UDD Death Probabilities Across Partial Years

Death Probabilities Within an Integer Age Year

An annual mortality probability gives the chance of death during a whole year of age. Under the uniform distribution of deaths assumption (UDD), that annual probability can also be used to calculate death probabilities over shorter intervals within the year. The key distinction is whether the probability is measured from exact age \(x\), or conditional on the person having survived to a later fractional age.

This tutorial considers only subintervals within the same integer age year: an interval beginning at age \(x+s\) and ending at age \(x+t\), where \(0\leq s<t\leq 1\). Let \(q_x={}_1q_x\) be the probability that a life aged \(x\) dies before age \(x+1\). Assume UDD applies during this year and \(q_x<1\), so survival to any fractional age in the year has positive probability.

Definition: Under UDD, deaths are spread uniformly over the interval from age \(x\) to age \(x+1\). For \(0\leq r\leq1\), the probability that a life aged \(x\) dies by age \(x+r\) is \({}_r q_x=rq_x\), and the probability of surviving to age \(x+r\) is \({}_r p_x=1-rq_x\). Thus the probability of death between ages \(x+s\) and \(x+t\), measured from age \(x\), is \((t-s)q_x\).

Unconditional Probability Over a Partial-Year Interval

“Unconditional” means that the probability is measured for a life initially aged \(x\), without requiring survival to the interval’s starting age. Since UDD assigns a proportion \(t-s\) of the year’s deaths to the interval from \(x+s\) to \(x+t\), the interval’s probability is the same proportion of the annual death probability.

P(s<T_x\leq t)={}_tq_x-{}_sq_x =tq_x-sq_x =(t-s)q_x,\qquad 0\leq s<t\leq1.

Here \(T_x\) is the future lifetime of a life currently aged \(x\), measured in years. The formula is an unconditional probability: its denominator is the original group of lives aged \(x\), not only those who reach age \(x+s\). Under UDD, deaths have no point mass at a particular fractional age, so including or excluding an interval endpoint does not change these probabilities.

Conditional Probability Given Survival to the Interval’s Start

If the question says the life is known to have survived to age \(x+s\), the relevant group is smaller. The probability of dying in the interval must therefore be divided by the probability of surviving to its start. For \(s<t\leq1\), this gives:

P(T_x\leq t\mid T_x>s) =\frac{P(s<T_x\leq t)}{P(T_x>s)} =\frac{(t-s)q_x}{1-sq_x}.

The numerator is the unconditional probability of dying in the specified interval, measured from age \(x\). The denominator is the probability of reaching the interval’s starting age. This conditional probability is not generally equal to \((t-s)q_x\), because the latter is still measured against the original group.

Why the Interval Formula Holds

The result can be seen directly from the life-table interpretation of UDD. Let \(\ell_x\) be the number alive at exact age \(x\), and let \(d_x=\ell_x-\ell_{x+1}\) be the number dying during the year. Since \(q_x=d_x/\ell_x\), UDD places a fraction \(r\) of those deaths in the first \(r\) years of the age interval. The number alive at age \(x+r\) is therefore \(\ell_x-rd_x\).

\ell_{x+r}=\ell_x-rd_x =\ell_x(1-rq_x),\qquad 0\leq r\leq1.

The deaths between \(x+s\) and \(x+t\) are the lives present at \(x+s\) who are no longer present at \(x+t\). Subtracting the two interpolated survivor counts gives the interval count. Dividing by the original number \(\ell_x\) produces the unconditional probability; dividing instead by the number alive at \(x+s\) produces the conditional probability.

\ell_{x+s}-\ell_{x+t} =\ell_x(1-sq_x)-\ell_x(1-tq_x) =\ell_x(t-s)q_x.

Consequently, the interval contains \(\ell_x(t-s)q_x\) deaths. Relative to \(\ell_x\), this is \((t-s)q_x\). Relative to the \(\ell_x(1-sq_x)\) lives who reach \(x+s\), it is \((t-s)q_x/(1-sq_x)\). This count-based derivation also shows why the conditional denominator must reflect survival to the interval’s start.

Worked Calculations

Worked Example: Unconditional and Conditional Interval Probabilities

A hypothetical life table gives \(q_x=0.20\) for the year of age from \(x\) to \(x+1\). Under UDD, calculate (a) the probability that a life aged \(x\) dies between ages \(x+0.25\) and \(x+0.70\), and (b) the probability of death in that interval given survival to age \(x+0.25\). No interest assumption is needed because this is a mortality-probability calculation.

Method. First calculate the interval probability from the original age \(x\). Then divide by the probability of surviving to the interval’s start for the conditional result.

(a) Unconditional probability. The interval length is \(0.70-0.25=0.45\) years. Its probability measured from age \(x\) is:

P(0.25<T_x\leq0.70) =(0.45)(0.20)=0.0900.

(b) Conditional probability. The probability of surviving to age \(x+0.25\) is:

{}_ {0.25}p_x=1-(0.25)(0.20)=1-0.0500=0.9500.

Therefore:

P(T_x\leq0.70\mid T_x>0.25) =\frac{0.0900}{0.9500} =0.0947368\ldots \approx 0.0947.

Check. Survival to age \(x+0.70\) is \(1-(0.70)(0.20)=0.8600\). The decrease in survival from age \(x+0.25\) to \(x+0.70\) is \(0.9500-0.8600=0.0900\), matching the unconditional interval probability. The conditional answer exceeds the unconditional answer because it uses the smaller group who survived to \(x+0.25\). The displayed probabilities are rounded to four decimal places.

Worked Example: Using Life-Table Counts

A hypothetical table has \(\ell_x=2{,}500\) and \(d_x=300\). Assume UDD during the year. Find the unconditional probability of death between ages \(x+0.20\) and \(x+0.65\), and then find the probability of death in that interval given survival to age \(x+0.20\).

Method. Convert the table entries to \(q_x\), use UDD to allocate deaths to the interval, and compare the interval deaths with the appropriate number of lives.

The annual death probability is:

q_x=\frac{d_x}{\ell_x} =\frac{300}{2{,}500}=0.12.

The interval is \(0.65-0.20=0.45\) years long. Under UDD, the number of deaths in it is:

$$ 0.45d_x=0.45(300)=135. $$

Relative to the original \(2{,}500\) lives, the unconditional probability is:

\frac{135}{2{,}500}=0.0540.

To find the conditional probability, first calculate the number alive at age \(x+0.20\):

\ell_{x+0.20} =\ell_x-0.20d_x =2{,}500-0.20(300) =2{,}440.

Of these \(2{,}440\) lives, \(135\) die before age \(x+0.65\). Thus:

P(T_x\leq0.65\mid T_x>0.20) =\frac{135}{2{,}440} =0.0553278\ldots \approx 0.0553.

Check. The conditional probability also follows from \((0.45)(0.12)/(1-(0.20)(0.12))=0.0540/0.9760=0.0553278\ldots\). Both approaches agree. The displayed probabilities are rounded to four decimal places.

Worked Example: Conditioning at a Later Fractional Age

For a hypothetical insured population, suppose \(q_x=0.25\) under UDD. For a life known to be alive at age \(x+0.60\), calculate the probability of death by age \(x+0.90\). Also calculate the probability of surviving to age \(x+0.90\), conditional on being alive at \(x+0.60\).

Method. The requested death probability is conditional on survival to \(x+0.60\), so use the interval probability in the numerator and survival to \(x+0.60\) in the denominator.

The interval length is \(0.90-0.60=0.30\) years. Its unconditional probability from age \(x\) is:

P(0.60<T_x\leq0.90) =(0.30)(0.25)=0.0750.

The probability of surviving from age \(x\) to \(x+0.60\) is:

$$ P(T_x>0.60)=1-(0.60)(0.25)=0.8500. $$

The conditional probability of death in the interval is therefore:

P(T_x\leq0.90\mid T_x>0.60) =\frac{0.0750}{0.8500} =0.0882353\ldots \approx 0.0882.

Conditional survival to age \(x+0.90\) is the complement:

1-0.0882353\ldots =0.9117647\ldots \approx 0.9118.

Check. The survival probabilities from age \(x\) are \(0.8500\) at \(x+0.60\) and \(1-(0.90)(0.25)=0.7750\) at \(x+0.90\). Their ratio is \(0.7750/0.8500=0.9117647\ldots\), as required. The conditional death and survival probabilities sum to \(1.0000\). Results are rounded to four decimal places.

Common Mistake: Using the Wrong Denominator

A frequent error is to report \((t-s)q_x\) as the conditional probability after being told that the life survived to age \(x+s\). That expression is correct only for the unconditional probability from age \(x\). For example, in the first worked example, using \(0.45(0.20)=0.0900\) as the conditional answer is wrong. It counts deaths per original life aged \(x\), rather than per life known to have reached \(x+0.25\). The correction is to divide by \(1-0.25(0.20)=0.9500\), giving \(0.0900/0.9500=0.0947\).

A second check is to keep the interval length separate from the starting fraction. The numerator uses \(t-s\), while the survival denominator uses \(s\). Substituting \(t\) for the interval length, or using \(1-tq_x\) as the conditioning probability, mixes up the event endpoint and the conditioning event.

Key takeaway: Under UDD, the unconditional probability of death between fractional ages \(x+s\) and \(x+t\) is \((t-s)q_x\). Given survival to \(x+s\), divide that interval probability by \(1-sq_x\). Always identify the group in the denominator before calculating.

Check Your Understanding

Assume UDD in each question. Give probabilities to four decimal places where needed.

A life has \(q_x=0.16\). Find the unconditional probability of death between ages \(x+0.20\) and \(x+0.75\). (Answer: 0.0880)
For \(q_x=0.10\), find the probability of death between ages \(x+0.35\) and \(x+0.80\), conditional on survival to age \(x+0.35\). (Answer: 0.0466)
A hypothetical table has \(\ell_x=1{,}800\) and \(d_x=270\). Find the number of deaths between ages \(x+0.10\) and \(x+0.60\), and the conditional probability of death in that interval given survival to \(x+0.10\). (Answer: 135 deaths; 0.0761)
Let \(q_x=0.22\). Find the probability of death between ages \(x+0.30\) and \(x+0.90\), conditional on survival to age \(x+0.30\). (Answer: 0.1413)
Under UDD, the conditional probability of death between \(x+0.25\) and \(x+0.75\), given survival to \(x+0.25\), is \(0.10\). Determine \(q_x\). (Answer: 0.1905)

What You'll Learn