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Could you please help me? In book Rüdiger Seydel "Tools for Computational Finance" in Chapter 4.5 "American Options as Free Boundary Problem" it is provided the following explanation for case $S>S_f$ , where $S_f$ is contact point:
If $S>S_f$, then early-exercise causes immediate loss, because $-V+K-S<0$. Well, we have options so in case of early exercise we should borrow $1$ risky asset and seal it with price $K$. So our current payoff will be $K$. Why here is $-V+K-S$ ?
Thank you.

Nate Eldredge
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Mikhail Norshteyn
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1 Answers1

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By the definition of $S_f$ (eqn 4.22), we have $V^{am}_P(S, t) > (K - S)^+$ whenever $S > S_f(t)$. Rearranging, $-V^{am}_P + (K - S)^+ < 0$, so we have

$-V^{am}_P + K - S \leq -V^{am}_P + (K - S)^+ < 0$

Daniel McLaury
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  • I understand, why $−V_{P}^{am}+K−S<0$, but how this variable $−V_{P}^{am}+K−S$ is connected to exercise? I mean, we have bought option at some arbitrary moment $\hat{t}$ for price $V(S(\hat{t}),\hat{t})=\hat{V}$, and we exercise at another time moment $t$ when $S>S_{f}$, so we get payoff $K-S(t)$. So cashflow will be $-\hat{V} +K -S(t)$ and its not always less than zero (only $-V(S(t),t) +K -S(t) < 0$). Thank you. I hope you understood my question. – Mikhail Norshteyn May 30 '13 at 00:02