## 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.

### Delta

Delta is the first derivative of the option price with respect to the underlying price (e.g., stock price). We first compute the Delta of a Call option, the price of which is given by:

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

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

We notice that the second component on the RHS (right hand side) has $$n \left( d_{1}\right)$$, whereas the third component has $$n \left( d_{1} -\sigma \sqrt{\tau} \right)$$, and we thus need to express both in common terms. We have, by definition:

$$n \left( d_{1}\right)=\frac{1}{ \sqrt{2\pi}} {\ e^{- \frac{d_{1}^{2}}{2}}}$$

Then

$$n \left( d_{1}-\sigma \sqrt{\tau}\right)=\frac{1}{ \sqrt{2 \pi}} e^{- \frac{1}{2} \left(d_{1} - \sigma \sqrt{\tau}\right)^{2}}$$ $$= \frac{1}{ \sqrt{2 \pi}} e^{- \frac{d_{1}^{2}}{2} - \frac{\sigma^{2} \tau}{2} + d_{1} \sigma \sqrt{\tau} }$$ $$= \frac{1}{ \sqrt{2 \pi}} e^{- \frac{d_{1}^{2}}{2}} e^{-\frac{\sigma^{2} \tau}{2} + d_{1} \sigma \sqrt{\tau} }$$ $$= n \left( d_{1}\right) e^{-\frac{\sigma^{2} \tau}{2} + d_{1} \sigma \sqrt{\tau} }$$ $$= n \left( d_{1}\right) e^{-\frac{\sigma^{2} \tau}{2} + \frac{1}{\sigma \sqrt{\tau}} \left(\ln{\left (\frac{S}{K} \right ) + \left(r_{d} - r_{f} + \frac{\sigma^{2}}{2}\right) \tau}\right) \sigma \sqrt{\tau} }$$ $$= n \left( d_{1}\right) e^{-\frac{\sigma^{2} \tau}{2} + \ln{\left (\frac{S}{K} \right ) + \left(r_{d} - r_{f} + \frac{\sigma^{2}}{2}\right) \tau} }$$ $$= n \left( d_{1}\right) e^{ \ln{\left (\frac{S}{K} \right ) + \left(r_{d} - r_{f}\right) \tau} }$$ $$= n \left( d_{1}\right) e^{ \ln{\left (\frac{S}{K} \right )}} e^{ \left (r_{d} - r_{f}\right) \tau}$$ $$= n \left( d_{1}\right) \frac{S}{K} e^{ \left (r_{d} - r_{f}\right) \tau}$$

We now substitute the above expression for $$n \left( d_{1} -\sigma \sqrt{\tau} \right)$$ to get:

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

We can now rapidly apply the above process to compute the Delta of a Put option, the price of which is given by

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

Where we have used $$n(-x)=n(x)$$ because the function $$n(x)$$ is symmetric around 0 (visualise the bell shaped standard normal distribution curve).