1.关于未来的不确定性，麦克斯·普尔曼正好生活在两个时期，\(t=\)\(0,1。\)让\(c{t}\in\mathbb{R}\)表示他在时期\(t的消费。\)Max对两个时期消费流的偏好（以\(t=0\)计算）由函数表示 \[ U\left(c_{0}, c_{1}\right)=u\left(c_{0}\right)+\delta E u\left(c_{1}\right) \] 其中，\(\delta\)是一个贴现因子，\(u（\cdot）\)是一个递增且严格凹的效用函数，\(E\)运算符表示他对周期t=1中的事件的期望（在\(t=0\)）。为简单起见，还可以假设消费的边际效用是凸的，即，\(u^{\prime \prime}>0\)
假设最初没有不确定性。设\(w{0}\geq 0\)为Max在第0阶段的收入，设\(w{1}\geq 0\)为Max在第1阶段的收入。\(Max\)可以储蓄或借贷。让\(s\in\mathbb{R}\)表示他的储蓄（注意，如果他借款，\(s\)可能是负数），让\(\rho\)表示储蓄的总回报（即，\(\rho=1+R\)，其中，\(R\)是利率）。因此，他在第0阶段的消耗是\(w{1}-\)，在第1阶段的消耗是\(w{1}+\rho s\)在整个练习中假设内部解。
1)写出Max选择储蓄\(s^{*}\)为正的充要条件。
3)现在假设Max在第一阶段的收入中面临不确定性。具体地说，假设他的第一阶段收入由\(w{1}+\tilde{x}，\)给出，其中\(w{1}\geq 0\)和随机变量\(\tilde{x}\)的预期值为\(E（\tilde{x}）=0。\)让\(s^{**}\)表示Max在这种情况下的新的最优储蓄。显示\(s^{**}>s^{*}\)。[提示：假设\(s^{**}=s^{*}\)，并使用Jensen不等式比较一阶条件。
solution
1)效用最大化：
\(\begin{aligned} \max : & U=\mu\left(c_0)+\delta u ( c_{1}\right) \\ &\left.=u\left(w_{0}-s\right)+\delta \mu (w_{1}+\rho s\right) \end{aligned}\)
由于 \(\frac{d u}{d s}=-u^{\prime}\left(c_{0}\right)+\delta \rho u^{\prime}\left(c_{1}\right)\)
\(\frac{d^{2} u}{d s^{2}}=u^{\prime \prime}\left(c_{0}\right)+\delta \rho^{2} u^{n}\left(c_{1}\right)<0\)
故 \(s^{*}>0\)的充要条件是：（正实根）
\(u(0)>0 \quad 即 \quad \delta \rho u^{\prime}\left(w_{1}\right)>u^{\prime}\left(w_{0}\right)\)
2）若 \(w_{1}=0: \quad c_{0}=w_{0}-s, \quad c_{1}=\rho s\)
\(\begin{aligned} \frac{d u}{d s} &\left.=-u^{\prime}\left(c_{0}\right)+\delta \rho u^{\prime} (c_{1}\right) \\ &=-u^{\prime}\left(w_{0}-s\right)+\delta \rho u^{\prime}(\rho s)=0 \end{aligned}\)
将上式去全微分：
\(u^{\prime \prime}\left(c_{0}\right) d s+\delta u^{\prime}\left(c_{1}\right) d \rho+\delta \rho u^{\prime}\left(c_{1}\right) d \rho s=0\)
\(\Rightarrow \frac{d s}{d p}=-\frac{\left.\delta u^{\prime}\left(c_{1}\right)+\delta \rho s u^{\prime \prime} c c_{1}\right)}{u^{\prime \prime}\left(c_{0}\right)+\delta \rho^{2} u^{\prime \prime}\left(c_{1}\right)}>0\)
\(\Rightarrow \quad \delta u^{\prime}\left(c_{1}\right)+\delta c_{1} u^{\prime \prime}\left(c_{1}\right)>0\)
\(\Rightarrow \quad-\frac{c_{1} u^{\prime \prime}\left(c_{1}\right)}{u^{\prime}\left(c_{1}\right)}<1\)
\(\Rightarrow \quad R_{R}\left(c_{1}\right) < 1\)
3)期望效用最大化：
\(\max : Eu=\mu\left(c_{0}\right)+\delta E u\left(u_{1}\right)\)
\(\left(\tilde{w}_{1}=w_{1}+\bar{x}\right)=u\left(w_{0}-s\right)+\delta E u\left(\tilde{w}_{1}+\rho s\right)\)
\(\left(w_{1}^{i}=w_{1}+x^{i}\right)=u\left(w_{0}-s\right)+\delta \sum_{i=1}^{N} \pi_{i} u\left(w_{1}^{i}+\rho s\right)\)
假设1时刻存在N种状态，发生概率为 \(\pi_{i}, \sum_{i=1}^{N} T_{i}=1\)
\(\begin{aligned} Foc: \frac{d Eu}{d s} &=-u^{\prime}\left(w_{0}-s^{* *}\right)+\delta \sum_{i=1}^{N} \pi_{i} u^{\prime}\left(w_{1}^{i}+\rho s^{* *}\right) \\ &=-u^{\prime}\left(w_{0}-s^{* *}\right)+\delta E u^{\prime}\left(\tilde{w}_{1}+\rho s^{* *}\right) \\ &=0 \end{aligned}\)
由1）知： \(-u^{\prime}\left(w_{0}-s^{*}\right)+\delta u^{\prime}\left(w_{1}+\rho s^{*}\right)=0\)
由于 \(\begin{aligned} E u^{\prime}\left(\tilde{w}_{1}+\rho S^{* *}\right) &=\sum_{i=1}^{N} \pi_{i} u^{\prime}\left(w_{1}^{i}+\rho s^{* *}\right) \\ &>u^{\prime}\left[\sum_{i=1}^N \pi_{i}\left(w_{1}^{i}+\rho s^{* *}\right)\right] \\ &=u^{\prime}\left[E\left( \tilde{w}_{1}+\rho s^{* *}\right)\right] \\ &=u^{\prime}\left(w_{1}+\rho s^{* *}\right) \end{aligned}\)
故 \(\begin{aligned} 0=\frac{d E u}{d s} \mid_{s=s^{* *}} &>-u^{\prime}\left(w_{0}-s^{* *}\right)+u^{\prime}\left(w_{1}+\rho_{S}^{* *}\right) \\ &=\frac{d u}{d s} \mid _{s=s^{* *}} \end{aligned}\)
则\(s^{*{*}}>s^{*}\)
求竞争性市场均衡条件下的产品价格以及每个消费者的情说。
交易后，人们的效用水平上升了还是下降了? 为什么?
帕累托最优的资源分配方案，即契约曲线表达式。
A 的效用函数变成 \(\mathrm{U}\left(\mathrm{X}_{\mathrm{A}}, \mathrm{Y}_{\mathrm{A}}\right)=\beta \ln \mathrm{X}_{\mathrm{A}}+\beta \ln \mathrm{Y}_{\mathrm{A}},\) 那么 (1)\(\sim(4)\) 的答案是否会发生变化? 为什么?
solution：
1）不妨设 \(p_{y}=1 ,p=p_{x} / p_{y}\)
A,B效用最大化：
\(m a x: \quad u^{A}=x_{A} y_{A}\) st : \(\quad p \cdot x_{A}+y_{A}=p e_{A}^{x}+e_{A}^{y}\)
\(\begin{aligned} \max : & u^{B}=\ln x_{B}+\alpha \ln y_{B} \\ \ st: & p \cdot x_{B}+y_{B}=p e_{B}^{x}+e_{B}^{y} \end{aligned}\)
拉格朗日函数：
\(\left\{\begin{array}{l}\mathcal{L}_{A}=x_{A} y_{A}+\lambda\left[p e_{A}^{x}+e_{A}^{y}-p \cdot x_{A}-y_{A}\right] \\ \mathcal{L}_{B}=\ln x_{B}+\alpha \ln y_{B}+\lambda\left[p e_{B}^{x}+e_{B}^{y}-p x_{B}-y_{B}\right]\end{array}\right.\)
FOC:\(\left\{\begin{array}{l}\frac{\partial \mathcal{L}_{A}}{\partial x_{A}}=y_{A}-\lambda p=0 \\ \frac{\partial \mathcal{L}_{B}}{\partial y_{A}}=x_{A}-\lambda=0\end{array}\right.\)
FOC:\(\left\{\begin{array}{l}\frac{\partial \mathcal{L}_{B}}{\partial x_{B}}=\frac{1}{x_{B}}-\lambda p=0 \\ \frac{\partial \mathcal{L}_{B}}{\partial y_{B}}=\frac{\alpha}{y_{B}}-\lambda=0\end{array}\right.\)
解得：\(\left\{\begin{array}{l}x_{A}=\frac{p e_{A}^{x}+e_{A}^{y}}{2 p} \\ Y_{A}=\frac{p e_{A}^{x}+e_{A}^{y}}{2}\end{array} \quad\left\{\begin{array}{l}x_{B}=\frac{p e_{B}^{x}+e_{B}^{y}}{(\alpha+1) p} \\ Y_{B}=\frac{\alpha\left(p e_B^{x}+e_{B}^{Y}\right)}{(\alpha+1)}\end{array}\right.\right.\)
竞争性均衡： \(Y_{A}+Y_{B}=E^{Y}\)
解得：\(p=\frac{(\alpha+1) e_{A}^{Y}+2 e_{B}^{Y}}{(\alpha+1) e_{A}^{x}+2 \alpha e_{B}^{x}}\)
2）当 \(\alpha=1\)时， \(p \equiv \frac{E^{y}}{E^{x}}\)
初始禀赋的移动不改变均衡价格
当\(\alpha\neq 1\)是， \(p=\frac{(\alpha+1) e_{A}^{Y}+2 e_{B}^{Y}}{(\alpha+1) e_{A}^{x}+2 \alpha e_{B}^{x}}\)
随 $$的变化而变化
x,y商品的总禀赋不变，x,y市场的总需求恒等于总禀赋。
3）交易后人们的效用上升
又福利经济学第一定理知，竞争性均衡为帕累托有效，故效用会有所增加，至少刽比原来差
4）契约曲线：
\(\max : x_{A} y_{A}\)
st: \(\left\{\begin{array}{l}\bar{u}_{B}=\ln x_{B}+\alpha \ln y_{B} \\ x_{A}+x_{B}=E^{x} \\ y_{A}+y_{B}=E^{Y}\end{array}\right.\)
\(\mathcal{L}=x_{A} y_{A}+\lambda\left[\bar{u}_{B}-\ln \left(E^{x}-x_{A}\right)-\alpha \ln \left(E^{y}-y_{A}\right)\right]\)
FOC:\(\left\{\begin{array}{l}\frac{\partial \mathcal{L}}{\partial x_{A}}=y_{A}+\frac{\lambda}{E^{x}-x_{A}}=0 \\ \frac{\partial \mathcal{L}}{\partial y_{A}}=x_{A}+\frac{\partial \lambda}{E^{y}-y_ A}=0\end{array}\right.\)
解得： \(x_{A} \cdot E^{x}-2 y_{A} E^{y}+(\alpha-1) x_{A} y_{A}=0\) \(\left( 0 \leq x_{A} \leq E^{x}\right)\)
5）若 \(u\left(x_{A} , y_{A}\right)=\beta \ln x_{A}+\beta \ln y_{A} \quad(\beta>0)\)
则以上结果不变，因为 \(u\left(x_{A}, y_{A}\right)=x_{A} y_{A}\)仅仅是进行了正单调变换。
3.考虑下面的Stackelberg竞争。有两家公司。公司1是Stackelberg leader并首先选择其数量\(q{1}\geq 0\)。公司2是follower并在观察公司l的选择后选择其数量\(q{2}\geq 0\)。假设市场需求由\(P\ left（q{1}，q{2}\ right）=a-q{1}-q{2}\)给出，其中\(a>0\) 。我们允许 对于负价格。让\(c{i}>0\)为公司\(i\)的边际生产成本。假设\(a>\max\left\{c{1}，c{2}\right\}\)
1）如果企业1选择\(q{1}\leq a-c{2}\)，那么企业2的最优数量是多少？、
2）如果企业1选择$q{1}>a-c{2}，$2的最优数量是多少？
假设\(c{1}>c{2}\)和\(2 c{1}-c{2}<a.\)企业1的最佳产出水平是多少？
\(假设\)c{1}<c{2}$和$2 c{2}-c{1}>a。厂商1的最佳产出水平是多少？
solution
企业2利润最大化：
\(\max : \pi_{2}=\left(a-c_{2}-q_{1}-q_{2}\right) q_{2}\)
\(Foc: \frac{\partial \pi_{2}}{\partial q_{2}}=a-c_{2}-q_{1}-2 q_{2}=0\)
则企业2的反应函数为：
\(q_{2}=\left\{\begin{array}{cc}\frac{a-c_{2}-q_{1}}{2} & q_{1} \leqslant a-c_{2} \\ 0 & q_{1}>a-c_{2}\end{array}\right.\)
企业1利润最大化：
若 \(q_{1}>a-c_{2}, \quad q_{2}=0\)，企业1垄断生产。
\(\max : \pi_{1}^{m}=\left(a-c_{1}-q_{1}\right) q_{1}\)
\(Foc: \frac{d \pi_{1}^m}{d q_{1}}=a-c_{1}-2 q_{1}=0\)
解得： \(q_{1}^{m}=\frac{a-c_{1}}{2}>a-c_{2}\)
即\(a<2 c_{2}-c_{1}\)
由于：\(a>\max \left\{c_{1}, c_{2}\right\}\)
则：\(2 c_{2}-c_{1}>\max \left\{c_{1}, c_{2}\right\}\)
即\(c_{2}>c_{1}\)
综上：\(c_{2}>c_{1}\) 且 \(a<2 c_{2}-c_{1}\)时
\(q_{1}=\frac{a-c_{2}}{2}, \quad q_{2}=0\)
II)若\(q_{1} \leq a-c_{2}\)，斯塔克伯格竞争：
此时
\(a \geqslant 2 c_{2}-c_{1}\) 或 \(c_{2} \leq c_{1}\)
\(\begin{aligned} \max : & \pi_{1}^{s}=\left[a-c_{1}-q_{1}-q_{2}\left(q_{1}\right)\right] q_{1} \\=& \frac{1}{2}\left(a-2 c_{1}+c_{2}-q_{1}\right) q_{1} \end{aligned}\)
若\(a<2 c_{1}-c_{2} ; q_{1}=0 , q_{2}=\frac{a-c_{2}}{2}\)角点解
若 \(\Rightarrow 2 c_{1}-c_{2}>\max \left\{c_{1}, c_{2}\right\}\) \(\Rightarrow c_{1}>c_{2}\)
若 \(a>2 c_{1}-c_{2} ; q_{1}=\frac{a-2 c_{1}+c_{2}}{2}, q_{2}=\frac{a+2 c_{1}-3 c_{2}}{4}\)
此时 \(\begin{aligned} a-c_{2} & \geqslant q_{1} \\ \Rightarrow a \geqslant & 3 c_{2}-2 c_{1} \end{aligned}\)
由于 \(\left(2 c_{1}-c_{2}\right)-\left(3c_2-2 c_{1})=4 ( c_{1}-c_{2}\right)\)
若\(c_{1}>c_{2}\)，则上是成立
若\(c_{1}<c_{2}\left\{\begin{array}{l}a \geqslant 3 c_{2}-2 c_{1} ，则上是成立\\ a<3 c_{2}-2 c_{1}\end{array}\right.\)
\(c_{1}<c_{2}\left\{\begin{array}{l}a \geqslant 3 c_{2}-2 c_{1} \\ a<3 c_{2}-2 c_{1},q_1=a-c_2-q_2=0\end{array}\right.\)
设事件\(A: a<2 c_{2}-c_{1} ; B: \quad c_{2}>c_{1}\)
1）AB \(q_{1}=\frac{a-c_{2}}{2}, q_{2}=0\)
2)A-B \(q_{1}=0 , q_{2}=\frac{a-c_{2}}{2}\)
3)1-AB \(q_{1}=\frac{a-2 c_{1}+c_{2}}{2} ; q_{2}=\frac{a+2 c_{1}-3 c_{2}}{4}\)
4)B-A
\(\left\{\begin{array}{l}a \geqslant 3 c_{2}-2 c_{1}: q_{1}=\frac{a-2 c_{1}+c_{2}}{2} ; \quad q_{2}=\frac{a+2 c_{1}-3 c_{2}}{4} \\ a<3 c_{2}-2 c_{1}: q_{1}=a-c_{2} \quad q_{2}=0\end{array}\right.\)
图示