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Hazard Rate Derivation

  1. Jan 10, 2012 #1
    F(t) = 1-Exp -((t-γ)/n))^β

    f(t) = dF(t)/dt = Exp/n[(t-x)/n]^β-1

    h(t) = f(t)/1-F(t)

    h(t)= β(t-y)^β-1/n^β



    The final answer: h(t)= β(t-y)β-1/n^β, Is the correct answer.

    But, I can't for the life of me work out why. Have I made a mistake in the f(t) derivation.

    Can anyone help?
     
  2. jcsd
  3. Jan 10, 2012 #2

    lanedance

    User Avatar
    Homework Helper

    its a bit hard to read, but you derivative doesn't look right

    so lets start with the chain rule
    [tex]
    \frac{d}{dt} g(h(t))= g'(h(t))h'(t)[/tex]

    applying to our case
    [tex]
    F(t) = 1-e^{-(\frac{t-γ}{n})^\beta}
    [/tex]

    so let
    [tex]
    g(x) = 1-e^{x}
    [/tex]
    [tex]
    g'(x) = e^{x}
    [/tex]
    [tex]
    h(t) = -(\frac{t-γ}{n})^\beta
    [/tex]
    [tex]
    h'(t) = -\frac{d}{dt}(\frac{t-γ}{n})^\beta
    [/tex]


    which gives
    [tex]
    f(t) = \frac{F(t)}{dt} = e^{-(\frac{t-γ}{n})^\beta}(-\frac{d}{dt}(\frac{t-γ}{n})^\beta)
    [/tex]
     
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