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I'm stuck in showing that the following function is increasing over the domain $\left[0,x_0\right]$:

$$\Pi\left(z\right) = \int_{0}^{\phi\left(z\right)}\int_{x}^{\bar{x}}\left(2y-b\left(x\right)-x\right)dG\left(y\right)dG\left(x\right)\\ + \int_{0}^{\phi\left(z\right)}\int_{0}^{x}\left(y-b\left(x\right)\right)dG\left(y\right)dG\left(x\right)\\ + \int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\int_{0}^{\bar{x}}\left(y-x\right)dG\left(y\right)dG\left(x\right)dG\left(t\right)$$

Primitives of the problem:

$X,Y\sim G$ with $supp\left(G\right)=\left[0,\bar{x}\right]$

$\phi\left(.\right)=b^{-1}\left(.\right)$

$\phi\left(x_0\right)=E\left[X\right]$

$z\in\left[0,x_0\right]$ with $x_0<\bar{x}$

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:

$\phi\left(z\right)\leq\phi\left(x_0\right)\leq\phi\left(\bar{x}\right)=\bar{x}$

$\phi\left(z\right)>z$ ; $\phi\left(x_0\right)>x_0$

$\phi\left(x_0\right)=E\left[Y\right]\geq\phi\left(z\right)\geq z$ $\Rightarrow E\left[Y\right]\geq z$

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:

$$\frac{\partial\Pi}{\partial z} = g\left(\phi\left(z\right)\right)\frac{\partial\phi\left(z\right)}{\partial z}\left[\int_{\phi\left(z\right)}^{\bar{x}}\left(2y-z-\phi\left(z\right)\right)dG\left(y\right)+\int_{0}^{\phi\left(z\right)}\left(y-z\right)dG\left(y\right)\right.\\ \left.-\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\left(\phi\left(x_0\right)-x\right)dG\left(x\right)+\phi\left(z\right)\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}g\left(t\right)dt-\phi\left(x_0\right)\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}g\left(t\right)dt\right]$$

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:

$$\small\int_{\phi\left(z\right)}^{\bar{x}}\left(2y-z-\phi\left(z\right)\right)dG\left(y\right)+\int_{0}^{\phi\left(z\right)}\left(y-z\right)dG\left(y\right)-\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\left(2\phi\left(x_0\right)-x-\phi\left(z\right)\right)dG\left(x\right)$$ is $>0$.

We can simplify the expression to get:

$$\small \int_{\phi\left(z\right)}^{\bar{x}}\left(2x-z-\phi\left(z\right)\right)dG\left(x\right)+\int_{0}^{\bar{x}}\left(x-z\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\bar{x}}\left(x-z\right)dG\left(x\right)-\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\left(2\phi\left(x_0\right)-x-\phi\left(z\right)\right)dG\left(x\right)$$ to be finally left with:

$$\left(E\left[X\right]-z\right)+\left[\int_{\phi\left(z\right)}^{\bar{x}}\left(x-\phi\left(z\right)\right)dG\left(x\right)\right]-\left[\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\left(2\phi\left(x_0\right)-x-\phi\left(z\right)\right)dG\left(x\right)\right]$$

Someone to help me for this? At least, to find some conditions on the CDF $G$ that could ensure $\Pi(z)$ is indeed increasing over $\left[0,x_0\right]$? Maybe, using the derivatives path is not the appropriate one (I guess).

Many thanks!

EDIT: Since we have that $E[X]>z$ so that maybe it suffices to show that: $$ \left[\int_{\phi\left(z\right)}^{\bar{x}}\left(x-\phi\left(z\right)\right)dG\left(x\right)\right]-\left[\int_{\phi\left(z\right)}^{\phi\left(x_0\right)}\left(2\phi\left(x_0\right)-x-\phi\left(z\right)\right)dG\left(x\right)\right]>0 $$

No? Any suggestions please?

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