0
$\begingroup$

Using the integral form of a call option price, I can derive the Dupire equation (no dividend) for the local volatility:

$$\sigma(S,T)^2=\frac{\frac{\partial C}{\partial T}+rK\frac{\partial C}{\partial K}}{\frac{1}{2}K^2\frac{\partial^2 C}{\partial K^2}}$$

However I have seen that this can be expressed in terms of the implied volatility, when assuming the Black-Scholes Model as:

$$\sigma(S,T)^2=\frac{\hat{\sigma}^2+2\hat{\sigma}(T-t)\frac{\partial \hat{\sigma}}{\partial T}+2r\hat{\sigma}K(T-t)\frac{\partial \hat{\sigma}}{\partial K}}{\left (1+Kd_1\sqrt{T-t}\frac{\partial \hat{\sigma}}{\partial K}\right )^2+\hat{\sigma}(T-t)K^2\left ( \frac{\partial^2\hat{\sigma}}{\partial K^2}-d_1\left ( \frac{\partial \hat{\sigma}}{\partial K}\right )^2\sqrt{T-t}\right )}$$

Now I've tried to get to this by using the chain rule: $$\frac{\partial C}{\partial T}=\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial T}$$ and similarly for $K$. However I cannot seem to get an expression that's quadratic in $\hat{\sigma}$ for the numerator. Am I on the right lines?

$\endgroup$
1
$\begingroup$

You are missing terms. The partial derivatives in the first formula must take into account the explicit dependence of the option price on expiry $T$ and strike $K$ and implicit dependence when using an implied volatility $\hat{\sigma}(T,K)$ that also depends on $T$ and $K$.

Using the ordinary derivative notation in the first formula to avoid confusion, we would expand as follows

$$\frac{d}{dT} C(S,K,T, r, \hat{\sigma}(T,K)) = \frac{\partial C}{\partial T} + \frac{\partial C}{\partial \hat{\sigma}} \frac{\partial \hat{\sigma}}{\partial T}$$

$\endgroup$
  • $\begingroup$ Thanks for your answer, to clarify, in the second derivative would we also have to take into account that the partial derivative $\frac{\partial C}{\partial T}$ also has a dependence on the implied vol. i.e.$$\frac{d^2 C}{dK^2}=\frac{\partial}{\partial K}\left ( \frac{\partial C}{\partial K}+\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial K}\right ) +\frac{\partial}{\partial \hat{\sigma}}\left (\frac{\partial C}{\partial K}+\frac{\partial C}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial K}\right )\frac{\partial \hat{\sigma}}{\partial K}$$ $\endgroup$ – George1811 Apr 7 '17 at 10:35
  • $\begingroup$ @George1811: That is correct. $\endgroup$ – RRL Apr 7 '17 at 16:30
  • $\begingroup$ However, the formula you are trying to derive does not look correct. What is the source? $\endgroup$ – RRL Apr 7 '17 at 16:33

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.