Expected value of functions of order statistics

Question

Let $F$ be a CDF over $[\underline{y},\bar{y}]$ and $g(\cdot)$ is some strictly increasing function. I know the following has at least one solution. \begin{equation} 2F(y)\int_{\underline{y}}^{y}g(x)\frac{dF(x)}{F(y)}=\int_{\underline{y}}^{y}g(x)\frac{dF(x)^2}{F(y)^2}\end{equation} Can we guarantee that the solution is unique? (assume $g(\underline{y})>0$)

Answer 1

The solution is not unique, in general. just take $F$ to be uniform on $[0,1]$. Write the equation as $$h(y)=f(y)-g(y)$$

The second derivative of $h$ must be strictly positive or negative in order to guarantee the uniqueness. One gets however (for $\int xdx^2$, i had $dx^2=dy$, $x=\sqrt{y}$)

$$h(y)=(3/2)y^5-y^{3/2}-(3/2)y^2c^3+c^{3/2}$$ and $$h^{''}(y;c)=30y^3-(3/4)y^{-1/2}-3c^3$$

Since $c=0$, because $F$ is uniform on $[0,1]$, we have $$h^{''}(y)=30y^3-(3/4)y^{-1/2}$$ which is neither positive nor negative for all $y$.

https://www.wolframalpha.com/input/?i=Plot+30x%5E3-(3%2F4)x%5E(-1%2F2)++from+0+to+0.5