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Suppose $V^+(S,t;K)$ is the value of a American option with strike $K$ before the exercise, and $V^-(S,t;K)$ is the value after exercise. Then how to understand the inequality $$V^+(S,t;K)\geq V^-(S,t;K) + (S - K)^+$$

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    $\begingroup$ Does an option exist after exercise? I would love to exercise an option and still own the option. Nice feature. $\endgroup$ – FinanceGuyThatCantCode May 23 '17 at 15:39
  • $\begingroup$ Life ends but trading never stops ;). Even if all outstanding options are exercised ahead of an event (such as a dividend) there will be buyers and sellers of options for strike K and maturity T after the event and the market will reconstitute itself at a new price $V^-$. $\endgroup$ – noob2 May 23 '17 at 17:20
  • $\begingroup$ If you're talking about a specific option, then when it's exercised it's gone. Unless you have some fancy otc product. If you're talking about the value of another American option on the same underlying that starts as soon as you exercise, with the same strike and expiry, then it will have the exact same value as the option you just exercised, because it's the same thing. $\endgroup$ – will May 23 '17 at 23:27
  • $\begingroup$ @will Is the following answer right? Regard the value after exercise $V^-(S,t;K)$ is $0?$ $\endgroup$ – A.Oreo May 24 '17 at 1:21
  • $\begingroup$ @noob2 Why we never consider the value after exercise usually? This case appears in the optimal static hedge of American call by vanilla call under the non-linear model. Could I understand as when we exercise the American option before the maturity of vanilla call, the vanilla call still exist i.e the hedge portfolio still have the value, so we should define the value of American call after the exercise? $\endgroup$ – A.Oreo May 24 '17 at 1:31
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After exercise, the option ceases to exist. It has no value.

$V^−(S,t;K)=0$

$V^+(S,t;K)=(S−K)^+$

If an option was exercised and

$V^+(S,t;K)>(S−K)$

then there was still time value in the option and it should not have been exercised.

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