<jupyter_start><jupyter_text>Financial Economics HW_04**11510691 程$% color% Aquamarine, black, blue, brown, cyan, darkgray, gray, green, lightgray, lime, magenta, olive, orange, pink, purple, red, teal, violet, white, yellow\DeclareMathOperator*{\argmin}{argmin}\DeclareMathOperator*{\argmax}{argmax}\DeclareMathOperator*{\plim}{plim}\newcommand{\ffrac}{\displaystyle \frac}\newcommand{\d}[1]{\displaystyle{1}}\newcommand{\space}{\text{ }}\newcommand{\bspace}{\;\;\;\;}\newcommand{\bbspace}{\;\;\;\;\;\;\;\;}\newcommand{\QQQ}{\boxed{?\:}}\newcommand{\void}{\left.\right.}\newcommand{\Tran}[1]{{1}^{\mathrm{T}}}\newcommand{\CB}[1]{\left\{ 1 \right\}}\newcommand{\SB}[1]{\left[ 1 \right]}\newcommand{\P}[1]{\left( 1 \right)}\newcommand{\abs}[1]{\left| 1 \right|}\newcommand{\norm}[1]{\left\| 1 \right\|}\newcommand{\given}[1]{\left. 1 \right|}\newcommand{\using}[1]{\stackrel{\mathrm{1}}{=}}\newcommand{\asim}{\overset{\text{a}}{\sim}}\newcommand{\RR}{\mathbb{R}}\newcommand{\EE}{\mathbb{E}}\newcommand{\II}{\mathbb{I}}\newcommand{\NN}{\mathbb{N}}\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\QQ}{\mathbb{Q}}\newcommand{\PP}{\mathbb{P}}\newcommand{\AcA}{\mathcal{A}}\newcommand{\FcF}{\mathcal{F}}\newcommand{\AsA}{\mathscr{A}}\newcommand{\FsF}{\mathscr{F}}\newcommand{\dd}{\mathrm{d}}\newcommand{\I}[1]{\mathrm{I}\left( 1 \right)}\newcommand{\N}[1]{\mathcal{N}\left( 1 \right)}\newcommand{\Exp}[1]{\mathrm{E}\left[ 1 \right]}\newcommand{\Var}[1]{\mathrm{Var}\left[ 1 \right]}\newcommand{\Avar}[1]{\mathrm{Avar}\left[ 1 \right]}\newcommand{\Cov}[1]{\mathrm{Cov}\left( 1 \right)}\newcommand{\Corr}[1]{\mathrm{Corr}\left( 1 \right)}\newcommand{\ExpH}{\mathrm{E}}\newcommand{\VarH}{\mathrm{Var}}\newcommand{\AVarH}{\mathrm{Avar}}\newcommand{\CovH}{\mathrm{Cov}}\newcommand{\CorrH}{\mathrm{Corr}}\newcommand{\ow}{\text{otherwise}}\newcommand{\wp}{\text{with probability }}\newcommand{\FSD}{\text{FSD}}\newcommand{\SSD}{\text{SSD}}$远星$^\dagger$**  Question 12.1$\P 1$$$\begin{align}v&= \Exp{-e^{-a\tilde w}}\\&= -\exp\CB{-a\Exp{\tilde w} + \frac{1}{2} a^2 \Var{\tilde w}}\\&= -\exp\CB{-a w\Tran z \P{\iota + \bar r} + \ffrac{1}{2} a^2 w^2 \Tran z \Sigma z}\end{align}$$$\P 2$$\bspace$Obviously, $v= \Exp{-e^{-a\tilde w}} = v\P{\Exp{\tilde w},\sigma\SB{\tilde w}} \equiv v\P{\bar w, \sigma_w}$$\P 3$$$\begin{align}\partial_1 v &= \ffrac{\partial v}{\partial \bar w} = -av > 0\\\partial_2 v &= \ffrac{\partial v}{\partial \sigma_w} = a^2 \sigma_w v < 0\\\partial_{11}^2 v &= \ffrac{\partial^2 v}{\partial \bar w^2} = a^2 v < 0\\\partial_{22}^2 v &= \ffrac{\partial^2 v}{\partial \sigma_w^2} = \P{a^4 \sigma_w^2 + a^2}v < 0\\\end{align}$$$\bspace$And the convexity: $\P{\partial_{11}^2 v} \cdot \P{\partial_{22}^2 v} - \P{\partial_{12}^2 v}^2 = \P{a^6\sigma_w^2 + a^4}v^2 - a^6\sigma_w^2v^2 = a^4v^2>0$  Question 12.3$\P 1$$\bspace$Suppose the portfolio to be $z = \SB{z_1;z_2;\cdots;z_N}$, then the variance is$$\sigma_p^2 = \Var{\sum_{i=1}^N z_i \tilde r_i} = \sigma^2 \sum_{i=1}^N z_i^2 + c\sum_{i\neq j}z_iz_j$$$$\ffrac{\partial \sigma_p^2}{z_i} = 2\sigma^2 z_i + c\sum_{i\neq j}z_j = \P{2\sigma^2-1}z_i + c\sum_i z_i = 0$$$\bspace$Thus $z_i \equiv \ffrac{c\sum_i z_i}{1-2\sigma^2} \equiv \ffrac{1}{N}$, $\forall\: i$$\P 2$$\bspace$Plug in the result we have: $\sigma_p^2 = \ffrac{\sigma^2}{N} + \ffrac{N\P{N-1}}{N^2}c \to c$ as $N \to \infty$$\P 3$$\bspace$As $N \to\infty$, the variance of one single securities will shrink to $0$, while the covariance of any two different securities will keep.  Question 12.4$\P 1$$\bspace$Suppose $\bar r_1 \neq \bar r_2$, and given expected return rate $\bar r_p$, we have $z\bar r_1 + \P{1-z} \bar r_2 = \bar r_p$ so that $z = \ffrac{\bar r_p - \bar r_2}{\bar r_1 - \bar r_2}$. Then the variance is$$\begin{align}\sigma_p^2 &= z^2 \sigma_1^2 + \P{1-z}^2 \sigma_2^2 + 2z\P{1-z}\sigma_{12}\\&= \ffrac{1}{\P{\bar r_1 - \bar r_2}^2}\P{\P{\bar r_p - \bar r_2}^2 \sigma_1^2 + \P{\bar r_1 - \bar r_p}^2 \sigma_2^2 + 2\P{\bar r_p - \bar r_2}\P{\bar r_1 - \bar r_p}\sigma_{12}}\\&= \ffrac{1}{\P{\bar r_1 - \bar r_2}^2}\P{\P{\sigma_1^2 + \sigma_2^2 -2\sigma_{12}}\bar r_p^2 - \P{2\sigma_1^2\bar r_2 + 2\sigma_2^2 \bar r_1 -2\P{\bar r_1 + \bar r_2}\sigma_{12}}\bar r_p + \P{\sigma_1^2\bar r_2^2 + \sigma_2^2 \bar r_1^2 - 2\bar r_1 \bar r_2 \sigma_{12}}}\end{align}$$$\bspace$Then the points $\P{\bar r_p,\sigma_p}$ constitute the mean variance effective frontier.$\P 2$$\bspace$If we let $\bar r_p = \bar r_1$, then the frontier portfolio is just security $1$, while letting $\bar r_p = \bar r_2$ leads to security $2$.$\P 3$$$\begin{align} \rho_{12} = -1 \Rightarrow \sigma_p &= \sqrt{\ffrac{\P{\sigma_1^2 + \sigma_2^2 +2\sigma_1\sigma_2}\bar r_p^2 - \P{2\sigma_1^2\bar r_2 + 2\sigma_2^2 \bar r_1 +2\P{\bar r_1 + \bar r_2}\sigma_1\sigma_2}\bar r_p + \P{\sigma_1^2\bar r_2^2 + \sigma_2^2 \bar r_1^2 + 2\bar r_1 \bar r_2 \sigma_1\sigma_2}}{\P{\bar r_1 - \bar r_2}^2}}\\&= \abs{\ffrac{\P{\sigma_1 + \sigma_2} \bar r_p - \P{\sigma_1 \bar r_2 + \sigma_2 \bar r_1}}{\bar r_1 - \bar r_2}} = \abs{\ffrac{\P{\bar r_p - \bar r_2}\sigma_1 + \P{\bar r_p - \bar r_1}\sigma_2}{\bar r_1 - \bar r_2}}\\\rho_{12} = 1 \Rightarrow \sigma_p &= \abs{\ffrac{\P{\sigma_1 - \sigma_2} \bar r_p - \P{\sigma_1 \bar r_2 - \sigma_2 \bar r_1}}{\bar r_1 - \bar r_2}} = \abs{\ffrac{\P{\bar r_p - \bar r_2}\sigma_1 - \P{\bar r_p - \bar r_1}\sigma_2}{\bar r_1 - \bar r_2}}\\\rho = 0 \Rightarrow \sigma_p &= \sqrt{\ffrac{\P{\sigma_1^2 + \sigma_2^2}\bar r_p^2 - \P{2\sigma_1^2\bar r_2 + 2\sigma_2^2 \bar r_1}\bar r_p + \P{\sigma_1^2\bar r_2^2 + \sigma_2^2 \bar r_1^2}}{\P{\bar r_1 - \bar r_2}^2}} = \sqrt{\ffrac{\P{\bar r_p - \bar r_2}^2\sigma_1^2 + \P{\bar r_p - \bar r_1}^2\sigma_2^2}{\P{\bar r_1 - \bar r_2}^2}}\end{align}$$| $\rho = -1$ | $\rho = 1$   | $\rho = 0$    || :---------: | :-----------:| :-----------: || ![_auto_0](attachment:_auto_0) | ![_auto_2](attachment:_auto_2) | ![_auto_3](attachment:_auto_3) |  Question 12.11$\bspace$Since $\Sigma$ is positive definite, then $\Tran{\bar\eta}\Sigma^{-1}\bar\eta \iff \bar\eta = 0$ and thus $\bar r = r_F \iota$  Question 13.1$\P 1$$\bspace$Suppose that the price vector of the stocks is $S$ and for agent $k$ with portfolio $\theta_k$, we have his demand for stocks:$$\begin{array}{cl}\d{\max_{\theta_k}} &\Exp{-e^{-\tilde w_{\void_k}}} = -\exp\CB{-\Exp{\tilde w_k} + \ffrac{1}{2}\Var{\tilde w_k}}\\s.t. & \begin{align}\tilde w_k &= \P{e_{k,0}+\Tran S \bar \theta_k -c_{k,0}}\P{1+r_F} + \Tran{\theta_k}\P{\tilde v - \P{1+r_F}S}+ \Tran{\bar \theta_k}S\P{1+r_F}+y_k\P{\tilde m + \tilde \varepsilon}\\&= \P{e_{k,0}+\Tran S \bar \theta_k -c_{k,0}} + \Tran{\theta_k} \tilde v + \Tran{\P{\bar \theta_k - \theta_k }}S + y_k \P{\Tran \iota \tilde v + \tilde \varepsilon}\equiv w_k + \Tran{\theta_k} \tilde v + \Tran{\P{\bar \theta_k - \theta_k }}S + y_k \P{\Tran \iota \tilde v + \tilde \varepsilon}\end{align}\end{array}$$$\bspace$Then to maximize $\Exp{-e^{-\tilde w_{\void_k}}}$ is equivalent to minimize $$\begin{align}&\ffrac{1}{2}\Var{\tilde w_k} - \Exp{\tilde w_k}\\=\;& \ffrac{1}{2}\Var{ \P{\Tran{\theta_k} + y_k \Tran\iota}\tilde v + y_k\tilde\varepsilon} - \P{w_k + \Tran{\theta_k} \bar v + \Tran{\P{\bar \theta_k - \theta_k }}S + y_k \Tran\iota\bar v}\\=\;& \ffrac{1}{2}\Tran{\P{\theta_k + y_k\iota}}\Sigma\P{\theta_k + y_k\iota} + \ffrac{\sigma^2y_k^2}{2} - w_k - \Tran{\theta_k} \bar v - \Tran{\P{\bar \theta_k - \theta_k }}S - y_k \Tran\iota\bar v\end{align}$$$\bspace$Take the derivative on $\theta_k$, since $\Sigma = \Sigma^{-1}$, we have$$0 = \P{\Tran\theta_k + y_k \Tran\iota}\Sigma + \Tran{\P{S - \bar v}} \Rightarrow \theta_k = \Sigma^{-1}\P{\bar v - S} - y_k\iota$$$\P 2$$\bspace$To reach balance, $\sum\theta_k = \sum\bar\theta_k = \iota$ so that$$K\cdot \Sigma^{-1}\P{\bar v - S} - \P{\sum y_k}\iota = \iota \Rightarrow S = \bar v - \ffrac{\P{\sum y_k +1}\Sigma \iota}{K}$$$\P 3$$\bspace$We now prove under this context $\bar r_n -r_F = \beta_{n,M}\P{\bar r_M - r_F}$ holds, where $\bar r_n$ is the mean value of $\tilde r_n = {\ffrac{\tilde v_n}{S_n}}-1$, $\bar r_M$ is the mean value of $\tilde r_M = {\ffrac{\Tran{\iota} \tilde v}{w_M}}-1$, for security $n$. $w_M = \Tran\iota S$. Then$$\begin{align}\bar r_n - r_F &= \bar r_n = \ffrac{\bar v_n - S_n}{S_n}\\&= \ffrac{\P{\sum y_k +1}\P{\Sigma \iota}_n}{K S_n}\\\bar r_M - r_F &= \bar r_M\\&= \ffrac{\Tran\iota \bar v}{w_M} - 1\end{align}$$$\bspace$Then the $\beta_{n,M} = \ffrac{\Cov{\tilde r_M, \tilde r_n}}{\Var{\tilde r_M}} = \Big.\ffrac{\Cov{\Tran\iota \tilde v,{\tilde v}_n}}{w_M S_n}\Big/\ffrac{\Var{\Tran\iota\tilde v}}{w_M^2} = \ffrac{w_M\P{\Sigma\iota}_n}{S_n \Tran\iota\Sigma\iota}$. Now we check$$\begin{align}\ffrac{\bar r_n - r_F}{\beta_{n,M}}&= \Big.\ffrac{\P{\sum y_k +1}\P{\Sigma \iota}_n}{K S_n}\Big/ \ffrac{w_M\P{\Sigma\iota}_n}{S_n \Tran\iota\Sigma\iota}\\&= \ffrac{1}{w_M}\ffrac{\P{\sum y_k +1}{\Tran\iota\Sigma \iota}}{K}\\&= \ffrac{1}{w_M}\P{\Tran\iota \bar v - \Tran\iota S} = \ffrac{\Tran\iota \bar v}{w_M} - 1\end{align}$$$\P 4$$\bspace$The marginal utility of agent $k$ at time period $1$ is now$$\begin{align}\partial_w U_k\P{\tilde w_{k,1}} &= \exp\CB{-\tilde w_k}\\&= \exp\CB{w_k+\Tran{\bar\theta_k}S + y_k \P{\tilde \varepsilon + \Tran\iota S} + \Tran{\P{\tilde v-S}}\Sigma^{-1}\P{\bar v - S}}\end{align}$$$\bspace$Thus the Pareto dominance require $\forall \: k$, $y_k$ is a constant and thus $y_k\tilde \varepsilon$. And this is automatically satisfied given $\tilde\varepsilon = 0$  Question 13.2$\P 1$$\bspace$Now we need to minimize$$\begin{align}&\ffrac{1}{2}\Var{\tilde w_k} - \Exp{\tilde w_k}\\=\;& \ffrac{1}{2}\Var{ \P{\Tran{\theta_k} + y_k \Tran\iota}\tilde v + y_k\tilde\varepsilon} - \P{w_k + \Tran{\theta_k} \bar v + \Tran{\P{\bar \theta_k - \theta_k }}S + y_k \Tran\iota\bar v}\\=\;& \ffrac{1}{2}\Tran{\P{\theta_k + y_k\iota}}\Sigma\P{\theta_k + y_k\iota} + \ffrac{\sigma^2y_k^2}{2} - w_k - \Tran{\theta_k} \bar v - \Tran{\P{\bar \theta_k - \theta_k }}S - y_k \Tran\iota\bar v\\&+ \P{\Tran\theta_k + y_k\Tran\iota}y_k c\end{align}$$$\bspace$Take the derivative on $\theta_k$ then we have$$0 = \P{\Tran\theta_k + y_k \Tran\iota}\Sigma + \Tran{\P{S - \bar v}} + y_k \Tran c \Rightarrow \theta_k = \Sigma^{-1}\P{\bar v - S - y_k c} - y_k\iota$$$\bspace$To see it's separation, we rewrite this as $\Sigma^{-1}\P{\bar v - S} - y_k\P{\iota + \Sigma^{-1}c}$ and these are the two components, thus $K=3$, with the last one the monetary fund.$\P 2$$\bspace$Similarly, we have $$K\cdot \Sigma^{-1}\P{\bar v - S} - \P{\sum y_k}\P{\iota + \Sigma^{-1}c} = \iota \Rightarrow S = \bar v - \ffrac{\P{\sum y_k +1}\Sigma \iota + c\sum y_k }{K}$$$\P 3$$\bspace$Now we still have $\beta_{n,M} = \ffrac{w_M\P{\Sigma\iota}_n}{S_n \Tran\iota\Sigma\iota}$. Consider$$\begin{align}\ffrac{\bar r_n - r_F}{\bar r_M - r_F} &= \ffrac{w_M}{S_n}\ffrac{\bar v_n -S_n}{\Tran\iota \bar v - w_M}\\&= \ffrac{w_M}{S_n}\ffrac{\ffrac{\P{\sum y_k +1}\P{\Sigma \iota}_n + c_n\sum y_k}{K}}{\ffrac{\P{\sum y_k +1}\Tran\iota\Sigma \iota + \Tran\iota c\sum y_k }{K}}\\&= \ffrac{w_M}{S_n}\ffrac{\P{\sum y_k +1}\P{\Sigma \iota}_n + c_n\sum y_k}{\P{\sum y_k +1}\Tran\iota\Sigma \iota + \Tran\iota c\sum y_k} \neq \beta_{n,M}\end{align}$$$\bspace$Thus CAPM model fails. And the reason is that now there's a covariance between the market risk and sigle stock risk and that violates the assumption. The expected return of stock $n$ is$$\bar r_n = \ffrac{\P{\sum y_k +1}\P{\Sigma \iota}_n + c_n\sum y_k}{K\bar v_n - \P{\sum y_k +1}\P{\Sigma \iota}_n + c_n\sum y_k}$$$\bspace$Various $\bar r_n$ is from- various $c_n$- various $\bar v_n$- various $\P{\Sigma\iota}_n$$\P 4$$$\begin{align}\partial_w U_k\P{\tilde w_{k,1}} &= \exp\CB{-\tilde w_k}\\&= \exp\CB{w_k+\Tran{\bar\theta_k}S + y_k \P{\tilde \varepsilon + \Tran\iota S} + \Tran{\P{\tilde v-S}}\Sigma^{-1}\P{\bar v - S - y_k c}}\\&= \exp\CB{w_k+\Tran{\bar\theta_k}S + y_k \P{\tilde \varepsilon + \Tran\iota S - \Tran{\P{\tilde v - S}}\Sigma^{-1}c} + \Tran{\P{\tilde v-S}}\Sigma^{-1}\P{\bar v - S}}\end{align}$$$\bspace$Thus the Pareto dominance requires $\forall\: k$, $y_k$ is a common constant, or $\tilde \varepsilon + \Tran\iota S - \Tran{\P{\tilde v - S}}\Sigma^{-1}c = 0$.  Question 13.3There's a ***fatal typo*** is the question, and I could ***NOT confirm*** it until the *last* ask... so I refuse to present it here.$\P 1$$\bspace$For agent $k$, in whichever country, his optimization problem is$$\begin{array}{cl}\d{\max_{\theta_{k,a}\:,\:\theta_{k,b}}} & \Exp{-e^{-\alpha \tilde w_{\void_k}}} = -\exp\CB{-\alpha\Exp{\tilde w_k} + \ffrac{1}{2}\alpha^2\Var{\tilde w_k}}\\s.t. & \tilde w_k = \P{w_k - \theta_{k,a}p_a - \theta_{k,b}p_b}\P{1+r_F}+ \Tran \theta_{k,a}\tilde v_a + \Tran\theta_{k,b}\tilde v_b\end{array}$$$\bspace$That is to say we need to minimize$$\begin{align}&\ffrac{1}{2}\alpha^2\Var{\tilde w_k} - \alpha\Exp{\tilde w_k}\\=\;& \ffrac{1}{2}\alpha^2\P{\Tran\theta_{k,a}\P{\Sigma_a+\sigma_w^2I}\theta_{k,a} + \Tran\theta_{k,b}\P{\Sigma_b+\sigma_w^2I}\theta_{k,b} + 2 \Tran\theta_{k,a}\sigma_w^2I\theta_{k,b}}\\&- \alpha\P{w_k - \Tran\theta_{k,a}\cdot p_a - \Tran\theta_{k,b}\cdot p_b}\P{1+r_F} - \alpha\P{\Tran\theta_{k,a} + \Tran\theta_{k,b}}\bar v\end{align}$$$\bspace$So then we take the derivatives on $\theta_{k,a}$ and $\theta_{k,b}$ which give$$\begin{cases}\ffrac{1}{2}\alpha^2\P{2 \Tran\theta_{k,a}\P{\Sigma_a+\sigma_w^2I} + 2\Tran\theta_{k,b}\sigma_w^2I}-\alpha\P{\Tran{\bar v}-\Tran p_a\P{1+r_F}} = 0\\\ffrac{1}{2}\alpha^2\P{2 \Tran\theta_{k,b}\P{\Sigma_b+\sigma_w^2I} + 2\Tran\theta_{k,a}\sigma_w^2I}-\alpha\P{\Tran{\bar v}-\Tran p_b\P{1+r_F}} = 0\end{cases}$$$$\Rightarrow\left\{\begin{align}\theta_{k,a} &= \ffrac{{\P{-\sigma_w^2\P{\Sigma_a+\Sigma_b} - \Sigma_b\Sigma_a}}^{-1}}{\alpha}\P{\sigma_w^2\bar v - \P{\Sigma_b+\sigma_w^2I}\bar v + \Sigma_bp_a\P{1+r_F}}\\\theta_{k,b} &= \ffrac{{\P{-\sigma_w^2\P{\Sigma_b+\Sigma_a} - \Sigma_a\Sigma_b}}^{-1}}{\alpha}\P{\sigma_w^2\bar v - \P{\Sigma_a+\sigma_w^2I}\bar v + \Sigma_ap_b\P{1+r_F}}\end{align}\right.$$$\P 2$$\bspace$In country $a$, $\sum_{K_a}\theta_{k,a} + \sum_{K_b}\theta_{k,a} = \iota$, thus$$\left\{\begin{align}\P{K_a + K_b}\P{\P{\Sigma_a+\sigma_w^2I} \theta_{k,a} + \sigma_w^2\theta_{k,b}} = \P{\Sigma_a+2\sigma_w^2I}\iota = \ffrac{K_a + K_b}{\alpha}\P{\bar v - p_a\P{1+r_F}}\\\P{K_a + K_b}\P{\P{\Sigma_b+\sigma_w^2I} \theta_{k,b} + \sigma_w^2\theta_{k,a}} = \P{\Sigma_b+2\sigma_w^2I}\iota = \ffrac{K_a + K_b}{\alpha}\P{\bar v - p_b\P{1+r_F}}\end{align}\right.$$$$\Rightarrow \left\{\begin{align}p_a &= \ffrac{1}{1+r_F}\P{\bar v -\ffrac{\alpha}{K_a + K_b}\P{\Sigma_a + 2\sigma_w^2I}\iota}\\p_b &= \ffrac{1}{1+r_F}\P{\bar v -\ffrac{\alpha}{K_a + K_b}\P{\Sigma_b + 2\sigma_w^2I}\iota}\end{align}\right.$$$\P 3$$\bspace$The risk premium of $n$th asset in country $a$ is$$\begin{align}\bar r_n - r_F &= \ffrac{\bar v_n}{\P{p_a}_n} -1-r_F\\&= \ffrac{\bar v_n}{\ffrac{1}{1+r_F}\P{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}}-\P{1+r_F}\\&= \ffrac{\P{\ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n + \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}\P{1+r_F}}{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}\end{align}$$$\bspace$And similarly, the risk premium of $m$th asset in country $b$ is$$\bar r_m - r_F = \ffrac{\P{\ffrac{\alpha}{K_a + K_b}\P{\Sigma_b\iota}_n + \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}\P{1+r_F}}{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_b\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}$$$\P 4$$\bspace$We first write $\tilde r_{wm} = \ffrac{\Tran\iota \P{\tilde v_a + \tilde v_b}}{\Tran\iota\P{p_a+p_b}} - 1$, then the risk premium or the market portfolio is$$\begin{align}\bar r_{wm} - r_F &= \ffrac{2\Tran\iota\bar v}{\Tran\iota\P{p_a+p_b}}-\P{1+r_F}\\&= \P{1+r_F}\P{\ffrac{2\Tran\iota\bar v}{2\Tran\iota\bar v - \ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}-1}\\&= \P{1+r_F}\ffrac{\ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}{2\Tran\iota\bar v - \ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}\end{align}$$$\bspace$Thus, for $n$th asset in country $a$,$$\ffrac{\bar r_n - r_F}{\bar r_{wm} - r_F} = \ffrac{\ffrac{\P{\ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n + \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}\P{1+r_F}}{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}}{\P{1+r_F}\ffrac{\ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}{2\Tran\iota\bar v - \ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}} = \ffrac{\ffrac{\ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n + \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}}{\ffrac{\ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}{2\Tran\iota\bar v - \ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}}$$$\bspace$And now the $\beta$$$\begin{align}\beta_{n,wm} &= \ffrac{\Cov{\tilde r_n,\tilde r_{wm}}}{\Var{\tilde r_{wm}}}\\&= \ffrac{\Cov{\ffrac{\tilde v_n}{\P{p_a}_n}-1,\ffrac{\Tran\iota \P{\tilde v_a + \tilde v_b}}{\Tran\iota\P{p_a+p_b}} - 1}}{\Var{\ffrac{\Tran\iota \P{\tilde v_a + \tilde v_b}}{\Tran\iota\P{p_a+p_b}} - 1}}\\&= \ffrac{\ffrac{1}{\P{p_a}_n\Tran\iota\P{p_a+p_b}}}{\ffrac{1}{\P{\Tran\iota\P{p_a+p_b}}^2}} \times \ffrac{\Cov{\tilde v_n,\Tran\iota\P{p_a+p_b}}}{\Var{\Tran\iota\P{p_a+p_b}}}\\&= \ffrac{\Tran\iota\ffrac{1}{1+r_F}\P{ 2\bar v - \ffrac{\alpha}{K_a+K_b}\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}}{\ffrac{1}{1+r_F}\P{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}}\cdot\ffrac{2\sigma_w^2+\P{\Sigma_a\iota}_n}{\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota} \\&= \ffrac{2\Tran\iota\bar v - \ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}{\bar v_n - \ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n - \ffrac{2\alpha}{K_a + K_b}\sigma_w^2} \cdot \ffrac{\ffrac{\alpha}{K_a + K_b}\P{\Sigma_a\iota}_n + \ffrac{2\alpha}{K_a + K_b}\sigma_w^2}{\ffrac{\alpha}{K_a+K_b}\Tran\iota\P{4\sigma_w^2I+\Sigma_a+\Sigma_b}\iota}\\&= \ffrac{\bar r_n - r_F}{\bar r_{wm} - r_F}\end{align}$$$\bspace$That's it. And similar for the $m$th asset in country $b$. Thus CAPM model holds. 委屈。。。  Question 13.5$\P 1$$\bspace$To eliminate the chance to arbitrage, $\forall\: \omega$, $\phi_\omega > 0$ and $\sum\phi_\omega \leq 1$.$\P 2$$\bspace$For state $\omega$, the interest rate is $\tilde r = \ffrac{w_\omega}{\phi_\omega}-1$. Then the mean value and the variance are $$\bar r = \sum_{\omega}\ffrac{\pi_\omega w_\omega}{\phi_\omega}-1,\bspace\Var{\tilde r} = \sum_\omega \pi_\omega\P{\ffrac{w_\omega}{\phi_\omega}}^2 - \P{\sum_{\omega}\ffrac{\pi_\omega w_\omega}{\phi_\omega}}^2$$$\bspace$And the covariance of portfolio $w$ and $w'$ is$$\Cov{\tilde r,\tilde r'} = \Exp{\tilde r\cdot\tilde r'} - \bar r \cdot\bar r' = \sum_\omega \pi_\omega\ffrac{w_\omega w_\omega'}{\phi_\omega^2} - \P{\sum_\omega \pi_\omega\ffrac{w_\omega}{\phi_\omega}}\P{\sum_\omega \pi_\omega\ffrac{w_\omega'}{\phi_\omega}}$$$\P3$$\bspace$The payoff, for all states $\omega$, is $\ffrac{w_\omega^a}{\phi_\omega}\times1 = \ffrac{1}{\sum\phi_\omega}$. And the price of that state-$\omega$ security is $\sum w_\omega^a = 1$, thus, it's riskless portfolio.$\bspace$To prove these two build up the frontier, consider the optimization problem$$\begin{array}{cl}\d{\min_{w_\omega}} & \ffrac{1}{2}\sum\nolimits_\omega {\pi_\omega} \P{\ffrac{w_\omega}{\phi_\omega}}^2\\s.t. & \sum_\omega \pi_\omega \ffrac{w_\omega}{\phi_\omega}-1 = \bar r_p \\& \sum_\omega w_\omega = 1\end{array}$$$\bspace$Then we write the corresponding Largrangian equation$$L = \ffrac{1}{2}\sum\nolimits_\omega {\pi_\omega} \P{\ffrac{w_\omega}{\phi_\omega}}^2 + \lambda_r \P{\bar r_p +1 - \pi_\omega \ffrac{w_\omega}{\phi_\omega} } + \lambda_1 \P{1-\sum\nolimits_\omega w_\omega}$$$\bspace$where the F.O.C. is $\pi_\omega \ffrac{w_\omega}{\phi_\omega^2} = \lambda_r \ffrac{\pi_\omega}{\phi_\omega} + \lambda_1$, which leads to$$w_\omega = \lambda_r \phi_\omega + \lambda_1 \phi_\omega q_\omega \Rightarrow \left\{\begin{align}\sum_{\omega}w_\omega = 1 &= \lambda_r\cdot\P{\sum\phi_\omega} + \lambda_1 \cdot\P{\sum\phi_\omega q_\omega}\\\sum_\omega \ffrac{w_\omega}{q_\omega} = 1 + \bar r_p &= \lambda_r\cdot\P{\sum\ffrac{\phi_\omega}{q_\omega}} + \lambda_1\cdot\P{\sum\phi_\omega} = \lambda_r + \lambda_1\cdot\P{\sum\phi_\omega}\end{align}\right.$$$$\Rightarrow\begin{cases}\lambda_r = \ffrac{\P{1+\bar r_p}\sum\phi_\omega q_\omega -\sum \phi_\omega}{\P{\sum\phi_\omega q_\omega} -\P{\sum \phi_\omega}^2}\\\lambda_1 = \ffrac{\P{1+\bar r_p}\sum\phi_\omega -1}{\P{\sum\phi_\omega q_\omega} -\P{\sum \phi_\omega}^2}\end{cases}$$$$\begin{align}w_\omega &= \lambda_r\phi_\omega + \ffrac{\lambda_1\phi_\omega^2}{\pi_\omega}\\&= \lambda_r\phi_\omega +\lambda_1\phi_\omega q_\omega\\&= \delta \cdot \ffrac{\phi_\omega}{\sum \phi_\omega} + \P{1-\delta}\cdot\ffrac{\phi_\omega q_\omega}{\sum\phi_\omega q_\omega}\end{align}$$$\bspace$where $\delta = \lambda_r \cdot \sum\phi_\omega = \ffrac{\P{1+\bar r_p}\sum\phi_\omega q_\omega\sum\phi_\omega -\P{\sum \phi_\omega}^2}{\P{\sum\phi_\omega q_\omega} -\P{\sum \phi_\omega}^2}$. That's it. ***$\\[35em]$<jupyter_code>import numpy as np
import matplotlib.pyplot as plt
sigma1 = 10
sigma2 = 5
bar_r1 = 11
bar_r2 = 10
X = np.linspace(0,5,1001)
Y_m1_u = ((bar_r1-bar_r2)*X+(sigma1*bar_r2+sigma2*bar_r1))/(sigma1+sigma2)
Y_m1_d = (-(bar_r1-bar_r2)*X+(sigma1*bar_r2+sigma2*bar_r1))/(sigma1+sigma2)

Y_1_u = ((bar_r1-bar_r2)*X+(sigma1*bar_r2-sigma2*bar_r1))/(sigma1-sigma2)
Y_1_d = (-(bar_r1-bar_r2)*X+(sigma1*bar_r2-sigma2*bar_r1))/(sigma1-sigma2)

I = np.ones(1001)
Y_0_u = (2*(sigma1**2)*bar_r2*I + 2*(sigma2**2)*bar_r1 * I +2*(bar_r1-bar_r2)*np.sqrt(X*X*(sigma1**2)+X*X*(sigma2**2) - I*(sigma1**2*sigma2**2)))/(2*(sigma1**2+sigma2**2))
Y_0_d = (2*(sigma1**2)*bar_r2*I + 2*(sigma2**2)*bar_r1 * I -2*(bar_r1-bar_r2)*np.sqrt(X*X*(sigma1**2)+X*X*(sigma2**2) - I*(sigma1**2*sigma2**2)))/(2*(sigma1**2+sigma2**2))
plt.plot(X,Y_m1_u)
plt.plot(X,Y_m1_d)
plt.plot(X,Y_1_u)
plt.plot(X,Y_1_d)
plt.plot(X,Y_0_u)
plt.plot(X,Y_0_d)<jupyter_output><empty_output>
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