Calculating Black–Scholes Vega from the Normal Density

Summary

The discussion explains how to implement the Black–Scholes vega formula by clarifying the notation for the derivative of the standard normal cumulative distribution function. By the fundamental theorem of calculus, that derivative is the standard normal probability density evaluated at the same input, so vega uses the density at d1 multiplied by spot price and the square root of time to expiry.

A second response gives an equivalent expression using strike price, the discounted strike, and the density evaluated at d2. These forms provide alternatives for implementation and cross-checking. The note is limited to the stated Black–Scholes expressions; it does not discuss discrete dividends, alternative pricing models, numerical approximations, or conventions for scaling vega across volatility units.

Key ideas

• The derivative of the standard normal cumulative distribution function is its probability density function.
• Black–Scholes vega is spot price times the normal density at d1 times the square root of time to expiry.
• An equivalent vega expression uses discounted strike and the normal density at d2.
• The formulas provide a direct implementation path but assume the stated Black–Scholes setup.

Tags

# Understanding Vega calculation in black Scholes model


# Understanding Vega calculation in black Scholes model












I am attempting to calculate the Greeks, and I understand their derivation. However when it comes to actually implementing Vega I am a little lost. Vega is defined analytically as:

$$ SN'(d_1)\sqrt{T-t} $$

It seems that it requires the first derivative of a normal distribution. I thought my probability knowledge was ok but I can't seem to find anything about its behavior, or how to implement the first derivative of a normal distribution programmatically.

Can anyone help me understand this?

Thank you!

## Answer by user16651 (score 2, accepted)

https://quant.stackexchange.com/a/27606

You know $$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-u^2}{2}}du$$ then $$N'(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$ by the Fundamental Theorem of Calculus. Therefore $$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$

## Answer by jaehyukchoi49 (score 0)

https://quant.stackexchange.com/a/71236

FYI, there is one more equivalent expression for the Black-Scholes vega: $$ \mathcal{V} = S_t N'(d_1) \sqrt{\tau} = \color{red}{K e^{-r\tau} N'(d_2) \sqrt{\tau}}. $$ See another answer and this question.

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