Central exposed to risk, and the two mortality rates it produces

CS2 · Risk Modelling & Survival Analysis · 12 min read

1Why a denominator is the hard part

You want to estimate mortality at age 65. You know how many deaths occurred. The difficulty is the denominator: how many lives were there to die?

That question has no single answer, because lives do not sit still. During one calendar year a policyholder may turn 65 in March, so only nine months of their year belongs to age 65. Another may take out a policy in May, so the insurer only observed them for eight months. A third may die in September, so they were only exposed for eight months — and nobody can be exposed after they are dead. A fourth may lapse the policy in June and walk out of the data entirely while still very much alive.

Counting heads cannot cope with any of that. What you need is time: the total amount of life-time spent being aged 65 and under observation. That quantity is the central exposed to risk, written Ecx, and it is measured in years, not people.

Every bit of difficulty in this topic is in that definition, and almost every mark lost in it comes from including a stretch of time that should not be there or leaving one out.

2The principle of correspondence

There is one rule, and it decides every borderline case:

A life must be included in the exposure for exactly those periods during which, had it died, that death would have been counted in the numerator.

Say it the other way round and it becomes a working test. Take any stretch of time and ask: if this life had died then, would I have a death certificate in my data? If yes, the stretch is exposure. If no, it is not.

That single test resolves the awkward cases. A policy that lapsed in June contributes exposure to June and nothing after, because a death in November would never have reached the insurer's records. A life that turned 66 in March contributes to age 65 up to March only, because a death in July would be counted against age 66. A death in September contributes exposure right up to the death, because that death is in the numerator.

A question that looks like it is about arithmetic is usually about this principle, and an answer that states it explicitly before computing anything is answering what was asked.

3Two ways to compute it

Exactly, from the file

If you have each life's dates, add up each one's own exposed period and total them. This is exact, it needs no assumptions, and on a modern policy file it is what software does. It is also what a question with eight lives in a table wants, and the method is worth stating plainly: for each life, find the start and end of the period during which a death would have counted, take the length in years, and add.

By census, from head counts

Historically, and still in questions, you do not have the file — you have a count of how many lives were aged x at each 1 January. Write Px(t) for the number aged x and under observation at time t. Then the exposure is the area under that count:

and with counts only at year ends, the trapezium rule approximates it:

In words: average the head count at the start and the end, and treat that average as having applied all year.

This is an approximation and it is worth being clear about what it assumes: that the count moves roughly linearly between censuses. On a large, stable portfolio that is close to true and the error is negligible. On a small portfolio, or one growing fast, or one where the age definition makes people arrive and leave in bursts, it is not. The worked example below computes the exposure both ways on the same eight lives and the two answers differ by nearly 10% — a useful corrective to the idea that the census formula is simply "the formula".

The age definition has to match

One more trap, and it is the reason real exposure calculations go wrong. "Aged 65" can mean age last birthday, age nearest birthday, or age next birthday, and the deaths data and the census data may be classified differently. If deaths are recorded by age last birthday and the census is taken by age nearest, the two are counting different groups of people and the ratio of them estimates nothing. The principle of correspondence again: the numerator and the denominator must refer to the same lives over the same periods, and adjusting one of them to match the other is part of the calculation, not a technicality.

4The two rates, and how they differ

With a numerator Dx (deaths) and a denominator, there are two different quantities you might want.

The force of mortality

Divide deaths by the central exposed to risk and you estimate the force of mortality, assumed constant across the year of age:

The x + ½ subscript is not decoration. An estimate built from a whole year of exposure is an estimate of the force somewhere in the middle of that year, and quoting it as μx misplaces it by six months. Under the constant-force assumption this estimator is also the maximum likelihood estimator, which is why it is the one used for graduation.

The rate of mortality

If you want qx, the probability that a life aged exactly x dies within the year, you need an initial exposed to risk: every life who was alive at age x gets a full year of credit, whatever happened afterwards, because the probability is a probability for a whole year.

In words: the central exposure already counts each life up to the moment it died; add back the rest of the year for each death, and on average that is half a year each.

The ½ is the assumption that deaths are spread uniformly across the year of age. It is only an approximation, but it is a good one at ordinary ages, and the adjustment is small: with two deaths it adds one year to a denominator of several hundred.

The two rates agree, and that is the check

Under a constant force across the year, the exact relation is

So you can get q̂x two ways — from the initial exposure directly, or from the force through this exponential. At light mortality they agree to several decimal places, as the worked example shows, and seeing them agree is the strongest check available that the exposure has been built correctly. When they disagree beyond rounding, the exposure is wrong, not the formula.

A worked example

Two parts. First the exposure for a small file, computed two ways to show the method and the size of the census approximation's error; then a portfolio-scale calculation producing both mortality rates. Every figure below is invented for this example and is not from any portfolio, investigation or published table.

1. 1Part one: eight lives, age 65, calendar year 2026

   Age here means age last birthday, and all dates fall on the first of a month, so each life's exposure is a whole number of twelfths of a year.

   Life	In the age-65 band from	Until	Why it ended	Months
   1	1 Jan	31 Dec	still 65 at the year end	12
   2	1 Jan	1 Apr	turned 66	3
   3	1 Jul	31 Dec	turned 65 in July	6
   4	1 Jan	1 Sep	died	8
   5	1 May	31 Dec	new policy, already 65	8
   6	1 Oct	31 Dec	turned 65 in October	3
   7	1 Jan	1 Jun	policy lapsed	5
   8	1 Mar	1 Nov	turned 65 in March, died in November	8

   Check each line against the principle of correspondence. Life 7 stops in June because a death in November would never have been reported. Life 4 runs to the death and no further. Life 2 stops at the birthday because a later death would be counted at age 66. Life 8 both starts and stops inside the year.

2. 2Route one: add up by life

   12 + 3 + 6 + 8 + 8 + 3 + 5 + 8 = 53 months
   E^c₆₅ = 53/12 = 4.4166667 years

3. 3Route two: add up by month

   Count how many of the eight lives are in the band in each month and total the counts. January: lives 1, 2, 4 and 7, so four. March: lives 1, 2, 4, 7 and 8, so five. And so on.

   Month	J	F	M	A	M	J	J	A	S	O	N	D
   In the band	4	4	5	4	5	4	5	5	4	5	4	4

   4+4+5+4+5+4+5+5+4+5+4+4 = 53 months
   E^c₆₅ = 53/12 = 4.4166667 years

   The same 53. Adding a rectangle's area by rows or by columns must give the same answer, and if the two totals differ, one of the eight exposure periods has been misread — which is exactly why it is worth doing both on a file small enough to allow it.

4. 4What the census approximation would have said

   At 1 January, the lives in the band are 1, 2, 4 and 7, so P65(0) = 4. At 31 December they are 1, 3, 5 and 6, so P65(1) = 4.

   E^c₆₅ ≈ ½(4 + 4) = 4.0000000 years

   Against the exact 4.4167 — low by 0.4167 years, or 9.4%. The two census counts happen to be equal, so the trapezium rule reports a flat population all year, while in truth the count rose to five in four separate months. On eight lives the approximation is poor; on four hundred, the ups and downs average out and it is fine. That is the honest statement of when to trust it.

5. 5Part two: the whole portfolio

   The investigation covers a much larger block. The counts of lives aged 65 last birthday and under observation were 412 at 1 January 2026 and 429 at 1 January 2027, and two deaths at age 65 last birthday were recorded during 2026. The deaths and the census use the same age definition, so the principle of correspondence is satisfied.

6. 6The central exposed to risk

   E^c₆₅ ≈ ½(412 + 429) = 420.5 years

7. 7The force of mortality

   μ̂_65.5 = 2 / 420.5 = 0.00475624

   Quoted at age 65½, not 65. About 0.48% a year, which is the right order of magnitude for light mortality at that age and a first sign that nothing has gone badly wrong.

8. 8The rate of mortality, route one: through the force

   q̂₆₅ = 1 − exp(−0.00475624) = 1 − 0.99525505 = 0.00474495

9. 9The rate of mortality, route two: through the initial exposure

   E₆₅ = 420.5 + ½(2) = 421.5 years
   q̂₆₅ = 2 / 421.5 = 0.00474496

10. 10The two routes

    0.00474495 and 0.00474496. They agree to seven decimal places, differing by one part in five hundred thousand. Two genuinely different calculations — one exponentiating a force, the other adjusting a denominator by half a death each — and they land on the same number because at this level of mortality the uniform-deaths assumption and the constant-force assumption are describing almost the same thing. If your two figures differ in the third decimal place, the exposure is wrong; the formulae are not the problem.

11. 11Reading the numbers

    The force is 0.00475624 and the rate is 0.00474496. The force is the larger, and it always will be: a probability is capped at 1 while a rate is not, so 1 − e−μ < μ for every positive μ. The gap here is about 0.2% of the figure. At a force of 0.5 it would be 21%, which is why the two must not be used interchangeably at older ages even though they look identical at 65.

    One last caution on the estimate itself, and it is the most important sentence here. Two deaths is two deaths. Treating the count as Poisson, a 95% interval for it runs from about 0.24 to 7.22 deaths, so the force of mortality could be anywhere from

    0.2422 / 420.5 = 0.000576   to   7.2247 / 420.5 = 0.017181

    on this data alone — a range of about thirty-fold around a point estimate of 0.00476. Note that the usual normal interval is no help at all at this count: ±1.96√2/420.5 gives a lower end of −0.0018, and a negative force of mortality is not a possible answer. That is the signal to use the Poisson interval rather than the reflex one.

    So the exposure has been computed to seven decimal places and the mortality rate is not known to within a factor of five. Both statements are true at once, and keeping them apart is the point: exposure is arithmetic and can be exact, while the rate is an estimate from a handful of deaths. Graduation across many ages exists precisely to borrow the strength from neighbouring ages that this single figure cannot supply.

Where this fits in CS2

This page explains one idea from CS2 Risk Modelling & Survival Analysis. The full CS2 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.

Other CS2 topics explained

• The survival function and the hazard rate, explained
• The Kaplan-Meier estimator, and why censoring needs it
• The Cox proportional hazards model, and what a hazard ratio means
• Markov chains: transition matrices, Chapman-Kolmogorov and the stationary distribution
• Bühlmann credibility: how much to trust one risk's own experience
• Extreme value theory: block maxima, thresholds and the generalised Pareto distribution
• Copulas and Sklar's theorem: modelling dependence without touching the margins

Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.