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\title{Advanced Microeconomics Homework 2}
\date\today
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\section*{Problem 1}
(a)
Alex:
(i) Interior solution:\\
$\max \limits_{(x_1^{A},x_2^{A})} u^{A}=(x_1^{A}-1)^{\frac{1}{3}}(x_2^{A}-1)^{\frac{2}{3}}$\\
$s.t.\quad p_1x_1^{A}+p_2x_2^{A}\leq p_1e_1^{A}+p_2e_2^{A}=p_1+4p_2 $\\
$\Longrightarrow \max \limits_{(x_1^{A},x_2^{A})}(x_1^{A}-1)^{\frac{1}{3}}(x_2^{A}-1)^{\frac{2}{3}}-\lambda (p_1x_1^{A}+p_2x_2^{A}-p_1-4p_2) $\\
$FOC: \quad \frac{1}{3}(x_1^{A}-1)^{-\frac{2}{3}}(x_2^{A}-1)^{\frac{2}{3}}-\lambda p_1=0$\\
$\quad \frac{2}{3}(x_1^{A}-1)^{\frac{1}{3}}(x_2^{A}-1)^{-\frac{1}{3}}-\lambda p_2=0$\\
$\Longrightarrow \frac{p_2}{p_1}=\frac{2(x_1^{A}-1)}{x_2^{A}-1}$\\
$p_2(x_2^{A}-1)=2p_1(x_1^{A}-1)$\\
together with $\quad p_1x_1^{A}+p_2x_2^{A}=p_1+4p_2 $\\
$\Longrightarrow x_1^{A}=1+\frac{p_2}{p_1}, \quad x_2^{A}=3$\\
(ii) Corner solution:\\
If $x_1^A=0,\quad p_2x_2^A=p_1+4p_2\Longrightarrow x_2^A=\frac{p_1}{p_2}+4$;\\
If $x_2^A=0,\quad p_1x_1^A=p_1+4p_2\Longrightarrow x_1^A=4\frac{p_2}{p_1}+1$.\\
Bob:
(i) Interior solution:\\
$\max \limits_{(x_1^{B},x_2^{B})} u^{B}=(x_1^{B}-1)^{\frac{2}{3}}(x_2^{B}-1)^{\frac{1}{3}}$\\
$s.t.\quad p_1x_1^{B}+p_2x_2^{B}\leq p_1e_1^{B}+p_2e_2^{B}=4p_1+p_2 $\\
$\Longrightarrow \max \limits_{(x_1^{B},x_2^{B})}(x_1^{B}-1)^{\frac{2}{3}}(x_2^{B}-1)^{\frac{1}{3}}-\lambda (p_1x_1^{B}+p_2x_2^{B}-4p_1-p_2) $\\
$FOC: \quad \frac{2}{3}(x_1^{B}-1)^{-\frac{1}{3}}(x_2^{B}-1)^{\frac{1}{3}}-\lambda p_1=0$\\
$\quad \frac{1}{3}(x_1^{B}-1)^{\frac{2}{3}}(x_2^{B}-1)^{-\frac{2}{3}}-\lambda p_2=0$\\
$\Longrightarrow \frac{p_2}{p_1}=\frac{x_1^{B}-1}{2(x_2^{B}-1)}$\\
$2p_2(x_2^{B}-1)=p_1(x_1^{B}-1)$\\
together with $\quad p_1x_1^{B}+p_2x_2^{B}=4p_1+p_2 $\\
$\Longrightarrow x_1^{B}=3, \quad x_2^{B}=1+\frac{p_1}{p_2}$\\
(ii) Corner solution:\\
If $x_1^B=0,\quad p_2x_2^B=4p_1+p_2\Longrightarrow x_2^B=4\frac{p_1}{p_2}+1$;\\
If $x_2^B=0,\quad p_1x_1^B=4p_1+p_2\Longrightarrow x_1^B=\frac{p_2}{p_1}+4$.\\
(b) Market clear:\\
$x_1^{A}+x_1^{B}=e_1^{A}+e_1^{B}=5 $\\
$x_2^{A}+x_2^{B}=e_2^{A}+e_2^{B}=5 $\\
(i) Interior solution:\\
$1+\frac{p_2}{p_1}+3=5$ $\Longrightarrow \frac{p_2}{p_1}=1$\\
(ii) Corner solution:\\
If $x_1^A=0, x_2^A=\frac{p_1}{p_2}+4;\quad x_1^B=\frac{p_2}{p_1}+4,x_2^B=0$, then $ 0+\frac{p_2}{p_1}+4=5 \Longrightarrow \frac{p_2}{p_1}=1$;\\
If $x_1^A=4\frac{p_2}{p_1}+1, x_2^A=0;\quad x_1^B=0,x_2^B=4\frac{p_1}{p_2}+1$, then $ 4\frac{p_2}{p_1}+1+0=5 \Longrightarrow \frac{p_2}{p_1}=1$.\\
To sum up, $\frac{p_2}{p_1}=1$.\\
(c) $p_1=p_2$,\\
(i) Interior solution:\\
$(x_1^{A},x_2^{A})=(2,3),\quad u^A(2,3)=2^{\frac{2}{3}}$;\\
$(x_1^{B},x_2^{B})=(3,2),\quad u^B(3,2)=2^{\frac{2}{3}}$.\\
(ii) Corner solution:\\
$(x_1^{A},x_2^{A})=(0,5),\quad u^A(0,5)=-4^{\frac{2}{3}}<u^A(2,3)$;\\
$(x_1^{B},x_2^{B})=(5,0),\quad u^B(5,0)=-4^{\frac{2}{3}}<u^B(3,2)$.\\
These are not equilibrium allocation.\\
$(x_1^{A},x_2^{A})=(5,0),\quad u^A(5,0)=2^{\frac{2}{3}}=u^A(2,3)$;\\
$(x_1^{B},x_2^{B})=(0,5),\quad u^B(0,5)=2^{\frac{2}{3}}=u^B(3,2)$.\\
Therefore, the equilibrium allocation is $(x_1^{A},x_2^{A})=(2,3)$ ,$(x_1^{B},x_2^{B})=(3,2)$; and $(x_1^{A},x_2^{A})=(5,0)$,$(x_1^{B},x_2^{B})=(0,5)$.\\
\section*{Problem 2}
(a)
Alex:
(i) Interior solution:\\
$\max \limits_{(x_1^{A},x_2^{A})} u^{A}=(x_1^{A}-2)^{\frac{1}{3}}(x_2^{A}-2)^{\frac{2}{3}}$\\
$s.t.\quad p_1x_1^{A}+p_2x_2^{A}\leq p_1e_1^{A}+p_2e_2^{A}=p_1+4p_2 $\\
$\Longrightarrow \max \limits_{(x_1^{A},x_2^{A})}(x_1^{A}-2)^{\frac{1}{3}}(x_2^{A}-2)^{\frac{2}{3}}-\lambda (p_1x_1^{A}+p_2x_2^{A}-p_1-4p_2) $\\
$FOC: \quad \frac{1}{3}(x_1^{A}-2)^{-\frac{2}{3}}(x_2^{A}-2)^{\frac{2}{3}}-\lambda p_1=0$\\
$\quad \frac{2}{3}(x_1^{A}-2)^{\frac{1}{3}}(x_2^{A}-2)^{-\frac{1}{3}}-\lambda p_2=0$\\
$\Longrightarrow \frac{p_2}{p_1}=\frac{2(x_1^{A}-2)}{x_2^{A}-2}$\\
$p_2(x_2^{A}-2)=2p_1(x_1^{A}-2)$\\
together with $\quad p_1x_1^{A}+p_2x_2^{A}=p_1+4p_2 $\\
$\Longrightarrow x_1^{A}=\frac{5}{3}+\frac{2}{3}\frac{p_2}{p_1}, \quad x_2^{A}=\frac{10}{3}-\frac{2}{3}\frac{p_1}{p_2}$\\
(ii) Corner solution:\\
If $x_1^A=0,\quad p_2x_2^A=p_1+4p_2\Longrightarrow x_2^A=\frac{p_1}{p_2}+4$;\\
If $x_2^A=0,\quad p_1x_1^A=p_1+4p_2\Longrightarrow x_1^A=4\frac{p_2}{p_1}+1$.\\
Bob:
(i) Interior solution:\\
$\max \limits_{(x_1^{B},x_2^{B})} u^{B}=(x_1^{B}-2)^{\frac{2}{3}}(x_2^{B}-2)^{\frac{1}{3}}$\\
$s.t.\quad p_1x_1^{B}+p_2x_2^{B}\leq p_1e_1^{B}+p_2e_2^{B}=4p_1+p_2 $\\
$\Longrightarrow \max \limits_{(x_1^{B},x_2^{B})}(x_1^{B}-2)^{\frac{2}{3}}(x_2^{B}-2)^{\frac{1}{3}}-\lambda (p_1x_1^{B}+p_2x_2^{B}-4p_1-p_2) $\\
$FOC: \quad \frac{2}{3}(x_1^{B}-2)^{-\frac{1}{3}}(x_2^{B}-2)^{\frac{1}{3}}-\lambda p_1=0$\\
$\quad \frac{1}{3}(x_1^{B}-2)^{\frac{2}{3}}(x_2^{B}-2)^{-\frac{2}{3}}-\lambda p_2=0$\\
$\Longrightarrow \frac{p_2}{p_1}=\frac{x_1^{B}-2}{2(x_2^{B}-2)}$\\
$2p_2(x_2^{B}-2)=p_1(x_1^{B}-2)$\\
together with $\quad p_1x_1^{B}+p_2x_2^{B}=4p_1+p_2 $\\
$\Longrightarrow x_1^{B}=\frac{10}{3}-\frac{2}{3}\frac{p_2}{p_1}, \quad x_2^{B}=\frac{5}{3}+\frac{2}{3}\frac{p_1}{p_2}$\\
(ii) Corner solution:\\
If $x_1^B=0,\quad p_2x_2^B=4p_1+p_2\Longrightarrow x_2^B=4\frac{p_1}{p_2}+1$;\\
If $x_2^B=0,\quad p_1x_1^B=4p_1+p_2\Longrightarrow x_1^B=\frac{p_2}{p_1}+4$.\\
(b) Market clear:\\
$x_1^{A}+x_1^{B}=e_1^{A}+e_1^{B}=5 $\\
$x_2^{A}+x_2^{B}=e_2^{A}+e_2^{B}=5 $\\
(i) Interior solution:\\
$ \frac{5}{3}+\frac{2}{3}\frac{p_2}{p_1}+\frac{10}{3}-\frac{2}{3}\frac{p_2}{p_1}=5$\\
$\frac{10}{3}-\frac{2}{3}\frac{p_1}{p_2}+\frac{5}{3}+\frac{2}{3}\frac{p_1}{p_2}=5$ always true.\\
In this case, the economy has a continuum of equilibria.\\
(ii) Corner solution:\\
If $x_1^A=0, x_2^A=\frac{p_1}{p_2}+4;\quad x_1^B=\frac{p_2}{p_1}+4,x_2^B=0$, then $ 0+\frac{p_2}{p_1}+4=5 \Longrightarrow \frac{p_2}{p_1}=1$;\\
If $x_1^A=4\frac{p_2}{p_1}+1, x_2^A=0;\quad x_1^B=0,x_2^B=4\frac{p_1}{p_2}+1$, then $ 4\frac{p_2}{p_1}+1+0=5 \Longrightarrow \frac{p_2}{p_1}=1$.\\
To sum up, $\frac{p_2}{p_1}=1$.\\
In this case, the economy does not have a continuum of equilibria.\\
\section*{Problem 3}
\begin{proof}
$x^i(p)$ is the solution, then $\forall h^i$, $u^i(h^i)\leq u^i(x^i(p))$.\\
According to the Walrasian Equilibrium, $x^i(p)-e^i\leq0$.\\
Suppose $x^i(p)-e^i<0$, because of local non-satiation,
$\exists y^i, \forall \epsilon >0,\textbar y^i-x^i(p)\textbar < \epsilon \quad s.t. \quad y^i-e^i<0 \quad and \quad u^i(y^i)>u^i(x^i(p)) $, contradiction.\\
Therefore $x^i(p)-e^i=0$ . $\because z^i(p)=x^i(p)-e^i$ $\therefore z^i(p)\cdot p=0$.
\end{proof}
\section*{Problem 4}
(a)\begin{proof}
local non-satiation,
$\forall x^i, \forall \epsilon >0, \exists y^i, \textbar y^i-x^i\textbar < \epsilon \quad s.t. \quad u^i(y^i)>u^i(x^i) $.\\
Alex's endowment:$e^A=(0.5,0.5)$, utility $u^A(e^A)=0$\\
If $y^A=(0.6,0.6)$, then $\textbar y^A-e^A\textbar < 0.1 $, but $u^A(y^A)=0=u^A(e^A)$, violates the local non-satiation property.\\
\end{proof}
(b)\begin{proof}
Walrasian Equilibrium,
$\max \limits_{x^i}u^i(x^i)\quad s.t.\quad a^i+b^i\leq 1.$\\
No allocation can increase Bob's utility, $u^B(e^B)=0.5$.\\
If $u^B(x^B)>0.5$, then $\min(a^B, b^B)>0.5$,which means $a^B>0.5\quad and \quad b^B>0.5$, therefore $a^B+b^B> 1$, contradiction.\\
No allocation can increase Alex's utility, $u^A(e^A)=0$.\\
If $u^A(x^A)>0$, then $\min( \lfloor a^A \rfloor, \lfloor b^A \rfloor)>0$,which means $a^A\geq 1 \quad and \quad b^A\geq 1 $, therefore $a^A+b^A\geq 2$, contradiction.\\
The initial endowment is a Walrasian Equilibrium.
\end{proof}
(c) The initial endowment is not Pareto efficient.
If $ (a^A, b^A)=(0,0)$ and $(a^B, b^B)=(1,1)$, then
$u^A(a^A, b^A)=0=u^A(e^A)$, and $u^B(a^B, b^B)=1>0.5=u^B(e^B)$, which means this allocation is Pareto better than the initial endowment.
\section*{Problem 5}
(a)\begin{proof}
find the Walrasian Equilibrium:
Alex:
$\max \limits_{(a_{Alex},b_{Alex})} u_{Alex}=a_{Alex}\cdot b_{Alex}-b_{Bob}$\\
$s.t.\quad a_{Alex}+b_{Alex}\leq 2 $ \\
$\Longrightarrow a_{Alex}=1 ,\quad b_{Alex}=1$\\
Bob:
$\max \limits_{(a_{Bob},b_{Bob})} u_{Bob}=a_{Bob}\cdot b_{Bob}$\\
$s.t.\quad a_{Bob}+b_{Bob}\leq 2 $ \\
$\Longrightarrow a_{Bob}=1 ,\quad b_{Bob}=1$\\
Market clear: $a_{Alex}+a_{Bob}=2,\quad b_{Alex}+b_{Bob}=2$.\\
$u_{Alex}(e_A)=0,\quad u_{Bob}(e_B)=1$.
Therefore the initial endowment is a Walrasian Equilibrium.
\end{proof}
(b) The initial allocation is not Pareto efficient. We can find a Pareto better allocation, in which Alex has $(a,b)$ and according to market clear, Bob has $((2-a),(2-b))$.\\
$u_{Alex}(a,b)=ab-(2-b)\geq 0=u_{Alex}(e_A)$ and $u_{Bob}(2-a,2-b)=(2-a)(2-b)\geq 1=u_{Bob}(e_B)$. Inequality is strict for at least one agent. \\
$\Longrightarrow \frac{2}{a+1}\leq b \leq 2-\frac{1}{2-a},\quad when \quad a \neq 2$. Inequality is strict for at least one side.\\
$\Longrightarrow \frac{2}{a+1}< 2-\frac{1}{2-a}.$
$\Longrightarrow \frac{1}{2}<a<1$.\\
Choose $a=\frac{3}{4}$, then $\frac{8}{7}\leq b\leq \frac{6}{5}$, then choose $b=\frac{41}{35}$. In this case, $u_{Alex}=\frac{1}{20}>0$ and $u_{Bob}=\frac{29}{28}>1$, Pareto better than the initial endowment.
\section*{Problem 6}
(a) Yes. For example, Bob only eats apples and only has bananas as endowment. The price of apples remains unchanged but the price of bananas decreases after the free trade with another country. Then Bob is worse off.
(b) Yes. It is possible that $u^i(x^i)\geq u^i(\tilde{x}^i)$ for all $i\leq I$, and at least one inequality is strict.
\begin{proof}
Suppose $ \exists i \leq I,\quad u^i(\tilde{x}^i)>u^i(x^i)$, then $p\cdot \tilde{x}^i>p\cdot e^i$. (Otherwise $\tilde{x}^i$ is in agent i's budget set, the agent should choose $\tilde{x}^i$ instead of $x^i$.)\\
Suppose $ \forall i \leq I,\quad u^i(\tilde{x}^i)\geq u^i(x^i)$. Claim: $p\cdot \tilde{x}^i\geq p\cdot e^i$.\\
Suppose $p\cdot \tilde{x}^i< p\cdot e^i$.\\
$\exists \epsilon >0,\quad \forall y^i, \quad s.t. \quad \textbar y^i-\tilde{x}^i \textbar< \epsilon$. Because $u^i$ is locally non-satiated, $u^i(y^i)>u^i(\tilde{x}^i)\geq u^i(x^i)$. Then $y^i$ instead of $x^i$ is in equilibrium.\\
Therefore $\forall i\leq I,\quad \sum_{i}p\tilde{x}^i>\sum_{i}pe^i$, which means $\tilde{x}^i$ is not feasible.\\
$\Longrightarrow u^i(x^i)\geq u^i(\tilde{x}^i)$ for all $i\leq I$, and at least one inequality is strict.
\end{proof}
\section*{Problem 7}
(a)\begin{proof}
$c(q,w)=\min \limits_{z:f(z)\leq q} w\cdot z $. Denote $z^*$ is the solution, then $c(q,w)=w\cdot z^*$, and $f(z^*)\leq q$ is binding.
$z^*=f^{-1}(q)=g(q)$.\\
$\Longrightarrow c(q,w)=w\cdot z^*=w\cdot f^{-1}(q)=w\cdot g(q)$.\\
$\frac{dg}{dq}=\frac{1}{\frac{df}{dz}}$
$\Longrightarrow \frac{d^2g}{dq^2}=\frac{d}{dq}（\frac{dg}{dq}）=\frac{d}{dq}（\frac{1}{\frac{df}{dz}}）=\frac{d}{dz}（\frac{1}{\frac{df}{dz}}）\frac{dz}{dq}=-\frac{\frac{d^2f}{dz^2}}{(\frac{df}{dz})^2}\frac{dz}{df}=-\frac{\frac{d^2f}{dz^2}}{(\frac{df}{dz})^3}$.\\
$\because f $ is production function, $\therefore \frac{df}{dz}\geq 0$\\
$\because f $ is concave, $\therefore \frac{d^2f}{dz^2}\leq 0$\\
$\Longrightarrow \frac{d^2c}{dq^2}=w\cdot \frac{d^2g}{dq^2}=w\cdot (-\frac{\frac{d^2f}{dz^2}}{(\frac{df}{dz})^3})\geq 0$. That is, $c(q,w)$ is convex in q.
\end{proof}
(b)Yes.\\
(c)Yes.
$f(z_1,z_2)=\max(z_1,z_2)=q$, $c(q,w)=\min \limits_{z:f(z)\geq q} w_1z_1+w_2z_2$.\\
If $z_1\geq z_2$, then $f(z_1,z_2)=z_1=q$, $c(q,w)=w_1q$\\
If $z_1< z_2$, then $f(z_1,z_2)=z_2=q$, $c(q,w)=w_2q$\\
$\Longrightarrow c(q,w)=qw_1\cdot 1_{w_1\leq w_2}+qw_2\cdot 1_{w_2<w_1}$, which is convex.
\section*{Problem 8}
(a)\begin{proof}
$\pi(p)=\sup \limits_{y}p\cdot y=p\cdot y(p)$\\
According to Hotelling's Lemma, \\
$\nabla \pi (p)=y(p)$, which is $\frac{\partial \pi}{\partial p}+\frac{\partial \pi}{\partial w_1}+\frac{\partial \pi}{\partial w_2}+\frac{\partial \pi}{\partial w_3}=y(p,w_1,w_2,w_3)$\\
$\pi(p)=p\cdot y(p) \Longrightarrow \nabla \pi (p)=y(p)+p\cdot \nabla y(p) \Longrightarrow p\cdot \nabla y(p)=0$ \\
$\Longrightarrow \frac{\partial y}{\partial p}\cdot p+\frac{\partial y}{\partial w_1}\cdot w_1+\frac{\partial y}{\partial w_2}\cdot w_2+\frac{\partial y}{\partial w_3}\cdot w_3=0$
\end{proof}
(b)
$y(p,w_1,w_2,w_3)=(c\cdot \frac{p^3}{w_1w_2w_3},-\frac{p^4}{w_1^2w_2w_3},-\frac{p^4}{w_1w_2^2w_3},-\frac{p^4}{w_1w_2w_3^2})$\\
$\because \frac{\partial y}{\partial p}\cdot p+\frac{\partial y}{\partial w_1}\cdot w_1+\frac{\partial y}{\partial w_2}\cdot w_2+\frac{\partial y}{\partial w_3}\cdot w_3=0$\\
$\therefore 3c\cdot \frac{p^2}{w_1w_2w_3}\cdot p+(-4)\cdot \frac{p^3}{w_1^2w_2w_3} \cdot w_1+(-4)\cdot \frac{p^3}{w_1w_2^2w_3}\cdot w_2+(-4)\cdot \frac{p^4}{w_1w_2w_3^2}\cdot w_3=0 \Longrightarrow c=4$.
\section*{Problem 9}
(a)\begin{proof}
input vector $y$, input price vector $w$, output price vector $p$, output function $f(y)$.\\
non-decreasing returns to scale: $\forall t>1,\quad f(ty)\geq tf(y)$\\
$\because \pi_1=pf(y)-wy,\quad \pi_2=pf(ty)-wty\geq tpf(y)-twy=t\pi_1,\\
\therefore pf(y)-wy$ is non-decreasing in y.\\
If $\pi_1=pf(y)-wy >0$, then profit maximization means the firm's profit is infinity;\\
If $\pi_1=pf(y)-wy \leq 0$, then profit maximization means the firm's profit is 0.
\end{proof}
(b)\begin{proof}
$c(q,w)=\min \limits_{z:f(z)\leq q} w\cdot z $. Denote $z^*$ is the solution, then $c(q,w)=w\cdot z^*$, and $f(z^*)\leq q$ is binding.
$z^*=f^{-1}(q)$ \\
$\therefore c(q,w)=w\cdot f^{-1}(q) ,\therefore c(\alpha q,w)=w\cdot f^{-1}(\alpha q)$\\
$\alpha \in [0,1], \quad f(\alpha z)\geq \alpha \cdot f(z)
\Longrightarrow f(\alpha z)\geq \alpha \cdot q \\
\Longrightarrow \alpha z \geq f^{-1}(\alpha q) \Longrightarrow \alpha f^{-1}(q)\geq f^{-1}(\alpha q) \\
\Longrightarrow w\cdot \alpha f^{-1}(q)\geq w\cdot f^{-1}(\alpha q),\quad \forall w>0 \\
\Longrightarrow \alpha c(q,w)\geq c(\alpha q,w)
\Longrightarrow \frac{c(q,w)}{q} \geq \frac{c(\alpha q,w)}{\alpha q} $.\\
$\because \alpha \in [0,1],\therefore \frac{c(q,w)}{q} $ is non-decreasing in q.
\end{proof}
\end{document}
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