已知：alpha_1leftmu_1-r_fright-alpha_1^2leftpi_1^ sigma_1^2-pi_2^ theta_1 sigma_1 sigma_2right=0 alpha_2leftmu_2-r_fright-alpha_2^2leftpi_2^ sigma_2^2-pi_1^ theta_2 sigma_1 sigma_2right=0 求解pi_1^和pi_2^

将第一个式子乘以$\pi_2^$，第二个式子乘以$\pi_1^$，然后相加，得到： $$ \alpha_1\pi_2^\left(\mu_1-r_f\right)-\alpha_1^2\pi_2^\left(\pi_1^* \sigma_1^2-\pi_2^* \theta_1 \sigma_1 \sigma_2\right)+\alpha_2\pi_1^\left(\mu_2-r_f\right)-\alpha_2^2\pi_1^\left(\pi_2^* \sigma_2^2-\pi_1^* \theta_2 \sigma_1 \sigma_2\right)=0 $$ 整理得： $$ \pi_1^\left(\alpha_2^2\theta_2\sigma_1\sigma_2-\alpha_2\left(\mu_2-r_f\right)\right)=\pi_2^\left(\alpha_1^2\theta_1\sigma_1\sigma_2-\alpha_1\left(\mu_1-r_f\right)\right) $$ 再将第一个式子乘以$\theta_1$，第二个式子乘以$\theta_2$，然后相减，得到： $$ \pi_1^\left(\alpha_2^2\theta_2\theta_1\sigma_1^2-\alpha_2\theta_1\left(\mu_2-r_f\right)\sigma_1\sigma_2\right)-\pi_2^\left(\alpha_1^2\theta_1\theta_2\sigma_1^2-\alpha_1\theta_2\left(\mu_1-r_f\right)\sigma_1\sigma_2\right)=0 $$ 整理得： $$ \pi_2^=\frac{\alpha_2\theta_1\left(\mu_1-r_f\right)\sigma_1\sigma_2+\alpha_1\theta_2\left(\mu_2-r_f\right)\sigma_1\sigma_2}{\alpha_1^2\theta_1\theta_2\sigma_1^2+\alpha_2^2\theta_1\theta_2\sigma_2^2} $$ 将$\pi_2^$代入前面的式子，得到： $$ \pi_1^*=\frac{\alpha_2\theta_1\left(\mu_1-r_f\right)\sigma_1\sigma_2+\alpha_1\theta_2\left(\mu_2-r_f\right)\sigma_1\sigma_2}{\alpha_1^2\theta_1^2\sigma_1^2+\alpha_2^2\theta_2^2\sigma_2^2-\alpha_1\alpha_2\theta_1\theta_2\sigma_1\sigma_2} $