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


$X$ = exercise price

$S$ = underlying stock price

$T$ = time to maturity

$t$ = time to today

$B(\tau)$ = optimal exercise boundary

  • $\begingroup$ 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. $\endgroup$ – Theoretical Economist Jan 28 '17 at 1:31
  • $\begingroup$ if you never seen this notation before its highly unlikely you can help with the question but i might be wrong $\endgroup$ – user10699 Jan 28 '17 at 10:02

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:

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


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

simply means that the extrinsic value of the option (i.e., the value in excess of $X-S> 0, \forall S<X)$ tends towards $0$ as $t \to T$.

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 $t=T$:

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

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.


$T$ 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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