Proof of relationship between hazard rate, probability density, survival function

Question

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)?

Answer 1

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 $S(t) = \exp(-H(t))$].

Answer 2

$$h(t) = \frac{f(t)}{S(t)}\ $$ $$ = \frac{f(t)}{1-F(t)}$$ $$= \frac{f(t)}{1- \int^t_0{f(s) ds}}$$

Integrate both sides: $$\int^t_0 h(s) ds = \int^t_0 \frac{f(s)}{1- \int^t_0{f(s)ds}}ds $$ $$= -\ln [1- \int^t_0{f(s)ds}]^t_0+ c $$ $$1- \int^t_0{f(s)ds} = \exp [-\int^t_0 h(s) ds]$$ Differentiate both sides: $$-f(t) = -h(t) \exp[-\int^t_0 h(s) ds]$$ $$f(t) = h(t) \exp[-\int^t_0 h(s) ds]$$

Since $$h(t) = \frac{f(t)}{S(t)}$$

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

Replace $f(t)$ by $h(t) \exp[-\int^t_0 h(s) ds]$ , $$S(t) = \frac{h(t) \exp[-\int^t_0 h(s) ds]}{h(t)}$$ Therefore, $$S(t) = \exp[-\int^t_0 h(s) ds]$$

Answer 3

We prove the following equation: $$ S(t)=\exp\{-\int_{0}^{t}h(u)du\} $$ proof:

We first prove $$ f(t)=-\frac{dS(t)}{dt} $$ proof:

$$ f(t)=\frac{dF(t)}{dt}=\frac{dP(T<t)}{dt}=\frac{d(1-S(t))}{dt}=-\frac{dS(t)}{dt}\ \blacksquare $$ And we know $$ h(t)=\frac{f(t)}{S(t)} $$ Substitute $f(t)$ into $h(t)$ we get $$ h(t)=\frac{-\frac{dS(t)}{dt}}{S(t)} $$ then continue our main proof. By integrate the both side of the above equation, we have $$ \int_0^th(u)du=\int_0^t\frac{-\frac{dS(t)}{dt}}{S(t)}dt=\int_0^t-S(t)^{-1}dS(t)\\ =-[\log S(t)-\log S(0)]=-\log S(t) $$ Then we get the result $$ S(t)=\exp\{-\int_0^th(u)du\}\ \blacksquare $$