K2−D2infπ2T−1exp−α21−γzT−11+rT−1+π2T−1μ2T−1−rT−1+π∗1T−1μ1T−1−rT−1+12α221−γ2π22T−1σ22T−1+π∗1T−12σ21T−1+2π∗1T−1π2T−1ρT−1σ1T−1σ2T−1对该式子求解由一阶最优化条件找到最优策略pi_2 T-1^

根据一阶最优化条件，我们需要对式子求偏导数，并令其等于0，即：

∂(K2−D2infπ2,T−1,T−1exp{−α2(1−γ)[zT−1(1+rT−1)+π2,T−1(μ2,T−1−rT−1)+π∗1,T−1(μ1,T−1−rT−1)]+12α22(1−γ)2[π22,T−1σ22,T−1+(π∗1,T−1)2σ21,T−1+2π∗1,T−1π2,T−1ρT−1σ1,T−1σ2,T−1]})/∂π2,T−1 = 0

化简上式，得到：

-K2 + D2inf + α2(1-γ)exp{-α2(1-γ)[zT-1(1+rT-1)+π2,T-1(μ2,T-1-rT-1)+π∗1,T-1(μ1,T-1-rT-1)]+12α22(1-γ)2[π22,T-1σ22,T-1+(π∗1,T-1)2σ21,T-1+2π∗1,T-1π2,T-1ρT-1σ1,T-1σ2,T-1]}π2,T-1 - α22(1-γ)2σ22,T-1π2,T-1 - α22(1-γ)2ρT-1σ1,T-1π∗1,T-1 = 0

移项，得到：

π2,T-1 = (K2 - D2inf - α22(1-γ)2σ22,T-1π2,T-1 - α22(1-γ)2ρT-1σ1,T-1π∗1,T-1)/α2(1-γ)exp{-α2(1-γ)[zT-1(1+rT-1)+π2,T-1(μ2,T-1-rT-1)+π∗1,T-1(μ1,T-1-rT-1)]+12α22(1-γ)2[π22,T-1σ22,T-1+(π∗1,T-1)2σ21,T-1+2π∗1,T-1π2,T-1ρT-1σ1,T-1σ2,T-1]}

因此，最优策略为：

pi_{2, T-1}^* = (K2 - D2inf - α22(1-γ)2σ22,T-1pi_{2, T-1}^* - α22(1-γ)2ρT-1σ1,T-1pi_{1, T-1}^)/α2(1-γ)exp{-α2(1-γ)[zT-1(1+rT-1)+pi_{2, T-1}^(μ2,T-1-rT-1)+pi_{1, T-1}^(μ1,T-1-rT-1)]+12α22(1-γ)2[pi_{2, T-1}^2σ22,T-1+(pi_{1, T-1}^)2σ21,T-1+2pi_{1, T-1}^*pi_{2, T-1}^*ρT-1σ1,T-1σ2,T-1]