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 $\displaystyle \mathbb{R}^+$.

I wish to prove that

$\displaystyle \frac{F(x)}{f(x)}f'(x) - 2f((x)<0$ (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 $\displaystyle r(x)=\frac{f(x)}{(1-F(x))}$ is weakly increasing. This, of course, leads to $\displaystyle r'(x) \geq 0$.

So that we can write

$\displaystyle f'(x)(1-F(x)) + f^2(x) \geq 0 \Rightarrow F(x)f'(x) - f^2(x) \leq f'(x).$ (2)

Now (1) can also be written $\displaystyle F(x)f'(x) - f^2(x) < f^2(x)$. (3)

(3) is true if f is IFR and if $\displaystyle f'(x)<f^2(x)$.

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

If $\displaystyle f'(x) \geq f^2(x)$, f'(x) is always positive because f is a pdf. So f is increasing over $\displaystyle \mathbb{R}^+$.

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

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 $\displaystyle f'(x)<f^2(x)$ 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.