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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!

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    $\begingroup$ It is just $\frac{1}{\sqrt 2 \pi} e^{-0.5 d^2_1}$, the "bell curve" or normal density. $\endgroup$
    – nbbo2
    Jun 14, 2016 at 5:06
  • $\begingroup$ Since @noob2 was first if you want to post that as an answer I will mark this resolved. $\endgroup$
    – user20664
    Jun 14, 2016 at 8:23
  • $\begingroup$ I would simply add that the function $N(x) = \mathbb{P}(X \leq x)$ for any given random variable $X$ and $x \in \mathcal{X}$ is called the cumulative distribution function (cdf) of $X$, while the function $N'(x) = n(x)$ is called the probability density function of $X$ (pdf). So what you are looking at is not the first derivative of a normal distribution, but rather the normal pdf itself (bell-shaped). Also by definition, $N(x) = \int_{-\infty}^x n(u) du$ for a random variable with support $\mathcal{X}=\mathbb{R}$. $\endgroup$
    – Quantuple
    Jun 14, 2016 at 9:19
  • $\begingroup$ Mine was just a hint. The full answer is below. $\endgroup$
    – nbbo2
    Jun 14, 2016 at 13:29

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You know $$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-u^2}{2}}du$$ then $$N'(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}$$ therefore $$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$

enter image description here

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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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