Hazard Rate

Study guides: Exam ASTAM, Exam MAS-I

The Hazard Rate $\mu(t)$ (also called the force of mortality in life contingencies, or the failure rate in reliability) is the instantaneous rate of failure at time $t$ given survival to $t$ .

$\mu(t) = \frac{f(t)}{S(t)}$

$= -\frac{d}{dt}\ln S(t)$

$S(t) = \exp\!\left(-\int_0^t \mu(s)\,ds\right)$

Also written $h(t)$ or $\lambda(t)$ depending on context; in life tables it is $\mu_x$
A constant hazard rate $\mu(t) = \mu$ corresponds to the Exponential distribution: $S(t) = e^{-\mu t}$
The cumulative hazard function $H(t) = \int_0^t \mu(s)\,ds = -\ln S(t)$
An increasing hazard rate indicates aging (e.g., Weibull with shape $> 1$ )
A decreasing hazard rate indicates infant-mortality effects (Weibull with shape $< 1$ )

Gompertz–Makeham model (common in life insurance): $\mu_x = A + Bc^x, \quad A \geq 0,\, B > 0,\, c > 1$

Computing the Hazard Rate from a Survival Function The survival function is $S(t) = e^{-0.05t^2}$ . Find the hazard rate $\mu(t)$ .

Answer $\mu(t) = -\frac{d}{dt}\ln S(t) = -\frac{d}{dt}(-0.05t^2) = 0.10t$ The hazard rate increases linearly with time, indicating an aging population.

Recovering the Survival Function from a Hazard Rate A force of mortality is $\mu(t) = 0.02 + 0.001t$ . Find $S(5)$ .

Answer $H(5) = \int_0^5 (0.02 + 0.001t)\,dt = \left[0.02t + 0.0005t^2\right]_0^5 = 0.10 + 0.0125 = 0.1125$ $S(5) = e^{-0.1125} \approx 0.8936$