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

  1. Jan 10, 2008 #1
    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
     
  2. jcsd
  3. Jan 11, 2008 #2

    HallsofIvy

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    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, [itex]\lambda(t)[/itex], is given by
    [tex]\lambda = -\frac{S'(t)}{S(t)}[/tex]
    Directly in terms of F, then, since S'(t)= (1- F(t))'= -F'(t),
    [tex]\lambda = \frac{F'(t)}{1- F(t)}[/tex]

    If f(t) is the density function, f(t)= F'(t), then
    [tex]\lambda = \frac{f(t)}{1- F(t)} [/tex]

    Alternatively, we can define the "cumulative hazard function", [itex]\Lambda(t)= -log(1- F(t))[/itex] and then the hazard function is the derivative: [itex]\lambda(t)= d \Lambda(t)/dt[/itex]
    In any case, finding [itex]\lambda[/itex] 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)= [itex]\int_0^t 1 dx= t[/itex] and the hazard function is given by [itex]\lambda(t)= f(t)/(1- F(t))= 1/(1- t)[/itex]. Alternatively, the cumulative hazard function is [itex]\Lamba(t)= -log(F(t))= -log(1-t)[/itex] and the hazard function is the derivative of that: [itex]\lambda(t)= d(-log(1-t))/dt= 1/(1-t)[/itex].

    Going the other way, if we were given [itex]\lambda(t)= 1/(1- t)[/itex], then [itex]\lambda(t)= F'/(1- F)= 1/(1- t)[/itex] so finding [itex]\lambda(t)[/itex] requires solving a differential equation: [itex]F'= \lambda(t)(1- F)= (1- F)/(1- t)[/itex]. 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.
     
  4. Jan 11, 2008 #3
    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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