Risk Modelling and Survival Analysis · Concepts of survival models

Life Table Functions and Notation: lx, dx, qx and px

Updated 11 October 2026 · Fact-checked

A life table lists lx, the expected number of survivors to age x from a starting group, and dx, the deaths between x and x+1. From these you get tpx = lx+t ÷ lx, tqx = 1 − tpx and t|uqx = (lx+t − lx+t+u) ÷ lx. Always convert to l values first.

Understand Life Table Functions and Notation

A life table follows a group of lives from a starting age. The starting group is called the radix, for example l0 = 1,00,000 births or l20 = 1,00,000 lives at age 20. The function lx is the expected number of lives still alive at exact age x. It is not a count of real people. It can be a decimal.

The function dx is the expected number of deaths between exact ages x and x+1. So dx = lx − lx+1. More generally, the deaths between ages x and x+n number lx − lx+n, which equals dx + dx+1 + … + dx+n−1.

The probabilities come from ratios of l values. tpx is the probability that a life aged x survives to age x+t. tqx is the probability that the life dies before x+t. When t = 1 you drop the prefix and write px and qx. Since a life either survives or dies, tpx + tqx = 1.

The deferred mortality probability t|uqx is the probability that a life aged x survives t years and then dies in the next u years. It means: survive to x+t, then die between x+t and x+t+u. If u = 1 you write t|qx. Think of it as a timeline with two cuts. The first cut at x+t is survival. The second cut at x+t+u is death before that point.

Every one of these probabilities is a ratio with lx in the denominator, because we condition on being alive at age x. Once you see that, you never need to memorise separate formulae. You just count survivors at the right ages and divide by lx.

Key rules to remember

Survival probability: tpx = lx+t ÷ lx; Probability that a life aged x survives to age x+t. Denominator is always the starting age.
Death probability: tqx = 1 − tpx = (lx − lx+t) ÷ lx; Probability of dying within t years of age x.
Annual deaths: dx = lx − lx+1 = lx × qx; Expected deaths between exact ages x and x+1.
One-year probabilities: px = lx+1 ÷ lx and qx = dx ÷ lx; These are tpx and tqx with t = 1.
Deferred mortality: t|uqx = tpx × uqx+t = (lx+t − lx+t+u) ÷ lx; Survive t years, then die within the next u years.
Deferred mortality as a difference: t|uqx = t+uqx − tqx = tpx − t+upx; Useful when you are given survival or death probabilities rather than l values.
Multiplication rule for survival: t+upx = tpx × upx+t; Survive to x+t first, then survive u more years from age x+t.
Sum of deaths: lx − lx+n = dx + dx+1 + … + dx+n−1; Use when you are given dx values instead of lx values.

How to solve Life Table Functions and Notation questions

Use this method for any question that asks for a probability from a life table or in actuarial notation.

1Read the notation and write down in words what the event is, for example 'survive 5 years then die within 3 years'.
2Mark the key ages on a timeline: the starting age x, then x+t, then x+t+u if needed.
3Write the probability as a ratio of l values. The denominator is lx at the starting age. The numerator counts the lives in the event.
4If you are given probabilities instead of l values, use tpx = product of one-year px values, and t|uqx = tpx × uqx+t.
5If you are given dx values, rebuild the l values by adding up deaths, or sum the dx values over the age range.
6Substitute the numbers and compute carefully. Keep full decimals until the end.
7Check that your answer lies between 0 and 1 and makes sense. A deferred probability must be smaller than tpx.

Quickest way: Timeline and l-ratio shortcut

When to use it: Use this in MCQs and in the first part of written questions when a life table is given.

Draw a line with ages x, x+t and x+t+u.
Write the survivors needed: survival to x+t is lx+t. Death in the next u years is lx+t − lx+t+u.
Put the answer over lx and divide.
For tqx, use (lx − lx+t) ÷ lx directly instead of 1 − tpx. It saves a step.
For t|uqx, remember the numerator is the difference of the two l values at the cuts.

Common mistakes in Life Table Functions and Notation

Dividing by lx+t instead of lx when computing t|uqx.
Students confuse it with uqx+t, which does use lx+t as the denominator.
Fix: Ask which age the life is known to be alive at. For t|uqx it is age x, so divide by lx. Only use lx+t if the question says 'given alive at x+t'.
Reading t|uqx as the probability of dying between ages t and u.
The notation looks like two ages, but t and u are durations, not ages.
Fix: Read it as: defer t years, then a window of u years. The window runs from x+t to x+t+u.
Using lx − lx+1 for the deaths in the wrong age interval.
Index slips when reading the table, such as using dx+1 for dx.
Fix: Write dx = lx − lx+1 beside your working and underline the ages you pick from the table.
Treating lx as an actual head count and rounding it to a whole number.
The word 'lives' suggests integers.
Fix: lx is an expected value. Keep the decimals given in the table and do not round until the final answer.
Writing t+upx = tpx + upx or tpx × upx.
Students forget the second survival is conditional on reaching age x+t.
Fix: Use t+upx = tpx × upx+t. The second factor starts at age x+t.
Mixing up t|uqx with t+uqx.
Both involve the total period t+u.
Fix: t+uqx includes death in the first t years. t|uqx excludes it. So t|uqx = t+uqx − tqx.

Worked examples

Example 1

A life table gives l60 = 80,000, l65 = 72,000, l66 = 70,560 and l70 = 62,400. For a life aged 60, find (a) 5p60, (b) 5|1q60, and (c) 5|5q60.

Show the solution

(a) 5p60 = l65 ÷ l60 = 72,000 ÷ 80,000 = 0.9.
(b) 5|1q60 = (l65 − l66) ÷ l60 = (72,000 − 70,560) ÷ 80,000.
The numerator is 1,440, so 5|1q60 = 1,440 ÷ 80,000 = 0.018.
(c) 5|5q60 = (l65 − l70) ÷ l60 = (72,000 − 62,400) ÷ 80,000.
The numerator is 9,600, so 5|5q60 = 9,600 ÷ 80,000 = 0.12.

Answer: 5p60 = 0.9, 5|1q60 = 0.018, 5|5q60 = 0.12.

Example 2

For a life aged 40, q40 = 0.002, q41 = 0.003 and q42 = 0.004. Find (a) 3p40, (b) 2|q40, and (c) the expected number of deaths between ages 40 and 43 out of 1,00,000 lives alive at age 40.

Show the solution

(a) p40 = 0.998, p41 = 0.997, p42 = 0.996.
3p40 = 0.998 × 0.997 × 0.996.
0.998 × 0.997 = 0.995006. Then 0.995006 × 0.996 = 0.991026 (to 6 decimal places: 0.995006 × 0.996 = 0.995006 − 0.003980024 = 0.991025976).
So 3p40 ≈ 0.991026.
(b) 2|q40 = 2p40 × q42. First 2p40 = 0.995006.
Then 2|q40 = 0.995006 × 0.004 = 0.003980024, about 0.003980.
(c) The probability of dying within 3 years is 3q40 = 1 − 0.991025976 = 0.008974024.
Expected deaths = 1,00,000 × 0.008974024 ≈ 897.4.

Answer: 3p40 ≈ 0.991026, 2|q40 ≈ 0.003980, and expected deaths ≈ 897.4 out of 1,00,000.

Exam tips

Always write the event in words and draw the age timeline before using any formula. It prevents most notation errors.
In MCQs, check the denominator first. Wrong options are often built by using lx+t instead of lx.
Show the l-ratio form in written answers. Examiners give method marks even if arithmetic slips.
For computer-based Paper B, build the lx column in R or Excel once and use vector indexing to pull survival and deferred probabilities. State the radix and the age offset in your code comments.
Check that your final probability is between 0 and 1 and that t|uqx ≤ tpx.

Practice questions from Concepts of survival models

Life Table Functions and Notation: frequently asked questions

What is the difference between qx and dx?

qx is a probability: the chance a life aged x dies before x+1. dx is an expected number of deaths between x and x+1 in the life table group. They are linked by dx = lx × qx.

How do I calculate t|uqx from a life table?

Take lx+t minus lx+t+u and divide by lx. This is the probability of surviving t years and then dying within the next u years. Make sure the denominator is lx, not lx+t.

Is t|uqx the same as t+uqx?

No. t+uqx is the probability of dying within t+u years, including the first t years. t|uqx only counts deaths in the window after the first t years. So t|uqx = t+uqx − tqx.

What does the radix of a life table mean?

The radix is the starting value of lx, such as l0 = 1,00,000. It is only a scale. The probabilities do not depend on it, because they are ratios of l values.