Perform survival model and hazard rate calculations.

A reinsurer needs to price a long-tail liability cover and asks how long a policyholder is likely to survive past age 65. The answer is not a single number but a whole survival curve, and the hazard rate is the engine that drives it.

The four equivalent representations. A non-negative random variable $T$ (time to death, failure, or claim) is fully described by any one of $F$, $S$, $f$, or $h$. Knowing one gives the other three.

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

KEY: The hazard is not a probability. It is a rate with units of 1/time. The conditional probability of failure in a small interval $[t,t+\Delta ]$ given survival to $t$ is approximately $h(t)\Delta$, not $h(t)$ itself.

From hazard back to survival. Integrating the log-derivative relation:

$$S(t)=\exp \left(-{\int }_0^{t}h(s)ds\right)=e^{-H(t)}$$

So if you can write the hazard, you can build the survival curve, and vice versa.

Conditional survival. Given that $T$ has reached age $s$, the probability of surviving an additional $t$ units is

Common mistakes

• Treating the hazard as a probability. A hazard of 0.06 per year is not a 6% chance of death; it is an instantaneous rate. The one-year mortality is $1-e^{-\int h}$, not $h$.
• Forgetting to multiply by $S(t)$ when computing density from hazard. The density is $f(t)=h(t)S(t)$, not just $h(t)$. Skipping the survival factor inflates the density toward the tail.
• Integrating the hazard incorrectly for non-constant rates. Writing $S(t)=e^{-h(t)\cdot t}$ is wrong unless $h$ is constant. You must integrate ${\int }_0^{t}h(s)ds$.

Bottom line

• Survival function $S(t)=P(T>t)=1-F(t)$; strictly non-increasing, $S(0)=1$, $S(\infty )=0$.
• Hazard rate $h(t)=f(t)/S(t)=-\frac{d}{dt}\ln S(t)$; instantaneous failure rate given survival, with units of 1/time (not a probability, so $h(t)>1$ is allowed).
• Cumulative hazard $H(t)={\int }_0^{t}h(s)ds=-\ln S(t)$, so $S(t)=e^{-H(t)}$.
• Mean lifetime $E[T]={\int }_0^{\infty }S(t)dt$; mean residual life $e(t)={\int }_t^{\infty }S(u)du/S(t)$.

Exam shortcut

When given a hazard, immediately compute the cumulative hazard and exponentiate; that is your survival function for everything downstream. For Gompertz or Makeham, remember the $1/\ln c$ factor; without it your survival probability looks like $10^{-5}$ when it should be 0.9. For conditional survival problems, compute the two cumulative hazards and subtract: ${}_t{p}_s=\exp (-(H(s+t)-H(s)))$ avoids dividing two tiny exponentials.

The full lesson (about 1,827 words, 12 min read) adds 2 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• A4

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