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Thread: Characteristics of an IFR pdf

  1. #1
    Apr 2009

    Characteristics of an IFR pdf


    I am looking for help in solving a problem that crept up in my doctoral thesis.
    Let f be a pdf, twice differentiable and F its cdf. Both are defined over \mathbb{R}^+.

    I wish to prove that
    \frac{F(x)}{f(x)}f'(x) - 2f((x)<0 (1).
    I wish to prove it for all functions f which are IFR in the sense developed by Richard E. Barlow and Franck Proschan in their book Mathematical theory of reliability - Google Book Search.
    When a pdf is IFR, then r(x)=\frac{f(x)}{(1-F(x))} is weakly increasing. This, of course, leads to r'(x) \geq 0.
    So that we can write
    f'(x)(1-F(x)) + f^2(x) \geq 0 \Rightarrow F(x)f'(x) - f^2(x) \leq f'(x). (2)

    Now (1) can also be written F(x)f'(x) - f^2(x) < f^2(x). (3)
    (3) is true if f is IFR and if f'(x)<f^2(x).

    What happens when f'(x) \geq f^2(x)?

    If f'(x) \geq f^2(x), f'(x) is always positive because f is a pdf. So f is increasing over \mathbb{R}^+.
    If there exists a x_0 such that f'(x_0)=0, then f exhibits a maximum at that point. but f being a pdf, that means that f(x_0) \neq 0, which contradicts the hypothesis unless f is the null function, which we discard. So \forall x \in \mathbb{R}^+, f'(x) >0. Hence f is striclty increasing. But by definition of F, \int_0^{\infty} f(t)dt =1.

    My hunch is that this last equality contradicts the fact that f is strictly increasing and positive.
    Is this the case and can we then say that if f is IFR then f'(x)<f^2(x) and so (1) is true when f is IFR?

    If that is so, then I have have won my day.

    Any help on this problem would be very much appreciated!
    Thank you.
    Last edited by Xavier_B; Apr 9th 2009 at 01:13 AM.
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