Let $t \in [0,1]$ or $t \in (0,1)$ be distributed according to $F(t)$.
Now consider the following equation: \begin{equation} \frac{\int_{\underline{t}}^{\overline{t}}(\gamma-t(2\gamma-1))dF(t)}{\int_{0}^{\overline{t}}(\gamma-t(2\gamma-1))dF(t)}=\kappa \end{equation}
where $\kappa \in (0,1)$, $\underline{t}=\frac{(1-\gamma)d}{\gamma+(1-\gamma)d}$ and $\overline{t}=\frac{\gamma d}{1-\gamma +\gamma d}$. Here, $\gamma \in (\frac{1}{2},1)$ and $d>0$.
Note that $\underline{t}$ is decreasing in $\gamma$ (from $\frac{d}{1+d}$ to $0$), whereas $\overline{t}$ is increasing in $\gamma$ (from $\frac{d}{1+d}$ to $1$). Moreover, the left-hand side of the equation is close to zero for $\gamma$ close to $\frac{1}{2}$, whereas the left-hand side of the equation is close to one for $\gamma$ close to $1$. Therefore, the equation is true for at least one $\gamma$.
The question in my mind is, would the equation be true for unique $\gamma$. In other words, would there exist $\gamma^{*}$ such that the left-hand side of the equation is smaller than $\kappa$ for $\gamma<\gamma^{*}$ and is larger than $\kappa$ for $\gamma>\gamma^{*}$? I have been playing around with this equation numerically, assuming that $t \sim$ Beta$(\alpha,\beta)$, and so far, I have not been able to find a numerical example, where the equation is true for more than one $\gamma$.
To the extent that the interval $(\underline{t},\overline{t})$ is expanding as $\gamma$ increases, whereas $(0,\underline{t})$ is shrinking as $\gamma$ increases, I think it's plausible to expect the equation being true for unique $\gamma$. But I have no clue how I am going to show this..
I'd greatly appreciate any kind of comments on the problem. Thanks so much for your time!
John Kim
