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I know that the Black-Scholes equation for an American option should satisfy the following inequality: $$ V_t + \frac{1}{2}\sigma^2 S^2 V_{SS} + r S V_S - rV \leq 0. $$ Actually we have 3 possibilities:

  1. $>0 \Rightarrow$ arbitrage.
  2. $=0 \Rightarrow$ hold, no arbitrage.
  3. $<0 \Rightarrow$ one should exercise already.

Let us assume that I want to find the form of $V(S)$ for which: $$ V_t + \frac{1}{2}\sigma^2 S^2 V_{SS} + r S V_S - rV =0. $$ and I obtain: $V(S) = 0$.

My question is: What I can conclude from this result? I think that it means that there is no optimal strategy for this option. In other words, it means that we should not exercise it at any time.

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  • $\begingroup$ Not sure where you are going with this. The form of V(S) for which the equation is zero is called the Black Scholes equation for a European option ;) $\endgroup$ – Alex C May 7 '18 at 14:44
  • $\begingroup$ You are right. But I think it can be also applied for American options. $\endgroup$ – MathMen May 7 '18 at 15:11
  • $\begingroup$ $V(S)=0$ is a _ solution, but is not _the only solution. For example, you may check that $V(S)=S$ also satisfies your equation. Mathematically, there are countless solutions to your equation, most of which yields no elegant expressions. Thus to get a financially meaningful solution, you must consider its initial/terminal/boundary condition, e.g., $V(S,T)=\max\left\{0,S-K\right\}$ for the European call. $\endgroup$ – hypernova May 8 '18 at 12:31
  • $\begingroup$ I understand. But Let us assume that I obtained the result: $V(S) = 0$. It does not satisfy my boundary condition, which means that the initial equation has no solution. My question is: How to interpret this result? Can I say that there is no optimal time to exercise such option? $\endgroup$ – MathMen May 8 '18 at 16:47
  • $\begingroup$ @JonasAl-Hadad: That's an interesting question. I'm not sure if $V(S)=0$ could be interpreted. Since it does not satisfy the boundary condition (the option value at expiry), it doesn't provide information for the option value we expect. Thus I think it doesn't have a financial interpretation. Nevertheless, I happened to find this, which might be helpful for you. $\endgroup$ – hypernova May 9 '18 at 17:24

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