I am an economist and need some math help.
My question is simply prove or disprove that $b_1(v) \leq b_2(v)$, which are selections of two maximization problems.
I have two objective functions. $v \in [0,1]$ and $ 0 \leq b \leq v.$
\begin{gather} \Pi(b; v) =\int_{0}^{b} (v - 2b)h(z)dz + (-b)(1-H(b)) \\ \Pi(b; v)= \int_{0}^{b} (v-2b)g(z)dz + (-b)(1-G(b)) \end{gather}
Which yield two FOCs. (let's ignore corner solution case $b(v)=0$) \begin{gather} (v-b)h(b) - [1-H(b)] = 0 \\ (v-b)g(b) - [1-G(b)] = 0 \end{gather}
From the two FOCs, I can get optimal selections functions $b_1(v)$ and $b_2(v)$. (1 denotes above, 2 denotes below). I want to prove of disprove $b_1(v) \leq b_2(v)$, for all $v$.
I have several conditions.
- $G(b)>H(b)>b$ on (0,1)
- I know the explicit form of $b_2(v)$ \begin{eqnarray} b_2(v)= v-\int^{v}_{0}\frac{1-H(v)}{1-H(x)}dx \end{eqnarray}
- $G(b)=H(b_2^{-1}(b))$
- $\frac{g(b)}{1-G(b)}$, $\frac{h(b)}{1-H(b)}$ are monotone increasing. Hazard rates are montone increasing
- Assume $G(b)$ and $H(b)$ are concave if you need it. But I hope to have a result without this.
So far I think of one way to show $b_1(v) \leq b_2(v)$
for all $b^*$ such that \begin{gather} (v-b^*)h(b^*) - [1-H(b^*)] = 0 \end{gather} if $b^*$ implies \begin{gather} (v-b^*)g(b^*) - [1-G(b^*)] > 0 \end{gather}
then $b_2(v) \geq b_1(v)$ (if this right?)
This is equivalent to show $\frac{g(b)}{1-G(b)}$>$\frac{h(b)}{1-H(b)}$
Do you think 1-4 implies $\frac{g(b)}{1-G(b)}$>$\frac{h(b)}{1-H(b)}$ ?
or do you have an idea to prove or disporve $b_1(v) \leq b_2(v)$?