%This is my super simple Real Analysis Homework template
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage[]{amsthm} %lets us use \begin{proof}
\usepackage[]{amssymb} %gives us the character \varnothing
\usepackage[]{dsfont}
\title{Advanced Microeconomics Homework 1}
\author{}
\date\today
%This information doesn't actually show up on your document unless you use the maketitle command below
\begin{document}
\maketitle %This command prints the title based on information entered above
%Section and subsection automatically number unless you put the asterisk next to them.
\section*{Problem 1}
(a) \begin{proof}
a utility function u represents a preference relation $\succsim$.\\
If $x_n\longrightarrow x$, $y_n\longrightarrow y$, $x_n\succsim y_n$ $ \Longrightarrow$ $u(x_n)\geq u(y_n)$.\\
u is continuous $\Longrightarrow$ $u(x_n)\longrightarrow u(x)$, $u(y_n)\longrightarrow u(y)$ $\Longrightarrow$ $u(x)\geq u(y)$ $\Longrightarrow$ $x\succsim y$ $\Longrightarrow$ $\succsim$ is continuous.\\
\end{proof}
(b) The converse is false.\\
If $X= \mathds{R} $, and $x\succsim y$, if and only if $x\geq y$, then $\succsim $ is continuous, but any strictly increasing function could represent $\succsim $.\\
Construction: \\
$$ u(x)=\left\{
\begin{array}{rcl}
x-1 & {\quad if \quad x \geq 0}\\
2x+100 & {\quad if\quad x\textgreater 0}
\end{array}
\right.
$$
\section*{Problem 2}
(a)
\begin{proof}
There exists two sets of points $x_n$ and $y_n$ .\\
$x_n=\{(a_n,z_1):n=1,2,...,a_n\in \mathds{Q},z_1\in \mathds{R},a_n>a\},x_n\longrightarrow x$\\
$x=\{(a,z_1):a\in \mathds{Q},z_1\in \mathds{R}\}$\\
$y_n=\{(b_n,z_2):n=1,2,...,b_n\in \mathds{Q},z_2\in \mathds{R},b_n $y=\{(b,z_2):b\in \mathds{Q},z_2\in \mathds{R}\}$\\ $a=b, z_1 $\Longrightarrow$ $x_n\succsim y_n$ but $y\succsim x$,so $\succ$ is not a continuous preference relation. \end{proof} (b)\begin{proof} Firstly, define the utility function on the set of rational points in $\mathds{Q} \times \mathds{R}$ .Since rational points are countable, let $(x_1,y_1),(x_2,y_2),...$ denote the set of rational points in $\mathds{Q} \times \mathds{R}$, $x_n\in \mathds{Q}$ and $y_n\in (\mathds{Q}\cap \mathds{R})$\\ $$u(x)=\sum_{n=1}^{\infty}\frac{1}{2^n}\mathds{1}_{(x,y)\succsim(x_n,y_n)}$$\\ \\ Claim:$\forall (x,y),(x',y') \in \mathds{Q} \times \mathds{R}, u(x,y)\geq u(x',y') \rightleftharpoons (x,y) \succsim (x',y') $ .\\ \begin{itemize} \item Prove the claim: $\forall (x,y),(x',y') \in \mathds{Q} \times \mathds{R}, (x,y) \succsim (x',y')\Longrightarrow u(x,y)\geq u(x',y') $ .\\ $ (x,y) \succsim (x',y')\Longrightarrow$ If $(x',y') \succsim (x_n,y_n)$,then $(x,y) \succsim (x_n,y_n) $ (According to Transitivity) $\Longrightarrow \mathds{1}_{(x,y)\succsim(x_n,y_n)} \geq \mathds{1}_{(x',y')\succsim(x_n,y_n)}, \forall n $ $\Longrightarrow u(x,y)\geq u(x',y')$ \\ \item Prove the claim: $\forall (x,y),(x',y') \in \mathds{Q} \times \mathds{R},u(x,y)\geq u(x',y')\Longrightarrow (x,y) \succsim (x',y') $ .\\ Contrapositive: If $(x',y') \succ (x,y)$ ,then $ u(x',y')> u(x,y) $\\ (i)\quad If $x'=x=a\in \mathds{Q}, y'>y \in \mathds{R}$,according to the proof of Debreu's Representation Theorem, $\exists z\in \mathds{Q} ,s.t. y'\geq z >y$. (In fact, the set of y is continuous.)\\ then $\mathds{1}_{(x',y')\succsim(x_n,y_n)} \geq \mathds{1}_{(x,y)\succsim(x_n,y_n)}, \forall n$\\ $\mathds{1}_{(x',y')\succsim(a,z)}=1 > 0=\mathds{1}_{(x',y')\succsim(a,z)}$\\ $\Longrightarrow u(x',y')> u(x,y)$\\ (ii)\quad If $x'>x\in \mathds{Q},y',y\in \mathds{R},\exists b\in\mathds{Q},s.t.x'\geq b>x$\\ then $\mathds{1}_{(x',y')\succsim(x_n,y_n)} \geq \mathds{1}_{(x,y)\succsim(x_n,y_n)}, \forall n$\\ $\mathds{1}_{(x',y')\succsim(b,y_n)}=1 > 0=\mathds{1}_{(x',y')\succsim(b,y_n)}$\\ $\Longrightarrow u(x',y')> u(x,y)$\\ \end{itemize} \end{proof} (c)There is not a continuous utility function that represents $\succ$ \section*{Problem 3} (a)IIA: if $B\subseteq A ,C(A)\in B \Longrightarrow C(B)=C(A)$ \begin{proof} Suppose c satisfies the IIA condition.\\ If $c(x,y,w,z)=x$, then $c(w,x)=x$,contradict to $c(w,x)=w$\\ so $c(x,y,w,z)\neq x$\\ If $c(x,y,w,z)=y$, then $c(x,y,z)=y$,contradict to $c(x,y,z)=x$\\ so $c(x,y,w,z)\neq y$\\ If $c(x,y,w,z)=w$, then $c(y,z,w)=w$,contradict to $c(y,z,w)=y$\\ so $c(x,y,w,z)\neq w$\\ If $c(x,y,w,z)=z$, then $c(x,y,z)=z$ and $c(y,z,w)=z$,contradict to $c(x,y,z)=x$ and $c(y,z,w)=y$\\ so $c(x,y,w,z)\neq z$\\ To sum up, c violates the IIA condition. \end{proof} (b)\quad There are two directions to prove.\\ (i)\quad a directed cycle exists $\Longrightarrow$ c violates the IIA condition.\\ (ii)\quad c violates the IIA condition $\Longrightarrow$ a directed cycle exists.\\ \begin{proof} (i)\quad Suppose c satisfies IIA condition\\ If $C(A_1\cup A_2...\cup A_n)=x_i $, because $x_i\in A_i$ and $x_i\in A_{i+1} \Longrightarrow C(A_{i+1}=x_i)$\\ Because $C(A_{i+1}=x_{i+1}$ ,$x_i\approx x_{i+1}$ and there is a cycle but not directed cycle. Contradiction.\\ \\ (ii)\quad Contrapositive: no directed cycle existes $\Longrightarrow$ c satisfies IIA condition\\ If $x_i\in A_i$ and $x_i \notin A_k, \forall K\neq i$, then there is no directed cycle. \\ $C(A_i)=x_i, \forall i$.Without loss of generality, suppose $x_1\succsim x_2\succsim x_3\succsim ...\succsim x_n$, then $C(A_1\cup A_2...\cup A_n)=x_1 $, satisfies IIA, $C(A_2\cup A_3...\cup A_n)=x_2 $, satisfies IIA, ...the rest can be done in the same manner and c satisfies IIA. \end{proof} \section*{Problem 4} Independence Axiom: $\forall \mu,\mu',\mu''\in \Delta (x)$ and $\forall \alpha \in(0,1)$, $\mu \succsim \mu' \rightleftharpoons \alpha \mu + (1-\alpha)\mu'' \succsim \alpha \mu' + (1-\alpha)\mu''$ \begin{proof} Suppose the agent is not indifferent between all lotteries over X, there exists $\mu$ and $\mu'$,s.t. $\mu \succ \mu'$ $\Longrightarrow \sum_{i=1}^{n} \mu_i\cdot u(x_i) > \sum_{i=1}^{n} \mu'_i\cdot u(x_i) \Longrightarrow \alpha \mu + (1-\alpha)\mu'' \succ \alpha \mu' + (1-\alpha)\mu''\Longrightarrow \alpha \sum_{i=1}^{n} \mu_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i) > \alpha \sum_{i=1}^{n} \mu'_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i)$\\ Because $x_1\sim x_2\sim...\sim x_n $, $u(x_1)=u(x_2)=...=u(x_n)=\bar{u},\sum_{i=1}^{n} \mu_i=1 $\\ $\alpha \sum_{i=1}^{n} \mu_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i) = \alpha \sum_{i=1}^{n} \mu_i\cdot \bar{u} + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot \bar{u}=\bar{u}$ and $\alpha \sum_{i=1}^{n} \mu'_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i) = \alpha \sum_{i=1}^{n} \mu'_i\cdot \bar{u} + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot \bar{u}=\bar{u}$\\ Contradict to $\sum_{i=1}^{n} \mu_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i) > \alpha \sum_{i=1}^{n} \mu'_i\cdot u(x_i) + (1-\alpha)\sum_{i=1}^{n} \mu''_i\cdot u(x_i)$. Therefore, the agent is indifferent between all lotteries over X. \end{proof} \section*{Problem 5} \begin{proof} (a)\quad $\forall w, u(w)>\frac{1}{2}u(w+250)+\frac{1}{2}u(w-200)$\\ u is concave $\Longrightarrow u(w-200)>\frac{4}{5}u(w-250)+\frac{1}{5}u(w)$\\ $\Longrightarrow u(w+250)-u(w)<\frac{4}{5}[u(w)-u(w-250)]$\\ \\ (b)\quad $u(+\infty)-u(w)=[u(w+250)-u(w)]+[u(w+500)-u(w+250)]+[u(w+750)-u(w+500)]+...+[u(+\infty)-u(+\infty-250)]$\\ $u(w+250)-u(w)<\frac{4}{5}[u(w)-u(w-250)]\\ \Longrightarrow u(w+500)-u(w+250)<(\frac{4}{5})^2[u(w)-u(w-250)]\\ \Longrightarrow u(w+750)-u(w+500)<(\frac{4}{5})^3[u(w)-u(w-250)]\\ ...\\ \Longrightarrow u(w+250n)-u(w+250(n-1))<(\frac{4}{5})^n[u(w)-u(w-250)]\\ \Longrightarrow u(+\infty)-u(w) < [\frac{4}{5}+(\frac{4}{5})^2+(\frac{4}{5})^3+...+(\frac{4}{5})^n][u(w)-u(w-250)], n\longrightarrow +\infty \\ \Longrightarrow u(+\infty)-u(w) < 4[u(w)-u(w-250)].$\\ \\ (c)\quad u is concave $\Longrightarrow u(w-250)>\frac{3}{4}u(w)+\frac{1}{4}u(w-1000)\\ \Longrightarrow u(w)-u(w-1000)<4[u(w)-u(w-250)]$ \end{proof} \section*{Problem 6} \begin{proof} DARA: $A(x,u)=-\frac{u''(x)}{u'(x)}$ is decreasing in x.\\ DRRA: $xA(x,u)=-x\frac{u''(x)}{u'(x)}$ is decreasing in x.\\ \\(a)\quad Let $g(x)= -x\frac{u''(x)}{u'(x)} $\\ then $g'(x)=-(\frac{u''(x)}{u'(x)}+x\frac{u'''(x)u'(x)-[u''(x)]^2}{[u'(x)]^2})=-\frac{u'(x)u''(x)+xu'''(x)u'(x)-x[u''(x)]^2}{[u'(x)]^2} < 0$\\ $\Longrightarrow u'(x)u''(x)+xu'''(x)u'(x)-x[u''(x)]^2 >0\\ \Longrightarrow x[u'''(x)u'(x)-[u''(x)]^2]> -u'(x)u''(x)>0\\ x\in [0,+\infty) u'''(x)u'(x)-[u''(x)]^2 >0 $\\ Let $h(x)= -\frac{u''(x)}{u'(x)} $\\ then $h'(x)=-\frac{u'''(x)u'(x)-[u''(x)]^2}{[u'(x)]^2} < 0 $\\ $\Longrightarrow A(x,u)=-\frac{u''(x)}{u'(x)}$ is decreasing in x, (DARA).\\ \\ (b)\quad Define $c_w: u(c_w)=\int u(w+x)dF(x)$\\ DARA is equivalent to $w-c_w$ decreases ($c_w-w$ increases )in w for all F. $w'>w, v_1(x)=u(w+x), v_2(x)=u(w'+x)\\ \mathds{E}[u(w+x)]=\mathds{E}[v_1(x)]=\inf v_1(x)dF(x)=u(c_w)>u(w)$\\ u is increasing $\Longrightarrow c_w>w \Longrightarrow c_w-w>0, w'>w \Longrightarrow c_w'-w'>c_w-w>0 \Longrightarrow u(c_w')>u(w') \\ LHS=\inf v_2(x)dF(x)=\mathds{E}[v_2(x)]=\mathds{E}[u(w'+x)]\Longrightarrow \mathds{E}[u(w'+x)] > u(w')$\\ \\ (c)\quad Define $c_w: u(c_w)=\int u(wx)dF(x)$\\ DARA is equivalent to $\frac{w}{c_w}$ decreases ($\frac{c_w}{w}$ increases )in w for all F. $w'>w, v_1(x)=u(wx), v_2(x)=u(w'x)\\ \mathds{E}[u(wx)]=\mathds{E}[v_1(x)]=\inf v_1(x)dF(x)=u(c_w)>u(w)$\\ u is increasing $\Longrightarrow c_w>w , c_w,w\in (0,+\infty )\Longrightarrow \frac{c_w}{w}>1, w'>w \Longrightarrow \frac{c_w'}{w'}>\frac{c_w}{w}>1 \Longrightarrow u(c_w')>u(w') \\ LHS=\inf v_2(x)dF(x)=\mathds{E}[v_2(x)]=\mathds{E}[u(w'x)]\Longrightarrow \mathds{E}[u(w'x)] > u(w')$\\ \end{proof} \section*{Problem 7} (a)\quad No. Construction: $$ u(x)=\left\{ \begin{array}{rcl} x & {\quad if \quad x =1,2}\\ 2x & {\quad if\quad x=3,4,5} \end{array} \right. $$ In this case, $\mathds{E}(L)=5.4$,$\mathds{E}(L')=5.6$, FOSD $\Longrightarrow L'\succsim L$ \\ (b) \begin{proof} SOSD: $\forall x, \int{-\infty}^{x}F(t)dt \leq \int{-\infty}^{x}G(t)dt \Longrightarrow F \succsim G $\\ \begin{itemize} \item$x=1,CDF F_L(t)=0.2, \int_{-\infty}^{x}F_L(t)dt=0.2 \\ F_{L'}(t)=0.4, \int_{-\infty}^{x}F_{L'}(t)dt=0.4$ \item$x=2,CDF F_L(t)=0.4, \int_{-\infty}^{x}F_L(t)dt=0.6 \\ F_{L'}(t)=0.4, \int_{-\infty}^{x}F_{L'}(t)dt=0.8$ \item$x=3,CDF F_L(t)=0.6, \int_{-\infty}^{x}F_L(t)dt=1.2 \\ F_{L'}(t)=0.6, \int_{-\infty}^{x}F_{L'}(t)dt=1.4$ \item$x=4,CDF F_L(t)=0.8, \int_{-\infty}^{x}F_L(t)dt=2 \\ F_{L'}(t)=0.6, \int_{-\infty}^{x}F_{L'}(t)dt=2$ \item$x=5,CDF F_L(t)=1, \int_{-\infty}^{x}F_L(t)dt=3 \\ F_{L'}(t)=1, \int_{-\infty}^{x}F_{L'}(t)dt=3$ \end{itemize} satisfies $\forall x, \int_{-\infty}^{x}F(t)dt \leq \int_{-\infty}^{x}G(t)dt $, therefore a risk averse agent always prefer L to L'. \end{proof} (c)\quad No. a counter-example: $$ u(x)=\left\{ \begin{array}{rcl} x & {\quad if \quad x =1,2}\\ 0.99x & {\quad if\quad x=3,4,5} \end{array} \right. $$ In this case, $\mathds{E}(L)=2.976$,$\mathds{E}(L'')=3.172$, $\Longrightarrow L''\succsim L$ \section*{Problem 8} \begin{proof} $\max \limits_{\alpha \in (0,1)}\int_{0}^{+\infty}u(w-q\alpha -L+L\alpha)dF(L)$\\ take the first derivative of $\alpha :\int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dF(L)=0$\\ Let $W(\alpha,L)= (L-q)\cdot u'(w-q\alpha -L+L\alpha) \\ \frac{\partial W(\alpha,L)}{\partial \alpha}= (L-q)^2\cdot u''(w-q\alpha -L+L\alpha)$\\ because $(L-q)^2 \geq 0$ and $u''(w-q\alpha -L+L\alpha) \leq 0,\frac{\partial W(\alpha,L)}{\partial \alpha} \leq 0$, that is, $W(\alpha,L)$ non-increasing in $\alpha$.\\ (a)\quad because $W(\alpha,L)$ non-increasing in $\alpha,\alpha \in (0,1)$, $W(\alpha,L) \geq W(1,L)$.\\ $\alpha=1, W(1,L)=(L-q)\cdot u'(w-q) \Longrightarrow W(\alpha,L) \geq (L-q)\cdot u'(w-q)$\\ $\Longrightarrow \int_{0}^{+\infty}W(\alpha,L)dF(L) \geq \int_{0}^{+\infty}(L-q)\cdot u'(w-q)dF(L)\\ \int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dF(L) \geq u'(w-q)\int_{0}^{+\infty}(L-q) dF(L)$\\ when the policy is actuarially fair, that is $q=\mathds{E}_F[L]=\inf LdF(L)$, $RHS=0$, $\int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dF(L) \geq 0$ when and only when $\alpha=1$ ,the equal sign can be taken, that is the agent purchases full insurance.\\ \\ (b)\quad $q=\mathds{E}_F[L]=\inf LdF(L) \Longrightarrow \int_{0}^{+\infty}(L-q)dF(L)<0 $, because $u'(w-q)>0, W(1,L)=u'(w-q)\int_{0}^{+\infty}(L-q)dF(L) <0$\\ According to the first derivative of $\alpha$, the optimal choice of the agent is $\alpha$ s.t. $W(\alpha,L)=0$.\\ because $W(\alpha,L)$ non-increasing in $\alpha,\alpha \in (0,1)$, the optimal choice of the agent is $\alpha <1$.\\ \\ (c)\quad take the first derivative of $\alpha :\\ \int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dF(L)=0\\ \int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dG(L)=0$\\ Suppose $\alpha_F > \alpha_G $, then $W(\alpha_F,L) $\Longrightarrow \int_{0}^{+\infty}W(\alpha_F,L)dG(L) <\int_{0}^{+\infty}W(\alpha_G,L)dG(L) \\ \int_{0}^{+\infty}W(\alpha_F,L)dG(L)=\int_{0}^{+\infty}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L)=\int_{0}^{q}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L)+\int_{q}^{+\infty}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L) $\\ Because G dominates F in the likelihood-ratio order, $\frac{g(x)}{f(x)}$ is non-decreasing in x. If $L \leq q$,then $\frac{g(L)}{f(L)} \leq \frac{g(q)}{f(q)}$ and $W(\alpha, L)<0 \Longrightarrow \int_{0}^{q}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L) \geq \int_{0}^{q}W(\alpha_F,L)\frac{g(q)}{f(q)}dF(L) $;\\ If $L > q$,then $\frac{g(L)}{f(L)} \geq \frac{g(q)}{f(q)}$ and $W(\alpha, L)>0 \Longrightarrow \int_{q}^{+\infty}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L) \geq \int_{q}^{+\infty}W(\alpha_F,L)\frac{g(q)}{f(q)}dF(L)$\\ $\Longrightarrow \int_{0}^{q}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L)+\int_{q}^{+\infty}W(\alpha_F,L)\frac{g(L)}{f(L)}dF(L) \geq \int_{0}^{q}W(\alpha_F,L)\frac{g(q)}{f(q)}dF(L)+\int_{q}^{+\infty}W(\alpha_F,L)\frac{g(q)}{f(q)}dF(L)=\int_{0}^{+\infty}W(\alpha_F,L)\frac{g(q)}{f(q)}dF(L)=\frac{g(q)}{f(q)}\int_{0}^{+\infty}W(\alpha_F,L)dF(L) =0$(According to the the first derivative of $\alpha$)\\ therefore, $\int_{0}^{+\infty}W(\alpha_G,L)dG(L) > \int_{0}^{+\infty}W(\alpha_F,L)dG(L) \geq \frac{g(q)}{f(q)}\int_{0}^{+\infty}W(\alpha_F,L)dF(L) =0 $, contradict to the first derivative of $\alpha :\\ \int_{0}^{+\infty}(L-q)\cdot u'(w-q\alpha -L+L\alpha)dG(L)=0$ \end{proof} \end{document}