4
$\begingroup$

I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

The transformed equation is \begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

And as an additional question, for the following graph

plot,

you get the most payoff when you wait until t = 4 and S = $e^{0.5}$. Is this insight correct? Additionally, in the graph above, what is the implication? Since the payoff is greatest when time to go, $t$ is maximum, does this mean we should exercise the option early?

$\endgroup$
1
$\begingroup$

You seem to be quite in to PDEs and less into finance. A call option gives at a future time $T$ the payoff $$ max(S_T-K,0), $$ thus if $S_T$ is bigger than $K$ then this is what you get, else nothing. The PDE tells you how the value changes with respect to changes in $S$ and $t$.

Theory tells us that the price of this option today (at $t=0$) is given as the discounted expected pay-off.

You solve the PDE on a mesh. There the value must be biggest, when $S$ is biggest. In reallity (and in theory on the whole real line) the value is unbounded.

EDIT: Note that the equation above is the terminal value. This one is triavial, the task is to calculate the price of this option at time $t$, $0 < t < T$. Furthermore mathematical finance deals with hedging of such a position.

I am not that much into PDEs but the one that you state is the PDE of the Europena style call -right? In this case you can only exercise the option at $T$. For so called American type options you can exercise at any point in time $u, 0 < u \le T$. If the stock is not paying dividens then you will never exercise early as the price of holding the option is always bigger(see here). In this case the price of the American style option is the same as of the European. Read about these two styles first. The basic reference is John Hull: OPTIONS, FUTURES, AND OTHER DERIVATIVES. You find slides on the web.

$\endgroup$
  • $\begingroup$ Thanks for your reply. I'm solving a finance problem in a primarily PDE-related project. Nonetheless, I still very much need to understand the the financial implication of the model. Is is always the case that the value is biggest when S is biggest? $\endgroup$ – meraxes Dec 1 '15 at 19:07
  • $\begingroup$ In addition, tn the graph above, what is the implication? Since the payoff is greatest when time to go, t is max, does this mean we should exercise the option early? $\endgroup$ – meraxes Dec 1 '15 at 19:13
  • $\begingroup$ See my edit in the answer. $\endgroup$ – Ric Dec 2 '15 at 7:36
  • 1
    $\begingroup$ So the graph above is only relevant for American type options (where the stock pays dividens) - is this right? Thanks for the reference, by the way! $\endgroup$ – meraxes Dec 2 '15 at 8:51
  • $\begingroup$ No, please google the terms "American" means that it can be exercised anytime up to expiration - this is a property of the option. Paying dividends is a property of the stock. As I see it you don't have dividends in your equations. $\endgroup$ – Ric Dec 2 '15 at 10:37
3
$\begingroup$

The form of the boundary conditions comes from the transformation. If my transformation maps $t \to\tau$, $S \to x$ then the boundary conditions are mapped by $\tau \to t$ and $x \to S$. So if $x=log(S)$, then my terminal condition will be $(e^x-K)^+$

$\endgroup$

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.