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Deriving Black–Scholes Call Vega from the Strike and Spot Terms

Article Quant Q&A · Author: foshizzle
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Summary

The document checks whether differentiating the Black–Scholes call price with respect to volatility produces the familiar vega formula. The derivation starts from the two density-weighted terms involving the spot price and discounted strike, then factors out the standard normal density at d-plus. It uses the relationship between d-plus and d-minus to simplify the exponential ratio and combine the remaining terms.

The result is call vega equal to spot multiplied by the standard normal density at d-plus and the square root of time to expiry. An equivalent expression uses the discounted strike and the density at d-minus. The explanation assumes the standard Black–Scholes setup and the stated definitions of d-plus and d-minus; the document does not discuss dividends, alternative volatility conventions, or other model extensions.

Key ideas

  • Call vega is the derivative of the Black–Scholes call price with respect to volatility.
  • Factoring out the standard normal density at d-plus helps simplify the differentiated price expression.
  • The exponential ratio links the d-plus and d-minus density terms to spot and discounted strike.
  • Vega can be written using either spot with the density at d-plus or discounted strike with the density at d-minus.
  • The resulting formula includes the square root of time to expiry.

Tags

OptionsDerivatives pricingVolatility
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# Derive vega for Black-Scholes call from this formula?


# Derive vega for Black-Scholes call from this formula?












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

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

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

## Answer by Gordon (score 9, accepted)

https://quant.stackexchange.com/a/26155

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.

## Answer by jaehyukchoi49 (score 4)

https://quant.stackexchange.com/a/71139

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.