# Terminal Condition for American Put Option

In a recent book I read, the author mentioned the terminal condition

$$\mathop {\lim }\limits_{t \to T} V(S,t) = \max \left\{ {X - S,0} \right\}$$

This is intuitive to understand. Then he defines $$\tau \equiv T - t$$ and when $r>0$, the terminal condition above can be simplified as

$$\mathop {\lim }\limits_{\tau \to 0} V(S,\tau) = 0$$

in the range $${\Sigma _1} = \left\{ {(S,\tau )|B(\tau ) \le S < + \infty ,0 \le \tau \le T} \right\}$$

This is not so intuitive . How can the value of the option be equal to zero in this case?

@Update: The book is " Homotopy Analysis Method in Nonlinear Differential Equations" p. 432

$B(\tau )$= optimal exercise boundary

• This doesn't seem to make sense. Can you provide the reference? – LocalVolatility Jan 22 '17 at 22:40
• This makes no sense, it say the payoff is zero as the option approach maturity. Can you double check. – SmallChess Jan 23 '17 at 5:11
• Could you also provide a reference? – Bob Jansen Jan 25 '17 at 9:48
• Assuming $B(\tau)$ figures the optimal exercice boundary when the time to expiry is $\tau$ which is still not completely clear, i.e. $B(\tau)$ is such that: $$B(\tau) = \text{argmax}_B \left\{ (X-B) \geq \Bbb{E}^\Bbb{Q} \left[ e^{-r\tau} (X-S_T)^+ \mid S_{T-\tau}=B \right] \right\}$$ then because $B(\tau) \to X$ in the limit as $\tau \to 0$, your put option is OTM in the range $\Sigma_1$ when $\tau \to 0$. – Quantuple Jan 25 '17 at 11:46
• I think @Quantuple answers it. Basically, it means as your spot price is too high, the payoff is zero as it approaches to maturity. – SmallChess Jan 26 '17 at 6:27

• But $S-X$ is not the same as $\max(S-X,0)$ – noob2 Jan 23 '17 at 18:56