1
$\begingroup$

I'm having troubles with the transformation from the Black-Scholes PDE and transforming it to the diffusion equation. I read this other stackexchange post (Here) and I understand most of the process, except where they changed the initial condition.

I got \begin{equation} \begin{split} u(x,0) &= e^{r\tau}C(S,T)\\ &=e^{r\tau}\text{max}(S-K,0)\\ &=e^{r\tau}\text{max}(e^y-K,0)\\ &=e^{r\tau}\text{max}(e^{x-(r-\sigma^2/2)\tau)}-K,0)\\ &=\text{max}(e^{x+\sigma^2\tau/2}-e^{r\tau}K,0)\\ \end{split} \end{equation}

Which is different from their equation of: \begin{equation} u(x,0) = u_0(x) = \text{max}(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0) \end{equation}

Where $a=2r/\sigma^2$

I would comment on the other post, however I don't have enough 'reputation' and this is a very specific question that I can't find elsewhere. Apparently it's in the textbook referenced in the original post, but the particular page referenced isn't freely available.

$\endgroup$
  • $\begingroup$ In Dewynne's et al. derivation, there is an extra step (involving going from "v" to "u") that is not shown in the stackexchange post you linked... $\endgroup$ – noob2 Dec 19 '18 at 21:41
1
$\begingroup$

$$u(x,0)=e^{\frac{k-1}{2}x}v(x,0)$$

and $$v(x,0)=max(e^{x}-1,0)$$ Hence $$u(x,0)=max(e^{\frac{k+1}{2}x}-e^{\frac{k-1}{2}x},0)$$

$\endgroup$

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.