Given that the Black-Scholes formula for a European Call is given by:


$S$ is stock price, $K$ is strike price

When I take limit as $t\rightarrow T^-$, where $\sigma>0$, what are the cases that I have to take into considerations?


1 Answer 1


Look at the values of $d_1$ and $d_2$ when $t \rightarrow T$:

$$d_1 = \frac{\ln(S/K) + \left(r - D + \dfrac{1}{2}\sigma^2\right)\tau}{\sigma \sqrt{\tau}}$$


$$d_2 = d_1 - \sigma \sqrt{\tau}$$

with $\tau = T -t$.

Therefore, $t \rightarrow T$ is equivalent to $\tau \rightarrow 0$

  • Case 1: $S > K$

$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow + \infty$

$d_2 \sim d_1\rightarrow + \infty$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow (S-K)$$

as $N(d_i) \rightarrow 1$ for $i \in \{ 1,2\}$.

  • Case 2: $S < K$

$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow - \infty$

$d_2 \sim d_1\rightarrow - \infty$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow 0$$

as $N(d_i) \rightarrow 0$ for $i \in \{ 1,2\}$.

  • Case 3: $S = K$

$d_1 = \frac{\left(r - D + \dfrac{1}{2}\sigma^2\right)\sqrt{\tau}}{\sigma} \rightarrow 0$

$d_2 \rightarrow 0$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2)$$

$$C(S,\tau) = \left( S - K \right) \dfrac{1}{2} \rightarrow 0$$

as $N(d_i) \rightarrow \dfrac{1}{2}$ for $i \in \{ 1,2\}$ and $S = K$ by assumption.

Putting everything together, you see that the price of the call tends to its payoff


when the time $t$ is "infinitely" close to the maturity of the call $T$.

  • 1
    $\begingroup$ for Case I, shall we separate $S>K$ and $S=K$? Because when $S=K$, $\ln (S/K)=\ln 1=0$, hence the limit will be $0$ instead. Am I right? $\endgroup$
    – lrh09
    Aug 2, 2017 at 15:15
  • $\begingroup$ Good point. I have updated my answer with a 3rd case. $\endgroup$ Aug 2, 2017 at 15:41
  • $\begingroup$ By the way, Case 3: $S=K$, when $N(d_1),N(d_2)\rightarrow \frac{1}{2}$, isn't $C(S,t)\rightarrow \frac{1}{2}(S-K)$? Because the power $\tau=T-t\rightarrow 0$, the terms with $e$ approach $1$ $\endgroup$
    – lrh09
    Aug 2, 2017 at 16:04
  • $\begingroup$ Yes, but since $S = K$, I factored out by $S$ first. But your approach may be more intuitive indeed. I will update $\endgroup$ Aug 2, 2017 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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