I am reading a bit on survival analyses and most textbooks state that
$h(t)= \lim_{ \Delta t \rightarrow 0} \frac{P(t < T \leq t+\Delta t |T \geq t )}{ \Delta t} =\frac{f(t)}{1-F(t)} (1)$
where $h(t)$ is the hazard rate,
$f(t)=\lim_{\Delta t \rightarrow 0} \frac{P(t < T \leq t+\Delta t)}{ \Delta t}(2)$ the density function,
$F(t)=Pr(T<t) (3)$ and
$S(t)=Pr(T>t)=1-F(t) (4)$
Also they state that
$S(t)= e^{- \int_0^t h(s)ds } (5)$
Most textbooks (at least those I have) do not provide proof for either (1) or (5). I think I managed to get through (1) as follows
$h(t)= \lim_{ \Delta t \rightarrow 0} \frac{P(t < T \leq t+\Delta t |T \geq t )}{ \Delta t}=$
$\lim_{ \Delta t \rightarrow 0} \frac{P(T \geq t |t < T \leq t+\Delta t ) P(t < T \leq t+\Delta t)}{ P(T \geq t)\Delta t}$ which because of (2) and (4) becomes
$\lim_{ \Delta t \rightarrow 0} \frac{P(T \geq t |t < T \leq t+\Delta t )f(t)}{S(t)\Delta t}$
but $P(T \geq t |t < T \leq t+\Delta t )=1$ therefore $h(t)=\frac{f(t)}{1-F(t)}$
How does one prove (5)?
Best Answer
The derivative of $S$ is $$ \frac{\mathrm{d}S(t)}{\mathrm{dt}} = \frac{\mathrm{d}(1 - F(t))}{\mathrm{dt}} = - \frac{\mathrm{d}F(t)}{\mathrm{dt}} = -f(t) $$ Therefore, as mentioned by @StéphaneLaurent, we have $$ -\frac{\mathrm{d}\log(S(t))}{\mathrm{dt}} = \cfrac{-\frac{\mathrm{d}S(t)}{\mathrm{dt}}}{S(t)} = \frac{f(t)}{S(t)} = h(t) $$ where the last equality follows from (1).
Taking the integral both sides of the previous relation, we obtain $$ -\log(S(t)) = \int_0^t h(s) \, \mathrm{d}s $$ so that $$ S(t) = \exp \left\{- \int_0^t h(s) \, \mathrm{d}s\right\} $$
This is your equation (5). The integral part in the exponential is the integrated hazard, also called cumulative hazard $H(t)$ [so that $S(t) = \exp(-H(t))$].