## Black's Price and Greeks

We derive the formulae for the price and Greeks of a call and a put options under the Black's model assumptions:

### Delta

Delta is the first derivative of the option price wrt the underlying price (e.g., stock price). We first compute the Delta of an option , the price of which is given by:

$$Black= e^{-r\tau} \left[ \phi S N{\left( \phi d_{1} \right )}- \phi K N{\left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right )} \right]$$

Re-arranging to avoid repetition of $$e^{-r\tau}$$ term, and then differentiating both sides wrt S, we get

$$e^{r\tau} Black= \phi S N{\left( \phi d_{1} \right )}- \phi K N{\left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right )}$$ $$e^{r\tau} \frac{\partial Black}{\partial S}=\frac{\partial}{\partial S} \left( \phi S N{\left( \phi d_{1} \right)}- \phi K N{\left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right )} \right)$$ $$= \frac {\partial } {\partial S} \left( \phi S N{\left( \phi d_{1} \right)} \right) - \frac {\partial } {\partial S} \left( \phi K N{\left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right )} \right)$$ $$= \left( \phi N{\left( \phi d_{1} \right )}\frac {\partial} {\partial S} S+ \phi S \frac {\partial } {\partial S} N{\left( \phi d_{1} \right )} \right)- \phi K \frac {\partial } {\partial S} \left( N{\left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right )} \right)$$ $$= \left( \phi N{\left( \phi d_{1} \right )}+ \phi S n{\left (d_{1} \right)} \frac {\partial } {\partial S} \left( \phi d_{1} \right) \right)- \phi K n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \frac {\partial } {\partial S} \left( \phi d_{1} - \phi \sigma \sqrt{\tau} \right)$$ $$= \phi N{\left( \phi d_{1} \right )}+ \phi S n{\left (d_{1} \right)} \frac {\partial \left( \phi d_{1} \right) } {\partial S} - \phi K n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \frac {\partial \left( \phi d_{1} \right) } {\partial S}$$

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

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

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

$$e^{r\tau} \frac {\partial Black } {\partial S}= \phi N{\left( \phi d_{1} \right )}+ \phi S n{\left (d_{1} \right)} \frac {\partial \left( \phi d_{1} \right) } {\partial S} - \phi K n{\left (d_{1} - \sigma \sqrt{\tau} \right )} \frac {\partial \left( \phi d_{1} \right) } {\partial S}$$ $$= \phi N{\left( \phi d_{1} \right )}+ \phi S n{\left (d_{1} \right)} \frac {\partial \left( \phi d_{1} \right) } {\partial S} - \phi K n \left( d_{1}\right) \frac{S}{K} \frac {\partial \left( \phi d_{1} \right) } {\partial S}$$ $$= \phi N{\left( \phi d_{1} \right )}+ \phi S n{\left (d_{1} \right)} \frac {\partial \left( \phi d_{1} \right) } {\partial S} - \phi n \left( d_{1}\right) S \frac {\partial \left( \phi d_{1} \right) } {\partial S}$$ $$= \phi N{\left( \phi d_{1} \right )}$$ $$\frac {\partial Black } {\partial S}= e^{-r\tau} \phi N{\left( \phi d_{1} \right )}$$