According to Kim (1990, p.560) in "The Analytic Valuation of American Options". I understand the first minimum condition where K sets the lower bound of the optimal exercise boundary at expiry, but the second one is unclear to me,

$ \lim_{s \mapsto 0} B(s)=B(0)=K$ if $\delta \leq r $

$ \lim_{s \mapsto 0} B(s)=B(0)= (r/\delta )K $ if $\delta > r$

Update: $\delta$ = divdend rate, risk-free interest rate = $r$, optimal exercise boundary as a function of time $B(s)$ , exercise price $K$, in addition i found the following explanation,

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  • $\begingroup$ Could you please clarify what the notations are ? $\endgroup$ – Antoine Conze Jan 24 '18 at 13:11
  • $\begingroup$ updates notations $\endgroup$ – asdf Jan 24 '18 at 14:16

To clarify you must be talking about the optimal exercice boundary for the American Put. Consider an American put with maturity $T$ and let $B(t)$ be the optimal exercise boundary as a function of time $t$.

Let $dt$ be a small time step. Let $S$ be the stock price at time $T-dt$ and assume that it is optimal to early exercise at that point.

First you must have $S \leq K$ since it would not make sense to exercise early and get zero intrinsic value.


  • If you exercise immediately you get an intrinsic value of $K - S$
  • if you wait until maturity $T$ you get on average under the risk neutral measure $K - S(1+ (r - \delta) dt)$, which you have to discount back to $T-dt$ to obtain a continuation value of $$(K - S)(1+ (r - \delta) dt)(1-r dt) \approx K - S + (S \delta - Kr) dt $$

Since it is optimal to exercise early when the intrinsic value is above the continuation value, you must have $$ K - S \geq K - S + (S \delta - Kr) dt $$ and therefore $S \delta - Kr \leq 0$ or equivalently $S \leq K r/ \delta $

Therefore you would only early exercise if $S \leq \min(K, K r/ \delta )$.

Now the optimal boundary is the $\max$ of the $S$ that satisfy that inequality, therefore $B(T-dt) = \min(K, K r/ \delta )$, and $$ \lim_{dt \rightarrow 0} B(T-dt) = \min(K, K r/ \delta ) $$


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