Risk Modelling and Survival Analysis · Concepts of survival models

Expected Future Lifetime and Life Expectancy: Complete and Curtate

Updated 11 October 2026 · Fact-checked

Expectation of life is the average future lifetime of a life aged x. The complete expectation is e°x = ∫0^∞ t px dt, using exact time of death. The curtate expectation is ex = Σ(k≥1) k px, counting only whole years lived. They are linked by e°x ≈ ex + ½.

Understand Expected Future Lifetime and Life Expectancy

Let Tx be the future lifetime of a life aged x. It is a continuous random variable. Its mean is the complete expectation of life, written e°x = E[Tx]. It tells you the average number of years, including fractions of a year, that a life aged x will still live.

You rarely integrate the density directly. Use the survival function instead. Since t px = P(Tx > t), the mean of a non-negative random variable gives e°x = ∫0^∞ t px dt. This works because the area under the survival curve equals the mean. The same idea gives the variance and higher moments.

The curtate future lifetime Kx is the whole number of complete years lived from age x. So Kx = ⌊Tx⌋. It is discrete, taking values 0, 1, 2, ... The curtate expectation of life is ex = E[Kx]. Here Kx ≥ k exactly when Tx ≥ k, so P(Kx ≥ k) = k px. Summing these tail probabilities gives ex = Σ(k=1 to ∞) k px.

Tx = Kx + Sx, where Sx is the fraction of the final year lived. So e°x = ex + E[Sx]. If deaths are spread evenly within each year of age (uniform distribution of deaths), E[Sx] = ½, and e°x ≈ ex + ½. This is an approximation. It is exact only under that assumption, and it is not a general identity.

In a life table, k px = l(x+k) ÷ lx. So ex = Σ l(x+k) ÷ lx over k ≥ 1. The complete expectation can be estimated from a table using the trapezium rule, which gives the same ½ adjustment.

Key rules to remember

Complete expectation of life: e°x = E[Tx] = ∫0^∞ t px dt; Equivalent to ∫0^∞ t · t px · μ(x+t) dt. The survival form is usually quicker.
Curtate expectation of life: ex = E[Kx] = Σ(k=1 to ∞) k px; Sum starts at k = 1, not 0.
Curtate probability mass function: P(Kx = k) = k px × q(x+k) = k| qx; Probability of surviving k years then dying in the next year.
Life table form: ex = Σ(k≥1) l(x+k) ÷ lx; Use the table directly. Stop where l becomes zero.
Relationship under uniform distribution of deaths: e°x ≈ ex + ½; Exact if deaths are uniform within each year of age. Otherwise only an approximation.
Recursion for curtate expectation: ex = px × (1 + e(x+1)); Useful for working backwards from the oldest age.
Recursion for complete expectation: e°x = ∫0^1 t px dt + px × e°(x+1); The first term is the expected time lived in the first year.
Temporary complete expectation: e°x:n| = ∫0^n t px dt; Expected years lived within the next n years.

How to solve Expected Future Lifetime and Life Expectancy questions

Use this method for any question on expected future lifetime.

1Identify whether the question asks for the complete (e°x) or curtate (ex) expectation, or a temporary version.
2Write the survival probabilities you need: t px for the complete case, k px for the curtate case.
3If a mortality law is given, form t px from it, for example t px = exp(−∫ μ). If a life table is given, use k px = l(x+k) ÷ lx.
4Write the formula as an integral (complete) or a sum starting at k = 1 (curtate), with correct limits.
5Evaluate. For recursions, start from the oldest age and work backwards using ex = px(1 + e(x+1)).
6To switch between the two, use e°x ≈ ex + ½ and state that you assume uniform distribution of deaths.
7Check the answer is sensible: it should be positive, below the remaining maximum lifetime, and e°x should be a little above ex.
8State units (years) and the assumption used.

Quickest way: Table sum and half-year adjustment

When to use it: When you have a life table and need the complete expectation fast, or when the question gives ex and asks for e°x.

Add the l(x+k) values for k = 1, 2, ... down to the end of the table.
Divide the total by lx to get ex.
Add ½ to get e°x, and write that you assume deaths are uniform within each year.
For a simple law such as constant force μ, use e°x = 1 ÷ μ directly, with no integration.

Common mistakes in Expected Future Lifetime and Life Expectancy

Starting the curtate sum at k = 0.
Students copy the sum form of other expectations. 0 px = 1 would add 1 to the answer.
Fix: Remember ex = Σ from k = 1. Kx ≥ k has probability k px, and k = 0 adds nothing.
Stating e°x = ex + ½ as an exact result.
The rule is taught so often that the assumption is forgotten.
Fix: Write 'assuming deaths uniformly distributed over each year of age' and use ≈, unless the law is exactly uniform in each year.
Adding ½ the wrong way, as ex = e°x + ½.
Mixing up which measure counts partial years.
Fix: Complete includes the fraction of the final year, so it is larger: e°x ≈ ex + ½.
Integrating t · μ(x+t) instead of using t px.
Students forget the density is t px μ(x+t).
Fix: Use e°x = ∫ t px dt. If you use the density, include both t px and μ(x+t).
Using l(x+k) without dividing by lx.
Table values look like probabilities already.
Fix: Always form k px = l(x+k) ÷ lx before summing.
Wrong recursion: ex = px × e(x+1) with the 1 missing.
Forgetting the year lived by the survivor.
Fix: Survivors live one year plus their further expectation: ex = px(1 + e(x+1)).

Worked examples

Example 1

Mortality follows a constant force μ = 0.04 at all ages. Find the complete expectation of life e°x, and the curtate expectation ex.

Show the solution

Under constant force, t px = e^(−0.04t).
e°x = ∫0^∞ e^(−0.04t) dt = 1 ÷ 0.04 = 25.
Here k px = e^(−0.04k) = v^k with v = e^(−0.04) = 0.960789.
ex = Σ(k≥1) v^k = v ÷ (1 − v) = 0.960789 ÷ 0.039211.
Compute: 0.960789 ÷ 0.039211 = 24.5033 (approx.).
Check: e°x − ex = 0.4967, close to ½ but not exactly, since deaths are not exactly uniform here.

Answer: e°x = 25 years and ex ≈ 24.50 years.

Example 2

From a life table, l60 = 90,000, l61 = 88,200, l62 = 86,000, l63 = 83,000, l64 = 79,000 and l65 = 0 (assume l65 = 0 as the end of the table). Calculate the curtate expectation of life at age 60 and estimate the complete expectation, stating your assumption.

Show the solution

Sum l(60+k) for k = 1 to 5: 88,200 + 86,000 + 83,000 + 79,000 + 0.
88,200 + 86,000 = 174,200; + 83,000 = 257,200; + 79,000 = 336,200.
Divide by l60: e60 = 336,200 ÷ 90,000 = 3.7356 (approx.).
Assume deaths are uniformly distributed over each year of age, so e°60 ≈ e60 + ½.
e°60 ≈ 3.7356 + 0.5 = 4.2356.

Answer: Curtate expectation e60 ≈ 3.74 years. Complete expectation e°60 ≈ 4.24 years, assuming uniform distribution of deaths.

Exam tips

Read whether the question wants complete or curtate. Examiners often ask for both and then the difference.
Always state the uniform distribution of deaths assumption when you add ½. Marks are given for it.
For a constant force μ, e°x = 1 ÷ μ. Use this to check your integration.
In written answers, show the formula in survival-function form before substituting numbers.
Check the table sum ends where l becomes zero, and that you divided by lx.

Practice questions from Concepts of survival models

Expected Future Lifetime and Life Expectancy: frequently asked questions

What is the difference between complete and curtate expectation of life?

Complete expectation e°x is the mean of the exact future lifetime Tx, including fractions of a year. Curtate expectation ex is the mean of Kx, the whole number of years lived. So e°x is slightly larger.

Why does the curtate sum start at k = 1?

Kx takes the value k or more with probability k px. Summing P(Kx ≥ k) for k ≥ 1 gives E[Kx]. The k = 0 term would add 1 wrongly.

Is e°x = ex + 0.5 always true?

No. It holds approximately when deaths are uniformly distributed within each year of age. Under other assumptions, such as constant force within each year, the difference is slightly different.

How do I get e°x from a life table?

Compute ex by summing l(x+k) ÷ lx for k ≥ 1, then add ½ under the uniform deaths assumption. This is the same as applying the trapezium rule to the integral of t px.