If so, is there a derivation anywhere that shows this? I was told that this could be done in a class but I don't see how it's possible.


It's quite straightforward when you plug in the formulae for the greeks into the PDE.


$\Delta =\frac{\partial c_t}{\partial S_t}=\Phi(d_1)$

$\Gamma=\frac{\partial^2 c_t}{\partial S_t^2}=\frac{\phi(d1)}{S_t\sigma\sqrt{u}}$

$\Theta=\frac{\partial c_t}{\partial t}=-rKe^{ru}\Phi(d_2)-S_t\phi(d_1)\frac{\sigma}{2\sqrt{u}}$

The Black Scholes PDE:

\begin{eqnarray} rc_t&=&\Theta+rS_t\Delta + \frac{1}{2}S_t^2\sigma^2\Gamma\\ RHS&=&-rKe^{ru}\Phi(d_2)-S_t\phi(d_1)\frac{\sigma}{2\sqrt{u}}+rS_t\Phi(d_1)+\frac{1}{2}S_t^2\sigma^2 \frac{\phi(d1)}{S_t\sigma\sqrt{u}}\\ &=&-rKe^{ru}\Phi(d_2)-\frac{S_t\phi(d_1)\sigma}{2\sqrt{u}}+rS_t\Phi(d_1)+ \frac{S_t\sigma\phi(d1)}{2\sqrt{u}}\\ &=&rS_t\Phi(d_1)-rKe^{ru}\Phi(d_2)\\ &=&rc_t\\ &=&LHS \end{eqnarray}

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.