Tag Info

Hot answers tagged

2

Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= ...


1

let $\frac{\partial C}{\partial S}=\delta_c$ let $\frac{\partial^2 C}{\partial S^2}=\Gamma_c$ let $\frac{\partial C_0}{\partial S}=\delta_0$ let $\frac{\partial^2 C_0}{\partial S^2}=\Gamma_0$ we want $\frac{\partial V}{\partial S}=\frac{\partial C}{\partial S}=\delta_c$ and $\frac{\partial^2 V}{\partial S^2}=\frac{\partial^2 C}{\partial S^2}=\Gamma_c$ ...



Only top voted, non community-wiki answers of a minimum length are eligible