0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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 $X-S> 0, \forall S<X)$ 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.

$\endgroup$
0
$\begingroup$

$T$ 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 $\Sigma_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.