6
$\begingroup$

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

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

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

$\endgroup$

2 Answers 2

6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

The answer by @Gordon is pretty complete, but let me add one more point. Let $n(x) = N'(x)$ be the PDF of standard normal distribution.

In the derivation, note that $$ e^{d_+^2/2 - d_-^2/2} = \frac{n(d_-)}{n(d_+)} = \frac{S_0}{Ke^{-rt}}. $$ Thanks to this relation, there are two equivalent expressions for the Black-Scholes vega: $$ \frac{\partial C}{\partial \sigma} = S_0 n(d_+) \sqrt{t} = K e^{-rt} n(d_-) \sqrt{t}. $$ See Wikipedia.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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