beginaligned& K_2-D_2 inf _pi_2 T-1 exp left-alpha_21-gammaleftz_T-1left1+r_T-1right+pi_2 T-1leftmu_2 T-1-r_T-1rightrightright & left+pi_1 T-1^leftmu_1 T-1-r_T-1rightright+frac12 alpha_2^21-gamma^2lef

根据一阶最优化条件，我们需要对 $\pi_{2, T-1}$ 求偏导数，并令其等于零，即：

$$\frac{\partial}{\partial \pi_{2, T-1}}\left(K_2-D_2 \exp \left{-\alpha_2(1-\gamma)\left[z_{T-1}\left(1+r_{T-1}\right)+\pi_{2, T-1}\left(\mu_{2, T-1}-r_{T-1}\right)\right.\right.\right.$$ $$\left.\left.+\pi_{1, T-1}^\left(\mu_{1, T-1}-r_{T-1}\right)\right]+\frac{1}{2} \alpha_2^2(1-\gamma)^2\left[\pi_{2, T-1}^2 \sigma_{2, T-1}^2\right.\right.$$ $$\left.\left.+\left(\pi_{1, T-1}^\right)^2 \sigma_{1, T-1}^2+2 \pi_{1, T-1}^* \pi_{2, T-1} \rho_{T-1} \sigma_{1, T-1} \sigma_{2, T-1}\right]\right}\right) = 0$$

化简上式，得到：

$$-D_2 \exp \left{-\alpha_2(1-\gamma)\left[z_{T-1}\left(1+r_{T-1}\right)+\pi_{2, T-1}^\left(\mu_{2, T-1}-r_{T-1}\right)\right.\right.$$ $$\left.\left.+\pi_{1, T-1}^\left(\mu_{1, T-1}-r_{T-1}\right)\right]+\frac{1}{2} \alpha_2^2(1-\gamma)^2\left[\pi_{2, T-1}^* \sigma_{2, T-1}^2\right.\right.$$ $$\left.\left.+2 \pi_{1, T-1}^* \rho_{T-1} \sigma_{1, T-1} \sigma_{2, T-1}\right]\right}\left[\alpha_2(1-\gamma)\left(\mu_{2, T-1}-r_{T-1}\right)+\alpha_2^2(1-\gamma)^2 \pi_{2, T-1}^* \sigma_{2, T-1}^2\right] = 0$$

由于 $\exp(\cdot)$ 函数是正的，因此我们可以将上式化简为：

$$\alpha_2(1-\gamma)\left(\mu_{2, T-1}-r_{T-1}\right)+\alpha_2^2(1-\gamma)^2 \pi_{2, T-1}^* \sigma_{2, T-1}^2 = 0$$

解出 $\pi_{2, T-1}^*$，得到：

$$\pi_{2, T-1}^* = -\frac{\mu_{2, T-1}-r_{T-1}}{\alpha_2(1-\gamma)\sigma_{2, T-1}^2}$$

因此，最优策略为：

$$\pi_{2, T-1}^* = -\frac{\mu_{2, T-1}-r_{T-1}}{\alpha_2(1-\gamma)\sigma_{2, T-1}^2}$