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Risk Modelling and Survival Analysis · Concepts of survival models

Force of Mortality and Hazard Rate: Formula and Meaning

Updated 11 October 2026 · Fact-checked

The force of mortality μx is the instantaneous rate of death at exact age x, given survival to x. For lifetime X from birth, μx = f(x) ÷ S(x) = −d/dx ln S(x). For future lifetime T_x, μx = f_x(0) ÷ S_x(0). Integrating gives survival: tpx = exp(−∫ from 0 to t of μ(x+s) ds).

Understand Force of Mortality and Hazard Rate

Start with a simple idea. At age x you are alive. You want a measure of how fast death is arriving at that moment. A probability such as qx covers a whole year. It cannot describe a single instant. The force of mortality μx does that job.

Formally, μx = lim as h→0+ of (h q_x) ÷ h, where h q_x is the probability that a life aged x dies within h years. It is a rate, not a probability. It can exceed 1. It is never negative. It is also called the hazard rate or the transition intensity in a two-state model.

The link to the lifetime distribution comes from conditioning. For the future lifetime T_x, the density at t is f(t) = tpx · μ(x+t). So the density is the probability of surviving to x+t, multiplied by the rate of dying at that moment. Rearranging gives μ(x+t) = f(t) ÷ tpx, which is the density divided by the survival function.

Since μ(x+t) = −d/dt ln(tpx), integrating both sides from 0 to t gives tpx = exp(−∫ μ(x+s) ds). The integral inside is the cumulative hazard. If μ is constant, tpx = e^(−μt), and T_x is exponential.

The difference from qx matters. qx = 1 − px is the probability of dying within one year. It is bounded between 0 and 1. μx is the instantaneous rate. For small rates, qx ≈ μ(x+½), but this is an approximation, not an identity.

Key rules to remember

Definition of force of mortality
μx = lim(h→0+) (h q_x) ÷ h = f_x(0) ÷ S_x(0)
Here f_x and S_x refer to the future lifetime T_x. It is a rate, not a probability.
In terms of the survival function of age at death
μx = −S′(x) ÷ S(x) = −d/dx ln S(x)
S(x) = P(X > x) for lifetime X from birth, or l_x up to a constant: μx = −(1/l_x) · dl_x/dx.
Survival probability from force of mortality
tpx = exp(−∫ from 0 to t of μ(x+s) ds)
Holds for any t ≥ 0 where μ is integrable. For constant μ, tpx = e^(−μt).
Density of future lifetime
f_x(t) = tpx · μ(x+t)
Density of T_x. Also equals −d/dt (tpx).
Deferred death probability
t|u qx = tpx − (t+u)px = ∫ from t to t+u of spx · μ(x+s) ds
Probability of surviving t years then dying in the next u years.
Cumulative hazard
Λ_x(t) = ∫ from 0 to t of μ(x+s) ds, so tpx = e^(−Λ_x(t))
Useful when the integral is simple.

How to solve Force of Mortality and Hazard Rate questions

Most questions give you μ or a survival function and ask for a probability, a density or a rate. Follow these steps.

  1. 1Write down what is given: μ(x), S(x), tpx, or a density. State the age and the time interval clearly.
  2. 2Decide which quantity you need: a survival probability tpx, a death probability tqx, a density, or μ itself.
  3. 3If you have μ and need tpx, integrate μ(x+s) from 0 to t and apply tpx = exp(−integral).
  4. 4If you have tpx or S(x) and need μ, take −d/dt of ln(tpx), or use μ = −S′ ÷ S.
  5. 5For death probabilities, use tqx = 1 − tpx. For deferred or interval probabilities, subtract two survival probabilities.
  6. 6For a density, use f(t) = tpx · μ(x+t). Check it integrates to 1 if the question asks you to verify validity.
  7. 7Substitute numbers last. Keep the units of time consistent and state the answer to a sensible number of decimal places.

Quickest way: Constant or simple-form μ shortcut

When to use it: Use when μ is constant, piecewise constant or a simple polynomial over the interval.

  1. If μ is constant over the interval, write tpx = e^(−μt) directly.
  2. If μ is piecewise constant, split the interval and multiply the survival probabilities for each piece.
  3. If μ is a power such as μ(x+s) = k(x+s)^n, integrate to get [k/(n+1)]·[(x+t)^(n+1) − x^(n+1)] and exponentiate with a minus sign.
  4. For the reverse direction, compute ln of tpx and differentiate. Do not simplify before differentiating.

Common mistakes in Force of Mortality and Hazard Rate

  • Treating μx as a probability and saying it cannot exceed 1.

    The name sounds like a probability and qx is always between 0 and 1.

    Fix: Remember that μx is a rate per year. It is non-negative but can be above 1. qx and tpx are probabilities between 0 and 1, whereas μx and densities are rates or densities and are not bounded by 1.

  • Writing qx = μx.

    Students match the two as 'chance of death at age x'.

    Fix: qx = 1 − exp(−∫ from 0 to 1 of μ(x+s) ds). Even for constant μ, qx = 1 − e^(−μ), which is only approximately μ when μ is small.

  • Forgetting the minus sign in the exponent of tpx.

    Students memorise exp(∫ μ) and drop the sign.

    Fix: A positive hazard must reduce survival. Check that your tpx is between 0 and 1 and falls as t grows.

  • Using f(t) = μ(x+t) for the density.

    The density and hazard look similar.

    Fix: Density f = tpx · μ(x+t). The hazard is f ÷ tpx. They differ by the survival factor.

  • Integrating from x to t instead of from 0 to t with argument x+s.

    Confusion between age and duration.

    Fix: Either integrate μ(x+s) over s from 0 to t, or integrate μ(y) over y from x to x+t. Both are right. Mixing them is wrong.

  • Differentiating tpx with respect to x instead of t when finding μ(x+t).

    Both x and t appear in the expression.

    Fix: For μ(x+t) with x fixed, use −d/dt ln(tpx). If you differentiate in x, you must use the correct two-variable form, which is easier to avoid.

Worked examples

Example 1

The force of mortality is μ(x) = 0.002 for 40 ≤ x < 50 and μ(x) = 0.004 for x ≥ 50. Find the probability that a life aged 45 survives to age 55.

Show the solution
  1. Split the 10-year interval at age 50. From 45 to 50 is 5 years with μ = 0.002. From 50 to 55 is 5 years with μ = 0.004.
  2. Cumulative hazard = 5 × 0.002 + 5 × 0.004 = 0.010 + 0.020 = 0.030.
  3. 10p45 = exp(−0.030).
  4. exp(−0.030) = 0.97045.

Answer: 10p45 ≈ 0.9704

Example 2

For a life aged 30, the survival function from age 30 is tp30 = (1 − t/70)^(1/2) for 0 ≤ t < 70. Find the force of mortality at age 40 and the density of T30 at t = 10.

Show the solution
  1. Take logs: ln(tp30) = ½ ln(1 − t/70).
  2. Differentiate with respect to t: d/dt ln(tp30) = ½ × (−1/70) ÷ (1 − t/70) = −1 ÷ (140(1 − t/70)) = −1 ÷ (2(70 − t)).
  3. So μ(30+t) = 1 ÷ (2(70 − t)).
  4. At age 40, t = 10: μ(40) = 1 ÷ (2 × 60) = 1/120 ≈ 0.008333.
  5. Density f(10) = 10p30 × μ(40).
  6. 10p30 = (1 − 10/70)^(1/2) = (6/7)^(1/2) = 0.92582.
  7. f(10) = 0.92582 × 0.008333 = 0.007715.

Answer: μ40 = 1/120 ≈ 0.00833 and f(10) ≈ 0.00772

Exam tips

  • Show the integral and the exponential before substituting numbers. Method marks are given for the setup.
  • If a question says 'derive', start from the definition with h q_x and limits. Do not jump to the result.
  • For piecewise μ, draw a quick timeline of the age bands. It prevents missed intervals.
  • Check your final tpx lies between 0 and 1 and decreases with t. This catches sign errors fast.
  • In MCQs, watch for options that confuse μx with qx. Work out which quantity the question actually asks for.

Practice questions from Concepts of survival models

Force of Mortality and Hazard Rate: frequently asked questions

What is the force of mortality in simple words?

It is the instantaneous rate of dying at age x for someone who has survived to x. Think of it as a speedometer for death risk. It is measured per year and is not a probability.

How do I derive tpx from the force of mortality?

Start with μ(x+t) = −d/dt ln(tpx). Integrate both sides from 0 to t and use 0px = 1. This gives ln(tpx) = −∫ μ(x+s) ds, so tpx = exp(−∫ from 0 to t of μ(x+s) ds).

What is the difference between μx and qx?

qx is the probability of dying within one year of age x, so it lies between 0 and 1. μx is the instantaneous rate at exact age x and can be larger than 1. They are linked by qx = 1 − exp(−∫ from 0 to 1 of μ(x+s) ds).

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

Yes, for lifetime data they are the same thing. The hazard rate is the general statistical term. Force of mortality is the actuarial name, and transition intensity is used in multi-state models.