這些 SDE 是如何得出的？
誰能給我一個詳細的解釋下面的方程（3）和（4）是如何從（1）和（2）推導出來的？
$$ \begin{align*} \frac{dF_{t,T}}{F_{t,T}} &=\sigma e^{-\lambda(T-t)}dB_t, \tag{1}\ \ln(F_{t,T})&=\ln(F_{0,T})-1/2\int_{0}^{t}\sigma^2 e^{-2\lambda(T-s)}ds+\int_{0}^{t}\sigma e^{-\lambda(T-s)}dB_s.\tag{2} \end{align*} $$ 給定 $ \ln(S_t)=\ln(F_{t,t}) $ ， 我們有： $$ \begin{align*} \frac{dS_t}{S_t}=(\mu_t-\lambda \ln(S_t))dt+\sigma dB_t,\tag{3} \end{align*} $$ 在哪裡 $$ \begin{align*} \mu_t=\frac{\partial \ln(F_{0,t})}{\partial t} +\lambda \ln(F_{0,t})+\frac{1}{4}\sigma^2(1-e^{-2\lambda t}). \tag{4} \end{align*} $$ 或者與他們相關的任何事情都會有所幫助。
從 $ (2) $ ,
$$ \begin{align*} \ln S_t &=\ln F_{t, t} \ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} $$ 然後， $$ \begin{align*} \lambda e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s = \lambda \ln S_t - \lambda \ln F_{0, t} + \frac{\sigma^2}{4} \left(1-e^{-2\lambda t}\right). \end{align*} $$ 所以， $$ \begin{align*} d\ln S_t &= \left(\frac{\partial \ln F_{0, t}}{\partial t}-\frac{\sigma^2}{2}e^{-2\lambda t} - \lambda e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s\right)dt +\sigma dB_t\ &=\left[\frac{\partial \ln F_{0, t}}{\partial t}-\frac{\sigma^2}{2}e^{-2\lambda t}+\lambda \ln F_{0, t} - \frac{\sigma^2}{4} \left(1-e^{-2\lambda t}\right) -\lambda \ln S_t\right]dt + \sigma dB_t\ &=\left(\frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} -\frac{\sigma^2}{4} - \frac{\sigma^2}{4} e^{-2\lambda t} -\lambda \ln S_t\right)dt + \sigma dB_t. \end{align*} $$ 注意 $$ \begin{align*} d\langle \ln S, \ln S \rangle_t= \sigma^2 dt. \end{align*} $$ 根據伊藤引理， $$ \begin{align*} dS_t &= de^{\ln S_t}\ &= e^{\ln S_t} d \ln S_t + \frac{1}{2}e^{\ln S_t}d\langle \ln S, \ln S \rangle_t\ &=S_t d \ln S_t + \frac{1}{2} \sigma^2 S_t dt\ &= S_t\left[\left(\frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} -\frac{\sigma^2}{4} - \frac{\sigma^2}{4} e^{-2\lambda t} -\lambda \ln S_t\right)dt + \sigma dB_t + \frac{\sigma^2}{2} dt \right]\ &=S_t\big[\left(\mu_t - \lambda \ln S_t\right)dt + \sigma dB_t\big], \end{align*} $$ 在哪裡 $$ \begin{align*} \mu_t = \frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} +\frac{\sigma^2}{4}\left(1- e^{-2\lambda t}\right). \end{align*} $$