隨機波動 PDE 的 Gatheral 變數變化
這取自 Gatheral 的書“波動率表面”，他試圖從方程 2.3 到方程 2.4。
我們有以下 PDE，
$$ \frac{\partial V}{\partial t}+\frac{1}{2}vS^2\frac{\partial ^2 V}{\partial S^2} + \rho\eta vS\frac{\partial ^2 V}{\partial S\partial v}+\frac{1}{2}\eta^2v\frac{\partial ^2 V}{\partial v^2} + rS\frac{\partial V}{\partial S}-rV-\lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$ 通過使用變數的變化，
$$ x=\ln{\frac{Se^{r\tau}}{K}}, \tau = T-t $$ 表明它減少到
$$ -\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} - \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$ 我的工作如下…
我們有 $ C(x,v,\tau)=V(Ke^{x-r\tau}, v, T-\tau) $ . 偏導數是，
$$ \begin{aligned} \frac{\partial V}{\partial t} &= \frac{\partial V}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial V}{\partial t}\ &=\frac{\partial V}{\partial x}\frac{\partial x}{\partial S}\frac{\partial S}{\partial \tau}\frac{\partial \tau}{\partial t}+\frac{\partial V}{\partial \tau}\frac{\partial \tau}{\partial t}\ &=\frac{\partial V}{\partial x}\frac{1}{S}(-rS)(-1)-\frac{\partial V}{\partial \tau}\ &= r\frac{\partial V}{\partial x}-\frac{\partial V}{\partial \tau}\ \frac{\partial V}{\partial S} &= \frac{\partial V}{\partial x}\frac{\partial x}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial x} \ \frac{\partial^2 V}{\partial S^2}&= \frac{1}{S}\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial x}-\frac{1}{S^2}\frac{\partial V}{\partial x} \ &=\frac{1}{S^2}\left(\frac{\partial^2 V}{\partial x^2}-\frac{\partial V}{\partial x} \right)\ \frac{\partial V}{\partial S\partial v} &= \frac{\partial }{\partial S}\frac{\partial V}{\partial v}=\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial v} = \frac{1}{S}\frac{\partial V}{\partial x\partial v} \end{aligned} $$ 代入原始的 PDE，不幸的是，我得到了，
$$ r\frac{\partial C}{\partial x}-\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} +r\frac{\partial C}{\partial x}-rC- \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$ 我不知道如何擺脫 $ r\frac{\partial C}{\partial x} $ 和 $ rC $ 條款。
我想也許我留下了涉及他評論的關鍵步驟：“進一步，假設我們只考慮歐洲期權價格的到期 C 的未來價值，而不是它今天的價值，並定義 $ \tau = T − t $ 。”
有人可以幫忙嗎？謝謝！
首先請注意，您在金錢的定義中有一個錯字。它應該是
$$ \begin{equation} x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right). \end{equation} $$ 根據您引用的評論，我們然後定義
$$ \begin{equation} e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t). \end{equation} $$ 注意 $ C $ 是一個函式 $ \tau $ 並不是 $ t $ - 這在你的符號中似乎很奇怪。相應的偏導數由下式給出
$$ \begin{eqnarray} \frac{\partial V}{\partial t} & = & e^{-r \tau} \left( r C - r \frac{\partial C}{\partial x} - \frac{\partial C}{\partial \tau} \right), \ \frac{\partial V}{\partial S} & = & e^{-r \tau} \frac{1}{S} \frac{\partial C}{\partial x}, \ \frac{\partial^2 V}{\partial S^2} & = & e^{-r \tau} \frac{1}{S^2} \left( \frac{\partial^2 C}{\partial x^2} - \frac{\partial C}{\partial x} \right). \end{eqnarray} $$ 代入 Gatheral 書中的公式 (2.4)。我還建議您明確區分 $ C $ 和 $ V $ 並且沒有 $ V $ 在你的偏導數的兩邊。