Ytukyg

Problem
This is a follow-up of this question.

Let $G$ be a proper prior distribution for $thetain[0,1]$ and $ain(0,1)$ be a fixed value. I'd like to show that the following quantity is non-negative for any $a$ and for any integer $0leq xleq n-1$,
$$
g(x)=int_{a}^1 s^{x+1}(1-s)^{n-x-1}dG(s)int_{0}^{a} t^{x}(1-t)^{n-x}dG(t)-int_{a}^1 s^{x}(1-s)^{n-x}dG(s)int_{0}^{a} t^{x+1}(1-t)^{n-x-1}dG(t)
$$

I've simulated that this is true when $G$ is uniform or some beta distributions. However, I have some trouble showing this algebraically. I can lower-bound the first product of integrals using Cauchy-Schwarz and upper-bound terms in the second product of integrals by the product of expectations over $G$, noting that $s$ and $1-s$ are negatively correlated. However, none of these give me useful results for showing $g(x)geq 0$.

Here's what I have so far:

First let $f(s)=frac{s}{1-s}$ and $h(s)=s^x (1-s)^{n-x}$. Then by expressing $int_a^1 f(s)h(s)dG(s)=int_0^1 f(s)h(s)dG(s)-int_a^1 f(s)h(s)dG(s)$ (and similarly for $int_a^1 h(s) dG(s))$, we can rewrite
$$
g(x)=int_{0}^1 f(s)h(s)dG(s)int_{0}^{1} h(s)1(sleq a)dG(s)-int_0^1 h(s) dG(s)int_0^1 f(s)h(s)1(sleq a)dG(s)
$$
Now the first product can be bounded below by Cauchy-Schwarz
$$
int_{0}^1 f(s)h(s)dG(s)int_{0}^{1} h(s)1(sleq a)dG(s)geq int_0^1 f(s)^{1/2}h(s)1(sleq a) dG(s)=int_0^a f(s)^{1/2}h(s) dG(s)
$$
The second product can be bounded above by noting that on $(0,1)$, $h(s)$ attains the maximum at $s=x/n$ and $f(s)$ is increasing on $(0,a)$.
$$
int_0^1 h(s) dG(s)int_0^1 f(s)h(s)1(sleq a)dG(s)leq f(a)hleft(frac{x}{n}right)int_0^a h(s) dG(s)
$$
Putting everything together, I get
begin{eqnarray*}
g(x)&geq& int_0^a f(s)^{1/2}h(s) dG(s)-f(a)hleft(frac{x}{n}right)int_0^a h(s) dG(s)\
&=&int_0^a left[f(s)^{1/2}-f(a)hleft(frac{x}{n}right)right] h(s) dG(s)
end{eqnarray*}
Now on $sin(0,a)$, $f(s)^{1/2}-f(a)hleft(frac{x}{n}right)$ may not always be greater than 0, but I need to show that the integral is (if possible) - it may also be the case that my bounds are not tight enough.

Another thought:

Since
begin{eqnarray*}
g(x)&=&int_{0}^1 f(s)h(s)dG(s)int_{0}^{a} h(s)dG(s)-int_0^1 h(s) dG(s)int_0^a f(s)h(s)dG(s)\
&=&int_0^aleft(int_{0}^1 f(s)h(s)dG(s)- f(t)int_0^1 h(s) dG(s)right)h(t)dG(t)
end{eqnarray*}
then I just need to show that for $tin(0,a)$,
$$
int_{0}^1 f(s)h(s)dG(s)- f(t)int_0^1 h(s) dG(s)geq 0
$$
or after substitution
$$
frac{t}{1-t}leq frac{int_{0}^1 left(frac{s}{1-s}right)s^x(1-s)^{n-x}dG(s)}{int_0^1 s^x(1-s)^{n-x} dG(s)}hspace{3mm}forall tin(0,a)
$$
The LHS is just a particular value of $f(t)=frac{t}{1-t}$ on a constrained range $(0,a)$, while the RHS seems to be a weighted average of $f(t)$ over a larger range (i.e. $(0,1)$). Does this inequality always hold? Does it depend on the prior $G$?

The answer seems to be easier than initially thought:

begin{eqnarray*}
g(x)&=&int_{a}^1 f(s)h(s)dG(s)int_{0}^{a} h(t)dG(t)-int_a^1 h(s) dG(s)int_0^a f(t)h(t)dG(t)\
&=&int_0^aleft(int_{a}^1 f(s)h(s)dG(s)- f(t)int_a^1 h(s) dG(s)right)h(t)dG(t)\
&=&int_0^aleft(int_{a}^1 [f(s)-f(t)]h(s)dG(s)right)h(t)dG(t)
end{eqnarray*}
Since $f(cdot)$ is continuous and increasing on $(0,1)$, $f(s)geq f(t)$ since $sin(a,1)$ and $tin(0,a)$. Therefore, $g(x)geq 0$ as desired.