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 $h(t) = f(t)/S(t) = -\tfrac{d}{dt}\ln S(t)$; 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 2,660 words, 18 min read) adds 8 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• A4

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