Black-Scholes PDE 變換
摘自 Wilmott、Howison 和 Dewynne 的“金融衍生品數學”，第 5.4 節，第 76 頁。我如何開始進行轉換以得到無量綱方程？即我們從標準的 Black-Scholes PDE 開始：
$$ {\frac {\partial V}{\partial t}} +{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}} +rS{\frac {\partial V}{\partial S}}-rV=0 $$
並應用以下轉換： $$ S=Ee^x, t = T - \tau/\frac{1}{2}\sigma^2, V=Ev(x,\tau) $$ 獲得： $$ {\frac {\partial v}{\partial \tau}} = {\frac {\partial ^{2}v}{\partial x^{2}}} +(k-1){\frac {\partial v}{\partial x}}-kv $$ 在哪裡 $ k = r/\frac{1}{2}\sigma^2 $ .
我如何從這些轉變開始？假設我採取 $ rS{\frac {\partial V}{\partial S}} $ 術語和替代 $ S $ 和 $ V $ : $$ rS{\frac {\partial V}{\partial S}} = r \times Ee^x \times\frac {\partial Ev(x,\tau)}{\partial Ee^x } $$ 我該如何從這裡開始？
我認為通過注意到它可以簡化
$$ \dfrac{\partial}{\partial S} = \dfrac{\partial}{\partial (E e^x)} = \dfrac{\partial x}{\partial (E e^x)}\dfrac{\partial}{\partial x} = \left(\dfrac{\partial (E e^x)}{\partial x}\right)^{-1}\dfrac{\partial}{\partial x} = \dfrac{1}{E e^x}\dfrac{\partial}{\partial x} , $$
和
$$ \dfrac{\partial}{\partial t} = \dfrac{\partial \tau}{\partial t}\dfrac{\partial}{\partial \tau} = -\dfrac{2}{\sigma^2}\dfrac{\partial}{\partial \tau} . $$
使用這些表達式，只需替換 $ V $ 通過新變數，應該這樣做。