user2966729

I need to take a ciggarate for 3 minutes. I haven't smoke for hours. Coming right back. Really thankful.

So your point about the bounds is ...? I would like to hear about that.

I think the bounds you mention is automatically satisfied. If you draw an arbitrary distribution function on $[0,1]$, and take any point on the distribution function, you will see the facts.

You missed an $x$ in $(2x-1)f(x)$ in the second equality of $U'(x)$. $$U'(x)=(3x-1)F(x)+(2x-1)xf(x)+\int_x^1 F(t)dt.$$. But that doesn't matter. I see you point there.

Thank you so much for your help. Very good insights. I can assume $f$ is everywhere differentiable for less general results on this problem. Is there any other insights for differentiable assumption?

For $f=0$, the only negative part in $U'(x)$ is removed, see the second equation in my original post.

@Erick Wong, I think there are errors in your calculations. When $x\in[0,\frac{1}{4}]$, $U'(x)=2 - 8 x + 18 x^2$, which reaches a minimum at $x^*=\frac{2}{9}$, where $U'(x^*)=\frac{10}{9}$. But thanks for inspirations.

@Erick Wong, oh sorry and thanks, let me check.

@Erick Wong, you see, I have shown $U'(m)\geq 0$ for $F=m^a, a\geq 0$. When $a=1$, the distribution is uniform. So I guess there are some errors in your calculations.

@Math1000, I think the computation is correct. According to $\frac{d}{dx}\int_x^b h(t)dt = -h(x)$, when $h(t)=(1-t)f(t)$, we get $-(1-x)f(x)$.