<p>I don't think this boundary condition has anything to do with the optimal exercise price and it should therefore hold for both European and American style options. It is simply a terminal condition which allows us to rewrite the PDE in closed-form.</p>

<p>The relationship you show above in which:</p>

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

<p>$\to$</p>

<blockquote>
  <p>$\mathop {\lim }\limits_{\tau  \to 0} V(S,\tau) = 0$</p>
</blockquote>

<p>simply means that <strong><em>the extrinsic value of the option (i.e., the value in excess of $X-S&gt; 0, \forall S&lt;X)$ tends towards $0$ as $t \to T$.</em></strong></p>

<p>Dropping the "$max$" function is simply a restatement of the payoff condition in terms of the Heaviside function. The Heaviside function is essentially equivalent to the maximum function except that it enforces the following boundary at $t=T$:</p>

<p>${\displaystyle V(S,\,\tau)=0\quad \forall \;\;S&lt;X}$,</p>

<p>Though the difference seems subtle, it is important since $V_t$ need not be finite at S = 0, or even defined for that matter, which allows us to more easily perform the substitution of variables required to express the contingent pay-off in terms of the (heat) diffusion equation which is the general solution to the Black-Scholes model.</p>

<p>I hope that the intuition of this boundary condition is clearer from this explanation.</p>

CC BY-SA 3.0
2017-05-01T23:46:58.497Z
2017-05-01T23:46:58.497Z
David Addison
<p>$T$ is fixed so $\tau \to 0$ if and only if $t \to T$. Hence, it must be the case that</p>

<p>$$ \lim_{\tau \to 0} V(S,\tau) = \max \{ X-S,0 \} $$</p>

<p>However, when $(S,\tau) \in \Sigma_1$, it is optimal not to exercise the option. This means that as $\tau \to 0$, it must be the case that $X \le S$. Otherwise, you would not be in $\Sigma_1$.</p>

CC BY-SA 3.0
2017-01-29T13:22:56.257Z
2017-01-29T13:22:56.257Z
Theoretical Economist
<p>In a recent book I read the author mentioned the terminal condition: </p>

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

<p>This is intuitive to understand. </p>

<p>Then, the author defines: $$\tau  \equiv T - t$$</p>

<p>With this, the terminal condition above can be simplified to:</p>

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

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

<p>(note: in the space)   $${\Sigma _1} = \left\{ {(S,\tau )|B(\tau ) \le S &lt;  + \infty ,0 \le \tau  \le T} \right\}$$</p>

<p>Notations: </p>

<p>$X$ = exercise price </p>

<p>$S$ = underlying stock price</p>

<p>$T$ = time to maturity</p>

<p>$t$ = time to today</p>

<p>$B(\tau)$ = optimal exercise boundary</p>

CC BY-SA 3.0
2017-01-28T01:31:36.563Z
What is $B( \tau )$? I assume $S$ is the price of the underlying, $X$ is the strike price,and $T$ is the expiry date. You really ought to be explicit about what you mean -- the notation you use isn't necessarily standard and not everyone will have read the book you've been reading.
Theoretical Economist
CC BY-SA 3.0
2017-01-28T10:02:23.210Z
if you never seen this notation before its highly unlikely you can help with the question but i might be wrong
user10699
<asset-pricing>
Terminal Condition for American Put Option
user10699
# Terminal Condition for American Put Option

(Asked by: username_0 on 2017-01-22)

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, the author defines: $$\tau  \equiv T - t$$

With this, the terminal condition above can be simplified to:

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

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

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

Notations:

$X$ = exercise price

$S$ = underlying stock price

$T$ = time to maturity

$t$ = time to today

$B(\tau)$ = optimal exercise boundary

**Tags:** asset-pricing

> **Comments:**
> **username_1**: What is $B( \tau )$? I assume $S$ is the price of the underlying, $X$ is the strike price,and $T$ is the expiry date. You really ought to be explicit about what you mean -- the notation you use isn't necessarily standard and not everyone will have read the book you've been reading.  
> **username_0**: if you never seen this notation before its highly unlikely you can help with the question but i might be wrong  

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## 2 answers

### Answer by username_2 on 2017-05-01. Score: 1

I don't think this boundary condition has anything to do with the optimal exercise price and it should therefore hold for both European and American style options. It is simply a terminal condition which allows us to rewrite the PDE in closed-form.

The relationship you show above in which:

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

$\to$

> $\mathop {\lim }\limits\_{\tau \to 0} V(S,\tau) = 0$

simply means that ***the extrinsic value of the option (i.e., the value in excess of $X-S> 0, \forall S***

Dropping the "$max$" function is simply a restatement of the payoff condition in terms of the Heaviside function. The Heaviside function is essentially equivalent to the maximum function except that it enforces the following boundary at $t=T$:

${\displaystyle V(S,\,\tau)=0\quad \forall \;\;S

Though the difference seems subtle, it is important since $V\_t$ need not be finite at S = 0, or even defined for that matter, which allows us to more easily perform the substitution of variables required to express the contingent pay-off in terms of the (heat) diffusion equation which is the general solution to the Black-Scholes model.

I hope that the intuition of this boundary condition is clearer from this explanation.


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### Answer by username_1 on 2017-01-29. Score: 0

$T$ is fixed so $\tau \to 0$ if and only if $t \to T$. Hence, it must be the case that

$$ \lim\_{\tau \to 0} V(S,\tau) = \max \{ X-S,0 \} $$

However, when $(S,\tau) \in \Sigma\_1$, it is optimal not to exercise the option. This means that as $\tau \to 0$, it must be the case that $X \le S$. Otherwise, you would not be in $\Sigma\_1$.


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