, calculate the survival factor for each year segment and multiply the factors. Equivalently, use the ratio of the interpolated survivor count at the endpoint to the starting integer-age count. Do not apply a single year’s \(q_x\) to a segment that extends into a later age interval.

Why Linear Interpolation Follows from UDD

UDD says that, among the \(d_x\) deaths in the interval from age \(x\) to age \(x+1\), the proportion that have occurred by fractional time \(t\) is \(t\). Thus, by age \(x+t\), \(t d_x\) of the age-\(x\) group have died, leaving \(l_x-t d_x\) alive. Dividing by \(l_x\) gives the survival probability. Since \(q_x=d_x/l_x\), the same result is \(1-tq_x\).

\frac{l_x-t d_x}{l_x} =1-t\frac{d_x}{l_x} =1-tq_x.

This justification also shows why the formulas agree whether the table supplies \(l_x\) and \(d_x\), or \(l_x\) and \(q_x\): the inputs describe the same one-year deaths and starting population. The formulas require a valid life-table relationship, in particular \(0\le d_x\le l_x\) and \(q_x=d_x/l_x\). Rounding in a printed table can cause small discrepancies, so calculations should use the supplied values consistently and note any stated rounding.

Worked Calculations

Worked Example: Starting with \(l_x\) and \(d_x\)

A hypothetical life table has \(l_{42}=18{,}000\) and \(d_{42}=1{,}260\). Assume UDD between ages 42 and 43. Find \(q_{42}\), the interpolated number alive at age 42.3, the probability of surviving from age 42 to 42.3, and the probability of dying before age 42.3. No interest rate is involved; this is a mortality calculation.

Method. First convert the table entries to \(q_{42}\), then use the UDD interpolation for the fractional part \(t=0.3\).

One-year death probability. Divide deaths in the year by the number alive at its start:

q_{42}=\frac{d_{42}}{l_{42}} =\frac{1{,}260}{18{,}000} =0.07.

Interpolated survivors. By age 42.3, UDD allocates \(0.3\) of the year’s deaths to the elapsed part of the year:

l_{42.3}=l_{42}-0.3d_{42} =18{,}000-0.3(1{,}260) =18{,}000-378 =17{,}622.

Probabilities. The survival probability is the fraction of the original 18,000 still alive. The death probability is its complement:

{}_{0.3}p_{42}=\frac{17{,}622}{18{,}000}=0.979, \qquad {}_{0.3}q_{42}=1-0.979=0.021.

The check using \(q_{42}\) gives \(1-0.3(0.07)=0.979\), and \(0.3(0.07)=0.021\). The values agree. The common mistake this method avoids is treating \(d_{42}=1{,}260\) as a probability; it is a count, and must be divided by \(l_{42}\) to obtain \(q_{42}\).

Worked Example: Starting with \(q_x\) Across an Integer Age

For a hypothetical mortality table, \(q_x=0.08\) and \(q_{x+1}=0.15\). Assume UDD separately in each year. Calculate the probability of surviving from exact age \(x\) to age \(x+1.6\), and the probability of dying during that interval. No interest rate is involved.

Method. The interval contains a full year at age \(x\), followed by \(0.6\) of a year at age \(x+1\). Calculate the survival in each segment and multiply.

First year. The probability of surviving from \(x\) to \(x+1\) is \(1-q_x=1-0.08=0.92\).

Next \(0.6\) year. Under UDD, the probability of surviving that fractional segment, conditional on reaching age \(x+1\), is:

{}_{0.6}p_{x+1}=1-0.6q_{x+1} =1-0.6(0.15) =1-0.09 =0.91.

Combine the segments. Reaching age \(x+1.6\) requires surviving both segments:

{}_{1.6}p_x=(0.92)(0.91)=0.8372.

Therefore, the probability of dying before age \(x+1.6\) is \(1-0.8372=0.1628\). As a check, the probability of dying in the first year is \(0.08\); the probability of surviving that year and then dying in the \(0.6\)-year segment is \(0.92(0.09)=0.0828\). Their sum is \(0.08+0.0828=0.1628\). The common mistake avoided is using \(q_x\) for the \(0.6\)-year segment at age \(x+1\); that segment uses \(q_{x+1}\).

Worked Example: Checking \(l_x\), \(d_x\), and \(q_x\) Together

A hypothetical table gives \(l_x=12{,}000\), \(d_x=600\), and \(d_{x+1}=855\). It also reports \(q_{x+1}=0.075\). Assume UDD in both years. Find the interpolated number alive at age \(x+1.4\) and the probability of surviving from age \(x\) to \(x+1.4\). No interest rate is involved.

Method. Derive the missing one-year probabilities and the integer-age survivor count, then interpolate within the second year. Finally, compare the survivor-count ratio with the product of the segment survival probabilities.

Age \(x\) inputs. The one-year probability at age \(x\) is:

q_x=\frac{d_x}{l_x} =\frac{600}{12{,}000} =0.05.

Thus \(l_{x+1}=l_x-d_x=12{,}000-600=11{,}400\). At age \(x+1\), the supplied values are consistent because \(855/11{,}400=0.075=q_{x+1}\).

Interpolate within the second year. Here the fractional part is \(t=0.4\), so:

l_{x+1.4} =l_{x+1}-0.4d_{x+1} =11{,}400-0.4(855) =11{,}400-342 =11{,}058.

Calculate and check survival. The survivor-count ratio is \(11{,}058/12{,}000=0.9215\). Using the year-by-year factors gives the same result:

(1-q_x)(1-0.4q_{x+1}) =(1-0.05)(1-0.4(0.075)) =(0.95)(0.97) =0.9215.

The probability of dying before age \(x+1.4\) is the complement, \(1-0.9215=0.0785\). The common mistake avoided is interpolating from \(l_x\) using \(d_{x+1}\); \(d_{x+1}\) belongs to the year that starts at age \(x+1\), so its interpolation starts from \(l_{x+1}=11{,}400\).

Practical Checks and Common Traps

A reliable calculation keeps the age interval and the form of the input visible at every step. Before interpolating, verify which integer age starts the relevant year. Then check that the fractional part \(t\) is between 0 and 1, that \(d_x\) is a count, and that \(q_x\) is a probability. The interpolated count should lie between \(l_x\) and \(l_{x+1}\), and survival and death probabilities for the same interval should sum to 1.

One tempting error is to use a different interpolation rule without noticing. For example, if \(q_x=0.20\) and \(t=0.4\), UDD gives:

{}_{{0.4}}p_x=1-0.4(0.20)=0.92.

A calculation such as \((1-q_x)^{0.4}=0.8^{0.4}\) does not implement UDD. It uses a different fractional-age assumption and therefore is not a substitute for the linear UDD formula. Another error is to carry \(q_x\) across an integer-age boundary: the next segment must use the life-table entry for the next age.

Key takeaway: Under UDD, \(l_{x+t}=l_x-t d_x=l_x(1-tq_x)\) for \(0\le t\le1\), and \({}_t p_x=1-tq_x\). Convert the supplied life-table inputs first, use the entry for the correct integer-age year, and multiply survival factors when an interval crosses an integer age.

Check Your Understanding

For each problem, assume UDD within every age-year mentioned. No interest rate is involved.

A hypothetical table has \(l_{50}=12{,}000\) and \(d_{50}=900\). Find \(q_{50}\), \(l_{50.6}\), and the probability of surviving from age 50 to age 50.6. (Answer: \(q_{50}=0.075\), \(l_{50.6}=11{,}460\), \({}_{0.6}p_{50}=0.955\).)
A table gives \(q_x=0.12\) and \(q_{x+1}=0.05\). Find the probability of surviving from exact age \(x\) to age \(x+1.7\), then find the probability of dying during the interval. (Answer: survival \(=0.8492\); death \(=0.1508\).)
A table has \(l_x=5{,}000\) and \(d_x=350\). Find \(q_x\) and the interpolated number alive at age \(x+0.2\). (Answer: \(q_x=0.07\), \(l_{x+0.2}=4{,}930\).)
A table gives \(l_x=9{,}000\), \(q_x=0.08\), and \(q_{x+1}=0.10\). Find \(d_x\), the number expected alive at age \(x+1.5\), and the probability of surviving from age \(x\) to age \(x+1.5\). (Answer: \(d_x=720\), \(l_{x+1.5}=7{,}866\), survival \(=0.874\).)
A table gives \(l_x=20{,}000\), \(d_x=1{,}600\), \(d_{x+1}=1{,}840\), and \(q_{x+1}=0.10\). Check that the age-\(x+1\) entries agree, then find \(l_{x+1.25}\) and the probability of surviving from age \(x\) to age \(x+1.25\). (Answer: \(l_{x+1}=18{,}400\), \(d_{x+1}/l_{x+1}=0.10\), \(l_{x+1.25}=17{,}940\), survival \(=0.897\).)

What You'll Learn