Let $z^*(F):= \arg\max_z (1-F(z)) \times g(E_F[x|x>z]) + F(z) \times g(E_F[x|x\leq z])$, where $F$ is the CDF of a continuous distribution, and $g'>0, g''>0$.
Could someone help me prove/verify that if $F_1$ first order stochastic dominates $F_2$, then $z^*(F_1)>z^*(F_2)$? If it is easier, let $g(x):=x^{1/(1-\alpha)}$, where $0<\alpha<1$.
Thank you!
I figured the answer out! :)
Assume $g(x)=x^{1/(1-\alpha)}$, where $0<\alpha<1$.
Proof: Step 1: By taking the derivative of the maximand, we get $z^*= \Phi(z^*;F)$, where $\Phi(z;F):=\alpha \frac{(E_F[x|x>z])^{1/(1-\alpha}-(E_F[x|x\leq z])^{1/(1-\alpha}}{(E_F[x|x>z])^{\alpha/(1-\alpha}-(E_F[x|x\leq z])^{\alpha/(1-\alpha}}.$ It thus suffices to show that $\Phi(z;F_1)>\Phi(z;F_2)$ for all $z$.
Step 2: Since $F1$ fosd $F2$, it follows that $E_{F_1}[x|x>z]>E_{F_2}[x|x>z]$ and $E_{F_1}[x|x\leq z]>E_{F_2}[x|x\leq z]$
Step 3: Lemma: if $g$ is such that $g',g''>0$, and if $x_1>y_1,x_2>y_2$ and $x_1>x_2$, $y_1>y_2$, then $\frac{g(x_1)-g(y_1)}{x_1-y_1}>\frac{g(x_2)-g(y_2)}{x_2-y_2}$. Proof is a bit lengthy but straightforward.
The three steps complete the proof.