# Characteristics of an IFR pdf

• April 9th 2009, 12:01 AM
Xavier_B
Characteristics of an IFR pdf
Hi,

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).

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) 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.