CS2 · explained

The survival function and the hazard rate, explained

Four functions describe the same lifetime, and CS2 expects you to move between them without hesitating. This page sets out what the survival function, the density, the hazard rate and the integrated hazard each mean, and shows that any one of them determines the other three.

CS2 · Risk Modelling & Survival Analysis · 11 min read

After this page you can

Move between the survival function, the density, the hazard rate and the integrated hazard, and compute a survival probability from any one of them

The idea, before any symbols

Suppose you are watching a group of people, or policies, or machines, and waiting for something to happen to each of them. There are two quite different questions you might ask, and the whole of survival modelling turns on keeping them apart.

The first is how many are still going. Out of everyone who started, what fraction has not yet had the event by the time five years have passed? That is a proportion, it starts at 1 and falls towards 0, and it is what most people mean when they say "survival".

The second is how dangerous it is right now for somebody who has made it this far. Not "what fraction will fail", but "of those still standing at this instant, at what rate are they failing?" That is a different quantity altogether. It need not fall as time passes; for human mortality after about age thirty it rises steadily, and for a newly installed machine it often falls.

The survival function answers the first question and the hazard rate answers the second. Almost every mistake in this part of CS2 is one being used where the other is meant.

Why the hazard rate has to exist

You might reasonably ask why a second function is needed when the first already contains all the information. The answer is that the hazard rate is the one that can be modelled.

Nobody has a theory of what fraction of a population will be alive at age 73. People do have theories about the force acting on a life at an instant: that it rises roughly exponentially with age in adulthood, that it is raised by a multiplicative factor by smoking, that it jumps at the moment of surgery and then decays. Those are statements about the hazard rate, and they are statements you can write down, fit to data and test. The survival function is then whatever those statements imply. That is the direction the whole subject runs in: assume something about the hazard, derive the survival.

The second reason is censoring, which the next topic is about. A life still alive when a study ends tells you nothing about when it will die, but it does tell you it was exposed to the hazard for the whole time it was watched. The hazard rate is the quantity that can use that life; a raw proportion cannot.

The four functions

Let T be the future lifetime, measured from whatever starting point the question defines — birth, entry to a study, the start of a policy, a diagnosis.

The survival function S(t) is the probability that the event has not happened by time t.

S(t) = P(T > t)

In words: the proportion of the original group still going at time t. It equals 1 at t = 0, never increases, and tends to 0.

The density f(t) is the rate at which events are happening among the original group.

f(t) = −d/dt S(t)

In words: how steeply the survival curve is falling. The minus sign is there only because S is decreasing and a density cannot be negative.

The hazard rate h(t), called the force of mortality and written μx when the time scale is age, is the rate at which events happen among those still at risk.

h(t) = f(t) / S(t)

In words: take the rate of events in the whole original group and divide by the fraction still eligible to have one. That division is the entire difference between f and h, and it is why h can rise while f falls: the numerator is shrinking but the denominator is shrinking faster.

The integrated hazard H(t), also called the cumulative hazard, is the hazard added up over the whole period.

H(t) = ∫0t h(s) ds

In words: the total amount of danger the life has been through by time t. It is not a probability and can exceed 1.

The one identity that ties them together

Everything above collapses into a single relation, and it is the most useful line in the topic:

S(t) = exp(−H(t)) = exp(−∫0t h(s) ds)

It is worth seeing where it comes from, because one pass makes it impossible to misremember. Since h = f/S and f = −S′, we have h(t) = −S′(t)/S(t), which is −d/dt ln S(t). Integrate both sides from 0 to t, use S(0) = 1 so that ln S(0) = 0, and exponentiate. Three lines.

The consequence is the thing to carry: any one of the four functions determines the other three. Give a question a hazard and it has given you a survival function. Give it a survival function and it has given you a hazard. There is never a need to ask which one the question "really" supplied.

The named models, and what each one assumes

A parametric survival model is a choice of shape for h(t). Three appear repeatedly.

Constant hazard: the exponential model

h(t) = λ, so H(t) = λt and S(t) = e−λt. The danger does not change with time. This is the simplest possible model and it has one famous property: it has no memory. A life that has survived ten years faces exactly the hazard a new life faces. For human mortality over any long span that is plainly wrong, which is precisely why it is useful as a baseline to measure a better model against.

Power-law hazard: the Weibull model

h(t) = λγtγ−1, giving H(t) = λtγ and S(t) = exp(−λtγ). One extra parameter buys a direction of travel: γ above 1 gives a hazard that rises, γ below 1 one that falls, and γ exactly 1 returns the exponential. It is the standard model wherever the hazard changes smoothly but you have no theory for how.

Exponentially rising hazard: the Gompertz model

μx = Bcx with c > 1. The force of mortality multiplies by the same factor every year. This is the empirical shape of adult human mortality and it is the model the worked example below uses. Integrating it gives

H = ∫0t Bcx+s ds = Bcx(ct − 1) / ln c

The ln c on the bottom is the signature of this model and is where the algebra usually goes wrong. Adding a constant term, μx = A + Bcx, gives the Makeham model; the A accounts for deaths that have nothing to do with ageing, and it integrates to a plain At.

Two checks that take ten seconds

The bound. S(t) must lie between 0 and 1, must be 1 at t = 0, and must never rise as t grows. An answer outside that has an error you can find without re-reading the algebra.

The light-mortality sanity test. When the hazard is small, S(t) ≈ 1 − H(t), because e−H ≈ 1 − H. So an integrated hazard of 0.08 should give a survival probability near 0.92. If it has given you 0.6, one of the two is wrong.

Questions students ask

Is the hazard rate the same thing as qx?

No, and the difference matters. qx is a probability over a whole year and can never exceed 1. The force of mortality is an instantaneous rate and has no upper limit. They are close in value when mortality is light, which is what makes the confusion hard to spot. Under a constant force across the year, the exact relation is qx = 1 − e−μ, and the worked example on exposed to risk shows how little the two differ at ordinary rates.

Why write the survival function as an exponential rather than just quoting it?

Because the exponential form is what makes the later topics work. Proportional hazards multiplies h by a constant, and under the exponential form that becomes raising S to a power — one line instead of a new derivation. Multi-state models add hazards together, and adding inside an exponential is multiplying outside it. Every convenience later in the paper comes from this one representation.

Can the hazard rate be zero, or infinite?

Zero, yes: a period during which the event cannot happen has zero hazard and contributes nothing to H, so the survival function is flat there. Infinite, only in the limit — a hazard that becomes infinite at some finite age forces the survival function to zero at that age, which is how a model with a maximum attainable age is built.

The risk of dying does not hold still, it compounds

00.010.020.03607080AgeForce of mortalityif it never changed2.4 times the age 60 figure
  • the real force of mortality, multiplying by 1.09 for every year of age
  • what holding it flat at the age 60 level would assume
The force of mortality multiplies by 1.09 for every year of age, so it is 2.4 times as large at 70 as at 60, and about 5.6 times by 80. Hold it flat at its age 60 level, as the simplest model does, and you would say 94.9 of every 100 sixty year olds reach 70. The honest answer is 92.0.

A worked example

A life aged exactly 60 is modelled with a Gompertz force of mortality, μx = Bcx, with B = 0.00003 and c = 1.09. Both parameters are invented for this example. They are not from any published mortality table and are not a mortality assumption anybody should use.

Find the probability that the life survives to 70, three independent ways, and then the probability that it dies between 70 and 80.

  1. The force of mortality at the two ends

    ln c = ln 1.09 = 0.0861777
    c60 = 176.0312920, so μ60 = 0.00003 × 176.0312920 = 0.00528094
    c70 = 416.7300862, so μ70 = 0.00003 × 416.7300862 = 0.01250190

    The force has multiplied by 2.3674 over the decade, which is 1.0910 — the defining property of the model, and a first check that the two figures are consistent.

  2. Route one: the closed form

    H = μ60(c10 − 1) / ln c
    c10 = 2.3673637, so c10 − 1 = 1.3673637
    1.3673637 / 0.0861777 = 15.8667927
    H = 0.00528094 × 15.8667927 = 0.0837916

  3. Route two: as a difference of forces

    Because dμ/dx = μ ln c, integrating the Gompertz force between two ages is the change in the force divided by ln c. Different numbers, same integral:

    H = (μ70 − μ60) / ln c = (0.01250190 − 0.00528094) / 0.0861777
        = 0.00722096 / 0.0861777 = 0.0837915

  4. Route three: numerically

    Splitting the ten years into two million strips and adding μ60+s ds across them gives 0.0837916. The three agree to six decimal places — 0.0837916, 0.0837915, 0.0837916 — and the middle one differs in the seventh only because the two forces were each rounded to eight places before being subtracted. That is rounding, not a mistake, and telling the two apart is most of what checking an answer is. If the first two had disagreed in the fourth decimal place, the third route is the tie-breaker.

  5. The survival probability

    10p60 = exp(−0.0837916) = 0.9196229

    So about 92.0% of such lives reach 70. The light-mortality check holds: H is 0.0838 and 1 − 0.0838 = 0.9162, close to 0.9196 and a little below it, which is the right side for a small positive H.

  6. What a constant force would have said

    Hold the force at its age-60 value for the whole decade and the answer becomes exp(−10 × 0.00528094) = 0.9485608. That is 94.9% against 92.0% — the exponential model overstates ten-year survival by nearly three percentage points, because it ignores the fact that the force more than doubles on the way. On one life that is a rounding difference; on a block of annuities it is a reserving error.

  7. The density and the hazard at age 70, as a consistency check

    The density of the future lifetime at t = 10 is f(10) = 10p60 × μ70:

    f(10) = 0.9196229 × 0.01250190 = 0.0114970

    and dividing it straight back by the survival probability returns 0.0114970 / 0.9196229 = 0.0125019, which is μ70 again. That is h = f/S closing on itself, and it is the cheapest proof that the three quantities belong to the same model.

  8. Dying between 70 and 80

    First the twenty-year integrated hazard:

    c20 = 5.6044108, so H(20) = 0.00528094 × (5.6044108 − 1) / 0.0861777 = 0.2821567
    20p60 = exp(−0.2821567) = 0.7541555
    P(die between 70 and 80) = 10p60 − 20p60 = 0.9196229 − 0.7541555 = 0.1654674

    Check it as a product instead of a difference — survive to 70, then die in the following ten years:

    10p70 = 20p60 / 10p60 = 0.7541555 / 0.9196229 = 0.8200704
    0.9196229 × (1 − 0.8200704) = 0.9196229 × 0.1799296 = 0.1654674

    And check 10p70 independently, by running the Gompertz formula from age 70 instead of dividing: μ70(c10 − 1)/ln c = 0.01250190 × 15.8667927 = 0.1983651, and exp(−0.1983651) = 0.8200704. The same figure to seven places, reached without using the age-60 answer at all.

  9. Reading the three numbers together

    Survival to 70 is 0.9196, survival to 80 is 0.7542, and the chance of dying in the decade between them is 0.1655. The integrated hazard over the first decade is 0.0838 and over the second it is 0.2821567 − 0.0837916 = 0.1983651, nearly two and a half times as much — which is what an exponentially rising force means in plain terms, and the reason a level premium charged across both decades has to build a reserve in the first.

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

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.