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

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)?

• Have you noted that $h(t)$ is the derivative of $- \log S(t)$ ? Commented May 3, 2013 at 18:05
• Yeah I don't get that either... Commented May 3, 2013 at 18:14
• In your proof of (1), you should first argue that the 2nd probability in the numerator is 1, and then apply (2) and (4). Commented May 3, 2013 at 18:32
• Why is the order important? Commented May 3, 2013 at 18:52
• If you keep your ordering, you should argue that the limit as $\Delta t \rightarrow 0$ (rather than the proba itself) equals $1$. Anyway, this is a detail... Commented May 3, 2013 at 19:21

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

• Could you please be a bit more explicit at $$-\frac{\mathrm{d}\log(S(t))}{\mathrm{dt}} = \cfrac{-\frac{\mathrm{d}S(t)}{\mathrm{dt}}}{S(t)}$$ Commented May 3, 2013 at 19:37
• This is the chaine rule. We have $\frac{\mathrm{d}\, \log(x)}{\mathrm{d}x} = \frac{1}{x}$ so that $$\cfrac{\mathrm{d}\, \log(f(x))}{\mathrm{d}x} = \cfrac{\frac{\mathrm{d}\,f(x)}{\mathrm{d}x}}{x}$$ Commented May 4, 2013 at 4:19
• Should the x in the right hand side of the last equation be f(x)?,i.e.To differentiate y = log S(t). Let u = S(t) therefore $$\frac{du}{dt} =dS(t)/dt = S'(t)$$. Additionally, we have $y = log S(t) = log(u)$ and so $$\frac{dy}{du} = \frac{1}{u} = \frac{1}{S(t)}$$. By the chain rule, so $$\frac{dy}{dt} = \frac{dy}{du} \frac{du}{dt} = \frac{1}{S(t)} S'(t) = \frac{S'(t)}{S(t)}$$ Commented Jan 5, 2017 at 5:27
• @user1420372: Yes, you are right. It should have been f(x). Commented Jan 5, 2017 at 6:50

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

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 $$S(t)=\exp\{-\int_0^th(u)du\}\ \blacksquare$$