How Do You Convert Between HT(t), FT(t), and fT(t) in Probability Calculations?

[estado3]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

 

[HallsofIvy]

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.

 

[estado3]

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?