求解关于 π_(T-1) 的一阶最优化条件

为了对 'π_(T-1)' 求一阶最优化条件，我们需要先对该式子取关于 'π_(T-1)' 的偏导数：

\begin{aligned} \frac{\partial}{\partial \pi_{T-1}} &\left[D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \mathrm{e}^{-\gamma q_{T-1} \beta_{T-1}^2 / 2}\right] \&= -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} r_{T-1} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} \mu_{T-1} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} \frac{\sigma_{T-1}^2}{2} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \beta_{T-1}^2 \mathrm{e}^{-\gamma q_{T-1} \beta_{T-1}^2 / 2} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} \alpha_{T-1} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} q_{T-1} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \beta_{T-1} \end{aligned}

现在我们需要令上述式子等于 0，得到最优化条件：

\begin{aligned} & -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} r_{T-1} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} \mu_{T-1} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma \pi_{T-1} \mu_{T-1}} \mathrm{e}^{-\gamma \pi_{T-1} \sigma_{T-1}^2 / 2} \frac{\sigma_{T-1}^2}{2} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \beta_{T-1}^2 \mathrm{e}^{-\gamma q_{T-1} \beta_{T-1}^2 / 2} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} \alpha_{T-1} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \&\quad -\gamma D \mathrm{e}^{-\gamma\left(w_{T-1}-\pi_{T-1} r_{T-1}\right)} \mathrm{e}^{-\gamma c_{T-1}} \mathrm{e}^{-\gamma \delta\left(q_{T-1}\right)} q_{T-1} \mathrm{e}^{-\gamma q_{T-1} \alpha_{T-1}} \beta_{T-1} = 0 \&\end{aligned}

这是一个非线性方程，一般需要使用数值方法求解。