Risk Modelling and Survival Analysis · Transition intensities dependent on age (exact or census)

Two-State Model and Force of Mortality Explained

Q: What is the force of mortality in simple terms?

It is the instantaneous rate of death at age x for a person alive at age x. For a short interval h, the chance of dying is about μx × h. It is measured per year and is not capped at 1.

Q: How do I derive the survival function from the force of mortality?

Integrate the force from age x to x+t to get the cumulative hazard. Then tpx = exp(−cumulative hazard). This comes from solving d/dt tpx = −tpx μ(x+t) with 0px = 1.

Q: What is the difference between μx and qx?

The force μx is an instantaneous rate at exact age x. The quantity qx is the probability of death within one year of age x. They are linked by qx = 1 − exp(−∫ μ(x+s) ds) from 0 to 1.

Q: Is the force of mortality the same as the hazard rate?

Yes. In survival analysis, the hazard rate of the future lifetime random variable is the force of mortality. The cumulative hazard is the integral of the force.

Updated 11 October 2026 · Fact-checked

The two-state model has two states, alive and dead, with one-way movement from alive to dead. The force of mortality μx is the instantaneous rate of death at age x, given survival to x. You find survival by integrating it: tpx = exp(−∫ μ(x+s) ds) from 0 to t.

Understand Two-State Model and Force of Mortality

The simplest survival model has two states: alive and dead. A life starts alive and can only move one way, to dead. There is no return. The only thing that matters is when the move happens.

The model is described by a transition intensity from alive to dead. In this setting it is called the force of mortality, written μx at age x. It measures how fast death is arriving at age x for someone who is alive at age x. It is a rate, not a probability. It has units of 'per year' and can be larger than 1.

Formally, μx = lim (h→0+) of P(dies within h years | alive at x) ÷ h. So for a small h, the probability that a life aged x dies within h years is approximately μx × h. This is where the idea of a rate comes from. In the IAI notation the model assumes the probability of death in a short interval is hμ(x+t) + o(h), where o(h) is a term that goes to zero faster than h.

The force of mortality links to everything else. If T is the future lifetime of a life aged x, then its density is fx(t) = tpx × μ(x+t), and its survival function is tpx = exp(−∫ μ(x+s) ds). So if you know μ, you know the whole lifetime distribution. If you know the survival function, you can recover μ by differentiating its log.

Do not confuse μx with qx. The quantity qx is the probability that a life aged x dies before age x+1. It is a probability between 0 and 1. The force μx is an instantaneous rate at exact age x. They are close when mortality is low and changes slowly, but they are not equal. Under a constant force μ over the year, qx = 1 − e^(−μ).

Key rules to remember

Definition of force of mortality: μx = lim (h→0+) [P(T ≤ h | alive at x)] ÷ h = lim (h→0+) hqx ÷ h; Here T is the future lifetime of a life aged x. It is an instantaneous rate.
Short-interval approximation: hqx ≈ h × μx for small h; Gives the two-state model assumption: P(death in (t, t+h)) = h μ(x+t) + o(h).
Force from survival function: μ(x+t) = −d/dt ln(tpx) = −(d/dt tpx) ÷ tpx; Use for deriving μ from a given survival function.
Survival function from force: tpx = exp(−∫ from 0 to t of μ(x+s) ds); The integral is the cumulative hazard. Note the limits and the age shift x+s.
Density of future lifetime: fx(t) = tpx × μ(x+t); Also fx(t) = −d/dt tpx.
Probability of death in a period: t|uqx = tpx × uq(x+t) = ∫ from t to t+u of spx × μ(x+s) ds; Deferred probability: survive t years, then die within the next u years.
Constant force: If μ(x+s) = μ for 0 ≤ s ≤ t, then tpx = e^(−μt); Future lifetime is then exponential with mean 1/μ.

How to solve Two-State Model and Force of Mortality questions

Most exam questions give you either μ or the survival function and ask for a probability, a density, or the other function. Use this routine.

1Identify what is given: μ(x+t), tpx, the density, or a life table value. Note the age x carefully.
2Write the target in standard notation, such as tpx, tqx, or t|uqx.
3If you are given μ, integrate it from 0 to t, using μ(x+s) with the age shift, to get the cumulative hazard. Then tpx = exp(−cumulative hazard).
4If you are given tpx or the survival function, take the log and differentiate with respect to t, then change sign, to get μ(x+t).
5For a probability of death in a range, use the difference of survival probabilities, or integrate spx × μ(x+s) over the range. Use the multiplication rule s+tpx = spx × tp(x+s) for conditional questions.
6Substitute numbers last. Keep exact forms such as e^(−0.3) until the final step, then give the answer to the requested accuracy.
7Check the answer: probabilities must lie in [0, 1], tpx must decrease in t, and μ must be non-negative.

Quickest way: Cumulative hazard shortcut

When to use it: Use when μ is given as a simple function or is piecewise constant, and you need survival or death probabilities.

Write tpx = exp(−H) where H = ∫ μ(x+s) ds from 0 to t.
If μ is constant over a stretch, H is just μ × length of that stretch.
For piecewise constant μ, add the pieces: H = μ1 t1 + μ2 t2 + ...
For tqx, compute 1 − tpx. For a deferred probability, use tpx × (1 − upx+t).
For a Gompertz or Makeham form, use the known integral rather than integrating again, but check the form of the given μ first.

Common mistakes in Two-State Model and Force of Mortality

Treating μx as the probability of death within a year.
Both measure mortality at age x and have similar values when mortality is low.
Fix: Remember μx is a rate per year, and qx = 1 − exp(−∫ μ(x+s) ds) over 0 to 1. Only for small, slowly changing μ is qx ≈ μx.
Integrating μ(s) from 0 to t instead of μ(x+s).
Students forget the life starts at age x, not age 0.
Fix: Always write the integrand as μ(x+s) with s running from 0 to t. Or integrate μ(u) from x to x+t.
Forgetting the minus sign in the exponent.
The formula is memorised without understanding that survival must fall.
Fix: Use tpx = exp(−∫ μ). If your tpx rises with t, you have lost the sign.
Using the density as tpx × μx instead of tpx × μ(x+t).
The age at which the force is evaluated is confused.
Fix: The density at time t uses the force at age x+t, because death occurs then.
Assuming tqx = t × μx for large t.
The small-h approximation hqx ≈ hμx is applied outside its range.
Fix: Use the approximation only for small h. Otherwise use 1 − exp(−∫ μ).
Mixing up t|uqx with tqx or uqx.
The deferred notation is read quickly.
Fix: Read t|uqx as: survive t years, then die within the next u years. Use tpx × uq(x+t) or tpx − t+upx.

Worked examples

Example 1

The force of mortality for a life aged 40 is μ(40+s) = 0.002 + 0.0001 s for s ≥ 0. Calculate 5p40 and 5q40 to 4 decimal places.

Show the solution

Cumulative hazard H = ∫ from 0 to 5 of (0.002 + 0.0001 s) ds.
H = 0.002 × 5 + 0.0001 × 5² ÷ 2 = 0.010 + 0.00125 = 0.01125.
5p40 = exp(−0.01125). Since e^(−0.01125) ≈ 1 − 0.01125 + 0.0000633 − 0.0000002 = 0.988813, 5p40 ≈ 0.9888.
5q40 = 1 − 0.988813 = 0.011187, so 5q40 ≈ 0.0112.

Answer: 5p40 ≈ 0.9888 and 5q40 ≈ 0.0112.

Example 2

A lifetime has survival function S(t) = (1 − t/100)^(1/2) for 0 ≤ t < 100, with S(t) = t p0. Find the force of mortality at age 36 and the probability that a life aged 36 survives 28 more years.

Show the solution

ln S(t) = (1/2) ln(1 − t/100).
Differentiate: d/dt ln S(t) = (1/2) × (−1/100) ÷ (1 − t/100) = −1 ÷ (2(100 − t)).
So μt = −d/dt ln S(t) = 1 ÷ (2(100 − t)).
At age 36: μ36 = 1 ÷ (2 × 64) = 1/128 = 0.0078125.
28p36 = S(64) ÷ S(36) = (0.36)^(1/2) ÷ (0.64)^(1/2) = 0.6 ÷ 0.8 = 0.75.

Answer: μ36 = 1/128 ≈ 0.0078, and 28p36 = 0.75.

Exam tips

Write the notation and the integration limits explicitly. Marks are given for the setup even if the arithmetic slips.
In MCQs, check whether the question asks for a rate or a probability. A value above 1 can only be a force, never a probability.
For piecewise μ, split the integral at the age boundaries and add the cumulative hazards. Do not average the forces.
In derivation questions, state the two-state assumption, that P(death in (t, t+h)) = h μ(x+t) + o(h), then derive the differential equation d/dt tpx = −tpx μ(x+t).
In the computer-based paper, check R or Excel results against a hand estimate such as exp(−μt) to catch a wrong age shift.

Practice questions from Transition intensities dependent on age (exact or census)

Two-State Model and Force of Mortality: frequently asked questions

What is the force of mortality in simple terms?

It is the instantaneous rate of death at age x for a person alive at age x. For a short interval h, the chance of dying is about μx × h. It is measured per year and is not capped at 1.

How do I derive the survival function from the force of mortality?

Integrate the force from age x to x+t to get the cumulative hazard. Then tpx = exp(−cumulative hazard). This comes from solving d/dt tpx = −tpx μ(x+t) with 0px = 1.

What is the difference between μx and qx?

The force μx is an instantaneous rate at exact age x. The quantity qx is the probability of death within one year of age x. They are linked by qx = 1 − exp(−∫ μ(x+s) ds) from 0 to 1.

Is the force of mortality the same as the hazard rate?

Yes. In survival analysis, the hazard rate of the future lifetime random variable is the force of mortality. The cumulative hazard is the integral of the force.