Distribution satisfying Increasing failure rate

Question

Suppose we have distribution $f(x)$ satisfying increasing failure rate, i.e., $\frac{f(x)}{1-F(x)}$ increases as $x$.

If $a<b$, then can we prove that $\frac{F(b)-F(a)}{f(b)-f(a)}$ increases as $a$ increases of as $b$ decreases?

Answer 1

NO, as illustrated by the following counterexample:

Consider the probability density function $$f(x)=\begin{cases}x^2, &\text{if }x\in [0,\sqrt[3]{3}]\\0, &\text{otherwise} \end{cases},$$ whose cdf is $$F(x)=\begin{cases}0, &\text{if }x< 0\\\frac{x^3}{3}, &\text{if }0\le x<\sqrt[3]{3}\\1, &\text{if }x\ge \sqrt[3]{3}\end{cases}.$$ Then on $[0,\sqrt[3]{3}]$, we have $$\frac{f(x)}{1-F(x)}=\frac{x^2}{1-x^3/3},$$ which is clearly increasing. However, for all $0\le a<b\le \sqrt[3]{3}$, $$G(a,b):=\frac{F(b)-F(a)}{f(b)-f(a)}=\frac 13\,\frac{b^3-a^3}{b^2-a^2}=\frac 13\,\frac{b^2+ab+a^2}{b+a}.$$ Clearly, $$\frac{\partial G}{\partial a}\propto b(b+a)-(b^2+ab+a^2)=-a^2\le 0,$$ $$\frac{\partial G}{\partial b}\propto (2b+a)(b+a)-(b^2+ab+a^2)=b^2+2ab>0.$$ In this example, $G(a,b)$ increases as $a$ decreases and as $b$ increases, which is totally to the opposite of what you claim.