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How to find limit of hazard function

  1. Mar 1, 2013 #1
    lim(∫λu du)

    when limit T tend to infinity and ∫ between 0 to t
     
  2. jcsd
  3. Mar 1, 2013 #2

    mathman

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    You need to clarify. T, t, λ.
     
  4. Mar 2, 2013 #3
    FR=λ(t)
    λ(t)>= 0
    whay
    lim∫λ(u)du→∞
    not:lim(t → ∞)
    ∫ btween(0,t)

    plz i have test
    thanks
     
  5. Mar 2, 2013 #4

    HallsofIvy

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    That makes it even more non-understandable. At first you had [itex]\int \lambda u du[/itex] but then you say "[itex]FR= \lambda(t)[/itex]. You hadn't mentioned "FR" before and now it looks like [itex]\lambda[/itex] is a function of t (by which I guess you mean "T", the upper limit in the original integral.

    What you first posted, [itex]\lim_{T\to\infty}\int_0^T \lambda u du[/itex] is fairly easy.
    [tex]\lambda\int_0^T udu= \lambda T^2/2[/tex].
    That limit, as T goes to infinity, does not exist. (Or the limit is "infinity" which just says it does not exist for a particular reason- it grows without bound.)

    The second one you posted depends upon exactly what function of T [itex]\lambda[/itex] is.
     
  6. Mar 2, 2013 #5

    chiro

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    Hey rawand and welcome to the forums.

    Just out of curiosity, does T refer to the resolution of the hazard function? (Higher t means more values between 0 and t)
     
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