求解最优策略 pi_{2, T-1}^* - 一阶最优化条件应用

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：/n/n$$/n//frac{//partial}{//partial //pi_{2, T-1}}//left(K_2-D_2 //exp //left//{-//alpha_2(1-//gamma)//left[z_{T-1}//left(1+r_{T-1}//right)+//pi_{2, T-1}//left(//mu_{2, T-1}-r_{T-1}//right)//right.//right.//right./n$$/n$$/n//left.//left.+//pi_{1, T-1}^//left(//mu_{1, T-1}-r_{T-1}//right)//right]+//frac{1}{2} //alpha_2^2(1-//gamma)^2//left[//pi_{2, T-1}^2 //sigma_{2, T-1}^2//right.//right./n$$/n$$/n//left.//left.+//left(//pi_{1, T-1}^//right)^2 //sigma_{1, T-1}^2+2 //pi_{1, T-1}^* //pi_{2, T-1} //rho_{T-1} //sigma_{1, T-1} //sigma_{2, T-1}//right]//right//}//right) = 0/n$$/n/n化简上式，得到：/n/n$$-D_2 //exp //left//{-//alpha_2(1-//gamma)//left[z_{T-1}//left(1+r_{T-1}//right)+//pi_{2, T-1}^//left(//mu_{2, T-1}-r_{T-1}//right)//right.//right.//right./n$$/n$$/n//left.//left.+//pi_{1, T-1}^//left(//mu_{1, T-1}-r_{T-1}//right)//right]+//frac{1}{2} //alpha_2^2(1-//gamma)^2//left[//pi_{2, T-1}^* //sigma_{2, T-1}^2//right.//right./n$$/n$$/n//left.//left.+2 //pi_{1, T-1}^* //rho_{T-1} //sigma_{1, T-1} //sigma_{2, T-1}//right]//right//}//right/}/n$$/n$$/n//left[//alpha_2(1-//gamma)//left(//mu_{2, T-1}-r_{T-1}//right)+//alpha_2^2(1-//gamma)^2 //pi_{2, T-1}^* //sigma_{2, T-1}^2//right] = 0/n$$/n/n由于 'exp(·)' 函数是正的，因此我们可以将上式化简为：/n/n$$/n//alpha_2(1-//gamma)//left(//mu_{2, T-1}-r_{T-1}//right)+//alpha_2^2(1-//gamma)^2 //pi_{2, T-1}^* //sigma_{2, T-1}^2 = 0/n$$/n/n解出 'pi_{2, T-1}^'，得到：/n/n$$/n//pi_{2, T-1}^ = -//frac{//mu_{2, T-1}-r_{T-1}}{//alpha_2(1-//gamma)//sigma_{2, T-1}^2}/n$$/n/n因此，最优策略为：/n/n$$/n//pi_{2, T-1}^* = -//frac{//mu_{2, T-1}-r_{T-1}}{//alpha_2(1-//gamma)//sigma_{2, T-1}^2}/n$$