3
$\begingroup$

Is it possible to get the right formula for vega of a call option under the black scholes model from this formula?

$$\frac{\partial{C}}{\partial{\sigma}}=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)$$

$d_-=\frac{\ln{\frac{S_0}{k}}+(r-\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$ $d_+=\frac{\ln{\frac{S_0}{k}}+(r+\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$

$\endgroup$
4
$\begingroup$

Note that, \begin{align*} \frac{\partial{C}}{\partial{\sigma}} &=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)\\ &=\frac{1}{\sqrt{2\pi}}e^{\frac{-d_+^2}{2}}\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{d_+^2}{2} - \frac{d_-^2}{2}} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{1}{2}(d_+-d_-)(d_++d_-)} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{1}{2}\sigma \sqrt{t}\, \frac{2\ln \frac{S_0}{K} +2rt}{\sigma \sqrt{t}}} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{S_0d_+}{\sigma} \right]\\ &=S_0 N'(d_+)\sqrt{t}, \end{align*} which is the Black-Scholes vega formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.