Getting from the hazard function to the cumulative and density func and vice versa

1. The problem statement, all variables and given/known data

Looking for a step by step online guide/tutorial/worked example showing equations for getting to the hazard function from the density function, the cumulative distribution function from the hazard function, and vice versa

2. Relevant equations

HT(t) = hazard function
FT(t) = Cumulative distribution function
fT(t) = Density function

3. The attempt at a solution
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 Recognitions: Gold Member Science Advisor Staff Emeritus Distribution function and density function I know but I had to look up "Hazard function". According to Wikipedia, http://en.wikipedia.org/wiki/Survival_analysis If we have the cumulative distribution function, [/itex]F(t)= Pr(T\le t)[/itex], we define S(t)= 1- T(t). Then the Hazard function, $\lambda(t)$, is given by $$\lambda = -\frac{S'(t)}{S(t)}$$ Directly in terms of F, then, since S'(t)= (1- F(t))'= -F'(t), $$\lambda = \frac{F'(t)}{1- F(t)}$$ If f(t) is the density function, f(t)= F'(t), then $$\lambda = \frac{f(t)}{1- F(t)}$$ Alternatively, we can define the "cumulative hazard function", $\Lambda(t)= -log(1- F(t))$ and then the hazard function is the derivative: $\lambda(t)= d \Lambda(t)/dt$ In any case, finding $\lambda$ involves solving a first order differential equation. To take a simple example, the uniform distribution from 0 to 1, the density function is a constant, f(x)= 1, so F(x)= $\int_0^t 1 dx= t$ and the hazard function is given by $\lambda(t)= f(t)/(1- F(t))= 1/(1- t)$. Alternatively, the cumulative hazard function is $\Lamba(t)= -log(F(t))= -log(1-t)$ and the hazard function is the derivative of that: $\lambda(t)= d(-log(1-t))/dt= 1/(1-t)$. Going the other way, if we were given $\lambda(t)= 1/(1- t)$, then $\lambda(t)= F'/(1- F)= 1/(1- t)$ so finding $\lambda(t)$ requires solving a differential equation: $F'= \lambda(t)(1- F)= (1- F)/(1- t)$. That's a "separable" differential equation: dF/(1- F)= dt/(1-t) . Integrating, log(1- F(t))= log(1- t)+ C1 so 1- F(t)= C2(1- t). In order that F(0)= 0, we must have C2= 1 so 1- F(t)= 1- t and F(t)= t as before.
 thanks for the link, I am wondering at what stage we use the hazard function, I understand the p.d.f is used for a moment in time, while the c.d.f is used for a time period i.e 0< = T, at what stage do we need the hazard function?

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