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).