I'm stuck in showing that the following function is increasing over the domain $\left[0,\hat{b}\right]$:
\begin{eqnarray} \Pi\left(z\right) & = & \int_{0}^{\phi\left(z\right)}\int_{x}^{\bar{x}}\left(2y-b\left(x\right)-x\right)dG\left(y\right)dG\left(x\right)\nonumber \\ & + & \int_{0}^{\phi\left(z\right)}\int_{0}^{x}\left(y-b\left(x\right)\right)dG\left(y\right)dG\left(x\right)\\ & + & \int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\int_{0}^{\bar{x}}\left(y-x\right)dG\left(y\right)dG\left(x\right)dG\left(t\right)\nonumber \end{eqnarray}
Primitives of the problem:
$X,Y\sim G$ with $\operatorname{supp}\left(G\right)=\left[0,\bar{x}\right]$
$\phi\left(.\right)=b^{-1}\left(.\right)$
$\phi\left(\hat{b}\right)=E\left[X\right]$
$z\in\left[0,\hat{b}\right]$ with $\hat{b}<\bar{x}$.
Function $b\left(x\right)$ is concave over $\left[0,\bar{x}\right]$ with: $b\left(\bar{x}\right)=\bar{x}$, $b(0)=0$ and $b\left(x\right)\leq x$, $\forall x\in[0,\bar{x}]$. Thus, $\phi\left(.\right)$ is convex over $\left[0,\bar{x}\right]$ with $\phi\left(x\right)\geq x,$ $\forall x\in\left[0,\bar{x}\right]$ and $\phi\left(\bar{x}\right)=\bar{x}$.
We have the following:
$\phi\left(z\right)\leq\phi\left(\hat{b}\right)\leq\phi\left(\bar{x}\right)=\bar{x}$
$\phi\left(z\right)>z$ et $\phi\left(\hat{b}\right)>\hat{b}$
$\phi\left(\hat{b}\right)=E\left[Y\right]\geq\phi\left(z\right)\geq z$ $\Rightarrow E\left[Y\right]\geq z$.
I've tried to show that $\Pi'(z)>0$, but I get blocked at the end and I'm not quite sure that it is the good way to use... Taking the derivative at $z$, I get:
\begin{eqnarray*} \frac{\partial\Pi}{\partial z} & = & g\left(\phi\left(z\right)\right)\frac{\partial\phi\left(z\right)}{\partial z}\left[\int_{\phi\left(z\right)}^{\bar{x}}\left(2y-z-\phi\left(z\right)\right)dG\left(y\right)+\int_{0}^{\phi\left(z\right)}\left(y-z\right)dG\left(y\right)\right.\\ & & \left.-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(\phi\left(\hat{b}\right)-x\right)dG\left(x\right)+\phi\left(z\right)\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}g\left(t\right)dt-\phi\left(\hat{b}\right)\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}g\left(t\right)dt\right] \end{eqnarray*}
We have that $g\left(\phi\left(z\right)\right)\frac{\partial\phi\left(z\right)}{\partial z}>0$ so we are left with showing that:
$$\small \int_{\phi\left(z\right)}^{\bar{x}}\left(2y-z-\phi\left(z\right)\right)dG\left(y\right)+\int_{0}^{\phi\left(z\right)}\left(y-z\right)dG\left(y\right)-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right) \gtrless 0 $$
We can simplify the expression to get:
\begin{eqnarray*}\scriptsize \int_{\phi\left(z\right)}^{\bar{x}}\left(2x-z-\phi\left(z\right)\right)dG\left(x\right)+\int_{0}^{\bar{x}}\left(x-z\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\bar{x}}\left(x-z\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right) & \gtrless & 0\\ \scriptsize E\left[X\right]-z-\int_{\phi\left(z\right)}^{\bar{x}}\left(x-z\right)dG\left(x\right)+\int_{\phi\left(z\right)}^{\bar{x}}\left(2x-\phi\left(z\right)\right)dG\left(x\right)-z\int_{0}^{\bar{x}}dG\left(x\right)-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right) & \gtrless & 0\\ \scriptsize E\left[X\right]-z-\int_{\phi\left(z\right)}^{\bar{x}}xdG\left(x\right)+\int_{\phi\left(z\right)}^{\bar{x}}\left(2x-\phi\left(z\right)\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right) & \gtrless & 0\\ E\left[X\right]-z+\int_{\phi\left(z\right)}^{\bar{x}}\left(x-\phi\left(z\right)\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right) & \gtrless & 0 \end{eqnarray*}
to be finally left with: \begin{eqnarray*} \left(E\left[X\right]-z\right)+\left[\int_{\phi\left(z\right)}^{\bar{x}}\left(x-\phi\left(z\right)\right)dG\left(x\right)\right]-\left[\int_{\phi\left(z\right)}^{\phi\left(\hat{b}\right)}\left(2\phi\left(\hat{b}\right)-x-\phi\left(z\right)\right)dG\left(x\right)\right] & \gtrless & 0 \end{eqnarray*}
Could anyone help me for this? At least, to find some conditions on the CDF $G$ that could ensure $\Pi(z)$ is indeed increasing over $\left[0,\hat{b}\right]$? Maybe, using the derivatives path was not the appropriate way to do it (I guess).
Many thanks!
