關於將Black Scholes微分方程轉換為熱方程的問題
我正在閱讀一本關於將 Black Scholes 方程轉換為熱方程的書，對於那些我有疑問的人，我用粗體突出顯示，非常感謝你的建議。
讓 $ S $ , $ T $ , $ V $ 分別表示標的資產價格、期限和期權價格。下面是轉換過程：
讓 $ y=lnS $ 自從 $ (S=e^y) $ 和 $ \tau_t=T-t $ ，然後 $ \frac{\partial V}{\partial t}=-\frac{\partial V}{\partial \tau_t} $ , $ \frac{\partial V}{\partial S}=\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial y} $ 和 $ \frac{\partial^2 V}{\partial S^2}=\frac{\partial }{\partial S}(\frac{\partial V}{\partial S})=\frac{\partial }{\partial S}(\frac{1}{S}\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S}\frac{\partial }{\partial S}(\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} $ ,
這是我的第一個疑問：為什麼 $ \frac{1}{S}\frac{\partial V}{\partial S}(\frac{\partial V}{\partial y})=\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} $ 持有？
布萊克斯科爾斯方程 $ \frac{\partial V}{\partial t} + rS \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}-rV = 0 $
可以轉換為
$ -\frac{\partial V}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial V}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial y^2}-rV = 0 $
讓 $ u=e^{r\tau_t}V $ ,
方程變為
$ -\frac{\partial u}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial u}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2} = 0 $
最後，讓
$ x=y+(r-\frac{1}{2}\sigma^2)\tau_t=lnS+(r-\frac{1}{2}\sigma^2)\tau_t $
和
$ \tau=\tau_t $ ， 然後 $ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial x} $
和
$ \frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x} $ ,
這是我的第二個疑問：為什麼 $ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial x} $ 和 $ \frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x} $ 抓住？
你問題的第一部分：
- $ \frac{\partial y}{\partial S} = \frac{\partial ln S}{\partial S} = \frac{1}{S} $
- $ \frac{\partial^2 V}{\partial S \partial y} = \frac{\partial}{\partial y} \frac{\partial V}{\partial S} = \frac{\partial}{\partial y} (\frac{\partial y}{\partial S}\frac{\partial V}{\partial y})= \frac{\partial}{\partial y} (\frac{1}{ S}\frac{\partial V}{\partial y}) = \frac{-1}{S^2} \frac{\partial S}{\partial y}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} = \frac{-1}{S}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} $
- $ \frac{\partial^2 V}{\partial S^2} = \frac{\partial}{\partial S} (\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}) = \ \frac{\partial^2 V}{\partial S \partial y} \frac{\partial y}{\partial S} + \frac{\partial V}{\partial y} \frac{\partial^2 y}{\partial S^2} = \ \frac{\partial^2 V}{\partial S \partial y} \frac{1}{S} - \frac{1}{S^2}\frac{\partial V}{\partial y} = \ \frac{-1}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} -\frac{1}{S^2}\frac{\partial V}{\partial y} = \ \frac{-2}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} $
第二部分的關鍵是 $ \frac{\partial x}{\partial y} $ 是 1。