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?

  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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: $\frac{\partial V}{\partial B} = 0$? $\endgroup$
    – MMM
    May 3 '18 at 15:12
  • $\begingroup$ +1, It's a really nice calculation. Not sure why no one else upvoted it. I like the randomisation approach + maximise wrt $B$ very much! $\endgroup$
    – Kevin
    Aug 20 at 0:05

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