会计与财务管理

由题意得，要求均衡消费和证券价格，即求使效用函数

u(c_0,c_1,c_2)=ln(c_0)+\frac{1}{2}ln(c_1)+\frac{1}{2}ln(c_2)\\ s.t. \begin{equation}\left\{\begin{array}{lr} c_0\leq1-s_1\theta_1-s_2\theta_2 \\ c_1\leq1+\theta_1+2\theta_2\\ c_2\leq2+\theta_1 \end{array}\right.\end{equation}

$c_0,c_1,c_2,s_1,s_2$ 。构造拉格朗日函数：

L(c,\theta)=ln(c_0)+\frac{1}{2}ln(c_1)+\frac{1}{2}ln(c_2)+\lambda(1-s_1\theta_1-s_2\theta_2-c_0)+\mu(1+\theta_1+2\theta_2-c_1)+\psi(2+\theta_1-c_2)

令：

\begin{equation}\left\{\begin{array}{lr} \frac{\partial L}{\partial c_0}=\frac{1}{c_0}-\lambda=0 \\ \frac{\partial L}{\partial c_1}=\frac{1}{2c_1}-\mu=0 \\ \frac{\partial L}{\partial c_2}=\frac{1}{2c_2}-\psi=0 \\ \frac{\partial L}{\partial \theta_1}=-s_1\lambda+\mu+\psi=0 \\ \frac{\partial L}{\partial \theta_2}=-s_2\lambda+2\mu=0 \\ \lambda(1-s_1\theta_1-s_2\theta_2-c_0)=0 \\ \mu(1+\theta_1+2\theta_2-c_1)=0 \\ \psi(2+\theta_1-c_2)=0 \\ \end{array}\right.\end{equation}

整理得：

\begin{equation}\left\{\begin{array}{lr} c_0=\frac{2(1+\theta_1+2\theta_2)(2+\theta_1)}{2(1+\theta_1+2\theta_2)(2+\theta_1)-[(3+2\theta_1+2\theta_2)\theta_1+2(2+\theta_1)\theta_2)]}\\ c_1=1+\theta_1+2\theta_2\\ c_2=2+\theta_1\\ s_1=\frac{c_0(c_1+c_2)}{2c_1c_2}\\ s_2=\frac{c_0}{c_1}\\ \lambda=\frac{1}{c_0}\\ \mu=\frac{1}{2c_1}\\ \psi=\frac{1}{2c_2} \end{array}\right.\end{equation}

$\theta_1=\theta_2=0$ ，代入上式解之得：

\lambda=1,\mu=\frac{1}{2},\psi=\frac{1}{4},s_1=\frac{3}{4},s_2=1,c_0=1,c_1=1,c_2=2

$(s_1^*,s_2^*)=(\frac{3}{4},1)$ $(c_0^*,c_1^*,c_2^*)=(1,1,2)$