用 \(\mathrm{p}\) 和 \(\mathrm{q}\) 表示买者和卖者的利润函数。
p 和 q 在什么条件下使买者和卖者的利润和最大?
（3）若卖方选择交易价格，求最优条件下各自的利润。
solution：
1）买者的利润函数：
\(\pi^{b}=\int_{0}^{q}(210-q) d q-p q=(210-p) q-\frac{1}{2} q^{2}\)
卖者的利润函数 \(\pi^{s}=(p-10) q\)
2）利润函数之和最大化：
\(\max : \pi=\pi^{b}+\pi^{s}=200 q-\frac{1}{2} q^{2}\)
\(Fo{c}: \frac{d \pi}{d q}=200-q=0\)
解得：
\(q^{*}=200, \quad p^{*}=10\)
3）卖方选择交易价格：
\(\max : \pi^{s}=(p-10) \cdot q\)
st: \(\pi^{b}=(210-p) q-\frac{1}{2} q^{2}=0\)
\(\Rightarrow \max _{p}: \pi^{s}=(p-10)(420-2 p)\)
\(\Rightarrow\left\{\begin{array}{l}p^{*}=110 \\ q^{*}=200\end{array}\right.\)
\(\Rightarrow\left\{\begin{array}{l}\pi^{s}=20000 \\ \pi_{b}=0\end{array}\right.\)
4）买方选择交易价格：
\(\max : \pi^{b}=(210-p) \cdot q-\frac{1}{2} q^{2}\)
st: \(\left\{\begin{array}{l}\pi^{s}=(p-10) q=0 \\ p \geqslant 10\end{array}\right.\)
\(\Rightarrow p^{*}=10 \quad q^{*}=200\)
\(\Rightarrow\left\{\begin{array}{l}\pi_{b}=20000 \\ \pi_{s}=0\end{array}\right.\)
假定委托人与代理人之间签订一个线性合约: \(\boldsymbol{w}=s+b \boldsymbol{y},\) 代理人会采取什么行动? 代理人的行动“ \(\boldsymbol{\alpha} "\) 会如何随 \(\boldsymbol{b}\) 而发生变化? 代理人的行动会如何随 \(\boldsymbol{k}\) 而发生变动?
现在假定代理人的效用函数形式为 \[ u(x)=-e^{-r x} \] 又假定代理人的努力成本函数为 \(\boldsymbol{C}(\boldsymbol{\alpha})=\frac{\mathbf{1}}{\mathbf{2}} {\alpha}^{2}\)
证明，最优线性契约中的激励系数 \(\boldsymbol{b}^{*}\) 必满足 \[ b^{*}=\frac{k^{2}}{k^{2}+r \sigma^{2}} \]
solution：
1）线性契约：
不妨假设代理人风险中性且
\(c^{\prime}(\alpha)>0 . c^{\prime \prime}(\alpha)>0\)
\(\max _{\alpha}: E(w)=5+b k \alpha-c(\alpha)\)
st: \(s+b k \alpha-c(\alpha) \geqslant 0\)
Foc: \(\quad c^{\prime}\left(\alpha^{*}\right)=b k\)
\(\Rightarrow c^{\prime \prime}\left(\alpha^{*}\right) d \alpha^{*}=b d k+k d b\)
\(\begin{aligned} \Rightarrow \frac{\partial \alpha^{*}}{\partial b} &=\frac{k}{c^{\prime \prime}\left(\alpha^{*}\right)}>0 \\ \frac{\partial \alpha^{*}}{\partial k} &=\frac{b}{c^{\prime \prime}\left(\alpha^{*}\right)}>0 \end{aligned}\)
故 \(\alpha^{*}\)随k，b的上升而上升
假设非负约束满足
2）若 \(u(x)=-e^{-r x} , c(\alpha)=\frac{1}{2} \alpha^{2}\)
委托人收益最大化：设为风险中性
\(\max : E \pi=(1-b) k \alpha-s\)
st: \(\max : \pi_{1}=E U(w)-c(\alpha)\)
\(=s+b k \alpha-\frac{1}{2} r b^{2} \sigma^{2}-\frac{1}{2} \alpha^{2}\)
\(s+b k \alpha-\frac{1}{2} r b^{2} \sigma^{2}-\frac{1}{2} \alpha^{2} \geqslant 0\)
对IC约束简化：
\(\alpha^{*}=b k\)
对IR约束简化： \(s^{*}=\frac{1}{2} b^{2}\left(r \sigma^{2}-k^{2}\right)\)
委托人收益最大化：
\(\max _{b}: E \pi=(1-b) b k^{2}-\frac{1}{2} b^{2}\left(r \sigma^{2}-k^{2}\right)\)
\(Foc: \frac{d E \pi}{d b}=(1-2 b) k^{2}-b\left(r \sigma^{2}-k^{2}\right)\)
\(\Rightarrow \quad b^{*}=\frac{k^{2}}{k^{2}+r \sigma^{2}}\)
solution：
1）父母效用最大化：
\(\max : \pi_p=U_{2}\left(\gamma_{2}-L\right)+2 v_{1}\left(\gamma_{1}+L\right)\)
\(Fo{c}: \frac{d \pi_p}{d L}=-U_{2}^{\prime}\left(r_{2}-L\right)+\alpha U_{1}^{\prime}\left(\gamma_{1}+L\right)=0\)
\(\Rightarrow \quad \partial U_{1}^{\prime}\left(x_{1}+L\right)=U_{2}^{\prime}\left(Y_{2}-L\right)\) \((*)\)
2）孩子效用最大化： \(\max : \pi_{c}=U_{1}\left[\gamma_{1}(r)+L(r)\right]\)
\(Fo{c}: \quad \frac{d \pi_c}{d r}=U_{1}^{\prime}\left(\gamma_{1}+L\right) \cdot\left(\gamma_{1}^{\prime}(r)+L^{\prime}(r))=0\right.\)
\(\Rightarrow \quad \gamma_{1}^{\prime}(r)+L^{\prime}(r)=0 \quad(* *)\)
3）*式对r求导得：
\(\partial U_{1}^{\prime \prime}\left(\gamma_{1}+L\right) \cdot\left(\gamma_{1}^{\prime}(r)+L^{\prime}(r)\right)=U_{2}^{\prime \prime}\left(\gamma_{2}-L\right)\left(\gamma_{2}^{\prime}(r)-L^{\prime}(r)\right)\)
由**式可知：
\(\gamma_{1}^{\prime}(r)+L^{\prime}(r)=0\)
\(\Rightarrow \gamma_{2}^{\prime}(r)-L^{\prime}(r)=0\)
\(\Rightarrow \quad \gamma_{1}^{\prime}(r)+\gamma_{2}^{\prime}(r)=0\)
\(\Rightarrow\) 盖子的最终目标是选择r最大化 \(r_{1}(r)+r_{2}(r)\)