## Black Scholes Greeks

We derive the formulae for the Price and Greeks (derivatives with respect to inputs) of the European options under the Black-Scholes assumptions.

### Vega

We now differentiate the Black-Scholes Call Option price formula with respect to $$\sigma$$, the volatility of the underlying, in order to derive a formula for the Vega:

$$\frac {\partial BS Call Price} {\partial \sigma} =\frac {\partial } {\partial \sigma} \left( S e^{- r_{f} \tau} N{\left (d_{1} \right )}- K e^{- r_{d} \tau} N{\left (d_{1} - \sigma \sqrt{\tau} \right )} \right)$$ $$= \frac {\partial } {\partial \sigma} \left( S e^{- r_{f} \tau} N{\left (d_{1} \right )}\right) -\frac {\partial } {\partial \sigma} \left( K e^{- r_{d} \tau} N{\left (d_{1} - \sigma \sqrt{\tau} \right )} \right)$$ $$= S e^{- r_{f} \tau} \frac {\partial } {\partial \sigma} \left( N{\left (d_{1} \right )} \right)-K e^{- r_{d} \tau} \frac {\partial } {\partial \sigma} \left( N{\left (d_{1} - \sigma \sqrt{\tau} \right )} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left( d_{1} \right) - K e^{- r_{d} \tau} n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \frac {\partial } {\partial \sigma} \left(d_{1} - \sigma \sqrt{\tau} \right)$$

To simplify, we substitute the expression for $$n \left( d_{1} -\sigma \sqrt{\tau} \right)$$ from the Delta Section:

$$n \left( d_{1}-\sigma \sqrt{\tau}\right)=n \left( d_{1}\right) \frac{S}{K} e^{ \left (r_{d} - r_{f}\right) \tau}$$

Thus,

$$\frac {\partial BS Call Price} {\partial \sigma} = S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left( d_{1} \right) - K e^{- r_{d} \tau} n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \frac {\partial } {\partial \sigma} \left(d_{1} - \sigma \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left( d_{1} \right) - K e^{- r_{d} \tau} { n \left( d_{1}\right) \frac{S}{K} e^{ \left (r_{d} - r_{f}\right) \tau} } \frac {\partial } {\partial \sigma} \left(d_{1} - \sigma \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left( d_{1} \right) - S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left(d_{1} - \sigma \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial } {\partial \sigma} \left( d_{1} \right) - S e^{- r_{f} \tau} n{\left (d_{1} \right)} \left( \frac {\partial } {\partial \sigma} \left( d_{1} \right) - \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} \sqrt{\tau} n{\left (d_{1} \right)}$$

And for a Put option, we get:

$$\frac {\partial BS Put Price} {\partial \sigma}= \frac {\partial } {\partial \sigma} \left(- S e^{- r_{f} \tau} N{\left (-d_{1} \right )}+ K e^{- r_{d} \tau} N{\left (-d_{1} + \sigma \sqrt{\tau} \right )} \right)$$ $$=\frac {\partial } {\partial \sigma} \left( -S e^{- r_{f} \tau} N{\left (-d_{1} \right )}\right) +\frac {\partial } {\partial \sigma} \left( K e^{- r_{d} \tau} N{\left (-d_{1} + \sigma \sqrt{\tau} \right )} \right)$$ $$= -S e^{- r_{f} \tau} \frac {\partial } {\partial \sigma} N{\left (-d_{1} \right )} + K e^{- r_{d} \tau} \frac {\partial } {\partial \sigma} N{\left (-d_{1} + \sigma \sqrt{\tau} \right )}$$ $$= - S e^{- r_{f} \tau} n{\left (-d_{1} \right)} \frac {\partial } {\partial \sigma} \left( -d_{1} \right) + K e^{- r_{d} \tau} n{\left (-d_{1} + \sigma \sqrt{\tau} \right )} \frac {\partial } {\partial \sigma} \left(-d_{1} + \sigma \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (-d_{1} \right)} \frac {\partial \left( d_{1} \right) } {\partial \sigma} - K e^{- r_{d} \tau} n{\left (-d_{1} + \sigma \sqrt{\tau} \right )} \frac {\partial \left(d_{1} - \sigma \sqrt{\tau} \right) } {\partial \sigma}$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial \left( d_{1} \right) } {\partial \sigma} - K e^{- r_{d} \tau} n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \left( \frac {\partial \left( d_{1} \right)} {\partial \sigma} - \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial \left( d_{1} \right) } {\partial \sigma} - K e^{- r_{d} \tau}n \left( d_{1}\right) \frac{S}{K} e^{ \left (r_{d} - r_{f}\right) \tau} \left( \frac {\partial \left( d_{1} \right)} {\partial \sigma} - \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial \left( d_{1} \right) } {\partial \sigma} - S e^{- r_{f} \tau} n{\left (d_{1} \right)} \frac {\partial \left( d_{1} \right) } {\partial \sigma} \left( \frac {\partial \left( d_{1} \right)} {\partial \sigma} - \sqrt{\tau} \right)$$ $$= S e^{- r_{f} \tau} n{\left (d_{1} \right)} \sqrt{\tau}= S e^{- r_{f} \tau} \sqrt{\tau} n{\left (d_{1} \right)}$$

Where again we have used $$n(-x)=n(x)$$ because the function $$n(x)$$ is symmetric around 0.