TITLE: Distribution and second order differential inequality
QUESTION [9 upvotes]: I would like to solve the following $2^{\mathrm{nd}}$ order differential inequality
$$
\theta_F'(x) = \frac{2F'(x)^2 - F(x)F''(x) + F''(x)}{F'(x)^2} < 0
$$
for some subinterval $I \subset [0,+\infty)$, subject to the condition that $F$ is a cumulative distribution function, that is,
$$
\lim_{x \to -\infty} F(x) = 0, \qquad \lim_{x \to +\infty} F(x) = 1.
$$
Is this something feasible? By feasible I mean, can I hope for a closed form solution for $F$ given these conditions?
I have tried solving when $\theta'_F(x) = 0$ but I'm not sure how this could be useful.
This ODE comes from differentiating
$$
\theta_F(x) = x - \frac{1-F(x)}{f(x)} = x - \frac{1-F(x)}{F'(x)}.
$$
Basically, I am looking for a general form for distributions $F$ such that $\theta_F(x)$ is decreasing on some subinterval of $x \in [0,+\infty)$.
Any help would be greatly appreciated.

REPLY [4 votes]: For every $F$ let $H_F=1/(1-F)$ then $$(F')^2\cdot\theta'_F=H_F''\cdot(1-F)^3,$$ hence the condition that $\theta'_F\lt0$ on the interval $I$ is equivalent to the fact that the function $H_F$ is strictly concave on $I$. 
Every CDF $F$ such that $H_F$ is strictly concave on $I$ solves the inequality, thus, every $F=1-1/h$ with $h\gt1$ càdlàg, increasing, and concave on $I$, solves the inequality.
Examples: The PDF $f_{a,c}$ defined by $$f_{a,c}(x)=\dfrac{ac\mathbf 1_{x\geqslant0}}{(1+cx)^{a+1}},$$ for every $x$ solves this on every $I\subseteq(0,\infty)$, for every parameters $c\gt0$ and $a$ in $(0,1)$. Likewise for the PDF $f_{a,b,c}$ defined by $$f_{a,b,c}(x)=(\mathrm e^{a\xi}-b)\frac{a\mathrm e^{ax}\mathbf 1_{x\gt\xi}}{(\mathrm e^{ax}-b)^2},$$
for every $(a,b,\xi)$ such that $a\gt0$ and $\mathrm e^{a\xi}\gt b$, on every interval $I\subseteq(\xi,\infty)$.