Distribution and second order differential inequality 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.
 A: 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 $I\subseteq(\xi,\infty)$.
A: Too long for a comment. 
Notice that 
$$\frac{d\theta}{dx}=\frac{2(F'(x))^2-F(x)F''(x)+F''(x)}{(F'(x))^2}=2+F''(x)\cdot\frac{1-F(x)}{(F'(x))^2}$$
if $$\frac{d\theta}{dx}<0$$
over some interval then a necessary condition would be 
$$F''(x)<0$$
on that interval. Let this interval be $[a,b)\subseteq(0,\infty)$ i.e. 
$$\frac{d\theta}{dx}<0$$
for all $x\in[a,b)$. Then 
$$2+F''(x)\cdot\frac{1-F(x)}{(F'(x))^2}<0\Leftrightarrow \frac{F''(x)}{F'(x)}<-2\frac{F'(x)}{1-F(x)}$$
Integrating both sides 
$$\int^{y}_{a}\frac{F''(x)}{F'(x)}\,dx<\int^{y}_{a}-2\frac{F'(x)}{1-F(x)}\,dx$$
$$\Leftrightarrow$$
$$\int^{y}_{a}\frac{d(F'(x))}{F'(x)}<\int^{y}_{a}2\frac{d(1-F(x))}
{1-F(x)}$$
$$\Leftrightarrow$$
$$\ln|F'(x)|\Big|^{y}_{a}<2\ln|1-F(x)|\Big|^{y}_{a}$$
where $a\geq0$ and $y\in(a,b)$. Then the above result can be rewritten as
$$\ln(F'(x))\Big|^{y}_{a}<\ln(1-F(x))^2\Big|^{y}_{a}$$
$$\Leftrightarrow$$
$$\ln\Big[\frac{F'(y)}{F'(a)}\Big]<\ln\Big[\frac{(1-F(y))^2}{(1-F(a))^2}\Big]$$
$$\Leftrightarrow$$
$$\frac{F'(y)}{F'(a)}<\frac{(1-F(y))^2}{(1-F(a))^2}$$
$$\Leftrightarrow$$
$$\frac{F'(y)}{(1-F(y))^2}<\frac{F'(a)}{(1-F(a))^2}$$
which holds for all $y\in(a,b)$. Since $y>a$ integrating both sides of the above inequality over the interval $[a+\epsilon,z)$ where $\epsilon>0$ and $z\leq b$ yields
$$\int^{z}_{a+\epsilon}\frac{F'(y)}{(1-F(y))^2}\,dy<\int^{z}_{a+\epsilon}\frac{F'(a)}{(1-F(a))^2},dy$$
$$\Leftrightarrow$$
$$\frac{1}{(1-F(z))}-\frac{1}{(1-F(a+\epsilon))}<\frac{F'(a)}{(1-F(a))^2}(z-a-\epsilon)$$
Letting $\epsilon\to0$ yields
$$\frac{1}{(1-F(z))}-\frac{1}{(1-F(a))}<\frac{F'(a)}{(1-F(a))^2}(z-a)$$
Denoting by $$G(z)=\frac{1}{(1-F(z))}$$
then the last expression is equivalent to
$$G(z)-G(a)<G '(a)(z-a)\Leftrightarrow G(z)<G '(a)(z-a)+G(a)$$$$\Leftrightarrow F(z)<1-\frac{1}{G '(a)(z-a)+G(a)}$$
Since $F(z)$ is a cdf then $F(z)\geq F(a)$ for all $z\geq a$ therefore any such cdf must satisfy 
$$1-\frac{1}{G(a)}=F(a)\leq F(z)<1-\frac{1}{G '(a)(z-a)+G(a)}$$
A: Subinterval the OP says. Then we have a very simple example
that fits the bill, unless I am missing something (such as differentiable everywhere) . Put:
$$
F(x) = \left\{ \begin{array}{lll} 0 & \mbox{for} & x \le 0 \\
\sqrt{x} & \mbox{for} & 0 \le x \le 1 \\ 1 & \mbox{for} & 1 \le x
\end{array} \right. \qquad \Longrightarrow \quad
\theta_F'(x) < 0 \quad \mbox{for} \quad I = \left( 0, \frac{1}{9} \right)
$$
