Monotonicity of $\int^x_0\frac{F(y)}{F(x)}dy$ Let $F$ be a continuous and strictly increasing CDF with support $[0,1]$.
Can we show that $\int^x_0\frac{F(y)}{F(x)}dy$ is increasing in $x$? Or can we find a CDF that we can use it as a counterexample? 
I've tried some well-known CDF's with a finite support, but I had monotonicity with all of the CDF's.
Any tips or help?
 A: The answer is: No, your integral is not always increasing. Consider the following strictly increasing, continuous CDF:
\begin{align}
F:\Bbb R&\to[0,1] \\
x&\mapsto 
\begin{cases}
0, &x\le 0  \\
\frac{49}{10}x, & 0\le x \le \frac1{10} \\
\frac{39}{80}+\frac{x}{40}, & \frac1{10}\le x\le \frac9{10} \\
\frac{49}{10}x-\frac{39}{10}, & \frac9{10}\le x\le 1 \\
1, & 1\le x
\end{cases}.
\end{align}
Plot of $F$:

Then we have 
\begin{gather}
\int_0^{\frac9{10}} F = \frac{849}{2000}, \\
\int_0^1 F = \frac12,
\end{gather}
and thus 
\begin{equation}
\frac{\displaystyle\int_0^{\frac9{10}} F}{F\left(\frac9{10}\right)}=\frac{\frac{849}{2000}}{\frac12}=\frac{849}{1000}>\frac12=\frac{\displaystyle\int_0^{1} F}{F\left(1\right)}.
\end{equation}
Hence, $x\mapsto \dfrac{\int_0^x F}{F(x)}$ is not increasing (in fact, it decreases strictly on $]\frac9{10},1[$).
Remark. Since you can approximate any continuous function on $[0,1]$ arbitrary well (w.r.t. $\|\cdot\|_\infty$-norm) by smooth functions, it is also possible to construct a smooth counter-example.
A: Thought I had an idea that might work; maybe someone else can see how to finish it.
Let $G=F^{-1}$; then making the change of variables $y=G(t)$,
$$
\int_0^x \frac{F(y)}{F(x)} \,dy = \int_0^{F(x)} \frac{t}{F(x)} G'(t) \,dt.
$$
Since $w=F(x)$ is increasing, it suffices to determine whether
$$
\frac1w \int_0^w tG'(t) \,dt
$$
is an increasing function of $w$. Integrating by parts,
$$
\frac1w \int_0^w tG'(t) \,dt = \frac1w \bigg( tG(t)\bigg|_0^w - \int_0^w G(t)\,dt \bigg) = G(w) -\frac1w \int_0^w G(t)\,dt ...
$$
