2
$\begingroup$

Using a standard PDE approach to price an American perpetual put option I obtain that the price of such option has the following form: $$ V(S) = A S + B S^{-2r/\sigma^2}. $$ And then I need to find a proper $A$ and $B$ coefficients to have the final solution. Finally I receive: $$ V(S) = \frac{K\sigma^2}{2r + \sigma^2}\left(\frac{S}{K} \frac{2r + \sigma^2}{2r}\right)^{-2r/\sigma^2}, \quad S \geq S^{*} = \frac{K}{1+\frac{\sigma^2}{2r}}. $$

This result is taken f.e. from 'Paul Wilmott on Quantitative Finance' book.

My question is:

Why I can not use the same technique to price American perpetual call option? When I apply the same method I obtain that my price has a form: $$ V(S) = A S. $$ But I am not able to derive that the coefficient $A$ should be equal to $1$.

Can anybody explain me where is the key issue of this problem?

$\endgroup$
  • $\begingroup$ One of the boundary conditions in the derivation of the Black-Scholes formula is that $\lim_{S\to\infty}\lvert C - S\rvert = 0$. The same boundary condition should hold here as well. This dictates $A = 1$. $\endgroup$ – Calculon Apr 2 '18 at 20:27
2
$\begingroup$

I suggest you first value a perpetual up and out call with a barrier B above max of strike K and initial spot and a rebate paid at first barrier hit equal to B - K. Then maximize this value over B. Continuing to assume no dividends, I believe you will find that the optimal B is infinite and that the up and out call value converges to spot. I haven’t actually done the calculation but it seems like a worthwhile exercise.

$\endgroup$
0
$\begingroup$

As suggested by Peter, you start by assuming a given policy to exercise when the spot price hits the level $B > K$ for the first time. Then the value matching condition implies

\begin{equation} V(B) = A S = B - K \qquad \Leftrightarrow \qquad A = \frac{B - K}{B}. \end{equation}

Thus for $S \leq B$, we have

\begin{equation} V(S) = \left( 1 - \frac{K}{B} \right) S. \end{equation}

Taking the derivative w.r.t. $B$ yields

\begin{equation} \frac{\partial V}{\partial B} = \frac{K S}{B^2} \end{equation}

Since this is strictly positive, it follows that the $B^* = \infty$ and thus $A^* = 1$. The only exception is when $S = 0$. In this case the option is worthless no matter what exercise policy you employ.

$\endgroup$
  • $\begingroup$ I do not understand this sentence: "Since this is strictly positive, it follows that the $B^{*}=\infty$". I mean you assumed that $\frac{\partial V}{\partial B} = 0$, hence we have to set $B^{*} = \infty.$ But why we have to have this equality: $\frac{\partial V}{\partial B} = 0$? $\endgroup$ – MathMen May 3 '18 at 15:12

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.