Cash or Nothing options Greeks under Black Scholes

We derive the formulae for the Greeks, derivatives with respect to inputs, of a digital or a binary option that pays one unit of cash or nothing.

Theta

And in the final section, we derive formula for Theta, which is the first derivative of the option price with respect to t. Note that in our representation so far \( \tau=(T-t) \) so the dependence on t comes through \( \tau \).

$$\frac {\partial Price} {\partial \tau} =\frac {\partial } {\partial \tau} e^{- r_{d} \tau} N \left (\phi d_{2} \right ) $$ $$=e^{- r_{d} \tau}\frac{\partial}{\partial \tau} \left( N\left( \phi d_{2} \right ) \right) + N\left( \phi d_{2} \right ) \frac{\partial}{\partial \tau} \left( e^{- r_{d} \tau} \right)$$ $$=\phi e^{- r_{d} \tau} n{\left (\phi d_{2} \right )} \frac{\partial}{\partial \tau} d_{2} - r_{d} e^{- r_{d} \tau} N{\left (\phi d_{2} \right )}$$

Now, by definition:

$$ \frac{\partial}{\partial \tau} \left( d_{2} \right)=\frac{\partial}{\partial \tau} \left( \frac{1}{\sigma \sqrt{\tau}} \left(\ln{\left (\frac{S}{K} \right ) + \left(r_{d} - r_{f} - \frac{\sigma^{2}}{2}\right) \tau}\right) \right)$$ $$=\frac{\partial}{\partial \tau} \left( \frac{1}{\sigma \sqrt{\tau}}\ln\left (\frac{S}{K} \right )+ \frac{1}{\sigma \sqrt{\tau}} \left(r_{d} - r_{f} - \frac{\sigma^{2}}{2}\right) \tau \right)$$ $$=\frac{\partial}{\partial \tau} \left( \frac{1}{\sigma \sqrt{\tau}}\ln\left (\frac{S}{K} \right )+ \frac{1}{\sigma } \left(r_{d} - r_{f} - \frac{\sigma^{2}}{2}\right) \sqrt{\tau} \right)$$ $$= - \frac{1}{2\sigma \sqrt[3]{\tau}}\ln\left (\frac{S}{K} \right )+ \frac{1}{2\sigma \sqrt{\tau} } \left(r_{d} - r_{f} - \frac{\sigma^{2}}{2}\right) $$ $$=-\frac{1}{2\tau}\frac{1}{\sigma \sqrt{\tau}} \left(\ln{\left (\frac{S}{K} \right ) - \left(r_{d} - r_{f} - \frac{\sigma^{2}}{2}\right) \tau}\right)$$

Thus

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

Now

$$\frac {\partial Price} {\partial t}=\frac {\partial Price} {\partial \tau}\frac {\partial \tau} {\partial t}=\frac {\partial Price} {\partial \tau}\frac {\partial \left(T-t \right)} {\partial t}=-\frac {\partial Price} {\partial \tau}$$

Hence

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